Class 11 NCERT Solutions- Chapter 3 Trigonometric Function – Exercise 3.3 | Set 2

Last Updated : 07 Apr, 2021

Question 15: cot 4x (sin 5x + sin 3x) = cot x (sin 5x – sin 3x)

Solution:

Taking LHS in consideration, we get

= cot 4x (sin 5x + sin 3x)

= $\frac{cos \hspace{0.1cm}4x}{sin \hspace{0.1cm}4x}$ (sin 5x + sin 3x)

Using the identity,

sin A + sin B = 2 sin $\mathbf{(\frac{A+B}{2})}$ cos $\mathbf{(\frac{A-B}{2})}$

= $\frac{cos \hspace{0.1cm}4x}{sin \hspace{0.1cm}4x}$ (2 sin $(\frac{5x+3x}{2})$ cos $(\frac{5x-3x}{2})$ )

= $\frac{cos \hspace{0.1cm}4x}{sin \hspace{0.1cm}4x}$ (2 sin $(\frac{8x}{2})$ cos $(\frac{2x}{2})$ )

= $\frac{cos \hspace{0.1cm}4x}{sin \hspace{0.1cm}4x}$ (2 sin 4x cos x)

= 2 cos 4x cos x

Now, taking RHS in consideration, we get

= cot x (sin 5x – sin 3x)

= $\frac{cos \hspace{0.1cm}x}{sin \hspace{0.1cm}x}$ (sin 5x – sin 3x)

Using the identity,

sin A – sin B = 2 cos $\mathbf{(\frac{A+B}{2})}$ sin $\mathbf{(\frac{A-B}{2})}$

= $\frac{cos \hspace{0.1cm}x}{sin \hspace{0.1cm}x}$ (2 cos $(\frac{5x+3x}{2})$ sin $(\frac{5x-3x}{2})$ )

= $\frac{cos \hspace{0.1cm}x}{sin \hspace{0.1cm}x}$ (2 cos $(\frac{2x}{2})$ sin $(\frac{2x}{2})$ )

= $\frac{cos \hspace{0.1cm}x}{sin \hspace{0.1cm}x}$ (2 cos 4x sin x)

= 2 cos 4x cos x

Hence, LHS = RHS

Question 16: $\frac{cos \hspace{0.1cm}9x - cos\hspace{0.1cm} 5x}{sin \hspace{0.1cm}17x - sin \hspace{0.1cm}3x} = -\frac{sin \hspace{0.1cm}2x}{cos \hspace{0.1cm}10x}$

Solution:

Taking LHS in consideration, we get

= $\frac{cos \hspace{0.1cm}9x - cos\hspace{0.1cm} 5x}{sin \hspace{0.1cm}17x - sin \hspace{0.1cm}3x}$

Using the identity,

cos A – cos B = 2 sin $\mathbf{(\frac{A+B}{2})}$ sin $\mathbf{(\frac{B-A}{2})}$

sin A – sin B = 2 cos $\mathbf{(\frac{A+B}{2})}$ sin $\mathbf{(\frac{A-B}{2})}$

= $\frac{2 sin \hspace{0.1cm}(\frac{9x+5x}{2})\hspace{0.1cm} sin\hspace{0.1cm}(\frac{5x-9x}{2})}{2 cos \hspace{0.1cm}(\frac{17x+3x}{2}) \hspace{0.1cm} sin \hspace{0.1cm}(\frac{17x-3x}{2})}$

= $\frac{2 sin \hspace{0.1cm}(\frac{14x}{2})\hspace{0.1cm} sin\hspace{0.1cm}(\frac{-4x}{2})}{2 cos \hspace{0.1cm}(\frac{20x}{2}) \hspace{0.1cm} sin \hspace{0.1cm}(\frac{14x}{2})}$

= $\frac{2 sin \hspace{0.1cm}(7x)\hspace{0.1cm} sin\hspace{0.1cm}(-2x)}{2 cos \hspace{0.1cm}(10x) \hspace{0.1cm} sin \hspace{0.1cm}(7x)}$

= $-\frac{sin\hspace{0.1cm}(2x)}{cos \hspace{0.1cm}(10x)}$

Hence, LHS = RHS

Question 17: $\frac{sin \hspace{0.1cm}5x + sin \hspace{0.1cm}3x}{cos \hspace{0.1cm}5x + cos \hspace{0.1cm}3x}$ = tan 4x

