Pre calculus Grade 11 Learner's Module Senior High School

DEPED COPY

Precalculus

Learner’s Material

Department of Education
Republic of the Philippines

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DEPED COPY Precalculus
Learner’s Material
First Edition 2016

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DEPED COPY Table of Contents
To the Precalculus Learners 1
DepEd Curriculum Guide for Precalculus 2
Unit 1: Analytic Geometry 6
Lesson 1.1: Introduction to Conic Sections and Circles . . . . . . . . 7
1.1.1: An Overview of Conic Sections . . . . . . . . . . . . . . . . . . . . . . . . . . . 7
1.1.2: Denition and Equation of a Circle. . . . . . . . . . . . . . . . . . . . . . . 8
1.1.3: More Properties of Circles. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10
1.1.4: Situational Problems Involving Circles. . . . . . . . . . . . . . . . . . . . 12
Lesson 1.2: Parabolas. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19
1.2.1: Denition and Equation of a Parabola. . . . . . . . . . . . . . . . . . . . 19
1.2.2: More Properties of Parabolas. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23
1.2.3: Situational Problems Involving Parabolas . . . . . . . . . . . . . . . . 26
Lesson 1.3: Ellipses . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33
1.3.1: Denition and Equation of an Ellipse. . . . . . . . . . . . . . . . . . . . . 33
1.3.2: More Properties of Ellipses . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36
1.3.3: Situational Problems Involving Ellipses. . . . . . . . . . . . . . . . . . . 40
Lesson 1.4: Hyperbolas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46
1.4.1: Denition and Equation of a Hyperbola . . . . . . . . . . . . . . . . . . 46
1.4.2: More Properties of Hyperbolas . . . . . . . . . . . . . . . . . . . . . . . . . . . 50
1.4.3: Situational Problems Involving Hyperbolas . . . . . . . . . . . . . . . 54
Lesson 1.5: More Problems on Conic Sections . . . . . . . . . . . . . . . . 60
1.5.1: Identifying the Conic Section by Inspection. . . . . . . . . . . . . . . 60
1.5.2: Problems Involving Dierent Conic Sections . . . . . . . . . . . . . . 62
iiiAll rights reserved. No part of this material may be reproduced or transmitted in any form or by any means -
electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2016.

DEPED COPY Lesson 1.6: Systems of Nonlinear Equations . . . . . . . . . . . . . . . . . . 67
1.6.1: Review of Techniques in Solving Systems of Linear
Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 68
1.6.2: Solving Systems of Equations Using Substitution . . . . . . . . . 69
1.6.3: Solving Systems of Equations Using Elimination. . . . . . . . . . 70
1.6.4: Applications of Systems of Nonlinear Equations . . . . . . . . . . 73
Unit 2: Mathematical Induction 80
Lesson 2.1: Review of Sequences and Series . . . . . . . . . . . . . . . . . . . 81
Lesson 2.2: Sigma Notation. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 86
2.2.1: Writing and Evaluating Sums in Sigma Notation . . . . . . . . . 87
2.2.2: Properties of Sigma Notation. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 89
Lesson 2.3: Principle of Mathematical Induction . . . . . . . . . . . . . . 96
2.3.1: Proving Summation Identities . . . . . . . . . . . . . . . . . . . . . . . . . . . . 97
2.3.2: Proving Divisibility Statements. . . . . . . . . . . . . . . . . . . . . . . . . . .101
?
2.3.3: Proving Inequalities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 102
Lesson 2.4: The Binomial Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . 108
2.4.1: Pascal's Triangle and the Concept of Combination. . . . . . . . 109
2.4.2: The Binomial Theorem. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 111
2.4.3: Terms of a Binomial Expansion . . . . . . . . . . . . . . . . . . . . . . . . . . 114
?
2.4.4: Approximation and Combination Identities . . . . . . . . . . . . . . . 116
Unit 3: Trigonometry 123
Lesson 3.1: Angles in a Unit Circle . . . . . . . . . . . . . . . . . . . . . . . . . . . 124
3.1.1: Angle Measure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 124
3.1.2: Coterminal Angles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 128
3.1.3: Arc Length and Area of a Sector . . . . . . . . . . . . . . . . . . . . . . . . . 129
Lesson 3.2: Circular Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 135
3.2.1: Circular Functions on Real Numbers . . . . . . . . . . . . . . . . . . . . . 136
3.2.2: Reference Angle. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 139All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means -
electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2016.

DEPED COPY Lesson 3.3: Graphs of Circular Functions and Situational
Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 144
3.3.1: Graphs ofy= sinxandy= cosx. . . . . . . . . . . . . . . . . . . . . . . . 145
3.3.2: Graphs ofy=asinbxandy=acosbx. . . . . . . . . . . . . . . . . . . 147
3.3.3: Graphs ofy=asinb(xc) +dand
y=acosb(xc) +d. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 151
3.3.4: Graphs of Cosecant and Secant Functions . . . . . . . . . . . . . . . . 154
3.3.5: Graphs of Tangent and Cotangent Functions . . . . . . . . . . . . . 158
3.3.6: Simple Harmonic Motion. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 160
Lesson 3.4: Fundamental Trigonometric Identities. . . . . . . . . . . . .171
3.4.1: Domain of an Expression or Equation . . . . . . . . . . . . . . . . . . . . 171
3.4.2: Identity and Conditional Equation . . . . . . . . . . . . . . . . . . . . . . . 173
3.4.3: The Fundamental Trigonometric Identities . . . . . . . . . . . . . . . 174
3.4.4: Proving Trigonometric Identities . . . . . . . . . . . . . . . . . . . . . . . . . 176
Lesson 3.5: Sum and Dierence Identities. . . . . . . . . . . . . . . . . . . . .181
3.5.1: The Cosine Dierence and Sum Identities . . . . . . . . . . . . . . . . 181
3.5.2: The Cofunction Identities and the Sine Sum and
Dierence Identities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 183
3.5.3: The Tangent Sum and Dierence Identities . . . . . . . . . . . . . . . 186
Lesson 3.6: Double-Angle and Half-Angle Identities. . . . . . . . . . .192
3.6.1: Double-Angle Identities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 192
3.6.2: Half-Angle Identities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 195
Lesson 3.7: Inverse Trigonometric Functions . . . . . . . . . . . . . . . . . . 201
3.7.1: Inverse Sine Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 202
3.7.2: Inverse Cosine Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 205
3.7.3: Inverse Tangent Function and the Remaining Inverse
Trigonometric Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 208
Lesson 3.8: Trigonometric Equations. . . . . . . . . . . . . . . . . . . . . . . . . .220
3.8.1: Solutions of a Trigonometric Equation. . . . . . . . . . . . . . . . . . . . 221
3.8.2: Equations with One Term . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 224
3.8.3: Equations with Two or More Terms . . . . . . . . . . . . . . . . . . . . . . 227All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means -
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DEPED COPY Lesson 3.9: Polar Coordinate System . . . . . . . . . . . . . . . . . . . . . . . . . 236
3.9.1: Polar Coordinates of a Point . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 237
3.9.2: From Polar to Rectangular, and Vice Versa. . . . . . . . . . . . . . . 241
3.9.3: Basic Polar Graphs and Applications . . . . . . . . . . . . . . . . . . . . . 244
Answers to Odd-Numbered Exercises in Supplementary Problems
and All Exercises in Topic Tests 255
References 290All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means -
electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2016.

DEPED COPY To the Precalculus Learners
The Precalculus course bridges basic mathematics and calculus. This course
completes your foundational knowledge on algebra, geometry, and trigonometry.
It provides you with conceptual understanding and computational skills that are
prerequisites for Basic Calculus and future STEM courses.
Based on the Curriculum Guide for Precalculus of the Department of Edu-
cation (see pages 2-5), the primary aim of this Learning Manual is to give you
an adequate stand-alone material that can be used for the Grade 11 Precalculus
course.
The Manual is divided into three units: analytic geometry, summation no-
tation and mathematical induction, and trigonometry. Each unit is composed
of lessons that bring together related learning competencies in the unit. Each
lesson is further divided into sub-lessons that focus on one or two competencies
for eective learning.
At the end of each lesson, more examples are given inSolved Examplesto
reinforce the ideas and skills being developed in the lesson. You have the oppor-
tunity to check your understanding of the lesson by solving theSupplementary
Problems. Finally, two sets of Topic Testare included to prepare you for the
exam.
Answers, solutions, or hints to odd-numbered items in the Supplementary
Problems and all items in the Topic Tests are provided at the end of the Manual
to guide you while solving them. We hope that you will use this feature of the
Manual responsibly.
Some items are marked with a star. A starred sub-lesson means the discussion
and accomplishment of the sub-lesson are optional. This will be decided by your
teacher. On the other hand, a starred example or exercise means the use of
calculator is required.
We hope that you will nd this Learning Manual helpful and convenient to
use. We encourage you to carefully study this Manual and solve the exercises
yourselves with the guidance of your teacher. Although great eort has been
put into this Manual for technical correctness and precision, any mistake found
and reported to the Team is a gain for other students. Thank you for your
cooperation.
The Precalculus LM TeamAll rights reserved. No part of this material may be reproduced or transmitted in any form or by any means -
electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2016.

DEPED COPY K to 12 BASIC EDUCATION CURRICULUM
SENIOR HIGH SCHOOL – SCIENCE, TECHNOLOGY, ENGINEERING AND MATHEMATICS (STEM) SPECIALIZED SUBJECT

K to 12 Senior High School STEM Specialized Subject – Pre-Calculus December 2013 Page 1 of 4

Grade: 11 Semester: First Semester
Core Subject Title: Pre-Calculus No. of Hours/ Semester: 80 hours/ semester
Pre-requisite (if needed):

Subject Description: At the end of the course, the students must be able to apply concepts and solve problems involving conic sections, systems of nonlinear equations,
series and mathematical induction, circular and trigonometric functions, trigonometric identities, and polar coordinate system.

CONTENT
CONTENT
STANDARDS
PERFORMANCE
STANDARDS
LEARNING COMPETENCIES CODE
Analytic
Geometry

The learners
demonstrate an
understanding
of...

key concepts of
conic sections and
systems of
nonlinear
equations

The learners shall be able
to...

model situations
appropriately and solve
problems accurately using
conic sections and systems
of nonlinear equations

The learners...

1. illustrate the different types of conic sections: parabola, ellipse,
circle, hyperbola, and degenerate cases.***

STEM_PC11AG-Ia-1
2. define a circle. STEM_PC11AG-Ia-2
3. determine the standard form of equation of a circle STEM_PC11AG-Ia-3
4. graph a circle in a rectangular coordinate system STEM_PC11AG-Ia-4
5. define a parabola STEM_PC11AG-Ia-5
6. determine the standard form of equation of a parabola STEM_PC11AG-Ib-1
7. graph a parabola in a rectangular coordinate system STEM_PC11AG-Ib-2
8. define an ellipse STEM_PC11AG-Ic-1
9. determine the standard form of equation of an ellipse STEM_PC11AG-Ic-2
10. graph an ellipse in a rectangular coordinate system STEM_PC11AG-Ic-3
11. define a hyperbola STEM_PC11AG-Id-1
12. determine the standard form of equation of a hyperbola STEM_PC11AG-Id-2

2All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means -
electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2016.

DEPED COPY K to 12 BASIC EDUCATION CURRICULUM
SENIOR HIGH SCHOOL – SCIENCE, TECHNOLOGY, ENGINEERING AND MATHEMATICS (STEM) SPECIALIZED SUBJECT

K to 12 Senior High School STEM Specialized Subject – Pre-Calculus December 2013 Page 2 of 4

CONTENT
CONTENT
STANDARDS
PERFORMANCE
STANDARDS
LEARNING COMPETENCIES CODE

13. graph a hyperbola in a rectangular coordinate system STEM_PC11AG-Id-3
14. recognize the equation and important characteristics of the
different types of conic sections
STEM_PC11AG-Ie-1
15. solves situational problems involving conic sections STEM_PC11AG-Ie-2
16. illustrate systems of nonlinear equations STEM_PC11AG-If-1
17. determine the solutions of systems of nonlinear equations using
techniques such as substitution, elimination, and graphing***
STEM_PC11AG-If-g-1
18. solve situational problems involving systems
of nonlinear equations
STEM_PC11AG-Ig-2
Series and
Mathematical
Induction

key concepts of
series and
mathematical
induction and the
Binomial
Theorem.

keenly observe and
investigate patterns, and
formulate appropriate
mathematical statements
and prove them using
mathematical induction
and/or Binomial Theorem.
1. illustrate a series
STEM_PC11SMI -Ih-1
2. differentiate a series from a sequence STEM_PC11SMI -Ih-2
3. use the sigma notation to represent a series STEM_PC11SMI -Ih-3
4. illustrate the Principle of Mathematical Induction STEM_PC11SMI -Ih-4
5. apply mathematical induction in proving identities STEM_PC11SMI -Ih-i-1
6. illustrate Pascal’s Triangle in the expansion of &#3627408485;+&#3627408486;
??????
for small
positive integral values of ??????
STEM_PC11SMI -Ii-2
7. prove the Binomial Theorem STEM_PC11SMI -Ii-3
8. determine any term of &#3627408485;+&#3627408486;
??????
, where ?????? is a positive integer,
without expanding
STEM_PC11SMI -Ij-1
9. solve problems using mathematical induction and the Binomial
Theorem
STEM_PC11SMI -Ij-2

3All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means -
electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2016.

DEPED COPY K to 12 BASIC EDUCATION CURRICULUM
SENIOR HIGH SCHOOL – SCIENCE, TECHNOLOGY, ENGINEERING AND MATHEMATICS (STEM) SPECIALIZED SUBJECT

K to 12 Senior High School STEM Specialized Subject – Pre-Calculus December 2013 Page 3 of 4
CONTENT
CONTENT
STANDARDS
PERFORMANCE
STANDARDS
LEARNING COMPETENCIES CODE
Trigonometry

key concepts of
circular functions,
trigonometric
identities, inverse
trigonometric
functions, and
the polar
coordinate
system
1. formulate and solve
accurately situational
problems involving
circular functions
1. illustrate the unit circle and the relationship between the linear
and angular measures of a central angle in a unit circle

STEM_PC11T-IIa-1
2. convert degree measure to radian measure and vice versa STEM_PC11T-IIa-2
3. illustrate angles in standard position and coterminal angles STEM_PC11T-IIa-3
4. illustrate the different circular functions STEM_PC11T-IIb-1
5. uses reference angles to find exact values of circular functions STEM_PC11T-IIb-2
6. determine the domain and range of the different circular functions STEM_PC11T-IIc-1
7. graph the six circular functions (a) amplitude, (b) period, and (c)
phase shift
STEM_PC11T-IIc-d-1
8. solve problems involving circular functions STEM_PC11T-IId-2
2. apply appropriate
trigonometric identities in
solving situational
problems
9. determine whether an equation is an identity or a conditional
equation
STEM_PC11T-IIe-1
10. derive the fundamental trigonometric identities STEM_PC11T-IIe-2
11. derive trigonometric identities involving sum and difference of
angles
STEM_PC11T-IIe-3
12. derive the double and half-angle formulas STEM_PC11T-IIf-1
13. simplify trigonometric expressions STEM_PC11T-IIf-2
14. prove other trigonometric identities STEM_PC11T-IIf-g-1
15. solve situational problems involving trigonometric identities STEM_PC11T-IIg-2
3. formulate and solve
accurately situational
problems involving
appropriate trigonometric
functions
16. illustrate the domain and range of the inverse trigonometric
functions.
STEM_PC11T-IIh-1
17. evaluate an inverse trigonometric expression. STEM_PC11T-IIh-2
18. solve trigonometric equations. STEM_PC11T-IIh-i-1
19. solve situational problems involving inverse trigonometric
functions and trigonometric equations
STEM_PC11T-IIi-2
4. formulate and solve
accurately situational
problems involving the
polar coordinate system
20. locate points in polar coordinate system STEM_PC11T-IIj-1
21. convert the coordinates of a point from rectangular to polar
systems and vice versa
STEM_PC11T-IIj-2
22. solve situational problems involving polar coordinate system STEM_PC11T-IIj-3

***Suggestion for ICT-enhanced lesson when available and where appropriate
4All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means -
electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2016.

DEPED COPY K to 12 BASIC EDUCATION CURRICULUM
SENIOR HIGH SCHOOL – SCIENCE, TECHNOLOGY, ENGINEERING AND MATHEMATICS (STEM) SPECIALIZED SUBJECT

K to 12 Senior High School STEM Specialized Subject – Pre-Calculus December 2013 Page 4 of 4

Code Book Legend

Sample: STEM_PC11AG-Ia-1

DOMAIN/ COMPONENT CODE
Analytic Geometry AG
Series and Mathematical Induction SMI
Trigonometry T

LEGEND SAMPLE
First Entry
Learning Area and
Strand/ Subject or
Specialization
Science, Technology,
Engineering and Mathematics
Pre-Calculus
STEM_PC11AG
Grade Level Grade 11
Uppercase
Letter/s
Domain/Content/
Component/ Topic
Analytic Geometry
-
Roman Numeral
*Zero if no specific
quarter
Quarter First Quarter I
Lowercase
Letter/s
*Put a hyphen (-) in
between letters to
indicate more than a
specific week
Week Week one a
-
Arabic Number Competency
illustrate the different types
of conic sections: parabola,
ellipse, circle, hyperbola,
and degenerate cases
1
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electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2016.

DEPED COPY Unit 1
Analytic Geometry
San Juanico Bridge, by Morten Nrbe, 21 June 2009,
https://commons.wikimedia.o
rg/wiki/File%3ASan
JuanicoBridge2.JPG. Public Domain.
Stretc
hing from Samar to Leyte with a total length of more than two kilome-
ters, the San Juanico Bridge has been serving as one of the main thoroughfares
of economic and social development in the country since its completion in 1973.
Adding picturesque eect on the whole architecture, geometric structures are
subtly built to serve other purposes. The arch-shaped support on the main span
of the bridge helps maximize its strength to withstand mechanical resonance and
aeroelastic utter brought about by heavy vehicles and passing winds.
6All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means -
electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2016.

DEPED COPY Lesson 1.1. Introduction to Conic Sections and Circles
Learning Outcomes of the Lesson
At the end of the lesson, the student is able to:
(1) illustrate the dierent types of conic sections: parabola, ellipse, circle, hyper-
bola, and degenerate cases;
(2) dene a circle;
(3) determine the standard form of equation of a circle;
(4) graph a circle in a rectangular coordinate system; and
(5) solve situational problems involving conic sections (circles).
Lesson Outline
(1) Introduction of the four conic sections, along with the degenerate conics
(2) Denition of a circle
(3) Derivation of the standard equation of a circle
(4) Graphing circles
(5) Solving situational problems involving circles
Introduction
We present the conic sections, a particular class of curves which sometimes
appear in nature and which have applications in other elds. In this lesson, we
rst illustrate how each of these curves is obtained from the intersection of a
plane and a cone, and then discuss the rst of their kind, circles. The other conic
sections will be covered in the next lessons.
1.1.1. An Overview of Conic Sections
We introduce the conic sections (or conics), a particular class of curves which
oftentimes appear in nature and which have applications in other elds. One
of the rst shapes we learned, a circle, is a conic. When you throw a ball, the
trajectory it takes is a parabola. The orbit taken by each planet around the sun
is an ellipse. Properties of hyperbolas have been used in the design of certain
telescopes and navigation systems. We will discuss circles in this lesson, leaving
parabolas, ellipses, and hyperbolas for subsequent lessons.
Circle (Figure 1.1) - when the plane is horizontal
Ellipse (Figure 1.1) - when the (tilted) plane intersects only one cone to form
a bounded curve
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DEPED COPY Parabola (Figure 1.2) - when the plane intersects only one cone to form an
unbounded curve
Hyperbola (Figure 1.3) - when the plane (not necessarily vertical) intersects
both cones to form two unbounded curves (each called a branch of the hyper-
bola)
Figure 1.1 Figure 1.2 Figure 1.3
We can draw these conic sections (also called conics) on a rectangular co-
ordinate plane and nd their equations. To be able to do this, we will presentequivalent denitions of these conic sections in subsequent sections, and use these
to nd the equations.
There are other ways for a plane and the cones to intersect, to form what are
referred to asdegenerate conics: a point, one line, and two lines. See Figures 1.4,
1.5 and 1.6.
Figure 1.4 Figure 1.5 Figure 1.6
1.1.2. Denition and Equation of a Circle
A circle may also be considered a special kind of ellipse (for the special case when
the tilted plane is horizontal). As we get to know more about a circle, we will
also be able to distinguish more between these two conics.
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DEPED COPY See Figure 1.7, with the pointC(3;1) shown. From the gure, the distance
ofA(2; 1) fromCisAC= 5. By the distance formula, the distance ofB(6;5)
fromCisBC=
p
(63)
2
+ (51)
2
= 5. There are other pointsPsuch that
PC= 5. The collection of all such points which are 5 units away fromC, forms
a circle.
Figure 1.7 Figure 1.8
LetCbe a given point. The set of all pointsPhaving the same
distance fromCis called acircle. The point Cis called thecenterof
the circle, and the common distance itsradius.
The termradiusis both used to refer to a segment from the centerCto a
pointPon the circle, and the length of this segment.
See Figure 1.8, where a circle is drawn. It has centerC(h; k) and radiusr >0.
A pointP(x; y) is on the circle if and only ifPC=r. For any such point then,
its coordinates should satisfy the following.
PC=r
p
(xh)
2
+ (yk)
2
=r
(xh)
2
+ (yk)
2
=r
2
This is thestandard equationof the circle with centerC(h; k) and radiusr. If
the center is the origin, thenh= 0 andk= 0. The standard equation is then
x
2
+y
2
=r
2
.
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DEPED COPY Example 1.1.1.In each item, give the
standard equation of the circle satisfy-
ing the given conditions.
(1) center at the origin, radius 4
(2) center (4; 3), radius
p
7
(3) circle in Figure 1.7(4) circleAin Figure 1.9
(5) circleBin Figure 1.9
(6) center (5;6), tangent to they-
axis
Figure 1.9
(7)
center (5;6), tangent to thex-axis
(8) It has a diameter with endpointsA(1; 4) andB(4;2).
Solution.(1)x
2
+y
2
= 16
(2) (x + 4)
2
+ (y3)
2
= 7
(3) The center is (3;1) and the radius is 5, so the equation is (x3)
2
+(y1)
2
=
25.
(4) By inspection, the center is (2;1) and the radius is 4. The equation is
(x+ 2)
2
+ (y+ 1)
2
= 16.
(5) Similarly by inspection, we have (x 3)
2
+ (y2)
2
= 9.
(6) The center is 5 units away from they-axis, so the radius isr= 5 (you can
make a sketch to see why). The equation is (x 5)
2
+ (y+ 6)
2
= 25.
(7) Similarly, since the center is 6 units away from thex-axis, the equation is
(x5)
2
+ (y+ 6)
2
= 36.
(8) The centerCis the midpoint ofAandB:C=

1+4
2
;
4+2
2

=

3
2
;3

.
The
radius is thenr=AC=
q

1
3
2

2
+
(43)
2
=
q
29
4
.
The circle has
equation

x
3
2

2
+
(y3)
2
=
29
4
. 2
1.1.3.
More Properties of Circles
After expanding, the standard equation

x
3
2

2
+
(y3)
2
=
29
4
can
be rewritten as
x
2
+y
2
3x6y+ 4 = 0;
an equation of the circle ingeneral form.
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DEPED COPY If the equation of a circle is given in the general form
Ax
2
+Ay
2
+Cx+Dy+E= 0; A6= 0;
or
x
2
+y
2
+Cx+Dy+E= 0;
we can determine the standard form by completing the square in both variables.
Completing the square in an expression likex
2
+ 14x means determining
the term to be added that will produce a perfect polynomial square. Since the
coecient ofx
2
is already 1, we take half the coecient ofxand square it, and
we get 49. Indeed,x
2
+ 14x + 49 = (x + 7)
2
is a perfect square. To complete
the square in, say, 3x
2
+ 18x, we factor the coecient ofx
2
from the expression:
3(x
2
+ 6x), then add 9 inside. When completing a square in an equation, any
extra term introduced on one side should also be added to the other side.
Example 1.1.2.Identify the center and radius of the circle with the given equa-
tion in each item. Sketch its graph, and indicate the center.
(1)x
2
+y
2
6x= 7
(2)x
2
+y
2
14x+ 2y=14
(3) 16x
2
+ 16y
2
+ 96x 40y= 315
Solution.The rst step is to rewrite each equation in standard form by complet-
ing the square inxand iny. From the standard equation, we can determine the
center and radius.
(1)
x
2
6x+y
2
= 7
x
2
6x+ 9 +y
2
= 7 + 9
(x3)
2
+y
2
= 16
Center (3;0),r= 4, Figure 1.10
(2)
x
2
14x+y
2
+ 2y=14
x
2
14x+ 49 +y
2
+ 2y+ 1 =14 + 49 + 1
(x7)
2
+ (y+ 1)
2
= 36
Center (7;1),r= 6, Figure 1.11
(3)
16x
2
+ 96x + 16y
2
40y= 315
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DEPED COPY 16(x
2
+ 6x) + 16

y
2

5
2
y

= 315
16(x
2
+ 6x+ 9) + 16

y
2

5
2
y+
25
16

= 315 + 16(9) + 16

25
16

16(x+ 3)
2
+ 16

y
5
4

2
= 484
(x+ 3)
2
+

y
5
4

2
=
484
16
=
121
4
=

11
2

2
Center

3;
5
4

,r= 5:5, Figure 1.12. 2
Figure 1.10 Figure 1.11 Figure 1.12
In
the standard equation (x h)
2
+ (yk)
2
=r
2
, both the two squared
terms on the left side have coecient 1. This is the reason why in the preceding
example, we divided by 16 at the last equation.
1.1.4. Situational Problems Involving Circles
Let us now take a look at some situational problems involving circles.
?
Example 1.1.3.A street with two lanes, each 10 ft wide, goes through a
semicircular tunnel with radius 12 ft. How high is the tunnel at the edge of each
lane? Round o to 2 decimal places.
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DEPED COPY Solution.We draw a coordinate system with origin at the middle of the highway,
as shown. Because of the given radius, the tunnel's boundary is on the circle
x
2
+y
2
= 12
2
. PointPis the point on the arc just above the edge of a lane, so
itsx-coordinate is 10. We need itsy-coordinate. We then solve 10
2
+y
2
= 12
2
fory >0, giving usy= 2
p
116:63 ft. 2
Example 1.1.4.A piece of a broken plate was dug up in an archaeological site.
It was put on top of a grid, as shown in Figure 1.13, with the arc of the plate
passing throughA(7; 0),B(1;4) andC(7;2). Find its center, and the standard
equation of the circle describing the boundary of the plate.
Figure 1.13
Figure 1.14
Solution.W
e rst determine the center. It is the intersection of the perpendicular
bisectors ofABandBC(see Figure 1.14). Recall that, in a circle, the perpen-
dicular bisector of any chord passes through the center. Since the midpointM
ofABis

7+1
2
;
0+4
2

=
(3;2), andmAB=
40
1+7
=
1
2
,
the perpendicular bisector
ofABhas equationy2 =2(x+ 3), or equivalently,y=2x4.
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DEPED COPY Since the midpointNofBCis

1+7
2
;
4+2
2

= (4;3), andmBC=
24
71
=
1
3
,
the perpendicular bisector ofBChas equationy3 = 3(x 4), or equivalently,
y= 3x9.
The intersection of the two linesy= 2x4 andy= 3x9 is (1;6) (by
solving a system of linear equations). We can take the radius as the distance of
this point from any ofA,BorC(it's most convenient to useBin this case). We
then getr= 10. The standard equation is thus (x 1)
2
+ (y+ 6)
2
= 100.2
More Solved Examples
1. In each item, give the standard equation of the circle satisying the given con-
ditions.
(a) center at the origin, contains (0;3)
(b) center (1; 5), diameter 8
(c) circleAin Figure 1.15
(d) circleBin Figure 1.15
(e) circleCin Figure 1.15
(f) center (2; 3), tangent to they-
axis
(g) center (2; 3), tangent to thex-
axis
(h) contains the points (2;0) and
(8;0), radius 5
Figure 1.15
Solution:
(a) The radius is 3, so the equation isx
2
+y
2
= 9.
(b) The radius is 8=2 = 4, so the equation is (x1)
2
+ (y5)
2
= 16.
(c) The center is (2;2) and the radius is 2,
so the equation is (x +2)
2
+(y2)
2
= 4.
(d) The center is (2;3) and the radius is 1,
so the equation is (x 2)
2
+(y3)
2
= 1.
(e) The center is (1;1) and by the
Pythagorean Theorem, the radius (see
Figure 1.16) is
p
2
2
+ 2
2
=
p
8, so the
equation is (x 1)
2
+ (x+ 1)
2
= 8.
Figure 1.16
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DEPED COPY (f) The radius is 3, so the equation is (x + 2)
2
+ (y+ 3)
2
= 9.
(g) The radius is 2, so the equation is (x + 2)
2
+ (y+ 3)
2
= 4.
(h) The distance between (2; 0) and (8;0) is 10; since the radius is 5, these
two points are endpoints of a diameter. Then the circle has center at
(3;0) and radius 5, so its equation is (x3)
2
+y
2
= 25.
2. Identify the center and radius of the circle with the given equation in each
item. Sketch its graph, and indicate the center.
(a)x
2
+y
2
+ 8y= 33
(b) 4x
2
+ 4y
2
16x+ 40y + 67 = 0
(c) 4x
2
+ 12x + 4y
2
+ 16y 11 = 0
Solution:
(a)
x
2
+y
2
+ 8y= 33
x
2
+y
2
+ 8y+ 16 = 33 + 16
x
2
+ (y+ 4)
2
= 49
Center (0;4), radius 7, see Figure 1.17.
(b)
4x
2
+ 4y
2
16x+ 40y + 67 = 0
x
2
4x+y
2
+ 10y =
67
4
x
2
4x+ 4 +y
2
+ 10y + 25 =
67
4
+ 4 + 25
(x2)
2
+ (y+ 5)
2
=
49
4
=

7
2

2
Center (2;5), radius 3.5, see Figure 1.18.
(c)
4x
2
+ 12x + 4y
2
+ 16y 11 = 0
x
2
+ 3x+y
2
+ 4y=
11
4
x
2
+ 3x+
9
4
+y
2
+ 4y+ 4 =
11
4
+
9
4
+ 4

x+
3
2

2
+ (y+ 2)
2
= 9
Center

3
2
;2

, radius 3, see Figure 1.19.
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DEPED COPY Figure 1.17 Figure 1.18 Figure 1.19
3. A circular play area with radius 3 m is
to be partitioned into two sections using
a straight fence as shown in Figure 1.20.
How long should the fence be?
Solution:To determine the length of the
fence, we need to determine the coordi-
nates of its endpoints. From Figure 1.20,
the endpoints havexcoordinate1 and
are on the circlex
2
+y
2
= 9. Then
1 +y
2
= 9, ory=2
p
2. Therefore,
the length of the fence is 4
p
25:66 m.
Figure 1.20
4. A Cartesian coordinate system was used to identify locations on a circu-
lar track. As shown in Figure 1.21, the circular track contains the points
A(2; 4),B(2;3),C(5;2). Find the total length of the track.
Figure 1.21 Figure 1.22
Solution:The segmentABis vertical and has midpoint (2;0:5), so its
perpendicular bisector has equationy=0:5. On the other hand, the segment
BChas slope1=7 and midpoint (1:5;2:5), so its perpendicular bisector has
equationy2:5 = 7(x1:5), or 7x y8 = 0.
The center of the circle is the intersection ofy=0:5 and 7xy8 = 0;
that is, the center is at

15
14
;
1
2

.
The radius of the circle is the distance from the center to any of the pointsA,
B, orC; by the distance formula, the radius is
r
2125
98
=
5
14
p
170. Therefore,
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DEPED COPY the total length of the track (its circumference), is
2
5
14
p
17029:26 units:
Supplementary Problems 1.1
Identify the center and radius of the circle with the given equation in each item.
Sketch its graph, and indicate the center.
1.x
2
+y
2
=
1
4
2. 5x
2
+ 5y
2
= 125
3. (x+ 4)
2
+

y
3
4

2
= 1
4.x
2
4x+y
2
4y8 = 0
5.x
2
+y
2
14x+ 12y = 36
6.x
2
+ 10x +y
2
16y11 = 0
7. 9x
2
+ 36x + 9y
2
+ 72y + 155 = 0
8. 9x
2
+ 9y
2
6x+ 24y = 19
9. 16x
2
+ 80x + 16y
2
112y+ 247 = 0
Find the standard equation of the circle which satises the given conditions.
10. center at the origin, radius 5
p
3
11. center at (17;5), radius 12
12. center at (8;4), contains (4; 2)
13. center at (15;7), tangent to thex-axis
14. center at (15;7), tangent to they-axis
15. center at (15;7), tangent to the liney=10
16. center at (15;7), tangent to the linex= 8
17. has a diameter with endpoints (3; 1) and (7;6)
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DEPED COPY 18. has a diameter with endpoints

9
2
;4

and

3
2
;2

19. concentric withx
2
+ 20x +y
2
14y+ 145 = 0, diameter 12
20. concentric withx
2
2x+y
2
2y23 = 0 and has 1=5 the area
21. concentric withx
2
+ 4x+y
2
6y+ 9 = 0 and has the same circumference as
x
2
+ 14x +y
2
+ 10y + 62 = 0
22. contains the points (3; 3), (7;1), (0;2)
23. contains the points (1; 4), (1; 2), (4;3)
24. center at (3;2) and tangent to the line 2x 3y= 1
25. center at (5;1) and tangent to the linex+y+ 10 = 0
26. has center withx-coordinate 4 and tangent to the linex+ 3y= 9 at (3; 4)
27. A stadium is shaped as in Figure 1.23, where its left and right ends are circular
arcs both with center atC. What is the length of the stadium 50 m from one
of the straight sides?
Figure 1.23
28.
A waterway in a theme park has a
semicircular cross section with di-
ameter 11 ft. The boats that are
going to be used in this waterway
have rectangular cross sections and
are found to submerge 1 ft into the
water. If the waterway is to be
lled with water 4.5 ft deep, what is
the maximum possible width of the
boats?
Figure 1.24
4
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DEPED COPY Lesson 1.2. Parabolas
Learning Outcomes of the Lesson
At the end of the lesson, the student is able to:
(1) dene a parabola;
(2) determine the standard form of equation of a parabola;
(3) graph a parabola in a rectangular coordinate system; and
(4) solve situational problems involving conic sections (parabolas).
Lesson Outline
(1) Denition of a parabola
(2) Derivation of the standard equation of a parabola
(3) Graphing parabolas
(4) Solving situational problems involving parabolas
Introduction
A parabola is one of the conic sections. We have already seen parabolas which
open upward or downward, as graphs of quadratic functions. Here, we will see
parabolas opening to the left or right. Applications of parabolas are presented
at the end.
1.2.1. Denition and Equation of a Parabola
Consider the pointF(0;2) and the line`having equationy=2, as shown in
Figure 1.25. What are the distances ofA(4;2) fromFand from`? (The latter
is taken as the distance ofAfromA`, the point on`closest toA). How about
the distances ofB(8;8) fromFand from`(fromB`)?
AF= 4 and AA`= 4
BF=
p
(80)
2
+ (82)
2
= 10 and BB`= 10
There are other pointsPsuch thatPF=PP`(whereP`is the closest point on
line`). The collection of all such points forms a shape called aparabola.
LetFbe a given point, and`a given line not containingF. The set of
all pointsPsuch that its distances fromFand from`are the same, is
called aparabola. The point Fis itsfocusand the line`itsdirectrix.
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DEPED COPY Figure 1.25
Figure 1.26
Consider
a parabola with focusF(0; c) and directrix`having equationy=c.
See Figure 1.26. The focus and directrix arecunits above and below, respectively,
the origin. LetP(x; y) be a point on the parabola soPF=PP`, whereP`is the
point on`closest toP. The pointPhas to be on the same side of the directrix
as the focus (ifPwas below, it would be closer to`than it is fromF).
PF=PP`
p
x
2
+
y c)
2
=y(c) =y+c
x
2
+y
2
2cy+c
2
=y
2
+ 2cy+c
2
x
2
= 4cy
The vertexVis the point midway between the focus and the directrix. This
equation,x
2
= 4cy, is then the standard equation of a parabola opening upward
with vertexV(0;0).
Suppose the focus isF(0;c) and the directrix isy=c. In this case, a
pointPon the resulting parabola would be below the directrix (just like the
focus). Instead of opening upward, it will open downward. Consequently,PF=
p
x
2
+
y +c)
2
andPP`=cy(you may draw a version of Figure 1.26 for
this case). Computations similar to the one done above will lead to the equation
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DEPED COPY x
2
=4cy.
We collect here the features of the graph of a parabola with standard equation
x
2
= 4cyorx
2
=4cy, wherec >0.
(1)vertex:
originV(0;0)
If the parabola opens upward, the vertex is the lowest point. If the
parabola opens downward, the vertex is the highest point.
(2)directrix: the liney=cory=c
The directrix iscunits below or above the vertex.
(3)focus:F(0; c) orF(0;c)
The focus iscunits above or below the vertex.
Any point on the parabola has the same distance from the focus as it
has from the directrix.
(4)axis of symmetry:x= 0 (they-axis)
This line divides the parabola into two parts which are mirror images
of each other.
Example 1.2.1.Determine the focus and directrix of the parabola with the
given equation. Sketch the graph, and indicate the focus, directrix, vertex, and
axis of symmetry.
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DEPED COPY (1)x
2
= 12y (2)x
2
=6y
Solution.(1) The vertex isV(0;0) and the parabola opens upward. From 4c =
12,c= 3. The focus,c= 3 units above the vertex, isF(0;3). The directrix,
3 units below the vertex, isy=3. The axis of symmetry isx= 0.
(2) The vertex isV(0
;0) and the parabola opens downward. From 4c = 6,c=
3
2
.
The
focus,c=
3
2
units
below the vertex, isF

0;
3
2

.
The directrix,
3
2
units
ab
ove the vertex, isy=
3
2
.
The axis of symmetry isx= 0.
Example 1.2.2.What
is the standard equation of the parabola in Figure 1.25?
Solution.From the gure, we deduce thatc= 2. The equation is thusx
2
=
8y. 2
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DEPED COPY 1.2.2. More Properties of Parabolas
The parabolas we considered so far are \vertical" and have their vertices at the
origin. Some parabolas open instead horizontally (to the left or right), and some
have vertices not at the origin. Their standard equations and properties are given
in the box. The corresponding computations are more involved, but are similar
to the one above, and so are not shown anymore.
In all four cases below, we assume thatc >0. The vertex isV(h; k), and it
lies between the focusFand the directrix`. The focusFiscunits away from
the vertexV, and the directrix iscunits away from the vertex. Recall that, for
any point on the parabola, its distance from the focus is the same as its distance
from the directrix.
(xh
)
2
= 4c(yk) (yk)
2
= 4c(xh)
(xh
)
2
=4c(y k) (yk)
2
=4c(x h)
directrix`
: horizontal directrix`: vertical
axis of symmetry:x
=h, vertical axis of symmetry:y=k, horizontal
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DEPED COPY Note the following observations:
The equations are in terms ofxhandyk: the vertex coordinates are
subtracted from the corresponding variable. Thus, replacing bothhandk
with 0 would yield the case where the vertex is the origin. For instance, this
replacement applied to (x h)
2
= 4c(yk) (parabola opening upward) would
yieldx
2
= 4cy, the rst standard equation we encountered (parabola opening
upward, vertex at the origin).
If thex-part is squared, the parabola is \vertical"; if they-part is squared,
the parabola is \horizontal." In a horizontal parabola, the focus is on the left
or right of the vertex, and the directrix is vertical.
If the coecient of the linear (non-squared) part is positive, the parabola
opens upward or to the right; if negative, downward or to the left.
Example 1.2.3.Figure 1.27 shows the graph of parabola, with only its focus
and vertex indicated. Find its standard equation. What is its directrix and its
axis of symmetry?
Solution.The vertex isV(5;4) and the focus isF(3;4). From these, we
deduce the following:h= 5,k=4,c= 2 (the distance of the focus from the
vertex). Since the parabola opens to the left, we use the template (yk)
2
=
4c(x h). Our equation is
(y+ 4)
2
=8(x5):
Its directrix isc= 2 units to the right ofV, which isx= 7. Its axis is the
horizontal line throughV:y=4.
Figure 1.27
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DEPED COPY The standard equation (y + 4)
2
=8(x5) from the preceding example can
be rewritten asy
2
+ 8x+ 8y24 = 0, an equation of the parabola in general
form.
If the equation is given in the general formAx
2
+Cx+Dy+E= 0 (AandC
are nonzero) orBy
2
+Cx+Dy+E= 0 (BandCare nonzero), we can determine
the standard form by completing the square in both variables.
Example 1.2.4.Determine the vertex, focus, directrix, and axis of symmetry
of the parabola with the given equation. Sketch the parabola, and include these
points and lines.
(1)y
2
5x+ 12y =16
(2) 5x
2
+ 30x + 24y = 51
Solution.(1) We complete the square ony, and movexto the other side.
y
2
+ 12y = 5x16
y
2
+ 12y + 36 = 5x16 + 36 = 5x+ 20
(y+ 6)
2
= 5(x + 4)
The parabola opens to the right. It has vertexV(4;6). From 4c = 5, we
getc=
5
4
= 1:25. The focus isc= 1:25 units to the right ofV:F(2:75; 6).
The (vertical) directrix isc= 1:25 units to the left ofV:x=5:25. The
(horizontal) axis is throughV:y=6.
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DEPED COPY (2) We complete the square onx, and moveyto the other side.
5x
2
+ 30x =24y+ 51
5(x
2
+ 6x+ 9) =24y+ 51 + 5(9)
5(x+ 3)
2
=24y+ 96 =24(y 4)
(x+ 3)
2
=
24
5
(y4)
In the last line, we divided by 5 for the squared part not to have any coe-
cient. The parabola opens downward. It has vertexV(3;4).
From 4c =
24
5
, we getc=
6
5
= 1:2. The focus isc= 1:2 units belowV:
F(3;2:8). The (horizontal) directrix isc= 1:2 units aboveV:y= 5:2. The
(vertical) axis is throughV:x=3.
Example 1.2.5.A
parabola has focusF(7;9) and directrixy= 3. Find its
standard equation.
Solution.The directrix is horizontal, and the focus is above it. The parabola
then opens upward and its standard equation has the form (xh)
2
= 4c(yk).
Since the distance from the focus to the directrix is 2c = 93 = 6, thenc= 3.
Thus, the vertex isV(7;6), the point 3 units belowF. The standard equation is
then (x 7)
2
= 12(y6). 2
1.2.3. Situational Problems Involving Parabolas
Let us now solve some situational problems involving parabolas.
Example 1.2.6.A satellite dish has a shape called a paraboloid, where each
cross-section is a parabola. Since radio signals (parallel to the axis) will bounce
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DEPED COPY o the surface of the dish to the focus, the receiver should be placed at the focus.
How far should the receiver be from the vertex, if the dish is 12 ft across, and 4:5
ft deep at the vertex?
Solution.The
second gure above shows a cross-section of the satellite dish drawn
on a rectangular coordinate system, with the vertex at the origin. From the problem, we deduce that (6; 4:5) is a point on the parabola. We need the distance
of the focus from the vertex, i.e., the value ofcinx
2
= 4cy.
x
2
= 4cy
6
2
= 4c(4:5)
c=
6
2
44
:5
= 2
Thus, the receiver should be 2 ft away from the vertex. 2
Example 1.2.7.The cable of a suspension bridge hangs in the shape of a
parabola. The towers supporting the cable are 400 ft apart and 150 ft high. If the cable, at its lowest, is 30 ft above the bridge at its midpoint, how high is the cable 50 ft away (horizontally) from either tower?
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DEPED COPY Solution.Refer to the gure above, where the parabolic cable is drawn with
its vertex on they-axis 30 ft above the origin. We may write its equation as
(x0)
2
=a(y30); since we don't need the focal distance, we use the simpler
variableain place of 4c. Since the towers are 150 ft high and 400 ft apart, we
deduce from the gure that (200;150) is a point on the parabola.
x
2
=a(y30)
200
2
=a(15030)
a=
200
2
120
=
1000
3
The parabola has equationx
2
=
1000
3
(y30), or equivalently,
y= 0:003x
2
+ 30. For the two points on the parabola 50 ft away from the
towers,x= 150 orx=150. Ifx= 150, then
y= 0:003(150
2
) + 30 = 97:5:
Thus, the cable is 97:5 ft high 50 ft away from either tower. (As expected, we
get the same answer fromx=150.) 2
More Solved Examples
For Examples 1 and 2, determine the focus and directrix of the parabola with the
given equation. Sketch the graph, and indicate the focus, directrix, and vertex.
1.y
2
= 20x
Solution:
Vertex:V(0;0), opens to the right
4c= 20)c= 5
Focus:F(5;0), Directrix:x=5
See Figure 1.28.
2. 3x
2
=12y
Solution:3x
2
=12y ,x
2
=4y
Vertex:V(0;0), opens downward
4c= 4)c= 1
Focus:F(0;1), Directrix:y= 1
See Figure 1.29.
Figure 1.28 Figure 1.29
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DEPED COPY 3. Determine the standard equation of the
parabola in Figure 1.30 given only its
vertex and focus. Then determine its di-
rectix and axis of symmetry.
Solution:
V

3
2
;4

; F(4;4)
c=
5
2
)4c= 10
Parabola opens to the left
Equation: (y 4)
2
=10

x+
3
2

Directrix:x= 1, Axis:y= 4
Figure 1.30
4. Determine the standard equation of the
parabola in Figure 1.31 given only its
vertex and diretrix. Then determine its
focus and axis of symmetry.
Solution:
V

5;
13
2

, directrix:y=
15
2
c= 1)4c= 4
Parabola opens downward
Equation:

y
13
2

2
=4 (x5)
Focus:

5;
11
2

, Axis:x= 5
Figure 1.31
For Examples 5 and 6, determine the vertex, focus, directrix, and axis of sym-
metry of the parabola with the given equation. Sketch the parabola, and include
these points and lines.
5.x
2
6x2y+ 9 = 0
Solution:
x
2
6x= 2y9
x
2
6x+ 9 = 2y
(x3)
2
= 2y
V(3;0), parabola opens upward
4c= 2)c=
1
2
,F

3;
1
2

,
directrix:y=
1
2
, axis:x= 3
See Figure 1.32.
Figure 1.32
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DEPED COPY 6. 3y
2
+ 8x+ 24y + 40 = 0
Solution:
3y
2
+ 24y =8x40
3(y
2
+ 8y) =8x40
3(y
2
+ 8y+ 16) =8x40 + 48
3(y+ 4)
2
=8x+ 8
(y+ 4)
2
=
8
3
(x1)
V(1;4), parabola opens to the left
4c=
8
3
)c=
2
3
,F

