Class 11 RD Sharma Solutions – Chapter 8 Transformation Formulae – Exercise 8.2 | Set 1

Question 1. Express each of the following as the product of sines and cosines:

(i) sin 12θ + sin 4θ

Solution:

We know, sin A + sin B = 2 sin (A+B)/2 cos (A–B)/2

sin 12θ + sin 4θ = 2 sin (12θ + 4θ)/2 cos (12θ – 4θ)/2

= 2 sin 16θ/2 cos 8θ/2

= 2 sin 8θ cos 4θ

(ii) sin 5θ – sin θ

Solution:

We know, sin A – sin B = 2 cos (A+B)/2 sin (A–B)/2

sin 5θ – sin θ = 2 cos (5θ + θ)/2 sin (5θ – θ)/2

= 2 cos 6θ/2 sin 4θ/2

= 2 cos 3θ sin 2θ

(iii) cos 12θ + cos 8θ

Solution:

We know, cos A + cos B = 2 cos (A+B)/2 cos (A–B)/2

cos 12θ + cos 8θ = 2 cos (12θ + 8θ)/2 cos (12θ – 8θ)/2

= 2 cos 20θ/2 cos 4θ/2

= 2 cos 10θ cos 2θ

(iv) cos 12θ – cos 4θ

Solution:

We know, cos A – cos B = –2 sin (A+B)/2 sin (A–B)/2

cos 12θ – cos 4θ = –2 sin (12θ + 4θ)/2 sin (12θ – 4θ)/2

= –2 sin 16θ/2 sin 8θ/2

= –2 sin 8θ sin 4θ

(v) sin 2θ + cos 4θ

Solution:

sin 2θ + cos 4θ = sin 2θ + sin (90^o – 4θ)

We know, sin A + sin B = 2 sin (A+B)/2 cos (A–B)/2

sin 2θ + sin (90^o – 4θ) = 2 sin (2x + 90^o – 4θ)/2 cos (2θ – 90^o + 4θ)/2

= 2 sin (90^o – 2θ)/2 cos (6θ – 90^o)/2

= 2 sin (45^o – θ) cos (3θ – 45^o)

= 2 sin (45^o – θ) cos [–(45^o – 3θ)]

= 2 sin (45^o – θ) cos (45^o – 3θ)

= 2 sin (π/4 – θ) cos (π/4 – 3θ)

Question 2. Prove that :

(i) sin 38° + sin 22° = sin 82°

Solution:

Given, L.H.S. = sin 38° + sin 22°.

We know, sin A + sin B = 2 sin (A+B)/2 cos (A–B)/2

sin 38° + sin 22° = 2 sin (38^o + 22^o)/2 cos (38^o – 22^o)/2

= 2 sin 60^o/2 cos 16^o/2

= 2 sin 30^o cos 8^o

= 2 × (1/2) × cos 8^o

= cos 8^o

= cos (90° – 82°)

= sin 82°

= R.H.S.

Hence proved.

(ii) cos 100° + cos 20° = cos 40°

Solution:

Given, L.H.S. = cos 100° + cos 20°.

We know, cos A + cos B = 2 cos (A+B)/2 cos (A–B)/2

cos 100° + cos 20° = 2 cos (100^o + 20^o)/2 cos (100^o – 20^o)/2

= 2 cos 120^o/2 cos 80^o/2

= 2 cos 60^o cos 40^o

= 2 × (1/2) × cos 40^o

= cos 40^o

= R.H.S.

Hence Proved.

(iii) sin 50° + sin 10° = cos 20°

Solution:

Given, L.H.S. = sin 50° + sin 10°.

We know, sin A + sin B = 2 sin (A+B)/2 cos (A–B)/2.

sin 50° + sin 10° = 2 sin (50^o + 10^o)/2 cos (50^o – 10^o)/2

= 2 sin 60^o/2 cos 40^o/2

= 2 sin 30^o cos 20^o

= 2 × (1/2) × cos 20^o

= cos 20^o

= R.H.S

Hence Proved.

(iv) sin 23° + sin 37° = cos 7°

Solution:

Given L.H.S. = sin 23° + sin 37°.

