# Class 11 Maths NCERT Solutions for Chapter 3 Trigonometric Functions Exercise 3.3

### Trigonometric Functions Exercise 3.3 Solutions

1. sin2 (Ï€/6) + cos2 (Ï€/3) – tan2 (Ï€/4) = -1/2

Solution

L.H.S. = sin2 (Ï€/6) + cos2 (Ï€/3) – tan2 (Ï€/4)

2. Prove that 2sin2 (Ï€/6) + cosec2 (7Ï€/6) cos2 (Ï€/3) = 3/2

Solution

3. Prove that cot2 (Ï€/6) + cosec (5Ï€/6) + 3 tan2 (Ï€/6) = 6
Solution

4. Prove that 2sin2 (3Ï€/4) + 2cos2 (Ï€/4) + 2sec2 (Ï€/3) = 10
Solution

= 1 + 1 + 8 = 10
= R.H.S

5. Find the value of:
(i) sin 75°
(ii) tan 15°
Solution

(i) sin 75° = sin (45° + 30°)
= sin 45° cos 30° + cos 45° sin 30°
[sin (x + y) = sin x cos y + cos x sin y]

(ii) tan 15° = tan(45° - 30°)

6. Prove that : cos(Ï€/4 - x) cos (Ï€/4 - y) - sin(Ï€/4 - x) sin(Ï€/4 - y) = sin(x + y)
Solution

7. Prove that : tan(Ï€/4 + x)/tan(Ï€/4 - x) = [(1+ tan x)/(1 - tan x)2
Solution

8. Prove that  [cos(Ï€ + x) cos (-x)]/[sin(Ï€ - x) cos(Ï€/2 + x)] = cot2 x
Solution

9. cos(3Ï€/2 + x) cos(2Ï€ + x) [cot(3Ï€/2 - x) + cot(2Ï€ + x)] = 1
Solution
L.H.S = cos(3Ï€/2 + x) cos(2Ï€ + x) [cot(3Ï€/2 - x) + cot(2Ï€ + x)]
= sin x cos x[tan x + cot x]

= 1 = R.H.S.

10. Prove that sin (n + 1)x sin (n + 2)x + cos (n + 1)x cos (n + 2)x = cos x
Solution
L.H.S. = sin (n + 1)x sin (n + 2)x + cos (n + 1)x cos (n + 2)x
= (1/2)[2sin(n + 1)x sin(n + 2)x + 2cos(n + 1)x cos(n + 2)x]

= cos(-x) = cos x = R.H.S.

11. Prove that cos(3Ï€/4 + x) - cos(3Ï€/4 - x) = -√2 sin x
Solution

= -2sin(3Ï€/4) sin x
= -2sin(Ï€ - Ï€/4) sin x
= -2 sin (Ï€/4) sin x
= -2 × 1/√2 × sin x
= -√2 sin x
= R.H.S.

12. Prove that sin2 6x – sin2 4x = sin 2x sin 10x
Solution
It is known that sin A + sin B = 2 sin (A + B)/2 cos (A - B)/2 , sin A - sin B = 2 cos (A + B)/2 sin (a - B)/2  .
∴ L.H.S. = sin2 6x - sin2 4x
= (sin 6x + sin 4x)(sin 6x - sin 4x) =
= (2 sin 5x cos x) (2 cos 5x sin x)
= (2 sin 5x cos 5x) (2 sin x cos x)
= sin 10x sin 2x
= R.H.S.

13. Prove that cos2 2x – cos2 6x = sin 4x sin 8x
Solution
It is known that cos A + cos B = [2 cos(A+B)/2][cos(A-B)/2], cos A - cos B = -[2 sin(A+B)/2][sin (A-B)/2].
∴ L.H.S. = cos2 2x – cos2 6x
= (cos 2x + cos 6x) (cos 2– 6x)

= [2 cos 4x cos(-2x)][-2 sin 4x sin(-2x)]
= [2 cos 4x cos 2x] [–2 sin 4(–sin 2x)]
= (2 sin 4x cos 4x) (2 sin 2x cos 2x)
= sin 8x sin 4x
= R.H.S.

