1. POSITIVE ANGLE The angle measured in an anticlockwise direction. NEGATIVE ANGLE The angle measured in a clockwise direction. y x y x       TRIGONOMETRIC FUNCTION

2. y x 1 st QUADRANT 2 nd QUADRANT 3 rd QUADRANT 4 th QUADRANT SIN, COS, TAN ADD SUGAR TO COFFEE sin, cosine and tangent have positive values cosine has positive value tangent has positive value sin has positive value

3. y x REFERENCE ANGLE     1 st QUADRANT 2 nd QUADRANT 3 rd QUADRANT 4 th QUADRANT

4. EXAMPLE Calculate the followings: (a)sin 120  (b)cosine -120  y x (a) y x sin 120  = sin ( 180   120  ) = sin 60  cosine (  120  ) = cosine (  180  (  120  ) ) = cosine ( 60  ) =  cosine 60  = 0.8660 =  0.5 (b)

5. EXAMPLE Calculate the followings: (a)sin 845  (b)tan ( -860  ) y x (a) sin 845  = sin ( 845   720  ) = sin 125  = 0.8192 = sin (180   125  ) = sin 55  y x (b) tan (  860  ) = tan (  860   (  720  )) = tan (  140  ) = 0.8391 = tan (180   140  ) = tan ( 40  )

6. DEFINITION OF SINE COSINE AND TANGENT y x  P  m,n  1 1 0 n m 1  sin  = cos  = tan  =

7. EXAMPLE  P  -0.75,-0.9  1 1 0 1  y x  Q  -0.8,0.5   Given the points P (-0.75, -0.9) and Q (-0.8, 0.5) on a unit circle as shown in the diagram. Find the values of (a)cos  (b)sin  (c)tan  (d)tan  (e)cos  (a)cos  = - 0.8 (b)sin  = - 0.9 (c)tan  = 1.2 (d)tan  = - 0.625 (e)cos  = - 0.75 Do the exercises in SP 2

8. DEFINITIONS OF SECANT, COSECANT AND COTANGENT The signs of cot , cosec  and sec  follow the signs of tan , sin  and cos  in the respective quadrant.

9. EXAMPLE 1 1 0 y x  Q  -0.78,0.6  Given the points Q (-0.78, 0.6) is on a unit circle as shown in the diagram. Find the values of (a)cosec 143.13  (b)sec 143.13  (c)cot 143.13  (d)tan 143.13  (e)cos 143.13  143.13  cosec 143.13 (a) = = = == Determine the value of the reference angle

10. sec 143.13 (b) = = = == cot 143.13 (c) = = = == TRY (d) and (e) Do the exercises in SP 3

11. EXAMPLE Determine the value for each of the following trigonometric functions. (a)cosec 140  (b)cot - ⅔  y x  140  cosec 140  (a) = = = == y x  -⅔ -⅔  cot - ⅔  (b) = = = == Do the exercises in SP 4

12. 2 1 2 1 60  30  SPECIAL ANGLES 30 , 45 , 60  2

13. 1 1 45  22

14. EXAMPLE Without using a calculator, determine the value for each of the following trigonometric functions. (a)cot 240  (b)tan -225  y x (a) 240 

15. y x (b) -225  Do the exercises in SP 5

16. SOLVING TRIGONOMETRIC EQUATIONS EXAMPLE Solve the following trigonometric equations for 0     360 . (a)sin   - 0.6532(b)cos 2  = - 0.6824 (a) sin   - 0.6532 -ve shows the quadrant   sin -1 0.6532   40.78  // 40  47’ referenc e angle y x 40.78    180   40.78  // 180   40  47’, 360  - 40.78  // 360  - 40  47’.   220.78  // 220  47’, 319.22  // 319  13’.

17. (b) cos 2  = - 0.6824 0   2   720  2   cos -1 0.6824 2   46.97  // 46  59’ reference angle y x 46.97  2   180  - 46.97 , 180  + 46.97 , 2   133.03 , 226.97 , 360  + 133.03 , 360  + 226.97  586.97 . 493.03 ,   66.52 , 113.49 , 246.52 , 293.49 .

18. 12   ½x+12  192  EXAMPLE Solve 4 cos (½x + 12  ) + 1  3.626 for 0   x  360 . 4 cos (½x + 12  ) + 1  3.626 SOLUTION 4 cos (½x + 12  )  3.626 - 1 cos (½x + 12  )  ½x + 12   cos -1 0.6565 ½x + 12   ½x  48.97  - 12  ½x  36.97  x  2 ( 36.97  ) x  73.94  48.97  Do the exercises in SP 6

19. SOLUTION FOR SP 7, N0 1 (a) y x  5 13 12 0     180  (b) (i) sin  = (ii) tan  = (iii) cosec  = (iv) sec  = Do the exercises in SP 7 90   

20. EXAMPLE Solve 6 tan x – 3 cot x = 7 for 0   x  360 . SOLUTION 6 tan x – 3 cot x = 7 6 tan x – 3- 7 = 0 6 tan 2 x – 3- 7tan x = 0 6 tan 2 x – 7tan x -3 = 0

21. EXAMPLE Solve 2 cos x=sec x for 0   x  360 . SOLUTION