We know sin A + sin B = 2 sin (A+B)/2 cos (A–B)/2

sin 23° + sin 37° = 2 sin (23^o + 37^o)/2 cos (23^o – 37^o)/2

= 2 sin 60^o/2 cos (–14^o/2)

= 2 sin 30^o cos (–7^o)

= 2 × (1/2) × cos 7^o

= cos 7^o

= R.H.S.

Hence Proved.

(v) sin 105° + cos 105° = cos 45°

Solution:

Given, L.H.S. = sin 105° + cos 105°

sin 105° + cos 105° = sin 105^o + sin (90^o – 105^o)

= sin 105^o + sin (–15^o)

= sin 105^o – sin 15^o

We know, sin A – sin B = 2 cos (A+B)/2 sin (A–B)/2

sin 105^o – sin 15^o = 2 cos (105^o + 15^o)/2 sin (105^o – 15^o)/2

= 2 cos 120^o/2 sin 90^o/2

= 2 cos 60^o sin 45^o

= 2 × (1/2) × (1/√2)

= 1/√2

= cos 45^o

= R.H.S.

Hence Proved.

(vi) sin 40° + sin 20° = cos 10°

Solution:

Given, L.H.S. = sin 40° + sin 20°.

We know, sin A + sin B = 2 sin (A+B)/2 cos (A–B)/2

sin 40° + sin 20° = 2 sin (40^o + 20^o)/2 cos (40^o – 20^o)/2

= 2 sin 60^o/2 cos 20^o/2

= 2 sin 30^o cos 10^o

= 2 × (1/2) × cos 10^o

= cos 10^o

= R.H.S.

Hence Proved.

Question 3. Prove that:

(i) cos 55° + cos 65° + cos 175° = 0

Solution:

Given, L.H.S. = cos 55° + cos 65° + cos 175°.

We know, cos A + cos B = 2 cos (A+B)/2 cos (A–B)/2

cos 55° + cos 65° + cos 175° = 2 cos (55^o + 65^o)/2 cos (55^o – 65^o) + cos (180^o – 5^o)

= 2 cos 120^o/2 cos (–10o)/2 – cos 5^o

= 2 cos 60° cos (–5°) – cos 5°

= 2 × (1/2) × cos 5^o – cos 5^o

=cos 5^o – cos 5^o

= 0

= R.H.S.

Hence Proved.

(ii) sin 50° – sin 70° + sin 10° = 0

Solution:

Given, L.H.S. = sin 50° – sin 70° + sin 10°.

We know, sin A – sin B = 2 cos (A+B)/2 sin (A–B)/2

sin 50° – sin 70° + sin 10° = 2 cos (50^o + 70^o)/2 sin (50^o – 70^o) + sin 10^o

= 2 cos 120^o/2 sin (–20^o)/2 + sin 10^o

= 2 cos 60^o (–sin 10^o) + sin 10^o

= 2 × (1/2) × (–sin 10^o) + sin 10^o

= 0

= R.H.S.

Hence Proved.

(iii) cos 80° + cos 40° – cos 20° = 0

Solution:

Given L.H.S. = cos 80° + cos 40° – cos 20°.

We know, cos A + cos B = 2 cos (A+B)/2 cos (A–B)/2

cos 80° + cos 40° – cos 20° = 2 cos (80^o + 40^o)/2 cos (80^o – 40^o) – cos 20^o

= 2 cos 120^o/2 cos 40^o/2 – cos 20^o

= 2 cos 60° cos 20^o – cos 20°

= 2 × (1/2) × cos 20^o – cos 20^o

= 0

= R.H.S.

Hence Proved.

(iv) cos 20° + cos 100° + cos 140° = 0

Solution:

Given, L.H.S. = cos 20° + cos 100° + cos 140°.

We know, cos A + cos B = 2 cos (A+B)/2 cos (A–B)/2.

cos 20° + cos 100° + cos 140° = 2 cos (20^o + 100^o)/2 cos (20^o – 100^o) + cos (80^o – 40^o)

= 2 cos 120^o/2 cos (–80^o)/2 – cos 40^o

= 2 cos 60° cos (–40°) – cos 40°

= 2 × (1/2) × cos 40^o – cos 40^o

= 0

= R.H.S.

Hence Proved.