1
3
;4

,
directrix:x=
5
3
, axis:y=4
See Figure 1.33.
Figure 1.33
7. A parabola has focusF(11; 8) and directrixx=17. Find its standard
equation.
Solution:Since the focus is 6 units to the right of the directrix, the parabola
opens to the right with 2c = 6. Thenc= 3 andV(14; 8). Hence, the
equation is (y 8)
2
= 12(x+ 14).
8. A ashlight is shaped like a
paraboloid and the light source
is placed at the focus so that
the light bounces o parallel to
the axis of symmetry; this is
done to maximize illumination.
A particular ashlight has its
light source located 1 cm from
the base and is 6 cm deep; see
Figure 1.34. What is the width
of the ashlight's opening?
Figure 1.34
Solution:Let the base (the vertex) of the ashlight be the pointV(0;0).
Then the light source (the focus) is atF(0;1); soc= 1. Hence, the parabola's
equation isx
2
= 4y. To get the width of the opening, we need thexcoordinates
of the points on the parabola withycoordinate 6.
x
2
= 4(6))x=2
p
6
Therefore, the width of the opening is 22
p
6 = 4
p
69:8 cm.
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DEPED COPY 9. An object thrown from a height of 2
m above the ground follows a parabolic
path until the object falls to the ground;
see Figure 1.35. If the object reaches
a maximum height (measured from the
ground) of 7 m after travelling a hor-
izontal distance of 4 m, determine the
horizontal distance between the object's
initial and nal positions.
Figure 1.35
Solution:LetV(0;7) be the parabola's vertex, which corresponds to the high-
est point reached by the object. Then the parabola's equation is of the form
x
2
=4c(y 7) and the object's starting point is at (4;2). Then
(4)
2
=4c(27))c=
16
20
=
4
5
:
Hence, the equation of the parabola isx
2
=
16
5
(y7). When the object hits
the ground, theycoordinate is 0 and
x
2
=
16
5
(07) =
112
5
)x=4
r
7
5
:
Since this point is to the right of the vertex, we choosex= +4
r
7
5
. Therefore,
the total distance travelled is 4
r
7
5
(4)8:73 m.
Supplementary Problems 1.2
Determine the vertex, focus, directrix, and axis of symmetry of the parabola with
the given equation. Sketch the graph, and include these points and lines.
1.y
2
=36x
2. 5x
2
= 100y
3.y
2
+ 4x14y=53
4.y
2
2x+ 2y1 = 0
5. 2x
2
12x+ 28y = 38
6. (3x 2)
2
= 84y 112
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DEPED COPY Find the standard equation of the parabola which satises the given conditions.
7. vertex (7;11), focus (16; 11)
8. vertex (10;5), directrixy=1
9. focus

10;
23
2

, directrixy=
11
2
10. focus

3
2
;3

, directrixx=
37
2
11. axis of symmetryy= 9, directrixx= 24, vertex on the line 3y5x= 7
12. vertex (0; 7), vertical axis of symmetry, through the pointP(4;5)
13. vertex (3;8), horizontal axis of symmetry, through the pointP(5;12)
14. A satellite dish shaped like a paraboloid has its receiver located at the focus.
How far is the receiver from the vertex if the dish is 10 ft across and 3 ft deep
at the center?
15. A ashlight shaped like a paraboloid has its light source at the focus located
1.5 cm from the base and is 10 cm wide at its opening. How deep is the
ashlight at its center?
16. The ends of a rope are held in place at the top of two posts, 9 m apart and
each one 8 m high. If the rope assumes a parabolic shape and touches the
ground midway between the two posts, how high is the rope 2 m from one of
the posts?
17. Radiation is focused to an unhealthy area in a patient's body using a parabolic
reector, positioned in such a way that the target area is at the focus. If the
reector is 30 cm wide and 15 cm deep at the center, how far should the base
of the reector be from the target area?
18. A rectangular object 25 m wide is to pass under a parabolic arch that has a
width of 32 m at the base and a height of 24 m at the center. If the vertex
of the parabola is at the top of the arch, what maximum height should the
rectangular object have?
4
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DEPED COPY Lesson 1.3. Ellipses
Learning Outcomes of the Lesson
At the end of the lesson, the student is able to:
(1) dene an ellipse;
(2) determine the standard form of equation of an ellipse;
(3) graph an ellipse in a rectangular coordinate system; and
(4) solve situational problems involving conic sections (ellipses).
Lesson Outline
(1) Denition of an ellipse
(2) Derivation of the standard equation of an ellipse
(3) Graphing ellipses
(4) Solving situational problems involving ellipses
Introduction
Unlike circle and parabola, an ellipse is one of the conic sections that most stu-
dents have not encountered formally before. Its shape is a bounded curve which
looks like a attened circle. The orbits of the planets in our solar system around
the sun happen to be elliptical in shape. Also, just like parabolas, ellipses have
reective properties that have been used in the construction of certain structures.
These applications and more will be encountered in this lesson.
1.3.1. Denition and Equation of an Ellipse
Consider the pointsF1(3;0) andF2(3;0), as shown in Figure 1.36. What is the
sum of the distances ofA(4;2:4) fromF1and fromF2? How about the sum of
the distances ofB(andC(0;4)) fromF1and fromF2?
AF1+AF2= 7:4 + 2:6 = 10
BF1+BF2= 3:8 + 6:2 = 10
CF1+CF2= 5 + 5 = 10
There are other pointsPsuch thatPF1+PF2= 10. The collection of all such
points forms a shape called anellipse.
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DEPED COPY Figure 1.36
Figure 1.37
LetF1andF2be two distinct points. The set of all pointsP, whose
distances fromF1and fromF2add up to a certain constant, is called
anellipse. The points F1andF2are called thefociof the ellipse.
Given are two points on thex-axis,F1(c;0) andF2(c;0), the foci, bothc
units away from their center (0;0). See Figure 1.37. LetP(x; y) be a point on
the ellipse. Let the common sum of the distances be 2a(the coecient 2 will
make computations simpler). Thus, we havePF1+PF2= 2a.
PF1= 2aPF2
p
(x+c)
2
+y
2
= 2a
p
(xc)
2
+y
2
x
2
+ 2cx+c
2
+y
2
= 4a
2
4a
p
(xc)
2
+y
2
+x
2
2cx+c
2
+y
2
a
p
(xc)
2
+y
2
=a
2
cx
a
2

x
2
2cx+c
2
+y
2

=a
4
2a
2
cx+c
2
x
2
(a
2
c
2
)x
2
+a
2
y
2
=a
4
a
2
c
2
=a
2
(a
2
c
2
)
b
2
x
2
+a
2
y
2
=a
2
b
2
by lettingb=
p
a
2
c
2
, soa > b
x
2
a
2
+
y
2
b
2
= 1
When we letb=
p
a
2
c
2
, we assumeda > c. To see why this is true, look at
4PF1F2in Figure 1.37. By the Triangle Inequality,PF1+PF2> F1F2, which
implies 2a > 2c, soa > c.
We collect here the features of the graph of an ellipse with standard equation
x
2
a
2
+
y
2
b
2
= 1, wherea > b. Let c=
p
a
2
b
2
.
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DEPED COPY (1)c
: origin (0;0)
(2)foci:F1(c;0) andF2(c;0)
Each focus iscunits away from the center.
For any point on the ellipse, the sum of its distances from the foci is 2a.
(3)vertices:V1(a;0) andV2(a;0)
The vertices are points on the ellipse, collinear with the center and foci.
Ify= 0, thenx=a. Each vertex isaunits away from the center.
The segmentV1V2is called themajor axis. Its length is 2a. It divides
the ellipse into two congruent parts.
(4)covertices:W1(0;b) andW2(0; b)
The segment through the center, perpendicular to the major axis, is the
minor axis. It meets the ellipse at the covertices. It divides the ellipse
into two congruent parts.
Ifx= 0, theny=b. Each covertex isbunits away from the center.
The minor axisW1W2is 2bunits long. Sincea > b, the major axis is
longer than the minor axis.
Example 1.3.1.Give the coordinates of the foci, vertices, and covertices of the
ellipse with equation
x
2
25
+
y
2
9
=
1:
Sketch the graph, and include these points.
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DEPED COPY Solution.Witha
2
= 25 andb
2
= 9, we havea= 5,b= 3, andc=
p
a
2
b
2
= 4.
foci:F1(4;0); F2(4;0) vertices:V1(5;0); V2(5;0)
covertices:W1(0;3); W 2(0;3)
Example 1.3.2.Find
the (standard) equation of the ellipse whose foci are
F1(3;0) andF2(3;0), such that for any point on it, the sum of its distances
from the foci is 10. See Figure 1.36.
Solution.We have 2a = 10 andc= 3, soa= 5 andb=
p
a
2
c
2
=

equation is
x
2
25
+
y
2
16
=
1: 2
1.3.2. More Properties of Ellipses
The ellipses we have considered so far are \horizontal" and have the origin as their
centers. Some ellipses have their foci aligned vertically, and some have centers
not at the origin. Their standard equations and properties are given in the box.
The derivations are more involved, but are similar to the one above, and so are
not shown anymore.
In all four cases below,a > bandc=
p
a
2
b
2
.
F1andF2arec
units away from the center. The verticesV1andV2areaunits away from the
center, the major axis has length 2a, the coverticesW1andW2arebunits away
from the center, and the minor axis has length 2b. Recall that, for any point on
the ellipse, the sum of its distances from the foci is 2a.
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DEPED COPY Center Corresponding Graphs
(0;0)
x
2
a
2
+
y
2
b
2
=
1,a > b
x
2
b
2
+
y
2
a
2
=
1,b > a
(h; k)
(xh
)
2
a
2
+
(
yk)
2
b
2
=
1
(xh)
2
b
2
+
(
yk)
2
a
2
=
1
a > b b
> a
major axis: horizontal major axis: vertical
minor axis: vertical minor axis: horizontal
In the standard equation, if thex
-part has the bigger denominator, the ellipse
is horizontal. If they-part has the bigger denominator, the ellipse is vertical.
Example 1.3.3.Give the coordinates of the center, foci, vertices, and covertices
of the ellipse with the given equation. Sketch the graph, and include these points.
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DEPED COPY (1)
(x+ 3)
2
24
+
(y5)
2
49
= 1
(2) 9x
2
+ 16y
2
126
x+ 64y = 71
Solution.(1) Froma
2
= 49 andb
2
= 24, we havea= 7,b= 2
p
64:9, and
c=
p
a
2
b
2
= 5. The ellipse is vertical.
center: (3; 5)
foci:F1(3;0),F2(3;10)
vertices:V1(3;2),V2(3;12)
covertices:W1(32
p
6;5)(7:9; 5)
W2(3 + 2
p
6;5)(1:9;5)
(2) We rst change the given equation to standard form.
9(
x
2
14x) + 16(y
2
+ 4y) = 71
9(x
2
14x+ 49) + 16(y
2
+ 4y+ 4) = 71 + 9(49) + 16(4)
9(x7)
2
+ 16(y + 2)
2
= 576
(x7)
2
64
+
(
y+ 2)
2
36
=
1
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DEPED COPY We havea= 8 andb= 6. Thus,c=
p
a
2
b
2
= 2
p
75:3. The ellipse is
horizontal.
center: (7; 2)
foci:F1(72
p
7;2)(1:7;2)
F2(7 + 2
p
7;2)(12:3; 2)
vertices:V1(1;2),V2(15;2)
covertices:W1(7;8),W2(7;4)
Example 1.3.4.The
foci of an ellipse are (3;6) and (3; 2). For any point
on the ellipse, the sum of its distances from the foci is 14. Find the standard
equation of the ellipse.
Solution.The midpoint (3; 2) of the foci is the center of the ellipse. The
ellipse is vertical (because the foci are vertically aligned) andc= 4. From the
given sum, 2a = 14 soa= 7. Also,b=
p
a
2
c
2
=
p
33. The equation is
(
x+ 3)
2
33
+
(
y+ 2)
2
49
=
1. 2
Example 1.3.5.An ellipse has vertices (2
p
61;
5) and (2 +
p
61;
5), and
its minor axis is 12 units long. Find its standard equation and its foci.
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DEPED COPY Solution.The midpoint (2; 5) of the vertices is the center of the ellipse, which is
horizontal. Each vertex isa=
p
61 units away from the center. From the length of
the minor axis, 2b = 12 sob= 6. The standard equation is
(x2)
2
61
+
(y+ 5)
2
36
=
1. Each focus isc=
p
a
2
b
2
= 5 units away from (2;5), so their coordinates
are (3; 5) and (7; 5). 2
1.3.3. Situational Problems Involving Ellipses
Let us now apply the concept of ellipse to some situational problems.
?
Example 1.3.6.A tunnel has the shape of a semiellipse that is 15 ft high at
the center, and 36 ft across at the base. At most how high should a passing truck
be, if it is 12 ft wide, for it to be able to t through the tunnel? Round o your
answer to two decimal places.
Solution.Refer
to the gure above. If we draw the semiellipse on a rectangular
coordinate system, with its center at the origin, an equation of the ellipse which contains it, is
x
2
18
2
+
y
2
15
2
=
1:
To maximize its height, the corners of the truck, as shown in the gure, would have to just touch the ellipse. Since the truck is 12 ft wide, let the point (6; n) be the corner of the truck in the rst quadrant, wheren >0, is the (maximum)
height of the truck. Since this point is on the ellipse, it should t the equation.
Thus, we have
6
2
18
2
+
n
2
15
2
=
1
n= 10
p
214
:14 ft 2
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DEPED COPY Example 1.3.7.The orbit of a planet has the shape of an ellipse, and on one
of the foci is the star around which it revolves. The planet is closest to the star
when it is at one vertex. It is farthest from the star when it is at the other vertex.
Suppose the closest and farthest distances of the planet from this star, are 420
million kilometers and 580 million kilometers, respectively. Find the equation of
the ellipse, in standard form, with center at the origin and the star at thex-axis.
Assume all units are in millions of kilometers.
Solution.In
the gure above, the orbit is drawn as a horizontal ellipse with
center at the origin. From the planet's distances from the star, at its closest and farthest points, it follows that the major axis is 2a= 420 + 580 = 1000
(million kilometers), soa= 500. If we place the star at the positivex-axis,
then it isc= 500420 = 80 units away from the center. Therefore, we get
b
2
=a
2
c
2
= 500
2
80
2
= 243600. The equation then is
x
2
250000
+
y
2
243600
=
1:
The star could have been placed on the negativex-axis, and the answer would
still be the same. 2
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DEPED COPY More Solved Examples
1. Give the coordinates of the foci, vertices,
and covertices of the ellipse with equa-
tion
x
2
169
+
y
2
144
= 1. Then sketch the
graph and include these points.
Solution:The ellipse is horizontal.
a
2
= 169)a= 13,b
2
= 144)b= 12,
c=
p
169144 = 5
Foci:F1(5;0),F2(5;0)
Vertices:V1(13; 0),V2(13;0)
Covertices:W1(0;12),W2(0;12)
See Figure 1.38.
Figure 1.38
2.
Find the standard equation of the ellipse whose foci areF1(0;8) andF2(0;8),
such that for any point on it, the sum of its distances from the foci is 34.
Solution:The ellipse is vertical and has center at (0; 0).
2a= 34)a= 17
c= 8)b=
p
17
2
8
2
=

The equation is
x
2
225
+
y
2
289
=
1.
For Examples 3 and 4, give the coordinates of the center, foci, vertices, and
covertices of the ellipse with the given equation. Sketch the graph, and include
these points.
3.
(x7)
2
64
+
(
y+ 2)
2
25
=
1
Solution:The ellipse is horizontal.
a
2
= 64)a= 8,b
2
= 25)b= 5
c=
p
6425

p
396
:24
center: (7; 2)
foci:F1(7
p
39;
2)(0:76; 2)
F2(7 +
p
39;
2)(13:24; 2)
vertices:V1(1;2); V2(15;2)
covertices:W1(7;7); W 2(7;3)
See Figure 1.39.
Figure 1.39
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DEPED COPY 4. 16x
2
+ 96x + 7y
2
+ 14y + 39 = 0
Solution:
16x
2
+ 96x + 7y
2
+ 14y =39
16(x
2
+ 6x+ 9) + 7(y
2
+ 2y+ 1) =39 + 151
16(x+ 3)
2
+ 7(y + 1)
2
= 112
(x+ 3)
2
7
+
(y+ 1)
2
16
= 1
The ellipse is vertical.
a
2
= 16)a= 4,b
2
= 7)b=
p
7
2:65
c=
p
167 = 3
center: (3; 1)
foci:F1(3;4),F2(3;2)
vertices:V1(3;5); V2(3;3)
covertices:W1(3
p
7;1)(5:65; 1)
W2(3 +
p
7;1)(0:35; 1)
See Figure 1.40.
Figure 1.40
5. The covertices of an ellipse are (5;6) and (5; 8). For any point on the ellipse,
the sum of its distances from the foci is 12. Find the standard equation of the
ellipse.
Solution:The ellipse is horizontal with center at the midpoint (5;7) of the
covertices. Also, 2a = 12 soa= 6 whileb= 1. The equation is
(x5)
2
36
+
(y7)
2
1
= 1.
6. An ellipse has foci (4
p
15;3) and (4 +
p
15;3), and its major axis is 10
units long. Find its standard equation and its vertices.
Solution:The ellipse is horizontal with center at the midpoint (4; 3) of the
foci; alsoc=
p
15. Since the length of the major axis is 10, 2a= 10 and
a= 5. Thusb=
p
5
2
15 =
p
10. Therefore, the equation of the ellipse is
(x+ 4)
2
25
+
(y3)
2
10
= 1 and its vertices are (9;3) and (1; 3).
7. A whispering gallery is an enclosure or room where whispers can be clearly
heard in some parts of the gallery. Such a gallery can be constructed by
making its ceiling in the shape of a semi-ellipse; in this case, a whisper from
one focus can be clearly heard at the other focus. If an elliptical whispering
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DEPED COPY gallery is 90 feet long and the foci are 50 feet apart, how high is the gallery at
its center?
Solution:W
e set up a Cartesian coordinate system by assigning the center of
the semiellipse as the origin. The point on the ceiling right above the center is a covertex of the ellipse. Since 2a= 90 and 2c= 50; thenb
2
= 45
2
25
2
= 1400.
The height is given byb=
p
140037
:4 ft.
8. A spheroid (or oblate spheroid) is the surface obtained by rotating an ellipse
around its minor axis. The bowl in Figure 1.41 is in the shape of the lower half of a spheroid; that is, its horizontal cross sections are circles while its vertical cross sections that pass through the center are semiellipses. If this bowl is 10
in wide at the opening and
p
10 in deep at the center, how deep does a circular
co
ver with diameter 9 in go into the bowl?
Figure 1.41
Solution:W
e set up a Cartesian coordinate system by assigning the center
of the semiellipse as the origin. Thena= 5,b=
p
10, and the equation of
the
ellipse is
x
2
25
+
y
2
10
=
1. We want they-coordinate of the points on the
ellipse that hasx=4:5. This coordinate isy=
q
10

1
x
2
25

1:38.
Therefore, the cover will go 1.38 inches into the bowl.
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DEPED COPY Supplementary Problems 1.3
Give the coordinates of the center, foci, vertices, and covertices of the ellipse with
the given equation. Sketch the graph, and include these points.
1.
x
2
8
+
y
2
4
= 1
2.
x
2
16
+
(y2)
2
25
= 1
3. (x1)
2
+ (2y 2)
2
= 4
4.
(x+ 5)
2
49
+
(y2)
2
121
= 1
5. 16x
2
224x+ 25y
2
+ 250y 191 = 0
6. 25x
2
200x+ 16y
2
160y= 800
Find the standard equation of the ellipse which satises the given conditions.
7. foci (2
p
33;8) and (2 +
p
33;8), the sum of the distances of any point from
the foci is 14
8. center (3;7), vertical major axis of length 20, minor axis of length 12
9. foci (21;10) and (3; 10), contains the point (9;15)
10. a vertex at (3;18) and a covertex at (12;7), major axis is either hori-
zontal or vertical
11. a focus at (9; 15) and a covertex at (1; 10), with vertical major axis
12. A 40-ft wide tunnel has the shape of a semiellipse that is 5 ft high a distance
of 2 ft from either end. How high is the tunnel at its center?
13. The moon's orbit is an ellipse with Earth as one focus. If the maximum
distance from the moon to Earth is 405 500 km and the minimum distance is
363 300 km, nd the equation of the ellipse in a Cartesian coordinate system
where Earth is at the origin. Assume that the ellipse has horizontal major
axis and that the minimum distance is achieved when the moon is to the right
of Earth. Use 100 km as one unit.
14. Two friends visit a whispering gallery (in the shape of a semiellipsoid) where
they stand 100 m apart to be at the foci. If one of them is 6 m from the
nearest wall, how high is the gallery at its center?
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DEPED COPY 15. A jogging path is in the shape of an ellipse. If it is 120 ft long and 40 ft wide,
what is the width of the track 15 ft from either vertex?
16. Radiation is focused to an unhealthy area in a patient's body using a semiel-
liptic reector, positioned in such a way that the target area is at one focus
while the source of radiation is at the other. If the reector is 100 cm wide
and 30 cm high at the center, how far should the radiation source and the
target area be from the ends of the reector?
4
Lesson 1.4.Hyperbolas
Learning Outcomes of the Lesson
At the end of the lesson, the student is able to:
(1) dene a hyperbola;
(2) determine the standard form of equation of a hyperbola;
(3) graph a hyperbola in a rectangular coordinate system; and
(4) solve situational problems involving conic sections (hyperbolas).
Lesson Outline
(1) Denition of a hyperbola
(2) Derivation of the standard equation of a hyperbola
(3) Graphing hyperbolas
(4) Solving situational problems involving hyperbolas
Introduction
Just like ellipse, a hyperbola is one of the conic sections that most students
have not encountered formally before. Its graph consists of two unbounded
branches which extend in opposite directions. It is a misconception that each
branch is a parabola. This is not true, as parabolas and hyperbolas have very
dierent features. An application of hyperbolas in basic location and navigation
schemes are presented in an example and some exercises.
1.4.1. Denition and Equation of a Hyperbola
Consider the pointsF1(5;0) andF2(5;0) as shown in Figure 1.42. What is the
absolute value of the dierence of the distances ofA(3:75; 3) fromF1and from
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DEPED COPY F2? How about the absolute value of the dierence of the distances ofB

5;
16
3

fromF1and fromF2?
jAF1AF2j=j9:253:25j= 6
jBF1BF2j=

16
3

34
3

= 6
There are other pointsPsuch thatjPF1PF2j= 6. The collection of all such
points forms a shape called ahyperbola, which consists of two disjoint branches.
For pointsPon the left branch,PF2PF1= 6; for those on the right branch,
PF1PF2= 6.
Figure 1.42
Figure 1.43
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DEPED COPY LetF1andF2be two distinct points. The set of all pointsP, whose
distances fromF1and fromF2dier by a certain constant, is called a
hyperbola. The points F1andF2are called thefociof the hyperbola.
In Figure 1.43, given are two points on thex-axis,F1(c;0) andF2(c;0), the
foci, bothcunits away from their midpoint (0;0). This midpoint is thecenter
of the hyperbola. LetP(x; y) be a point on the hyperbola, and let the absolute
value of the dierence of the distances ofPfromF1andF2, be 2a (the coecient
2 will make computations simpler). Thus,jPF1PF2j= 2a, and so

p
(x+c)
2
+y
2

p
(xc)
2
+y
2
= 2a:
Algebraic manipulations allow us to rewrite this into the much simpler
x
2
a
2

y
2
b
2
= 1;whereb=
p
c
2
a
2
:
When we letb=
p
c
2
a
2
, we assumedc > a. To see why this is true, suppose
thatPis closer toF2, soPF1PF2= 2a. Refer to Figure 1.43. Suppose also
thatPis not on thex-axis, so4PF1F2is formed. From the triangle inequality,
F1F2+PF2> PF1. Thus, 2c > PF 1PF2= 2a, soc > a.
Now we present a derivation. For now, assumePis closer toF2soPF1> PF2,
andPF1PF2= 2a.
PF1= 2a+PF2
p
(x+c)
2
+y
2
= 2a+
p
(xc)
2
+y
2
p
(x+c)
2
+y
2

2
=

2a+
p
(xc)
2
+y
2

2
cxa
2
=a
p
(xc)
2
+y
2
(cxa
2
)
2
=

a
p
(xc)
2
+y
2

2
(c
2
a
2
)x
2
a
2
y
2
=a
2
(c
2
a
2
)
b
2
x
2
a
2
y
2
=a
2
b
2
by lettingb=
p
c
2
a
2
>0
x
2
a
2

y
2
b
2
= 1
We collect here the features of the graph of a hyperbola with standard equa-
tion
x
2
a
2

y
2
b
2
= 1:
Letc=
p
a
2
+b
2
.
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DEPED COPY Figure 1.44 Figure 1.45
(1)c
enter: origin (0;0)
(2)foci:F1(c;0) andF2(c;0)
Each focus iscunits away from the center.
For any point on the hyperbola, the absolute value of the dierence of
its distances from the foci is 2a.
(3)vertices:V1(a;0) andV2(a;0)
The vertices are points on the hyperbola, collinear with the center and
foci.
Ify= 0, thenx=a. Each vertex isaunits away from the center.
The segmentV1V2is called thetransverse axis. Its length is 2a.
(4)asymptotes:y=
b
a
xandy=
b
a
x
, the lines`1and`2in Figure 1.45
The asymptotes of the hyperbola are two lines passing through the cen- ter which serve as a guide in graphing the hyperbola: each branch of the hyperbola gets closer and closer to the asymptotes, in the direction towards which the branch extends. (We need the concept of limits from calculus to explain this.)
An aid in determining the equations of the asymptotes: in the standard equation, replace 1 by 0, and in the resulting equation
x
2
a
2
y
2
b
2=
0, solve
fory.
To help us sketch the asymptotes, we point out that the asymptotes `1and`2are the extended diagonals of theauxiliary rectangledrawn
in Figure 1.45. This rectangle has sides 2aand 2bwith its diagonals
intersecting at the centerC. Two sides are congruent and parallel to
the transverse axisV1V2. The other two sides are congruent and parallel
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DEPED COPY to theconjugate axis, the segment shown which is perpendicular to the
transverse axis at the center, and has length 2b.
Example 1.4.1.Determine the foci, vertices, and asymptotes of the hyperbola
with equation
x
2
9

y
2
7
= 1:
Sketch the graph, and include these points and lines, the transverse and conjugate
axes, and the auxiliary rectangle.
Solution.Witha
2
= 9 andb
2
= 7, we have
a= 3,b=
p
7, andc=
p
a
2
+b
2
= 4.
foci:F1(4;0) andF2(4;0)
vertices:V1(3;0) andV2(3;0)
asymptotes:y=
p
7
3
xandy=
p
7
3
x
The graph is shown at the right. The conju-
gate axis drawn has its endpointsb=
p
7
2:7 units above and below the center.2
Example 1.4.2.Find
the (standard) equation of the hyperbola whose foci are
F1(5;0) andF2(5;0), such that for any point on it, the absolute value of the
dierence of its distances from the foci is 6. See Figure 1.42.
Solution.We have 2a= 6 andc= 5, soa= 3 andb=
p
c
2
a
2
=

hyperbola then has equation
x
2
9

y
2
16
=
1. 2
1.4.2. More Properties of Hyperbolas
The hyperbolas we considered so far are \horizontal" and have the origin as their
centers. Some hyperbolas have their foci aligned vertically, and some have centers
not at the origin. Their standard equations and properties are given in the box.
The derivations are more involved, but are similar to the one above, and so are
not shown anymore.
In all four cases below, we letc=
p
a
2
+b
2
.
F1andF2arecunits
away from the centerC. The verticesV1andV2areaunits away from the center.
The transverse axisV1V2has length 2a. The conjugate axis has length 2b and is
perpendicular to the transverse axis. The transverse and conjugate axes bisect each other at their intersection point,C. Each branch of a hyperbola gets closer
and closer to the asymptotes, in the direction towards which the branch extends. The equations of the asymptotes can be determined by replacing 1 in the standard equation by 0. The asymptotes can be drawn as the extended diagonals of the auxiliary rectangle determined by the transverse and conjugate axes. Recall that,
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DEPED COPY for any point on the hyperbola, the absolute value of the dierence of its distances
from the foci is 2a.
Center Corresponding Hyperbola
(0;0)
x
2
a
2

y
2
b
2
=
1
y
2
a
2

x
2
b
2
=
1
(h; k)
(xh
)
2
a
2

(
yk)
2
b
2
=
1
(yk)
2
a
2

(
xh)
2
b
2
=
1
transverse axis: horizontal transverse axis: vertical
conjugate axis: vertical conjugate axis: horizontal
In the standard equation, aside from being positive, there are no other re-
strictions
onaandb. In fact,aandbcan even be equal. The orientation of the
hyperbola is determined by the variable appearing in the rst term (the positive term): the corresponding axis is where the two branches will open. For example, if the variable in the rst term isx, the hyperbola is \horizontal": the transverse
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DEPED COPY axis is horizontal, and the branches open to the left and right in the direction of
thex-axis.
Example 1.4.3.Give the coordinates of the center, foci, vertices, and asymp-
totes of the hyperbola with the given equation. Sketch the graph, and include
these points and lines, the transverse and conjugate axes, and the auxiliary rect-
angle.
(1)
(y+ 2)
2
25

(x7)
2
9
= 1
(2) 4x
2
5y
2
+ 32x + 30y = 1
Solution.(1) Froma
2
= 25 andb
2
= 9, we havea= 5,b= 3, andc=
p
a
2
+b
2
=
p
345:8. The hyperbola is vertical. To determine the asymp-
totes, we write
(y+2)
2
25

(x7)
2
9
= 0, which is equivalent toy+ 2 =
5
3
(x7).
We can then solve this fory.
center:C(7;2)
foci:F1(7;2
p
34)(7;7:8) andF2(7;2 +
p
34)(7;3:8)
vertices:V1(7;7) andV2(7;3)
asymptotes:y=
5
3
x
41
3
andy=
5
3
x+
29
3
The conjugate axis drawn has its endpointsb= 3 units to the left and right
of the center.
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DEPED COPY (2) We rst change the given equation to standard form.
4(x
2
+ 8x)5(y
2
6y) = 1
4(x
2
+ 8x+ 16)5(y
2
6y+ 9) = 1 + 4(16)5(9)
4(x+ 4)
2
5(y3)
2
= 20
(x+ 4)
2
5

(y3)
2
4
= 1
We havea=
p
52:2 andb= 2. Thus,c=
p
a
2
+b
2
= 3. The hyperbola
is horizontal. To determine the asymptotes, we write
(x+4)
2
5

(y3)
2
4
= 0
which is equivalent toy3 =
2
p
5
(x+ 4), and solve fory.
center:C(4;3)
foci:F1(7;3) andF2(1;3)
vertices:V1(4
p
5;3)(6:2; 3) andV2(4 +
p
5;3)(1:8; 3)
asymptotes:y=
2
p
5
x+
8
p
5
+ 3 andy=
2
p
5
x
8
p
5
+ 3
The conjugate axis drawn has its endpointsb= 2 units above and below
the center.
Example 1.4.4.The
foci of a hyperbola are (5;3) and (9; 3). For any point
on the hyperbola, the absolute value of the dierence of its of its distances from
the foci is 10. Find the standard equation of the hyperbola.
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DEPED COPY Solution.The midpoint (2; 3) of the foci is the center of the hyperbola. Each
focus isc= 7 units away from the center. From the given dierence, 2a= 10 so
a= 5. Also,b
2
=c
2
a
2
= 24. The hyperbola is horizontal (because the foci are
horizontally aligned), so the equation is
(x2)
2
25

(y+ 3)
2
24
= 1: 2
Example 1.4.5.A hyperbola has vertices (4; 5) and (4; 9), and one of its
foci is (4; 2
p
65). Find its standard equation.
Solution.The midpoint (4; 2) of the vertices is the center of the hyperbola,
which is vertical (because the vertices are vertically aligned). Each vertex is
a= 7 units away from the center. The given focus isc=
p
65 units away from
the center. Thus,b
2
=c
2
a
2
= 16, and the standard equation is
(y2)
2
49

(x+ 4)
2
16
= 1: 2
1.4.3. Situational Problems Involving Hyperbolas
Let us now give an example on an application of hyperbolas.
Example 1.4.6.An explosion was heard by two stations 1200 m apart, located
atF1(600; 0) andF2(600;0). If the explosion was heard inF1two seconds before
it was heard inF2, identify the possible locations of the explosion. Use 340 m/s
as the speed of sound.
Solution.Using the given speed of sound, we can deduce that the sound traveled
340(2) = 680 m farther in reachingF2than in reachingF1. This is then the
dierence of the distances of the explosion from the two stations. Thus, the
explosion is on a hyperbola with foci areF1andF2, on the branch closer toF1.
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DEPED COPY We havec= 600 and 2a= 680, soa= 340 andb
2
=c
2
a
2
= 244400.
The explosion could therefore be anywhere on the left branch of the hyperbola
x
2
115600

y
2
244400
= 1. 2
More Solved Examples
1. Determine the foci, vertices, and asymptotes of the hyperbola with equation
x
2
16

y
2
33
= 1. Sketch the graph, and include these points and lines, the
transverse and conjugate axes, and the auxiliary rectangle.
Solution:The hyperbola is horizontal.
a
2
= 16)a= 4,
b
2
= 33)b=
p
33,
c=
p
16 + 33 = 7
center: (0; 0)
foci:F1(7;0),F2(7;0)
vertices:V1(4;0),V2(4;0)
asymptotes:y=
p
33
4
x,y=
p
33
4
x
The conjugate axis has endpoints
(0;
p
33) and (0;
p
33). See Figure
1.46.
Figure 1.46
2. Find the standard equation of the hyperbola whose foci areF1(0;10) and
F2(0;10), such that for any point on it, the absolute value of the dierence of
its distances from the foci is 12.
Solution:The hyperbola is vertical and has center at (0; 0). We have 2a = 12,
soa= 6; also,c= 10. Thenb=
p
10
2
6
2
= 8. The equation is
y
2
36

x
2
64
= 1.
For Examples 3 and 4, give the coordinates of the center, foci, vertices, and
asymptotes of the hyperbola with the given equation. Sketch the graph, and in-
clude these points and lines, the transverse and conjugate axes, and the auxiliary
rectangle.
3.
(y+ 6)
2
25

(x4)
2
39
= 1
Solution:The hyperbola is vertical.
a
2
= 25)a= 5,b
2
= 39)b=
p
39,c=
p
25 + 39 = 8
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DEPED COPY center: (4; 6)
foci:F1(4;14),F2(4;2)
vertices:V1(4;11),V2(4;1)
asymptotes:
(y+ 6)
2
25

(x4)
2
39
= 0
,y+ 6 =
5
p
39
(x4)
The conjugate axis has endpointsb=
p
39 units to the left and to the right of
the center. See Figure 1.47.
Figure 1.47
4. 9x
2
+ 126x 16y
2
96y+ 153 = 0
Solution:
9x
2
+ 126x 16y
2
96y=153
9(x
2
+ 14x + 49)16(y
2
+ 6y+ 9) =153 + 9(49)16(9)
9(x+ 7)
2
16(y+ 3)
2
= 144
(x+ 7)
2
16

(y+ 3)
2
9
= 1
The hyperbola is horizontal.
a
2
= 16)a= 4; b
2
= 9)b= 3; c=
p
16 + 9 = 5
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DEPED COPY center: (7; 3)
foci:F1(12; 3),F2(2;3)
vertices:V1(11; 3),V2(3;3)
asymptotes:
(x+ 7)
2
16

(y+ 3)
2
9
= 0,y+ 3 =
3
4
(x+ 7)
The conjugate axis have endpoints (7; 6) and (7;0). See Figure 1.48.
Figure 1.48
5. The foci of a hyperbola are (17;3) and (3;3). For any point on the
hyperbola, the absolute value of the dierence of its distances from the foci is
14. Find the standard equation of the hyperbola.
Solution:The hyperbola is horizontal with center at the midpoint (7;3) of
the foci. Also, 2a = 14 soa= 7 whilec= 10. Thenb
2
= 10
2
7
2
= 51. The
equation is
(x+ 7)
2
49

(y+ 3)
2
51
= 1.
6. The auxiliary rectangle of a hyperbola has vertices (24; 15), (24; 9), (10;9),
and (10;15). Find the equation of the hyperbola if its conjugate axis is hor-
izontal.Solution:The hyperbola is vertical. Using the auxiliary rectangle's dimen-
sions, we see that the length of the transverse axis is 2a = 24 while the
length of the conjugate axis is 2b = 34. Thus,a= 12 andb= 17. The
hyperbola's vertices are the midpoints (7; 15) and (7;9) of the bottom
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DEPED COPY and top sides, respectively, of the auxiliary rectangle. Then the hyperbola's
center is (7;3), which is the midpoint of the vertices. The equation is
(y+ 3)
2
144

(x+ 7)
2
289
= 1.
7. Two LORAN (long range navigation) stationsAandBare situated along a
straight shore, whereAis 200 miles west ofB. These stations transmit radio
signals at a speed 186 miles per millisecond. The captain of a ship travellingon the open sea intends to enter a harbor that is located 40 miles east ofstationA.
Due to the its location, the harbor experiences a time dierence in receivingthe signals from both stations. The captain navigates the ship into the harborby following a path where the ship experiences the same time dierence as the
harbor.
(a) What time dierence between station signals should the captain be look-
ing for in order the ship to make a successful entry into the harbor?
(b) If the desired time dierence is achieved, determine the location of the
ship if it is 75 miles oshore.
Solution:
(a) LetHrepresent the harbor on the shoreline. Note thatBHAH= 160
40 = 120. The time dierence on the harbor is given by 1201860:645
milliseconds. This is the time dierence needed to be maintained in order
to for the ship to enter the harbor.
(b) Situate the stationsAandBon the Cartesian plane so thatA(100; 0)
andB(100;0). LetPrepresent the ship on the sea, which has coordinates
(h;75). SincePBPA= 120, then it should follow thath <0. More-
over,Pshould lie on the left branch of the hyperbola whose equation is
given by
x
2
a
2

y
2
b
2
= 1
where 2a = 120)a= 60, andb=
p
c
2
a
2
=
p
100
2
60
2
= 80.
Therefore,
h
2
60
2

75
2
80
2
= 1
h=
s

1 +
75
2
80
2

60
2
82:24
This means that the ship is around 17.76 miles to the east of stationA.
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DEPED COPY Supplementary Problems 1.4
Give the center, foci, vertices, and asymptotes of the hyperbola with the given
equation. Sketch the graph and the auxiliary rectangle, then include these points
and lines.
1.
x
2
100

y
2
81
= 1
2.yx=
1
y+x
3. 4x
2
15(y5)
2
= 60
4.
(y6)
2
64

(x8)
2
36
= 1
5. 9y
2
+ 54y 6x
2
36x27 = 0
6. 16x
2
+ 64x 105y
2
+ 840y 3296 = 0
Find the standard equation of the hyperbola which satises the given conditions.
7. foci (7; 17) and (7; 17), the absolute value of the dierence of the distances
of any point from the foci is 24
8. foci (3;2) and (15; 2), a vertex at (9; 2)
9. center (10; 4), one corner of auxiliary rectangle at (1; 12), with horizontal
transverse axis
10. asymptotesy=
71
3

4
3
xandy=
4
3
x
17
3
and a vertex at (17;9)
11. asymptotesy=
5
12
x+
19
3
andy=
5
12
x+
29
3
and a focus at (4;5)
12. horizontal conjugate axis, one corner of auxiliary rectangle at (3;8), and an
asymptote 4x + 3y= 12
13. two corners of auxiliary rectangle at (2; 3) and (16; 1), and horizontal trans-
verse axis
14. Two radio stations are located 150 miles apart, where stationAis west of sta-
tionB. Radio signals are being transmitted simultaneously by both stations,
travelling at a rate of 0.2 miles/sec. A plane travelling at 60 miles above
ground level has just passed by stationBand is headed towards the other
station. If the signal fromBarrives at the plane 480sec before the signal
sent fromA, determine the location of the plane.
4
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DEPED COPY Lesson 1.5. More Problems on Conic Sections
Learning Outcomes of the Lesson
At the end of the lesson, the student is able to:
(1) recognize the equation and important characteristics of the dierent types of
conic sections; and
(2) solve situational problems involving conic sections.
Lesson Outline
(1) Conic sections with associated equations in general form
(2) Problems involving characteristics of various conic sections
(3) Solving situational problems involving conic sections
Introduction
In this lesson, we will identify the conic section from a given equation. We
will analyze the properties of the identied conic section. We will also look at
problems that use the properties of the dierent conic sections. This will allow
us to synthesize what has been covered so far.
1.5.1. Identifying the Conic Section by Inspection
The equation of a circle may be written in standard form
Ax
2
+Ay
2
+Cx+Dy+E= 0;
that is, the coecients ofx
2
andy
2
are the same. However, it does not follow
that if the coecients ofx
2
andy
2
are the same, the graph is a circle.
General Equation Standard Equation graph
(A) 2x
2
+ 2y
2
2x+ 6y+ 5 = 0

x
1
2

2
+

y+
3
2

2
= 0 point
(B)x
2
+y
2
6x8y+ 50 = 0 (x3)
2
+ (y4)
2
=25 empty set
For a circle with equation (x h)
2
+ (yk)
2
=r
2
, we haver
2
>0. This is
not the case for the standard equations of (A) and (B).
In (A), because the sum of two squares can only be 0 if and only if each square
is 0, it follows thatx
1
2
= 0 andy+
3
2
= 0. The graph is thus the single point

1
2
;
3
2

.
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DEPED COPY In (B), no real values ofxandycan make the nonnegative left side equal to
the negative right side. The graph is then the empty set.
Let us recall the general form of the equations of the other conic sections. We
may write the equations of conic sections we discussed in the general form
Ax
2
+By
2
+Cx+Dy+E= 0:
Some terms may vanish, depending on the kind of conic section.
(1)Circle:bothx
2
andy
2
appear, and their coecients are the same
Ax
2
+Ay
2
+Cx+Dy+E= 0
Example: 18x
2
+ 18y
2
24x+ 48y 5 = 0
Degenerate cases: a point, and the empty set
(2)Parabola:exactly one ofx
2
ory
2
appears
Ax
2
+Cx+Dy+E= 0 (D6= 0, opens upward or downward)
By
2
+Cx+Dy+E= 0 (C6= 0, opens to the right or left)
Examples: 3x
2
12x+ 2y+ 26 = 0 (opens downward)
2y
2
+ 3x+ 12y 15 = 0 (opens to the right)
(3)Ellipse:bothx
2
andy
2
appear, and their coecientsAandBhave the same
sign and are unequal
Examples: 2x
2
+ 5y
2
+ 8x10y7 = 0 (horizontal major axis)
4x
2
+y
2
16x6y+ 21 = 0 (vertical major axis)
IfA=B, we will classify the conic as a circle, instead of an ellipse.
Degenerate cases: a point, and the empty set
(4)Hyperbola:bothx
2
andy
2
appear, and their coecientsAandBhave dif-
ferent signs
Examples: 5x
2
3y
2
20x18y22 = 0 (horizontal transverse axis)
4x
2
+y
2
+ 24x + 4y36 = 0 (vertical transverse axis)
Degenerate case: two intersecting lines
The following examples will show the possible degenerate conic (a point, two
intersecting lines, or the empty set) as the graph of an equation following a similar
pattern as the non-degenerate cases.
(1) 4x
2
+ 9y
2
16x+ 18y + 25 = 0 =)
(x2)
2
3
2
+
(y+ 1)
2
2
2
= 0
=)one point: (2; 1)
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DEPED COPY (2) 4x
2
+ 9y
2
16x+ 18y + 61 = 0 =)
(x2)
2
3
2
+
(y+ 1)
2
2
2
=1
=)empty set
(3) 4x
2
9y
2
16x18y+ 7 = 0 =)
(x2)
2
3
2

(y+ 1)
2
2
2
= 0
=)two lines:y+ 1 =
2
3
(x2)
A Note on Identifying a Conic Section
by Its General Equation
It is only after transforming a given general equation to standard
form that we can identify its graph either as one of the degenerate
conic sections (a point, two intersecting lines, or the empty set) or as
one of the non-degenerate conic sections (circle, parabola, ellipse, or
hyperbola).
1.5.2. Problems Involving Dierent Conic Sections
The following examples require us to use the properties of dierent conic sections
at the same time.
Example 1.5.1.A circle has center at the focus of the parabolay
2
+16x+4y=
44, and is tangent to the directrix of this parabola. Find its standard equation.
Solution.The standard equation of the parabola is (y+ 2)
2
=16(x 3). Its
vertex isV(3;2). Since 4c = 16 orc= 4, its focus isF(1;2) and its directrix
isx= 7. The circle has center at (1; 2) and radius 8, which is the distance
fromFto the directrix. Thus, the equation of the circle is
(x+ 1)
2
+ (y+ 2)
2
= 64: 2
Example 1.5.2.The vertices and foci of 5x
2
4y
2
+ 50x + 16y + 29 = 0 are,
respectively, the foci and vertices of an ellipse. Find the standard equation of
this ellipse.
Solution.We rst write the equation of the hyperbola in standard form:
(x+ 5)
2
16

(y2)
2
20
= 1:
For this hyperbola, using the notationsah,bh, andchto refer toa,b, andcof
the standard equation of the hyperbola, respectively, we haveah= 4,bh= 2
p
5,
ch=
p
a
2
h
+b
2
h
= 6, so we have the following points:
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DEPED COPY center: (5; 2)
vertices: (9; 2) and (1; 2)
foci: (11; 2) and (1; 2).
It means that, for the ellipse, we have these points:
center: (5; 2)
vertices: (11; 2) and (1; 2)
foci: (9; 2) and (1;2).
In this case,ce= 4 andae= 6, so thatbe=
p
a
2
ec
2
e=
p
20. The standard
equation of the ellipse is
(x+ 5)
2
36
+
(y2)
2
20
= 1: 2
More Solved Examples
1. Identify the graph of each of the following equations.
(a) 4x
2
8x49y
2
+ 196y388 = 0
(b)x
2
+ 5x+y
2
y+ 7 = 0
(c)y
2
48x+ 6y=729
(d) 49x
2
+ 196x + 100y
2
+ 1400y +
196 = 0
(e) 36x
2
+360x+64y
2
512y+1924 =
0
(f)x
2
+y
2
18y19 = 0
(g)5x
2
+ 60x + 7y
2
+ 84y + 72 = 0
(h)x
2
16x+ 20y = 136
Solution:
(a) Since the coecients ofx
2
andy
2
have opposite signs, the graph is a
hyperbola or a pair of intersecting lines. Completing the squares, we get
4x
2
8x49y
2
+ 196y 388 = 0
4(x
2
2x)49(y
2
4y) = 388
4(x
2
2x+ 1)49(y
2
4y+ 4) = 388 + 4(1)49(4)
(x1)
2
49

(y2)
2
4
= 1:
Thus, the graph is a hyperbola.
(b) Sincex
2
andy
2
have equal coecients, the graph is a circle, a point, or
the empty set. Completing the squares, we get
x
2
+ 5x+y
2
y+ 7 = 0
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DEPED COPY x
2
+ 5x+
25
4
+y
2
y+
1
4
=7 +
25
4
+
1
4

x+
5
2

2
+

y
1
2

2
=
1
2
:
Since the right hand side is negative, the graph is the empty set.
(c) By inspection, the graph is a parabola.
(d) Since the coecients ofx
2
andy
2
are not equal but have the same sign,
the graph is an ellipse, a point, or the empty set. Completing the squares,
we get
49x
2
+ 196x + 100y
2
+ 1400y + 196 = 0
49(x
2
+ 4x) + 100(y
2
+ 14y) =196
49(x
2
+ 4x+ 4) + 100(y
2
+ 14y + 49) =196 + 49(4) + 100(49)
(x+ 2)
2
100
+
(y+ 7)
2
49
= 1:
Thus, the graph is an ellipse.
(e) Since the coecients ofx
2
andy
2
are not equal but have the same sign,
the graph is an ellipse, a point, or the empty set. Completing the squares,we get
36x
2
+ 360x + 64y
2
512y+ 1924 = 0
36(x
2
+ 10x) + 64(y
2
8y) =1924
36(x
2
+ 10x + 25) + 64(y
2
8y+ 16) =1924 + 36(25) + 64(16)
(x+ 5)
2
64
+
(y4)
2
36
= 0:
Since the right-hand side is 0, the graph is a single point (the point is(5;4)).
(f) Sincex
2
andy
2
have equal coecients, the graph is a circle, a point, or
the empty set. Completing the squares, we get
x
2
+y
2
18y19 = 0
x
2
+y
2
18y+ 81 = 19 + 81
x
2
+ (y9)
2
= 100:
Thus, the graph is a circle.
(g) Since the coecients ofx
2
andy
2
have opposite signs, the graph is a
hyperbola or a pair of intersecting lines. Completing the squares, we get
5x
2
+ 60x + 7y
2
+ 84y + 72 = 0
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DEPED COPY 5(x
2
12x) + 7(y
2
+ 12y) =72
5(x
2
12x+ 36) + 7(y
2
+ 12y + 36) =725(36) + 7(36)
(x6)
2
7

(y+ 6)
2
5
= 0:
Since the right hand side is 0, the graph is a pair of intersecting lines;
these arey+ 6 =
r
5
7
(x6).
(h) By inspection, the graph is a parabola.
2. The center of a circle is the vertex of the parabolay
2
+ 24x 12y+ 132 = 0.
If the circle intersects the parabola's directrix at a point wherey= 11, nd
the equation of the circle.
Solution:
y
2
12y=24x 132
y
2
12y+ 36 =24x 132 + 36
(y6)
2
=24x 96
(y6)
2
=24(x + 4)
The vertex of the parabola is (4; 6) and its directrix isx= 2. Thus, the
circle has center (4; 6) and contains the point (2; 11). Then its radius is
p
(42)
2
+ (611)
2
=
p
61. Therefore, the equation of the circle is (x+
4)
2
+ (y6)
2
= 61.
3. The vertices of the hyperbola with equation 9x
2
72x16y
2
128y256 = 0
are the foci of an ellipse that contains the point (8;10). Find the standard
equation of the ellipse.Solution:
9x
2
72x16y
2
128y256 = 0
9(x
2
8x)16(y
2
+ 8y) = 256
9(x
2
8x+ 16)16(y
2
+ 8y+ 16) = 256 + 9(16)16(16)
(x4)
2
16

(y+ 4)
2
9
= 1
The vertices of the hyperbola are (0;4) and (8; 4). Since these are the foci
of the ellipse, the ellipse is horizontal with centerC(4;4); also, the focal
distance of the ellipse isc= 4. The sum of the distances of the point (8; 10)
from the foci is
p
(80)
2
+ (10 (4))
2
+
p
(88)
2
+ (10 (4))
2
= 16:
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DEPED COPY This sum is constant for any point on the ellipse; so 2a= 16 anda= 8. Then
b
2
= 8
2
4
2
= 48. Therefore, the equation of the ellipse is
(x4)
2
64
+
(y+ 4)
2
48
= 1:
Supplementary Problems 1.5
For items 1 to 8, identify the graph of each of the following equations.
1. 9x
2
+ 72x 64y
2
+ 128y + 80 = 0
2. 49x
2
490x+ 36y
2
+ 504y + 1225 = 0
3.y
2
+ 56x 18y+ 417 = 0
4.x
2
+ 20x +y
2
20y+ 200 = 0
5.x
2
10x48y+ 265 = 0
6.144x
2
1152x + 25y
2
150y5679 = 0
7.x
2
+ 4x+ 16y
2
128y+ 292 = 0
8.x
2
6x+y
2
+ 14y + 38 = 0
9. An ellipse has equation 100x
2
1000x + 36y
2
144y956 = 0. Find the
standard equations of all circles whose center is a focus of the ellipse and
which contains at least one of the ellipse's vertices.
10. Find all parabolas whose focus is a focus of the hyperbolax
2
2x3y
2
2 = 0
and whose directrix contains the top side of the hyperbola's auxiliary rectangle.
11. Find the equation of the circle that contains all corners of the auxiliary rect-
angle of the hyperbolax
2
18x+y
2
+ 10y 81 = 0.
12. Find the equations of all horizontal parabolas whose focus is the center of the
ellipse 9x
2
+ 17y
2
170y+ 272 = 0 and whose directrix is tangent to the same
ellipse.
13. Find all values ofr6= 1 so that the graph of
(r1)x
2
+ 14(r 1)x+ (r1)y
2
6(r1)y= 6057r
is
(a) a circle,
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DEPED COPY (b) a point,
(c) the empty set.
14. Find all values ofm6=7;0 so that the graph of
2mx
2
16mx+my
2
+ 7y
2
= 2m
2
18m
is
(a) a circle.
(b) a horizontal ellipse.
(c) a vertical ellipse.
(d) a hyperbola.
(e) the empty set.
4
Lesson 1.6.Systems of Nonlinear Equations
Learning Outcomes of the Lesson
At the end of the lesson, the student is able to:
(1) illustrate systems of nonlinear equations;
(2) determine the solutions of systems of nonlinear equations using techniques
such as substitution, elimination, and graphing; and
(3) solve situational problems involving systems of nonlinear equations.
Lesson Outline
(1) Review systems of linear equations
(2) Solving a system involving one linear and one quadratic equation
(3) Solving a system involving two quadratic equations
(4) Applications of systems of nonlinear equations
Introduction
After recalling the techniques used in solving systems of linear equations in
Grade 8, we extend these methods to solving a system of equations to systems
in which the equations are not necessarily linear. In this lesson, the equations
are restricted to linear and quadratic types, although it is possible to adapt the
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electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2016.

