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

Chapter 3 Trigonometric Functions Exercise 3.3 NCERT Solutions for Class 11 is helpful in clearing your concepts about the chapter and solving your doubts easily. Class 11 Maths NCERT Solutions will be useful in building your basic concepts and preparing for higher classes. The NCERT solutions provided here are prepared by Studyrankers subject matter experts that will guide whenever you got problem while solving a question.

1. Prove that:
Sin2 Ï€/6 + cos2 Ï€/3 –tan2 Ï€/4 = -1/2h

L.HS. = sin2 Ï€/6 + cos2 Ï€/3 – tan2 Ï€/4
= (sin Ï€/6)2 + (cos Ï€/3)2 – (tan Ï€/4)2
=(1/2)2 + (1/2)2 – 12
[sin Ï€/6 = ½, cos Ï€/3 = 1/2, tan Ï€/4 = 1]
= 1/4 + 3/4 - 1 = 1/2 - 1 = -1/2 = R.H.S

2. 2sin2 Ï€/6 + cosec2 7 Ï€/6 cos2 Ï€/3 = 3/2

L.H.S = 2sin2 Ï€/6 + cosec2 Ï€/3 cos2 7Ï€/6
= 2(sin Ï€/6)2 + (cosec 7Ï€/6)2 (cos Ï€/3)2
= 2.(1/2)2 + [cosec(Ï€ + Ï€/6)]2 (1/2)2
= 2 × 1/4+ (-cosec Ï€/6)2 (1/4)
[cosec (Ï€ +Î¸ ) = - cosec]
= 1/2 + (2)2 1/4= 1/2 + 1 = 3/2 = R.H.S

3. CotÏ€/6 + cosec 5Ï€/6 + 3tan2 Ï€/6 = 6

L.H.S = cot2 Ï€/6 + cosec 5Ï€/6 + 3tan2 Ï€/6
= (cot Ï€/6)2 + cosec(Ï€ – Ï€/6) + 3(tan Ï€/6)2
=(√3)2 + cosec Ï€/6 + 3 (1/√3)2
= 3 + 2 + 3 × 1/3 = 3 + 2 + 1 = 6 = R.H.S.

4. 2sin2 3Ï€/4 + 2cos2 Ï€/4 + 2sec2 Ï€/3 = 10

L.H.S = 2sin2 3Ï€/4 + 2cos2 Ï€/4 + 2sec2 Ï€/3
= 2(sin 3Ï€/4)2 + 2(cos Ï€/4)2 + 2(sec Ï€/3)2
= 2(sin [Ï€ – Ï€/4])2 + 2 × (1/2)2 + 2(22)
= 2(sin Ï€/4)2 + 2 × 1/2 + 8
= 2 × (1/2)2 + 1 + 8 = 2 × 1/2 + 1 +8
= 1 + 1 + 8 = 10 = R.H.S

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

(i) We have
sin 75° = sin(45° + 30°)
=sin45° cos 30° + cos 45° sin30°
= (1/√2)(√3/2) + (1/√2)(1/2)
= √3/2√2 + 1/2√2 = (√3 + 1)/2√2
= {(√3 + 1)√2}/4 = 1/4 (√2 + √6)
(ii) similar tan 15° = tan(45° - 30°)
= (tan 45° - tan 30°)/(1 + tan 45° tan 30°)
= (1 - 1/√3)/{1 + (1)(1/√3) = {(√3 -1)/√3}/{(√3 + 1)/√3}
= (√3 - 1)/(√3 + 1) = (√3 - 1)/√3 + 1) × (√3 - 1)/(√3 - 1)
= {(√3)2 - 2√3 + (1)2}/(3 - 1)
= (3 - 2√3 + 1)/2 = (4 -2√3)/2
={2(2 - √3)}/2 = 2 -√3

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

L.H.S = cos (Ï€/4 – x) cos(Ï€/4 – y) – sin(Ï€/4 – x) sin(Ï€/4 – y)
Let Ï€/4 – x = A, Ï€/4 – y = B
L.H.S = cos A cosB – sin A sin B
= cos (A + B)
= cos(Ï€/4 – x + Ï€/4 – y)
= cos(Ï€/2 – (x + y))
= sin( x + y) = R.H.S

7. Prove that {tan(Ï€/4 + x)}/{tan(Ï€/4 – x)} = {(1 + tan x)/(1 – tan x)}2

L.H.S = {tan(Ï€/4 + x)}/{tan(Ï€/4 – x)} = {(1 + tan x)/(1 – tan x)}/{(1 – tan x)/(1 + tan x)}
= {(1 + tan x)/( 1 – tan x)}2 = R.H.S