(v) sin 5π/18 – cos 4π/9 = √3 sin π/9

Solution:

Given, L.H.S. = sin 5π/18 – cos 4π/9

= sin 5π/18 – sin (π/2 – 4π/9)

= sin 5π/18 – sin (9π – 8π)/18

= sin 5π/18 – sin π/18

We know, sin A – sin B = 2 cos (A+B)/2 sin (A– B)/2

= 2 cos (6π/36) sin (4π/36)

= 2 cos π/6 sin π/9

= 2 cos 30^o sin π/9

= 2 × (√3/2) × sin π/9

= √3 sin π/9

= R.H.S.

Hence Proved.

(vi) cos π/12 – sin π/12 = 1/√2

Solution:

Given, cos π/12 – sin π/12 = sin (π/2 – π/12) – sin π/12

= sin (6π – 5π)/12 – sin π/12

= sin 5π/12 – sin π/12

We know, sin A – sin B = 2 cos (A+B)/2 sin (A–B)/2

= 2 cos (6π/24) sin (4π/24)

= 2 cos π/4 sin π/6

= 2 cos 45^o sin 30^o

= 2 × (1/√2) × (1/2)

= 1/√2

= R.H.S.

Hence Proved.

(vii) sin 80° – cos 70° = cos 50°

Solution:

We have, sin 80° = cos 50° + cos 70^o

Here, R.H.S. = cos 50° + cos 70^o

We know,

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

cos 50° + cos 70^o = 2 cos (50^o + 70^o)/2 cos (50^o – 70^o)/2

= 2 cos 120^o/2 cos (–20^o)/2

= 2 cos 60^o cos (–10^o)

= 2 × (1/2) × cos 10^o

= cos 10^o

= cos (90° – 80°)

= sin 80°

= L.H.S.

Hence Proved.

(viii) sin 51° + cos 81° = cos 21°

Solution:

Given, L.H.S. = sin 51° + cos 81°

= sin 51^o + sin (90^o – 81^o)

= sin 51^o + sin 9^o

We know, sin A + sin B = 2 sin (A+B)/2 cos (A–B)/2

sin 51^o + sin 9^o = 2 sin (51^o + 9^o)/2 cos (51^o – 9^o)/2

= 2 sin 60^o/2 cos 42^o/2

= 2 sin 30^o cos 21^o

= 2 × (1/2) × cos 21^o

= cos 21^o

= R.H.S.

Hence Proved.

Question 4. Prove that:

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

Solution:

Given, L.H.S. = cos (3π/4 + x) – cos (3π/4 – x)

We know, cos A – cos B = –2 sin (A+B)/2 sin (A–B)/2

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

= –2 sin (6π/4)/2 sin 2x/2

= –2 sin 6π/8 sin x

= –2 sin 3π/4 sin x

= –2 sin (π – π/4) sin x

= –2 sin π/4 sin x

= –2 × (1/√2) × sin x

= –√2 sin x

= R.H.S.

Hence proved.

(ii) cos (π/4 + x) + cos (π/4 – x) = √2 cos x

Solution:

Given, L.H.S. = cos (π/4 + x) + cos (π/4 – x)

We know, cos A + cos B = 2 cos (A+B)/2 cos (A–B)/2

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

= 2 cos (2π/4)/2 cos 2x/2

= 2 cos 2π/8 cos x

= 2 sin π/4 cos x

= 2 × (1/√2) × cos x

= √2 cos x

= R.H.S.

Hence proved.

Question 5. Prove that:

(i) sin 65^o + cos 65^o = √2 cos 20^o

Solution:

Given L.H.S. = sin 65^o + cos 65^o

= sin 65^o + sin (90^o – 65^o)

= sin 65^o + sin 25^o

We know, sin A + sin B = 2 sin (A+B)/2 cos (A–B)/2

sin 65^o + sin 25^o = 2 sin (65^o + 25^o)/2 cos (65^o – 25^o)/2

= 2 sin 90^o/2 cos 40^o/2

= 2 sin 45^o cos 20^o

= 2 × (1/√2) × cos 20^o

= √2 cos 20^o

= R.H.S.

Hence proved.