DEPED COPY methodology to systems with other types of equations. We focus on quadratic
equations for two reasons: to include a graphical representation of the solution
and to ensure that either a solution is obtained or it is determined that there is
no solution. The latter is possible because of the quadratic formula.
1.6.1. Review of Techniques in Solving Systems of Linear
Equations
Recall the methods we used to solve systems of linear equations.There were three
methods used: substitution, elimination, and graphical.
Example 1.6.1.Use the substitution method to solve the system, and sketch
the graphs in one Cartesian plane showing the point of intersection.
8
<
:
4x+y= 6
5x+ 3y= 4
Solution.Isolate the variableyin the rst equation, and then substitute into the
second equation.
4x+y= 6
=)y= 64x
5x+ 3y= 4
5x+ 3(64x) = 4
7x+ 18 = 4
x= 2
y= 64(2) =2
Example 1.6.2.Use
the elimination method to solve the system, and sketch the
graphs in one Cartesian plane showing the point of intersection.
8
<
:
2x+ 7 = 3y
4x+ 7y= 12
Solution.We eliminate rst the variablex. Rewrite the rst equation wherein
only the constant term is on the right-hand side of the equation, then multiply
it by2, and then add the resulting equation to the second equation.
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DEPED COPY 2x3y=7
(2)(2x 3y) = (2)(7)
4x+ 6y= 14
4x+ 6y= 14
4x+ 7y= 12
13y= 26
y= 2
x=
1
2
1.6.2. Solving Systems of Equations Using Substitution
W
e begin our extension with a system involving one linear equation and one
quadratic equation. In this case, it is always possible to use substitution by
solving the linear equation for one of the variables.
Example 1.6.3.Solve the following system, and sketch the graphs in one Carte-
sian plane.
8
<
:
xy+ 2 = 0
y1 =x
2
Solution.We solve foryin terms ofxin the rst equation, and substitute this
expression to the second equation.
xy+ 2 = 0 =)y=x+ 2
y1 =x
2
(x+ 2)1 =x
2
x
2
x1 = 0
x=
1
p
5
2
x=
1
+
p
5
2
=)y=
1
+
p
5
2
+
2 =
5 +
p
5
2
x=
1
p
5
2
=)y=
1
p
5
2
+
2 =
5
p
5
2
Solutions:

1
+
p
5
2
;
5
+
p
5
2
!
and

1
p
5
2
;
5
p
5
2
!
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DEPED COPY The rst equation represents a line withx-intercept2 andy-intercept 2,
while the second equation represents a parabola with vertex at (0;1) and which
opens upward.
1.6.3. Solving Systems of Equations Using Elimination
Elimination
method is also useful in systems of nonlinear equations. Sometimes,
some systems need both techniques (substitution and elimination) to solve them.
Example 1.6.4.Solve the following system:
8
<
:
y
2
4x6y= 11
4(3x) = (y3)
2
:
Solution 1.We expand the second equation, and eliminate the variablexby
adding the equations.
4(3x) = (y 3)
2
=)124x=y
2
6y+ 9 =)y
2
+ 4x6y= 3
8
<
:
y
2
4x6y= 11
y
2
+ 4x6y= 3
Adding these equations, we get
2y
2
12y= 14 =)y
2
6y7 = 0 =)(y7)(y+1) = 0 =)y= 7 ory=1:
Solving forxin the second equation, we have
x= 3
(y3)
2
4
:
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DEPED COPY y= 7 =)x=1 and y=1 =)x=1
Solutions: (1; 7) and (1;1) 2
The graphs of the equations in the preceding example with the points of
intersection are shown below.
Usually, the general form is more convenient to use in solving systems of
equations.
However, sometimes the solution can be simplied by writing the
equations in standard form. Moreover, the standard form is best for graphing.
Let us again solve the previous example in a dierent way.
Solution 2.By completing the square, we can change the rst equation into stan-
dard form:
y
2
4x6y= 11 =)4(x+ 5) = (y3)
2
:
8
<
:
4(x+ 5) = (y3)
2
4(3x) = (y3)
2
Using substitution or the transitive property of equality, we get
4(x+ 5) = 4(3x) =)x=1:
Substituting this value ofxinto the second equation, we have
4[3(1)] = (y 3)
2
=)16 = (y 3)
2
=)y= 7 ory=1:
The solutions are (1; 7) and (1; 1), same as Solution 1. 2
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DEPED COPY Example 1.6.5.Solve the system and graph the curves:
8
<
:
(x3)
2
+ (y5)
2
= 10
x
2
+ (y+ 1)
2
= 25:
Solution.Expanding both equations, we obtain
8
<
:
x
2
+y
2
6x10y+ 24 = 0
x
2
+y
2
+ 2y24 = 0:
Subtracting these two equations, we get
6x12y+ 48 = 0 =)x+ 2y8 = 0
x= 82y:
We can substitutex= 82yto either the rst equation or the second equation.
For convenience, we choose the second equation.
x
2
+y
2
+ 2y24 = 0
(82y)
2
+y
2
+ 2y24 = 0
y
2
6y+ 8 = 0
y= 2 ory= 4
y= 2 =)x= 82(2) = 4 and y= 4 =)x= 82(4) = 0
The solutions are (4;2) and (0; 4).
The graphs of both equations are circles. One has center (3; 5) and radius
p
10, while the other has center (0;1) and radius 5. The graphs with the points
of intersection are show below.
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DEPED COPY 1.6.4. Applications of Systems of Nonlinear Equations
Let us apply systems of equations to a problem involving modern-day television
sets.
?
Example 1.6.6.The screen size of television sets is given in inches. This
indicates the length of the diagonal. Screens of the same size can come in dierent
shapes. Wide-screen TV's usually have screens with aspect ratio 16 : 9, indicating
the ratio of the width to the height. Older TV models often have aspect ratio
4 : 3. A 40-inch LED TV has screen aspect ratio 16 : 9. Find the length and the
width of the screen.
Solution.Letwrepresent the width andhthe height of the screen. Then, by
Pythagorean Theorem, we have the system
8
<
:
w
2
+h
2
= 40
2
=)w
2
+h
2
= 1600
w
h
=
16
9
=)h=
9w
16
w
2
+h
2
= 1600 =) w
2
+

9w
16

2
= 1600
337w
2
256
= 1600
w=
r
409 600
337
34:86
h=
19x
16

19(34:86)
16
= 19:61
Therefore, a 40-inch TV with aspect ratio 16 : 9 is about 35:86 inches wide and
19:61 inches high. 2
More Solved Examples
Solve the system and graph the curves.
1.
8
<
:
x
2
y
2
= 21
x+y= 7
Solution:We can writeyin terms ofxusing the second equation asy= 7x.
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electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2016.

DEPED COPY Substituting this into the rst equation, we have
x
2
(7x)
2
= 21
14x70 = 0
x= 5:
Thus, the point of intersection is (5; 2).
2.
8
<
:
x
2
+y
2
x+ 6y+ 5 = 0
x+y+ 1 = 0
Solution:We can writeyin terms ofxusing the second equation asy=
(x+ 1).
Substituting this into the rst equation, we have
x
2
+ ((x + 1))
2
x+ 6((x + 1)) + 5 = 0
2x
2
5x= 0
x(2x5) = 0;
which yieldsx= 0 andx=
5
2
. Thus, the points of intersection are (0;1)
and

5
2
;
7
2

.
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DEPED COPY 3.
8
<
:
(y2)
2
= 4(x 4)
(y4)
2
=x5
Solution:We can rewrite the rst equation as
x4 =
(y2)
2
4
;
which can be substituted into the second equation by rewriting it as
(y4)
2
= (x4)1 =
(y2)
2
4
1
which upon expansion yields
3y
2
28y+ 64 = 0:
This equation has rootsy= 16=3 and y= 4, giving us the points (5; 4) and

61
9
;
16
3

.
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DEPED COPY 4.
8
<
:
x=
1
2
(y+ 5)
2
2
y
2
+ 10y + (x2)
2
=9
Solution:We can rewrite the rst equation as
(y+ 5)
2
= 2x+ 4;
which can be substituted into the second equation by completing the square
to get
(y
2
+ 10y + 25) + (x 2)
2
=9 + 25
(y+ 5)
2
+ (x2)
2
= 16
(2x+ 4) + (x2)
2
= 16
x
2
2x8 = 0;
This equation has rootsx=2 andx= 4, giving us the points (2;5),
(4;5
p
12), and

4;5 +
p
12

.
Find the system of equations that represents the given problem and solve.
5. The dierence of two numbers is 12, and the sum of their squares is 144. Find
the numbers.
Solution:Ifxandyare the two numbers, then we have the resulting system
(
xy= 12
x
2
+y
2
= 144;
where the rst equation yieldsx=y+ 12. Combining this with the second
equation yields (y + 12)
2
+y
2
= 144 or equivalently 2y(y+ 12) = 0, giving us
the ordered pairs (12;0) and (0; 12).
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DEPED COPY Supplementary Problems 1.6
1. Solve the system and graph the curves:
(a)
8
<
:
x
2
+ 3xy+ 2 = 0
y5x= 1
(b)
8
<
:
(y2)
2
= 9(x + 2)
9x
2
+ 4y
2
+ 18x 16y= 0
(c)
8
<
:
(x+ 1)
2
+ 2(y 4)
2
= 12
y
2
8y= 4x16
(d)
8
<
:
x
2
2x4y
2
+ 8y2 = 0
5x
2
10x+ 12y
2
+ 24y 58 = 0
(e)
8
<
:
x
2
+y
2
= 2
xy= 4
2. Ram is speeding along a highway when he sees a police motorbike parked on
the side of the road right next to him. He immediately starts slowing down,
but the police motorbike accelerates to catch up with him. It is assumed that
the two vehicles are going in the same direction in parallel paths.
The distance that Ram has traveled in meterstseconds after he starts to
slow down is given byd(t) = 150 + 75t1:2t
2
. The distance that the police
motorbike travels can be modeled by the equationd(t) = 4t
2
. How long will
it take for the police motorbike to catch up to Ram?
3. The square of a certain number exceeds twice the square of another number
by
1
8
. Also, the sum of their squares is
5
16
. Find possible pairs of numbers
that satisfy these conditions.
4. Solve the system of equations
8
<
:
x
2
+y
2
= 41
xy= 20
5. Determine the value(s) ofksuch that the circlex
2
+ (y6)
2
= 36 and the
parabolax
2
= 4kywill intersect only at the origin.
4
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DEPED COPY Topic Test 1 for Unit 1
1. Identify the graph of each of the following equations.
(a)x
2
x+y
2
+ 3y
3
2
= 0
(b)x
2
+ 4x14y= 52
(c) 3x
2
42x4y
2
24y+ 99 = 0
(d) 7x
2
112x+ 2y
2
+ 448 = 0
2. Determine and sketch the conic with the given equation. Identify the impor-
tant parts of the conic and include them in the graph.
(a) 25x
2
+ 7y
2
175 = 0
(b)64x
2
+ 128x + 36y
2
+ 288y 1792 = 0
3. Find the equation of the conic with the given properties.
(a) parabola; vertex at (1;3); directrixx=7
(b) hyperbola; asymptotesy=
12
5
x
1
5
andy=
12
5
x
49
5
; one vertex at
(3;5)
4. Solve the following system of equations:
8
<
:
(x1)
2
+ (y+ 1)
2
= 5
y= 2(x 1)
2
8
5. A doorway is in the shape of a rectangle capped by a semi-ellipse. If the
rectangle is 1 m wide and 2 m high while the ellipse is 0.3 m high at the
center, can a cabinet that is 2.26 m high, 0.5 m wide, and 2 m long be pushed
through the doorway? Assume that the cabinet cannot be laid down on its
side.
6. A point moves so that its distance from the point (0;1) is twice its distance
from the linex= 3. Derive the equation (in standard form) of the curve that
is traced by the point, and identify the curve.
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DEPED COPY Topic Test 2 for Unit 1
1. Identify the graph of each of the following equations.
(a)y
2
+ 8x10y= 15
(b)x
2
+ 10x +y
2
+ 18y + 110 = 0
(c) 9x
2
+ 36x + 4y
2
8y+ 4 = 0
(d)11x
2
+ 132x + 17y
2
136y124 = 0
2. Determine and sketch the conic with the given equation. Identify the impor-
tant parts of the conic and include them in the graph.
(a)x
2
y
2
= 64 (b) 4x
2
+ 24x + 49y
2
196y+ 36
3. Find the equation of the conic with the given properties.
(a) parabola; directrixy=2; focus at (7;12)
(b) ellipse; vertical or horizontal major axis; one vertex at (5;12); one
covertex at (1; 3)
4. Solve the following system of equations:
8
<
:
9x
2
4y
2
+ 54x + 45 = 0
(x+ 3)
2
= 4y + 4
5. Nikko goes to his garden to water his plants. He holds the water hose 3 feet
above the ground, with the hoses opening as the vertex and the water ow
following a parabolic path. The water strikes the ground a horizontal distance
of 2 feet from where the opening is located. If he were to stand on a 1.5 feet
stool, how much further would the water strike the ground?
6. A point moves so that its distance from the point (2;0) is two-thirds its dis-
tance from the liney= 5. Derive the equation (in standard form) of the curve
that is traced by the point, and identify the curve.
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DEPED COPY Unit 2
Mathematical Induction
Batad Rice Terraces in Ifugao, by Ericmontalban, 30 September 2012,
https://commons.wikimedia.o
rg/wiki/File%3ABatad
riceterracesinIfugao.jpg.

Listed as one of the United Nations Educational, Scientic and Cultural
Organization (UNESCO) World Heritage sites since 1995, the two-millennium-
old Rice Terraces of the Philippine Cordilleras by the Ifugaos is a living testimony
of mankind's creative engineering to adapt to physically-challenging environment
in nature. One of the ve clusters of terraces inscribed in the UNESCO list is
the majestic Batad terrace cluster (shown above), which is characterized by its
amphitheater-like, semicircular terraces with a village at its base.
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DEPED COPY Lesson 2.1. Review of Sequences and Series
Learning Outcomes of the Lesson
At the end of the lesson, the student is able to:
(1) illustrate a series; and
(2) dierentiate a series from a sequence.
Lesson Outline
(1) Sequences and series
(2) Dierent types of sequences and series (Fibonacci sequence, arithmetic and
geometric sequence and series, and harmonic series)
(3) Dierence between sequence and series
Introduction
In this lesson, we will review the denitions and dierent types of sequences
and series.
Lesson Proper
Recall the following denitions:
Asequenceis a function whose domain is the set of positive integers
or the setf1;2;3; : : : ; ng.
Aseriesrepresents the sum of the terms of a sequence.
If a sequence is nite, we will refer to the sum of the terms of the
sequence as the series associated with the sequence. If the sequence has
innitely many terms, the sum is dened more precisely in calculus.
A sequence is a list of numbers (separated by commas), while a series is a
sum of numbers (separated by \+" or \" sign). As an illustration, 1;
1
2
;
1
3
;
1
4
is a sequence, and 1
1
2
+
1
3

1
4
=
7
12
is its associated series.
The sequence withnth termanis usually denoted byfang, and the associated
series is given by
S=a1+a2+a3+ +an:
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DEPED COPY Example 2.1.1.Determine the rst ve terms of each dened sequence, and
give their associated series.
(1)f2ng
(2)f1 + 2n+ 3n
2
g
(3)f(1)
n
g
(4)f1 + 2 + 3 + +ng
Solution.We denote thenth term of a sequence byan, andS=a1+a2+a3+
a4+a5.
(1)an= 2n
First ve terms:a1= 21 = 1,a2= 22 = 0,a3=1,a4=2,a5=3
Associated series:S=a1+a2+a3+a4+a5= 1 + 0123 =5
(2)an= 1 + 2n+ 3n
2
First ve terms:a1= 1 + 21 + 31
2
= 6,a2= 17,a3= 34,a4= 57,a5= 86
Associated series:S= 6 + 17 + 34 + 57 + 86 = 200
(3)an= (1)
n
First ve terms:a1= (1)
1
=1,a2= (1)
2
= 1,a3=1,a4= 1,
a5=1
Associated series:S=1 + 11 + 11 =1
(4)an= 1 + 2 + 3 + +n
First ve terms:a1= 1,a2= 1+2 = 3,a3= 1+2+3 = 6,a4= 1+2+3+4 =
10,a5= 1 + 2 + 3 + 4 + 5 = 15
Associated series:S= 1 + 3 + 6 + 10 + 15 = 35 2
The sequencefangdened byan=an1+an2forn3, wherea1=
a2= 1, is called aFibonacci sequence. It terms are 1; 1;2;3;5;8;13; : : :.
Anarithmetic se
quenceis a sequence in which each term after the rst
is obtained by adding a constant (called thecommon dierence) to the
preceding term.
If thenth term of an arithmetic sequence isanand the common dierence is
d, then
an=a1+ (n1)d:
The associated arithmetic series withnterms is given by
Sn=
n(a1+an)
2
=
n[2a1+ (
n1)d]
2
:
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DEPED COPY Ageometric sequenceis a sequence in which each term after the rst
is obtained by multiplying the preceding term by a constant (called
thecommon ratio).
If thenth term of a geometric sequence isanand the common ratio isr, then
an=a1r
n1
:
The associated geometric series withnterms is given by
Sn=
8
>
<
>
:
na1 ifr= 1
a1(1r
n
)
(1r)
ifr6= 1.
The proof of this sum formula is an example in Lesson 2.3.
When1< r <1, the innite geometric series
a1+a1r+a1r
2
+ +a1r
n1
+
has a sum, and is given by
S=
a1
1r
:
Iffangis an arithmetic sequence, then the sequence withnth term
bn=
1
an
is aharmonic sequence.
More Solved Examples
1. How many terms are there in an arithmetic sequence with rst term 5, common
dierence3, and last term76?
Solution:a1= 5,d=3,an=76. Findn.
an= 5 + (n1)(3) = 76
n1 =
765
3
= 27; )n= 28
2. List the rst three terms of the arithmetic sequence if the 25th term is 35 and
the 30th term is 5.
Solution:a24=a1+ 24d = 35 anda30=a1+ 29d = 5
Eliminatinga1by subtraction, 5d =30, ord=6
This implies thata1= 179, and the rst three terms are 179, 173, 167.
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DEPED COPY 3. Find the sum of all positive three-digit odd integers.
Solution:Findsnifa1= 101 = 1 + 50(2),an= 999 = 1 + 499(2).
There are 450 terms froma1toan, hencen= 450.
sn=
450(101 + 999)
2
= 247 500
4. The seventh term of a geometric sequence is6 and the tenth term is 162.
Find the fth term.
Solution:a7=a1r
6
=6 anda10=a1r
9
= 162.
Eliminatinga1by division:
a10
a7
=r
3
=
162
6
=27. Thusr=3
Sincea5r
2
=a7,a5=
6
9
=
2
3
.
5. Insert three numbers (called geometric means) between 6 and 32=27, so that
the ve numbers form a geometric sequence.Solution:Ifa1= 6 and there are three terms betweena1and 32=27, then
a5= 32=27.
a5= 6(r)
4
=
32
27
)r
4
=
16
81
)r=
2
3
One possible set of three numbers is 4; 8=3;16=9, the other is4;8=3;16=9.
6. A ball dropped from the top of a building 180 m high always rebounds three-
fourths the distance it has fallen. How far (up and down) will the ball have
traveled when it hits the ground for the 6th time?
Solution:a1= 180,r= 3=4,n= 6
s6=
180
h
1

3
4

6
i
1
3
4
The ball traveled 2s 61801003:71 meters.
7. The Cantor set is formed as follows. Divide a segment of one unit into three
equal parts. Remove the middle one-third of the segment. From each of the
two remaining segments, remove the middle third. From each of the remaining
segments, remove the middle third. This process is continued indenitely. Find
the total length of the segments removed.
Solution:Letanrepresent the total length removed in thenth iteration. Hence
a1= 1=3,a2= 2=9,a3= 4=27, and so on.
This meansr=
2
3
. The sum to innity iss=
1
3
1
2
3
= 1.
The total length of the segments removed is 1 unit.
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DEPED COPY 8. The 7th term of an arithmetic sequence is 25. Its rst, third, and 21st term
form a geometric sequence. Find the rst term and the common dierence of
the sequence.
Solution:a7=a1+ 6d= 25)a1= 256d
a3
a1
=
a21
a3
, ora1a21=a
2
3.
(256d) (256d+ 20d) = (25 6d+ 2d)
2
d= 0 andan= 25 for alln, ord= 4 anda1= 1.
9. Letfangbe an arithmetic sequence andfbngan arithmetic sequence of positive
integers. Prove that the sequence withn
th
termabnis arithmetic.
Solution:Let the common dierence offangbedand offbngbec.
abn+1abn= [a1+ (bn+11)d][a1+ (bn1)d] =bn+1bn=c
This proves that the dierence between any two consecutive terms offabngis
a constant independent ofn.
10. Letfangbe a geometric sequence. Prove thatfa
3
ngis a geometric sequence.
Solution:Letrbe the common ratio offang.
a
3
n= (a1r
n1
)
3
=a
3
1(r
3
)
n1
.
Thusfa
3
ngis a geometric sequence with rst terma
3
1and common ratior
3
.
11. Iffangis a sequence such that its rst three terms form both an arithmetic
and a geometric sequence, what can be concluded aboutfang?
Solution:There is a real numberrsuch thata2=a1randa3=a1r
2
.
Sincea1,a2anda3form an arithmetic sequence, thena2a1=a3a2, or
a32a2+a1= 0.
a32a2+a1=a1r
2
2a1r+a1=a
1
(r1)
2
= 0)a1= 0 orr= 1.
Ifa1= 0, thena1=a2=a3= 0. Ifr= 1, thena1=a2=a3.
In all cases,a1=a2=a3.
Supplementary Problems 2.1
1. Find the 5th term of the arithmetic sequence whose 3rd term is 35 and whose
10th term is 77.
2. Suppose that the fourth term of a geometric sequence is
2
9
and the sixth term
is
8
81
. Find the rst term and the common ratio.
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DEPED COPY 3. The partial sum in the arithmetic series with rst term 17 and a common
dierence 3 is 30705. How many terms are in the series?
4. An arithmetic sequencea1; a2; : : : ; a100has a sum of 15,000. Find the rst
term and the common dierence if the sum of the terms in the sequence
a3; a6; a9; : : : ; a99is 5016.
5. The sum of an innite geometric series is 108, while the sum of the rst 3
terms is 112. Determine the rst term of this series.
6. Evaluate the innite series
3
2
2
0
5
1
+
3
3
2
1
5
2
+ +
3
k+1
2
k1
5
k
+ .
7. Letn= 0:123 = 0:123123: : :be a nonterminating repeating decimal. Find
a rational number that is equal tonby expressingnas an innite geometric
series. Simplify your answer.
8. An arithmetic sequence whose rst term is 2 has the property that its sec-
ond, third, and seventh terms are consecutive terms of a geometric sequence.Determine all possible second terms of the arithmetic sequence.
9. Eighty loaves of bread are to be divided among 4 people so that the amounts
they receive form an arithmetic progression. The rst two together receiveone-third of what the last two receive. How many loaves does each personreceive?
10. Givenaandb, suppose that three numbers are inserted between them so
that the ve numbers form a geometric sequence. If the product of the threeinserted numbers betweenaandbis 27, show thatab= 9.
11. For what values ofnwill the innite series (2n1) + (2n1)
2
+: : :+
(2n1)
i
+: : :have a nite value?
4
Lesson 2.2.Sigma Notation
Learning Outcomes of the Lesson
At the end of the lesson, the student is able to use the sigma notation to
represent a series.
Lesson Outline
(1) Denition of and writing in sigma notation
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DEPED COPY (2) Evaluate sums written in sigma notation
(3) Properties of sigma notation
(4) Calculating sums using the properties of sigma notation
Introduction
The sigma notation is a shorthand for writing sums. In this lesson, we will
see the power of this notation in computing sums of numbers as well as algebraic
expressions.
2.2.1. Writing and Evaluating Sums in Sigma Notation
Mathematicians use the sigma notation to denote a sum. The uppercase Greek
letter (sigma) is used to indicate a \sum." The notation consists of several
components or parts.
Letf(i) be an expression involving an integeri. The expression
f(m) +f(m+ 1) +f(m+ 2) + +f(n)
can be compactly written insigma notation, and we write it as
n
X
i=m
f(i);
which is read he summation off(i) fromi=mton." Here,m
andnare integers withmn,f(i) is aterm(orsummand) of the
summation, and the letteriis theindex, mthelower bound, andn
theupper bound.
Example 2.2.1.Expand each summation, and simplify if possible.
(1)
4
X
i=2
(2i+ 3)
(2)
5
X
i=0
2
i
(3)
n
X
i=1
ai
(4)
6
X
n=1
p
n
n+ 1
Solution.We apply the denition of sigma notation.
(1)
4
X
i=2
(2i+ 3) = [2(2) + 3] + [2(3) + 3] + [2(4) + 3] = 27
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DEPED COPY (2)
5
X
i=0
2
i
= 2
0
+ 2
1
+ 2
2
+ 2
3
+ 2
4
+ 2
5
= 63
(3)
n
X
i=1
ai=a1+a2+a3+ +an
(4)
6
X
n=1
p
n
n+ 1
=
1
2
+
p
2
3
+
p
3
4
+
2
5
+
p
5
6
+
p
6
7
2
Example 2.2.2.Write each expression in sigma notation.
(1) 1 +
1
2
+
1
3
+
1
4
+ +
1
100
(2)1 + 23 + 45 + 67 + 89 + 25
(3)a2+a4+a6+a8+ +a20
(4) 1 +
1
2
+
1
4
+
1
8
+
1
16
+
1
32
+
1
64
+
1
128
Solution.(1) 1 +
1
2
+
1
3
+
1
4
+ +
1
100
=
100
X
n=1
1
n
(2)1 + 23 + 45 + 25
= (1)
1
1 + (1)
2
2 + (1)
3
3 + (1)
4
4
+ (1)
5
5 + + (1)
25
25
=
25
X
j=1
(1)
j
j
(3)a2+a4+a6+a8+ +a20
=a2(1)+a2(2)+a2(3)+a2(4)+ +a2(10)
=
10
X
i=1
a2i
(4) 1 +
1
2
+
1
4
+
1
8
+
1
16
+
1
32
+
1
64
+
1
128
=
7
X
k=0
1
2
k
2
The sigma notation of a sum expression is not necessarily unique. For ex-
ample, the last item in the preceding example can also be expressed in sigma
notation as follows:
1 +
1
2
+
1
4
+
1
8
+
1
16
+
1
32
+
1
64
+
1
128
=
8
X
k=1
1
2
k1
:
However, this last sigma notation is equivalent to the one given in the example.
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DEPED COPY 2.2.2. Properties of Sigma Notation
We start with nding a formula for the sum of
n
X
i=1
i= 1 + 2 + 3 + +n
in terms ofn.
The sum can be evaluated in dierent ways. One informal but simple approach
is pictorial.
n
X
i
=1
i= 1 + 2 + 3 + +n=
n(n+ 1)
2
Another
way is to use the formula for an arithmetic series witha1= 1 and
an=n:
S=
n(a1+an)
2
=
n
(n+ 1)
2
:
W
e now derive some useful summation facts. They are based on the axioms
of arithmetic addition and multiplication.
n
X
i=m
cf(i) =c
n
X
i=m
f(i),cany real number.
Proof.
n
X
i=m
cf(i) =cf(m) +cf(m+ 1) +cf(m+ 2) + +cf(n)
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DEPED COPY =c[f(m) +f(m+ 1) + +f(n)]
=c
n
X
i=m
f(i) 2
n
X
i=m
[f(i) +g(i)] =
n
X
i=m
f(i) +
n
X
i=m
g(i)
Proof.
n
X
i=m
[f(i) +g(i)]
= [f(m) +g(m)] + + [f(n) +g(n)]
= [f(m) + +f(n)] + [g (m) + +g(n)]
=
n
X
i=m
f(i) +
n
X
i=m
g(i) 2
n
X
i=m
c=c(nm+ 1)
Proof.
n
X
i=m
c=c+c+c+ +c
| {z }
nm+1 terms
=c(nm+ 1) 2
A special case of the above result which you might encounter more often is
the following:
n
X
i=1
c=cn:
Telescoping Sum
n
X
i=m
[f(i+ 1)f(i)] =f(n+ 1)f(m)
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DEPED COPY Proof.
n
X
i=m

f(i+ 1)f(i)

= [f(m+ 1)f(m)] + [f (m+ 2)f(m+ 1)]
+ [f(m+ 3)f(m+ 2)] + + [f(n+ 1)f(n)]
Note that the terms,f(m+1); f(m+2); : : : ; f(n), all cancel out. Hence, we have
n
X
i=m
[f(i+ 1)f(i)] =f(n+ 1)f(m): 2
Example 2.2.3.Evaluate:
30
X
i=1
(4i5).
Solution.
30
X
i=1
(4i5) =
30
X
i=1
4i
30
X
i=1
5
= 4
30
X
i=1
i
30
X
i=1
5
= 4
(30)(31)
2
5(30)
= 1710 2
Example 2.2.4.Evaluate:
1
12
+
1
23
+
1
34
+ +
1
99100
:
Solution.
1
12
+
1
23
+
1
34
+ +
1
99100
=
99
X
i=1
1
i(i+ 1)
=
99
X
i=1
i+ 1i
i(i+ 1)
=
99
X
i=1

i+ 1
i(i+ 1)

i
i(i+ 1)

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DEPED COPY =
99
X
i=1

1
i

1
i+ 1

=
99
X
i=1

1
i+ 1

1
i

Usingf(i) =
1
i
and the telescoping-sum property, we get
99
X
i=1
1
i(i+ 1)
=

1
100

1
1

=
99
100
: 2
Example 2.2.5.Derive a formula for
n
X
i=1
i
2
using a telescoping sum with terms
f(i) =i
3
.
Solution.The telescoping sum property implies that
n
X
i=1

i
3
(i1)
3

=n
3
0
3
=n
3
:
On the other hand, using expansion and the other properties of summation,
we have
n
X
i=1

i
3
(i1)
3

=
n
X
i=1
(i
3
i
3
+ 3i
2
3i+ 1)
= 3
n
X
i=1
i
2
3
n
X
i=1
i+
n
X
i=1
1
= 3
n
X
i=1
i
2
3
n(n+ 1)
2
+n:
Equating the two results above, we obtain
3
n
X
i=1
i
2

3n(n+ 1)
2
+n=n
3
6
n
X
i=1
i
2
3n(n+ 1) + 2n= 2n
3
6
n
X
i=1
i
2
= 2n
3
2n+ 3n(n+ 1)
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DEPED COPY = 2n(n
2
1) + 3n (n+ 1)
= 2n(n1)(n+ 1) + 3n(n+ 1)
=n(n+ 1)[2(n1) + 3]
=n(n+ 1)(2n + 1):
Finally, after dividing both sides of the equation by 6, we obtain the desired
formula
n
X
i=1
i
2
=
n(n+ 1)(2n + 1)
6
: 2
More Solved Examples
1. Expand the following sums and simplify if possible:
(a)
5
X
i=1
(i
2
i+ 1)
(b)
6
X
i=3
i
2

i+ 1
2

2
(c)
5
X
i=0
x
3i
y
153i
(d)
9
X
i=1
x
2i+1
(i+ 1)
2
(e)
1
X
i=1
3
i+2
2
i+1
Solution:
(a)
5
X
i=1
(i
2
i+ 1) = (1
2
1 + 1) + (2
2
2 + 1) +: : :+ (5
2
5 + 1) = 45 or
5(5 + 1)(2(5) + 1)
6

5(5 + 1)
2
+ 5 = 45
(b)
6
X
i=3
i
2

i+ 1
2

2
= 3
2

4
2

2
+: : :+ 6
2

7
2

2
= 802
(c)
5
X
i=0
x
3i
y
153i
=x
3(0)
y
153(0)
+: : :+x
3(5)
y
153(5)
=xy
15
+x
3
y
12
+x
6
y
9
+
x
9
y
6
+x
12
y
3
+x
15
y
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DEPED COPY (d)
9
X
i=1
x
2i+1
(i+ 1)
2
=
x
3
(1 + 1)
2
+
x
5
(2 + 1)
2
+: : :+
x
19
(9 + 1)
2
= 4x
3
+ 9x
5
+ 16x
7
+
: : :+ 100x
19
(e)
1
X
i=1
3
i+2
2
i+1
= 27
1
X
i=1

2
3

i+1
, which is an innite geometric series with
jrj=
2
3
<1 anda1=
4
9
, giving us
1
X
i=1
3
i+2
2
i+1
= 27
4=9
1(2=3)
= 36.
2. Evaluate
20
X
i=1
[2(i1) + 2].
Solution:
20
X
i=1
(2(i1) + 2) =
20
X
i=1
2i= 4
(20)(21)
2
= 840.
3. Find a formula for
1
1(3)
+
1
2(4)
+
1
3(5)
+ +
1
n(n+ 2)
given any positive
integern.
Solution:We have
1
1(3)
+
1
2(4)
+
1
3(5)
+ +
1
n(n+ 2)
=
n
X
i=1
1
i(i+ 2)
. Rewriting
yields
1
i(i+ 2)
=
(21) + (i i)
i(i+ 2)
=
1
i

1
i+ 2

1
i(i+ 2)
or equivalently
1
i+ 2
=
1
2

1
i

1
i+ 2

:
Expanding the sum term by term,
n
X
i=1
1
i(i+ 2)
=
n
X
i=1
1
2

1
i

1
i+ 2

=

1
1
3

+

1
2

1
4

+

1
3

1
5

+

1
n2

1
n

+

1
n1

1
n+ 1

+

1
n

1
n+ 2

= 1 +
1
2

1
n+ 1

1
n+ 2
=
n(3n+ 2)
2(n+ 1)(n + 2)
:
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DEPED COPY 4. Determine the value ofNsuch that
N
X
i=0
1
i
2
+ 3i+ 2
=
97
98
.
Solution:Rewrite the sum as
N
X
i=1
1
i
2
+ 3i+ 2
=
N
X
i=1
1
(i+ 1)(i + 2)
=
N
X
i=1

1
i+ 1

1
i+ 2

Setf(i) =
1
i+ 1
and use telescoping sums to get
N
X
i=1
1
i
2
+ 3i+ 2
=
N
X
i=1

1
i+ 2

1
i+ 1

=

1
N+ 2

1
1

=
N+ 1
N+ 2
:
Since we want the sum to be equal to
97
98
,N= 96.
Supplementary Problems 2.2
1. Expand the following sums:
(a)
10
X
i=3
p
3
i
2
(b)
5
X
i=1
x
2i
2
i
(c)
5
X
i=2
(1)
i
x
i1
2. Write the following in sigma notation.
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DEPED COPY (a) (x + 5)(x+ 3)
2
+ (x+ 1)
3
(x1)
4
(b)
1
3
3
+
2
2
4
3
+
3
2
5
3
+: : :+
10
2
11
3
(c)a3+a6+a9+: : :+a81
3. Evaluate the following sums
(a)
150
X
i=1
(4i+ 2)
(b)
120
X
i=3
i(i5)
(c)
50
X
i=1
(2i1)(2i+ 1)
4. If
50
X
i=1
f(i) = 20 and
50
X
i=1
g(i) = 30, what is the value of
50
X
i=1
g(i) + 3f(i)
p
2
?
5. Ifs=
200
X
i=1

(i1)
2
i
2

, express
200
X
i=1
iin terms ofs.
6. Ifs=
n
X
i=1
aiandt=
n
X
i=1
bi, does it follow that
n
X
i=1
ai
bi
=
s
t
?
4
Lesson 2.3.Principle of Mathematical Induction
Learning Outcomes of the Lesson
At the end of the lesson, the student is able to:
(1) illustrate the Principle of Mathematical Induction; and
(2) apply mathematical induction in proving identities.
Lesson Outline
(1) State the Principle of Mathematical Induction
(2) Prove summation identities using mathematical induction
(3) Prove divisibility statements using mathematical induction
(4) Prove inequalities using mathematical induction
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DEPED COPY Introduction
We have derived and used formulas for the terms of arithmetic and geometric
sequences and series. These formulas and many other theorems involving positive
integers can be proven with the use of a technique calledmathematical induction.
2.3.1. Proving Summation Identities
The Principle of Mathematical Induction
LetP(n) be a property or statement about an integern. Suppose
that the following conditions can be proven:
(1)P(n0) is true (that is, the statement is true whenn=n0).
(2) IfP(k) is true for some integerkn0, thenP(k+ 1) is true
(that is, if the statement is true forn=k, then it is also true for
n=k+ 1).
Then the statementP(n) is true for all integersnn0.
The Principle of Mathematical Induction is often compared to climbing an
innite staircase. First, you need to be able to climb up to the rst step. Second,
if you are on any step (n=k), you must be able to climb up to the next step
(n=k+ 1). If you can do these two things, then you will be able to climb up
the innite staircase.
Part 1 Part 2
Another
analogy of the Principle of Mathematical Induction that is used is
toppling an innite line of standing dominoes. You need to give the rst domino a push so that it falls down. Also, the dominoes must be arranged so that if the
kth domino falls down, the next domino will also fall down. These two conditions
will ensure that the entire line of dominoes will fall down.
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DEPED COPY Standing Domino Tiles, by Nara Cute, 16 October 2015,
https://commons.wikimedia.o
rg/wiki/File:Wallpaper
kartudomino.png.
Public Domain.
There are many mathematical results that can be proven using mathematical
induction. In this lesson, we will focus on three main categories: summation
identities, divisibility statements, and inequalities.
Let us now take a look at some examples on the use of mathematical induction
in proving summation identities.
Example 2.3.1.Using mathematical induction, prove that
1 + 2 + 3 + +n=
n(n+ 1)
2
for
all positive integersn.
Solution.We need to establish the two conditions stated in the Principle of Math-
ematical Induction.
Part 1.
Prove that the identity is true forn= 1.
The left-hand side of the equation consists of one term equal to 1. The right-
hand side becomes
1(1 + 1)
2
=
2
2
=
1:
Hence, the formula is true forn= 1.
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DEPED COPY Part 2.Assume that the formula is true forn=k1:
1 + 2 + 3 + +k=
k(k+ 1)
2
:
We want to show that the formula is true forn=k+ 1; that is,
1 + 2 + 3 + +k+ (k+ 1) =
(k+ 1)(k + 1 + 1)
2
:
Using the formula forn=kand addingk+ 1 to both sides of the equation,
we get
1 + 2 + 3 + +k+ (k+ 1) =
k(k+ 1)
2
+ (k+ 1)
=
k(k+ 1) + 2(k + 1)
2
=
(k+ 1)(k + 2)
2
=
(k+ 1) [(k + 1) + 1]
2
We have proven the two conditions required by the Principle of Mathematical
Induction. Therefore, the formula is true for all positive integersn. 2
Example 2.3.2.Use mathematical induction to prove the formula for the sum
of a geometric series withnterms:
Sn=
a1(1r
n
)
1r
; r6= 1:
Solution.Letanbe thenth term of a geometric series. From Lesson 2.1, we know
thatan=a1r
n1
.
Part 1.Prove that the formula is true forn= 1.
a1(1r
1
)
1r
=a1=S1
The formula is true forn= 1.
Part 2.Assume that the formula is true forn=k1:Sk=
a1(1r
k
)
1r
. We
want to prove that it is also true forn=k+ 1; that is,
Sk+1=
a1(1r
k+1
)
1r
:
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DEPED COPY We know that
Sk+1=a1+a2+ +ak
| {z }
Sk
+ak+1
=Sk+ak+1
=
a1

1r
k

1r
+a1r
k
=
a1

1r
k

+a1r
k
(1r)
1r
=
a1

1r
k
+r
k
r
k+1

1r
=
a1

1r
k+1

1r
By the Principle of Mathematical Induction, we have proven that
Sn=
a1(1r
n
)
1r
for all positive integersn. 2
Example 2.3.3.Using mathematical induction, prove that
1
2
+ 2
2
+ 3
2
+ +n
2
=
n(n+ 1)(2n + 1)
6
for all positive integersn.
Solution.We again establish the two conditions stated in the Principle of Math-
ematical Induction.
Part 1
1(1 + 1)(21 + 1)
6
=
123
6
= 1 = 1
2
The formula is true forn= 1.
Part 2
Assume: 1
2
+ 2
2
+ 3
2
+ +k
2
=
k(k+ 1)(2k + 1)
6
.
Prove: 1
2
+ 2
2
+ 3
2
+ +k
2
+ (k+ 1)
2
=
(k+ 1)(k + 2) [2(k + 1) + 1]
6
=
(k+ 1)(k + 2)(2k + 3)
6
:
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DEPED COPY 1
2
+ 2
2
+ 3
2
+ +k
2
+ (k+ 1)
2
=
k(k+ 1)(2k + 1)
6
+ (k+ 1)
2
=
k(k+ 1)(2k + 1) + 6(k+ 1)
2
6
=
(k+ 1) [k (2k+ 1) + 6(k + 1)]
6
=
(k+ 1) (2k
2
+ 7k+ 6)
6
=
(k+ 1)(k + 2)(2k + 3)
6
Therefore, by the Principle of Mathematical Induction,
1
2
+ 2
2
+ 3
2
+ +n
2
=
n(n+ 1)(2n + 1)
6
for all positive integersn. 2
2.3.2. Proving Divisibility Statements
We now prove some divisibility statements using mathematical induction.
Example 2.3.4.Use mathematical induction to prove that, for every positive
integern, 7
n
1 is divisible by 6.
Solution.Similar to what we did in the previous session, we establish the two
conditions stated in the Principle of Mathematical Induction.
Part 1
7
1
1 = 6 = 61
7
1
1 is divisible by 6.
Part 2
Assume: 7
k
1 is divisible by 6.
To show: 7
k+1
1 is divisible by 6.
7
k+1
1 = 77
k
1 = 67
k
+ 7
k
1 = 67
k
+ (7
k
1)
By denition of divisibility, 67
k
is divisible by 6. Also, by the hypothesis
(assumption), 7
k
1 is divisible by 6. Hence, their sum (which is equal to
7
k+1
1) is also divisible by 6.
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DEPED COPY Therefore, by the Principle of Math Induction, 7
n
1 is divisible by 6 for all
positive integersn. 2
Note that 7
0
1 = 11 = 0 = 60 is also divisible by 6. Hence, a stronger
and more precise result in the preceding example is: 7
n
1 is divisible by 6 for
every nonnegative integern. It does not make sense to substitute negative values
ofnsince this will result in non-integer values for 7
n
1.
Example 2.3.5.Use mathematical induction to prove that, for every nonnega-
tive integern,n
3
n+ 3 is divisible by 3.
Solution.We again establish the two conditions in the Principle of Mathematical
Induction.
Part 1Note that claim of the statement is that it is true for every nonnegative
integern. This means that Part 1 should prove that the statement is true for
n= 0.
0
3
0 + 3 = 3 = 3(1)
0
3
0 + 3 is divisible by 3.
Part 2.We assume thatk
3
k+ 3 is divisible by 3. By denition of divisibility,
we can writek
3
k+ 3 = 3afor some integera.
To show: (k+ 1)
3
(k+ 1) + 3 is divisible by 3.
(k+ 1)
3
(k+ 1) + 3 =k
3
+ 3k
2
+ 2k+ 3
= (k
3
k+ 3) + 3k
2
+ 3k
= 3a+ 3k
2
+ 3k
= 3(a +k
2
+k)
Sincea+k
2
+kis also an integer, by denition of divisibility, (k+1)
3
(k+1)+3
is divisible by 3.
Therefore, by the Principle of Math Induction,n
3
n+ 3 is divisible by 3 for
all positive integersn. 2
?
2.3.3. Proving Inequalities
Finally, we now apply the Principle of Mathematical Induction in proving some
inequalities involving integers.
Example 2.3.6.Use mathematical induction to prove that 2
n
>2nfor every
integern3.
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DEPED COPY Solution.Just like the previous example, we establish the two conditions in the
Principle of Mathematical Induction.
Part 1
2
3
= 8>6 = 2(3)
This conrms that 2
3
>2(3).
Part 2
Assume: 2
k
>2k, wherekis an integer withk3
To show: 2
k+1
>2(k+ 1) = 2k+ 2
We compare the components of the assumption and the inequality we need to
prove. On the left-hand side, the expression is doubled. On the right-hand side,
the expression is increased by 2. We choose which operation we want to apply to
both sides of the assumed inequality.
Alternative 1.We double both sides.
Since 2
k
>2k, by the multiplication property of inequality, we have 22
k
>
22k.
2
k+1
>2(2k) = 2k+ 2k >2k+ 2 ifk3.
Hence, 2
k+1
>2(k+ 1).
Alternative 2.We increase both sides by 2.
Since 2
k
>2k, by the addition property of inequality, we have 2
k
+2>2k+2.
2(k+ 1) = 2k+ 2<2
k
+ 2<2
k
+ 2
k
ifk3.
The right-most expression above, 2
k
+ 2
k
, is equal to 2

2
k

= 2
k+1
.
Hence, 2(k + 1)<2
k+1
.
Therefore, by the Principle of Math Induction, 2
n
>2nfor every integer
n3. 2
We test the above inequality for integers less than 3.
2
0
= 1>0 = 2(0) True
2
1
= 2 = 2(1) False
2
2
= 4 = 2(2) False
The inequality is not always true for nonnegative integers less than 3. This
illustrates the necessity of Part 1 of the proof to establish the result. However,
the result above can be modied to: 2
n
2nfor all nonnegative integersn.
Before we discuss the next example, we review the factorial notation. Recall
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DEPED COPY that 0! = 1 and, for every positive integern,n! = 123 n. The factorial also
satises the property that (n + 1)! = (n+ 1)n!.
Example 2.3.7.Use mathematical induction to prove that 3
n
<(n+ 2)! for
every positive integern. Can you rene or improve the result?
Solution.We proceed with the usual two-part proof.
Part 1
3
1
= 3<6 = 3! = (1 + 2)! =)3
1
<(1 + 2)!
Thus, the desired inequality is true forn= 1.
Part 2
Assume: 3
k
<(k+ 2)!
To show: 3
k+1
<(k+ 3)!
Given that 3
k
<(k+ 2)!, we multiply both sides of the inequality by 3 and
obtain
3

3
k

<3 [(k+ 2)!]:
This implies that
3

3
k

<3 [(k+ 2)!]<(k+ 3) [(k + 2)!];sincek >0,
and so
3
k+1
<(k+ 3)!:
Therefore, by the Principle of Math Induction, we conclude that 3
n
<(n+2)!
for every positive integern.
The left-hand side of the inequality is dened for any integern. The right-
hand side makes sense only ifn+ 20, orn 2.
Whenn=2: 3
2
=
1
9
<1 = 0! = (2 + 2)!
Whenn=1: 3
1
=
1
3
<1 = 1! = (1 + 2)!
Whenn= 0: 3
0
= 1<2 = 2! = (0 + 2)!
Therefore, 3
n
<(n+ 2)! for any integern 2. 2
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DEPED COPY More Solved Examples
Use mathematical induction to prove the given statements below.
1. 23 + 23
2
+: : :+ 23
n1
= 3
n
3 forn1
Solution:
Part 1.
23 = 6 = 3
2
3.
The formula is true forn= 1.
Part 2.
Assume:P= 23 + 23
2
+: : :+ 23
k1
= 3
k
3.
To show: 23 + 23
2
+: : :+ 23
k
= 3
k+1
3.
23 + 23
2
+: : :+ 23
k
=P+ 23
k
= 3
k
3 + 23
k
= 33
k
3
= 3
k+1
3:
2. 1 + 4 + 4
2
+: : :+ 4
n1
=
1
3
(4
n
1) forn1
Solution:Part 1.
1 =
1
3
(4
1
1).
The formula is true forn= 1.
Part 2.
Assume:P= 1 + 4 + 4
2
+: : :+ 4
k1
=
1
3

4
k
1

.
To show: 1 + 4 + 4
2
+: : :+ 4
k
=
1
3

4
k+1
1

.
1 + 4 + 4
2
+: : :+ 4
k
=P+ 4
k
=
1
3

4
k
1

+ 4
k
=
4
3
4
k

1
3
=
1
3

4
k+1
1

:
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DEPED COPY 3.