8. Prove that {cos(Ï€ + x)cos(-x)}/{sin( Ï€ -x)cos(Ï€/2 + x)} = cot2x

L.H.S = {cos( Ï€ + x)cos(-x)}/{sin(Ï€ -x)cos(Ï€/2 + x)}
= {(-cosx)cosx/{sinx(-sinx)} = -cos2x/-sin2x
= cot2x = R.H.S

9. Prove that,
Cos(3Ï€/2 + x)cos(2Ï€ + x)[cot(3Ï€/2 – x) + cos(2Ï€ + x)] = 1

L.H.S = cos(3Ï€/2 + x) cos(2Ï€ + x)
[cot(3Ï€/2 – x) + cot(2Ï€ + x)
Now,
Cos(3Ï€/2 + x) = sin x, cos(2Ï€ + x) = cos x and
Cot(3Ï€/2 - x) = tanx, cot(2Ï€ + x) = cot x
L.H.S = sin x. cos x[tan x + cot x]
= sin x.cos x[sinx/cos x + cosx/sinx]
= sin x. cos x[(sin2 x + cos2x)/cosxsinx]
= (sin x. cosx)1/cosxsinx = 1
[∵ sin2x + cos2x = 1]

10. Prove that sin( n + 1)x sin( n + 2)x + cos(n + 1)x cos(n + 1)x cos(n + 2)x = cosx.

L.H.S = sin( n+ 1)x sin(n + 2)x + cos( n+1)x cos( n + 2)x
Write A = (n + 1)x
B = ( n + 2)x
L.H.S = sin A sin B + cos A cos B
= cos (A – B)
=cos ((n + 1)x – (n + 2)x)
= cos( x – 2x)
= cos(-x) [ cos(-Î¸) = cosÎ¸ ∀ Î¸ ∈ R]
= cos x = R.H.S

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

L.H.S = cos(3Ï€/4 + x) – cos(3Ï€/4 – x),
= -2sin ((3Ï€/4 + x + 3Ï€/4 - x)/2)sin((3Ï€/4 + x – 3Ï€/4 + x)/2)
= - 2sin ((3Ï€/2))/2 sin(2x/2) = -2sin 3Ï€/4 sin x
= -2sin(Ï€ - Ï€/4)sin x = -2 sin Ï€/4 sin x
= -2 × 1/√2 sin x = - √2sin x = R.H.S

12. Prove that
Sin26x –sin24x = sin2x sin10x

L.H.S = sin2 6x – sin24x
= (1 – cos 12x)/2 – (1 – cos 8x)/2
= (cos 8x – cos12x)/2
= (2 sin 10x. sin 2x)/2
=sin 10x . sin 2x = R.H.S Proved

13. Prove that
Cos22x – cos26x = sin 4x sin 8x

L.H.S = cos22x – cos26x
= (cos 2x + cos 6x)(cos 2x – cos 6x)

= [(2cos 4x cos(-2x))(-2 sin 4x sin (-2x))]
=(2 cos 4x xos 2x)(2 sin 4x sin 2x)
[ cos(-Î¸) = Î¸ , sin(-Î¸) = -sinÎ¸]
(2 sin 4x cos 4x)(2 sin 2x cos 2x)
= sin 8x sin 4x = R.H.S
[ sin 2A = 2 sin A cos A]

14. Prove that
Sin26x –sin24x = sin2x sin10x

L.H.S = sin2x + 2 sin 4x + sin 6x
= (sin 6x + sin 2x) + 2sin4x
(sin C + sin D = 2 sin (C + D)/2 cos (C – D)/2)
= 2sin (6x + 2x)/2 cos (6x – 2x)/2 + 2 sin 4x
= 2 sin 4x [cos 2x + 1]
[ cos 2x + 1 = 2 cos2x – 1 + 1 = 2 cos2x]
= 2 sin 4x cos2x
= 4 sin 4x cos2x = R.H.S.