(ii) sin 47^o + cos 77^o = cos 17^o

Solution:

Given, L.H.S. = sin 47^o + cos 77^o

= sin 47^o + sin (90^o – 77^o)

= sin 47^o + sin 13^o

We know, sin A + sin B = 2 sin (A+B)/2 cos (A–B)/2

sin 47^o + sin 13^o = 2 sin (47^o + 13^o)/2 cos (47^o – 13^o)/2

= 2 sin 60^o/2 cos 34^o/2

= 2 sin 30^o cos 17^o

= 2 × (1/2) × cos 17^o

= cos 17^o

= R.H.S.

Hence proved.

Question 6. Prove that:

(i) cos 3A + cos 5A + cos 7A + cos 15A = 4 cos 4A cos 5A cos 6A

Solution:

Given, L.H.S. = cos 3A + cos 5A + cos 7A + cos 15A

= (cos 5A + cos 3A) + (cos 15A + cos 7A)

We know, cos A + cos B = 2 cos (A+B)/2 cos (A–B)/2

= (cos 5A + cos 3A) + (cos 15A + cos 7A)

= [2 cos (5A+3A)/2 cos (5A–3A)/2] + [2 cos (15A+7A)/2 cos (15A–7A)/2]

= [2 cos 8A/2 cos 2A/2] + [2 cos 22A/2 cos 8A/2]

= [2 cos 4A cos A] + [2 cos 11A cos 4A]

= 2 cos 4A (cos 11A + cos A)

= 2 cos 4A [2 cos (11A+A)/2 cos (11A-A)/2]

= 2 cos 4A [2 cos 12A/2 cos 10A/2]

= 2 cos 4A [2 cos 6A cos 5A]

= 4 cos 4A cos 5A cos 6A

= R.H.S.

Hence proved.

(ii) cos A + cos 3A + cos 5A + cos 7A = 4 cos A cos 2A cos 4A

Solution:

Given L.H.S. = cos A + cos 3A + cos 5A + cos 7A

= (cos 3A + cos A) + (cos 7A + cos 5A)

We know, cos A + cos B = 2 cos (A+B)/2 cos (A–B)/2

= (cos 3A + cos A) + (cos 7A + cos 5A)

= [2 cos (3A+A)/2 cos (3A–A)/2] + [2 cos (7A+5A)/2 cos (7A–5A)/2]

= [2 cos 4A/2 cos 2A/2] + [2 cos 12A/2 cos 2A/2]

= [2 cos 2A cos A] + [2 cos 6A cos A]

= 2 cos A (cos 6A + cos 2A)

= 2 cos A [2 cos (6A+2A)/2 cos (6A–2A)/2]

= 2 cos A [2 cos 8A/2 cos 4A/2]

= 2 cos A [2 cos 4A cos 2A]

= 4 cos A cos 2A cos 4A

= R.H.S.

Hence proved.

(iii) sin A + sin 2A + sin 4A + sin 5A = 4 cos A/2 cos 3A/2 sin 3A

Solution:

Given, L.H.S. = sin A + sin 2A + sin 4A + sin 5A

= (sin 2A + sin A) + (sin 5A + sin 4A)

We know, sin A + sin B = 2 sin (A+B)/2 cos (A–B)/2

= (sin 2A + sin A) + (sin 5A + sin 4A)

= [2 sin (2A+A)/2 cos (2A–A)/2] + [2 sin (5A+4A)/2 cos (5A–4A)/2]

= [2 sin 3A/2 cos A/2] + [2 sin 9A/2 cos A/2]

= 2 cos A/2 (sin 9A/2 + sin 3A/2)

= 2 cos A/2 [2 sin (9A/2 + 3A/2)/2 cos (9A/2 – 3A/2)/2]

= 2 cos A/2 [2 sin ((9A+3A)/2)/2 cos ((9A–3A)/2)/2]

= 2 cos A/2 [2 sin 12A/4 cos 6A/4]

= 2 cos A/2 [2 sin 3A cos 3A/2]

= 4 cos A/2 cos 3A/2 sin 3A

= R.H.S.

Hence proved.

(iv) sin 3A + sin 2A – sin A = 4 sin A cos A/2 cos 3A/2

Solution:

Given, L.H.S. = sin 3A + sin 2A – sin A

= (sin 3A – sin A) + sin 2A

We know, sin A – sin B = 2 cos (A+B)/2 sin (A–B)/2

= (sin 3A – sin A) + sin 2A

= 2 cos (3A + A)/2 sin (3A – A)/2 + sin 2A

= 2 cos 4A/2 sin 2A/2 + sin 2A

= 2 cos 2A sin A + 2 sin A cos A

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

= 2 sin A [2 cos (2A+A)/2 cos (2A-A)/2]

= 2 sin A [2 cos 3A/2 cos A/2]

= 4 sin A cos A/2 cos 3A/2

= R.H.S.

