Trigonometry-1.ppt

trigo

1. © Boardworks Ltd 20051of 47
These icons indicate that teacher’s notes or useful web addresses are available in the Notes Page.
This icon indicates the slide contains activities created in Flash. These activities are not editable.
For more detailed instructions, see the Getting Startedpresentation.
© Boardworks Ltd 20051of 47
AS-Level Maths:
Core 2
for Edexcel
C2.4 Trigonometry 1

2. © Boardworks Ltd 20052of 47
Contents
© Boardworks Ltd 20052of 47
The sine, cosine and tangent of any angle
The graphs of sin θ, cos θand tan θ
Exact values of trigonometric functions
Trigonometric equations
Trigonometric identities
Examination-style questions
The sine, cosine and tangent of any angle

3. © Boardworks Ltd 20053of 47
θ
O
P
P
O
S
I
T
E
H
Y
P
O
T
E
N
U
S
E
A D J A C E N T
The three trigonometric ratios
Sin θ=
Opposite
Hypotenuse
S O H
Cos θ=
Adjacent
Hypotenuse
C A H
Tan θ=
Opposite
Adjacent
T O A
Remember:S O HC A HT O A
The three trigonometric ratios, sine, cosineand tangent, can
be defined using the ratios of the sides of a right-angled
triangle as follows:

4. © Boardworks Ltd 20054of 47
x
y
O
P(x, y)
r
The sine, cosine and tangent of any angle
These definitions are limited because a right-angled triangle
cannot contain any angles greater than 90°.
To extend the three trigonometric ratios to include angles
greater than 90°and less than 0°we consider the rotation of a
straight line OP of fixed length rabout the origin O of a
coordinate grid.
Angles are then measured
anticlockwisefrom the
positive x-axis.
For any angle θthere is an
associated acute angle α
between the line OP and
the x-axis.
α
θ

5. © Boardworks Ltd 20055of 47
The sine, cosine and tangent of any angle
The three trigonometric ratios are then given by:sin =
y
r
 cos =
x
r
 tan =
y
x

The xand ycoordinates can be positive or negative, while ris
always positive.
This means that the sign of the required ratio will depend on
the sign of the x-coordinate and the y-coordinate of the point P.

6. © Boardworks Ltd 20056of 47
The sine, cosine and tangent of any angle
The relationship between θmeasured from the positive x-axis
and the associated acute angle αdepends on the quadrant
that θfalls into.
If we take rto be 1 unit long then these ratios can be written
as:sin = =
1
y
y cos = =
1
x
x tan =
y
x
  sin
tan =
cos



For example, if θis between 90°and 180°it will fall into the
second quadrant and αwill be equal to (180 –θ)°.

7. © Boardworks Ltd 20057of 47
The sine of any angle
If the point P is taken to revolve about a unit circle then sin θis
given by the y-coordinate of the point P.

8. © Boardworks Ltd 20058of 47
The cosine of any angle
If the point P is taken to revolve about a unit circle then cos θ
is given by the x-coordinate of the point P.

9. © Boardworks Ltd 20059of 47
The tangentof any angle
Tan θis given by the y-coordinate of the point P divided by the
x-coordinate.

10. © Boardworks Ltd 200510of 47
The tangentof any angle
Tan θcan also be given by the length of tangent from the point
P to the x-axis.

11. © Boardworks Ltd 200511of 47
3
rd
quadrant
2
nd
quadrant 1
st
quadrant
4
th
quadrant
Tangent is positiveT
Sine is positiveS All are positiveA
Remember CAST
We can use CAST to remember in which quadrant each of
the three ratios are positive.
Cosine is positiveC

12. © Boardworks Ltd 200512of 47
The sine, cosine and tangent of any angle
The sin, cos and tan of angles in the first quadrant are positive.
In the second quadrant:sin θ= sin α
cos θ= –cos α
tan θ= –tan α
In the third quadrant:sin θ= –sin α
cos θ= –cos α
tan θ= tan α
In the fourth quadrant:sin θ= –sin α
cos θ= cos α
tan θ= –tan α
where αis the
associated
acute angle.