1
1
2
2

1
1
3
2

1
1
(n1)
2

1
1
n
2

=
n+ 1
2n
forn2.
Solution:
Part 1.
1
1
2
2
=
3
4
=
2 + 1
2(2)
.
The formula is true forn= 2.
Part 2.
Assume:P=

1
1
2
2

1
1
3
2

1
1
(k1)
2

1
1
k
2

=
k+ 1
2k
.
To show:

1
1
2
2

1
1
k
2

1
1
(k+ 1)
2

=
k+ 2
2(k+ 1)
.

1
1
2
2

1
1
(k+ 1)
2

=P

1
1
(k+ 1)
2

=
k+ 1
2k

(k
2
+ 2k+ 1)1
(k+ 1)
2
=
k+ 1
2k

k(k+ 2)
(k+ 1)
2
=
k+ 2
2(k+ 1)
:
4. Prove that 4
n+1
+ 5
2n1
is divisible by 21 for all integersn1.
Solution:
Part 1.
4
1+1
+ 5
2(1)1
= 21.
The number is divisible by 21 forn= 1.
Part 2.
Assume: 4
k+1
+ 5
2k1
is divisible by 21.
Prove: 4
k+2
+ 5
2(k+1)1
is divisible by 21.
4
k+2
+ 5
2(k+1)1
= 44
k+1
+ 255
2k1
= 4

4
k+1
+ 5
2k1

+ 215
2k1
215
2k1
is divisible by 21 and by the hypothesis (assumption), 4
k+1
+5
2k1
is
divisible by 21. Hence, their sum which is equal to 4
k+2
+ 5
2(k+1)1
is divisible
by 21.
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DEPED COPY 5.n
2
>2n+ 3 forn4.
Solution:
Part 1.
2
4
= 16>7 = 2(2) + 3 The inequality is true forn= 4.
Part 2
Assume:k
2
>2k+ 3
Prove: (k+ 1)
2
>2(k+ 1) + 3
We expand (k + 1)
2
and use the inequality in the hypothesis to get
(k+ 1)
2
=k
2
+ 2k+ 1>(2k+ 3) + 2k+ 1 = 4(k + 1)>2(k+ 1) + 3 ifk >0.
Therefore, by the principle of math induction,n
2
>2n+ 3 forn4.
6. Prove that 2
n+3
<(n+ 3)! forn4.
Solution:
Part 1.
2
4+3
= 2
7
<123 7 = (4 + 3)! The inequality is true forn= 1.
Part 2
Assume: 2
k+3
<(k+ 3)!
Prove: 2
k+4
<(k+ 4)!
Given that 2
k+3
<(k+ 3)!, we multiply both sides of the inequality by 2 and
obtain
2

2
k+3

<2 [(k+ 3)!].
This implies that
2
k+4
<2 [(k+ 3)!]<(k+ 4) [(k + 3)!], ifk >0.
Therefore, by the principle of math induction, 2
k+3
<(k+3)! for every positive
integern.
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DEPED COPY Supplementary Problems 2.3
Prove the following by mathematical induction:
1.
1
2
+
2
2
2
+
3
2
3
+ +
n
2
n
= 2
n+ 2
2
n
forn1
2.
n
X
i=1
(i+ 1) =
n(n+ 3)
2
3. 1(1!) + 2(2!) +: : :+n(n!) = (n + 1)!1.
4. The sum of the rstnodd numbers is equal ton
2
.
5.

1
1
2

1
1
3

1
1
4

: : :

1
1
n

=
1
2n
:
6.
n
X
i=1
(1)
i
i
2
=
(1)
n
n(n+ 1)
2
7. 4
3n+1
+ 2
3n+1
+ 1 is divisible by 7
8. 11
n+2
+ 12
2n+1
is divisible by 133
9. 5
2n+1
2
n+2
+ 3
n+2
2
2n+1
is divisible by 19
10. 11
n
6 is divisible by 5
11.
10
n
3
+
5
3
+ 4
n+2
is divisible by 3
12.n
2
<2
n
forn5.
13.
1
1
3
+
1
2
3
+
1
3
3
+: : :+
1
n
3
2
1
n
forn1.
14. The sequencean=
p
2an1; a1=
p
2 is increasing; that is,an< an+1.
4
Lesson 2.4.The Binomial Theorem
Learning Outcomes of the Lesson
At the end of the lesson, the student is able to:
(1) illustrate Pascal's Triangle in the expansion of (x+y)
n
for small positive
integral values ofn;
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DEPED COPY (2) prove the Binomial Theorem;
(3) determine any term in (x+y)
n
, wherenis a positive integer, without ex-
panding; and
(4) solve problems using mathematical induction and the Binomial Theorem.
Lesson Outline
(1) Expand (x +y)
n
for small values ofnusing Pascal's Triangle
(2) Review the denition of and formula for combination
(3) State and prove the Binomial Theorem
(4) Compute all or specied terms of a binomial expansion
(5) Prove some combination identities using the Binomial Theorem
Introduction
In this lesson, we study two ways to expand (a +b)
n
, wherenis a positive
integer. The rst, which uses Pascal's Triangle, is applicable ifnis not too big,
and if we want to determine all the terms in the expansion. The second method
gives a general formula for the expansion of (a +b)
n
for any positive integern.
This formula is useful especially whennis large because it avoids the process of
going through all the coecients for lower values ofnobtained through Pascal's
Triangle.
2.4.1. Pascal's Triangle and the Concept of Combination
Consider the following powers ofa+b:
(a+b)
1
=a+b
(a+b)
2
=a
2
+ 2ab+b
2
(a+b)
3
=a
3
+ 3a
2
b+ 3ab
2
+b
3
(a+b)
4
=a
4
+ 4a
3
b+ 6a
2
b
2
+ 4ab
3
+b
4
(a+b)
5
=a
5
+ 5a
4
b+ 10a
3
b
2
+ 10a
2
b
3
+ 5ab
4
+b
5
We now list down the coecients of each expansion in a triangular array as
follows:
n= 1 : 1 1
n= 2 : 1 2 1
n= 3 : 1 3 3 1
n= 4 : 1 4 6 4 1
n= 5 : 1 5 10 10 5 1
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DEPED COPY The preceding triangular array of numbers is part of what is called thePas-
cal's Triangle, named after the French mathematician, Blaise Pascal (1623-1662).
Some properties of the Triangle are the following:
(1) Each row begins and ends with 1.
(2) Each row hasn+ 1 numbers.
(3) The second and second to the last number of each row correspond to the
row number.
(4) There is symmetry of the numbers in each row.
(5) The number of entries in a row is one more than the row number (or one
more than the number of entries in the preceding row).
(6) Every middle number after rst row is the sum of the two numbers above
it.
It is the last statement which is useful in constructing the succeeding rows of the
triangle.
Example 2.4.1.Use Pascal's Triangle to expand the expression (2x3y)
5
.
Solution.We use the coecients in the fth row of the Pascal's Triangle.
(2x3y)
5
= (2x)
5
+ 5(2x)
4
(3y) + 10(2x)
3
(3y)
2
+ 10(2x)
2
(3y)
3
+ 5(2x)(3y )
4
+ (3y )
5
= 32x
5
240x
4
y+ 720x
3
y
2
1080x
2
y
3
+ 810xy
4
243y
5
2
Example 2.4.2.Use Pascal's Triangle to expand (a+b)
8
.
Solution.We start with the sixth row (or any row of the Pascal's Triangle that
we remember).
n= 6 : 1 6 15 20 15 6 1
n= 7 : 1 7 21 35 35 21 7 1
n= 8 : 1 8 28 56 70 56 28 8 1
Therefore, we get
(a+b)
8
=a
8
+ 8a
7
b+ 28a
6
b
2
+ 56a
5
b
3
+ 70a
4
b
4
+ 56a
3
b
5
+ 28a
2
b
6
+ 8ab
7
+b
8
2
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DEPED COPY We observe that, for eachn, the expansion of (a +b)
n
starts witha
n
and the exponent ofain the succeeding terms decreases by 1, while
the exponent ofbincreases by 1. This observation will be shown to
be true in general.
Let us review the concept of combination. Recall thatC(n; k) or

n
k

counts
the number of ways of choosingkobjects from a set ofnobjects. It is also useful
to know some properties ofC(n; k):
(1)C(n;0) =C(n; n) = 1,
(2)C(n;1) =C(n; n1) =n, and
(3)C(n; k) =C(n; nk).
These properties can explain some of the observations we made on the num-
bers in the Pascal's Triangle. Also recall the general formula for the number of
combinations ofnobjects takenkat a time:
C(n; k) =

n
k

=
n!
k!(nk)!
;
where 0! = 1 and, for every positive integern,n! = 123 n.
Example 2.4.3.Compute

5
3

and

8
5

.
Solution.
5
3

=
5!
(53)!3!
=
5!
2!3!
= 10

85

=
8!
(85)!5!
=
10!
3!5!
= 56 2
You may observe that the value of

5
3

and the fourth coecient in the fth
row of Pascal's Triangle are the same. In the same manner,

8
5

is equal to the
sixth coecient in the expansion of (a +b)
8
(see Example 2.4.2). These observed
equalities are not coincidental, and they are, in fact, the essence embodied in the
Binomial Theorem, as you will see in the succeeding sessions.
2.4.2. The Binomial Theorem
As the powerngets larger, the more laborious it would be to use Pascal's Triangle
(and impractical to use long multiplication) to expand (a+b)
n
. For example,
using Pascal's Triangle, we need to compute row by row up to the thirtieth row
111All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means -
electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2016.

DEPED COPY to know the coecients of (a +b)
30
. It is, therefore, delightful to know that it is
possible to compute the terms of a binomial expansion of degreenwithout going
through the expansion of all the powers less thann.
We now explain how the concept of combination is used in the expansion of
(a+b)
n
.
(a+b)
n
= (a+b)(a+b)(a+b) (a+b)
| {z }
nfactors
When the distributive law is applied, the expansion of (a +b)
n
consists of
terms of the forma
m
b
i
, where 0m; in. This term is obtained by choosing
aformof the factors andbfor the rest of the factors. Hence,m+i=n, or
m=ni. This means that the number of times the terma
ni
b
i
will appear
in the expansion of (a +b)
n
equals the number of ways of choosing (ni) ori
factors from thenfactors, which is exactlyC(n; i). Therefore, we have
(a+b)
n
=
n
X
i=0

n
i

a
ni
b
i
:
To explain the reasoning above, consider the casen= 3.
(a+b)
3
= (a+b)(a+b)(a+b)
=aaa+aab+aba+abb+baa+bab+bba+bbb
=a
3
+ 3a
2
b+ 3ab
2
+b
3
That is, each term in the expansion is obtained by choosing eitheraorbin each
factor. The terma
3
is obtained whenais chosen each time, whilea
2
bis obtained
whenais selected 2 times, or equivalently,bis selected exactly once.
We will give another proof of this result using mathematical induction. But
rst, we need to prove a result about combinations.
Pascal's Identity
Ifnandkare positive integers withkn, then

n+ 1
k

=

n
k

+

n
k1

:
Proof.The result follows from the combination formula.

n
k

+

n
k1

=
n!
k!(nk)!
+
n!
(k1)!(nk+ 1)!
=
n!(nk+ 1) +n!(k)
k!(nk+ 1)!
112All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means -
electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2016.

DEPED COPY =
n!(nk+ 1 +k)
k!(n+ 1k)!
=
n!(n+ 1)
k!(n+ 1k)!
=
(n+ 1)!
k!(n+ 1k)!
=

n+ 1
k

2
Pascal's identity explains the method of constructing Pascal's Triangle, in
which an entry is obtained by adding the two numbers above it. This identity
is also an essential part of the second proof of the Binomial Theorem, which we
now state.
The Binomial Theorem
For any positive integern,
(a+b)
n
=
n
X
i=0

n
i

a
ni
b
i
:
Proof.We use mathematical induction.
Part 1
1
X
i=0

1
i

a
1i
b
i
=

1
0

a
1
b
0
+

1
1

a
0
b
1
=a+b
Hence, the formula is true forn= 1.
Part 2.Assume that
(a+b)
k
=
k
X
i=0

k
i

a
ki
b
i
:
We want to show that
(a+b)
k+1
=
k+1
X
i=0

k+ 1
i

a
k+1i
b
i
:
(a+b)
k+1
= (a+b)(a+b)
k
= (a+b)
k
X
i=0

k
i

a
ki
b
i
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electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2016.

DEPED COPY =a
k
X
i=0

k
i

a
ki
b
i
+b
k
X
i=0

k
i

a
ki
b
i
=
k
X
i=0

k
i

a
ki+1
b
i
+
k
X
i=0

k
i

a
ki
b
i+1
=

k
0

a
k+1
b
0
+
k
X
i=1

k
i

a
k+1i
b
i
+

k
0

a
k
b
1
+

k
1

a
k1
b
2
+

k
2

a
k2
b
3
+ +

k
k1

a
1
b
k
+

k
k

a
0
b
k+1
=a
k+1
+
k
X
i=1

k
i

a
k+1i
b
i
+
k
X
i=1

k
i1

a
k+1i
b
i
+b
k+1
=

k+ 1
0

a
k+1
b
0
+
k
X
i=1

k
i

+

k
i1

a
k+1i
b
i
+

k+ 1
k+ 1

a
0
b
k+1
=
k+1
X
i=0

k+ 1
i

a
k+1i
b
i
The last expression above follows from Pascal's Identity.
Therefore, by the Principle of Mathematical Induction,
(a+b)
n
=
n
X
i=1

n
i

a
ni
b
i
for any positive integern. 2
2.4.3. Terms of a Binomial Expansion
We now apply the Binomial Theorem in dierent examples.
Example 2.4.4.Use the Binomial Theorem to expand (x +y)
6
.
114All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means -
electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2016.

DEPED COPY Solution.
(x+y)
6
=
6
X
k=0

6
k

x
6k
y
k
=

6
0

x
6
y
0
+

6
1

x
5
y
1
+

6
2

x
4
y
2
+

6
3

x
3
y
3
+

6
4

x
2
y
4
+

6
5

x
1
y
5
+

6
6

x
0
y
6
=x
6
+ 6x
5
y+ 15x
4
y
2
+ 20x
3
y
3
+ 15x
2
y
2
+ 6xy
5
+y
6
2
Since the expansion of (a +b)
n
begins withk= 0 and ends withk=n, the
expansion hasn+ 1 terms. The rst term in the expansion is

n
0

a
n
=a
n
, the
second term is

n
1

a
n1
b=na
n=1
b, the second to the last term is

n
n1

ab
n1
=
nab
n1
, and the last term is

n
n

b
n
=b
n
.
Thekth term of the expansion is

n
k1

a
nk+1
b
k1
. Ifnis even, there is a
middle term, which is the

n
2
+ 1

th term. Ifnis odd, there are two middle
terms, the

n+1
2

th and

n+1
2
+ 1

th terms.
The general term is often represented by

n
k

a
nk
b
k
. Notice that, in any term,
the sum of the exponents ofaandbisn. The combination

n
k

is the coecient
of the term involvingb
k
. This allows us to compute any particular term without
needing to expand (a +b)
n
and without listing all the other terms.
Example 2.4.5.Find the fth term in the expansion of

2x
p
y

20
.
Solution.The fth term in the expansion of a fth power corresponds tok= 4.

20
4

(2x)
204
(
p
y)
4
= 4845

65536x
16

y
2
= 317521920x
16
y
2
2
Example 2.4.6.Find the middle term in the expansion of

x
2
+ 3y

6
.
Solution.Since there are seven terms in the expansion, the middle term is the
fourth term (k = 3), which is

6
3

x
2

3
(3y)
3
= 20

x
3
8

27y
3

=
135x
3
y
3
2
: 2
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electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2016.

DEPED COPY Example 2.4.7.Find the term involvingx(with exponent 1) in the expansion
of

x
2

2y
x

8
.
Solution.The general term in the expansion is

8
k

x
2

8k

2y
x

k
=

8
k

x
162k

(2)
k
y
k
x
k
=

8
k

(2)
k
x
162kk
y
k
=

8
k

(2)
k
x
163k
y
k
:
The term involvesxif the exponent ofxis 1, which means 163k= 1, or
k= 5. Hence, the term is

8
5

(2)
5
xy
5
=1792xy
5
: 2
?
2.4.4. Approximation and Combination Identities
We continue applying the Binomial Theorem.
?
Example 2.4.8.(1) Approximate (0:8)
8
by using the rst three terms in the
expansion of (10:2)
8
. Compare your answer with the calculator value.
(2) Use 5 terms in the binomial expansion to approximate (0:8)
8
. Is there an
improvement in the approximation?
Solution.
(0:8)
8
= (10:2)
8
=
8
X
k=0

8
k

(1)
8k
(0:2)
k
=
8
X
k=0

8
k

(0:2)
k
(1)
2
X
k=0

8
k

(0:2)
k
=

8
0

+

8
1

(0:2) +

8
2

(0:2)
2
= 11:6 + 1:12 = 0:52
The calculator value is 0:16777216, so the error is 0:35222784.
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electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2016.

DEPED COPY (2)
4
X
k=0

8
k

(0:2)
k
=

8
0

+

8
1

(0:2) +

8
2

(0:2)
2
+

8
3

(0:2)
3
+

8
4

(0:2)
4
= 0:520:448 + 0:112 = 0:184
The error is 0:01622784, which is an improvement on the previous estimate.
2
Example 2.4.9.Use the Binomial Theorem to prove that, for any positive in-
tegern,
n
X
k=0

n
k

= 2
n
:
Solution.Seta=b= 1 in the expansion of (a +b)
n
. Then
2
n
= (1 + 1)
n
=
n
X
k=0

n
k

(1)
nk
(1)
k
=
n
X
k=0

n
k

: 2
Example 2.4.10.Use the Binomial Theorem to prove that

100
0

+

100
2

+

100
4

+ +

100
100

=

100
1

+

100
3

+

100
5

+ +

100
99

Solution.Leta= 1 andb=1 in the expansion of (a+b)
100
. Then

1 + (1)

100
=
100
X
k=0

100
k

(1)
100k
(1)
k
:
0 =

100
0

+

100
1

(1) +

100
2

(1)
2
+

100
3

(1)
3
+ +

100
99

(1)
99
+

100
100

(1)
100
Ifkis even, then (1)
k
= 1. Ifkis odd, then (1)
k
=1. Hence, we have
0 =

100
0

100
1

+

100
2

100
3

+

100
99

+

100
100

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electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2016.

DEPED COPY Therefore, after transposing the negative terms to other side of the equation, we
obtain

100
0

+

100
2

+

100
4

+ +

100
100

=

100
1

+

100
3

+

100
5

+ +

100
99

2
More Solved Examples
1. Use the Binomial Theorem to expand (2x
4
3y
2
)
5
.
Solution:

2x
4
3y
2

5
=
5
X
k=0

2x
4

5k
3y
2

k
= 32x
20
240x
16
y
2
+720x
12
y
4

1080x
8
y
6
+ 810x
4
y
8
243y
10
2. Determine the 20th term in the expansion of (x
3
3y)
28
.
Solution:We see thatk= 19 should yield the 20th term, yielding3
19

28
19

x
27
y
19
.
3. Find the term containing
x
2
y
2
in the expansion of

x
y

y
2
2x
2

20
.
Solution:Settinga=
x
y
; b=
y
2
2x
2
, the (k + 1)th term in the binomial
expansion is (1)
k

20
k

x
y

20k
y
2
2x
2

k
=
(1)
k
2
k

n
k

x
203k
y
3k20
. To get
x
2
y
2
, we get 203k= 2)k= 6, yielding
1
2
6

20
6

x
2
y
2
.
4. Determine the term not involvingxin the expansion of

x
3
+
2
x
5

16
.
Solution:Settinga=x
3
; b=
2
x
5
, the (k +1)th term in the binomial expansion
is

16
k

x
3

16k

2
x
5

k
= 2
k

16
k

x
488k
. To get the term withoutx, we get
488k= 0)k= 6, yielding 2
6

16
6

.
5. Determine the coecient ofx
9
in the expansion of (1 + 2x)
10
.
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electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2016.

DEPED COPY Solution:Settinga= 1; b= 2x, the (k + 1)th term in the binomial expansion
of the rst factor is

10
k

(1)
10k
(2x)
k
= 2
k

10
k

x
k
. To getx
9
, we setk= 9,
yielding 2
9

10
9

x
9
.
6. Prove that
n
X
i=0
(1)
k

n
k

3
nk
= 2
n
.
Solution:Seta= 3; b=1.
7. If
p
3 +
p
2

5
is written in the forma
p
3 +b
p
2 wherea; bare integers, what
isa+b?
Solution:We have
p
3 +
p
2

5
=
5
X
k=0

5
k

p
3

5kp
2

k
. Note that if
5kis odd (or equivalently,kis even), the term has a factor of
p
3, while
the rest have a factor of
p
2. Thus,a=

5
0

+

5
2

+

5
4

= 16 and
b=

5
1

+

5
3

+

5
5

= 16 yieldsa+b= 32.
Supplementary Problems 2.4
1. Use the Binomial Theorem to expand the following:
(a) (2x 3y)
5
(b)
p
x
3

2
x
2

4
(c) (1 +
p
x)
4
2. Without expanding completely, nd the indicated value(s) in the expansion of
the following:
(a) (2 +x)
9
, two middle terms
(b)

p
2
+
2
q

10
, 3rd term
(c) (x
2
+y
4
)
21
, last 2 terms
(d)

1
p
x

20
, middle term
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electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2016.

DEPED COPY (e)

2y
4
x
3
+
x
5
4y

15
, term not involvingy
(f)

1
2x
2
x
2

13
, term involvingx
2
(g) (12x)
6
, coecient ofx
3
(h)

2y
7=3

1
2y
5=3

30
, coecient of
1
y
2
(i) (
p
x3)
8
, coecient ofx
7=2
(j) (
p
x+ 2)
6
, coecient ofx
3=2
3. Approximate (2:1)
10
by using the rst 5 terms in the expansion of (2 + 0:1)
20
.
Compare your answer with the calculator result.
4. In the expansion of (4x + 3)
34
, thekth value and the (k+ 1)st terms have
equal coecients. What is the value ofk?
5. Determine the value of

19
0

19
1

3 +

19
2

3
2

19
3

3
3
+: : :+

19
18

3
18

19
19

3
19
4
120All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means -
electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2016.

DEPED COPY Topic Test 1 for Unit 2
1. Determine if the given sequence is arithmetic, geometric, or neither by writing
A, G, or O, respectively.
(a)
1
3
;
1
2
;
3
4
;
9
16
;
27
32
; : : :
(b)
1
2
;
1
7
;
1
12
;
1
17
;
1
21
; : : :
(c) 0;3;8;15;24; : : :
2. Three numbers form an arithmetic sequence, the common dierence being 5.
If the last number is increased by 1, the second by 2, and the rst by 4, the
resulting numbers form a geometric sequence. Find the numbers.
3. Evaluate the sum
50
X
i=1
2i
3
+ 9i
2
+ 13i + 6
i
2
+ 3i+ 2
.
4. Find the indicated terms in the expansion of the given expression.
(a)

x
2

1
2

8
, term involvingx
8
(b) (n
3
3m)
28
, 20th term
5. Prove the statement below for all positive integersnby mathematical induc-
tion.
1
13
+
1
35
+ +
1
(2n1)(2n + 1)
=
n
2n+ 1
6. On his 20th birthday, Ian deposited an amount of 10,000 pesos to a time-
deposit scheme with a yearly interest of 4%. Ian decides not to withdraw anyamount of money or earnings and vows to keep it in the same time-depositscheme year after year. Show that the new amounts in Ian's time-depositaccount in each succeeding birthday represent a geometric sequence, and use
this to determine the value of the money during Ian's 60th birthday.
121All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means -
electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2016.

DEPED COPY Topic Test 2 for Unit 2
1. Determine if the given sequence is arithmetic, geometric, or neither by writing
A, G, or O, respectively.
(a)
2
3
;
8
15
;
32
75
;
128
375
;
512
1875
; : : :
(b)
1
2
;
2
3
;
3
4
;
4
5
;
5
6
; : : :
(c) 3;
11
2
;8;
21
2
;13; : : :
2. The sum of the rst two terms of an arithmetic sequence is 9 and the sum of
the rst three terms is also 9. How many terms must be taken to give a sum
of126?
3. Evaluate the following sums.
(a)
50
X
i=1
(2i+ 1)(i 3) (b)
30
X
i=1
r
i
2
2i+ 1
4
4. Find the term not involvingxin the expansion of

x
3
+
1
x

8
.
5. Prove that the following statements are true for all positive integersnby
mathematical induction.
(a) 1 + 4 + 7 +: : :+ (3n 2) =
n(3n1)
2
(b) 3
n
+ 7
n1
+ 8 is divisible by 12.
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electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2016.

DEPED COPY Unit 3
Trigonometry
Puerto Princesa Subterranean River National Park, by Giovanni G. Navata, 12 November 2010,
https://commons.wikimedia.org/wiki/File%3AUndergroundRiver.jpg.

Named as one of the New Seven Wonders of Nature in 2012 by the New7Wonders
Foundation, the Puerto Princesa Subterranean River National Park is world-
famous for its limestone karst mountain landscape with an underground river.
The Park was also listed as UNESCO World Heritage Site in 1999. The under-
ground river stretches about 8.2 km long, making it one of the world's longest
rivers of its kind.
123All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means -
electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2016.

DEPED COPY Lesson 3.1. Angles in a Unit Circle
Learning Outcomes of the Lesson
At the end of the lesson, the student is able to:
(1) illustrate the unit circle and the relationship between the linear and angular
measures of arcs in a unit circle.
(2) convert degree measure to radian measure, and vice versa.
(3) illustrate angles in standard position and coterminal angles.
Lesson Outline
(1) Linear and angular measure of arcs
(2) Conversion of degree to radian, and vice versa
(3) Arc length and area of the sector
(4) Angle in standard position and coterminal angles
Introduction
Angles are being used in several elds like engineering, medical imaging, elec-
tronics, astronomy, geography and many more. Added to that, surveyors, pilots,
landscapers, designers, soldiers, and people in many other professions heavily use
angles and trigonometry to accomplish a variety of practical tasks. In this les-
son, we will deal with the basics of angle measures together with arc length and
sectors.
3.1.1. Angle Measure
An angle is formed by rotating a ray about its endpoint. In the gure shown
below, theinitial sideof\AOBisOA, while itsterminal sideisOB. An angle
is said to bepositiveif the ray rotates in a counterclockwise direction, and the
angle isnegativeif it rotates in a clockwise direction.
124All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means -
electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2016.

DEPED COPY An angle is instandard positionif it is drawn in thexy-plane with its vertex
at the origin and its initial side on the positivex-axis. The angles,, andin
the following gure are angles in standard position.
To measure angles, we use degrees, minutes, seconds, and radians.
A central angle of a circle measures onedegree, written 1

, if it inter-
cepts
1
360
of the circumference of the circle. Oneminute, written 1
0
, is
1
60
of 1

, while onesecond, written 1
00
, is
1
60
of 1
0
.
For example, in degrees, minutes, and seconds,
10

30
0
18
00
= 10

30 +
18
60

0
= 10

30:3
0
=

10 +
30:3
60

= 10:505

and
79:251

= 79

(0:25160)
0
= 79

15:06
0
= 79

15
0
(0:0660)
00
= 79

15
0
3:6
00
:
Recall that the unit circle is the circle with center at the origin and radius 1
unit.
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DEPED COPY A central angle of the unit circle that intercepts an arc of the circle
with length 1 unit is said to have a measure of oneradian, written 1
rad. See Figure 3.1.
Figure 3.1
In
trigonometry, as it was studied in Grade 9, the degree measure is often used.
On the other hand, in some elds of mathematics like calculus, radian measure of
angles is preferred. Radian measure allows us to treat the trigonometric functions
as functions with the set of real numbers as domains, rather than angles.
Example 3.1.1.In the following gure, identify the terminal side of an angle in
standard position with given measure.
(1) degree measure: 135

,135

,90

, 405

(2) radian measure:

4
rad,
3

4
rad,
3

2
rad,

2
rad
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DEPED COPY Solution.(1) 135

:
!
OC;135

:
!
OD;90

:
!
OE; and 405

:
!
OB
(2) radian measure:

4
rad:
!OB;
3
4
rad:
!OD;
3
2
rad:
!OE; and

2
rad:
!
OE 2
Since a unit circle has circumference 2 , a central angle that measures 360

has measure equivalent to 2 radians. Thus, we obtain the following conversion
rules.
Converting degree to radian, and vice versa
1. To convert a degree measure to radian, multiply it by

180
.
2. To convert a radian measure to degree, multiply it by
180

.
Figure 3.2 shows some special angles in standard position with the indicated
terminal sides. The degree and radian measures are also given.
Figure 3.2
Example 3.1.2.Express 75

and 240

in radians.
Solution.
75

180

=
5
12
=)75

=
5
12
rad
240

180

=
4
3
=)240

=
4
3
rad 2
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DEPED COPY Example 3.1.3.Express

8
rad and
11
6
rad in degrees.
Solution.

8

180

= 22:5 =)

8
rad = 22:5

11
6

180

= 330 =)
11
6
rad = 330

2
3.1.2. Coterminal Angles
Two angles in standard position that have a common terminal side are called
coterminal angles. Observe that the degree measures of coterminal angles dier
by multiples of 360

.
Two angles are coterminal if and only if their degree measures dier
by 360k , wherek2Z.
Similarly, two angles are coterminal if and only if their radian mea-
sures dier by 2k , wherek2Z.
As a quick illustration, to nd one coterminal angle with an angle that mea-
sures 410

, just subtract 360

, resulting in 50

. See Figure 3.3.
Figure 3.3
Example 3.1.4.Find the angle coterminal with380

that has measure
(1) between 0

and 360

, and
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DEPED COPY (2) between360

and 0

.
Solution.A negative angle moves in a clockwise direction, and the angle380

lies in Quadrant IV.
(1)380

+ 2360

= 340

(2)380

+ 360

=20

2
3.1.3. Arc Length and Area of a Sector
In a circle, a central angle whose radian measure issubtends an arc that is the
fraction

2
of the circumference of the circle. Thus, in a circle of radiusr(see
Figure 3.4), the lengthsof an arc that subtends the angleis
s=

2
circumference of circle =

2
(2r) =r:
Figure 3.4
In a circle of radiusr, the lengthsof an arc intercepted by a central
angle with measureradians is given by
s=r:
Example 3.1.5.Find the length of an arc of a circle with radius 10 m that
subtends a central angle of 30

.
Solution.Since the given central angle is in degrees, we have to convert it into
radian measure. Then apply the formula for an arc length.
30

180

=

6
rad
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DEPED COPY s= 10

6

=
5
3
m 2
Example 3.1.6.A central anglein a circle of radius 4 m is subtended by an
arc of length 6 m. Find the measure ofin radians.
Solution.
=
s
r
=
6
4
=
3
2
rad 2
A sector of a circle is the portion of the interior of a circle bounded by the
initial and terminal sides of a central angle and its intercepted arc. It is like a
\slice of pizza." Note that an angle with measure 2radians will dene a sector
that corresponds to the whole \pizza." Therefore, if a central angle of a sector
has measureradians, then the sector makes up the fraction

2
of a complete
circle. See Figure 3.5. Since the area of a complete circle with radiusrisr
2
, we
have
Area of a sector =

2
(r
2
) =
1
2
r
2
:
Figure 3.5
In a circle of radiusr, the areaAof a sector with a central angle
measuringradians is
A=
1
2
r
2
:
Example 3.1.7.Find the area of a sector of a circle with central angle 60

if
the radius of the circle is 3 m.
Solution.First, we have to convert 60

into radians. Then apply the formula for
computing the area of a sector.
60

180

=

3
rad
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DEPED COPY A=
1
2
(3
2
)

3
=
3
2
m
2
2
Example 3.1.8.A sprinkler on a golf course fairway is set to spray water over
a distance of 70 feet and rotates through an angle of 120

. Find the area of the
fairway watered by the sprinkler.
Solution.
120

180

=
2
3
rad
A=
1
2
(70
2
)
2
3
=
4900
3
5131 ft
2
2
More Solved Examples
1. Find the equivalent degree measure of
5
48
radians.
Solution:
5
48
rad =
5
48

180

=

75
4

2. Find the equivalent angle measure in degrees and in radians of an angle tracing
2
3
5
revolutions.
Solution:One revolution around a circle is equivalent to tracing 360

.
2
3
5
rev = 2
3
5
rev

360
1 rev

= 936

936

= 936

180

=
26
5
rad
3. Find the smallest positive angle coterminal with2016

.
Solution:Add 6 complete revolutions or 6(360

) = 2160

to the given angle
(or keep on adding 360

until you get a positive angle).
2016

+ 2160

= 144

4. Find the largest negative angle coterminal with
137
5
.
Solution:Subtract 14 complete revolutions or 14(2) = 28 to the given angle
(or keep on subtracting 2 until you get a negative angle).
137
5
28=
3
5
rad
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DEPED COPY 5. Find the length of the arc of a circle with radius 15 cm that subtends a central
angle of 84

.
Solution:
84

= 84

180

=
7
15
rad
s= 15

7
15

= 7cm
6. A central anglein a circle of radius 12 inches is subtended by an arc of length
27 inches. Find the measure ofin degrees.
Solution:
s=r=)=
s
r
=
12
27
=
9
4
rad
9
4
rad =
9
4

180

=

405

7. Find the area of a sector of a circle with central angle of 108

if the radius of
the circle is 15 cm.
Solution:
108

= 108

180

=
3
5
rad
A=
1
2
(15)
2
3
5
=
135
2
cm
2
8. Given isosceles right triangleABCwithACas the hypotenuse (as shown
below), a circle with center atAand radiusABintersectsACatD. What is
the ratio of the area of sectorBADto the area of the regionBCD?
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DEPED COPY Solution:Letrbe the radius of the circle; that is,r=AB.
\A=

4
rad =)Area of sectorBAD=
1
2
r
2

4

=
r
2
8
Area of regionBCD= Area of4ABCArea of sectorBAD=
4r
2
r
2
8
area of sectorBAD
area of the regionBCD
=
r
2
8
4r
2
r
2
8
=

4
Supplementary Problems 3.1
1. How many degrees is 1
1
5
of a complete revolution?
2. How many radians is
11
5
of a complete revolution?
3. What is the length of an arc of a circle with radius 4 cm that subtends a
central angle of 216

?
4. Find the length of an arc of a circle with radius
6

cm that subtends a central
angle of 99

.
5. What is the smallest positive angle coterminal with 2110

?
6. Find the largest negative angle coterminal with
107
6
.
7. Find the area of a sector of a circle with central angle of
7
6
if the diameter of
the circle is 9 cm.
8. Find the area of a sector of a circle with central angle of 108

if the radius of
the circle is 15 cm.
9. What is the radius of a circle in which a central angle of 150

determines a
sector of area 15 in
2
?
10. Find the radius of a circle in which a central angle of
5
4
determines a sector
of area 32 in
2
.
?
11. A central angle of a circle of radius 6 inches is subtended by an arc of length 6
inches. What is the central angle in degrees (rounded to two decimal places)?
?
12. An arc of length

5
cm subtends a central angleof a circle with radius
2
3
cm.
What isin degrees?
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DEPED COPY 13. Two overlapping circles of radii 1 cm are drawn such that each circle passes
through the center of the other. What is the perimeter of the entire region?
14. The length of arcABof a circle with center atOis equal to twice the length
of the radiusrof the circle. Find the area of sectorAOBin terms ofr.
15. The angle of a sector in a given circle is 20

and the area of the sector is equal
to 800 cm
2
. Find the arc length of the sector.
16. In Figure 3.6,AEandBCare arcs of two concentric circles with center atD.
IfAD= 2 cm,BD= 8 cm, and\ADE= 75

, nd the area of the region
AECB.
17. In Figure 3.7,ABandDEare diameters. IfAB= 12 cm and\AOD= 126

,
nd the area of the shaded region.
Figure 3.6 Figure 3.7 Figure 3.8
18.
A point moves outside an equilateral triangle of side 5 cm such that its distance
from the triangle is always 2 cm. See Figure 3.8. What is the length of one
complete path that the point traces?
Figure 3.9
19.
The segment of a circle is the region bounded by a chord and the arc subtended
by the chord. See Figure 3.9. Find the area of a segment of a circle with a central angle of 120

and a radius of 64 cm.
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DEPED COPY Figure 3.10
20.
In Figure 3.10, diameterABof circleOmeasures 12 cm and arcBCmeasures
120

. Find the area of the shaded region.
4
Lesson 3.2. Circular Functions
Learning Outcomes of the Lesson
At the end of the lesson, the student is able to:
(1) illustrate the dierent circular functions; and
(2) use reference angles to nd exact values of circular functions.
Lesson Outline
(1) Circular functions
(2) Reference angles
Introduction
We dene the six trigonometric function in such a way that the domain of
each function is the set of angles in standard position. The angles are measured
either in degrees or radians. In this lesson, we will modify these trigonometric
functions so that the domain will be real numbers rather than set of angles.
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DEPED COPY 3.2.1. Circular Functions on Real Numbers
Recall that the sine and cosine functions (and four others: tangent, cosecant,
secant, and cotangent) of angles measuring between 0

and 90

were dened in
the last quarter of Grade 9 as ratios of sides of a right triangle. It can be veried
that these denitions are special cases of the following denition.
Letbe an angle in standard position andP() =P(x; y) the point
on its terminal side on the unit circle. Dene
sin=y csc=
1
y
; y6= 0
cos=x sec=
1
x
; x6= 0
tan=
y
x
; x6= 0 cot =
x
y
; y6= 0
Example 3.2.1.Find the values of cos 135

, tan 135

, sin(60

), and sec(60

).
Solution.Refer to Figure 3.11(a).
(a) (b)
Figure 3.11
From properties of 45

-45

and 30

-60

right triangles (with hypotenuse 1
unit), we obtain the lengths of the legs as in Figure 3.11(b). Thus, the coordinatesofAandBare
A=

p
2
2
;
p
2
2
!
andB=

1
2
;
p
3
2
!
:
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DEPED COPY Therefore, we get
cos 135

=
p
2
2
;tan 135

=1;
sin(60

) =
p
3
2
;and sec(60

) = 2: 2
From the last example, we may then also say that
cos

4
rad

=
p
2
2
;sin

3
rad

=
p
3
2
;
and so on.
From the above denitions, we dene the same six functions on real numbers.
These functions are calledtrigonometric functions.
Letsbe any real number. Supposeis the angle in standard position
with measuresrad. Then we dene
sins= sin cscs= csc
coss= cos secs= sec
tans= tan cots= cot
From the last example, we then have
cos

4

= cos

4
rad

= cos 45

=
p
2
2
and
sin

3

= sin

3
rad

= sin(60

) =
p
3
2
:
In the same way, we have
tan 0 = tan(0 rad) = tan 0

= 0:
Example 3.2.2.Find the exact values of sin
3
2
, cos
3
2
, and tan
3
2
.
Solution.LetP

3
2

be the point on the unit circle and on the terminal side of
the angle in the standard position with measure
3
2
rad. ThenP

3
2

= (0;1),
and so
sin
3
2
=1; cos
3
2
= 0;
but tan
3
2
is undened. 2
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DEPED COPY Example 3.2.3.Supposesis a real number such that sins=
3
4
and coss >0.
Find coss.
Solution.We may considersas the angle with measuresrad. LetP(s) = (x; y )
be the point on the unit circle and on the terminal side of angles.
SinceP(s) is on the unit circle, we know thatx
2
+y
2
= 1. Since sins=y=

3
4
, we get
x
2
= 1y
2
= 1

3
4

2
=
7
16
=)x=
p
7
4
:
Since coss=x >0, we have coss=
p
7
4
. 2
LetP(
x1; y1) andQ(x; y) be points on the terminal side of an anglein
standard position, wherePis on the unit circle andQon the circle of radiusr
(not necessarily 1) with center also at the origin, as shown above. Observe that
we can use similar triangles to obtain
cos=x1=
x1
1
=
x
r
and
sin=y1=
y1
1
=
y
r
:
W
e may then further generalize the denitions of the six circular functions.
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DEPED COPY Letbe an angle in standard position,Q(x; y) any point on the ter-
minal side of, andr=
p
x
2
+y
2
>0. Then
sin=
y
r
csc=
r
y
; y6= 0
cos=
x
r
sec=
r
x
; x6= 0
tan=
y
x
; x6= 0 cot =
x
y
; y6= 0
We then have a second solution for Example 3.2.3 as follows. With sins=
3
4
and sins=
y
r
, we may choosey=3 andr= 4 (which is always positive). In
this case, we can solve forx, which is positive since coss=
x
4
is given to be
positive.
4 =
p
x
2
+ (3)
2
=)x=
p
7 =)coss=
p
7
4
3.2.2. Reference Angle
We observe that if1and2are coterminal angles, the values of the six circular
or trigonometric functions at1agree with the values at2. Therefore, in nding
the value of a circular function at a number, we can always reduceto a number
between 0 and 2. For example, sin
14
3
= sin

14
3
4

= sin
2
3
. Also, observe
from Figure 3.12 that sin
2
3
= sin

3
.
Figure 3.12
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DEPED COPY In general, if1,2,3, and4are as shown in Figure 3.13 withP(1) =
(x1; y1), then each of thex-coordinates ofP(2),P(3), andP(4) isx1, while
they-coordinate isy1. The correct sign is determined by the location of the
angle. Therefore, together with the correct sign, the value of a particular circular
function at an anglecan be determined by its value at an angle1with radian
measure between 0 and

2
. The angle1is called thereference angleof.
Figure 3.13
The signs of the coordinates ofP() depends on the quadrant or axis where
it terminates. It is important to know the sign of each circular function in each
quadrant. See Figure 3.14. It is not necessary to memorize the table, since the
sign of each function for each quadrant is easily determined from its denition.
We note that the signs of cosecant, secant, and cotangent are the same as sine,
cosine, and tangent, respectively.
Figure 3.14
Using the fact that the unit circle is symmetric with respect to thex-axis, the
y-axis, and the origin, we can identify the coordinates of all the points using the
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DEPED COPY coordinates of corresponding points in the Quadrant I, as shown in Figure 3.15
for the special angles.
Figure 3.15
Example 3.2.4.Use reference angle and appropriate sign to nd the exact value
of each expression.
(1) sin
11
6
and cos
11
6
(2) cos

7
6

(3) sin 150

(4) tan
8
3
Solution.(1) The reference angle of
11
6
is

6
, and it lies in Quadrant IV wherein
sine and cosine are negative and positive, respectively.
sin
11
6
=sin

6
=
1
2
cos
11
6
= cos

6
=
p
3
2
(2) The angle
7
6
lies in Quadrant II wherein cosine is negative, and its refer-
ence angle is

6
.
cos

7
6

=cos

6
=
p
3
2
(3) sin 150

= sin 30

=
1
2
(4) tan
8
3
=tan

3
=
sin

3
cos

3
=
p
3
2
1
2
=
p
3 2
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DEPED COPY More Solved Examples
1. IfP() is a point on the unit circle and=
17
3
, what are the coordinates of
P()?
Solution:
17
3
is coterminal with
5
3
which terminates in QIV. The reference
angle is

3
, thereforeP

17
3

=

1
2
;
p
3
2

.
2. IfP() is a point on the unit circle and=
5
6
, nd the values of the six
trigonometric functions of.
Solution:The angle
5
6
terminates in QIII, the reference angle is

6
, therefore
P

5
6

=

p
3
2
;
1
2

.
cos

5
6

=
p
3
2
sec

5
6

=
2
p
3
=
2
p
3
3
sin

5
6

=
1
2
csc

5
6

=2
tan

5
6

=
1
p
3
=
p
3
3
cot

5
6

=
p
3
1
=
p
3
3. Find the six trigonometric functions of the angleif the terminal side ofin
standard position passes through the point (5; 12).
Solution:x= 5,y=12,r=
p
(5)
2
+ (12)
2
= 13.
cos=
x
r
=
5
13
sec=
r
x
=
13
5
sin=
y
r
=
12
13
csc=
r
y
=
13
12
tan=
y
x
=
12
5
cot=
x
y
=
5
12
4. Given sec=
25
24
and
3
2
, nd sin+ cos.
Solution:r= 25,x=24,y=
p
(25)
2
(24)
2
=7.
Sinceis in QIII,y=7.
sin+ cos=
7
25
+
24
25
=
31
25
:
5. If tanA=
4
5
, determine
2 sinAcosA
3 cosA
.
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DEPED COPY Solution:
tanA=
4
5
=)
sinA
cosA
=
4
5
2 sinAcosA
3 cosA
=
2
3

sinA
cosA

1
3

cosA
cosA

=
2
3

4
5

1
3
(1) =
1
5
6. What is the reference angle of
29
6
? Find the value of tan

29
6

.
Solution:
29
6
is coterminal with
7
6
in QIII, so its reference angle is

6
.
tan

29
6

= tan

6

=
p
3
3
7. For what anglein the third quadrant is cos= sin
5
3
?
Solution:
sin
5
3
= cos
cos=
p
3
2
andin QIII =)=
7
6
Supplementary Problems 3.2
1. In what quadrant isP() located if=
33
4
?
2. In what quadrant isP() located if=
17
6
?
3. In what quadrant isP() located if sec >0 and cot <0?
4. In what quadrant isP() located if tan >0 and cos <0 ?
5. IfP() is a point on the unit circle and=
5
6
, what are the coordinates of
P()?
6. IfP() is a point on the unit circle and=
11
6
, what are the coordinates
ofP()?
7. If cos >0 and tan=
2
3
, nd
sec+tan
sectan
8. If tan=
3
5
andis in QIII, what is sec?
9. If csc= 2 and cos <0, nd sec.
10. Find the values of the other trigonometric functions ofif cot=
4
3
and
sin <0.
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electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2016.

