Analytic Trigonometry: Odd and Even Identities, Sum and Difference Formula,Double Angle Formula, Half Angle Formula

Trigonometry, presents the odd even identities, sum and difference formula and how it is derived, double angle formula and it's origin, half angle formula and how it was derived.

1. ANALYTIC TRIGONOMETRY

2. TRIGONOMETRIC IDENTITIES

3. WHAT IS AN IDENTITY ? Presentation title 3

4. Example of Identities Presentation title 4

5. Example of Conditional Equation Presentation title 5

6. Trigonometric Identities Presentation title 6

7. EXAMPLE Simplify Trigonometric Expressions: Presentation title 7

8. SOLUTION Presentation title 8

9. SOLUTION Presentation title 9

10. ESTABLISH IDENTITIES Presentation title 10

11. ESTABLISH IDENTITIES Presentation title 11

12. EXAMPLE Presentation title 12

13. EXAMPLE Presentation title 13

14. SOLUTION Presentation title 14

15. SOLUTION Presentation title 15

16. SOLUTION Presentation title 16

17. SOLUTION Presentation title 17

18. SOLUTION Presentation title 18

19. SOLUTION Presentation title 19

20. SUM AND DIFFERENCE FORMULA

21. Presentation title 21 sin(A + B) = sinA cosB + cosA sinB . sin(A - B) = sinA cosB - cosA sinB . cos(A + B) = cosA cosB - sinA sinB . cos(A - B) = cosA cosB + sinA sinB . tan(A + B) = ( tanA + tanB ) / (1 - tanA tanB ) tan(A - B) = ( tanA - tanB ) / (1 + tanA tanB ) Six Sum and Difference Formulas in Trigonometry

22. 22 Solve using the Sum Formula Verify the identity: = cot x cot y + 1   cot x cot y + 1 = Expand using cos(A - B) = cosA cosB + sinA sinB .   cot x cot y + 1 = + ][   cot x cot y + 1 = cot x cot y +   = +  

23. Presentation title 23 Solve using the Difference Formula Solve 3 sin(x-∏) = 3 in the interval [0, 2 ∏ ]   sin x cos ∏ - cos x sin ∏ = 1 sin x (-1) – cos x(0) = 1 -sin x = 1 First, get sin (x- ∏) by itself, divide both side by 3 Sin (x- ∏) = 1   Expand the left side using sin(A - B) = sinA cosB - cosA sinB Then sin x = -1 when x is  

24. Presentation title 24 Find all the solutions for  2cos 2 ⁡(x+ )=1  in the interval  [0,2 ) . Solution:   2cos 2 ⁡(x+ ) = 1 Divide each term by 2     cos 2 ⁡(x+ ) = Take the square root     cos⁡(x+ ) = =  

25. Presentation title 25 use the  cosine sum formula cos(A + B) = cosA cosB - sinA sinB   to expand and solve cos⁡(x+ ) =   cos⁡xcos – sinx sin =   cos⁡x (0)– sinx (1)=   – sinx (1)=   sinx = -   sin x = - is in Quadrants III and IV, so x = and  

26. Presentation title 26 find  sin⁡( + ) , use the  sine sum formula sin(A + B) = sinA cosB + cosA sinB   sin⁡( + ) = sin cos + cos sin   = (-1) ( )+ (0) ( ) = -  

27. 27 Solve for all values of x between [0,2 ) for 2tan 2 ⁡(x+ )+1=7.   2tan 2 (x + ) + 1 = 7 Transpose 1 to the right side   2tan 2 (x + ) = 6 Multiple each side by   tan 2 (x + ) = 3 Take the square root     tan (x + ) = ±   First, solve for tan (x + )   SOLUTION:

28. 28 Use tan sum formula tan(A + B) =   =   =   =   = =   =   tan (x + ) = ±  

29. 29 = transposing terms   =   =   =   tan x = is true when x = and   =  

30. 30 Find all solutions to  sin⁡(x+ )=sin⁡(x - ) , when  x  is between  [0,2 ) .   SOLUTION: E xpand all sides sin⁡(x+ ) = sin⁡(x -   sin ⁡x cos + cos ⁡x sin = sin⁡ x cos - cos x sin   sin ⁡x( )+ cos x ⁡( = sin⁡ x - cos x Set the two sides equal   + ⁡( = sin⁡ x - cos x Multiply all terms by 2   + = sin⁡ x - cos x Transpose to join similar terms  