Hence proved.

(v) cos 20^o cos 100^o + cos 100^o cos 140^o – cos 140^o cos 200^o = – 3/4

Solution:

Given L.H.S. = cos 20^o cos 100^o + cos 100^o cos 140^o – cos 140^o cos 200^o

= 1/2 [2 cos 100^o cos 20^o + 2 cos 140^o cos 100^o – 2 cos 200^o cos 140^o]

We know that, 2 cos A cos B = cos (A+B) + cos (A–B)

= 1/2 [cos (100^o + 20^o) + cos (100^o – 20^o) + cos (140^o + 100^o) + cos (140^o – 100^o) – cos (200^o + 140^o) – cos (200^o – 140^o)]]

= 1/2 [cos 120^o + cos 80^o + cos 240^o + cos 40^o – cos 340^o – cos 60^o]

= 1/2 [cos (90^o + 30^o) + cos 80^o + cos (180^o + 60^o) + cos 40^o – cos (360^o – 20^o) – cos 60^o]

= 1/2 [–sin 30^o + cos 80^o – cos 60^o + cos 40^o – cos 20^o – cos 60^o]

= 1/2 [–sin 30^o + cos 80^o + cos 40^o – cos 20^o – 2 cos 60^o]

= 1/2 [–sin 30^o + 2 cos (80^o+40^o)/2 cos (80^o–40^o)/2 – cos 20^o – 2 × 1/2]

= 1/2 [–sin 30^o + 2 cos 120^o/2 cos 40^o/2 – cos 20^o – 1]

= 1/2 [–sin 30^o + 2 cos 60^o cos 20^o – cos 20^o – 1]

= 1/2 [–1/2 + 2×(1/2)×cos 20^o – cos 20^o – 1]

= 1/2 [–1/2 + cos 20^o – cos 20^o – 1]

= 1/2 [–1/2 –1]

= 1/2 [–3/2]

= –3/4

= R.H.S.

Hence proved.

(vi) sin x/2 sin 7x/2 + sin 3x/2 sin 11x/2 = sin 2x sin 5x

Solution:

Given L.H.S. = sin x/2 sin 7x/2 + sin 3x/2 sin 11x/2

= 1/2 [2 sin 7x/2 sin x/2 + 2 sin 11x/2 sin 3x/2]

We know, 2 sin A sin B = cos (A–B) – cos (A+B)

= 1/2 [cos (7x/2 – x/2) – cos (7x/2 + x/2) + cos (11x/2 – 3x/2) – cos (11x/2 + 3x/2)]

= 1/2 [cos (7x–x)/2 – cos (7x+x)/2 + cos (11x–3x)/2 – cos (11x+3x)/2]

= 1/2 [cos 6x/2 – cos 8x/2 + cos 8x/2 – cos 14x/2]

= 1/2 [cos 3x – cos 7x]

= –1/2 [cos 7x – cos 3x]

= –1/2 [–2 sin (7x+3x)/2 sin (7x–3x)/2]

= –1/2 [–2 sin 10x/2 sin 4x/2]

= –1/2 [–2 sin 5x sin 2x]

= –2/–2 sin 5x sin 2x

= sin 2x sin 5x

= R.H.S.

Hence proved.

(vii) cos x cos x/2 – cos 3x cos 9x/2 = sin 4x sin 7x/2

Solution:

Given L.H.S. = cos x cos x/2 – cos 3x cos 9x/2

= 1/2 [2 cos x cos x/2 – 2 cos 9x/2 cos 3x]

We know, 2 cos A cos B = cos (A+B) + cos (A–B)

= 1/2 [cos (x + x/2) + cos (x – x/2) – cos (9x/2 + 3x) – cos (9x/2 – 3x)]

= 1/2 [cos (2x+x)/2 + cos (2x–x)/2 – cos (9x+6x)/2 – cos (9x–6x)/2]

= 1/2 [cos 3x/2 + cos x/2 – cos 15x/2 – cos 3x/2]

= 1/2 [cos x/2 – cos 15x/2]

= – 1/2 [cos 15x/2 – cos x/2]

= – 1/2 [–2 sin (15x/2 + x/2)/2 sin (15x/2 – x/2)/2]

= -1/2 [–2 sin (16x/2)/2 sin (14x/2)/2]

= -1/2 [–2 sin 16x/4 sin 7x/2]

= – 1/2 [–2 sin 4x sin 7x/2]

= –2/–2 [sin 4x sin 7x/2]

= sin 4x sin 7x/2

= R.H.S.