DEPED COPY 11. Find the values of the other trigonometric functions ofif csc=4 and
does not terminate in QIII.
12. The terminal side of an anglein standard position contains the point (7;1).
Find the values of the six trigonometric functions of.
13. The terminal side of an anglein standard position contains the point (2;4).
Find the values of the six trigonometric functions of.
14. If the terminal point of an arc of lengthlies on the line joining the origin
and the point (3;1), what is cos
2
sin
2
?
15. If the terminal point of an arc of lengthlies on the line joining the origin
and the point (2;6), what is sec
2
csc
2
?
16. Determine the reference angle of
35
4
, and nd cos
35
4
.
17. If
3
2
< <2, ndif cos= sin
2
3
.
18. Evaluate the sum of sin 30

+ sin 60

+ sin 90

+ + sin 510

+ sin 540

.
19. Iff(x) = sin 2x + cos 2x+ sec 2x+ csc 2x+ tan 2x+ cot 2x, what isf

7
8

?
20. Evaluate the sum of sec

6
+ sec
13
6
+ sec
25
6
+ + sec
109
6
.
4
Lesson 3.3.Graphs of Circular Functions and Situational
Problems
Learning Outcomes of the Lesson
At the end of the lesson, the student is able to:
(1) determine the domain and range of the dierent circular functions;
(2) graph the six circular functions with its amplitude, period, and phase shift;
and
(3) solve situational problems involving circular functions.
Lesson Outline
(1) Domain and range of circular functions
(2) Graphs of circular functions
(3) Amplitude, period, and phase shift
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DEPED COPY Introduction
There are many things that occur periodically. Phenomena like rotation of
the planets and comets, high and low tides, and yearly change of the seasons
follow a periodic pattern. In this lesson, we will graph the six circular functions
and we will see that they are periodic in nature.
3.3.1. Graphs ofy= sinxandy= cosx
Recall that, for a real numberx, sinx= sinfor an anglewith measurex
radians, and that sinis the second coordinate of the pointP() on the unit
circle. Since eachxcorresponds to an angle, we can conclude that
(1) sinxis dened for any real numberxor the domain of the sine function is
R, and
(2) the range of sine is the set of all real numbers between1 and 1 (inclusive).
From the denition, it also follows that sin(x+2) = sinxfor any real number
x. This means that the values of the sine function repeat every 2 units. In this
case, we say that the sine function is aperiodic functionwithperiod2.
Table 3.16 below shows the values ofy= sinx, wherexis the equivalent radian
measure of the special angles and their multiples from 0 to 2. As commented
above, these values determine the behavior of the function onR.
x0

6

4

3

2
2
3
3
4
5
6

y0
1
2
p
2
2
p
3
2
1
p
3
2
p
2
2
1
2
0
00:50:710:8710:870:710:5 0
x
7
6
5
4
4
3
3
2
5
3
7
4
11
6
2
y
1
2

p
2
2

p
3
2
1
p
3
2

p
2
2

1
2
0
0:5 0:71 0:87 10:87 0:71 0:5 0
Table 3.16
From the table, we can observe that asxincreases from 0 to

2
, sinxalso
increases from 0 to 1. Similarly, asxincreases from
3
2
to 2, sinxalso increases
from1 to 0. On the other hand, notice that asxincreases from

2
to, sinx
decreases from 1 to 0. Similarly, asxincreases fromto
3
2
, sinxdecreases from
0 to1.
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DEPED COPY To sketch the graph ofy= sinx, we plot the points presented in Table 3.16,
and join them with a smooth curve. See Figure 3.17. Since the graph repeats
every 2 units, Figure 3.18 shows periodic graph over a longer interval.
Figure 3.17
Figure 3.18
W
e can make observations about the cosine function that are similar to the
sine function.
y= cosxhas domainRand range [1;1].
y= cosxis periodic with period 2. The graph ofy= cosxis shown in
Figure 3.19.
Figure 3.19
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DEPED COPY From the graphs ofy= sinxandy= cosxin Figures 3.18 and 3.19, re-
spectively, we observe that sin(x) = sinxand cos(x) = cosxfor any real
numberx. In other words, the graphs ofy= cos(x) andy= cosxare the same,
while the graph ofy= sin(x) is the same as that ofy=sinx.
In general, if a functionfsatises the property thatf(x) = f(x) for allx
in its domain, we say that such function iseven. On the other hand, we say that
a functionfisoddiff(x) = f(x) for allxin its domain. For example, the
functionsx
2
and cosxare even, while the functionsx
3
3xand sinxare odd.
3.3.2. Graphs ofy=asinbxandy=acosbx
Using a table of values from 0 to 2, we can sketch the graph ofy= 3 sinx, and
compare it to the graph ofy= sinx. See Figure 3.20 wherein the solid curve
belongs toy= 3 sinx, while the dashed curve toy= sinx. For instance, ifx=

2
,
theny= 1 wheny= sinx, andy= 3 wheny= 3 sinx. The period,x-intercepts,
and domains are the same for both graphs, while they dier in the range. The
range ofy= 3 sinxis [3;3].
Figure 3.20
In
general, the graphs ofy=asinxandy=acosxwitha >0 have the same
shape as the graphs ofy= sinxandy= cosx, respectively. Ifa <0, there is a
reection across thex-axis.
In the graphs ofy=asinxandy=acosx, the numberjajis called
itsamplitude. It dictates the height of the curve. When jaj<1,
the graphs are shrunk vertically, and whenjaj>1, the graphs are
stretched vertically.
Now, in Table 3.21, we consider the values ofy= sin 2x on [0; 2].
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DEPED COPY x0

6

4

3

2
2
3
3
4
5
6

y0
p
3
2
1
p
3
2
0
p
3
2
1
p
3
2
0
00:8710:8700:87 1 0:87 0
x
7
6
5
4
4
3
3
2
5
3
7
4
11
6
2
y
p
3
2
1
p
3
2
0
p
3
2
1
p
3
2
0
0:87 10:87 00:87 10:87 0
Table 3.21
Figure 3.22
Figure
3.22 shows the graphs ofy= sin 2x (solid curve) andy= sinx(dashed
curve) over the interval [0;2]. Notice that, for sin 2x to generate periodic values
similar to [0;2] fory= sinx, we just need values ofxfrom 0 to. We then
expect the values of sin 2xto repeat everyunits thereafter. The period of
y= sin 2x is.
Ifb6
= 0, then bothy= sinbxandy= cosbxhave period given by
2
jbj
.
If 0<jbj<1, the graphs are stretched horizontally, and ifjbj>1, the
graphs are shrunk horizontally.
To sketch the graphs ofy=asinbxandy=acosbx,a; b6
= 0, we may proceed
with the following steps:
(1) Determine the amplitudejaj, and nd the period
2
jbj
. To draw one cycle
of the graph (that is, one complete graph for one period), we just need to
complete the graph from 0 to
2
jbj
.
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DEPED COPY (2) Divide the interval into four equal parts, and get ve division points:x1= 0,
x2,x3,x4, andx5=
2
jbj
, wherex3is the midpoint betweenx1andx5(that
is,
1
2
(x1+x5) =x3),x2is the midpoint betweenx1andx3, andx4is the
midpoint betweenx3andx5.
(3) Evaluate the function at each of the vex-values identied in Step 2. The
points will correspond to the highest point, lowest point, andx-intercepts
of the graph.
(4) Plot the points found in Step 3, and join them with a smooth curve similar
to the graph of the basic sine curve.
(5) Extend the graph to the right and to the left, as needed.
Example 3.3.1.Sketch the graph of one cycle ofy= 2 sin 4x.
Solution.(1) The period is
2
4
=

2
, and the amplitude is 2.
(2) Dividing the interval [0;

2
] into 4 equal parts, we get the followingx-
coordinates: 0,

8
,

4
,
3
8
, and

2
.
(3) Whenx= 0,

4
, and

2
, we gety= 0. On the other hand, whenx=

8
, we
havey= 2 (the amplitude), andy=2 whenx=
3
8
.
(4) Draw a smooth curve by connecting the points. There is no need to proceed
to Step 5 because the problem only asks for one cycle.
Example 3.3.2.Sk
etch the graph ofy=3 cos
x
2
.
Solution.(1)
The amplitude isj 3j= 3, and the period is
2
1
2
=
4.
(2) We divide the interval [0;4] into four equal parts, and we get the following
x-values: 0,, 2, 3, and 4.
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DEPED COPY (3) We havey= 0 whenx=and 3,y=3 whenx= 0 and 4, andy= 3
whenx= 2.
(4) We trace the points in Step 3 by a smooth curve.
(5) We extend the pattern in Step 4 to the left and to the right.
Example 3.3.3.Sk
etch the graph of two cycles ofy=
1
2
sin

2
x
3

.
Solution.Since
the sine function is odd, the graph ofy=
1
2
sin

2
x
3

is
the same
as that ofy=
1
2
sin
2
x
3
.
(1)
The amplitude is
1
2
,
and the period is
2
2
3
=
3.
(2) Dividing the interval [0;3] into four equal parts, we get thex-coordinates
of the ve important points:
0 + 3
2
=
3

2
;
0
+
3
2
2
=
3

4
;
3

2
+
3
2
=
9

4
:
(3)
We gety= 0 whenx= 0,
3
2
,
and 3 ,y=
1
2
when
3

4
,
andy=
1
2
when
9

4
.
(4)
We trace the points in Step 3 by a smooth curve.
(5) We extend the pattern in Step 4 by one more period to the right.
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DEPED COPY 3.3.3. Graphs ofy=asinb(xc) +dandy=acosb(xc) +d
We rst compare the graphs ofy= sinxandy= sin

x

3

using a table of
values and the 5-step procedure discussed earlier.
Asxruns from

3
to
7
3
, the value of the expressionx

3
runs from 0 to 2. So
for one cycle of the graph ofy= sin

x

3

, we then expect to have the graph of
y= sinxstarting fromx=

3
. This is conrmed by the values in Table 3.23. We
then apply a similar procedure to complete one cycle of the graph; that is, divide
the interval [

3
;
7
3
] into four equal parts, and then determine the key values of
xin sketching the graphs as discussed earlier. The one-cycle graph ofy= sinx
(dashed curve) and the corresponding one-cycle graph ofy= sin

x

3

(solid
curve) are shown in Figure 3.24.
x

3
5
6
4
3
11
6
7
3
x

3
0

2

3
2
2
sin

x

3

0101 0
Table 3.23
Figure 3.24
Observ
e that the graph ofy= sin

x

3

shifts

3
units
to the right of
y= sinx. Thus, they have the same period, amplitude, domain, and range.
The graphs of
y=asinb(xc) andy=acosb(xc)
have the same shape asy=asinbxandy=acosbx, respectively, but
shiftedcunits to the right whenc >0 and shiftedjcjunits to the left
ifc <0. The numbercis called thephase shiftof the sine or cosine
graph.
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DEPED COPY Example 3.3.4.In the same Cartesian plane, sketch one cycle of the graphs of
y= 3 sinxandy= 3 sin

x+

4

.
Solution.We have sketched the graph ofy= 3 sinxearlier at the start of the
lesson. We considery= 3 sin

x+

4

. We expect that it has the same shape as
that ofy= 3 sinx, but shifted some units.
Here, we havea= 3,b= 1, andc=

4
. From these constants, we get
the amplitude, the period, and the phase shift, and these are 3, 2, and

4
,
respectively.
One cycle starts atx=

4
and ends atx=

4
+ 2=
7
4
. We now compute
the important values ofx.

4
+
7
4
2
=
3
4
;

4
+
3
4
2
=

4
;
3
4
+
7
4
2
=
5
4
x

4

4
3
4
5
4
7
4
y= 3 sin

x+

4

0303 0
While the eect ofciny=asinb(xc) andy=acosb(xc) is
a horizontal shift of their graphs from the corresponding graphs of
y=asinbxandy=acosbx, the eect ofdin the equationsy=
asinb(xc) +dandy=acosb(xc) +dis a vertical shift. That is,
the graph ofy=asinb(xc)+dhas the same amplitude, period, and
phase shift as that ofy=asinb(xc), but shifteddunits upward
whend >0 andjdjunits downward whend <0.
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DEPED COPY Example 3.3.5.Sketch the graph of
y=2 cos 2

x

6

3:
Solution.Here,a=2,b= 2,c=

6
, andd=3. We rst sketch one cycle of
the graph ofy=2 cos 2

x

6

, and then extend this graph to the left and to
the right, and then move the resulting graph 3 units downward.
The graph ofy=2 cos 2

x

6

has amplitude 2, period, and phase shift

6
.
Start of one cycle:

6
End of the cycle:

6
+=
7
6

6
+
7
6
2
=
2
3
;

6
+
2
3
2
=
5
12
;
2
3
+
7
6
2
=
11
12
x

6
5
12
2
3
11
12
7
6
y=2 cos 2

x

6

2 0202
y=2 cos 2

x

6

353135
Before we end this sub-lesson, we make the following observation, which will
b
e used in the discussion on simple harmonic motion (Sub-Lesson 3.3.6).
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DEPED COPY Dierent Equations, The Same Graph
1. The graphs ofy= sinxandy= sin(x + 2k),kany integer, are
the same.
2. The graphs ofy= sinx,y=sin(x+),y= cos(x

2
), and
y=cos(x+

2
) are the same.
3. In general, the graphs of
y=asinb(xc) +d;
y=asin[b(x c) ++ 2k] +d;
y=acos[b(x c)

2
+ 2k] +d;
and
y=acos[b(x c) +

2
+ 2k] +d;
wherekis any integer, are all the same.
Similar observations are true for cosine.
3.3.4. Graphs of Cosecant and Secant Functions
We know that cscx=
1
sinx
if sinx6= 0. Using this relationship, we can sketch the
graph ofy= cscx.
First, we observe that the domain of the cosecant function is
fx2R: sinx6= 0g=fx2R:x6=k; k2Zg:
Table 3.25 shows the key numbers (that is, numbers wherey= sinxcrosses the
x-axis, attain its maximum and minimum values) and some neighboring points,
where \und" stands for \undened," while Figure 3.26 shows one cycle of the
graphs ofy= sinx(dashed curve) andy= cscx(solid curve). Notice the
asymptotes of the graphy= cscx.
x 0

6

2
5
6

7
6
3
2
11
6
2
y= sinx0
1
2
1
1
2
0
1
2
1
1
2
0
y= cscxund212und212und
Table 3.25
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DEPED COPY Figure 3.26
W
e could also sketch the graph of cscxdirectly from the graph ofy= sinx
by observing the following facts:
(1) If sinx= 1 (or1), then cscx= 1 (or1).
(2) At eachx-intercept ofy= sinx,y= cscxis undened; but a vertical
asymptote is formed because, when sinxis close to 0, the value of cscxwill
have a big magnitude with the same sign as sinx.
Refer to Figure 3.27 for the graphs ofy= sinx(dashed curve) andy= cscx
(solid curve) over a larger interval.
Figure 3.27
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DEPED COPY Like the sine and cosecant functions, the cosine and secant functions are also
reciprocals of each other. Therefore,y= secxhas domain
fx2R: cosx6
= 0g=fx2R:x6 =
k
2
; kodd integerg:
Similarly, the graph ofy= secxcan be obtained from the graph ofy= cosx.
These graphs are shown in Figure 3.28.
Figure 3.28
Example
3.3.6.Sketch the graph ofy= 2 csc
x
2
.
Solution.First,
we sketch the graph ofy= 2 sin
x
2
,
and use the technique dis-
cussed above to sketch the graph ofy= 2 csc
x
2
.
The vertical asymptotes ofy=
2 csc
x
2
are
thex-intercepts ofy= 2 sin
x
2
:
x=
0;2;4; : : :. After setting up the asymptotes, we now sketch the graph
ofy= 2 csc
x
2
as
shown below.
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DEPED COPY Example 3.3.7.Sk
etch the graph ofy= 2sec 2x.
Solution.Sketch the graph ofy=cos 2x(note that it has period), then sketch
the graph ofy=sec 2x(as illustrated above), and then move the resulting
graph 2 units upward to obtain the graph ofy= 2sec 2x.
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DEPED COPY 3.3.5. Graphs of Tangent and Cotangent Functions
We know that tanx=
sinx
cosx
, where cosx6
= 0. From this denition of the tangent
function, it follows that its domain is the same as that of the secant function,
which is
fx2R: cosx6
= 0g=fx2R:x6 =
k
2
; kodd integerg:
We note that tanx= 0 when sinx= 0 (that is, whenx=k,kany integer), and
that the graph ofy= tanxhas asymptotesx=
k
2
,kodd integer. Furthermore,
by recalling the signs of tangent from Quadrant I to Quadrant IV and its values,
we observe that the tangent function is periodic with period.
To sketch the graph ofy= tanx, it will be enough to know its one-cycle
graph on the open interval

2
;

2

. See Table 3.29 and Figure 3.30.
x

2

3

4

6
0
y= tanxund
p
31
p
3
3
0
x

6

4

3

2
y= tanx
p
3
3
1
p
3und
Table 3.29
Figure 3.30
In
the same manner, the domain ofy= cotx=
cosx
sinx
is
f
x2R: sinx6
= 0g=fx2R:x6 =k; k2Zg;
and its period is also. The graph ofy= cotxis shown in Figure 3.31.
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DEPED COPY Figure 3.31
In
general, to sketch the graphs ofy=atanbxandy=acotbx,a6
= 0 and
b >0, we may proceed with the following steps:
(1) Determine the period

b
.
Then we draw one cycle of the graph on

2b
;

2b

fory=atanbx
, and on

0;

b

fory=acotbx
.
(2) Determine the two adjacent vertical asymptotes. Fory=atanbx, these
vertical asymptotes are given byx=

2b
.
Fory=acotbx, the vertical
asymptotes are given byx= 0 andx=

b
.
(3)
Divide the interval formed by the vertical asymptotes in Step 2 into four
equal parts, and get three division points exclusively between the asymp-
totes.
(4) Evaluate the function at each of thesex-values identied in Step 3. The
points will correspond to the signs andx-intercept of the graph.
(5) Plot the points found in Step 3, and join them with a smooth curve ap-
proaching to the vertical asymptotes. Extend the graph to the right and to
the left, as needed.
Example 3.3.8.Sketch the graph ofy=
1
2
tan
2x.
Solution.The period of the function is

2
,
and the adjacent asymptotes arex=

4
;
3

4
;
: : :. Dividing the interval

4
;

4

in
to four equal parts, the keyx-values
are

8
,
0, and

8
.
x

8
0

8
y=
1
2
tan
2x

1
2
0
1
2
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DEPED COPY Example 3.3.9.Sk
etch the graph ofy= 2 cot
x
3
on
the interval (0; 3).
Solution.The period of the function is 3 , and the adjacent asymptotes arex= 0
andx= 3. We now divide the interval (0; 3) into four equal parts, and the
keyx-values are
3
4
,
3

2
,
and
9
4
.
x
3
4
3
2
9
4
y=

x
3
202
3.3.6. Simple Harmonic Motion
Rep
etitive or periodic behavior is common in nature. As an example, the time-
telling device known assundialis a result of the predictable rising and setting
of the sun everyday. It consists of a at plate and a gnomon. As the sun moves
across the sky, the gnomon casts a shadow on the plate, which is calibrated to
tell the time of the day.
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DEPED COPY Sundial, by liz west, 29 March 2007,
https://commons.wikimedia.o
rg/wiki/File:Sundial
2r.jpg.

Some motions are also periodic. When a weight is suspended on a spring,
pulled down, and released, the weight oscillates up and down. Neglecting resis-
tance, this oscillatory motion of the weight will continue on and on, and its height
is periodic with respect to time.
t=

t=
:8 sec
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DEPED COPY t=
:1 sec
t=

Periodic motions are usually modeled by either sine or cosine function, and are
calledsimple harmonic motions. Unimpeded movements of objects like oscilla-
tion, vibration, rotation, and motion due to water waves are real-life occurrences
that behave in simple harmonic motion.
Equations of Simple Harmonic Motion
The displacementy(directed height or length) of an object behaving
in a simple harmonic motion with respect to timetis given by one of
the following equations:
y=asinb(tc) +d
or
y=acosb(tc) +d:
In both equations, we have the following information:
amplitude =jaj=
1
2
(Mm) - the maximum displacement above
and below the rest position or central position or equilibrium, where
Mis the maximum height andmis the minimum height;
period =
2
jbj
- the time required to complete one cycle (from one
highest or lowest point to the next);
frequency =
jbj
2
- the number of cycles per unit of time;
c- responsible for the horizontal shift in time; and
d- responsible for the vertical shift in displacement.
Example 3.3.10.A weight is suspended from a spring and is moving up and
down in a simple harmonic motion. At start, the weight is pulled down 5 cm below
the resting position, and then released. After 8 seconds, the weight reaches its
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DEPED COPY highest location for the rst time. Find the equation of the motion.
Solution.We are given that the weight is located at its lowest position att= 0;
that is,y=5 whent= 0. Therefore, the equation isy=5 cosbt.
Because it took the weight 8 seconds from the lowest point to its immediate
highest point, half the period is 8 seconds.
1
2

2
b
= 8 =)b=

8
=)y=5 cos
t
8
2
?
Example 3.3.11.Suppose you ride a Ferris wheel. The lowest point of the
wheel is 3 meters o the ground, and its diameter is 20 m. After it started, the
Ferris wheel revolves at a constant speed, and it takes 32 seconds to bring you
back again to the riding point. After riding for 150 seconds, nd your approximate
height above the ground.
Solution.We ignore rst the xed value of 3 m o the ground, and assume that
the central position passes through the center of the wheel and is parallel to the
ground.
Lettbe the time (in seconds) elapsed that you have been riding the Ferris
wheel, andyis he directed distance of your location with respect to the assumed
central position at timet. Becausey=10 whent= 0, the appropriate model
isy=10 cosbtfort0.
Given that the Ferris wheel takes 32 seconds to move from the lowest point
to the next, the period is 32.
2
b
= 32 =)b=

16
=)y=10 cos
t
16
Whent= 150, we gety= 10 cos
150
16
3:83.
Bringing back the original condition given in the problem that the riding point
is 3 m o the ground, after riding for 150 seconds, you are approximately located
3:83 + 13 = 16:83 m o the ground. 2
In the last example, the central position or equilibrium may be vertically
shifted from the ground or sea level (the role of the constantd). In the same way,
the starting point may also be horizontally shifted (the role of the constantc).
Moreover, as observed in Sub-Lesson 3.3.3 (see page 154), to nd the function
that describes a particular simple harmonic motion, we can either choose
y=asinb(tc) +d
or
y=acosb(tc) +d;
and determine the appropriate values ofa,b,c, andd. In fact, we can assume
thataandbare positive numbers, andcis the smallest such nonnegative number.
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DEPED COPY Example 3.3.12.A signal buoy in Laguna Bay bobs up and down with the
heighthof its transmitter (in feet) above sea level modeled byh(t) =asinbt+d
at timet(in seconds). During a small squall, its height varies from 1 ft to 9 ft
above sea level, and it takes 3:5 seconds from one 9-ft height to the next. Find
the values of the constantsa,b, andd.
Solution.We solve the constants step by step.
The minimum and maximum values ofh(t) are 1 ft and 9 ft, respectively.
Thus, the amplitude isa=
1
2
(Mm) =
1
2
(91) = 4.
Because it takes 3:5 seconds from one 9-ft height to the next, the period is
3:5. Thus, we have
2
b
= 3:5, which givesb=
4
7
.
Because the lowest point is 1 ft above the sea level and the amplitude is 4,it follows thatd= 5. 2
Example 3.3.13.A variable star is a star whose brightness uctuates as ob-
served from Earth. The magnitude of visual brightness of one variable star rangesfrom 2:0 to 10:1, and it takes 332 days to observe one maximum brightness to
the next. Assuming that the visual brightness of the star can be modeled by theequationy=asinb(tc) +d,tin days, and puttingt= 0 at a time when the
star is at its maximum brightness, nd the constantsa,b,c, andd, wherea; b >0
andcthe least nonnegative number possible.
Solution.
a=
Mm
2
=
10:12:0
2
= 4:05
2
b
= 332 =)b=

166
d=a+m= 4:05 + 2:0 = 6 :05
For the (ordinary) sine function to start at the highest point att= 0, the least
possible horizontal movement to the right (positive value) is
3
2
units.
bc=
3
2
=)c=
3
2b
=
3
2

166
= 249 2
?
Example 3.3.14.The path of a fast-moving particle traces a circle with equa-
tion
(x+ 7)
2
+ (y5)
2
= 36:
It starts at point (1; 5), moves clockwise, and passes the point (7; 11) for the
rst time after traveling 6 microseconds. Where is the particle after traveling 15microseconds?
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DEPED COPY Solution.As described above, we may choose sine or cosine function. Here, we
choose the sine function to describe bothxandyin terms of timetin microsec-
onds; that is, we let
x=asinb(tc) +dandy=esinf(tg) +h;
where we appropriately choose the positive values fora,b,e, andf, and the least
nonnegative values forcandg.
The given circle has radius 6 and center (7; 5). Dening the central position
of the values ofxas the linex=7 and that of the values ofyas the liney= 5,
we geta=e= 6,d=7, andh= 5.
From the point (1;5) to the point (7; 11) (moving clockwise), the particle
has traveled three-fourths of the complete cycle; that is, three-fourths of the
period must be 2.
3
4

2
b
=
3
4

2
f
= 6 =)b=f=

4
As the particle starts at (1; 5) and moves clockwise, the values ofxstart
at its highest value (x=1) and move downward toward its central position
(x=7) and continue to its lowest value (x =13). Therefore, the graph of
asinbt+dhas to move
3
2b
= 6 units to the right, and so we getc= 6.
As to the value ofg, we observe the values ofystart at its central position
(y= 5) and go downward to its lowest value (y =1). Similar to the argument
used in determiningc, the graph ofy=esinft+hhas to move

b
= 4 units to
the right, implying thatg= 4.
Hence, We have the following equations ofxandyin terms oft:
x= 6 sin

4
(t6)7 andy= 6 sin

4
(t4) + 5:
Whent= 15, we get
x= 6 sin

4
(156)7 =7 + 3
p
2 2:76
and
y= 6 sin

4
(154) + 5 = 5 + 3
p
29:24:
That is, after traveling for 15 microseconds, the particle is located near the point(2:76; 9:24). 2
More Solved Examples
1. Find the period of the functiony= 4 sin

x
4

3.
Solution:y= 4 sin

x
4

3 =)y= 4 sin
1
3
(x)3 =)P=
2
1
3
= 6
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electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2016.

DEPED COPY 2. In the functiony= 3 tan(2kx ), the period is 4. Find the value ofkand
the phase shift of the graph of the function.
Solution:The period of the tangent function isP=

b
.
y= 3 tan(2kx) =)y= 3 tan 2k

x

2k

=)P=

2k
= 4:
1
2k
= 4 =)k=
1
8
and Phase shift =

2k
=

2

1
8
= 4
3. Sketch the graph of functiony=
1
2
sin
1
2

x+

6

+2 over one period. Determine
the domain and range of the function.Solution:The graph is a vertical translation ofy=
1
2
sin
1
2

x+

6

by 2 units
upward. The period of the given function is
2
1
2
= 4. One complete cycle may
start atx=

6
and end atx=

6
+ 4=
23
6
.
The critical points for the graph are
x=

6
; x=
5
6
; x=
11
6
; x=
17
6
;andx=
23
6
:
The domain of the function isRand
its range is

5
2
;
3
2

.
4.
Sketch the graph of the functiony=2 cos(x

2
)
+3 over two periods. Find
the domain and range of the function. Solution:The graph of the given function is a vertical translation ofy=
2 cos(x

2
)
by 3 units upward. The period of the function is 2. One
complete cycle may start and end atx=

2
andx=
5

2
,
respectively. The
next complete cycle starts atx=
5
2
and
ends atx=
9
2
.
critical
points:

2
;
;
3
2
;2
;
5
2
;3
;
7
2
;4
;
9
2
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DEPED COPY The domain of the given cosine function isR
, and its range is [1; 5].
5. Sketch the graph of the functiony=
1
4
tan

x

4

o
ver three periods. Find
the domain and range of the function.
Solution:The period of the function is. One complete cycle may start at
x=

4
and
end atx=
5
4
.
The domain of the function isf
xjx6 =
3
4
+k
; k2Zg, and its range isR.
6. Sketch the graph of the functiony=3 cot

1
2
x+

12

+
2 over three periods.
Find the domain and range of the function.
Solution:y=3 cot

1
2
x+

12

+
2 =3 cot
1
2

x+

6

+
2 =)P= 2
One complete cycle may start atx=

6
and
end atx=
11
6
.
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DEPED COPY The domain of the function isf
xjx6 =

6
+
2k; k2Zg, and its range isR.
7. The graph of the functiong(x) is the same as that off(x) = 3 sin

x

3

but
shifted
2 units downward and

2
units
to the right. What isg()?
Solution:The functionf(x) = 3 sin

x

3

when
shifted 2 units downward
and

2
units
to the right is
g(x) = 3 sin

x

3

2

2
= 3 sin

x
5
6

2
:
g() = 3 sin

5
6

2
=
1
2
8.
The graph of the functionh(x) is the same as that off(x) = 3 sin(2x3)+1
but shifted 3 units upward and

2
units
to the left. What ish(
5
6
)?
Solution:h
(x) = 3 sin

2

x+

2

3

+ 1 + 3 = 3 sin(2x2) + 4
h

5
6

=
3 sin

2

5
6

2

+ 4 =
83
p
3
2
9.
Sketch the graph ofy= 2 sec
1
2

x

4

o
ver two periods. Find the domain
and range of the function.
Solution:The period of the function is 4. One complete cycle may start at
x=

4
and
end atx=
17
4
.
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DEPED COPY The domain of the function isf
xjx6 =
5
4
+
2k; k2Zg, and its range is
(1; 2][[2;1).
10. Sketch the graph ofy=csc

x+

3

+
2 over two periods. Find the domain
and range of the function.
Solution:The period of the function is 2. One complete cycle may start and
end atx=

3
andx=
5

3
,
respectively.
The domain of the function isf
xjx6=

3
+k
; k2Zg, and its range is
(1; 1][[3;1).
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DEPED COPY Supplementary Problems 3.3
1. What is the period of the functiony=2 cos
1
4

x

2

?
2. The amplitude and period of the functiony= 4
a
2
cos

bx
3

are 3 and 4,
respectively. Findjaj+b.
3. In the functiony= 23cot
4
k
(x2), the period is 2. Find the value ofk.
4. What are the minimum and maximum values of the functiony= 3 sin
3
4

x+
2
3

5?
5. Given the functiony= 3 sin
3
4

x+
2
3

5, nd the value ofywhenx=
8
9
.
6. Given the functiony=2 cot
4
3

x

6

+ 3, nd the value ofywhenx=
7
6
.
7. Find the domain and range of the functiony=
2
3
sin
1
3

x
3
4

+ 2?
8. Find the range of the functiony= 3 sec

2x
3

.
9. Find the equation of the secant function whose graph is the graph ofy=
3 sec 2xshiftedunits to the right and 3 units downward.
10. Find the equation of the sine function whose graph is the graph ofy=
2 sin 2

x

4

+ 1 shifted

2
units to the left and 3 units upward.
11. Given the tangent functiony= 13 tan

2x
4

, nd the equations of all its
vertical asymptotes.
12. Given the cosecant functiony= csc

x
2

3

, nd the equations of all its
vertical asymptotes.
13. Sketch the graph over one period, and indicate the period, phase shift, domain,
and range for each.
(a)y= 2 sin
1
4

x+

4

1
(b)y= tan
1
2

2x+

3

2
(c)y=
1
2
csc
3
4
(2x)1
(d)y= sec
1
2

4x+
2
3

+ 2
14. A pointPin simple harmonic motion has a frequency of
1
2
oscillation per
minute and amplitude of 4 ft. Express the motion ofPby means of an
equation in the formd=asinbt.
?
15. A mass is attached to a spring, and then pulled and released 8 cm below its
resting position at the start. If the simple harmonic motion is modeled by
y=acos
1
10
(tc), wherea >0,cthe least nonnegative such number, andt
in seconds, nd the location of the mass 10 seconds later.
4
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DEPED COPY Lesson 3.4. Fundamental Trigonometric Identities
Learning Outcomes of the Lesson
At the end of the lesson, the student is able to:
(1) determine whether an equation is an identity or a conditional equation;
(2) derive the fundamental trigonometric identities;
(3) simplify trigonometric expressions using fundamental trigonometric identi-
ties; and
(4) prove other trigonometric identities using fundamental trigonometric identi-
ties.
Lesson Outline
(1) Domain of an equation
(2) Identity and conditional equation
(3) Fundamental trigonometric identities
(4) Proving trigonometric identities
Introduction
In previous lessons, we have dened trigonometric functions using the unit
circle and also investigated the graphs of the six trigonometric functions. This
lesson builds on the understanding of the dierent trigonometric functions by
discovery, deriving, and working with trigonometric identities.
3.4.1. Domain of an Expression or Equation
Consider the following expressions:
2x+ 1;
p
x
2
1;
x
x
2
3x4
;
x
p
x1
:
What are the real values of the variablexthat make the expressions dened in
the set of real numbers?
In the rst expression, every real value ofxwhen substituted to the expression
makes it dened in the set of real numbers; that is, the value of the expression is
real whenxis real.
In the second expression, not every real value ofxmakes the expression dened
inR. For example, whenx= 0, the expression becomes
p
1, which is not a real
number.
p
x
2
12R() x
2
10() x 1 orx1
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DEPED COPY Here, for
p
x
2
1 to be dened inR,xmust be in (1; 1][[1;1).
In the third expression, the values ofxthat make the denominator zero make
the entire expression undened.
x
2
3x4 = (x4)(x+ 1) = 0() x= 4 orx=1
Hence, the expression
x
x
2
3x4
is real whenx6= 4 andx6=1.
In the fourth expression, because the expression
p
x1 is in the denominator,
xmust be greater than 1. Although the value of the entire expression is 0 when
x= 0, we do not include 0 as allowed value ofxbecause part of the expression
is not real whenx= 0.
In the expressions above, the allowed values of the variablexconstitute the
domain of the expression.
Thedomainof an expression (or equation) is the set of all real values of
the variable for which every term (or part) of the expression (equation)
is dened inR.
In the expressions above, the domains of the rst, second, third, and fourth
expressions areR, (1; 1][[1;1),Rn f1; 4g, and (1; 1), respectively.
Example 3.4.1.Determine the domain of the expression/equation.
(a)
x
2
1
x
3
+ 2x
2
8x

p
x+ 1
1x
(b) tansincos 2
(c)x
2

p
1 +x
2
=
2
3
p
x
2
1
(d)z
cos
2
z
1 + sinz
= 4 sinz1
Solution.(a)x
3
+ 2x
2
8x=x(x+ 4)(x 2) = 0() x= 0; x=
4;orx= 2
p
x+ 12R() x+ 10() x 1
1x= 0() x= 1
Domain = [1; 1)n f4; 0;1;2g
= [1;0)[(0;1)[(1;2)[(2;1)
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DEPED COPY (b) tansincos 2=
sin
cos
sincos 2
cos= 0() =
k
2
; kodd integer
Domain =Rn f
k
2
jkodd integerg
(c) The expression 1+x
2
is always positive, and so
p
1 +x
2
is dened inR. On
the other hand, the expression
3
p
x
2
1 is also dened inR, but it cannot
be zero because it is in the denominator. Therefore,xshould not be1
and 1.
Domain =Rn f1; 1g
(d) 1 + sinz= 0() z=
3
2
+ 2k; k2Z
Domain =Rn f
3
2
+ 2kjk2Zg 2
3.4.2. Identity and Conditional Equation
Consider the following two groups of equations:
Group A Group B
(A1)x
2
1 = 0 (B1)x
2
1 = (x 1)(x+ 1)
(A2) (x + 7)
2
=x
2
+ 49 (B2) (x + 7)
2
=x
2
+ 14x + 49
(A3)
x
2
4
x2
= 2x1 (B3)
x
2
4
x2
=x+ 2
In each equation in Group A, some values of the variable that are in the
domain of the equation do not satisfy the equation (that is, do not make the
equation true). On the other hand, in each equation in Group B, every element
in the domain of the equation satises the given equation. The equations in
Group A are called conditional equations, while those in Group B are called
identities.
Anidentityis an equation that is true for all values of the variable
in the domain of the equation. An equation that is not an identity iscalled aconditional equation. (In other words, if some values of the
variable in the domain of the equation do not satisfy the equation,then the equation is a conditional equation.)
Example 3.4.2.Identify whether the given equation is an identity or a condi-
tional equation. For each conditional equation, provide a value of the variable inthe domain that does not satisfy the equation.
(1)x
3
2 =

x
3
p
2

x
2
+
3
p
2x+
3
p
4

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DEPED COPY (2) sin
2
= cos
2
+ 1
(3) sin= cos1
(4)
1
p
x
1 +
p
x
=
12
p
x+x
1x
Solution.(1) This is an identity because this is simply factoring of dierence of
two cubes.
(2) This is a conditional equation. If= 0, then the left-hand side of the equation
is 0, while the right-hand side is 2.
(3) This is also a conditional equation. If= 0, then both sides of the equation
are equal to 0. But if=, then the left-hand side of the equation is 0,
while the right-hand side is2.
(4) This is an identity because the right-hand side of the equation is obtained by
rationalizing the denominator of the left-hand side. 2
3.4.3. The Fundamental Trigonometric Identities
Recall that ifP(x; y) is the terminal point on the unit circle corresponding to,
then we have
sin=ycsc=
1
y
tan=
y
x
cos=xsec=
1
x
cot=
x
y
:
From the denitions, the following reciprocal and quotient identities immedi-
ately follow. Note that these identities hold ifis taken either as a real number
or as an angle.
Reciprocal Identities
csc=
1
sin
sec=
1
cos
cot=
1
tan
Quotient Identities
tan=
sin
cos
cot=
cos
sin
We can use these identities to simplify trigonometric expressions.
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DEPED COPY Example 3.4.3.Simplify:
(1)
tancos
sin
(2)
cos
cot
Solution.(1)
tancos
sin
=
sin
cos
cos
sin
= 1
(2)
cos
cot
=
cos
cos
sin
= sin 2
IfP(x; y) is the terminal point on the unit circle corresponding to, then
x
2
+y
2
= 1. Since sin=yand cos=x, we get
sin
2
+ cos
2
= 1:
By dividing both sides of this identity by cos
2
and sin
2
, respectively, we obtain
tan
2
+ 1 = sec
2
and 1 + cot
2
= csc
2
:
Pythagorean Identities
sin
2
+ cos
2
= 1
tan
2
+ 1 = sec
2
1 + cot
2
= csc
2

Example 3.4.4.Simplify:
(1) cos
2
+ cos
2
tan
2
(2)
1 + tan
2

1 + cot
2

Solution.(1) cos
2
+ cos
2
tan
2
= (cos
2
)(1 + tan
2
)
= cos
2
sec
2

= 1
(2)
1 + tan
2

1 + cot
2

=
sec
2

csc
2

=
1
cos
2

1
sin
2

=
sin
2

cos
2

= tan
2
2
In addition to the eight identities presented above, we also have the following
identities.
Even-Odd Identities
sin( ) =sin cos( ) = cos
tan( ) =tan
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DEPED COPY The rst two of the negative identities can be obtained from the graphs of
the sine and cosine functions, respectively. (Please review the discussion on page
147.) The third identity can be derived as follows:
tan( ) =
sin( )
cos( )
=
sin
cos
=tan:
The reciprocal, quotient, Pythagorean, and even-odd identities constitute
what we call thefundamental trigonometric identities.
We now solve Example 3.2.3 in a dierent way.
Example 3.4.5.If sin=
3
4
and cos >0. Find cos.
Solution.Using the identity sin
2
+ cos
2
= 1 with cos >0, we have
cos=
p
1sin
2
=
s
1

3
4

2
=
p
7
4
: 2
Example 3.4.6.If sec=
5
2
and tan <0, use the identities to nd the values
of the remaining trigonometric functions of.
Solution.Note thatlies in QIV.
cos=
1
sec
=
2
5
sin=
p
1cos
2
=
s
1

2
5

2
=
p
21
5
csc=
1
sin
=
5
p
21
21
tan=
sin
cos
=

p
21
5
2
5
=
p
21
2
cot=
1
tan
=
2
p
21
21
2
3.4.4. Proving Trigonometric Identities
We can use the eleven fundamental trigonometric identities to establish other
identities. For example, suppose we want to establish the identity
csccot=
sin
1 + cos
:
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electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2016.

DEPED COPY To verify that it is an identity, recall that we need to establish the truth of the
equation for all values of the variable in the domain of the equation. It is not
enough to verify its truth for some selected values of the variable. To prove it, we
use the fundamental trigonometric identities and valid algebraic manipulations
like performing the fundamental operations, factoring, canceling, and multiplying
the numerator and denominator by the same quantity.
Start on the expression on one side of the proposed identity (preferably the
complicated side), use and apply some of the fundamental trigonometric identities
and algebraic manipulations, and arrive at the expression on the other side of the
proposed identity.
Expression Explanation
csccot Start on one side.
=
1
sin

cos
sin
Apply some reciprocal and
quotient identities.
=
1cos
sin
Add the quotients.
=
1cos
sin

1 + cos
1 + cos
Multiply the numeratorand denominator by1 + cos.
=
1cos
2

(sin)(1 + cos)
Multiply.
=
sin
2

(sin)(1 + cos)
Apply a Pythagoreanidentity.
=
sin
1 + cos
Reduce to lowest terms.
Upon arriving at the expression of the other side, the identity has been estab-
lished. There is no unique technique to prove all identities, but familiarity withthe dierent techniques may help.
Example 3.4.7.Prove: secxcosx= sinxtanx.
Solution.
secxcosx=
1
cosx
cosx
=
1cos
2
x
cosx
=
sin
2
x
cosx
= sinx
sinx
cosx
= sinxtanx 2
Example 3.4.8.Prove:
1 + sin
1sin

1sin
1 + sin
= 4 sinsec
2

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DEPED COPY Solution.
1 + sin
1sin

1sin
1 + sin
=
(1 + sin)
2
(1sin)
2
(1sin)(1 + sin)
=
1 + 2 sin+ sin
2
1 + 2 sinsin
2

1sin
2

=
4 sin
cos
2

= 4 sinsec
2
2
More Solved Examples
1. Express each of the other circular functions ofin terms of cos.
Solution:
sec=
1
cos
sin
2
+ cos
2
= 1 =)sin
2
= 1cos
2
=)sin=
p
1cos
2

csc=
1
sin
=
1

p
1cos
2

cot=
cos
sin
=
cos

p
1cos
2

tan=
sin
cos
=

p
1cos
2

cos
2. If tan=a, express cos
2
in terms ofa.
Solution:
a=
sin
cos
=)a
2
=
sin
2

cos
2

=)a
2
=
1cos
2

cos
2

a
2
cos
2
= 1cos
2

a
2
cos
2
+ cos
2
= 1 =)cos
2
(a
2
+ 1) = 1 =)cos
2
=
1
a
2
+ 1
3. Givena= cosx, simplify and express sin
4
xcos
4
xin terms ofa.
Solution:sin
4
xcos
4
x= (sin
2
x+ cos
2
x)(sin
2
xcos
2
x)
= sin
2
xcos
2
x
= 1cos
2
xcos
2
x
= 12 cos
2
x= 12a
2
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DEPED COPY 4. Simplify (cscxsecx)
2
+ (cscx+ secx)
2
.
Solution:(cscxsecx)
2
+ (cscx+ secx)
2
= (csc
2
x2 cscxsecx+ sec
2
x) + (csc
2
x+ 2 cscxsecx+ sec
2
x)
= 2 csc
2
x+ 2 sec
2
x
=
2
sin
2
x
+
2
cos
2
x
=
2(cos
2
x+ sin
2
x)
sin
2
xcos
2
x
=
2
sin
2
xcos
2
x
= 2 csc
2
xsec
2
x
5. Verify the identity
csc
tan+ cot
= cos.
Solution:
csc
tan+ cot
=
1
sin
sin
cos
+
cos
sin
=
1
sin
sin
2
+ cos
2

cossin
=
1
sin

cossin
1
= cos
6. Establish the identity
csc+ cot1
cotcsc+ 1
=
1 + cos
sin
.
Solution:
csc+ cot1
cotcsc+ 1
=
csc+ cot1
cotcsc+ 1

csc+ cot
csc+ cot
=
(csc+ cot1)(csc+ cot)
(cotcsc)(csc+ cot) + (csc+ cot)
=
(csc+ cot1)(csc+ cot)
cot
2
csc
2
+ csc+ cot
=
(csc+ cot1)(csc+ cot)
1 + csc+ cot
= csc+ cot=
1
sin
+
cos
sin
=
1 + cos
sin
Supplementary Problems 3.4
1. Using fundamental identities, simplify the expression
tanxsinx
sinx
.
2. Using fundamental identities, simplify the expression
1
cscxcotx
.
3. Simplify sinA+
cos
2
A
1 + sinA
.
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DEPED COPY 4. Simplify (1cos
2
A)(1 + cot
2
A).
5. Express
cscx+ secx
cotx+ tanx
in terms of sine and cosine.
6. Express
tanxcotx
tanx+ cotx
in terms of sine and cosine.
7. Express
tanx+ sinx
cscx+ cotx
in terms cosine only.
8. Express
1
1 + tan
2
x
in terms sine only.
9. If cot=a, express sincosin terms ofa.
10. If sec=a >0 and sin >0, express sincosin terms ofa.
For numbers 11 - 20, establish the identities.
11.
csca+ 1
csca1
=
1 + sina
1sina
12.
1 + sina
1sina

1sina
1 + sina
= 4 tanaseca
13.
cosa
seca+ tana
= 1sina
14.
csca+ secatana
csc
2
a
= tanaseca
15.
1
1cosa
+
1
1 + cosa
= 2 csc
2
a
16.
sin
3
cos
3

sincos
= 1 + sincos
17.
tan
1tan
2

=
sincos
2 cos
2
1
18.
tan
2
+ sec+ 1
tan+ cot
= tan+ sin
19.
cotsinsec
seccsc
= cos
2
sin
2

20. tan
2
sec
2
sec
2
+ 1 = tan
4

4
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DEPED COPY Lesson 3.5. Sum and Dierence Identities
Learning Outcomes of the Lesson
At the end of the lesson, the student is able to:
(1) derive trigonometric identities involving sum and dierence of two angles;
(2) simplify trigonometric expressions using fundamental trigonometric identities
and sum and dierence identities;
(3) prove other trigonometric identities using fundamental trigonometric identi-
ties and sum and dierence identities; and
(4) solve situational problems involving trigonometric identities.
Lesson Outline
(1) The sum and dierence identities for cosine, sine, and tangent functions
(2) Cofunction identities
(3) More trigonometric identities
Introduction
In previous lesson, we introduced the concept of trigonometric identity, pre-
sented the fundamental identities, and proved some identities. In this lesson, we
derive the sum and dierence identities for cosine, sine, and tangent functions,
establish the cofunction identities, and prove more trigonometric identities.
3.5.1. The Cosine Dierence and Sum Identities
Letuandvbe any real numbers with 0< vu <2. Consider the unit circle
with pointsA= (1; 0),P1,P2,P3, anduandvwith corresponding angles as
shown below. ThenP1P2=AP3.
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DEPED COPY Recall thatP1=P(u) = (cosu;sinu),P2=P(v) = (cosv;sinv), andP3=
P(uv) = (cos(uv);sin(uv)), so that
P1P2=
p
(cosucosv)
2
+ (sinusinv)
2
;
while
AP3=
p
[cos(u v)1]
2
+ [sin(u v)0]
2
:
Equating these two expressions and expanding the squares, we get
(cosucosv)
2
+ (sinusinv)
2
= [cos(uv)1]
2
+ sin
2
(uv)
cos
2
u2 cosucosv+ cos
2
v+ sin
2
u2 sinusinv+ sin
2
v
= cos
2
(uv)2 cos(uv) + 1 + sin
2
(uv)
Applying the Pythagorean identity cos
2
+sin
2
= 1 and simplifying the resulting
equations, we obtain
(cos
2
u+ sin
2
u) + (cos
2
v+ sin
2
v)2 cosucosv2 sinusinv
= [cos
2
(uv) + sin
2
(uv)]2 cos(uv) + 1
1 + 12 cosucosv2 sinusinv= 12 cos(uv) + 1
cos(uv) = cosucosv+ sinusinv:
We have thus proved another identity.
Although we assumed at the start that 0< vu <2, but because
cos( ) = cos(one of the even-odd identities), this new identity is true for
any real numbersuandv. As before, the variables can take any real values or
angle measures.
Cosine Dierence Identity
cos(A B) = cosAcosB+ sinAsinB
ReplacingBwithB, and applying the even-odd identities, we immediately
get another identity.
Cosine Sum Identity
cos(A +B) = cosAcosBsinAsinB
Example 3.5.1.Find the exact values of cos 105

and cos

12
.
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electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2016.