13. © Boardworks Ltd 200513of 47
The sine, cosine and tangent of any angle
The value of the associated acute angle αcan be found using
a sketch of the four quadrants.
For angles between 0°and 360°it is worth remembering that:
when 0°< θ< 90°, α= θ
when 90°< θ< 180°,α= 180°–θ
when 180°< θ< 270°,α= θ–180°
when 270°< θ< 360°,α= 360°–θ
For example, if θ= 230°we have:
230°is in the third quadrant where only tan is positive and so:
α= 230°–180°=50°
sin 230°= –sin 50°
cos 230°= –cos 50°
tan 230°= tan 50°

14. © Boardworks Ltd 200514of 47
Contents
© Boardworks Ltd 200514of 47
The sine, cosine and tangent of any angle
The graphs of sin θ, cos θand tan θ
Exact values of trigonometric functions
Trigonometric equations
Trigonometric identities
Examination-style questions
The graphs of sin θ, cos θand tan θ

15. © Boardworks Ltd 200515of 47
The graph of y= sin θ

16. © Boardworks Ltd 200516of 47
The graph of y= sin θ
The graph of sine is said to be periodicsince it repeats itself
every 360°.
We can say that the periodof the graph y= sin θis 360°.
Other important features of the graph y= sin θinclude the
fact that:
It passes through the origin, since sin 0°= 0.
The maximum value is 1 and the minimum value is –1.
Therefore, the amplitudeof y= sin θis 1.
It has rotational symmetry about the origin. In other words, it
is an odd function and so sin (–θ) = –sin θ.

17. © Boardworks Ltd 200517of 47
The graph of y= cos θ

18. © Boardworks Ltd 200518of 47
The graph of y= cos θ
Like the graph of y= sin θthe graph of y= cos θis periodic
since it repeats itself every 360°.
We can say that the periodof the graph y= cos θis 360°.
Other important features of the graph y= cos θinclude the
fact that:
It passes through the point (0, 1), since cos 0°= 1.
The maximum value is 1 and the minimum value is –1.
Therefore, the amplitudeof y= cos θis 1.
It is symmetrical about the y-axis. In other words, it is an
evenfunction and so cos (–θ) = cos θ.
It the same as the graph of y= sin θtranslated left 90°. In
other words, cos θ= sin (90°–θ).

19. © Boardworks Ltd 200519of 47
The graph of y= tan θ

20. © Boardworks Ltd 200520of 47
The graph of y= tan θ
You have seen that the graph of y= tan θhas a different
shape to the graphs of y= sin θand y= cos θ.
Important features of the graph y= tan θinclude the fact that:
It passes through the point (0, 0), since tan 0°= 0.
It is symmetrical about the origin. In other words, it is an odd
function and so tan (–θ) = –tan θ.
It is periodic with a period of 180°.
tan θis not defined for θ= ±90°, ±270°,±450°,… that is, for
odd multiples of 90°.The graph y= tan θtherefore contains
asymptotes at these points.
Its amplitude is not defined, since it ranges from +∞ to –∞.

21. © Boardworks Ltd 200521of 47
Contents
© Boardworks Ltd 200521of 47
The sine, cosine and tangent of any angle
The graphs of sin θ, cos θand tan θ
Exact values of trigonometric functions
Trigonometric equations
Trigonometric identities
Examination-style questions
Exact values of trigonometric functions

22. © Boardworks Ltd 200522of 47
Sin, cos and tan of 45°
A right-angled isosceles triangle has two acute angels of 45°.
45°
45°
Suppose the equal sides are of
unit length.
1
1
Using Pythagoras’ theorem:
We can use this triangle to write exact values for sin, cos and
tan 45°:
cos 45°= tan 45°=12 2
The hypotenuse
22
11 1
2
sin 45°=1
2

