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8. Transformation Formulae
Exercise 8.1
1. Question
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
Answer
(i) We know, 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) We know, 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) We know, 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) We know, 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. Question
Prove that :
i. 
ii. 
iii.
Answer
i. We know, 2 sin A sin B = cos (A – B) – cos (A + B)
= cos 60° - cos 90°
Hence Proved
ii. We know, 2 cos A cos B = cos (A + B) + cos (A – B)
Page 2


8. Transformation Formulae
Exercise 8.1
1. Question
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
Answer
(i) We know, 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) We know, 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) We know, 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) We know, 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. Question
Prove that :
i. 
ii. 
iii.
Answer
i. We know, 2 sin A sin B = cos (A – B) – cos (A + B)
= cos 60° - cos 90°
Hence Proved
ii. We know, 2 cos A cos B = cos (A + B) + cos (A – B)
= cos 90° + cos 60°
Hence Proved
iii. We know, 2 sin A cos B = sin (A + B) + sin (A – B) 
= sin 90° + sin 60°
Hence Proved
3. Question
Show that:
i. 
ii. 
Answer
i. We know, 2 sin A cos B = sin (A + B) + sin (A – B)
{sin (-x) = - sin x}
{sin (180° - x) = sin x}
Page 3


8. Transformation Formulae
Exercise 8.1
1. Question
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
Answer
(i) We know, 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) We know, 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) We know, 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) We know, 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. Question
Prove that :
i. 
ii. 
iii.
Answer
i. We know, 2 sin A sin B = cos (A – B) – cos (A + B)
= cos 60° - cos 90°
Hence Proved
ii. We know, 2 cos A cos B = cos (A + B) + cos (A – B)
= cos 90° + cos 60°
Hence Proved
iii. We know, 2 sin A cos B = sin (A + B) + sin (A – B) 
= sin 90° + sin 60°
Hence Proved
3. Question
Show that:
i. 
ii. 
Answer
i. We know, 2 sin A cos B = sin (A + B) + sin (A – B)
{sin (-x) = - sin x}
{sin (180° - x) = sin x}
Hence Proved
ii. We know, 2 sin A cos B = sin (A + B) + sin (A – B)
Take L.H.S
{sin (-x) = - sin x}
= R.H.S.
Hence Proved
4. Question
Prove that:
Answer
Take L.H.S
{2 cos A cos B = cos (A + B) + cos (A – B)}
= 2 cos x {cos 120° + cos 2x}
= 2 cos x {cos (180° - 60°) + cos 2x}
{cos (180° - A) = - cos A}
= 2 cos x (cos 2x - cos 60°)
= 2 cos 2x cos x – 2 cos x cos 60°
= cos 3x + cos x – cos x
= cos 3x = R.H.S.
Hence Proved
5 A. Question
Prove that:
Answer
Take L.H.S
Multiplying & Dividing by 2:
Page 4