DEPED COPY Solution.
cos 105

= cos(60

+ 45

)
= cos 60

cos 45

sin 60

sin 45

=
1
2

p
2
2

p
3
2

p
2
2
=
p
2
p
6
4
cos

12
= cos

4

6

= cos

4
cos

6
+ sin

4
sin

6
=
p
2
2

p
3
2
+
p
2
2

1
2
=
p
6 +
p
2
4
2
Example 3.5.2.Given cos=
3
5
and sin=
12
13
, wherelies in QIV andin
QI, nd cos(+).
Solution.We will be needing sinand cos.
sin=
p
1cos
2
=
s
1

3
5

2
=
4
5
cos=
q
1sin
2
=
s
1

12
13

2
=
5
13
cos( +) = coscossinsin
=
3
5

5
13

4
5

12
13
=
63
65
2
3.5.2. The Cofunction Identities and the Sine Sum and Dierence
Identities
In the Cosine Dierence Identity, if we letA=

2
, we get
cos

2
B

= cos

2

cosB+ sin

2

sinB
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electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2016.

DEPED COPY = (0) cosB+ (1) sinB
= sinB:
From this identity, if we replaceBwith

2
B, we have
cos
h

2

2
B
i
= sin

2
B

cosB= sin

2
B

:
As for the tangent function, we have
tan

2
B

=
sin

2
B

cos

2
B

=
cosB
sinB
= cotB:
We have just derived another set of identities.
Cofunction Identities
cos

2
B

= sinB sin

2
B

= cosB
tan

2
B

= cotB
Using the rst two cofunction identities, we now derive the identity for sin(A+
B).
sin(A +B) = cos
h

2
(A+B)
i
= cos
h

2
A

B)
i
= cos

2
A

cosB+ sin

2
A

sinB
= sinAcosB+ cosAsinB
Sine Sum Identity
sin(A +B) = sinAcosB+ cosAsinB
In the last identity, replacingBwithBand applying the even-odd identities
yield
sin(A B) = sin[A + (B )]
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DEPED COPY = sinAcos(B ) + cosAsin(B )
= sinAcosBcosAsinB:
Sine Dierence Identity
sin(A B) = sinAcosBcosAsinB
Example 3.5.3.Find the exact value of sin

5
12

.
Solution.
sin

5
12

= sin

4
+

6

= sin

4

cos

6

+ cos

4

sin

6

=
p
2
2

p
3
2
+
p
2
2

1
2
=
p
6 +
p
2
4
2
Example 3.5.4.If sin=
3
13
and sin=
1
2
, where 0< <

2
and

2
< < ,
nd sin( +) and sin().
Solution.We rst compute cosand cos.
cos=
p
1sin
2
=
s
1

3
13

2
=
4
p
10
13
cos=
q
1sin
2
=
s
1

1
2

2
=
p
3
2
sin(+) = sincos+ cossin
=
3
13

p
3
2
!
+
4
p
10
13

1
2
=
4
p
103
p
3
26
sin() = sincoscossin
=
1
2

4
p
10
13

p
3
2
!
3
13
=
4
p
10 + 3
p
3
26
2
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electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2016.

DEPED COPY Example 3.5.5.Prove:
sin(x+y) = (1 + cotxtany) sinxcosy:
Solution.
(1 + cotxtany) sinxcosy= sinxcosy+ cotxtanysinxcosy
= sinxcosy+
cosx
sinx
siny
cosy
sinxcosy
= sinxcosy+ cosxsiny= sin(x+y) 2
3.5.3. The Tangent Sum and Dierence Identities
Recall that tanxis the ratio of sinxover cosx. When we replacexwithA+B,
we obtain
tan(A +B) =
sin(A +B)
cos(A +B)
:
Using the sum identities for sine and cosine, and then dividing the numerator
and denominator by cosAcosB, we have
tan(A +B) =
sinAcosB+ cosAsinB
cosAcosBsinAsinB
=
sinAcosB
cosAcosB
+
cosAsinB
cosAcosB
cosAcosB
cosAcosB

sinAsinB
cosAcosB
=
tanA+ tanB
1tanAtanB
:
We have just established thetangent sum identity.
In the above identity, if we replaceBwithBand use the even-odd identity
tan( ) =tan, we get
tan(A B) = tan[A + (B )] =
tanA+ tan(B )
1tanAtan(B )
=
tanAtanB
1 + tanAtanB
:
This is thetangent dierence identity.
Tangent Sum and Dierence Identities
tan(A +B) =
tanA+ tanB
1tanAtanB
tan(A B) =
tanAtanB
1 + tanAtanB
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DEPED COPY More Solved Examples
1. Given cosA=
3
5
, < A <
3
2
, and tanB=
7
24
,Bin QI, nd: (a) sin(A +B),
(b) cos(A +B), and (c) tan(A+B).
Solution:cosA=
3
5
andAin QIII =)sinA=
4
5
tanB=
7
24
andBin QIII =)sinB=
7
25
and cosB=
24
25
(a) sin(A +B) = sinAcosB+ cosAsinB
=

4
5

24
25

+

3
5

7
25

=
117
125
(b) cos(A+B) = cosAcosBsinAsinB
=

3
5

24
25

4
5

7
25

=
44
125
(c) tan(A +B) =
sin(A +B)
cos(A +B)
=
117
125
44
125
=
117
44
2. Find the exact value of cos
5
12
.
Solution:cos
5
12
= cos

2
3

4

= cos
2
3
cos

4
+ sin
2
3
sin

4
=

1
2

p
2
2
!
+
p
3
2
! p
2
2
!
=
p
6
p
2
4
3. IfA+B=

2
+ 2k,k2Z, prove that sinA= cosB.
Solution:sinA= sin

2
+ 2kB

= sin

2
+ 2k

cosBcos

2
+ 2k

sinB= cosB
4. Find the value of
(tan 10

)(tan 15

)(tan 20

)(tan 15

) (tan 65

)(tan 70

)(tan 75

)(tan 80

):
Solution:From the previous item, we know that sin= cos(90

). We
write each tangent in terms of sine and cosine.
(tan 10

)(tan 15

)(tan 20

)(tan 15

) (tan 65

)(tan 70

)(tan 75

)(tan 80

)
=

sin 10
cos 10

sin 15
cos 15

sin 20
cos 20

sin 70
cos 70

sin 75
cos 75

sin 80
cos 80

= 1
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electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2016.

DEPED COPY 5. IfA,BandCare the angles of a triangle and
(tanA)(tanB)(tanC) =
p
3
3
;
nd tanA+ tanB+ tanC.
Solution:
A+B+C= 180

=)tan(A +B+C) = tan 180

= 0
tanA+ tan(B +C)
1tanAtan(B +C)
= 0 =)tanA+ tan(B +C) = 0
tanA+
tanB+ tanC
1tanBtanC
= 0
tanAtanAtanBtanC+ tanB+ tanC
1tanBtanC
= 0
=)tanAtanAtanBtanC+ tanB+ tanC= 0
tanA+ tanB+ tanC= tanAtanBtanC=
p
3
3
6. Establish the identity
sin(A +B)
cos(A B)
=
tanA+ tanB
1 + tanAtanB
.
Solution:
sin(A +B)
cos(A B)
=
sinAcosB+ cosAsinB
cosAcosB+ sinAsinB
=
sinAcosB+ cosAsinB
cosAcosB+ sinAsinB

1
cosAcosB
1
cosAcosB
=
sinAcosB
cosAcosB
+
cosAsinB
cosAcosB
cosAcosB
cosAcosB
+
sinAsinB
cosAcosB
=
tanA+ tanB
1 + tanAtanB
Supplementary Problems 3.5
1. If
3
2
< <2, nd the radian measure ofif cos= sin
2
3
.
2. For what anglein QIV is sin= cos
4
3
?
3. IfA+B=

2
+k,k2Z, prove that tanA= cotB.
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electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2016.

DEPED COPY 4. What is the exact value of cot

5
12

?
5. What is the exact value of sin 105

cos 15

?
6. What is the exact value of tan 1875

?
7. Letandbe acute angles such that cot= 7 and csc=
p
10. Find
cos( +).
8. Given sin=
8
17
and sin=
1
2
, nd cos( +) if bothandare in QIV.
9. If 3 sinx= 2, nd sin(x ) + sin(x +).
10. Simplify: cos

x+

2

+ cos

2
x

.
11. Given sinA=
4
5
,

2
A, and cosB=
4
5
,Bnot in QI, nd: (a) sin(AB),
(b) cos(AB), and (c) tan(AB). Also, determine the quadrant in which
ABterminate.
12. Given cscA=
p
3,Ain QI, and secB=
p
2, sinB <0, nd: (a) sin(A B),
(b) cos(AB), and (c) tan(AB). Also, determine the quadrant in which
ABterminate.
13. Given sin=
4
5
and cos=
5
13
, nd sin( +) + sin( ).
14. Given sin=
2
3
,in QII, and cos=
3
4
, nd cos( +) + cos( ).
15. IfAandBare acute angles (in degrees) such that cscA=
p
17 and cscB=
p
34
3
, what isA+B?
16. If tan(x+y) =
1
3
and tany=
1
2
, what is tanx?
17. Evaluate:
tan

9
+ tan
23
36
1tan

9
tan
23
36
.
18. Establish the identity:
sin(A +B+C) = sinAcosBcosC+ cosAsinBcosC
+ cosAcosBsinCsinAsinBsinC:
19. Prove: sin 2 = 2 sincos.
20. Prove: cot 2=
cot
2
1
2 cot
.
4
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DEPED COPY Topic Test 1 for Unit 3
1. A central angle in a circle of radius 6 cm measures 37:5

. Find: (a) length of
the intercepted arc and (b) the area of the sector.
2. The point (1; 2) lies on the terminal side of the anglein standard position.
Find sin+ cos+ tan.
3. Given sinA=
12
13
, whereAis not in QI, and cscB=
5
3
, whereBis not in
QIII, nd: (a) cos(A B) and (b) tan(AB).
4. Find the exact value of
tan 57

+ tan 78

1tan 57

tan 78

.
5. If sinx=aand cosx0, express
cosxtanx+ sinx
tanx
in terms ofa.
6. Prove the identity cos
6
x+ sin
6
x= 3 cos
4
x3 cos
2
x1.
7. A regular hexagon of side length 1 unit is inscribed in a unit circle such that
two of its vertices are located on thex-axis. Determine the coordinates of the
hexagon.
8. Determine the amplitude, period and phase shift of the graph of
y= 2 sin

x
2
+

3

1;
and sketch its graph over one period. Find the range of the function.
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DEPED COPY Topic Test 2 for Unit 3
1. The area of a sector of a circle formed by a central angle of 30

isd

3
cm
2
.
Find the length of the intercepted arc.
2. The point (8; 6) lies on the terminal side of the anglein standard position.
Find (sin+ cos)
2
.
3. Given sinA=
8
17
and cosA >0, evaluate sin

2
A

+ cos

2
A

.
4. Find the exact value of sin 160

cos 35

sin 70

cos 55

.
5. Find the exact value of tan
7
12
.
6. Given cosA=
3
5
, whereAis not in QII, and tanB=
24
7
, whereBis not in
QI, nd: (a) sin(A +B) and (b) cot(A+B).
7. Establish the identity
tan
2
x
tanx+ tan
3
x
= sinxcosx.
8. If sinxcosx=
1
3
, nd
sinx
secx
.
9. Determine the period and phase shift of the graph ofy= tan

18

x
3

+ 2,
and sketch its graph over two periods.
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DEPED COPY Lesson 3.6. Double-Angle and Half-Angle Identities
Learning Outcomes of the Lesson
At the end of the lesson, the student is able to:
(1) derive the double-angle and half-angle identities;
(2) simplify trigonometric expressions using known identities;
(3) prove other trigonometric identities using known identities; and
(4) solve situational problems involving trigonometric identities.
Lesson Outline
(1) The double-angle and half-angle identities for cosine, sine, and tangent
(2) More trigonometric identities
Introduction
Trigonometric identities simplify the computations of trigonometric expres-
sions. In this lesson, we continue on establishing more trigonometric identities.
In particular, we derive the formulas forf(2) andf

1
2

, wherefis the sine,
cosine, or tangent function.
3.6.1. Double-Angle Identities
Recall the sum identities for sine and cosine.
sin(A +B) = sinAcosB+ cosAsinB
cos(A +B) = cosAcosBsinAsinB
WhenA=B, these identities becomes
sin 2A= sinAcosA+ cosAsinA= 2 sinAcosA
and
cos 2A= cosAcosAsinAsinA= cos
2
Asin
2
A:
Double-Angle Identities for Sine and Cosine
sin 2A= 2 sinAcosA cos 2A= cos
2
Asin
2
A
The double-identity for cosine has other forms. We use the Pythagorean
identity sin
2
+ cos
2
= 1.
cos 2A= cos
2
Asin
2
A
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DEPED COPY = cos
2
A(1cos
2
A)
= 2 cos
2
A1
cos 2A= cos
2
Asin
2
A
= (1sin
2
A)sin
2
A
= 12 sin
2
A
Other Double-Angle Identities for Cosine
cos 2A= 2 cos
2
A1 cos 2A= 12 sin
2
A
Example 3.6.1.Given sint=
3
5
and

2
< t < , nd sin 2t and cos 2t.
Solution.We rst nd costusing the Pythagorean identity. Sincetlies in QII,
we have
cost=
p
1sin
2
t=
s
1

3
5

2
=
4
5
:
sin 2t= 2 sintcost
= 2

3
5

4
5

=
24
25
cos 2t= 12 sin
2
t
= 12

3
5

2
=
7
25
2
In the last example, we may compute cos 2tusing one of the other two double-
angle identities for cosine. For the sake of answering the curious minds, we include
the computations here.
cos 2t= cos
2
tsin
2
t
=

4
5

2

3
5

2
=
7
25
cos 2t= 2 cos
2
t1
= 2

4
5

2
1
=
7
25
In the three cosine double-angle identities, which formula to use depends on
the convenience, what is given, and what is asked.
Example 3.6.2.Derive an identity for sin 3xin terms of sinx.
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DEPED COPY Solution.We use the sum identity for sine, the double-angle identities for sine
and cosine, and the Pythagorean identity.
sin 3x= sin(2x+x)
= sin 2x cosx+ cos 2xsinx
= (2 sinxcosx) cosx+ (12 sin
2
x) sinx
= 2 sinxcos
2
x+ sinx2 sin
3
x
= 2(sinx)(1sin
2
x) + sinx2 sin
3
x
= 3 sinx4 sin
3
x 2
For the double-angle formula for tangent, we recall the tangent sum identity:
tan(A +B) =
tanA+ tanB
1tanAtanB
:
WhenA=B, we obtain
tan(A +A) =
tanA+ tanA
1tanAtanA
=
2 tanA
1tan
2
A
:
Tangent Double-Angle Identity
tan 2A =
2 tanA
1tan
2
A
Example 3.6.3.If tan=
1
3
and sec >0, nd sin 2 , cos 2 , and tan 2 .
Solution.We can compute immediately tan 2 .
tan 2 =
2 tan
1tan
2

=
2

1
3

1

1
3

2
=
3
4
From the given information, we deduce thatlies in QIV. Using one Pythagorean
identity, we compute costhrough sec. (We may also use the technique dis-
cussed in Lesson 3.2 by solving forx,y, andr.) Then we proceed to nd cos 2.
sec=
p
1 + tan
2
=
s
1 +

1
3

2
=
p
10
3
cos=
1
sec
=
1
p
10
3
=
3
p
10
10
cos 2= 2 cos
2
1 = 2

3
p
10
10
!
2
1 =
4
5
tan 2 =
sin 2
cos 2
=)sin 2= tan 2 cos 2=
3
5
2
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electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2016.

DEPED COPY 3.6.2. Half-Angle Identities
Recall two of the three double-angle identities for cosine:
cos 2A= 2 cos
2
A1 and cos 2A= 12 sin
2
A:
From these identities, we obtain two useful identities expressing sin
2
Aand cos
2
A
in terms of cos 2Aas follows:
cos
2
A=
1 + cos 2A
2
and sin
2
A=
1cos 2A
2
:
Some Useful Identities
cos
2
A=
1 + cos 2A
2
sin
2
A=
1cos 2A
2
From these identities, replacingAwith
A
2
, we get
cos
2
A
2
=
1 + cos 2

A
2

2
=
1 + cosA
2
and
sin
2
A
2
=
1cos 2

A
2

2
=
1cosA
2
:
These are thehalf-angle identitiesfor sine and cosine.
Half-Angle Identities for Sine and Cosine
cos
2

A
2

=
1 + cosA
2
sin
2

A
2

=
1cosA
2
Because of the \square" in the formulas, we get
cos
A
2
=
r
1 + cosA
2
and sin
A
2
=
r
1cosA
2
:
The appropriate signs of cos
A
2
and sin
A
2
depend on which quadrant
A
2
lies.
Example 3.6.4.Find the exact values of sin 22:5

and cos 22:5

.
Solution.Clearly, 22:5

lies in QI (and so sin 22:5

and cos 22:5

are both posi-
tive), and 22:5

is the half-angle of 45

.
sin 22:5

=
r
1cos 45

2
=
s
1
p
2
2
2
=
p
2
p
2
2
cos 22:5

=
r
1 + cos 45

2
=
s
1 +
p
2
2
2
=
p
2 +
p
2
2
2
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DEPED COPY Example 3.6.5.Prove: cos
2

2

=
tan+ sin
2 tan
.
Solution.
cos
2

2

=
1 + cos
2
=
1 + cos
2

tan
tan

=
tan+ costan
2 tan
=
tan+ cos
sin
cos
2 tan
=
tan+ sin
2 tan
2
We now derive the rst version of the half-angle formula for tangent.
tan
A
2
=
sin
A
2
cos
A
2
=
sin
A
2
cos
A
2

2 sin
A
2
2 sin
A
2
!
=
2 sin
2

A
2

2 sin
A
2
cos
A
2
=
2
1cosA
2
sin

2
A
2

=
1cosA
sinA
There is another version of the tangent half-angle formula, and we can derive
it from the rst version.
tan
A
2
=
1cosA
sinA
=
1cosA
sinA

1 + cosA
1 + cosA

=
1cos
2
A
(sinA)(1 + cosA)
=
sin
2
A
(sinA)(1 + cosA)
=
sinA
1 + cosA
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DEPED COPY Tangent Half-Angle Identities
tan
A
2
=
1cosA
sinA
tan
A
2
=
sinA
1 + cosA
tan
A
2
=
sin
A
2
cos
A
2
tan
2

A
2

=
1cosA
1 + cosA
Example 3.6.6.Find the exact value of tan

12
.
Solution.
tan

12
=
1cos

6
sin

6
=
1
p
3
2
1
2
= 2
p
3 2
Example 3.6.7.If sin=
2
5
, cot >0, and 0 <2, nd sin

2
, cos

2
, and
tan

2
.
Solution.Since sin <0 and cot >0, we conclude the < <
3
2
. It follows
that

2
<

2
<
3
4
;
which means that

2
lies in QII.
cos=
p
1sin
2
=
s
1

2
5

2
=
p
21
5
sin

2
=
r
1cos
2
=
v
u
u
t
1

p
21
5

2
=
p
50 + 10
p
21
10
cos

2
=
r
1 + cos
2
=
v
u
u
t
1 +

p
21
5

2
=
p
5010
p
21
10
tan

2
=
1cos
sin
=
1

p
21
5

2
5
=
5 +
p
21
2
2
More Solved Examples
1. If cos=
5
13
with 0< < , nd sin 2 and cos 2.
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DEPED COPY Solution:In this problem, we use the Pythagorean identity sin
2
+cos
2
= 1.
Because cos=
5
13
, we must have
sin
2
= 1cos
2
= 1
25
169
=
144
169
:
Moreover, since 0< < , we take the square root of both sides of the above
equation to get
sin=
12
13
:
Now, using the double-angle identities we get
sin 2= 2 sincosand cos 2= cos
2
sin
2

= 2

12
13

5
13

=
25
169

144
169
=
120
169
=
119
169
:
2. Derive an identity for cos 3xin terms of cosx.
Solution:We use the sum identity for cosine, the double-angle identities for
sine and cosine, and the Pythagorean identity.
cos 3x= cos(2x +x)
= cos 2x cosxsin 2xsinx
= (2 cos
2
x1) cosx(2 sinxcosx) sinx
= 2 cos
3
xcosx2 sin
2
xcosx
= 2 cos
3
xcosx2(1cos
2
x) cosx
= 4 cos
3
x3 cosx:
3. Derive the identity for tan 3tin terms of tant.
Solution:Using the sum identity for tangent, we obtain
tan 3t = tan(2t+t) =
tan 2t + tant
1tan 2t tant
:
Now, using the tangent double-angle identity, we have
tan 3t =
2 tant
1tan
2
t
+ tant
1
2 tant
1tan
2
t
tant
:
Upon simplifying the terms on the right side of the equation, we nally obtain
tan 3t =
3 tanttan
3
t
13 tan
2
t
:
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DEPED COPY 4. Find the exact value of sin

12
.
Solution:To nd the value of sin

12
, we use the identity sin
21
2
y=
1cosy
2
.
Withy=

6
, we obtain
sin
2

12
= sin
2
1
2

6

=
1cos

6
2
=
1
p
3
2
2
=
2
p
3
4
:
Now, since 0<

12
< , sin

12
must be positive, and so
sin

12
=
p
2
p
3
2
:
5. Prove: sin
2

2

=
sec1
2 sec
.
Solution:
sin
2

2

=
1cos
2
=
1cos
2

sec
sec

=
seccossec
2 sec
=
seccos
1
cos
2 sec
=
sec1
2 sec
:
6. Use the half-angle identity to nd the exact value of tan 75

.
Solution:tan 75

= tan

1
2
150

=
1cos 150

sin 150
=
1+
p
3
2
1
2
= 2 +
p
3.
7. A ball is thrown following a projectile motion. It is known that the horizontal
distance (range) the ball can travel is given by
R=
v
2
0
g
sin 2;
whereRis the range (in feet),v0is the initial speed (in ft/s),is the angle
of elevation the ball is thrown, andg= 32ft=s
2
is the acceleration due to
gravity.
(a) Express the new range in terms of the original range when an angle
(0< 45

) is doubled?
(b) If a ball travels a horizontal distance of 20 ft when kicked at an angle
ofwith initial speed of 20
p
2 ft/s, nd the horizontal distance it can
travel when you double. Hint: Use the result of item (a)
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DEPED COPY Solution:
(a) LetR=
v
2
0
g
sin 2be the original range. When the angle is doubled, the
new range will becomeR
0
=
v
2
0
g
sin 4. Now, we solve sin 4 in terms of
the original range.
Note that sin 2 =
gR
v
2
0
. So, as a consequence of the fundamental identity,
we obtain
cos 2=
s
1
g
2
R
2
v
4
0
=
p
v
4
0g
2
R
2
v
2
0
:
Since sin 4 = 2 sin 2cos 2, it follows that
R
0
=
v
2
0
g
sin 4=
v
2
0
g

2
gR
v
2
0

p
v
4
0g
2
R
2
v
2
0
!
=
2R
p
v
4
0g
2
R
2
v
2
0
:
(b) Using the result in (a), ifis doubled, then the new range is given by
R
0
=
2R
p
v
4
0g
2
R
2
v
2
0
=
40
p
640000409600
800
= 24:
Therefore, the new horizontal distance is 24 ft.
Supplementary Problems 3.6
1. Letbe an angle in the rst quadrant and sin=
1
3
. Find
(a) sin 2
(b) cos 2
(c) tan 2
(d) sec 2
(e) csc 2
(f) cot 2
2. Find the approximate value of csc 46

and sec 46

, given that sin 23

0:3907:
3. If cost=
3
4
, what is cos 2t?
4. Derive a formula for sin 4xin terms of sinxand cosx.
5. Let

4
< x <0. Given that tan 2x =2, solve for tanx.
6. Obtain an identity for tan 4in terms of tan.
7. Solve for the exact value of cot 4if tan=
1
2
.
8. Use half-angle identities to nd the exact value of (a) sin
2
15

and (b) cos
2
15

.
9. Use half-angle identities to nd the exact value of (a) sin
25
8
and (b) cos
25
8
.
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DEPED COPY 10. Find the exact value of cos

8
.
11. Prove that
tan
1
2
y1
tan
1
2
y+ 1
=
sinycosy1
siny+ cosy+ 1
.
12. Verify that the following equation is an identity: cot
1
2
t= cott(sect+ 1).
13. Use half-angle identities to nd the exact value of (a) cos 105

and (b) tan 22:5

.
4
Lesson 3.7.Inverse Trigonometric Functions
Learning Outcomes of the Lesson
At the end of the lesson, the student is able to:
(1) graph the six basic inverse trigonometric functions;
(2) illustrate the domain and range of the inverse trigonometric functions;
(3) evaluate inverse trigonometric expressions; and
(4) solve situational problems involving inverse trigonometric functions.
Lesson Outline
(1) Denitions of the six inverse trigonometric functions
(2) Graphs of inverse trigonometric functions
(3) Domain and range of inverse trigonometric functions
(4) Evaluation of inverse trigonometric expressions
Introduction
In the previous lessons on functions (algebraic and trigonometric), we com-
puted for the value of a function at a number in its domain. Now, given a value
in the range of the function, we reverse this process by nding a number in the
domain whose function value is the given one. Observe that, in this process,
the function involved may or may not give a unique number in the domain. For
example, each of the functionsf(x) =x
2
andg(x) = cosxdo not give a unique
number in their respective domains for some values of each function. Given
f(x) = 1, the function givesx=1. Ifg(x) = 1, thenx= 2k, wherekis an
integer. Because of this possibility, in order for the reverse process to produce a
function, we restrict this process to one-to-one functions or at least restrict the
domain of a non-one-to-one function to make it one-to-one so that the process
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DEPED COPY works. Loosely speaking, a function that reverses what a given functionfdoes
is called itsinverse function, and is usually denoted by f
1
.
More formally, two functionsfandgareinverse functionsif
g(f(x)) =xfor anyxin the domain off,
and
f(g(x)) =xfor anyxin the domain ofg.
We denote the inverse function of a functionfbyf
1
. The graphs of a function
and its inverse function are symmetric with respect to the liney=x.
In this lesson, we rst restrict the domain of each trigonometric function
because each of them is not one-to-one. We then dene each respective inverse
function and evaluate the values of each inverse trigonometric function.
3.7.1. Inverse Sine Function
All the trigonometric functions that we consider are periodic over their entire
domains. This means that all trigonometric functions are not one-to-one if we
consider their whole domains, which implies that they have no inverses over those
sets. But there is a way to make each of the trigonometric functions one-to-one.
This is done by restricting their respective domains. The restrictions will give us
well-dened inverse trigonometric functions.
The domain of the sine function is the setRof real numbers, and its range is
the closed interval [1;1]. As observed in the previous lessons, the sine function
is not one-to-one, and the rst step is to restrict its domain (by agreeing what the
convention is) with the following conditions: (1) the sine function is one-to-one
in that restricted domain, and (2) the range remains the same.
The inverse of the (restricted) sine functionf(x) = sinx, where the
domain is restricted to the closed interval

2
;

2

, is called theinverse
sine functionorarcsine function, denoted byf
1
(x) = sin
1
xor
f
1
(x) = arcsinx. Here, the domain off
1
(x) = arcsinxis [1; 1],
and its range is

2
;

2

. Thus,
y= sin
1
xory= arcsinx
if and only if
siny=x;
where1x1 and

2
y

2
.
Throughout the lesson, we interchangeably use sin
1
xand arcsinxto mean
the inverse sine function.
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DEPED COPY Example 3.7.1.Find the exact value of each expression.
(1) sin
11
2
(2) arcsin(1)
(3) arcsin 0
(4) sin
1

1
2

Solution.(1) Let= sin
11
2
. This is equivalent to sin=
1
2
. This means that
we are looking for the numberin the closed interval

2
;

2

whose sine is
1
2
. We get=

6
. Thus, we have sin
11
2
=

6
.
(2) arcsin(1) =

2
because sin

2

=1 and

2
2

2
;

2

.
(3) arcsin 0 = 0
(4) sin
1

1
2

=

6
2
As emphasized in the last example, as long as1x1, sin
1
xis that
numbery2

2
;

2

such that siny=x. Ifjxj>1, then sin
1
xis not dened in
R.
We can sometimes nd the exact value of sin
1
x(that is, we can nd a value
in terms of), but if no such special value exists, then we leave it in the form
sin
1
x. For example, as shown above, sin
11
2
is equal to

6
. However, as studied
in Lesson 3.2, no special numbersatises sin=
2
3
, so we leave sin
12
3
as is.
Example 3.7.2.Find the exact value of each expression.
(1) sin

sin
11
2

(2) arcsin

sin

3

(3) arcsin(sin)
(4) sin

sin
1

1
2

Solution.(1) sin

sin
11
2

= sin

6
=
1
2
(2) arcsin

sin

3

= arcsin
p
3
2
=

3
(3) arcsin(sin) = arcsin 0 = 0
(4) sin

sin
1

1
2

= sin

6

=
1
2
2
From the last example, we have the following observations:
1. sin(arcsinx) =xfor anyx2[1;1]; and
2. arcsin(sin) =if and only if2

2
;

2

, and if62

2
;

2

, then
arcsin(sin) =', where'2

2
;

2

such that sin'= sin.
To sketch the graph ofy= sin
1
x, Table 3.32 presents the tables of values
fory= sinxandy= sin
1
x. Recall that the graphs ofy= sinxandy= sin
1
x
are symmetric with respect to the liney=x. This means that if a point (a; b) is
ony= sinx, then (b; a) is on y= sin
1
x.
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DEPED COPY y= sinx
x

2

3

4

6
0

6

4

3

2
y1
p
3
2

p
2
2

1
2
0
1
2
p
2
2
p
3
2
1
y= sin
1
x
x1
p
3
2

p
2
2

1
2
0
1
2
p
2
2
p
3
2
1
y

2

3

4

6
0

6

4

3

2
Table 3.32
The graph (solid thick curve) of the restricted sine functiony= sinxis shown
in Figure 3.33(a), while the graph of inverse sine functiony= arcsinxis shown
in Figure 3.33(b).
(a)y=
x
(b)y=

1
x
Figure 3.33
Example 3.7.3.Sketch the graph ofy= sin
1
(x+ 1).
Solution 1.In this solution, we use translation of graphs.
Becausey= sin
1
(x+ 1) is equivalent toy= sin
1
[x(1)], the graph of
y= sin
1
(x+ 1) is 1-unit to the left ofy= sin
1
x. The graph below shows
y= sin
1
(x+ 1) (solid line) andy= sin
1
x(dashed line).
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DEPED COPY Solution 2.In
this solution, we graph rst the corresponding sine function, and
then use the symmetry with respect toy=xto graph the inverse function.
y= sin
1
(x+ 1)()siny=x+ 1()x= siny1
The graph below shows the process of graphing ofy= sin
1
(x+ 1) fromy=
sinx1 with

2
x

2
,
and then reecting it with respect toy=x.
3.7.2. Inverse Cosine Function
The
development of the other inverse trigonometric functions is similar to that
of the inverse sine function.
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DEPED COPY y= cos
1
xory= arccosx
means
cosy=x;
where1x1 and 0y.
The graph (solid thick curve) of the restricted cosine functiony= cosxis
shown in Figure 3.34(a), while the graph of inverse cosine functiony= arccosx
is shown in Figure 3.34(b).
(a)y=
x
(b)y=

1
x
Figure 3.34
Example 3.7.4.Find the exact value of each expression.
(1) cos
1
0
(2) arccos

p
3
2

(3)
cos

cos
1

p
3
2

(4)
cos
1

cos
3
4

(5)
arccos

cos
7
6

(6)
sin

cos
1
p
2
2

Solution.(1)
cos
1
0 =

2
b
ecause cos

2
=
0 and

2
2[0
; ].
(2) arccos

p
3
2

=
5

6
(3)
cos

cos
1

p
3
2

=
p
3
2
b
ecause
p
3
2
2[
1;1]
(4) cos
1

cos
3
4

=
3

4
b
ecause
3
4
2[0
; ].
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DEPED COPY (5) arccos

cos
7
6

= arccos

p
3
2

=
5
6
(6) sin

cos
1
p
2
2

=
p
2
2
2
Example 3.7.5.Simplify: sin

arcsin
2
3
+ arccos
1
2

.
Solution.We know that arccos
1
2
=

3
. Using the Sine Sum Identity, we have
sin

arcsin
2
3
+ arccos
1
2

= sin

arcsin
2
3
+

3

= sin

arcsin
2
3

cos

3
+ cos

arcsin
2
3

sin

3
=
2
3

1
2
+ cos

arcsin
2
3

p
3
2
=
1
3
+
p
3
2
cos

arcsin
2
3

:
We compute cos

arcsin
2
3

. Let= arcsin
2
3
. By denition, sin=
2
3
, where
lies in QI. Using the Pythagorean identity, we have
cos

arcsin
2
3

= cos=
p
1sin
2
=
p
5
3
:
Going back to the original computations above, we have
sin

arcsin
2
3
+ arccos
1
2

=
1
3
+
p
3
2
cos

arcsin
2
3

=
1
3
+
p
3
2

p
5
3
=
2+
p
15
6
: 2
Example 3.7.6.Simplify: sin

2 cos
1

4
5

.
Solution.Let= cos
1

4
5

. Then cos=
4
5
. Because cos <0 and range
of inverse cosine function is [0; ], we know thatmust be within the interval

2
;

. Using the Pythagorean Identity, we get sin=
3
5
.
Using the Sine Double-Angle Identity, we have
sin

2 cos
1

4
5

= sin 2
= 2 sincos
= 2
3
5

4
5

=
24
25
: 2
Example 3.7.7.Sketch the graph ofy=
1
4
cos
1
(2x).
Solution.
y=
1
4
cos
1
(2x)()4y= cos
1
(2x)()x=
1
2
cos(4y )
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DEPED COPY We graph rsty=
1
2
cos(4x). The domain of this graph comes from the restriction
of cosine as follows:
04x=)0x

4
:
Then reect this graph with respect toy=x, and we nally obtain the graph of
y=
1
4
cos
1
(2x) (solid line).
In the last example, we may also use the following technique. In graphing
y=
1
4
cos

1
(2x), the horizontal length of cos
1
xis reduced to half, while the
vertical height is reduced to quarter. This comparison technique is shown in
the graph below with the graph ofy= cos
1
xin dashed line and the graph of
y=
1
4
cos

1
(2x) in solid line.
3.7.3. Inverse Tangent Function and the Remaining Inverse
T
rigonometric Functions
The inverse tangent function is similarly dened as inverse sine and inverse cosine functions.
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DEPED COPY y= tan
1
xory= arctanx
means
tany=x;
wherex2Rand

2
< y <

2
.
The graph (solid thick curve) of the restricted functiony= tanxis shown
in Figure 3.35(a), while the graph of inverse functiony= arctanxis shown in
Figure 3.35(b).
(a)y=
x
(b)y=

1
x
Figure 3.35
Example 3.7.8.Find the exact value of each expression.
(1) tan
1
1
(2) arctan

p
3

(3)
tan

tan
1

5
2

(4)
tan
1

tan

6

(5)
tan
1

tan
7
6

(6)
arctan

tan

19
6

Solution.Note
the range of arctan is the open interval

2
;

2

.
(1)
tan
1
1 =

4
(2)
arctan

p
3

=

3
(3)
tan

tan
1

5
2

=
5
2
(4)
tan
1

tan

6

=

6
b
ecause

6
2

2
;

2

.
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DEPED COPY (5) Here, note that
7
6
62

2
;

2

. Use the idea of reference angle, we know that
tan
7
6
= tan

6
.
tan
1

tan
7
6

= tan
1

tan

6

=

6
(6) Here, we cannot use the idea of reference angle, but the idea can help in a
way. The number (or angle)
19
6
is in QII, wherein tangent is negative, and
its reference angle is

6
.
arctan

tan

19
6

= arctan

tan

6

=

6
2
Example 3.7.9.Find the exact value of each expression.
(1) sin

2 tan
1

8
3

(2) tan

sin
13
5
tan
11
4

Solution.(1) Let= tan
1

8
3

. Then tan=
8
3
. Following the notations in
Lesson 3.2 and the denition of inverse tangent function, we know thatlies
in QIV, andx= 3 andy=8. We getr=
p
3
2
+ (8)
2
=
p
73.
Applying the Sine Double-Angle Identity (page 192) gives
sin

2 tan
1

8
3

= sin 2
= 2 sincos
= 2
y
r

x
r
= 2

8
p
73

3
p
73

=
48
73
:
(2) Using the Tangent Dierence Identity, we obtain
tan

sin
1
3
5
tan
1
1
4

=
tan

sin
13
5

tan

tan
11
4

1 + tan

sin
13
5

tan

tan
11
4

=
tan

sin
13
5

1
4
1 + tan

sin
13
5

1
4
:
We are left to compute tan

sin
13
5

. We proceed as in (1) above. Let
= sin
13
5
. Then sin=
3
5
. From the denition of inverse sine function and
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DEPED COPY the notations used in Lesson 3.2, we know thatlies in QI, andy= 3 and
r= 5. We getx=
p
5
2
3
2
= 4, so that tan=
y
x
=
3
4
.
tan

sin
1
3
5
tan
1
1
4

=
tan

sin
13
5

1
4
1 + tan

sin
13
5

1
4
=
3
4

1
4
1 +
3
4

1
4
=
8
19
2
?
Example 3.7.10.A student is
viewing a painting in a museum.
Standing 6 ft from the painting,
the eye level of the student is 5 ft
above the ground. If the paint-
ing is 10 ft tall, and its base is
4 ft above the ground, nd the
viewing angle subtended by the
painting at the eyes of the stu-
dent.
Solution.Letb
e the viewing angle, and let=+as shown below.
We observe that
tan=
1
6
and
tan=
9
6
:
Using
the Tangent Sum Identity, we have
tan= tan( +) =
tan+ tan
1tantan
=
1
6
+
9
6
1
1
6

9
6
=
20
9
:
Using a calculator, the viewing angle is=
tan
120
9
65
:8

. 2
We now dene the remaining inverse trigonometric functions.
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DEPED COPY Dene
cot
1
x=

2
tan
1
x:
It follows that the domain ofy= cot
1
xisRand its range is (0; ).
y= sec
1
xory= arcsecx
means
secy=x;
wherejxj 1 andy2

0;

2

[

;
3
2

.
Dene
csc
1
x=

2
sec
1
x:
This means that the domain ofy= csc
1
xis (1; 1][[1;1) and
its range is

;

2

[

0;

2

.
The graphs of these last three inverse trigonometric functions are shown in
Figures 3.36, 3.37, and 3.38, respectively.
(a)y=
x
(b)y=

1
x
Figure 3.36
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DEPED COPY (a)y=
x
(b)y=

1
x
Figure 3.37
(a)y=
x
(b)y=

1
x
Figure 3.38
Observe that the process in getting the value of an inverse function is the
same to all inverse functions. That is,y=f
1
(x) is the same asf(y) =x. We
need to remember the range of each inverse trigonometric function. Table 3.39
summarizes all the information about the six inverse trigonometric functions.
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DEPED COPY FunctionDomain Range Graph
sin
1
x [1;1]

2
;

2
Figure
3.33(b)
cos
1
x [1;1] [0; ]
Figure
3.34(b)
tan
1
x R

2
;

2
Figure
3.35(b)
cot
1
x R (0; )
Figure
3.36(b)
sec
1
xfx:jxj 1g

0;

2

[

;
3
2
Figure
3.37(b)
csc
1
xfx:jxj 1g

;

2

[

0;

2
Figure
3.38(b)
Table 3.39
Example 3.7.11.Find the exact value of each expression.
(1) sec
1
(2)
(2) csc
1

2
p
3
3

(3) cot
1

p
3

(4) sin

sec
1

3
2

csc
1

2
p
3
3

Solution.(1) sec
1
(2) =
4
3
because sec
4
3
=2 and
4
3
2

;
3
2

(2) csc
1

2
p
3
3

=
2
3
(3) cot
1

p
3

=
5
6
(4) From (2), we know that csc
1

2
p
3
3

=
2
3
. Let= sec
1

3
2

. Then
sec=
3
2
. From dened range of inverse secant function and the notations
in Lesson 3.2,lies in QIII, andr= 3 andx=2. Solving fory, we get
y=
p
3
2
(2)
2
=
p
5. It follows that sin=
p
5
3
and cos=
2
3
.
We now use the Sine Sum Identity.
sin

sec
1

3
2

csc
1

2
p
3
3
!!
= sin

2
3

= sin

+
2
3

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DEPED COPY = sincos
2
3
+ cossin
2
3
=

p
5
3
!

1
2

+

2
3

p
3
2
!
=
p
52
p
3
6
2
More Solved Examples
1. Find the exact values of the following, if they exist.
(a) sin
1
p
2
2
(b) arcsin

1
2

(c) sin
1
2
Solution:Note that the range off(x) = sin
1
xis [

2
;

2
]. Thus, if we let
y= sin
1
x, then we are looking fory2[

2
;

2
] such that siny=x. Hence,
(a) sin
1
p
2
2
=

4
,
(b) arcsin

1
2

=

6
, and
(c) sin
1
2 is undened because siny1.
2. Find the exact value of each expression.
(a) sin

sin
1
p
2
2

(b) cos

arcsin

1
2

(c) sin
1

sin
11
2

Solution:
(a) sin

sin
1
p
2
2

= sin(

4
) =
p
2
2
(b) cos

arcsin

1
2

= cos(

6
) =
p
3
2
(c) sin
1

sin
11
2

= sin
1
(1) =

2
3. Answer the following.
(a) What is the domain ofy= sin
1
2x?
(b) What is the range ofy= sin
1
2x?
(c) What is thexintercept ofy= sin
1
2x?
Solution:
(a) Consider the functionf() = sin
1
. The domain of sin
1
is [1; 1].
So,= 2x2[1;1]. Therefore, the domain of sin
1
2xis [1=2; 1=2].
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DEPED COPY (b) [=2; =2]
(c) (0;0)
4. From the concept of projectile motion, if an object is directed at an angle
(with2[0; =2]), then the range will beR=
v
2
0
g
sin 2(in feet) wherev0(in
ft=s) is the initial speed andg= 32ft=s
2
is the acceleration due to gravity.
At what angle shall the object be directed so that the range will be 100ft,
given that the initial speed isv0= 80ft=s?
Solution:From the formula of the range, we get
100 =
80
2
32
sin 2=)
1
2
= sin 2
Sincemust be from 0 to

2
(i.e. 02), this is equivalent to nding 2
such that 2= sin
11
2
. Hence,
2=

6
=)=

12
:
Therefore, the object must be directed at an angle of

12
rad(or 15

), to have
a projectile range of 100ft.
5. Find the exact values of the following, if they exist.
(a) cos
1
p
2
2
(b) cos

cos
1

1
2

(c) arccos(cos)
(d) arccos
Solution:
(a) cos
1
p
2
2
=

4
(b) cos

cos
1

1
2

= cos
2
3
=
1
2
(c) arccos(cos) = arccos(1) =
(d) Lety= arccos. Since cosy1, we haveyis undened because >3.
6. Simplify: (a) cos

cos
1
p
3
2
cos
11
3

Solution:We know that cos
1
p
3
2
=

6
. Let= cos
11
3
. Which is equivalent
to cos=
1
3
with 0. Using the Cosine Dierence Identity, we have
cos

cos
1
p
3
2
cos
1
1
3
!
= cos

6

= cos

6
cos+ sin

6
sin
=
p
3
2

1
3
+
1
2
sin:
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DEPED COPY Now, we solve for sinusing the Pythagorean Identity which gives
sin
2
= 1(1=3)
2
= 1(1=9) = 8=9:
Thus, sin=
2
p
2
3
because2[0; ]. Finally, we obtain
cos

cos
1
p
3
2
cos
1
1
3
!
=
p
3
2

1
3
+
1
2

2
p
2
3
=
p
3
6
+
2
p
2
6
=
p
3 + 2
p
2
6
:
7. Simplify: (a) cos

2 cos
12
5

; (b) sin

cos
12
5

Solution:Let= cos
12
5
. Which is equivalent to cos=
2
5
with 0
. Using the Double-Angle Identity for Cosine and one of the Fundamental
Idenity, we have
cos

2 cos
1
2
5

= cos(2 ) = 2 cos
2
1 = 2

2
5

2
1 =
8
25
1 =
17
25
and
sin

cos
1
2
5

= sin=
p
1cos
2
=
r
1
4
25
=
p
21
5
:
Here, sin0 because2[0; ].
8. Graph:y= 1 + cos
1
x
Solution:The graph can be obtained by translating the graph of the inverse
cosine function one unit upward.
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DEPED COPY 9. Find the exact value of each expression.
(a) arctan(tan
4
3
) (b) tan(tan
14
5
)
Solution:
(a) arctan(tan
4
3
) = arctan
p
3 ==3
(b) tan(tan
14
5
) =
4
5
10. Find the exact value of tan(tan
17
6
+ tan
11
2
).
Solution:
tan

tan
1
7
6
+ tan
1
1
2

=
tan(tan
17
6
) + tan(tan
11
2
)
1tan(tan
17
6
)tan(tan
11
2
)
=
7
6
+
1
2
1
7
6

1
2
= 4
11. Find the exact values of the following, if they exist.
(a) sec
1
p
2
(b) csc
1
1
(c) cot
1
p
3
3
(d) arcsec
1
(cot(

4
))
(e) cos(arccsc
1
2)
(f) arccot
1
(sin
20
3
)
Solution:
(a) sec
1
p
2 =

4
(b) csc
1
1 =

2
(c) cot
1
p
3
3
=

3
(d) arcsec(cot(

4
)) = arcsec(1) =
(e) cos(arccsc(2)) = cos

6
=
p
3
2
(f) arccot(sin
20
3
) = arccot
p
3
2
. Let= arccot
p
3
2
. Then,
cot=
p
3
2
=)tan=
2
p
3
=)= tan
1
2
p
3
(0:8571):
Here, we needed to use a calculator to solve for the approximate value,
since
2
p
3
is not a special value for tangent function.
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DEPED COPY Supplementary Problems 3.7
1. Find the exact value of the following.
(a) sin[sin
1
(1=2)]
(b) cos[cos
1
(
p
2=2)]
(c) tan[tan
1
(
p
3)]
(d) sin[arctan(
p
3)]
(e) cos[arccos(
p
2)]
(f) tan[arcsin(1=4)]
(g) cos[sin
1
(
p
3=2)]
2. Find the exact value of the following.
(a) sin
1
[sin(25=6)]
(b) arccos[cos(23=4)]
(c) tan
1
[tan(1)]
(d) arcsin[cos(13=4)]
(e) cos
1
[sec(23 )]
(f) arctan[sin(=12)]
3. Solve the exact value of the following.
(a) sin[2 cos
1
(4=5)]
(b) cos[2 sin
1
(5=13)]
(c) sin(sin
1
(3=5) + cos
1
(5=13))
(d) cos[sin
1
(1=2)cos
1
(8=17)]
4. Consider the functionf(x) = tan
1
(x+ 1). Do the following.
(a) Find the domain off.
(b) Find the range off.
(c) Find thexandyintercept off, if there are any.
(d) Graphf.
5. Evaluate and simplify the following, if they exist.
(a) arcsec(
p
2)
(b) arccsc(2)
(c) arccot
p
3
(d) [sec
1
(1)][cos
1
(1)]
(e) 2 cot
1
p
3 + 3 csc
1
2
(f) csc
1
0
6. Evaluate and simplify the following, if they exist.
(a)
cos(sec
1
3 + tan
1
2)
cos(tan
1
2)
(b) tan(2 arcsin(1=6))
(c) cos
2

sin
1
(1=2)
2

(d) arcsec(sin(100=3))
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DEPED COPY 7. A trough is in the shape of an inverted triangular prism whose cross section
has the shape of an inverted isosceles triangle (see Figure 3.40). If the length
of the base of the cross section is 2
p
3m:and the length of the trough is
100
p
3m:, nd the size of the vertex angle so that the volume is 900m
3
.
Hint:V=bhl=2:
Figure 3.40
4
Lesson 3.8. Trigonometric Equations
Learning Outcomes of the Lesson
At the end of the lesson, the student is able to:
(1) solve trigonometric equations; and
(2) solve situational problems involving trigonometric equations.
Lesson Outline
(1) Denition of a trigonometric equation
(2) Solution to a trigonometric equation
(3) Techniques of solving a trigonometric equation
Introduction
We have studied equations in Lesson 3.4. We dierentiated an identity from
a conditional equation. Recall that an identity is an equation that is true for all
220All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means -
electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2016.

