RD Sharma Solutions Class 11 Maths Chapter 8 Exercise 8.1 (Updated For 2024)

1. Express each of the following as the sum or difference of sines and cosines:
(i) 2 sin 3x cos x
(ii) 2 cos 3x sin 2x
(iii) 2 sin 4x sin 3x
(iv) 2 cos 7x cos 3x

Solution:

(i) 2 sin 3x cos x

By using the formula,

2 sin A cos B = sin (A + B) + sin (A – B)

2 sin 3x cos x = sin (3x + x) + sin (3x – x)

= sin (4x) + sin (2x)

= sin 4x + sin 2x

(ii) 2 cos 3x sin 2x

By using the formula,

2 cos A sin B = sin (A + B) – sin (A – B)

2 cos 3x sin 2x = sin (3x + 2x) – sin (3x – 2x)

= sin (5x) – sin (x)

= sin 5x – sin x

(iii) 2 sin 4x sin 3x

By using the formula,

2 sin A sin B = cos (A – B) – cos (A + B)

2 sin 4x sin 3x = cos (4x – 3x) – cos (4x + 3x)

= cos (x) – cos (7x)

= cos x – cos 7x

(iv) 2 cos 7x cos 3x

By using the formula,

2 cos A cos B = cos (A + B) + cos (A – B)

2 sin 3x cos x = cos (7x + 3x) + cos (7x – 3x)

= cos (10x) + cos (4x)

= cos 10x + cos 4x

2. Prove that:
(i) 2 sin 5π/12 sin π/12 = 1/2

(ii) 2 cos 5π/12 cos π/12 = 1/2

(iii) 2 sin 5π/12 cos π/12 = (√3 + 2)/2

Solution:

(i) 2 sin 5π/12 sin π/12 = 1/2

By using the formula,

2 sin A sin B = cos (A – B) – cos (A + B)

2 sin 5π/12 sin π/12 = cos (5π/12 – π/12) – cos (5π/12 + π/12)

= cos (4π/12) – cos (6π/12)

= cos (π/3) – cos (π/2)

= cos (180^o/3) – cos (180^o/2)

= cos 60° – cos 90°

= 1/2 – 0

= 1/2

Hence Proved.

(ii) 2 cos 5π/12 cos π/12 = 1/2

By using the formula,

2 cos A cos B = cos (A + B) + cos (A – B)

2 cos 5π/12 cos π/12 = cos (5π/12 + π/12) + cos (5π/12 – π/12)

= cos (6π/12) + cos (4π/12)

= cos (π/2) + cos (π/3)

= cos (180^o/2) + cos (180^o/3)

= cos 90° + cos 60°

= 0 + 1/2

= 1/2

Hence Proved.

(iii) 2 sin 5π/12 cos π/12 = (√3 + 2)/2

By using the formula,

2 sin A cos B = sin (A + B) + sin (A – B)

2 sin 5π/12 cos π/12 = sin (5π/12 + π/12) + sin (5π/12 – π/12)

= sin (6π/12) + sin (4π/12)

= sin (π/2) + sin (π/3)

= sin (180^o/2) + sin (180^o/3)

= sin 90° + sin 60°

= 1 + √3

= (2 + √3)/2

= (√3 + 2)/2

Hence Proved.

3. show that:

(i) sin 50^o cos 85^o = (1 – √2sin 35^o)/2√2

(ii) sin 25^o cos 115^o = 1/2 {sin 140^o – 1}

Solution:

(i) sin 50^o cos 85^o = (1 – √2sin 35^o)/2√2

By using the formula,

2 sin A cos B = sin (A + B) + sin (A – B)

sin A cos B = [sin (A + B) + sin (A – B)] / 2

sin 50^o cos 85^o = [sin(50^o + 85^o) + sin (50^o – 85^o)] / 2

= [sin (135^o) + sin (-35^o)] / 2

= [sin (135^o) – sin (35^o)] / 2 (since, sin (-x) = -sin x)

= [sin (180^o – 45^o) – sin 35^o] / 2

= [sin 45^o – sin 35^o] / 2

= [(1/√2) – sin 35^o] / 2

= [(1 – sin 35^o)/√2] / 2

= (1 – sin 35^o) / 2√2

Hence proved.

(ii) sin 25^o cos 115^o = 1/2 {sin 140^o – 1}

By using the formula,

2 sin A cos B = sin (A + B) + sin (A – B)

sin A cos B = [sin (A + B) + sin (A – B)] / 2

sin 20^o cos 115^o = [sin(25^o + 115^o) + sin (25^o – 115^o)] / 2

= [sin (140^o) + sin (-90^o)] / 2

= [sin (140^o) – sin (90^o)] / 2 (since, sin (-x) = -sin x)

= 1/2 {sin 140^o – 1}

Hence proved.

4. Prove that:

4 cos x cos (π/3 + x) cos (π/3 – x) = cos 3x

Solution:

Let us consider LHS:

4 cos x cos (π/3 + x) cos (π/3 – x) = 2 cos x (2 cos (π/3 + x) cos (π/3 – x))

By using the formula,

2 cos A cos B = cos (A + B) + cos (A – B)

2 cos x (2 cos (π/3+x) cos (π/3 – x)) = 2 cos x (cos (π/3+x + π/3-x) + cos (π/3+x – π/3+ x))

= 2 cos x (cos (2π/3) + cos (2x))

= 2 cos x {cos 120° + cos 2x}

= 2 cos x {cos (180° – 60°) + cos 2x}

= 2 cos x (cos 2x – cos 60°) (since, {cos (180° – A) = – cos A})

= 2 cos 2x cos x – 2 cos x cos 60°

= (cos (x + 2x) + cos (2x – x)) – (2cos x)/2

= cos 3x + cos x – cos x

= cos 3x

= RHS

Hence Proved.