31. 31 + = sin⁡ x - cos x Transpose to join similar terms   = - cos x -   = cos x (- – 1)   = cos x (- – 1)   =   . =   -  

32. Presentation title 32 EXERCISES

33. Presentation title 33 Double Angle Formulas of Sin, Cos and Tan sin 2A = 2 sin A cos A ( or )  = (2 tan A) / (1 + tan 2 A) cos 2A = cos 2 A - sin 2 A ( or ) = 2cos 2 A - 1 ( or ) = 1 - 2sin 2 A ( or ) = (1 - tan 2 A) / (1 + tan 2 A) tan 2A = (2 tan A) / (1 - tan 2 A)

34. Presentation title 34 Derivation of the Double Angle Formulas of sin, cos and tan Double Angle Formulas of Sin The sum formula of sine function is, sin (A + B) = sin A cos B + cos A sin B When A = B, the above formula becomes, sin (A + A) = sin A cos A + cos A sin A sin 2A = 2 sin A cos A

35. 35 Let us derive an alternate formula for sin 2A in terms of tan using the Pythagorean identity sec 2 A = 1 + tan 2 A. sin2 A =2sin A cos A multiply with   sin2 A =   =   =2tan A ⋅   =  

36. Presentation title 36 Double Angle Formulas of Cos The sum formula of cosine function is, cos (A + B) = cos A cos B - sin A sin B When A = B, the above formula becomes, cos (A + A) = cos A cos A - sin A sin A cos 2A = cos 2 A - sin 2 A

37. Presentation title 37 Let us use this as a base formula to derive two other formulas of cos 2A sin 2 A + cos 2 A = 1.  cos 2A = cos 2 A − (1 − cos 2 A) = cos 2 A − 1 + cos 2 A =2cos 2 A - 1 using the Pythagorean identity sin 2 A + cos 2 A = 1 transposing sin 2 A = 1 - cos 2 A substituting to the equation below: cos 2A = cos 2 A - sin 2 A (i) cos 2A = 2cos 2 A - 1 ( ii ) cos 2A = 1 - 2sin 2 A

38. Presentation title 38 Let us use this as a base formula to derive two other formulas of cos 2A sin 2 A + cos 2 A = 1.  cos 2A = (1 − sin 2 A) - sin 2 A =  1 - sin 2 A − + sin 2 A = 1 – 2sin 2 A using the Pythagorean identity sin 2 A + cos 2 A = 1 transposing cos 2 A = 1 - sin 2 A substituting to the equation below: cos 2A = cos 2 A - sin 2 A (i) cos 2A = 2cos 2 A - 1 ( ii ) cos 2A = 1 - 2sin 2 A

39. Presentation title 39 we will derive the formula of cos 2A in terms of tan using the base formula cos2 A =cos 2 A −sin 2 A =cos 2 A (1 -   =   =   =   double angle formulas of the cosine function are : cos 2A = cos 2 A - sin 2 A ( or ) = 2cos 2 A – 1 ( or ) = 1 - 2sin 2 A ( or ) =  

40. Presentation title 40 Double Angle Formulas of Tan tan (A + B) = (tan A + tan B) / (1 - tan A tan B) When A = B, the above formula becomes, tan (A + A) =   Thus, the double angle formula of tan function is, t an 2A =  

41. Presentation title 41 =   Examples Using Double Angle Formulas If tan A = 3 / 4, find the values of sin 2A, cos 2A, and tan 2A. Solution: Since the value of tan A is given, we use the double angle formulas for finding each of sin 2A, cos 2A, and tan 2A in terms of tan. a) sin2 A =   =   =  