Hence proved.

Question 7. Prove that:

(i) 

Solution:

We have,

L.H.S. = 

= 

= 

= 

= cot A

= R.H.S.

Hence proved.

(ii) 

Solution:

We have,

L.H.S. = 

= 

= 

= cot 8A

= R.H.S.

Hence proved.

(iii)

Solution:

We have,

L.H.S. =

=

=

=

= R.H.S.

Hence proved.

(iv)

Solution:

We have,

L.H.S. =

=

=

=

= R.H.S.

Hence proved.

(iv)

Solution:

We have,

L.H.S. =

=

=

=

= R.H.S.

Hence proved.

Question 8. Prove that:

(i) 

Solution:

We have,

L.H.S. = 

= 

= 

= 

= 

= 

= tan 3A

= R.H.S.

Hence proved.

(ii)

Solution:

We have,

L.H.S. =

=

=

=

=

= R.H.S.

Hence proved.

(iii) 

Solution:

We have,

L.H.S. = 

= 

= 

= 

= 

= cot 3A

= R.H.S.

Hence proved.

(iv) 

Solution:

We have,

L.H.S. = 

= 

= 

= 

= 

= 

= tan 6A

= R.H.S.

Hence proved.

(v) 

Solution:

We have,

L.H.S. = 

= 

= 

= 

= 

= 

= cot 6A

= R.H.S.

Hence proved.

(vi) 

Solution:

We have,

L.H.S. = 

Multiplying numerator and denominator by 2, we get

= 

= 

= 

= 

= 

= 

= 

= 

= 

= tan A

= R.H.S.

Hence proved.

(vii)

Solution:

We have,

L.H.S. =

=

=

=

=

= tan 8A

= R.H.S.

Hence proved.

(viii) 

Solution:

We have,

L.H.S. = 

On multiplying numerator and denominator by 2, we get

= 

= 

= 

= 

= 

= 

= 

= 

= tan 2A

= R.H.S.

Hence proved.

(ix) 

Solution:

We have,

L.H.S. = 

On multiplying numerator and denominator by 2, we get

= 

= 

= 

= 

= 

= 

= 

= tan 5A

= R.H.S.

Hence proved.

(x) 

Solution:

We have,

L.H.S. = 

= 

= 

= 

= 

= sin3A/sin5A

= R.H.S.

Hence proved.

(xi)

Solution:

We have,

L.H.S. =

=

=

=

=

=

=

= tan θ

= R.H.S.

Hence proved.

Question 9. Prove that:

(i) sin α + sin β + sin γ – sin (α + β + γ) = 4 sin (α + β)/2 sin (β + γ)/2 sin (α + γ)/2

Solution:

We have,

L.H.S. = sin α + sin β + sin γ – sin (α + β + γ)

=

=

=

=

=

=

= 4 sin (α + β)/2 sin (β + γ)/2 sin (α + γ)/2

= R.H.S.

Hence proved.

(ii) cos (A + B + C) + cos (A – B + C) + cos (A + B – C) + cos (–A + B + C) = 4 cos A cos B cos C

Solution:

We have,

L.H.S. = cos (A + B + C) + cos (A – B + C) + cos (A + B – C) + cos (–A + B + C)

=

=

= 2 cos (A + C) cos B + 2 cos B cos (A − C)

= 2 cos B [cos (A + C) + cos (A − C)]

= 2 cos B

= 2 cos B [2 cos A cos C]

= 4 cos A cos B cos C

= R.H.S.

Hence proved.

Question 10. If cos A + cos B = 1/2 and sin A + sin B = 1/4, prove that.

Solution:

We have,

cos A + cos B = 1/2

sin A + sin B = 1/4

=>

=>

=>

=>

=>

Hence proved.