8. Transformation Formulae
Exercise 8.1
1. Question
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
Answer
(i) We know, 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) We know, 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) We know, 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) We know, 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. Question
Prove that :
i. 
ii. 
iii.
Answer
i. We know, 2 sin A sin B = cos (A – B) – cos (A + B)
= cos 60° - cos 90°
Hence Proved
ii. We know, 2 cos A cos B = cos (A + B) + cos (A – B)
= cos 90° + cos 60°
Hence Proved
iii. We know, 2 sin A cos B = sin (A + B) + sin (A – B) 
= sin 90° + sin 60°
Hence Proved
3. Question
Show that:
i. 
ii. 
Answer
i. We know, 2 sin A cos B = sin (A + B) + sin (A – B)
{sin (-x) = - sin x}
{sin (180° - x) = sin x}
Hence Proved
ii. We know, 2 sin A cos B = sin (A + B) + sin (A – B)
Take L.H.S
{sin (-x) = - sin x}
= R.H.S.
Hence Proved
4. Question
Prove that:
Answer
Take L.H.S
{2 cos A cos B = cos (A + B) + cos (A – B)}
= 2 cos x {cos 120° + cos 2x}
= 2 cos x {cos (180° - 60°) + cos 2x}
{cos (180° - A) = - cos A}
= 2 cos x (cos 2x - cos 60°)
= 2 cos 2x cos x – 2 cos x cos 60°
= cos 3x + cos x – cos x
= cos 3x = R.H.S.
Hence Proved
5 A. Question
Prove that:
Answer
Take L.H.S
Multiplying & Dividing by 2:
{? 2 cos A cos B = cos (A + B) + cos (A – B)}
{? cos (-A) = cos A}
Multiplying & Dividing by 2:
{? 2 cos A cos B = cos (A + B) + cos (A – B)}
{? cos (180° - A) = - cos A}
= R.H.S
Hence Proved
5 B. Question
Prove that:
Answer
Take L.H.S
cos 40° cos 80° cos 160°
Multiplying & Dividing by 2:
{? 2 cos A cos B = cos (A + B) + cos (A – B)}
{? cos (180° - A) = - cos A & cos (180° + A) = - cos A}
Page 5


8. Transformation Formulae
Exercise 8.1
1. Question
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
Answer
(i) We know, 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) We know, 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) We know, 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) We know, 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. Question
Prove that :
i. 
ii. 
iii.
Answer
i. We know, 2 sin A sin B = cos (A – B) – cos (A + B)
= cos 60° - cos 90°
Hence Proved
ii. We know, 2 cos A cos B = cos (A + B) + cos (A – B)
= cos 90° + cos 60°
Hence Proved
iii. We know, 2 sin A cos B = sin (A + B) + sin (A – B) 
= sin 90° + sin 60°
Hence Proved
3. Question
Show that:
i. 
ii. 
Answer
i. We know, 2 sin A cos B = sin (A + B) + sin (A – B)
{sin (-x) = - sin x}
{sin (180° - x) = sin x}
Hence Proved
ii. We know, 2 sin A cos B = sin (A + B) + sin (A – B)
Take L.H.S
{sin (-x) = - sin x}
= R.H.S.
Hence Proved
4. Question
Prove that:
Answer
Take L.H.S
{2 cos A cos B = cos (A + B) + cos (A – B)}
= 2 cos x {cos 120° + cos 2x}
= 2 cos x {cos (180° - 60°) + cos 2x}
{cos (180° - A) = - cos A}
= 2 cos x (cos 2x - cos 60°)
= 2 cos 2x cos x – 2 cos x cos 60°
= cos 3x + cos x – cos x
= cos 3x = R.H.S.
Hence Proved
5 A. Question
Prove that:
Answer
Take L.H.S
Multiplying & Dividing by 2:
{? 2 cos A cos B = cos (A + B) + cos (A – B)}
{? cos (-A) = cos A}
Multiplying & Dividing by 2:
{? 2 cos A cos B = cos (A + B) + cos (A – B)}
{? cos (180° - A) = - cos A}
= R.H.S
Hence Proved
5 B. Question
Prove that:
Answer
Take L.H.S
cos 40° cos 80° cos 160°
Multiplying & Dividing by 2:
{? 2 cos A cos B = cos (A + B) + cos (A – B)}
{? cos (180° - A) = - cos A & cos (180° + A) = - cos A}
Multiplying & Dividing by 2:
{? 2 cos A cos B = cos (A + B) + cos (A – B)}
{? cos (180° - A) = - cos A}
= R.H.S
Hence Proved
5 C. Question
Prove that:
Answer
Take L.H.S
sin 20° sin 40° sin 80°
Multiplying & Dividing by 2:
{? 2 sin A sin B = cos (A – B) – cos (A + B)}
{? cos (-A) = cos A}
Multiplying & Dividing by 2:
{? 2 cos A sin B = sin (A + B) – sin (A – B)}
{? sin (-A) = - sin A}
{? sin (180° - A) = sin A}

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