Solution:

Taking LHS in consideration, we get

= $\frac{sin \hspace{0.1cm}5x + sin \hspace{0.1cm}3x}{cos \hspace{0.1cm}5x + cos \hspace{0.1cm}3x}$

Using the identity,

sin A + sin B = 2 sin $\mathbf{(\frac{A+B}{2})}$ cos $\mathbf{(\frac{A-B}{2})}$

cos A + cos B = 2 cos $\mathbf{(\frac{A+B}{2})}$ cos $\mathbf{(\frac{A-B}{2})}$

= $\frac{2 sin (\frac{5x+3x}{2}) cos (\frac{5x-3x}{2})}{2 cos (\frac{5x+3x}{2}) cos (\frac{5x-3x}{2})}$

= $\frac{2 sin (\frac{8x}{2}) cos (\frac{2x}{2})}{2 cos (\frac{8x}{2}) cos (\frac{2x}{2})}$

= $\frac{sin (4x) cos (x)}{cos (4x) cos (x)}$

= $\frac{sin (4x)}{cos (4x)}$

= tan 4x

Hence, LHS = RHS

Question 18: $\frac{sin\hspace{0.1cm} x - sin\hspace{0.1cm} y}{cos\hspace{0.1cm} x + cos\hspace{0.1cm} y} = tan(\frac{x-y}{2})$

Solution:

Taking LHS in consideration, we get

= $\frac{sin\hspace{0.1cm} x - sin\hspace{0.1cm} y}{cos\hspace{0.1cm} x + cos\hspace{0.1cm} y}$

Using the identity,

sin A – sin B = 2 cos $\mathbf{(\frac{A+B}{2})}$ sin $\mathbf{(\frac{A-B}{2})}$

cos A + cos B = 2 cos $\mathbf{(\frac{A+B}{2})}$ cos $\mathbf{(\frac{A-B}{2})}$

= $\frac{2 cos (\frac{x+y}{2}) sin (\frac{x-y}{2})}{2 cos (\frac{x+y}{2}) cos (\frac{x-y}{2})}$

= $\frac{sin (\frac{x-y}{2})}{cos (\frac{x-y}{2})}$

= $tan (\frac{x-y}{2})$

Hence, LHS = RHS

Question 19: $\frac{sin\hspace{0.1cm} x + sin \hspace{0.1cm}3x}{cos \hspace{0.1cm}x + cos \hspace{0.1cm}3x}$ = tan 2x

Solution:

Taking LHS in consideration, we get

= $\frac{sin \hspace{0.1cm}x + sin \hspace{0.1cm}3x}{cos \hspace{0.1cm}x + cos \hspace{0.1cm}3x}$

Using the identity,

sin A + sin B = 2 sin $\mathbf{(\frac{A+B}{2})}$ cos $\mathbf{(\frac{A-B}{2})}$

cos A + cos B = 2 cos $\mathbf{(\frac{A+B}{2})}$ cos $\mathbf{(\frac{A-B}{2})}$

= $\frac{2 sin (\frac{x+3x}{2}) cos (\frac{x-3x}{2})}{2 cos (\frac{x+3x}{2}) cos (\frac{x-3x}{2})}$

= $\frac{2 sin (\frac{4x}{2}) cos (\frac{-2x}{2})}{2 cos (\frac{4x}{2}) cos (\frac{-2x}{2})}$

= $\frac{sin (2x) cos (-x)}{cos (2x) cos (-x)}$

= $\frac{sin (2x) cos (x)}{cos (2x) cos (x)}$

= $\frac{sin (2x)}{cos (2x)}$

= tan 2x

Hence, LHS = RHS

Question 20: $\frac{sin \hspace{0.1cm}x - sin \hspace{0.1cm}3x}{sin^2 x - cos^2 x}$ = 2 sin x

Solution:

Taking LHS in consideration, we get

= $\frac{sin \hspace{0.1cm}x - sin \hspace{0.1cm}3x}{sin^2 x - cos^2 x}$

Using the identity,

sin A – sin B = 2 cos $\mathbf{(\frac{A+B}{2})}$ sin $\mathbf{(\frac{A-B}{2})}$

cos 2θ = cos² θ – sin² θ

= $\frac{2 cos (\frac{x+3x}{2}) sin (\frac{x-3x}{2})}{- cos 2x}$

= $\frac{2 cos (\frac{4x}{2}) sin (\frac{-2x}{2})}{- cos 2x}$

= $\frac{2 cos (2x) sin (-x)}{- cos 2x}$

= $\frac{- 2 sin (x)}{- 1}$

= 2 sin (x)