23. © Boardworks Ltd 200523of 47
2 2
2
60° 60°
60°
2
60°
30°
13
Sin, cos and tan of 30°
Suppose we have an equilateral triangle of side length 2.
We can use this triangle to write exact values for sin, cos and
tan 30°:
If we cut the triangle in half then we have
a right-angled triangle with acute angles
of 30°and 60°.
Using Pythagoras’ theorem:
The height of the triangle
22
21 3
sin 30°=1
2 cos 30°=3
2 tan 30°=1
3

24. © Boardworks Ltd 200524of 47
Sin, cos and tan of 60°
Suppose we have an equilateral triangle of side length 2.
We can also use this triangle to write exact values for sin, cos
and tan 60°:3
sin 60°=3
2 cos 60°=1
2 tan 60°=3
2 2
2
60° 60°
60°
2
60°
30°
13
If we cut the triangle in half then we have
a right-angled triangle with acute angles
of 30°and 60°.
Using Pythagoras’ theorem:
The height of the triangle
22
21

25. © Boardworks Ltd 200525of 47
Sin, cos and tan of 30°, 45°and 60°
The exact values of the sine, cosine and tangent of 30°, 45°
and 60°can be summarized as follows:
30°
sin
cos
tan
45° 60°
Use this table to write the exact value of cos 135°
cos 135°= –cos 45°=1
2 1
2 1
2 1
3 1
2 3
2 3
2 3 1 1
2

26. © Boardworks Ltd 200526of 47
Sin, cos and tan of 30°, 45°and 60°
Write the following ratios exactly:
1) cos 300°=
3) tan 240°=
5) cos –30°=
7) sin 210°=
2) tan 315°=
4) sin –330°=
6) tan –135°=
8) cos 315°=1
2 1
2 1
2 3
2 1
2 3
-1
1

27. © Boardworks Ltd 200527of 47
Contents
© Boardworks Ltd 200527of 47
The sine, cosine and tangent of any angle
The graphs of sin θ, cos θand tan θ
Exact values of trigonometric functions
Trigonometric equations
Trigonometric identities
Examination-style questions
Trigonometric equations

28. © Boardworks Ltd 200528of 47
Equations of the form sin θ= k
Equations of the form sin θ= k, where –1 ≤ k≤ 1, have an
infinite number of solutions.
If we use a calculator to find arcsin k(or sin
–1
k) the calculator
will give a value for θbetween –90°and 90°.
This is called the principal solutionof sin θ= k.
Other solutions in a given range can be found using the
graph of y= sin θor by considering the unit circle.
For example:
There is one and only one solution is this range.
Solve sin θ= 0.7 for –360°< θ< 360°
arcsin 0.7 = 44.4°(to 1 d.p.)

29. © Boardworks Ltd 200529of 47
44.4°
y= 0.7
y= sin θ
Equations of the form sin θ= k
Using the graph of y= sin θbetween –360°and 360°and the
line y= 0.7 we can locate the other solutions in this range.
So the solutions to sin θ= 0.7 for –360°< θ< 360°are:
θ= –315.6°, –224.4°, 44.4°, 135.6°(to 1 d.p)
135.6°–224.4°–315.6°

30. © Boardworks Ltd 200530of 47
Equations of the form sin θ= k

31. © Boardworks Ltd 200531of 47
44.4°
Equations of the form sin θ= k
We could also solve sin θ= 0.7 for –360°< θ< 360°by
considering angles in the first and second quadrants of a unit
circle where the sine ratio is positive.
Start by sketching the principal solution 44.4°in the first
quadrant.
Next, sketch the associated acute
angle in the second quadrant.
135.6°
–224.4° –315.6°
Moving anticlockwise from the
x-axis gives the second solution:
Moving clockwise from the
x-axis gives the third and fourth
solutions:
180°–44.4°= 135.6°
–(180°+ 44.4°) = –224.4°
–(360°–44.4°) = –315.6°

