1. Trigonometric Identities

2. Examples Prove that (1 – cos A)(1 + sec A)  sin A tan A L.H.S.
(1 – cos A)(1 + sec A) = 1 + sec A – cos A –Cos A sec A
= 1 + sec A – cos A - 1
= sec A – cos A
= sin A tan A
= R.H.S.

3. Examples
Prove that cot A + tan A  sec A cosec A
L.H.S.
= R.H.S.

4. Examples
R.H.S.
= R.H.S.

5. Solving equations Solve 2 tan2 x – 7 sec x + 8 = 0 for 0  x  360
2 (sec2x – 1) – 7 sec x + 8 = 0
2 sec2x – 2 – 7 sec x + 8 = 0
2 sec2x – 7 sec x + 6 = 0
(2 sec x – 3)(sec x – 2)= 0
sec x = 3/ or sec x = 2
cos x = 2/ or cos x = ½
x = or x = 60
or: x = 360 – or x =
complete solution: x = 48.2 or 60 or 300 or 311.8

6. Solving equations Solve 2 cos x = cot x for 0  x  360
2 cos x = cos x/ sin x
2 cos x sin x = cos x
2 cos x sin x – cos x= 0
cos x(2 sin x – 1)= 0
cos x = 0 or sin x = ½
cos x = 0  x = 90 or 270
sin x = ½  x = 30 or 330
complete solution: x = 30 or 90 or 270 or 30

7. Solving equations Solve 3 cot2 x – 10 cot x + 3 = 0 for 0  x  2
cot x = 1/3 or cot x = 3
 tan x = 3 or tan x = 1/3
tan x = 3  x = 1.24c or 4.39c
tan x = 1/3  x = 0.32c or 3.46c
complete solution: x = 0.32c or 1.24c or 3.46c or 4.39c

8. Solving equations Solve 5 cot2 x – 2 cosec x + 2 = 0 for 0  x  2
5(cosec2 x – 1) – 2 cosec x + 2 = 0
5cosec2 x – 5 – 2 cosec x + 2 = 0
5cosec2 x – 2 cosec x - 3 = 0
sin x = -5/3 not possible or sin x = 1  x = /2

9. Additional formulae sin (A + B) = sin A cos B + sin B cos A
cos (A + B) = cos A cos B - sin A sin B
cos (A - B) = cos A cos B + sin A sin B

10. Examples Find the exact value of sin 75
sin (A + B) = sin A cos B + sin B cos A
sin ( ) = sin 30 cos 45 + sin 45 cos 30

11. Examples Express cos (x + /3) in terms of cos x and sin x
cos (A + B) = cos A cos B - sin A sin B
cos (x + /3) = cos x cos /3 - sin /3 sin x

12. Examples
L.H.S.
= R.H.S.

13. Double angle formulae sin (A + B) = sin A cos B + sin B cos A
sin (A + A) = sin A cos A + sin A cos A
sin 2A = 2 sin A cos A
cos (A + B) = cos A cos B - sin A sin B
cos (A + A) = cos A cos A- sin A sin A
cos (A + A) = cos2A - sin2A
cos 2A = cos2A - sin2A
cos 2A = 2cos2A - 1
cos 2A = 1 – 2sin2A

14. Double angle formulae

15. Examples Given that cos A = 2/3, find the exact value of cos 2A.
cos 2A = 2cos2A - 1
Given that sin A = ¼ , find the exact value of sin 2A.
sin 2A = 2 sin A cos A A 4 1
15

16. Solving equations Solve cos 2A + 3 + 4 cos A = 0 for 0  x  2

17. Solving equations Solve sin 2A = sin A for -   x  
=2sin A cos A = sin A
=2 sin A cos A – sin A = 0
= sin A(2 cos A – 1) = 0
 sin A = 0 or cos A = ½
sin A = 0  A = -  or 0 or 
cos A = ½  A = - /3 or /3
Complete solution: A = -  or - /3 or 0 or /3 or 

18. Solving equations Solve tan 2A + 5 tan A = 0 for 0 x  2
tan A = 0  A = 0 or  or 2
7 – 5tan2 A = 0 tan A =  7/5  A = 0.97 , 2.27, 4.01 or 5.41c
Complete solution: A= 0.97 , 2.27, 4.01, 5.41c 0,  or 2

19. Harmonic form If a and b are positive
a sin x + b cos x can be written in the form R sin( x +  )
a sin x - b cos x can be written in the form R sin( x -  )
a cos x + b sin x can be written in the form R cos( x -  )
a cos x - b sin x can be written in the form R cos( x +  )

20. Examples Express 3 cos x + 4 sin x in the form R cos( x -  )
R cos( x -  ) = R cos x cos  + R sin x sin 
3 cos x + 4 sin x = R cos x cos  + R sin x sin 
R cos  = 3 [1] R sin  = [2]
[1]2 + [2]2 : R2 sin2 x + R2 cos2 x =
R2(sin2 x + cos2 x ) =
R2= = 25  R = 5
[2]  [1]: tan  = 4/3   = 53.1
3 cos x + 4 sin x = 5 cos( x  )

21. Examples Express 12 cos x + 5 sin x in the form R sin( x +  )
R sin( x +  ) = R sin x cos  + R cos x sin 
12 cos x + 5 sin x = R sin x cos  + R cos x sin 
R cos  = [1] R sin  = [2]
[1]2 + [2]2 : R2 cos2 x + R2 sin2 x =
R2(cos2 x + sin2 x ) =
R2= = 169  R = 13
[2]  [1]: tan  = 5/12   = 22.6
12 cos x + 5 sin x = 13 sin( x  )

22. Examples Express cos x - 3 sin x in the form R cos( x +  )
R cos( x +  ) = R cos x cos  - R sin x sin 
cos x - 3 sin x = R cos x cos  - R sin x sin 
R cos  = [1] R sin  =  [2]
[1]2 + [2]2 : R2 cos2 x + R2 sin2 x = 12 + (3 ) 2
R2(cos2 x + sin2 x ) =
R2= = 4  R = 2
[2]  [1]: tan  = 3   = 60
cos x + 3 sin x = 2 cos( x + 60 )

23. Solving equations Solve 7 sin x + 3 cos x = 6 for 0 x  2
R sin( x +  ) = R sin x cos  + R cos x sin 
7 sin x + 3 cos x = R sin x cos  + R cos x sin 
R cos  = 7 [1] R sin  = [2]
R2 =  R = 7.62
[2]  [1]: tan  = 3/7   = 0.405c (Radians)
7 sin x + 3 cos x = 7.62 sin( x )
7.62 sin( x ) = 6  x = sin-1(6/7.62)
x = or
x = 0.502c or c