42. Presentation title 42 Examples Using Double Angle Formulas If tan A = 3 / 4, find the values of sin 2A, cos 2A, and tan 2A. Solution: Since the value of tan A is given, we use the double angle formulas for finding each of sin 2A, cos 2A, and tan 2A in terms of tan. b ) cos2 A =   =   =  

43. Presentation title 43 Examples Using Double Angle Formulas If tan A = 3 / 4, find the values of sin 2A, cos 2A, and tan 2A. Solution: Since the value of tan A is given, we use the double angle formulas for finding each of sin 2A, cos 2A, and tan 2A in terms of tan. c ) tan2 A =   =   =   Answer:  sin 2A =  , cos 2A =  , and tan 2A =   

44. Presentation title 44 Prove the following identity  =tan 2 x   Solution: The double angle formula of cos i s cos 2A = 2cos 2 A - 1 (or) 1 - 2sin 2 A = =     =   =   =tan 2 x

45. Presentation title 45 Use the double angle formulas to derive the formula for sin 3x. Solution : sin (3x) = sin (2x + x ) ( using sin (A + B) = sin A cos B + cos A sin B) = ( sin 2x cos x) + cos2x sinx ( Using double angle formulas sin 2A = 2 sin A cos A cos 2A = 1 - 2sin 2 A = 2 sin x cos 2 x + sin x - 2 sin 3 x ( apply pythagorian ) = 2 sin x ( 1 - sin 2 x ) + sin x - 2 sin 3 x ( distributive property ) = 2 sin x - 2 sin 3 x + sin x - 2 sin 3 x = (2 sin x + sin x) + (-2 sin 3 x - 2 sin 3 x) = 3 sin x - 4 sin 3 x = ( 2 sin x cos x ) cos x + (1 - 2 sin 2 x) sin x ( Simplify ; distributive property )

46. Presentation title 46 Half Angle Formulas of Sin, Cos and Tan the values of the trigonometric functions (sin, cos , tan, cot, sec, cosec) for the angles like 0°, 30°, 45°, 60°, and 90° from the  trigonometric identities table. to know the exact values of sin 22.5°, tan 15°, etc , the half angle formulas are extremely useful half angle formulas are * Derived from the double angle formulas * helpful in proving several trigonometric identities

47. Presentation title 47 Half angle formula of sin:  sin = ±   cos = ±   tan = ± or   Half angle formula of cos :  Half angle formula of tan:  tan = ± or   tan = ± or   Half Angle Formulas of Sin, Cos and Tan

48. 48 Derivation of Half Angle Formulas of Sin, Cos and Tan From the double angle Formula: cos 2x = 1 -2 sin 2 x Replace x with   2 sin 2 ( ) = ; multiply by ½ sin 2 ( ) = sin ( ) = ±   cos 2 ( ) = 1 -2 sin 2 ( ) cos A = 1 -2 sin 2 ( ) Transposing  

49. 49 Derivation of Half Angle Formulas of Sin, Cos and Tan From the double angle Formula: cos 2x =  2 cos 2 x - 1 Replace x with   cos 2( ) = 2 cos 2 ( ) – 1 cos A = 2 cos 2 ( ) - 1 : transpose   2 cos 2 ( ) = 1 + cos A; multiply by ½ cos 2 ( ) = ( ) Cos ( ) = ±  

50. 50 Derivation of Half Angle Formulas of Sin, Cos and Tan From the double angle Formula: tan 2x = Replace x with   tan 2 ( ) = From the half angle formulas of sin and cos   tan = tan =  (1 - cos A) / sin A  

51. Presentation title 51 use the half angle formula.  1. Solve the trigonometric equation  sin 2 ⁡𝜃=2sin 2 ⁡ over the interval  [0,2𝜋) .   sin 2 𝜃 =2sin 2 ⁡ ⁡sin 2 𝜃 = 2 ( ⁡ ) Half angle identity   = Pythagorian identity = 0   = 0 Then = 0 or 1- = 0, w/c is =1   = 0,  

52. Presentation title 52 Solve  2cos 2 ⁡ =1  for  0≤𝑥<2𝜋   2 cos 2 ⁡ = 1 isolate cosine by multiplying it by 1/2   cos 2 ⁡ = then use half angle formula   = Cos x = 1-1 cos x = 0 when ⁡ =  