DEPED COPY values of the variable in the domain of the equation, while a conditional equation
is an equation that is not an identity.
In this lesson, we mostly study conditional trigonometric equations. Though
not explicitly, we have started it in the preceding lesson. For example, the equa-
tion sinx=
1
2
has the unique solutionx= sin
11
2
=

6
in the closed interval

2
;

2

. However, if we consider the entire domain (not the restricted domain)
of the sine function, which is the setRof real numbers, there are solutions (other
than

6
) of the equation sinx=
1
2
. This current lesson explores the techniques of
solving (conditional) trigonometric equations.
We divide the lesson into two groups of equations: the ones using a basic way
of solving, and those using more advanced techniques.
3.8.1. Solutions of a Trigonometric Equation
Any equation that involves trigonometric expressions is called atrigonometric
equation. Recall that a solutionor arootof an equation is a number in the domain
of the equation that, when substituted to the variable, makes the equation true.
The set of all solutions of an equation is called thesolution setof the equation.
Technically, the basic method to show that a particular number is a solution
of an equation is to substitute the number to the variable and see if the equation
becomes true. However, we may use our knowledge gained from the previous
lessons to do a quicker verication process by not doing the manual substitution
and checking. We use this technique in the example.
Example 3.8.1.Which numbers in the set

0;

6
;

4
;

3
;

2
;
2
3
;
3
4
;
5
6
; ;2

are
solutions to the following equations?
(1) sinx=
1
2
(2) tanx= 1
(3) 3 secx=2
p
3
(4)
p
3jcotxj= 1
(5) sec
2
xtan
2
x= 1
(6) sinx+ cosx= 0
(7) cos
2
x= cos 2x + sin
2
x
(8) sinx+ cos 2x= 0
(9) 2 sinx+ tanx2 cosx= 2
(10) sin
2
x+ cos
2
x= 2
(11) sin 2x= sinx
(12) 2 tanx+ 4 sinx= 2 + secx
Solution.Note that the choices (except 2 ) are numbers within the interval [0; ].
To quickly determine which numbers among the choices are solutions to a par-
ticular equation, we use some distinctive properties of the possible solutions.
(1) The sine function is positive on (0; ). From Lesson 3.2, we recall that

6
is
an obvious solution. We may imagine the graph ofy= sinx. We may also
use the idea of reference angle. Thus, among the choices, only

6
and
5
6
are
the only solutions of sinx=
1
2
.
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DEPED COPY (2) Since tanx= 1>0, any solution of the equation among the choices must
be in the interval

0;

2

(that is, in QI). Again, among the choices, the only
solution to tanx= 1 is

4
.
(3) Here, the given equation is equivalent to secx=
2
p
3
3
. Among the choices,
the only solution of the equation 3 secx=2
p
3 is
5
6
.
(4) Eliminating the absolute value sign, the given equation is equivalent to cotx=
p
3
3
or cotx=
p
3
3
. Among the choices, the only solution of cotx=
p
3
3
is

3
,
while the other equation has
2
3
. Thus, the only solutions of
p
3jcotxj= 1
from the given set are

3
and
2
3
.
(5) The given equation is one of the Pythagorean Identities (page 175). It means
that all numbers in the domain of the equation are solutions. The domain
of the equation isRn fx: cosx= 0g. Thus, all except

2
are solutions of
sec
2
xtan
2
x= 1.
(6) For the sum of sinxand cosxto be 0, they must have equal absolute values
but dierent signs. Among the choices, only
3
4
satises these properties, and
it is the only solution of sinx+ cosx= 0.
(7) This equations is one of the Double-Angle Identities for Cosine. This means
that all numbers in the domain of the equation are its solutions. Because thedomain of the given equation isR, all numbers in the given set are solutions
of cos
2
x= cos 2x + sin
2
x.
(8) We substitute each number in the choices to the expression on the left-side
of the equation, and select those numbers that give resulting values equal to1.
x= 0: sin 0 + cos 2(0) = 0 + 1 = 1
x=

6
: sin

6
+ cos 2(

6
) =
1
2
+
1
2
= 1
x=

4
: sin

4
+ cos 2(

4
) =
p
2
2
+ 0 =
p
2
2
x=

3
: sin

3
+ cos 2(

3
) =
p
3
2

1
2
=
p
31
2
x=

2
: sin

2
+ cos 2(

2
) = 11 = 0
x=
2
3
: sin
2
3
+ cos 2(
2
3
) =
p
3
2

1
2
=
p
31
2
x=
3
4
: sin
3
4
+ cos 2(
3
4
) =
p
2
2
+ 0 =
p
2
2
x=
5
6
: sin
5
6
+ cos 2(
5
6
) =
1
2
+
1
2
= 1
x=: sin+ cos 2= 0 + 1 = 1
x= 2: sin 2 + cos 2(2) = 0 + 1 = 1
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DEPED COPY From these values, the only solution of sinx+ cos 2x= 0 among the choices
is

2
.
(9) We again substitute the numbers in the given set one by one, and see which
resulting values are equal to 1.
x= 0: 2 sin 0 + tan 02 cos 0 =2
x=

6
: 2 sin

6
+ tan

6
2 cos

6
=
32
p
3
3
x=

4
: 2 sin

4
+ tan

4
2 cos

4
= 1
x=

3
: 2 sin

3
+ tan

3
2 cos

3
= 2
p
31
x=

2
: Since tan

2
is undened, this value ofxcannot be a solution of the
equation.
x=
2
3
: 2 sin
2
3
+ tan
2
3
2 cos
2
3
= 1
x=
3
4
: 2 sin
3
4
+ tan
3
4
2 cos
3
4
= 2
p
21
x=
5
6
: 2 sin
5
6
+ tan
5
6
2 cos
5
6
=
3+2
p
3
3
x=: 2 sin+ tan2 cos= 2
x= 2: 2 sin 2 + tan 22 cos 2 =2
Thus, the only solution of 2 sinx+tanx2 cosx= 2 from the given set is.
(10) This equation has no solution because one of the Pythagorean Identities says
sin
2
x+ cos
2
x= 1.
(11) We substitute each number in the given set to the expression of each side of
the equation, and see which resulting values are equal.
x= 0: sin 2(0) = 0; sin 0 = 0
x=

6
: sin 2(

6
) =
p
3
2
; sin

6
=
1
2
x=

4
: sin 2(

4
) = 1; sin

4
=
p
2
2
x=

3
: sin 2(

3
) =
p
3
2
; sin

3
=
p
3
2
x=

2
: sin 2(

2
) = 0; sin

2
= 1
x=
3
4
: sin 2(
3
4
) =1; sin
3
4
=
p
2
2
x=
5
6
: sin 2(
5
6
) =
p
3
2
; sin

3
=
1
2
x=: sin 2 = 0; sin= 0
x= 2: sin 2(2 ) = 0; sin 2= 0
Thus, among the numbers in the given set, the solutions of sin 2x= sinxare
0,

3
,, and 2.
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DEPED COPY (12) We employ the same technique used in the previous item.
x= 0 : 2 tan 0 + 4 sin 0 = 0
2 + sec 0 = 3
x=

6
: 2 tan

6
+ 4 sin

6
=
2
p
3+6
3
2 + sec

6
=
2
p
3+6
3
x=

4
: 2 tan

4
+ 4 sin

4
= 2
p
2 + 2
2 + sec

4
=
p
2 + 2
x=

3
: 2 tan

3
+ 4 sin

3
= 4
p
3
2 + sec

3
= 4
x=

2
: Both tan

2
and sec

2
are undened:
x=
2
3
: 2 tan
2
3
+ 4 sin
2
3
= 0
2 + sec
2
3
= 0
x=
3
4
: 2 tan
3
4
+ 4 sin
3
4
= 2
p
22
2 + sec
3
4
= 2
p
2
x=
5
6
: 2 tan
5
6
+ 4 sin
5
6
=
62
p
3
3
2 + sec
5
6
=
62
p
3
3
x=: 2 tan+ 4 sin= 0
2 + sec= 1
x= 2: 2 tan 2+ 4 sin 2= 0
2 + sec 2= 3
After checking the equal values, the solutions of 2 tanx+ 4 sinx= 2 + secx
among the given choices are

6
,
2
3
, and
5
6
. 2
3.8.2. Equations with One Term
From the preceding discussion, you may observe that there may be more solutions
of a given equation outside the given set. We now nd all solutions of a given
equation.
We will start with a group of equations having straightforward techniques
in nding their solutions. These simple techniques involve at least one of the
following ideas:
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DEPED COPY (1) equivalent equations (that is, equations that have the same solutions as the
original equation);
(2) periodicity of the trigonometric function involved;
(3) inverse trigonometric function;
(4) values of the trigonometric function involved on the interval [0; ] or [0; 2]
(depending on the periodicity of the function); and
(5) Zero-Factor Law:ab= 0 if and only ifa= 0 orb= 0.
To \solve an equation" means to nd all solutions of the equation. Here,
unless stated as angles measured in degrees, we mean solutions of the equation
that are real numbers (or equivalently, angles measured in radians).
Example 3.8.2.Solve the equation 2 cosx1 = 0.
Solution.The given equation is equivalent to
cosx=
1
2
:
On the interval [0;2], there are only two solutions of the last equation, and these
arex=

3
(this is in QI) andx=
5
3
(in QIV).
Because the period of cosine function is 2, the complete solutions of the
equation arex=

3
+k(2) andx=
5
3
+k(2) for all integersk. 2
In the preceding example, by saying that the \complete solutions arex=

3
+k(2) andx=
5
3
+k(2) for all integersk," we mean that any integral
value ofkwill produce a solution to the given equation. For example, when
k= 3,x=

3
+ 3(2) =
19
3
is a solution of the equation. Whenk=2,
x=
5
3
+ (2)(2 ) =
7
3
is another solution of 2 cosx1 = 0. The family of
solutionsx=

3
+k(2) can be equivalently enumerated asx=
19
3
+ 2k, while
the familyx=
5
3
+k(2) can also be stated asx=
7
3
+ 2k.
Example 3.8.3.Solve: (1 + cos)(tan1) = 0.
Solution.By the Zero-Factor Law, the given equation is equivalent to
1 + cos= 0 or tan 1 = 0
cos=1 tan= 1
=+ 2k; k2Z =

4
+k; k2Z:
Therefore, the solutions of the equation are=+ 2kand=

4
+kfor all
k2Z. 2
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DEPED COPY Example 3.8.4.Find all values ofxin the interval [2;2] that satisfy the
equation (sinx1)(sinx+ 1) = 0.
Solution.
sinx1 = 0 or sin x+ 1 = 0
sinx= 1 sinx=1
x=

2
or
3
2
x=
3
2
or

2
Solutions:

2
;
3
2
;
3
2
;

2
2
Example 3.8.5.Solve: cosx= 0:1.
Solution.There is no special number whose cosine is 0:1. However, because
0:12[1;1], there is a number whose cosine is 0:1. In fact, in any one-period
interval, with cosx= 0:1>0, we expect two solutions: one in QI and another in
QIV. We use the inverse cosine function.
From Lesson 3.7, one particular solution of cosx= 0:1 in QI isx= cos
1
0:1.
We can use this solution to get a particular solution in QIV, and this isx=
2cos
1
0:1, which is equivalent tox=cos
1
0:1.
From the above particular solutions, we can produce all solutions of cosx=
0:1, and these arex= cos
1
0:1+2kandx=cos
1
0:1+2kfor allk2Z.2
Example 3.8.6.Solve: 3 tan+ 5 = 0.
Solution.
3 tan+ 5 = 0 =)tan=
5
3
We expect only one solution in any one-period interval.
tan=
5
3
=)= tan
1

5
3

+k; k2Z 2
?
Example 3.8.7.The voltageV(in volts) coming from an electricity distribut-
ing company is uctuating according to the functionV(t) = 200 + 170 sin(120t)
at timetin seconds.
(1) Determine the rst time it takes to reach 300 volts.
(2) For what values oftdoes the voltage reach its maximum value?
Solution.(1) We solve for the least positive value oftsuch thatV(t) = 300.
200 + 170 sin(120t) = 300
sin(120t) =
100
170
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DEPED COPY 120t= sin
1
100
170
t=
sin
1100
170
120
0:00167 seconds
(2) The maximum value ofV(t) happens when and only when the maximum
value of sin(120t) is reached. We know that the maximum value of sin(120t)
is 1, and it follows that the maximum value ofV(t) is 370 volts. Thus, we
need to solve for all values oftsuch that sin(120t) = 1.
sin(120t) = 1
120t=

2
+ 2k; knonnegative integer
t=

2
+ 2k
120
t=
1
2
+ 2k
120
0:00417 + 0:017k
This means that the voltage is maximum whent0:00417+0:017k for each
nonnegative integerk. 2
3.8.3. Equations with Two or More Terms
We will now consider a group of equations having multi-step techniques of nding
their solutions. Coupled with the straightforward techniques we learned in the
preceding discussion, these more advanced techniques involve factoring of expres-
sions and trigonometric identities. The primary goal is to reduce a given equation
into equivalent one-term equations.
Example 3.8.8.Solve: 2 cosxtanx= 2 cosx.
Solution.
2 cosxtanx= 2 cosx
2 cosxtanx2 cosx= 0
(2 cosx)(tanx1) = 0
2 cosx= 0 or tan x1 = 0
cosx= 0 tan x= 1
x=

2
+ 2kor
x=
3
2
+ 2k,
k2Z
x=

4
+k,
k2Z
Solutions:

2
+ 2k;
3
2
+ 2k;

4
+k; k2Z 2
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DEPED COPY Example 3.8.9.Solve forx2[0;2): sin 2x= sinx.
Solution.
sin 2x= sinx
sin 2xsinx= 0
2 sinxcosxsinx= 0 Sine Double-Angle Identity
(sinx)(2 cosx1) = 0
sinx= 0 or 2 cos x1 = 0
x= 0 orx= cosx=
1
2
x=

3
orx=
5
3
Solutions: 0; ;

3
;
5
3
2
Tips in Solving Trigonometric Equations
(1) If the equation contains only one trigonometric term, isolate that
term, and solve for the variable.
(2) If the equation is quadratic in form, we may use factoring, nding
square roots, or the quadratic formula.
(3) Rewrite the equation to have 0 on one side, and then factor (if
appropriate) the expression on the other side.
(4) If the equation contains more than one trigonometric function,
try to express everything in terms of one trigonometric function.
Here, identities are useful.
(5) If half or multiple angles are present, express them in terms of a
trigonometric expression of a single angle, except when all angles
involved have the same multiplicity wherein, in this case, retain
the angle. Half-angle and double-angle identities are useful in
simplication.
Example 3.8.10.Solve forx2[0;2): 2 cos
2
x= 1 + sinx.
Solution.
2 cos
2
x= 1 + sinx
2(1sin
2
x) = 1 + sinxPythagorean Identity
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DEPED COPY 2 sin
2
x+ sinx1 = 0
(2 sinx1)(sinx+ 1) = 0 Factoring
2 sinx1 = 0 or sin x+ 1 = 0
sinx=
1
2
sinx=1
x=

6
orx=
5
6
x=
3
2
Solutions:

6
;
5
6
;
3
2
2
Example 3.8.11.Solve forx2[0;2) in the equation 3 cos
2
x+ 2 sinx= 2.
Solution.
3 cos
2
x+ 2 sinx= 2
3(1sin
2
x) + 2 sinx= 2 Pythagorean Identity
(3 sinx+ 1)(sinx1) = 0 Factoring
3 sinx+ 1 = 0 or sin x1 = 0
sinx=
1
3
sinx= 1
x= sin
1
(
1
3
) + 2
or
x=sin
1
(
1
3
)
x=

2
Solutions: 2 sin
1
(
1
3
)+; + sin
1
(
1
3
);

2
2
One part of the last solution needs further explanation. In the equation
sinx=
1
3
, we expect two solutions in the interval [0; 2): one in (;
3
2
) (which
is QIII), and another in (
3
2
;2) (which is QIV). Since no special number satises
sinx=
1
3
, we use inverse sine function. Because the range of sin
1
is [

2
;

2
], we
know that

2
<sin
1
(
1
3
)<0. From this value, to get the solution in (
3
2
;2),
we simply add 2 to this value, resulting tox= sin
1
(
1
3
) + 2. On the other
hand, to get the solution in (;
3
2
), we simply addsin
1
(
1
3
) to, resulting to
x=sin
1
(
1
3
).
Example 3.8.12.Solve: sin
2
x+ 5 cos
2x
2
= 2.
Solution.
sin
2
x+ 5 cos
2x
2
= 2
sin
2
x+ 5

1+cosx
2

= 2 Cosine Half-Angle Identity
2 sin
2
x+ 5 cosx+ 1 = 0
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DEPED COPY 2(1cos
2
x) + 5 cosx+ 1 = 0 Pythagorean Identity
2 cos
2
x5 cosx3 = 0
(2 cosx+ 1)(cosx3) = 0
2 cosx+ 1 = 0 or cos x3 = 0
cosx=
1
2
cosx= 3
x=
2
3
+ 2kor
x=
4
3
+ 2k,
k2Z
no solution
Solutions:
2
3
+ 2k;
4
3
+ 2k; k2Z 2
Example 3.8.13.Solve forx2[0;2) in the equation tan 2x 2 cosx= 0.
Solution.
tan 2x 2 cosx= 0
sin 2x
cos 2x
2 cosx= 0
sin 2x2 cosxcos 2x= 0
Apply the Double-Angle Identities for Sine and Cosine, and then factor.
2 sinxcosx2(cosx)(12 sin
2
x) = 0
(2 cosx)(2 sin
2
x+ sinx1) = 0
(2 cosx)(2 sinx1)(sinx+ 1) = 0
2 cosx= 0 or 2 sinx1 = 0 or sinx+ 1 = 0
cosx= 0 sinx=
1
2
sinx=1
x=

2
or
x=
3
2
x=

6
or
x=
5
6
x=
3
2
These values ofxshould be checked in the original equation because tan 2xmay
not be dened. Upon checking, this is not the case for each value ofxobtained.
The solutions are

2
,
3
2
,

6
,
5
6
, and
3
2
. 2
?
Example 3.8.14.A weight is suspended from a spring and vibrating vertically
according to the equation
f(t) = 20 cos

4
5

t
5
6

;
wheref(t) centimeters is the directed distance of the weight from its central
position attseconds, and the positive distance means above its central position.
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DEPED COPY (1) At what time is the displacement of the weight 5 cm below its central
position for the rst time?
(2) For what values oftdoes the weight reach its farthest point below its central
position?
Solution.(1) We nd the least positive value oftsuch thatf(t) =5.
20 cos

4
5

t
5
6

=5
cos

4
5

t
5
6

=
1
4
There are two families of solutions for this equation.

4
5

t
5
6

= cos
1

1
4

+ 2k,k2Z
t=
5
6
+
cos
1
(
1
4)+2k
4
5

In this family of solutions, the least positive value ofthappens when
k= 0, and this is
t=
5
6
+
cos
1

1
4

+ 2(0)
4
5

1:5589:

4
5

t
5
6

= 2cos
1

1
4

+ 2k,k2Z
t=
5
6
+
2cos
1
(
1
4)+2k
4
5

Here, the least positive value ofthappens whenk=1, and this is
t=
5
6
+
2cos
1

1
4

+ 2(1)
4
5

0:1078:
Therefore, the rst time that the displacement of the weight is 5 cm below
its central position is at about 0:1078 seconds.
(2) The minimum value off(t) happens when and only when the minimum
value of cos
4
5

t
5
6

is reached. The minimum value of cos
4
5

t
5
6

is
1, which implies that the farthest point the weight can reach below itscentral position is 20 cm. Thus, we need to solve for all values oftsuch that
cos
4
5

t
5
6

=1.
cos
4
5

t
5
6

=1
4
5

t
5
6

= cos
1
(1) + 2k; k 0
4
5

t
5
6

=+ 2k
t=
5
6
+
+2k
4
5

=
25
12
+
5
2
k
Therefore, the weight reaches its farthest point (which is 20 cm) below itscentral position att=
25
12
+
5
2
kfor every integerk0. 2
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DEPED COPY More Solved Examples
1. Give a particular solution of the following equation.
(a) sin
2
x1 = 0
(b) cotx=
p
3
(c) sec 3x =1
(d) csc
2
xcscx2 = 0
(e) cos
2
2x= sin
2
x
(f) 2 cosx3 = 0
Solution:
(a)x=

2
is a solution because sin
2
(

2
)1 = (1)
2
1 = 0.
(b) Note that cos

6
=
p
3
2
and sin

6
=
1
2
. Thus, cot

6
=
p
3. So,x=

6
is a
solution.
(c) Since sec=1 if and only if cos=1, a particular solution of the
equation in 3xis, that is, 3x =. Hence,x=

3
is a solution.
(d) Note that csc
3
2
=1. So, csc
23
2
= 1. As a consequence, csc
23
2

csc
3
2
2 = 1(1)2 = 0.
(e)x=

2
is a solution.
(f) Because cosxmust not be more than 1, then the equation has no solution.
2. What is the solution set of the following trigonometric equation sin
2
x+
cos
2
x= 1?
Solution:The equation is the Pythagorean Identity, meaning any element of
the domain of sinxand cosxsatises the equation. The domain of both sinx
and cosxisR. Therefore, the solution set of this trigonometric equation isR.
One may try the numbers

6
;0;and

4
for illustration.
(a)x=

6
sin
2
x+ cos
2
x= sin
2

6

+ cos
2

6

=

sin

6

2
+

cos

6

2
=

1
2

2
+
p
3
2
!
2
=
1
4
+
3
4
= 1:
(b)x= 0
sin
2
x+ cos
2
x= sin
2
0 + cos
2
0 = 0
2
+ 1
2
= 0 + 1 = 1:
(c)x=

4
sin
2
x+ cos
2
x= sin
2

4
+ cos
2

4
=
p
2
2
!
2
+
p
2
2
!
2
=
1
4
+
1
4
= 1:
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DEPED COPY 3. Find the solution set of the trigonometric equation tan
2
x+ 1 = sec
2
x:
Solution:Notice that this is a fundamental identity. Thus, the solution of
this equation is any number common to the domain of the tangent and secant
function. That is, the solution set is
fx2Rjcosx6= 0g=fx2Rjx6=

2
;
3
2
;
5
2
; :::g
=fx2Rjx6=
(2k+1)
2
; k2Zg:
4. Find all solutions of
p
3 tanx+ 1 = 0.
Solution:The equation is equivalent to tanx=
1
p
3
. This is true only if
x=
5
6
+kwherek2Z.
5. What are the solutions of
p
3 tanx+ 1 = 0, wherex2[0;2].
Solution:The solutions arex=
5
6
andx=
11
6
.
6. Determine all solutions of 4 cos
2
x1 = 0.
Solution:Note that the equation is quadratic in form, so we can apply tech-
niques in solving quadratic equations. For this case, we factor the expression
on the left and obtain, (2 cosx1)(2 cosx+ 1) = 0. Consequently, we have
cosx= 1=2 or cosx=1=2. The rst equation have solutions of the form
(=3 + 2k) or (5=3 + 2 k) wherek2Z, while the second equation have
solutions of the form (2=3 + 2 k) or (4=3 + 2 k). Combining the two solu-
tions, one observes that the solution set of the original equation may be given
by

3
;
2
3
;
4
3
;
5
3
;
7
3
; :::

:
We can write this in a more compact form as

k
3
:k6= 3j;wherej2Z

:
7. Find the solutions of 4 cos
2
x1 = 0 within the closed interval [0;2].
Solution:Similar to Example 6, the solution of the above equation is

k
3
:k6= 3j;wherej2Z

:
Since we are to nd solutions in [0;2], we takek= 1;2;4;and 5 to obtain
the solutions=3;2=3;4=3; and 5=3.
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DEPED COPY 8. Ifx2[0;2), solve the equation 2 sin
2
x=
p
3 sinx.
Solution:First, we write the equation as 2 sin
2
x
p
3 sinx= 0. Then, we
factor out sinxand get
sinx(2 sinx
p
3) = 0 =)sinx= 0 or sinx=
p
3=2:
The rst of these equations has solutionsx= 0 andx=, while the second
has solutionsx==3 and 2=3. The solutions of the original equation is the
union of the two, i.e., the solution set isf0; ;

3
;
2
3
g.
9. Solve 2 cos
2
x+ 5 cosx3 = 0, wherex2[0;2).
Solution:By factoring the left hand side of the given equation, we get (2 cosx
1)(cosx+ 3) = 0. This gives us two equations, namely
cosx=
1
2
and cosx=3:
First, we remark that the second equation does not have a solution because
cosxshould be more than or equal to -1. Hence, the solution of the rst
equation is the solution of the original equation. Thus, the solution set is
f

3
;
5
3
g.
10. Determine the solution set of the equation cos 2x = sinxon [0; 2).
Solution:Combining the equation cos 2x= sinxwith the cosine double-angle
identity cos 2x = 12 sin
2
x, we get
sinx= 12 sin
2
x:
This is equivalent to
2 sin
2
x+ sinx1 = 0 =)(2 sinx1)(sinx+ 1) = 0
=)sinx= 1=2 and sinx=1:
The solutions of the rst equation isx==6 andx= 5=6. The number
x= 3=2 is the solution of the second equation. Therefore, the solution set of
the original equation isf

6
;
5
6
;
3
2
g.
11. Solve cosx= cos 2x, for x2[0;2).
Solution:
cosx= cos 2x
)cosx= 2 cos
2
x1
) 0 = 2 cos
2
xcosx1
) 0 = (2 cosx+ 1)(cosx1):
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DEPED COPY The given trigonometric equation is equivalent to solving 2 cosx+ 1 = 0 and
cosx= 1. For 2 cosx+1 = 0 which is the same as cosx=1=2, the solutions
in the given interval arex= 2=3;4=3. For cosx= 1, the solution isx= 0.
Therefore, the solution set of the original equation isf0;
2
3
;
4
3
g.
12. A lighthouse at sea level is 34mifrom a boat. It is known that the top of
the lighthouse is 42:5 mifrom the boat and thatx=rcos, wherexis the
horizontal distance,ris the distance of the top of the lighthouse from the
boat, andis the angle of depression from the top of lighthouse. Find.
Solution:
x=rcos=)cos=
x
r
=
34
42:5
=
4
5
=)= cos
1
4
5
0:6435 (or 36:87

):
For this case, we used a calculator to nd the value of the unknown variable
since
4
5
is not a special value for cosine.
13. Three cities,A; B;andC, are positioned in a triangle as seen in the gure
below.
It is known that CityAis
140mifrom CityC, while CityBis 210mifrom
CityC. CitiesAandBare 70
p
7miapart.
Also, by the Cosine Law, we have
z
2
=x
2
+y
2
2xycos
wherex; y;andzare the respective distances of
BC;AC,AB,
=
m\ACB. Find.
Solution:Substituting the corresponding values ofx; y;andz, the problem is
now equivalent to solving the equation
34300 = 44100 + 1960058800 cos
) 29400 = 58800 cos
)
1
2
=
cos
)

3
=
:
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DEPED COPY Supplementary Problems 3.8
1. Find all solutions of the equation 2 cosxcosxsinx= 0.
2. Determine the solution set of the equation csc
2
x+ 1 = 0.
3. What are the solutions of sec
2
x+ secx2 = 0.
4. Find the solutions of the equation 4 sin
2
x1 = 0 on [0;2).
5. Find the values ofx2[0;2) for which csc 2x =
p
2.
6. What is the solution set of the equation sin= csc?
7. Solvet= sin
1
(cos 2t).
8. Letx2[0;2). Find the solutions of the equation cos
2
4x+ sin
2
2x= 1.
9. If a projectile, such as a bullet, is red into the air with an initial velocityvat
an angle of elevation, then the heighthof the projectile at timetis given by
h(t) =16t
2
+vtcosmeters. If the initial velocity is 109 meters per second,
at what angle should the bullet be red so that its height is 45 meters above
the oor in 2 seconds.
10. In a baseball eld, a pitcher throws the ball at a speed of 60km=hto the
catcher who is 100maway. When the ball leaves a starting point at an angle
of elevation of, the horizontal distance the ball travels is determined by
d=
v
2
32
sin, wheredis measured in meters and velocity in kilometers per
hour. At what angle of elevation (in degrees) is the ball thrown?
4
Lesson 3.9.Polar Coordinate System
Learning Outcomes of the Lesson
At the end of the lesson, the student is able to:
(1) locate points in polar coordinate system;
(2) convert the coordinates of a point from rectangular to polar system and vice
versa; and
(3) solve situational problems involving polar coordinate system.
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DEPED COPY Lesson Outline
(1) Polar coordinate system: pole and polar axis
(2) Polar coordinates of a point and its location
(3) Conversion from polar to rectangular coordinates, and vice versa
(4) Simple graphs and applications
Introduction
Two-dimensional coordinate systems are used to describe a point in a plane.
We previously used the Cartesian or rectangular coordinate system to locate a
point in the plane. That point is denoted by (x; y), wherexis the signed dis-
tance of the point from they-axis, andyis the signed distance of the point
from thex-axis. We sketched the graphs of equations (lines, circles, parabolas,
ellipses, and hyperbolas) and functions (polynomial, rational, exponential, log-
arithmic, trigonometric, and inverse trigonometric) in the Cartesian coordinate
plane. However, it is often convenient to locate a point based on its distance
from a xed point and its angle with respect to a xed ray. Not all equations
can be graphed easily using Cartesian coordinates. In this lesson, we also use
another coordinate system, which can be presented in dartboard-like plane as
shown below.
3.9.1. Polar Coordinates of a Point
W
e now introduce thepolar coordinate system. It is composed of a xed point
called thepole(which is the origin in the Cartesian coordinate system) and a
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DEPED COPY xed ray called thepolar axis(which is the nonnegativex-axis).
In the polar coordinate system, a point is described by the ordered pair (r
; ).
Theradial coordinaterrefers to the directed distance of the point from the pole.
Theangular coordinaterefers to a directed angle (usually in radians) from the
polar axis to the segment joining the point and the pole.
Because a point in polar coordinate system is described by an order pair of
radial
coordinate and angular coordinate, it will be more convenient to geomet-
rically present the system in apolar plane, which serves just like the Cartesian
plane. In the polar plane shown below, instead of rectangular grids in the Carte-
sian plane, we have concentric circles with common center at the pole to identify
easily the distance from the pole (radial coordinate) and angular rays emanating
from the pole to show the angles from the polar axis (angular coordinate).
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DEPED COPY Example 3.9.1.Plot the following points in one polar plane:A(3;

3
),B(1;
5
6
),
C(2;
7
6
),D(4;
19
12
),E(3;),F(4;
7
6
),G(2:5;
17
4
),H(4;
17
6
), andI(3;
5
3
).
Solution.
As seen in the last example, unlike in Cartesian plane where a point has a
unique
Cartesian coordinate representation, a point in polar plane have innitely
many polar coordinate representations. For example, the coordinates (3; 4) in
the Cartesian plane refer to exactly one point in the plane, and this particular
point has no rectangular coordinate representations other than (3; 4). However,
the coordinates (3;

3
)
in the polar plane also refer to exactly one point, but
this point has other polar coordinate representations. For example, the polar coordinates (3;
5
3
),
(3;
7
3
),
(3;
13
3
),
and (3;
19
3
)
all refer to the same point as
that of (3;

3
).
The polar coordinates (r; + 2k), wherek2Z, represent the same
point as that of (r; ).
In polar coordinate system, it is possible for the coordinates (r; ) to have
a negative value ofr. In this case, the point isjrjunits from the pole in the
opposite direction of the terminal side of, as shown in Figure 3.41.
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DEPED COPY Figure 3.41
Example
3.9.2.Plot the following points in one polar plane:A(3;
4
3
),B(
4;
11
6
),
C(
2;), andD(3:5;
7
4
).
Solution.As
described above, a polar point with negative radial coordinate lies
on the opposite ray of the terminal side of.
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DEPED COPY Points in Polar Coordinates
1. For any, the polar coordinates (0; ) represent the pole.
2. A point with polar coordinates (r; ) can also be represented by
(r; + 2k) or (r; ++ 2k) for any integerk.
3.9.2. From Polar to Rectangular, and Vice Versa
We now have two ways to describe points on a plane { whether to use the Carte-
sian coordinates (x; y) or the polar coordinates (r; ). We now derive the conver-
sion from one of these coordinate systems to the other.
We superimpose the Cartesian and polar planes, as shown in the following
diagram.
Figure 3.42
Suppose a pointPis represented by the polar coordinates (r; ). From Lesson
3.2 (in particular, the boxed denition on page 139), we know that
x=rcosand y=rsin:
Conversion from Polar to Rectangular Coordinates
(r; )!
8
<
:
x=rcos
y=rsin
!(x; y)
Given one polar coordinate representation (r; ), there is only one
rectangular coordinate representation (x; y) corresponding to it.
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DEPED COPY Example 3.9.3.Convert the polar coordinates (5; ) and (3;

6
) to Cartesian
coordinates.
Solution.
(5; )!
8
<
:
x= 5 cos=5
y= 5 sin= 0
!(5;0)
(3;

6
)!
8
<
:
x=3 cos

6
=
3
p
3
2
y=3 sin

6
=
3
2
!(
3
p
3
2
;
3
2
) 2
As explained on page 239 (right after Example 3.9.1), we expect that there
are innitely many polar coordinate representations that correspond to just one
given rectangular coordinate representation. Although we can actually determine
all of them, we only need to know one of them and we can chooser0.
Suppose a pointPis represented by the rectangular coordinates (x; y ). Re-
ferring back to Figure 3.42, the equation of the circle is
x
2
+y
2
=r
2
=) r=
p
x
2
+y
2
:
We now determine. Ifx=y= 0, thenr= 0 and the point is the pole. The
pole has coordinates (0; ), whereis any real number.
Ifx= 0 andy6= 0, then we may chooseto be either

2
or
3
2
(or their
equivalents) depending on whethery >0 ory <0, respectively.
Now, supposex6= 0. From the boxed denition again on page 139, we know
that
tan=
y
x
;
whereis an angle in standard position whose terminal side passes through the
point (x; y ).
Conversion from Rectangular to Polar Coordinates
(x; y) = (0;0)!(r; ) = (0; ); 2R
(0; y)
y6=0
!(r; ) =
8
<
:
(y;

2
) ify >0
(jyj;
3
2
) ify <0
(x;0)
x6=0
!(r; ) =
8
<
:
(x;0) ifx >0
(jxj; ) ifx <0
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DEPED COPY (x; y)
x6=0; y6=0
!(r; )
r=
p
x
2
+y
2
tan=
y
x
same quadrant as (x; y)
Given one rectangular coordinate representation (x; y), there are
many polar coordinate representations (r; ) corresponding to it. The
above computations just give one of them.
Example 3.9.4.Convert each Cartesian coordinates to polar coordinates (r; ),
wherer0.
(1) (4; 0)
(2) (4;4)
(3) (3;
p
3)
(4) (6;2)
(5) (3; 6)
(6) (12; 8)
Solution.(1) (4; 0)!(4; )
(2) The point (4; 4) is in QI.
r=
p
x
2
+y
2
=
p
4
2
+ 4
2
= 4
p
2
tan=
y
x
=
4
4
= 1 =)=

4
(4;4)!

4
p
2;

4

(3) (3;
p
3) in QIII
r=
q
(3)
2
+ (
p
3)
2
= 2
p
3
tan=

p
3
3
=
p
3
3
=)=
7
6
(3;
p
3)!

2
p
3;
7
6

(4) (6;2) in QIV
r=
p
6
2
+ (2)
2
= 2
p
10
tan=
2
6
=
1
3
=)= tan
1

1
3

(6;2)!

2
p
10;tan
1

1
3

(5) (3; 6) in QII
r=
p
(3)
2
+ 6
2
= 3
p
5
tan=
6
3
=2 =)=+ tan
1
(2)
(3;6)!

3
p
5; + tan
1
(2)

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DEPED COPY (6) (12; 8) in QIII
r=
p
(12)
2
+ (8)
2
= 4
p
13
tan=
8
12
=
2
3
=)=+ tan
12
3
(12; 8)!

4
p
13; + tan
12
3

2
3.9.3. Basic Polar Graphs and Applications
From the preceding session, we learned how to convert polar coordinates of a
point to rectangular and vice versa using the following conversion formulas:
r
2
=x
2
+y
2
;tan=
y
x
; x=rcos;andy=rsin:
Because a graph is composed of points, we can identify the graphs of some equa-tions in terms ofrand.
Graph of a Polar Equation
Thepolar graphof an equation involvingrandis the set of all
points with polar coordinates (r; ) that satisfy the equation.
As a quick illustration, the polar graph of the equationr= 22 sinconsists
of all points (r; ) that satisfy the equation. Some of these points are (2; 0), (1;

6
),
(0;

2
), (2; ), and (4;
3
2
).
Example 3.9.5.Identify the polar graph ofr= 2, and sketch its graph in the
polar plane.
Solution.Squaring the equation, we getr
2
= 4. Becauser
2
=x
2
+y
2
, we have
x
2
+y
2
= 4, which is a circle of radius 2 and with center at the origin. Therefore,
the graph ofr= 2 is a circle of radius 2 with center at the pole, as shown below.
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DEPED COPY In the previous example, instead of using the conversion formular
2
=x
2
+y
2
,
we may also identify the graph ofr= 2 by observing that its graph consists of
points (2; ) for all. In other words, the graph consists of all points with radial
distance 2 from the pole asrotates around the polar plane. Therefore, the
graph ofr= 2 is indeed a circle of radius 2 as shown.
Example 3.9.6.Identify and sketch the polar graph of=
5
4
.
Solution.The graph of=
5
4
consists of all points (r;
5
4
) forr2R. If
r >0, then points (r;
5
4
) determine a ray from the pole with angle
5
4
from
the polar axis. Ifr= 0, then (0;
5
4
) is the pole. Ifr <0, then the points
(r;
5
4
) determine a ray in opposite direction to that ofr >0. Therefore, the
graph of=
5
4
is a line passing through the pole and with angle
5
4
with
respect to the polar axis, as shown below.
Example 3.9.7.Iden
tify (and describe) the graph of the equationr= 4 sin.
Solution.
r= 4 sin
r
2
= 4rsin
x
2
+y
2
= 4y
x
2
+y
2
4y= 0
x
2
+ (y2)
2
= 4
Therefore, the graph ofr= 4 sinis a circle of radius 2 and with center at (2;

2
).
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DEPED COPY ?
Example
3.9.8.Sketch the graph ofr= 22 sin.
Solution.We construct a table of values.
x0

6

4

3

2
2
3
3
4
5
6

r210:590:2700:270:59 12
x
7
6
5
4
4
3
3
2
5
3
7
4
11
6
2
r33:413:73 43:733:41 32
This heart-shaped curve is called ac
ardioid. 2
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DEPED COPY ?
Example 3.9.9.The sound-pickup capability of a certain brand of microphone
is described by the polar equationr=4 cos, wherejrjgives the sensitivity of
the microphone to a sound coming from an angle(in radians).
(1) Identify and sketch the graph of the polar equation.
(2) Sound coming from what angle2[0; ] is the microphone most sensitive
to? Least sensitive?
Solution.(1) r=4 cos
r
2
=4rcos
x
2
+y
2
=4x
x
2
+ 4x+y
2
= 0
(x+ 2)
2
+y
2
= 4
This is a circle of radius 2 and with center at (2; ).
(2) We construct a table of values.
x0

6

4

3

2
2
3
3
4
5
6

r43:46 2:83 2022:833:464
From the table, the microphone is most sensitive to sounds coming from
angles=
0 and=, and least sensitive to sound coming from an angle
=

2
. 2
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DEPED COPY More Solved Examples
1. Locate in the polar plane the following polar points:M(1; =3); A(0; ); T(;0),
andH(4;5=3).
Solution:
2. Locate in the polar plane the following polar points:W(
1;7=4); X (2;=6); Y (4;5=6)
andZ(3;11=3).
Solution:
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DEPED COPY 3. Convert the following polar points to Cartesian coordinates.
(a) (5; 5=4) to Cartesian coordinates
(b) (2; 3=4) to Cartesian coordinates
(c) (; )
(d) (0; 10)
Solution:(a) Using the conversion formulas withr= 5 and= 5=4, we get
x=rcos= 5 cos(5=4) = 5
p
2=2
and
y=rsin= 5 sin(5=4) =5
p
2=2:
Therefore, (5;5=4) in Cartesian coordinate is(5
p
2=2;5
p
2=2).
(b) Using the conversion formulas withr=2 and= 5 = 3=4, we get
x=rcos=2 cos(3=4) = 2
p
2=2
and
y=rsin=2 sin(3=4) =2
p
2=2:
Therefore, (2; 3=4) in Cartesian coordinate is (2
p
2=2;2
p
2=2).
(c) Notice here thatis used in two dierent ways. First is, with numerical
value approximately equal to3:14, is used as a radius and second, as an
angle equivalent to 180

. That is, the point is in the negativex-axisunits
away from the origin. Hence, the Cartesian coordinate of (; ) is (; 0).
(d) Since the radius is 0, then the polar point (0; 10) is the origin with Cartesian
coordinate (0; 0).
4. Convert the following Cartesian points to polar coordinates.
(a) (5; 5)
(b) (3;
p
3)
(c) (5
p
3;15)
(d) (8;0)
Solution:(a) The point (x; y ) = (5; 5) is in the fourth quadrant. Using the
conversion formulas, we get
r=
p
x
2
+y
2
=
p
5
2
+ (5)
2
= 5
p
2
and
= tan
1
(y=x) = tan
1
(1) ==4:
Therefore, (5;5) in polar coordinate is (5
p
2;=4).
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DEPED COPY (b) Similarly, we use the conversion formulas to get
r=
q
(3)
2
+ (
p
3)
2
=
p
12 = 2
p
3
and
= tan
1
[
p
3=(3)] = =6:
Note that the point is in the second quadrant so we must use
pi
6
+. There-
fore, (3;
p
3) in polar coordinate is (2
p
3;5=6).
(c) The point is in the third quadrant.
r=
q
(5
p
3)
2
+ (15)
2
=
p
300 = 10
p
3
and
tan=15=(5
p
3))= 4=3:
Therefore, (5
p
3;15) in polar coordinate is (10
p
3;4=3).
(d) Using the conversion formula, one can show that the point (8;0) in polar
coordinate is also (8;0).
5. Identify (and describe) the graph of the equationr= 4 sin. Using a graphing
software, graph the following equations.
(a)r= 2 sin
(b)r=5
(c)= 2r
(d)r= 22 cos
Solution:(a)r= 2 sinis a circle with radius 1 centered at (1;

2
).
(b)r=5 is a standard circle with radius 5.
(c) Notice that asincreases, theralso increases. The graph of= 2ris a
spiral rotating counter-clockwise from the pole.
(d) The graph ofr= 22 cosis a cardioid.
(a) (b)
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DEPED COPY (c) (d)
6.
A boy is ying a kite with an angle of elevation of 60

from where he stands.
What is the direct distance of the kite from him, if the the kite is 6 ft above
the ground?
Solution:The problem can be illustrated as follows:
Here,r(in
is the angle of
depression. To solve forr, we apply the formulay=rsin. Thus,
r=y=sin= 6=sin(60

) = 6=(
p
3=2) = 12=
p
3 = 4
p
3:
Therefore,
the kite is 4
p
3 ft away from the boy.
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DEPED COPY Supplementary Problems 3.9
1. Give two more pairs of coordinates that describe the same point.
(a) (13; =3) (b) (0; 0) (c) (15; 15=4)
2. Locate the following points in the polar coordinate plane:
(a)P(3;)
(b)Q(3;7=4)
(c)R(5=2;5=2)
(d)S(8;23=6)
3. Transform the following to Cartesian coordinates:
(a) (3; )
(b) (3;7=4)
(c) (5=2;5=2)
(d) (8;23=6)
4. Transform the following to polar coordinates:
(a) (9; 40)
(b) (15; 20)
(c) (5=2;5=2)
(d) (14; 14)
5. Consider the equation in polar formr= 4 cos 2.
(a) Complete the table
0=6=4=3=22=3 3=4 5=6 7=6 5=4 4=3 3=2
r
(b) Plot the points obtained in part (a) in a polar coordinate system.
6. A helicopter is hovering 800 feet above a road. A truck driver observes the
helicopter at a horizontal distance of 600 feet. Find the angle of elevation of
the helicopter from the truck driver.
4
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DEPED COPY Topic Test 3 for Unit 3
1. Letbe an angle in QIII such that cos=
12
13
. Find the values of the six
trigonometric functions of 2 .
2. Prove that cot(2x) =
cot
2
x1
2 cotx
.
3. Using half-angle identities to nd the exact values of the following.
(a) tan 15