32. © Boardworks Ltd 200532of 47
Equations of the form cos θ= k and tan θ= k
Equations of the form cos θ= k, where –1 ≤ k≤ 1, and
tan θ= k, wherekis any real number, also have infinitely
many solutions. For example:
Solve tan θ= –1.5 for –360°<θ< 360°
Using a calculator, the principal solution is θ= –56.3°(to 1 d.p.)
–56.3°
–236.3°
123.7°
303.7°
Now look at angles in the second and
fourth quadrants of a unit circle where
the tangent ratio is negative.
This gives us four solutions in the
range –360°<θ< 360°:
θ= –236.3°,–56.3°,123.7°,303.7°

33. © Boardworks Ltd 200533of 47
Equations of the form cos θ= k

34. © Boardworks Ltd 200534of 47
Equations of the form tan θ= k

35. © Boardworks Ltd 200535of 47
Equations involving multiple or compound angles
Multiples anglesare angles of the form aθwhere ais a given
constant.
Compound anglesare angles of the form (θ+ b) where bis a
given constant.
When solving trigonometric equations involving these types of
angles, care should be taken to avoid ‘losing’ solutions.
Solve cos 2θ= 0.4 for –180°< θ< 180°
Start by changing the range to match the multiple angle:
Next, let x= 2θand solve the equation cos x= 0.4 in the range
–360°≤ x≤ 360°.
–180°< θ< 180°
–360°< 2θ< 360°

36. © Boardworks Ltd 200536of 47
Equations involving multiple or compound angles
Now, using a calculator: x= 66.4°(to 1 d.p.)
Using the unit circle to find the values
of xin the range –360°≤ x≤ 360°
gives:
66.4°
293.6°
–66.4°
–293.6°
x= 66.4°,
But x= 2θso:
θ= 33.2°, 146.8°, –33.2°, –146.8°
This is the complete solution set in the range –180°< θ< 180°.
293.6°,–66.4°,–293.6°

37. © Boardworks Ltd 200537of 47
Equations involving multiple or compound angles
Solve tan(θ+ 25°)= 0.8 for 0°< θ< 360°
Start by changing the range to match the compound angle:
Next, let x= θ + 25°and solve the equation tan x= 0.8 in the
range 25°< x< 385°.
0°< θ< 360°
25°< θ+ 25°< 385°
Using the unit circle to find the values
of xin the range 25°≤ x≤ 385°gives:
x= 38.7°, 218.7°(to 1 d.p.)
But x= θ+ 25°so:
θ= 13.7°, 193.7.7°(to 1 d.p.)
Using a calculator: x= 38.7°(to 1 d.p.)
38.7°
218.7°

38. © Boardworks Ltd 200538of 47
Trigonometric equations involving powers
Sometimes trigonometric equations involve powers of sin θ,
cos θand tan θ. For example:
Notice that (sin θ)
2
is usually written as sin
2
θ.
Solve 4sin
2
θ–1= 0 for –180°≤ θ≤ 180°2
4sin 1=0 2
4sin =1 2 1
4
sin = 1
2
sin = 1
2
When sin θ= ,1
2
When sin θ= –,
θ= 30°, 150°
θ= –30°, –150°
So the full solution set is θ= –150°, –30°, 30°, 150°

39. © Boardworks Ltd 200539of 47
Trigonometric equations involving powers
Treat this as a quadratic equation in cos θ.
Solve 3cos
2
θ–cos θ= 2 for 0°≤ θ≤ 360°
When cos θ= 1, θ= 0°, 360°
θ= 131.8°, 22.8.2°
So the full solution set is θ= 0°, 131.8°, 228.2°, 360°
3cos
2
θ–cosθ= 2
3cos
2
θ–cosθ–2 = 0
Factorizing:
(cos θ–1)(3cosθ+ 2) = 0
cos θ= 1 or cos θ= –2
3
When cos θ= –,2
3