Hence, LHS = RHS

Question 21: $\frac{cos\hspace{0.1cm} 4x + cos\hspace{0.1cm} 3x + cos \hspace{0.1cm}2x}{sin \hspace{0.1cm}4x + sin \hspace{0.1cm}3x + sin\hspace{0.1cm} 2x}$ = cot 3x

Solution:

Taking LHS in consideration, we get

= $\frac{(cos\hspace{0.1cm} 4x + cos\hspace{0.1cm} 2x) + cos \hspace{0.1cm}3x}{(sin \hspace{0.1cm}4x + sin \hspace{0.1cm}2x) + sin\hspace{0.1cm} 3x}$

Using the identity,

sin A + sin B = 2 sin $\mathbf{(\frac{A+B}{2})}$ cos $\mathbf{(\frac{A-B}{2})}$

cos A + cos B = 2 cos $\mathbf{(\frac{A+B}{2})}$ cos $\mathbf{(\frac{A-B}{2})}$

= $\frac{(2 cos (\frac{4x+2x}{2}) cos (\frac{4x-2x}{2})) + cos \hspace{0.1cm}3x}{(2 sin (\frac{4x+2x}{2}) cos (\frac{4x-2x}{2})) + sin\hspace{0.1cm} 3x}$

= $\frac{(2 cos (\frac{6x}{2}) cos (\frac{2x}{2})) + cos \hspace{0.1cm}3x}{(2 sin (\frac{6x}{2}) cos (\frac{2x}{2})) + sin\hspace{0.1cm} 3x}$

= $\frac{2 cos (3x) cos (x)+ cos \hspace{0.1cm}3x}{2 sin (3x) cos (x) + sin\hspace{0.1cm} 3x}$

Taking common, we have

= $\frac{cos (3x) [2 cos (x)+1]}{sin (3x)[2cos (x)+1]}$

= $\frac{cos \hspace{0.1cm}3x}{sin \hspace{0.1cm}3x}$

= cot 3x

Hence, LHS = RHS

Question 22: cot x cot 2x – cot 2x cot 3x – cot 3x cot x = 1

Solution:

Taking LHS in consideration, we get

= 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+x) (cot 2x + cot x)

Using the identity,

cot(A+B) = $\mathbf{\frac{cot A cot B - 1}{cot A + cot B}}$

= cot x cot 2x – $[\frac{cot 2x \hspace{0.1cm}cot x - 1}{cot 2x + cot x}]$ (cot 2x + cot x)

= cot x cot 2x – [cot 2x cot x – 1]

= cot x cot 2x – cot 2x cot x – 1

= 1

Hence, LHS = RHS

Question 23: tan 4x = $\frac{4\hspace{0.1cm}tan \hspace{0.1cm}x\hspace{0.1cm}(1-tan^2x)}{1-6tan^2x + tan^4x}$

Solution:

Taking LHS in consideration, we get

tan 4x = tan 2(2x)

Using the identity,

tan 2θ = $\mathbf{\frac{2 tan \theta}{1-tan^2\theta}}$

= $\frac{2 \hspace{0.1cm}tan (2x)}{1-tan^2(2x)}$

Again using the same identity, we get

= $\frac{2 (\frac{2 tan x}{1-tan^2x})}{1-(\frac{2 tan x}{1-tan^2x})^2}$

= $\frac{\frac{4 tan x}{1-tan^2x}}{1-(\frac{4 tan^2 x}{(1-tan^2x)^2})}$

= $\frac{\frac{4 tan x}{1-tan^2x}}{\frac{((1-tan^2x)^2 - 4 tan^2 x)}{(1-tan^2x)^2}}$

= $\frac{4 tan x (1-tan^2x)}{(1-tan^2x)^2 - 4 tan^2 x}$

= $\frac{4 \hspace{0.1cm}tan x (1-tan^2x)}{1+tan^4x-2tan^2x - 4 tan^2 x}$

= $\frac{4 \hspace{0.1cm}tan x (1-tan^2x)}{1-6tan^2x +tan^4x}$

Hence, LHS = RHS

Question 24: cos 4x = 1 – 8sin²x cos²x

Solution:

Taking LHS in consideration, we get

cos 4x = cos 2 (2x)

Using the identity,

cos 2θ = 1 – 2sin² θ

= 1 – 2sin² (2x)

= 1 – 2(2sin x cos x)²(As, sin 2θ = 2 sin θ cos θ)

= 1 – 2(4sin²x cos² x)

= 1 – 8sin² x cos² x

Hence, LHS = RHS

Question 25: cos 6x = 32 cos⁶x – 48cos⁴x + 18 cos²x – 1

Solution:

Taking LHS in consideration, we get

cos 6x = cos 3 (2x)

Using the identity,

cos 3θ = 4 cos³ θ – 3 cos θ

= 4 cos³ (2x) – 3 cos (2x)

= 4 cos³ (2x) – 3 cos (2x)

Now, Using the identity cos 2θ = 2cos² θ – 1

= 4 (2cos² x – 1)³ – 3 (2cos² x – 1)

Using algebraic identity,

(a-b)³ = a³ + b³ – 3a²b + 3ab²

= 4 [(2 cos² x) ³ – (1)³ – 3 (2 cos² x) ² + 3 (2 cos² x)(1)²] – 6cos² x + 3

= 4 [8cos⁶x – 1 – 12 cos⁴x + 6 cos²x] – 6 cos2x + 3

= 32 cos⁶x – 4 – 48 cos^4x + 24 cos² x – 6 cos²x + 3

= 32 cos⁶x – 48 cos⁴x + 18 cos²x – 1

Hence, LHS = RHS

Suggest improvement

Class 11 NCERT Solutions - Chapter 14 Mathematical Reasoning - Exercise 14.3

Class 11 RD Sharma Solutions - Chapter 17 Combinations- Exercise 17.3

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Class 11 NCERT Solutions- Chapter 3 Trigonometric Function – Exercise 3.3 | Set 2

Question 15: cot 4x (sin 5x + sin 3x) = cot x (sin 5x – sin 3x)

Question 16:

Question 17:= tan 4x

Question 18:

Question 19:= tan 2x

Question 20:= 2 sin x

Question 21:= cot 3x

Question 22: cot x cot 2x – cot 2x cot 3x – cot 3x cot x = 1

Question 23: tan 4x =

Question 24: cos 4x = 1 – 8sin2x cos2x

Question 25: cos 6x = 32 cos6x – 48cos4x + 18 cos2x – 1

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Question 16: $\frac{cos \hspace{0.1cm}9x - cos\hspace{0.1cm} 5x}{sin \hspace{0.1cm}17x - sin \hspace{0.1cm}3x} = -\frac{sin \hspace{0.1cm}2x}{cos \hspace{0.1cm}10x}$

Question 17: $\frac{sin \hspace{0.1cm}5x + sin \hspace{0.1cm}3x}{cos \hspace{0.1cm}5x + cos \hspace{0.1cm}3x}$ = tan 4x

Question 18: $\frac{sin\hspace{0.1cm} x - sin\hspace{0.1cm} y}{cos\hspace{0.1cm} x + cos\hspace{0.1cm} y} = tan(\frac{x-y}{2})$

Question 19: $\frac{sin\hspace{0.1cm} x + sin \hspace{0.1cm}3x}{cos \hspace{0.1cm}x + cos \hspace{0.1cm}3x}$ = tan 2x

Question 20: $\frac{sin \hspace{0.1cm}x - sin \hspace{0.1cm}3x}{sin^2 x - cos^2 x}$ = 2 sin x

Question 21: $\frac{cos\hspace{0.1cm} 4x + cos\hspace{0.1cm} 3x + cos \hspace{0.1cm}2x}{sin \hspace{0.1cm}4x + sin \hspace{0.1cm}3x + sin\hspace{0.1cm} 2x}$ = cot 3x

Question 23: tan 4x = $\frac{4\hspace{0.1cm}tan \hspace{0.1cm}x\hspace{0.1cm}(1-tan^2x)}{1-6tan^2x + tan^4x}$

Question 24: cos 4x = 1 – 8sin²x cos²x

Question 25: cos 6x = 32 cos⁶x – 48cos⁴x + 18 cos²x – 1