(b) tan 7:5

4. Find the exact value of the following.
(a) tan
1

cot
103
6

(b) cos

sin
1
40
41

5. Lety2[0;2). Find the solutions of the equation
sin
1
(cos
2
ycosy1) ==2:
6. Let2[0;2]. Find all the solutions of the equation
4 cos
2
sin= 3 sin:
7. Letr=22 sin. Complete the table and plot the points (r; ) in the same
polar coordinates.
0=6=4=3=22=3 3=4 5=6 7=6 5=4 4=3 3=2
r
8. Transform the following points from Cartesian to polar coordinates.
(a) (42; 56)
(b) (100; 100)
(c) (0; 7)
(d) (7; 0)
(e) (2; 2)
(f) (5; 12)
9. Transform the following points from polar to Cartesian coordinates.
(a) (3; =3)
(b) (45; 7=4)
(c) (1;)
(d) (5; 0)
(e) (2; 2)
(f) (9; 17=6)
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DEPED COPY Topic Test 4 for Unit 3
1. Letbe an angle in the 2
nd
quadrant such that cos=
7
25
. Find the
following.
(a) cos(2 ) (b) sin(2 ) (c) tan(2 )
2. Given that cos 48

0:6691. Find the approximate value of the following.
(a) cos
2
24

(b) sin
2
24

(c) tan
2
24

3. Using half-angle identities to nd the exact values of the following.
(a) tan(=12) (b) tan(=24)
4. Find the exact value of cos

cos
1
1
7
+ cos
1
3
5

.
5. Letx2[0;2). Find the solutions of the equation
4 sin
2
x+ (2
p
32
p
2) sinx
p
6 = 0:
6. Let2[0;2]. Find all the solutions of the equation
2 sin
2
(2)sin(2 )1 = 0:
7. Letr= 2 + 2 cos. Complete the table and plot the points (r; ) in the same
polar coordinates.
0=6=4=3=22=3 3=4 5=6 7=6 5=4 4=3 3=2
r
8. Transform the following points from Cartesian to polar coordinates.
(a) (21; 28)
(b) (100; 100)
(c) (0;5)
(d) (5;0)
(e) (; )
(f) (15; 8)
9. Transform the following points from polar to Cartesian coordinates.
(a) (4; =6)
(b) (100; 5=4)
(c) (1; )
(d) (5;0)
(e) (; )
(f) (15; 8=3)
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DEPED COPY 4
Answersto
Odd-Numbered Exercises
in Supplementary Problems
and
All Exercises in Topic Tests
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DEPED COPY 1. center (0; 0),r=
1
2
3. center

4;
3
4

,r=
1
5. center (7;
6),r= 11
256
Supplemen
tary Problems 1.1 (page 17)All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means -
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DEPED COPY 7. center (2;4),r=
5
3
9. center

5
2
;
7
2

,r=
7
4
11. (x 17)
2
+
(y5)
2
= 144
13. (x 15)
2
+ (y+ 7)
2
= 49
15. (x 15)
2
+ (y+ 7)
2
= 9
17. (x + 2)
2
+ (y3:5)
2
= 31:25
19. (x + 10)
2
+ (y7)
2
= 36
21. (x + 2)
2
+ (y3)
2
= 12
23. (x 2:5)
2
+ (y0:5)
2
= 14:5
25. (x + 5)
2
+ (y+ 1)
2
= 8
27. Set up a Cartesian coordinate system by assigningCas the origin. Then the
circle on the left end has radius 100 and has equationx
2
+y
2
= 10000. A
radius of the circle on the right end can be drawn fromCto the upper right
corner of the gure; this radius has length (by the Pythagorean theorem)
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DEPED COPY p
300
2
+ 100
2
=
p
100000. Then the circle on the right end has equationx
2
+
y
2
= 100000. We want the length of the segment aty= 50. In this case, the
left endpoint hasx=
p
1000050
2
=
p
7500 and the right endpoint has
x=
p
10000050
2
=
p
97500. Then the total length is
p
97500(
p
7500)
= 50
p
3 + 50
p
39 m398:85 m.
1. vertex (0;0), focus (9; 0), directrixx= 9, axisy= 0
3. vertex (
1;7), focus (2; 7), directrixx= 0, axisy= 7
258
Supplemen
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DEPED COPY 5. vertex (3;2), focus

3;
3
2

, directrixy=
11
2
, axisx= 3
7. (y11)
2
=
36(x7)
9. (x+ 10)
2
= 34(y3)
11. (y 9)
2
=80(x 4)
13. (y 8)
2
=8(x+ 3)
15.4:17 cm
17. 3:75 cm
1. center: (0; 0)
foci:F1(2;0),F2(2;0)
vertices:V1(2
p
2;0)
; V2(2
p
2;0)
co
vertices:W1(0;2); W 2(0;2)
3. cen
ter: (1; 1)
foci:F1(1
p
3;1),
F2(1
+
p
3;1)
v
ertices:V1(1;1); V2(3;1)
covertices:W1(1;0); W2(1;2)
259
Supplemen
tary Problems 1.3 (page 45)All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means -
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DEPED COPY 5. center: (7; 5)
foci:F1(1;5),F2(13;5)
vertices:V1(3;5); V2(17;5)
covertices:W1(7;13); W 2(7;3)
7.
(
x2)
2
49
+
(
y8)
2
16
=
1
9. The center is (9; 10) andc= 12. We see that the given point (9; 15) is
a covertex, sob= 5. Thena=
p
5
2
+

2
= 13. Therefore, the equation is
(x+ 9)
2
169
+
(
y10)
2
25
=
1.
11. Since the major axis is vertical, the center has the samexcoordinate as the
focus and the sameycoordinate as the covertex; that is, the center is (9; 10).
Thenc= 5; b= 10, anda
2
= 125. Therefore, the equation is
(x+ 9)
2
100
+
(
y10)
2
125
=
1.
13. Recall that the unit is 100 km. The vertices of the ellipse are at (3633; 0) and
(4055; 0). Then the center of the ellipse is at (211; 0). Thena= 3844
andc= 211. It follows thatb
2
= 14731815. The equation is
(x+ 211)
2
14776336
+
y
2
14731815
=
1.
15. Set up a coordinate system with the center of the ellipse at the origin. Then
a= 60 andb= 20. We want the length of the segment with endpoints (on the
ellipse) havingx= 45 (or45). Theycoordinates are given by
45
2
60
2+
y
2
20
2=
1,
ory=
q
20
2

1
45
2
60
2

13:23. Hence, the desired width is 26:46 ft.
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DEPED COPY 1. center: (0; 0)
foci:F1(
p
181;0),F2(
p
181;0)
vertices:V1(10; 0),V2(10;0)
asymptotes:y=
9
10
x
3. center: (0 ;5)
fo
ci:F1(
p
19;5),F2(
p
19;5)
v
ertices:V1(
p
15;5),V2(
p
15;5)
asymptotes:y5
=
2
p
15
x
5. cen
ter: (3; 3)
foci:F1(3;3
p
15),
F2(
3;3 +
p
15)
v
ertices:V1(3;3
p
6),
V2(
3;3 +
p
6)
asymptotes:y+
3 =
p
6
3
(
x+ 3)
261
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DEPED COPY 7.
y
2
144

(x+ 7)
2
145
= 1 9.
(x+ 10)
2
81

(y+ 4)
2
256
= 1
11. The intersection (4; 8) of the two asymptotes is the center of the hyperbola.
Then the hyperbola is vertical andc= 13. Since the slopes of the asymptotes
are
5
12
, we have
a
b
=
5
12
.
Sincec= 13, we havea
2
+b
2
= 169 andb=
p
169a
2
. It follows that
a
p
169a
2
=
5
12
=)a= 5 andb= 12 =)
(y8)
2
25

(x+ 4)
2
144
= 1:
13. The midpoint (9;1) of the two given corners is the center of the hyperbola.
Since the transverse axis is horizontal,a= 7 andb= 2. Therefore, the
equation is
(x9)
2
49

(y1)
2
4
= 1.
1. pair of intersecting lines
3. parabola
5. parabola
7. empty set
9. The standard equation of the ellipse is
(x5)
2
36
+
(y2)
2
100
= 1; so its foci
are (5; 10) and (5; 6) while its vertices are (5;12) and (5; 8). The equations
of the circles are (x5)
2
+ (y10)
2
= 4, (x5)
2
+ (y10)
2
= 324,
(x5)
2
+ (y+ 6)
2
= 4, and (x5)
2
+ (y+ 6)
2
= 324.
11. The standard equation of the hyperbola is
(y+ 5)
2
25

(x+ 9)
2
25
= 1. Its auxil-
iary rectangle has corners (14; 0);(4;0);(4;10); (14; 10). The equa-
tion of the circle is (x + 9)
2
+ (y+ 5)
2
= 50.
13. The equation simplies to
(x+ 7)
2
+ (y3)
2
=
r+ 2
r1
:
Its graph
(a) is a circle if
r+ 2
r1
>0; that is, whenr2(1; 2)[(1;+1).
(b) is a point if
r+ 2
r1
= 0; that is, whenr=2.
(c) is the empty set if
r+ 2
r1
<0; that is, whenr2(2;1).
262
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DEPED COPY 1. (a) (1;6)
(b)

4
3
;
9
20

;

4
3
;
89
20

(c) (1; 6)
;(1;2)
263
Supplemen
tary Problems 1.6 (page 77)All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means -
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DEPED COPY (d)

1;
3
2

;(1
p
15;1);(1 +
p
15;1)
(e) No solution
3. Let (x;
) be the ordered pair that satises the conditions. The resulting
system of equations is
8
>
<
>
:
x
2
= 2y
2
+
1
8
x
2
+y
2
=
5
16
Solving
yields

1
2
;
1
4

,

1
2
;
1
4

,

1
2
;
1
4

,
and

1
2
;
1
4

.
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DEPED COPY 5. We have the system
8
<
:
x
2
+ (y6)
2
= 36
x
2
= 4ky;
where the rst equation is a circle above thex-axis, tangent to thex-axis at
x= 0, and the second equation is a parabola facing up/down, depending on
k.
Substituting the second equation in the rst equation yieldsy
2
+(4k12)y= 0.
Note thaty= 0 is already a root.
We now consider two cases.
Ifk >0, the system might have one or two solutions. To ensure that the
solution is unique, we set the discriminant to be nonpositive: 4k120)
k3.
Ifk0, the system will always have a unique solution.
Thusk2(1; 0][(3;+1).
1. (a) Since the coecients ofx
2
andy
2
are equal, the graph is a circle, a point,
or the empty set. Completing the squares, we see that the equation is
equivalent to

x
1
2

2
+

y+
3
2

2
= 4:
Hence, the graph is a circle with center (0:5;1:5) and radius 2.
(b) By inspection, the graph is a parabola. Completing the squares, we see
that the equation is equivalent to (x + 2)
2
= 14(y+ 4). Hence, the graph
has vertex at (2;4) and is opening upward.
(c) Since the coecients ofx
2
andy
2
are of opposite signs, the graph is a
hyperbola or a pair of intersecting lines. Completing the squares, we seethat the equation is equivalent to
(x7)
2
4

(y+ 3)
2
3
= 1:
Hence, the graph is a horizontal hyperbola with center at (7;3).
(d) Since the coecients ofx
2
andy
2
have the same sign and are unequal,
the graph is an ellipse, a point, or the empty set. Completing the squares,we see that the equation is equivalent to
(x8)
2
2
+
y
2
7
= 0:
Hence, the graph is the point (8; 0).
265
Topic Test 1 for Unit 1 (page 78)All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means -
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DEPED COPY 2. (a) The equation is equivalent to
x
2
7
+
y
2
25
= 1. This is a vertical ellipse.
center: (0; 0)
foci:F1(0;3
p
2),F2(0;3
p
2)
vertices:V1(0;5),V2(0;5)
covertices:W1(
p
7;0); W2(
p
7;0)
(b) The equation is equivalent to
(
y+ 4)
2
64

(
x1)
2
36
=
1. This is a vertical
hyperbola.
center:C(1;4)
foci:F1(1;14),F2(1;6)
vertices:F1(1;12); F 2(1;6)
asymptotes:y+ 4 =
4
3
(
x1)
3. (a) The parabola opens to the right and has focal distancec=
6. Its equation
is (y3)
2
= 24(x+ 1).
(b) The intersection (2; 5) of the two asymptotes is the center of the
hyperbola. Then the hyperbola is horizontal anda= 5. Using the slopes
of the asymptotes, we have
b
a
=
12
4
.
It follows thatb= 12 and the
equation is
(x+ 2)
2
25

(
y+ 5)
2
144
=
1.
4. Multiplying the rst equation by 2, we get 2(x 1)
2
+ 2(y + 1)
2
= 10. By
subtracting the second equation from this new equation, we get the equation
2(y+ 1)
2
+ 8 = 10y. This has solutionsy= 0 andy=5=2.
Wheny= 0, the correspondingxvalues are 3 and1. Wheny=5=2,
the correspondingxvalues are
p
11
2
+
1. Therefore, the solutions are (1;0),
(3;0),

p
11
2
+
1;
5
2

.
266All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means -
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DEPED COPY 5. Set up a coordinate system by making the center of the door's base the origin.
Then the ellipse has center (0;2) witha= 1=2 andb= 0:3; then its is equation
x
2
0:5
2+
(y2)
2
0:3
2= 1.
To determine if the cabinet can be pushed through the doorway, we determine
the height of the doorway whenx= 0:25 (or0:25). We solve foryfrom the
equation
0:25
2
0:5
2+
(y2)
2
0:3
2= 1. Solving for theycoordinate, we see that the height
is2:2598 m. Hence, the cabinet cannot be pushed through the doorway.
6. Let (x; y) be the coordinates of the point. This point satises
p
x
2
+ (y+ 1)
2
= 2jx3j:
Manipulating this equation gives us
x
2
+ (y+ 1)
2
= 4(x
2
6y+ 9)
3(x
2
8x+ 16) + (y+ 1)
2
= 3648
3(x4)
2
+ (y+ 1)
2
=12
(x4)
2
4

(y+ 1)
2
12
= 1:
Therefore, the point traces a horizontal hyperbola with center at (4;1).
1. (a) By inspection, the graph is a parabola. Completing the squares, we see
that the equation is equivalent to (y 5)
2
=8(x5). Hence, the graph
has vertex at (5; 5) and is opening to the left.
(b) Since the coecients ofx
2
andy
2
are equal, the graph is a circle, a point,
or the empty set. Completing the squares, we see that the equation isequivalent to (x+ 5)
2
+ (y+ 9)
2
=4. Hence, the graph is the empty
set.
(c) Since the coecients ofx
2
andy
2
have the same sign and are unequal,
the graph is an ellipse, a point, or the empty set. Completing the squares,
we see that the equation is equivalent to
(x+ 2)
2
4
+
(y1)
2
9
= 1. Hence,
the graph is a vertical ellipse with center (2;1).
(d) Since the coecients ofx
2
andy
2
are of opposite signs, the graph is a
hyperbola or a pair of intersecting lines. Completing the squares, we
see that the equation is equivalent to
(y4)
2
11

(x6)
2
17
= 0. Hence,
the graph is a pair of intersecting lines given by the equationsy4 =

11
17
(x6).
267
Topic Test 2 for Unit 1 (page 79)All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means -
electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2016.

DEPED COPY 2. (a) The equation is equivalent to
x
2
64

y
2
64
= 1. This is a horizontal hyperbola.
center: (0; 0)
foci:F1(8
p
2;0),
F2(8
p
2;0)
vertices:V1(8;0),V2(8;0)
asymptotes:y=x
(b) The equation is equivalent to
(
x+ 3)
2
49
+
(
y2)
2
4
=
1. This is a horizontal
ellipse.
center:C(3;2)
foci:F1(33
p
5;2),F2(
3 + 3
p
5;2)
v
ertices:F1(10; 2); F2(4;2)
covertices:W1(3;0); W2(3;4)
3. (a) The parabola opens downward and has focal distancec=
5. Its equation
is (x7)
2
=20(y + 7).
(b) Since the ellipse has vertical or horizontal major axis, the center is at
either (1; 12) or (5; 3). Since the major axis is longer than the minor
axis, the center must be at (5; 3). Then the ellipse is vertical witha= 9
andb= 4. Its equation is
(x+ 5)
2
16
+
(
y3)
2
81
=
1:
268All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means -
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DEPED COPY 4. Completing the squares, we see that the rst equation is equivalent to 9(x+
3)
2
= 4y
2
+ 36. On the other hand, the second equation is equivalent to
9(x+ 3)
2
= 36y+ 36. Subtracting the second equation from the rst, we get
4y
2
36y= 0, which has solutionsy= 0 andy= 9.
Wheny= 0, the correspondingxvalues are5 and1. Wheny= 9, the
correspondingxvalues are32
p
10. Therefore, the solutions are (5; 0),
(1;0),

32
p
10;9

.
5. Set up a coordinate system so that the opening of the hose (the parabola's
vertex) is at (0; 3) and that the water ows towards the positivex-axis. Then
thex-axis (y = 0) corresponds to the ground; it follows the parabola passes
through the point (2; 0). Hence, the equation of the parabola isx
2
=
4
3
(y3).
If Nikko stands on a 1.5-ft stool and the vertex remains at (0;3), the line
y=1:5 will correspond to the ground. Hence, the water will strike the
ground wheny=1:5. This givesx=
q

4
3
(1:5 3) =
p
6. Therefore, the
water will travel
p
62 ft further before striking the ground.
6. Let (x; y) be the coordinates of the point. This point satises
p
(x2)
2
+y
2
=
2
3
jy5j:
Manipulating this equation gives us
(x2)
2
+y
2
=
4
9
(y
2
10y+ 25)
9(x2)
2
+ 5(y
2
+ 8y) = 100
9(x2)
2
+ 5(y + 4)
2
= 100 + 80
(x2)
2
20
+
(y+ 4)
2
36
= 1:
Therefore, the point traces a vertical ellipse with center at (2; 4).
1.a3=a1+(31)d= 35;a10=a1+(101)d= 77)d= 6; a1= 23)a5= 47.
3.sn=
n(2(17) + (n 1)3)
2
= 30705)3n
2
+ 31n 61410 = 0. Using the
quadratic formula and noting thatnmust be a whole number, we haven=
138.
5. We haves= 108 =
a1
1r
ands3= 112 =
a1(1r
3
)
1r
= 108 (1r
3
))r=

1
27
)a1= 144.
269
Supplementary Problems 2.1 (page 85)All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means -
electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2016.

DEPED COPY 7. Note that 0:123123: : := 0:123 + 0:000123 + 0:000000123 +: : := 0:123 +
0:123(0:001) + 0 :123(0:001)
2
+: : :, which is an innite geometric series with
r= 0:001. Thus, 0: 123 =
0:123
10:001
=
41
333
.
9. We haves4=
4 (2a 1+ (41)d)
2
= 80)2a1+ 3d= 40. Since the sum of the
rst two numbers are one-third of the sum of the last two numbers, we have
1
3
(a1+a2) =a3+a4)4a1+ 14d = 0. Combining yieldsd= 10, and thus
a1= 5; a2= 15; a3= 25; a 4= 35.
11. Note that this is a geometric series with common ratio 2n 1. Thus, the sum
will have a nite value ifj2n1j<1) 1 <2n1<1)0< n <1.
Thus,n2(0;1).
1. (a)
10
X
i=3
p
3
i
2
=
p
3
2

3
2
+: : :+
10
2

= 26
p
3
(b)
5
X
i=1
x
2i
2
i
=
x
2
2
+
x
4
4
+
x
6
8
+
x
8
16
+
x
10
32
(c)
5
X
i=2
(1)
i
x
i1
=xx
2
+x
3
x
4
3. (a)
150
X
i=1
(4i+ 2) = 4
150
X
i=1
i+
150
X
i=1
2 = 4
150(151)
2
+ 2(150) = 45;600
(b)
120
X
i=3
i(i5) =
120
X
i=1
(i
2
5i)1(15)2(25) =
120(121)(2(120) + 1)
6
+10 =
583;230
(c)
130
X
i=1
(2i3)(2i+3) =
130
X
i=1
(4i
2
9) = 4
130
X
i=1
i
2

130
X
i=1
9 =
130(131)(2(130) + 1)
6
+
9(130) = 741; 975
5.s=
200
X
i=1

(i1)
2
i
2

=
200
X
i=1
(12i))
200
X
i=1
i=
200s
2
270
Supplementary Problems 2.2 (page 95)All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means -
electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2016.

DEPED COPY 1.Part 1.
1
2
= 2
1 + 2
2
1
The formula is true forn= 1.
Part 2.
Assume:
k
X
i=1
i
2
i
= 2
k+ 2
2
k
.
To show:
k+1
X
i=1
i
2
i
= 2
k+ 3
2
k+1
.
k+1
X
i=1
i
2
i
=
k
X
i=1
i
2
i
+
k+ 1
2
k+1
= 2
k+ 2
2
k
+
k+ 3
2
k+1
= 2
k+ 3
2
k+1
:
3.Part 1.
1(1!) = (1 + 1)!1
The formula is true forn= 1.
Part 2.Assume:
k
X
i=1
i(i!) = (k+ 1)!1.
To show:
k+1
X
i=1
i(i!) = (k + 2)!1.
k+1
X
i=1
i(i!) =
k
X
i=1
i(i!) + (k + 1)[(k + 1)!]
= (k+ 1)!1 + (k+ 1)[(k + 1)!]
= (k+ 2)(k + 1)!1
= (k+ 2)!1:
271
Supplementary Problems 2.3 (page 108)All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means -
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DEPED COPY 5.Part 1.
1
1
2
=
1
2
=
1
2(1)
The formula is true forn= 2.
Part 2.
Assume:P=

1
1
2

1
1
3

1
1
k1

1
1
k

=
1
2k
.
To show:

1
1
2

1
1
k

1
1
k+ 1

=
1
2(k+ 1)
.

1
1
2

1
1
k+ 1

=P

1
1
k+ 1

=
1
2k

k
k+ 1
=
1
2(k+ 1)
:
7.Part 1.
4
3(1)+1
+ 2
3(1)+1
+ 1 = 273 = 7(39)
The number is divisible by 7 forn= 1.
Part 2.
Assume: 4
3k+1
+ 2
3k+1
+ 1 is divisible by 21.
Prove: 4
3(k+1)+1
+ 2
3(k+1)+1
+ 1 is divisible by 21.
4
3(k+1)+1
+2
3(k+1)+1
+1 = 644
3k+1
+82
3k+1
+1 = 564
3k+1
+8

4
3k+1
+ 2
3k+1
+ 1

7
9.Part 1.
5
2(1)+1
2
1+2
+ 3
1+2
2
2(1)+1
= 1216 = 19(64)
The number is divisible by 19 forn= 1.
Part 2.
Assume: 5
2k+1
2
k+2
+ 3
k+2
2
2k+1
is divisible by 19.
Prove: 5
2(k+1)+1
2
(k+1)+2
+ 3
(k+1)+2
2
2(k+1)+1
is divisible by 19.
5
2(k+1)+1
2
(k+1)+2
+ 3
(k+1)+2
2
2(k+1)+1
= 505
2k+1
2
k+2
+ 123
k+2
2
2k+1
=
12 (5
2n+1
2
n+2
+ 3
n+2
2
2n+1
) + 385
2n+1
2
n+2
272All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means -
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DEPED COPY 11.Part 1.
10
1
3
+
5
3
+ 4
1+2
= 1029 = 3(343)
The number is divisible by 3 forn= 1.
Part 2.
Assume:
10
k
3
+
5
3
+ 4
k+2
is divisible by 3.
Prove:
10
k+1
3
+
5
3
+ 4
k+3
is divisible by 3.
10
k+1
3
+
5
3
+4
k+3
= 10
10
k
3
+
5
3
+44
k+2
= 10

10
k
3
+
5
3
+ 4
k+2

64
n+2
9
5
3
13.Part 1.12
1
1
= 1
Part 2
Assume:
k
X
i=1
1
i
3
2
1
k
Prove:
k+1
X
i=1
1
i
3
2
1
k+ 1
k+1
X
i=1
1
i
3
2
1
k
+
1
(k+ 1)
3
= 2
(k+ 1)
3
k
(k+ 1)
3
. Note that 0<(k+ 1)
2
)
(k+ 1)
2
<(k+ 1)
3
k, thus 2
(k+ 1)
3
k
(k+ 1)
3
<2
(k+ 1)
2
(k+ 1)
3
= 2
1
k+ 1
.
1. (a) (2x3y)
5
= 32x
5
240x
4
y+ 720x
3
y
2
1080x
2
y
3
+ 810xy
4
243y
5
(b)
p
x
3

2
x
2

4
=
32
3
x
11=2
+ 16x
8
+
8
3
x
3
+
1
81
x
2

8
27
x
1=2
(c) (1 +
p
x)
4
= 4x
3=2
+x
2
+ 6x+ 4
p
x+ 1
3. Approximating yields (2:1)
10

4
X
k=0

10
k

2
10k
(0:1)
k
= 1667:904, which has
an approximate error of0:08.
5. In sigma notation we have
19
X
k=0

19
k

(3)
k
= (13)
19
= (2)
19
.
273
Supplementary Problems 2.4 (page 119)All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means -
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DEPED COPY 1. (a) G,r= 3=2 (b) O (c) O
2. First,a3=a2+ 5 =a1+ 10. Also,
a3+ 1
a2+ 2
=
a2+ 2
a1+ 4
. Thus,a1= 5,
a2= 10, anda3= 15.
3. We have
50
X
i=1
2i
3
+ 9i
2
+ 13i + 6
i
2
+ 3i+ 2
=
(2i+ 3)(i + 1)(i + 2)
(i+ 1)(i + 2)
= 2i+ 3.
Thus,
50
X
i=1
2i
3
+ 9i
2
+ 13i +

i
2
+ 3i+ 2
=
50
X
i=1
(2i+ 3) = 2700.
4. (a)

8
k

x
162k

1
2

k
=)162k= 8 =)k= 4
=)

8
4

x
8

1
2

4
=
35
8
x
8
(b)k= 19 =)

28
19

(n
3
)
2819
(3m )
19
=

28
19

3
19
n
27
m
19
5. Forn=k+ 1:
1
13
+
1
35
+ +
1
(2(k+ 1)1)(2(k + 1) + 1)
=
k
2k+ 1
+
1
(2k+ 1)(2k + 3)
=
2k
2
+ 3k+ 1
(2k+ 1)(2k+ 3)
=
k+ 1
2k+ 3
6.a1= 10;000,r= 1:04, 6020 = 40
s40= 10;000
1(1:04)
40
1(1:04)
499;675:83 pesos
1. (a) G,r= 4=5 (b) O (c) A,d= 5=2
2. We havea1+a2= 2a1+d= 9 anda1+a2+a3= 3a 1+ 2d= 9 yielding
a1= 9; d=9. Usingsn=126, we getn= 7.
3. (a)
50
X
i=1
(2i+ 1)(i 3) =
50
X
i=1
2i
2
5i3 = 79;472
(b)
30
X
i=1
r
i
2
2i+ 1
4
=
30
X
i=1
i1
2
=
435
2
274
Topic Test 1 for Unit 2 (page 121)
Topic Test 2 for Unit 2 (page 122)All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means -
electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2016.

DEPED COPY 4.

8
k

(x
3
)
8k

1
x

k
=

8
k

x
244k
=)244k= 0 =)k= 6 =)

8
6

= 28
5. (a) Forn=k+ 1:
1 + 4 + 7 + + (3(k + 1)2)
=
k(3k1)
2
+ (3k + 1) =
3k
2
+ 5k+ 2
2
=
(k+ 1)(3k + 2)
2
(b) Forn=k+ 1: 3
(n+1)
+ 7
(n+1)1
+ 8 = 7 (3
n
+ 7
n1
+ 8)43
n
68,
where 43
n
is divisible by 12 forn1, and 68 = 48 = 12(4).
1.
6
5
rev =
6
5
rev

360
1 rev

= 432

3. 216

= 216

180

=
6
5
rad;s= 4

6
5

=
24
5
cm
5. 2110

5(360

) = 310

7.=
7
6
;r=
9
2
cm;A=
1
2

9
2

2
7
6
=
189
16
cm
2
9.= 150

=
5
6
;A= 15 in
2
;r=
s
2(15)
5
6
=
6
p

=
6
p

in
11.r= 6 in;s= 6 in;=
s
r
=
6
6
= 1 rad; 1 rad = 1

180

57:30

13.
8
3
cm
15.= 20

=

9
rad;A= 800 cm
2
r=
v
u
u
t
2(800)

9
=
120
p

=
120
p

cm;s=

120
p

9

=
40
p

3
cm
17.r= 6 cm;= 54

=
3
10
Area of shaded region = 2area of sectorAOE= 2

1
2
(6)
2

3
10

=
54
5
cm
2
19.Asegment=AsectorAtriangle=
1
2

2
3

(6)
2

1
2
(3)(6
p
3) = (12 9
p
3) cm
2
275
Supplementary Problems 3.1 (page 133)All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means -
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DEPED COPY 1.
33
4
is coterminal with
33
4
8=

4
, and
33
4
terminates in QI.
3. The secant function is positive in QI and QIV. The cotangent function is
negative in QII and QIV. Therefore, the angleis in QIV.
5.
5
6
is in QII. The reference angle is

6
, and thereforeP

5
6

=

p
3
2
;
1
2
!
.
7. tan=
2
3
;cos >0 =)sec=
p
13
3
sec+ tan
sectan
=
p
13
3

2
3
p
13
3
+
2
3
=
174
p
13
9
9. csc= 2, cos <0;r= 2; y= 1; x=
p
3; sec=
r
x
=
2

p
3
=
2
p
3
3
11. csc=4 andnot in QIII =)in QIV
csc=
4
1
=)r= 4; y=1
x=
p
(4)
2
(1)
2
=
p
15,is in Quadrant IV,x=
p
15
cos=
x
r
=
p
15
4
sec=
r
x
=
4
p
15
15
sin=
y
r
=
1
4
csc=
r
y
=4 tan =
y
x
=
p
15
15
cot=
x
y
=
p
15
13.x=2; y= 4 =)r=
p
(2)
2
+ (4)
2
= 2
p
5
cos=
x
r
=
p
5
5
sec=
r
x
=
p
5 sin=
y
r
=
2
p
5
5
csc=
r
y
=
p
52 tan=
y
x
=2 cot =
x
y
=
1
2
15.x= 2; y=6; r=
p
(2)
2
+ (6)
2
= 2
p
10; sec=
p
10, csc=
p
10
3
sec
2
csc
2
= 10
10
9
=
80
9
17. cos= sin
2
3
=
p
3
2
and
3
2
< <2=)=
11
6
276
Supplementary Problems 3.2 (page 143)All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means -
electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2016.

DEPED COPY 19.f(x) = sin 2x + cos 2x+ sec 2x+ csc 2x+ tan 2x+ cot 2x
f

7
8

=
p
2
2
+
p
2
2
+
p
2
p
211 =2
1.P=
2
1
4
= 8
3.

4
k
= 2 =)k= 8
5.y= 3 sin
3
4

8
9
+
2
3

5 =
13
2
7. domain =R; range =

4
3
;
8
3

9.y= 3 sec 2(x)3
11. Asymptotes:x=
3
2
+ 2k; k2Z
13. (a)P= 8, phase shift =

4
, domain =R, range = [3; 1]
(b)P=,
phase shift =

6
,
domain =

xjx6
=

3
+k
; k2Z

, range =R
277
Supplemen
tary Problems 3.3 (page 170)All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means -
electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2016.

DEPED COPY (c)P=
4
3
, phase shift =

2
, domain =

xjx6
=

2
+
2k
3
; k2Z

, range =

1;
3
2

[

1
2
;1

(d)P=,
phase shift =

6
,
domain =

xjx6
=

12
+
k

2
;
k2Z

, range =
(1; 1][[3;1)
15.y=

1
10
(
t10); att= 10,y 4:32 (that is, the mass is located about
4:32 cm below the resting position)
278All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means -
electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2016.

DEPED COPY 1.
tanxsinx
sinx
=
sinx
cosx
sinx

sinx
sinx
=
sinx
cosx

1
sinx
1 = secx1
3. sinA+
cos
2
A
1 + sinA
=
sinA+ sin
2
Acos
2
A
1 + sinA
=
sinA+ 1
1 + sinA
= 1
5.
cscx+ secx
cotx+ tanx
=
1
sinx
+
1
cosx
cosx
sinx
+
sinx
cosx
=
cosx+ sinx
sinxcosx
cos
2
x+ sin
2
x
sinxcosx
= cosx+ sinx
7.
tanx+ sinx
cscx+ cotx
=
sinx
cosx
sinx
1
sinx
+
cosx
sinx
=
sinx+ sinxcosx
cosx
1 + cosx
sinx
=
sinx(1 + cosx)
cosx
1 + cosx
sinx
=
sin
2
x
cosx
=
1cos
2
x
cosx
9. sincos= sin
2

cos
sin
=
cot
csc
2

=
cot
1 + cot
2

=
a
1 +a
2
11.
csca+ 1
csca1
=
1
sina
+ 1
1
sina
1
=
1 + sina
sina
1sina
sina
=
1 + sina
1sina
13.
cosa
seca+ tana
=
cosa
1
cosa
+
sina
cosa
=
cos
2
a
1 + sina
=
1sin
2
a
1 + sina
= 1sina
15.
1
1cosa
+
1
1 + cosa
=
1 + cosa+ 1cosa
(1cosa)(1 + cosa)
=
2
1cos
2
a
=
2
sin
2
a
= 2 csc
2
a
17.
tan
1tan
2

=
sin
cos
1
sin
2

cos
2

=
sin
cos
cos
2
sin
2

sin
2

=
sin
cos

cos
2

cos
2
sin
2

=
sincos
cos
2
(1cos
2
)
=
sincos
2 cos
2
1
279
Supplementary Problems 3.4 (page 179)All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means -
electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2016.

DEPED COPY 19.
cotsinsec
seccsc
=
cos
sin

sin
cos
1
cos

1
sin
=
cos
2
sin
2

sincos
cossin= cos
2
sin
2

1. cos= sin
2
3
=
p
3
2
andin QIV =)=
11
6
3. tanA= tan

2
+kB

=
sin

2
+kB

cos

2
+kB

=
sin

2
+k

cosBcos

2
+k

sinB
cos

2
+k

cosB+ sin

2
+k

sinB
=
sin

2
+k

cosB
sin

2
+k

sinB
= cotB
5. sin 105

cos 15

= sin(90

+ 15

)cos 15

= cos 15

cos 15

= 0
7. cot= 7;csc=
p
10;andandare acute
=)cos=
7
p
2
10
;sin=
p
2
10
;sin=
p
10
10
;cos=
3
p
2
10
cos( +) = coscossinsin
=

7
p
2
10
!
3
p
2
10
!

p
2
10
! p
10
10
!
=
2
p
5
5
9. 3 sinx= 2 =)sinx=
2
3
sin(x) + sin(x +)
= sinxcoscosxsin+ sinxcos+ cosxsin
= 2 sinxcos= 2

2
3

(1) =
4
3
11. sinA=
4
5
andAin QII =)cosA=
3
5
cosB=
4
5
andBin QIV =)sinB=
3
5
(a) sin(A B) =

4
5

4
5

3
5

3
5

=
7
25
280
Supplementary Problems 3.5 (page 188)All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means -
electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2016.

DEPED COPY (b) cos(A B) =

3
5

4
5

+

4
5

3
5

=
24
25
(c) tan(A B) =
7
25

24
25
=
7
24
cos(A B)<0 and sin(AB)>0 =)ABin QII
13. Given: sin=
4
5
and cos=
5
13
sin(+) + sin( ) = sincos+ cossin+ sincoscossin
= 2 sincos= 2

4
5

5
13

=
8
13
15. cscA=
p
17; Ain QI =)tanA=
1
4
cscB=
p
34
3
; Bin QI =)tanB=
3
5
tan(A +B) =
1
4
+
3
5
1
1
4

3
5
= 1 =)A+B= 45

17.
tan

9
+ tan
23
36
1tan

9
tan
23
36
= tan

9
+
23
36

= tan
3
4
=1
19. sin 2 = sin(+) = sincos+ cossin= 2 sincos
1.r= 6 cm,= 37:5

= 37:5

180

=
5
24
rad
(a)s= 6

5
24

=
5
4
cm
(b)A=
1
2
(6)
2

5
24

=
15
4
cm
2
2.x=1,y=2,r=
p
(1)
2
+ (2)
2
=
p
5
sin+ cos+ tan=
2
p
5
5
+

p
5
5
+ 2 =
103
p
5
5
281
Topic Test 1 for Unit 3 (page 190)All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means -
electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2016.

DEPED COPY 3. sinA=
12
13
,Ais in QII =)cosA=
5
13
;tanA=
12
5
cosB=
5
3
,Bis in QIV =)sinB=
3
5
;cosB=
4
5
;tanA=
3
4
(a) cos(A B) = cosAcosB+ sinAsinB
=

5
13

4
5

+

12
13

3
5

=
56
65
(b) tan(A B) =
tanA+ tanB
1 + tanAtanB
=

12
5
+

3
4

1 +
12
5

3
4
=
33
56
4.
tan 57

+ tan 78

1tan 57

tan 78

= tan(57 + 78)

= tan 135

=1
5.
cosxtanx+ sinx
tanx
= cosx+
sinx
tanx
= 2 cosx= 2
p
1sin
2
x= 2
p
1a
2
6. cos
6
x+ sin
6
x= (cos
2
x)
3
+ (sin
2
x)
3
= (cos
2
x+ sin
2
x)(cos
4
xcos
2
xsin
2
x+ sin
4
x)
= cos
4
xcos
2
xsin
2
x+ sin
4
x
= cos
4
xcos
2
x(1cos
2
x) + (1cos
2
x)
2
= cos
4
xcos
2
x+ cos
4
x+ 12 cos
2
x+ cos
4
x
= 3 cos
4
x3 cos
2
x1
7. Connect the three diagonals of the hexagon. In doing this, the hexagon is
divided into 6 equilateral triangles. Hence,B

1
2
;
p
3
2
!
. Same coordinates
forC,EandF, except that they will just vary in signs depending on the
quadrant.
8.y= 2 sin

x
2
+

3

1 =)y= 2 sin
1
2

x+
2
3

1
P= 4, Phase Shift =
2
3
, Amplitude = 2, Range = [3;1]
282All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means -
electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2016.

DEPED COPY 1.Asector=

3
cm
2
,= 30

= 30

180

=

6
rad

3
=
1
2

6

r
2
=)r= 2 cm =)2

6

=

3
cm
2.x= 8,y=6,r=
p
(8)
2
+ (6)
2
= 10
(sin+ cos)
2
= sin
2
+ 2 sincos+ cos
2

= 1 + 2 sincos= 1 + 2

6
10

8
10

=
1
25
3. sinA=
8
17
sin

2
A

+ cos

2
A

= cosA+ sinA=
15
17
+
8
17
=
7
17
4. sin 160 cos 35sin 70 cos 55
= sin 20 cos 35cos 20 sin 35
= sin(2035) =sin(4530) =
p
2
p
6
4
5. tan
7
12
= tan

4
+

3

=
tan

4
+ tan

3
1tan

4
tan

3
=
1 +
p
3
1
p
3
=2
p
3
6. cosA=
3
5
,Ais in QIII =)sinA=
4
5
;tanA=
4
3
tanB=
24
7
,Bis in QIII =)sinB=
24
25
;cosB=
7
25
,
(a) sin(A +B) = sinAcosB+ cosAsinB
=

4
5

7
25

+

3
5

24
25

=
4
5
(b) cot(A +B) =
1tanAtanB
tanA+ tanB
=
1
4
3

24
7

4
3
+

24
7
=
3
4
7.
tan
2
x
tanx+ tan
3
x
=
tan
2
x
tanx(1 + tan
2
x)
=
tanx
sec
2
x
= sinxcosx
283
Topic Test 2 for Unit 3 (page 191)All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means -
electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2016.

DEPED COPY 8.
sinx
secx
= sinxcosx
sinxcosx=
1
3
(sinxcosx)
2
=

1
3

2
=)sin
2
x2 sinxcosx+ cos
2
x=
1
9
=)12 sinxcosx=
1
9
=) 2 sinxcosx=
8
9
=)sinxcosx=
4
9
9.y= tan

18

x
3

+ 2 =tan
1
3

x

6

+ 2
P= 3, phase shift =

6
p
2
9
(b)
cos 2=
7
9
(c)
tan 2 =
4
p
2
7
(d)
sec 2=
9
7
(e)
csc 2=
9
p
2
8
(f
) cot 2=
7
p
2
8
3.
cos(2t) =
1
8
5.
tanx=
1
p
5
2
7.
cot 4 = 1=(tan 4) =7=24
9. sin
25
8
=
2+
p
2
4
and
cos
25
8
=
2

p
2
4
11.
tan
1
2
y1
tan
1
2
y+
1
=
1cosy
siny
1
siny
1 + cosy
+
1
=
1cosysiny
siny
siny+
y
1 + cosy
=
1cos
2
ysinysinycosy
sin
2
y+
y+ sinycosy
=
sin
2
ysinysinycosy
sin
2
y+
y+ sinycosy
=
siny1cosy
siny+
y
:
284
Supplementary Problems 3.6 (page 200)
1. (a) sin 2 =4All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means -
electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2016.

DEPED COPY 13. (a) cos 105

=
p
2+
p
3
2
(b) tan 22:5

=
p
21
1. (a) sin[sin
1
(1=2)] = 1=2
(b) cos[cos
1
(
p
2=2)] =
p
2=2
(c) tan[tan
1
(
p
3)] =
p
3
(d) sin[arctan(
p
3)] =
p
3=2
(e) cos[arccos(
p
2)] does not exist
(f) tan[arcsin(1=4)] =
p
15=15
(g) cos(sin
1
p
3=2) = 1=2
3. (a) sin[2 cos
1
(4=5)] = 24=25
(b) cos[2 sin
1
(5=13)] = 119=169
(c) sin[sin
1
(3=5) + cos
1
(5=13)] = 33=65
(d) cos[sin
1
(1=2)cos
1
(8=17)] = (15 + 8
p
3)=34
5. (a) arcsec(
p
2) = 3=4
(b) arccsc(2) = =6
(c) arccot
p
3 ==6
(d) [sec
1
(1)][cos
1
(1)] ==
2
(e) 2 cot
1
p
3 + 3 csc
1
2 = 2(=6) + 3( =6) = 5=6
(f) csc
1
0 does not exist
7. Vertex angleshould be=3.
1. Solution set:f=2; 3=2;5=2;7=2; :::g =f(2k+ 1)=2 jk2Zg
3. Solution set:f2k=3 jk2Zg
5. Solution set:f=8; 3=8;9=8;11=8g
7. Solution set:f=2; =6g
9. The bullet should be red with an angle of= 60

.
285
Supplementary Problems 3.7 (page 219)
Supplementary Problems 3.8 (page 236)All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means -
electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2016.

DEPED COPY 1. (a) (13; 7=3), (13; 13=3)
(b) (0;2), (0; =4)
(c) (15; 7=4), (15; 23=4)
3. (a) (3; 0)
(b) (3
p
2=2;3
p
2=2)
(c) (0; 5=2)
(d) (4
p
3;4)
5. (a)r= 4 cos 2
0=6=4=3=22=3 3=4 5=6 7=6 5=4 4=3 3=2
r42024 2 0 242 0 2 4
(b)
169
(b)
sin(2 ) =
120
169
(c)
tan(2 ) =
120
119
(d)
sec(2 ) =
169
119
(e)
csc(2) =
169
120
(f
) cot(2 ) =
119
120
2.
Hint: Use the double-angle identity for tangent tan(2x) =
2 tanx
1tan
2
x
.
3.
(a) tan 15

= 2
p
3 (b)
tan 7:5

=
4
p
6
p
2
p
6
p
2
4.
(a) tan
1

cot
103
6

=

3
(b)
cos

sin
1
40
41

=
9
41
5.
Solution Set =

0;

2
;
3

2

286
Supplemen
tary Problems 3.9 (page 252)
Topic Test 3 for Unit 3 (page 253)
1. (a) cos(2) =119All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means -
electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2016.

DEPED COPY 6. Solution Set =

0; ;2;

6
;
5
6
;
7
6
;
11
6

7.r=22 sin
0=6 =4 =3 =2 2=3 3=4
r2 32
p
22
p
3 4 2
p
32
p
2
5=6 7=6 5=4 4=3 3=2
r3 2 1 2 +
p
22 +
p
3 0
8. (a) (r; )
= (70; tan
14
3
)
(b)
(r; ) = (100
p
2;

4
)
(c)
(r; ) = (7;

2
)
(d)
(r; ) = (7;0)
(e) (r; ) = (2
p
2;

4
)
(f
) (r; ) = (13; tan
112
5
)
9.
(a) (x; y) = (
3
2
;
3
p
3
2
)
(b)
(x; y) = (
45
p
2
2
;
45
p
2
2
)
(c)
(x; y) = (1; 0)
(d) (x; y) = (5;0)
(e) (x; y) = (2;0)
(f) (x; y) = (
9
p
3
2
;
9
2
)
287All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means -
electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2016.

DEPED COPY 625
(b) sin(2) =
336
625
(c) tan(2 ) =
336
527
2. (a) cos
2
24

0:8346 (b) sin
2
24

0:1655 (c) tan
2
24

0:1983
3. (a) tan(=12) = 2
p
3 (b) tan(=24) =
4
p
6
p
2
p
6
p
2
4. cos

cos
1
1
7
+ cos
1
3
5

=
316
p
3
35
5. Solution set =

4
;
3
4
;
4
3
;
5
3

6. Solution set =

5
8
;
7
8
;
13
8
;
15
8
;

4
;
5
4

7.r= 2 + 2 cos
0 =6 =4 =3 =2 2=3 3=4
r 4 2 +
p
32 +
p
2 3 2 12
p
2
5=6 7=6 5=4 4=3 3=2
r2
p
3 0 2
p
32
p
21 2
288
T
opic Test 4 for Unit 3 (page 254)
1. (a) cos(2) =527All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means -
electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2016.

DEPED COPY 8. (a) (r; ) = (35; tan
1
(
4
3
))
(b) (r; ) = (100
p
2;
5
4
)
(c) (r; ) = (5;

2
)
(d) (r; ) = (5; )
(e) (r; ) = (
p
2;

4
)
(f) (r; ) = (17; tan
1
(
8
15
))
9. (a) (x; y ) = (2
p
3;2)
(b) (x; y) = (50
p
2;50
p
2)
(c) (x; y) = (1; 0)
(d) (x; y) = (5;0)
(e) (x; y) = (;0)
(f) (x; y) = (
15
2
;
15
p
3
2
)
289All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means -
electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2016.

DEPED COPY References
[1] R.N. Aufmann, V.C. Barker, and R.D. Nation,College Trigonometry, Houghton
Miin Company, 2008.
[2] E.A. Cabral, M.L.A.N. De Las Pe~nas, E.P. De Lara-Tuprio, F.F. Francisco,
I.J.L. Garces, R.M. Marcelo, and J.F. Sarmiento,Precalculus, Ateneo de
Manila University Press, 2010.
[3] R. Larson,Precalculus with Limits, Brooks/Cole, Cengage Learning, 2014.
[4] L. Leithold,College Algebra and Trigonometry, Addison Wesley Longman
Inc., 1989, reprinted by Pearson Education Asia Pte. Ltd., 2002.
[5] M.L. Lial, J. Hornsby, and D.I. Schneider,College Algebra and Trigonometry
and Precalculus, Addison-Wesley Educational Publisher, Inc., 2001.
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290All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means -
electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2016.

Pre calculus Grade 11 Learner's Module Senior High School

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Pre calculus Grade 11 Learner&#39;s Module Senior High School

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Pre calculus Grade 11 Learner's Module Senior High School