40. © Boardworks Ltd 200540of 47
Contents
© Boardworks Ltd 200540of 47
The sine, cosine and tangent of any angle
The graphs of sin θ, cos θand tan θ
Exact values of trigonometric functions
Trigonometric equations
Trigonometric identities
Examination-style questions
Trigonometric identities

41. © Boardworks Ltd 200541of 47
Trigonometric identities
Two important identitiesthat must be learnt are:sin
tan (cos 0)
cos




sin
2
θ+ cos
2
θ≡1
An identity, unlike an equation, is true for everyvalue of the
given variable so, for example:
The symbol ≡means “is identically equal to” although an
equals sign can also be used.
sin
2
4°+ cos
2
4°≡1,sin
2
67°+ cos
2
67°≡1,sin
2
π+ cos
2
π≡1, etc.

42. © Boardworks Ltd 200542of 47
Trigonometric identities
We can prove these identities by considering a right-angled
triangle:
x
y
r
θy
r
sin = x
r
and cos = sin
=
cos


 y
r
x
r
= y
x
= tan as required.
Also:yx
rr

   
   
   
22
22
sin +cos = + 22
2
+
=
xy
r
But by Pythagoras’ theorem x
2
+ y
2
= r
2
so:2
22
2
sin +cos = =1 as required.
r
r


43. © Boardworks Ltd 200543of 47
71.6°
Trigonometric identities
One use of these identities is to simplify trigonometric
equations. For example:
Solve sin θ= 3 cos θfor 0°≤θ≤360°
Dividing through by cos θ:sin 3cos
=
cos cos

 tan =3
Using a calculator, the principal solution is θ= 71.6°(to 1 d.p.)
251.6°
So the solutions in the given range are:
θ= 71.6°,251.6°(to 1 d.p.)

44. © Boardworks Ltd 200544of 47
Trigonometric identities
Solve 2cos
2
θ–sin θ= 1 for 0 ≤ θ≤ 360°
We can use the identity cos
2
θ+ sin
2
θ = 1 to rewrite this
equation in terms of sin θ.
2(1 –sin
2
θ) –sin θ= 1
2 –2sin
2
θ–sin θ= 1
2sin
2
θ+ sin θ–1 = 0
(2sin θ–1)(sin θ+ 1) = 0
So: sin θ= 0.5 or sin θ= –1
If sin θ= 0.5, θ= 30°, 150°
If sin θ= –1, θ= 270°

45. © Boardworks Ltd 200545of 47
Contents
© Boardworks Ltd 200545of 47
The sine, cosine and tangent of any angle
The graphs of sin θ, cos θand tan θ
Exact values of trigonometric functions
Trigonometric equations
Trigonometric identities
Examination-style questions
Examination-style questions

46. © Boardworks Ltd 200546of 47
Examination-style question
Solve the equation
3 sin θ+ tan θ= 0
for θfor 0°≤θ≤360°
Rewriting the equation usingsin
tan :
cos



 sin
3sin + =0
cos


 3sin cos +sin =0   sin (3cos +1)=0
So sin θ= 0 or 3 cos θ+ 1 = 0
3 cos θ= –1
cos θ= –1
3

47. © Boardworks Ltd 200547of 47
Examination-style question
In the range 0°≤θ≤360°,
when sin θ= 0, θ = 0°, 180°, 360°.1
3
when cos θ= –:
cos
–1
–= 109.5°(to 1 d.p.)1
3
cos θis negative in the 2
nd
and 3
rd
quadrants so the second
solution in the range is:
250.5°
109.5°
θ= 250 .5°(to 1 d.p.)
So the complete solution set is:
θ= 0°, 180°, 360°, 109.5°, 250 .5°