15. Prove that,
Cot 4x[sin 5x + sin 3x] = cotx(sin 5x – sin 3x)

L.H.S = cot4x[sin 5x + sin 3x]
= cot 4x[2sin 4x cos x]
=cos4x/sin4x × 2sin 4x cosx = 2cos x cos 4x
= cosx/sinx × (sin5x – sin 3x)
[ ∵ 2cos A sin B = sin(A + B) – sin (A – B)]
= cotx (sin 5x – sin 3x) = R.H.S

16. Prove that (cos 9x – cox 5x)/(sin 17x – sin 3x) = - sin 2x/cos 10x

L.H.S = (cos 9x – cox 5x)/(sin 17x – sin 3x)
= {-2sin (9x + 5x/2) sin(9x – 5x/2)}/{2cos(17x + 3x/2)sin(17x – 3x)/2}

= (-sin 7x .sin 2x)/(cos 10x. sin 7x) = - sin 2x/cos 10x = R.H.S

17. Prove that (sin5x + sin 3x)/(cos5x + cos 3x) = tan 4x

L.H.S
= (sin5x + sin 3x)/(cos5x + cos 3x) = (2sin(5x + 3x/2)cos(5x – 3x/2))/(2cos(5x +3x/2)cos(5x -3x/2))
= sin4x/cos4x = tan 4x = R.H.S

18. Prove that (sin x – sin y)/(cos x + cos y) = tan (x – y)/2

L.H.S = (sin x – sin y)/(cos x + cos y) = (2cos (x + y/2)sin(x – y/2))/(2cos(x + y/2)cos(x – y/2))
= sin(x – y/2)/cos(x – y/2) = tan (x – y)/2 = R.H.S

19. Prove that (sin x + sin 3x)/(cos x + cos 3x) = tan2x

L.H.S = (sin x + sin 3x)/(cos x + cos 3x) = (sin 3x + sin x)/(cos3x + cosx)
= {2sin(3x + x/2) cos(3x – x/2)}/{2cos(3x + x/2)cos(3x – x/2)} = (sin2xcosx)/(cos2xcosx)
=sin2x/cos2x = tan 2x

20. Prove that (sinx – sin3x)/sin2x – cos2x = 2sin x

L.H.S = sinx – sin 3x)/sin2x – cos2x = -(sin 3x – sin x)/-(cos2x – sin2x)
= (sin 3x – sin x)/cos2x – sin2x = {2cos(3x + x/2)sin(3x –x/2)}/cos2x
= (2cos2x × sinx)/cos2x [ ∵ cos 2x = cos2x – sin2x]
= 2sin x

21. Prove that (cos 4x + cos 3x + cos 2x)/(sin4x + sin 3x + sin 2x) = cot 3x

L.H.S = (cos4x + cos3x + cos2x)/(sin4x + sin 3x + sin 2x)
= (cos 4x + cos 2x + cos 3x)/(sin 4x + sin2x + sin 3x)
= {2cos (4x + 2x/2)cos(4x – 2x/2) + cos3x}/{2 sin(4x +2x/2)cos(4x – 2x/2) + sin 3x}
= (2cos3xcosx + cos 3x)/(2sin 3x cos x + sin 3x)
= cos3x(2cosx + 1)/sin3x(2cosx + 1)
= cos3x/sin 3x = cot 3x = R.H.S

22. cot x cot 2x – cot2x cot 3x – cot 3x cot x = 1

L.H.S cot x cot 2x – cot 2x cot 3x – cot 3x cot x
We have 3x = x + 2x
Cot 3x = cot(x + 2x)
= (cot x cot 2x – 1)/(cot x + cot 2x)
By cross multiplication
cot 3x(cot x + cot 2x) = cot x cot 2x – 1
cot x cot 3x + cot 2x cot 3x = cos x cot 2x – 1
∴ cos x cot 2x – cot 2x cot3x – cot 3x cot x = 1

23. Prove that tan 4x = (4tan x( 1 – tan2x)/(1- 6tan2 x + tan4x)

L.H.S = tan 4x = tan2(2x)
= 2tan2x/1 - tan22x = 2.(2tan x/1 – tan2x)/1 – (2tan x/1 – tan2x)2

4tan x/(1 – tan2x) ( 1 – 6tan2x + tan4x) = RHS

24. Prove that cos 4x = 1 - 8 sin2x cos2x

L.H.S = cos4x = cos2(2x)
= 2cos22x – 1 [cos 2A = 2cos2 A -1]
= 2[2cos2 x – 1]2 – 1
= 2[4cos4x – 4cos2 x + 1] – 1
= 8cos4x – 8cos2x + 2 – 1
= 1 – 8cos2x(1 – cos2x)
= 1 – 8 cos2x sin2x = R.H.S

25. Prove that
Cos6x = 32cos6x – 48cosx4 + 18cos2x – 1