RD Sharma Solutions for Class 11 Chapter 8 - Transformation Formulae Exercise 8.1

In Exercise 8.1 of Chapter 8, we shall discuss problems based on formulae to transform the product into sum or difference. This exercise can be used as a model of reference by the students to improve their conceptual knowledge and understand the different ways used to solve the problems. Students can refer to RD Sharma Class 11 Solutions which are formulated by BYJUâ€™S experts in Maths to increase students confidence, which plays a crucial role in their board exams. The links to the solutions of this exercise can be accessed in the RD Sharma Class 11 Maths pdf, which is available in the below-mentioned links.

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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 â€“ B)

2 sin 3x cos x = sin (3x + x) + sin (3x â€“ 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) â€“ sin (A â€“ B)

2 cos 3x sin 2x = sin (3x + 2x) â€“ sin (3x â€“ 2x)

= sin (5x) â€“ sin (x)

= sin 5x â€“ sin x

(iii) 2 sin 4x sin 3x

By using the formula,

2 sin A sin B = cos (A â€“ B) â€“ cos (A + B)

2 sin 4x sin 3x = cos (4x â€“ 3x) â€“ cos (4x + 3x)

= cos (x) â€“ cos (7x)

= cos x â€“ cos 7x

(iv) 2 cos 7x cos 3x

By using the formula,

2 cos A cos B = cos (A + B) + cos (A â€“ B)

2 sin 3x cos x = cos (7x + 3x) + cos (7x â€“ 3x)

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

= cos 10x + cos 4x

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

(ii) 2 cos 5Ï€/12 cos Ï€/12 = 1/2

(iii) 2 sin 5Ï€/12 cos Ï€/12 = (âˆš3 + 2)/2

Solution:

(i) 2 sin 5Ï€/12 sin Ï€/12 = 1/2

By using the formula,

2 sin A sin B = cos (A â€“ B) â€“ cos (A + B)

2 sin 5Ï€/12 sin Ï€/12 = cos (5Ï€/12 – Ï€/12) â€“ cos (5Ï€/12 + Ï€/12)

= cos (4Ï€/12) â€“ cos (6Ï€/12)

= cos (Ï€/3) â€“ cos (Ï€/2)

= cos (180o/3) â€“ cos (180o/2)

= cos 60Â° – cos 90Â°

= 1/2 â€“ 0

= 1/2

Hence Proved.

(ii) 2 cos 5Ï€/12 cos Ï€/12 = 1/2

By using the formula,

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

2 cos 5Ï€/12 cos Ï€/12 = cos (5Ï€/12 + Ï€/12) + cos (5Ï€/12 – Ï€/12)

= cos (6Ï€/12) + cos (4Ï€/12)

= cos (Ï€/2) + cos (Ï€/3)

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

= cos 90Â° + cos 60Â°

= 0 + 1/2

= 1/2

Hence Proved.

(iii) 2 sin 5Ï€/12 cos Ï€/12 = (âˆš3 + 2)/2

By using the formula,

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

2 sin 5Ï€/12 cos Ï€/12 = sin (5Ï€/12 + Ï€/12) + sin (5Ï€/12 – Ï€/12)

= sin (6Ï€/12) + sin (4Ï€/12)

= sin (Ï€/2) + sin (Ï€/3)

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

= sin 90Â° + sin 60Â°

= 1 + âˆš3

= (2 + âˆš3)/2

= (âˆš3 + 2)/2

Hence Proved.

3. show that:

(i) sin 50o cos 85o = (1 – âˆš2sin 35o)/2âˆš2

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

Solution:

(i) sin 50o cos 85o = (1 – âˆš2sin 35o)/2âˆš2

By using the formula,

2 sin A cos B = sin (A + B) + sin (A â€“ B)

sin A cos B = [sin (A + B) + sin (A â€“ B)] / 2

sin 50o cos 85o = [sin(50o + 85o) + sin (50o â€“ 85o)] / 2

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

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

= [sin (180o â€“ 45o) â€“ sin 35o] / 2

= [sin 45o â€“ sin 35o] / 2

= [(1/âˆš2) â€“ sin 35o] / 2

= [(1 â€“ sin 35o)/âˆš2] / 2

= (1 â€“ sin 35o) / 2âˆš2

Hence proved.

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

By using the formula,

2 sin A cos B = sin (A + B) + sin (A â€“ B)

sin A cos B = [sin (A + B) + sin (A â€“ B)] / 2

sin 20o cos 115o = [sin(25o + 115o) + sin (25o â€“ 115o)] / 2

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

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

= 1/2 {sin 140o – 1}

Hence proved.

4. Prove that:

4 cos x cos (Ï€/3 + x) cos (Ï€/3 – x) = cos 3x

Solution:

Let us consider LHS:

4 cos x cos (Ï€/3 + x) cos (Ï€/3 – x) = 2 cos x (2 cos (Ï€/3 + x) cos (Ï€/3 – x))

By using the formula,

2 cos A cos B = cos (A + B) + cos (A â€“ B)

2 cos x (2 cos (Ï€/3+x) cos (Ï€/3 – x)) = 2 cos x (cos (Ï€/3+x + Ï€/3-x) + cos (Ï€/3+x – Ï€/3+ x))

= 2 cos x (cos (2Ï€/3) + cos (2x))

= 2 cos x {cos 120Â° + cos 2x}

= 2 cos x {cos (180Â° – 60Â°) + cos 2x}

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

= 2 cos 2x cos x â€“ 2 cos x cos 60Â°

= (cos (x + 2x) + cos (2x – x)) â€“ (2cos x)/2

= cos 3x + cos x â€“ cos x

= cos 3x

= RHS

Hence Proved.