14. Prove that sin 2x + 2sin 4x + sin 6x = 4cos2 x sin 4x
Solution
L.H.S = sin 2x + 2sin 4x + sin 6x
= [sin 2x + sin 6x]  + 2sin4x

[∵ sin A + sin B = 2sin (A + B)/2. cos(A - B)/2]
= 2 sin 4x cos (– 2x) + 2 sin 4x
= 2 sin 4x cos 2x + 2 sin 4x
= 2 sin 4x (cos 2x + 1)
= 2 sin 4x (2 cos2 x – 1 + 1)
= 2 sin 4x (2 cos2 x)
= 4cos2 x sin 4x
= R.H.S.

15. Prove that cot 4x (sin 5x + sin 3x) = cot x (sin 5x – sin 3x)
Solution
L.H.S = cot 4x (sin 5x + sin 3x)

= 2 cos 4x cos x
R.H.S. = cot x(sin 5x - sin 3x)

= (cos x/sin x) [2 cos 4x sin x]
= 2 cos 4x. cos x
= L.H.S = R.H.S

16. Prove that (cos 9x - cos 5x)/(sin 17x - sin 3x) = -sin 2x/cos 10x
Solution
It is known that
cos A - cos B = -2 sin(A+B)/2 .sin(A-B)/2, sin A - sin B = 2 cos(A+B)/2 .sin(A-B)/2

17. Prove that (sin 5x + sin 3x)/(cos 5x + cos 3x) = tan 4x
Solution
It is known that
cos A + cos B = 2cos(A+B)/2 .cos(A-B)/2, sin A + sin B = 2sin(A+B)/2 .cos(A-B)/2

20. Prove that (sin x - sin 3x)/(sin2 x - cos2 x) = 2 sin x
Solution
It is known that
sin A - sin B = 2cos(A+B)/2 .sin (A-B)/2, cos2A - sin2A = cos2A

= -2 ×(-sin x)
= 2sin x = R.H.S.

21. Prove that (cos 4x + cos 3x + cos 2x)/(sin 4x + sin 3x + sin 2x) = cot 3x
Solution

[∵ cos A + cos B = 2cos(A+B)/2 .cos(A-B)/2, sin A + sin B = 2sin(A+B)/2 .cos(A-B)/2]

22. Prove that cot x cot 2x – cot 2x cot 3x – cot 3x cot x = 1
Solution

L.H.S. = cot x cot 2x – cot 2x cot 3x – cot 3x cot x
= cot x cot 2x – cot 3x (cot 2x + cot x)
= cot x cot 2x – cot (2x) (cot 2x + cot x)

= cot x  cot 2x - (cot 2x cot x - )
= 1 = R.H.S.

23. Prove that tan 4x = [4 tan x(1 - tan2 x)]/(1 - 6 tan2 x + tan4 x)

Solution

It is known that tan 2A = 2 tan A/(1 - tan2 A
∴ L.H.S = tan 4x = tan 2(2x)

24. Prove that cos 4x = 1 – 8sinx cosx

Solution

L.H.S. = cos 4x
= cos 2(2x)
= 1 – 2 sin2 2x [cos 2A = 1 – 2 sin2 A]
= 1 – 2(2 sin x cos x)2 [sin2A = 2sin A cosA]
= 1 – 8 sin2x cos2x
= R.H.S.

25. Prove that: cos 6x = 32 cos6 x – 48 cos4 x + 18 cos2 – 1

Solution

L.H.S. = cos 6x
= cos 3(2x)
= 4 cos3 2x – 3 cos2x [cos 3A = 4 cos3 A – 3 cosA]
= 4 [(2 cos2 x – 1)3 – 3 (2 cos2 x – 1) [cos 2x = 2 cos2 x – 1]
= 4 [(2 cos2 x)3 – (1)3 – 3 (2 cos2 x)2 + 3 (2 cos2 x)] – 6cos2 x + 3
= 4 [8cos6x – 1 – 12 cos4x + 6 cos2x] – 6 cos2x + 3
= 32 cos6x – 4 – 48 cos4x + 24 cos2 x – 6 cos2x + 3
= 32 cos6x – 48 cos4x + 18 cos2x – 1
= R.H.S.