53. Presentation title 53 Solve  tan ⁡ = 4  for  0 ≤ a < 360º   tan⁡ = 4 use half angle formula   = 4 squaring both sides   = 16 cross multiply   17 cos a = -15 cos a = when a = 152º , 208º     cos a + = 1 - 16  

54. Presentation title 54 Solve  sin 22.5 º Sin 22.5 º = sin ⁡ = 1 use half angle formula   Sin 22.5 º = =   =   =   =  

55. Presentation title Find the exact value of cos 11 2.5 º cos 112.5 º = cos ⁡ = use half angle formula   cos 112.5 º = - =   cos 112.5 º = -   = -   -   = -  

56. Presentation title 56 tan = tan . use half angle formula   Find the exact value of tan   = = multiply with = = +   =   = =  

57. Presentation title 57 Use half angle identities to find the exact value of each expression. tan⁡15 º tan⁡22.5 º cot⁡75 º tan⁡67.5 º tan⁡157.5 º tan⁡112.5 º cos⁡105 º sin⁡112.5 º sec⁡15 º csc⁡22.5 º csc⁡75 º sec⁡67.5 º cot⁡157.5 º

58. Presentation title 58 Assignment/Exercises: Use the double angle formula and simplify sin 2x – sin x = 0 2. cos 2x – cos x = 0 3. Solve the equation sin = -1 for 0 ≤ x <  

59. Presentation title 59 - are mathematical equations that involve the trigonometric functions such as sine, cosine & tangent   T rigonometric E quations . - help us find values for the angles or sides that meet the specified criteria, and these angles are the solution of Trigonometric Equations - these equations create relationships between angles and sides of triangles. -Uses sin, cos & tan as the major trigonometric functions - require the use of  trigonometric identities  and specific angles

60. Presentation title 60   Solutions for T rigonometric E quations . Equations Solutions sin x = 0  x = n π cos x = 0 x = (n π + π /2) tan x = 0 x = n π sin x = 1 x = (2n π + π /2) = (4n+1) π /2 cos x = 1 x = 2n π sin x = sin θ x = n π + (-1) n θ, where θ ∈ [- π /2, π /2] Equations Solutions cos x = cos θ x = 2n π ± θ, where θ ∈ (0, π ] tan x = tan θ x = n π + θ, where θ ∈ (- π /2 , π /2] sin 2x = sin 2 θ x = n π ± θ cos 2x = cos 2 θ x = n π ± θ tan 2x = tan 2 θ x = n π ± θ cos x = cos θ x = 2n π ± θ, where θ ∈ (0, π ]

61. Presentation title 61   Problems Find the values of x 1. sin 2x – sin 4x + sin 6x = 0 Solution: sin 2x – sin 4x + sin 6x = 0 2sin4x.cos 2x – sin 4x = 0 sin 4x(2cos2x – 1) = 0 sin 4x = 0 or cos 2x = ½ 4x = n π or 2x = 2n π ± π/3 general solution: x = nπ/4 or nπ ± π/6

62. Presentation title 62 Solution: We already know that sin π/6 = 1/2 sin 5π/6 = sin (π – π/6) = sin π/6 = 1/2 Determine the primary solution to the equation sin x = 1/2 primary answers are x =π/6 and x = 5π/6.

63. Presentation title 63 Determine the answer to cos x = ½ . Solution: use the general solution of cos x = ½ . . From the wheel chart; cos  = ½ .   Equate the two equations ( i ) cos x = ½ .   (ii) cos = ½ . cos x = cos     x = 2n π + ( ), where n ∈ Z    As a result, cos x = ½ . has a generic solution of x = 2nπ + ( ), where n ∈ Z.   [With Cos θ = Cos α, the generic solution is θ = 2 n π + α, where n ∈ Z]

64. Presentation title 64 Determine the primary solutions to the trigonometric equation sin x =   Solution: sin x = , we know that   the primary solutions of sin x = are x = and .   sin = and sin (π – ) = sin = equating two equations sin = sin =  

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