1. 1 Trigonometry Review (I)Introduction By convention, angles are measured from the initial line or the x-axis with respect to the origin. If OP is rotated counter-clockwise from the x-axis, the angle so formed is positive. But if OP is rotated clockwise from the x-axis, the angle so formed is negative. O P x negative angle P O x positive angle

2. 2 (II)Degrees & Radians Angles are measured in degrees or radians. r r r 1c1c Given a circle with radius r, the angle subtended by an arc of length r measures 1 radian. Care with calculator! Make sure your calculator is set to radians when you are making radian calculations.

3. 3 (III)Definition of trigonometric ratios x y P(x, y)  r y x Note: Do not write cos  1  tan  1 

4. 4 1 0 11 Graph of y=sin x

5. 5 1 0 11 Graph of y=cos x

6. 6 0 Graph of y=tan x

7. 7 From the above definitions, the signs of sin , cos  & tan  in different quadrants can be obtained. These are represented in the following diagram: All +ve sin +ve tan +ve 1st 2nd 3rd 4th cos +ve

8. 8 What are special angles? (IV)Trigonometrical ratios of special angles Trigonometrical ratios of these angles are worth exploring

9. 9 1 sin 0°  0 sin 360°  0 sin 180°  0 sin 90°  1sin 270°  1 0   11

10. 10 cos 0°  1 cos 360°  1 cos 180°  1 cos 90°  0 cos 270°  1 0   11

11. 11 tan 180°  0 tan 0°  0 tan 90° is undefined tan 270° is undefined tan 360°  0 0  

12. 12 Using the equilateral triangle (of side length 2 units) shown on the right, the following exact values can be found.

13. 13 Complete the table. What do you observe?

14. 14

15. 15 2nd quadrant Important properties: 3rd quadrant 1st quadrant or 2     

16. 16 Important properties: 4th quadrant or    or 2      In the diagram,  is acute. However, these relationships are true for all sizes of 

17. 17 Complementary angles E.g.:30° & 60° are complementary angles. Two angles that sum up to 90° or radians are called complementary angles. are complementary angles. Recall:

18. 18 We say that sine & cosine are complementary functions. Also, tangent & cotangent are complementary functions. E.g.:

19. 19 E.g. 1: Simplify (i) sin 210  (ii) cos (iii) tan(– ) (iv) sin( ) sin(180°+30  )  (a) sin 210   Solution: 210° = 180°+30° 3 rd quadrant - sin 30  =

20. 20 (b) 4 th quadrant (c)

21. 21 sin (3  - x)  sin (2    - x)  sin (  - x)  sin x  0.6 cos (4  + x)  cos (2  + x)  0.8  cos x Soln : E.g. 2: If sin x = 0.6, cos x = 0.8, find (a) sin (3   x) (b) cos (4   x).

22. 22 (V)Basic Angle The basic angle is defined to be the positive, acute angle between the line OP & its projection on the x-axis. For any general angle, there is a basic angle associated with it. P O    P O   180°  or          Let  denotes the basic angle.

23. 23    360°   or   2     P O   P O    – 180° or    – 

24. 24 E.g.: P O       (1 st quadrant)

25. 25 E.g.: (2 nd quadrant)   P O   180°  or    

26. 26 E.g.: (3 rd quadrant)   P O    – 180° or    – 

27. 27 E.g.: (4 th quadrant)     360°   or   2    P O

28. 28 Principal Angle & Principal Range Example: sin θ = 0.5 Principal range Restricting y= sinθ inside the principal range makes it a one-one function, i.e. so that a unique θ= sin -1 y exists

29. 29 E.g. 3(a): sin. Solve for θ if Basic angle, α = Since sin is positive, it is in the 1 st or 2 nd quadrant Therefore Hence,

30. 30 E.g. 3(b): cos. Solve for θ if Since cos is negative, it is in the 2nd or 3rd quadrant Basic angle, α = 36.870 o ThereforeHence,

31. 31 r y x A O P(x, y) By Pythagoras’ Theorem, (VI)3 Important Identities sin 2 A  cos 2 A  1 Since and, Note: sin 2 A  (sin A) 2 cos 2 A  (cos A) 2

32. 32 (1) sin 2 A + cos 2 A  1(2) tan 2 A +1  sec 2 A(3) 1 + cot 2 A  csc 2 A tan 2 x = (tan x) 2 (VI)3 Important Identities Dividing (1) throughout by cos 2 A, Dividing (1) throughout by sin 2 A,

