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

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

Trigonometric Functions Exercise 3.3 Solutions

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

Solution

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


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

Solution


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

4. Prove that 2sin2 (3π/4) + 2cos2 (π/4) + 2sec2 (π/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(π/4 - x) cos (π/4 - y) - sin(π/4 - x) sin(π/4 - y) = sin(x + y) 
Solution

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

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

9. cos(3π/2 + x) cos(2π + x) [cot(3π/2 - x) + cot(2π + x)] = 1
Solution
L.H.S = cos(3π/2 + x) cos(2π + x) [cot(3π/2 - x) + cot(2π + 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(3π/4 + x) - cos(3π/4 - x) = -√2 sin x 
Solution

= -2sin(3π/4) sin x
= -2sin(π - π/4) sin x 
= -2 sin (π/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.

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