33. 33 (VII)Important Formulae (1)Compound Angle Formulae

34. 34 E.g. 4: It is given that tan A = 3. Find, without using calculator, (i)the exact value of tan , given that tan (  + A) = 5; (ii)the exact value of tan , given that sin (  + A) = 2 cos (  – A) Solution: (i) Given tan (  + A)  5 and tan A  3,

35. 35 Solution: sin  + cos  tan A = 2(cos  + sin  tan A) sin  + 3cos  = 2(cos  + 3sin  ) (ii) Given sin (  + A) = 2 cos (  – A) & tan A  3, 5sin  = cos  tan  = sin  cos A + cos  sin A = 2[ cos  cos A + sin  sin A ] (Divide by cos A on both sides)

36. 36 (2)Double Angle Formulae (i) sin 2A = 2 sin A cos A (ii) cos 2A = cos 2 A – sin 2 A = 2 cos 2 A – 1 = 1 – 2 sin 2 A (iii) Proof:

37. 37 (3)Triple Angle Formulae: (i) cos 3A = 4 cos 3 A – 3 cos A Proof: cos 3A = cos (2A + A) = cos 2A cos A – sin 2A sin A = ( 2cos 2 A  1)cos A – (2sin A cos A)sin A = 2cos 3 A  cos A – 2cos A sin 2 A = 2cos 3 A  cos A – 2cos A(1  cos 2 A) = 4cos 3 A  3cos A

38. 38 (ii) sin 3A = 3 sin A – 4 sin 3 A Proof: sin 3A = sin (2A + A) = sin 2A cos A + cos 2A sin A = (2sin A cos A )cos A + (1 – 2sin 2 A)sin A = 2sin A(1 – sin 2 A) + sin A – 2sin 3 A = 3sin A – 4sin 3 A

39. 39 E.g. 5: Given sin 2 A  & A is obtuse, find, without using calculators, the values of (i) cos 4A (ii) sin ½ A Solution: Since sin 2 A  But A is obtuse,sin A = A 5 33 4

40. 40 (i) A 5 33 4

41. 41 (ii) cos A = 1 – 2sin 2 ( )= 1 – 2sin 2 ( )sin ( ) = i.e. lies in the 1 st quadrant. So

42. 42 E.g. 6: Prove the following identities: (i) Solution: (i)LHS = = RHS cos 2A = cos 2 A – sin 2 A = 2 cos 2 A – 1 = 1 – 2 sin 2 A Recall:

43. 43 (ii)LHS = = RHS E.g. 6: Prove the following identities: (ii) Solution:

44. 44 LHS  E.g. 6: Prove the following identities: (iii) Solution:

45. 45  RHS

46. 46 LHS =  RHS E.g. 6: Prove the following identities: (iv) Solution:

47. 47 (5)The Factor Formulae (Sum or difference of similar trigo. functions) Recall compound angles formulae: ….  ….   +  : ….  ….     :  +  :    :

48. 48 By letting X = A + B and Y = A – B, we obtain the factor formulae:

49. 49 Solution: (i)LHS = cos  + cos 3  + cos 5  = cos 3  (4 cos 2  – 1) = RHS = cos 3  [ 2(2 cos 2  – 1) + 1 ] = (cos 5  + cos  ) + cos 3  = 2cos 3  cos 2  + cos 3  = cos 3  [2cos2  + 1] E.g. 8: Show that (i)

50. 50 (ii)LHS = = RHS E.g. 8: Show that (ii) Soln:

51. 51 (iii) LHS = sin  + sin 3  + sin 5  + sin 7  = (sin 3  + sin  ) + (sin 7  + sin 5  ) = 2sin 2  cos  + 2sin 6  cos  = 2cos  [ sin 6  + sin 2  ] (iii) sin  + sin 3  + sin 5  + sin 7  = 16 sin  cos 2  cos 2 2  E.g. 8: Show that Soln:

52. 52 = 16 sin  cos 2  cos 2 2  = RHS = 4 cos  cos 2  sin 4  = 4 cos  cos 2  [ 2 sin 2  cos 2  ] = 8 cos  cos 2 2  sin 2  = 8 cos  cos 2 2  ( 2 sin  cos  )