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

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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) $\frac{\sin A + \sin 3A}{\cos A - \cos 3A} = \cot A$

Solution:

We have,

L.H.S. = $\frac{\sin A + \sin 3A}{\cos A - \cos 3A}$

= $\frac{2\sin \left( \frac{A + 3A}{2} \right) \cos \left( \frac{A - 3A}{2} \right)}{2\sin \left( \frac{A + 3A}{2} \right) \sin \left( \frac{3A - A}{2} \right)}$

= $\frac{\sin 2A \cos \left( - A \right)}{\sin 2A \sin A}$

= $\frac{\sin 2A \cos A}{\sin 2A \sin A}$

= cot A

= R.H.S.

Hence proved.

(ii) $\frac{\sin 9A - \sin 7A}{\cos 7A - \cos 9A} = \cot 8A$

Solution:

We have,

L.H.S. = $\frac{\sin 9A - \sin 7A}{\cos 7A - \cos 9A}$

= $\frac{2\sin \left( \frac{9A - 7A}{2} \right) \cos \left( \frac{9A + 7A}{2} \right)}{2\sin \left( \frac{7A + 9A}{2} \right) \sin \left( \frac{9A - 7A}{2} \right)}$

= $\frac{\sin A \cos 8A}{\sin 8A \sin A}$

= cot 8A

= R.H.S.

Hence proved.

(iii) $\frac{sinA-sinB}{cosA+cosB}=tan\frac{A-B}{2}$

Solution:

We have,

L.H.S. = $\frac{sinA-sinB}{cosA+cosB}$

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

= $\frac{sin\frac{A-B}{2}}{cos\frac{A-B}{2}}$

= $tan\frac{A-B}{2}$

= R.H.S.

Hence proved.

(iv) $\frac{sinA+sinB}{sinA-sinB}=tan(\frac{A+B}{2})cot(\frac{A-B}{2})$

Solution:

We have,

L.H.S. = $\frac{sinA+sinB}{sinA-sinB}$

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

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

= $tan(\frac{A+B}{2})cot(\frac{A-B}{2})$

= R.H.S.

Hence proved.

(iv) $\frac{cosA+cosB}{cosB-cosA}=cot(\frac{A+B}{2})cot(\frac{A-B}{2})$

Solution:

We have,

L.H.S. = $\frac{cosA+cosB}{cosB-cosA}$

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

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

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

= R.H.S.

Hence proved.

Question 8. Prove that:

(i) $\frac{\sin A + \sin 3A + \sin 5A}{\cos A + \cos 3A + \cos 5A} = \tan 3A$

Solution:

We have,

L.H.S. = $\frac{\sin A + \sin 3A + \sin 5A}{\cos A + \cos 3A + \cos 5A}$

= $\frac{\sin A + \sin 5A + \sin 3A}{\cos A + \cos 5A + \cos 3A}$

= $\frac{2\sin \left( \frac{A + 5A}{2} \right) \cos \left( \frac{A - 5A}{2} \right) + \sin 3A}{2\cos \left( \frac{A + 5A}{2} \right) \cos \left( \frac{A - 5A}{2} \right) + \cos 3A}$

= $\frac{2 \sin 3A \cos \left( - 2A \right) + \sin 3A}{2\cos 3A \cos \left( - 2A \right) + \cos 3A}$

= $\frac{2\sin 3A \cos 2A + \sin 3A}{2\cos 3A \cos 2A + \cos 3A}$

= $\frac{\sin 3A \left[ 2\cos 2A + 1 \right]}{\cos 3A \left[ 2\cos 2A + 1 \right]}$

= tan 3A

= R.H.S.

Hence proved.

(ii) $\frac{cos3A+2cos5A+cos7A}{cosA+2cos3A+cos5A}=\frac{cos5A}{cos3A}$

Solution:

We have,

L.H.S. = $\frac{cos3A+2cos5A+cos7A}{cosA+2cos3A+cos5A}$

= $\frac{2cos\frac{10A}{2}cos\frac{4A}{2}+2cos5A}{cos\frac{6A}{2}cos\frac{4A}{2}+2cos3A}$

= $\frac{2cos5Acos2A+2cos5A}{2cos3Acos2A+2cos3A}$

= $\frac{2cos5A(cos2A+1)}{2cos3A(cos2A+1)}$

= $\frac{cos5A}{cos3A}$

= R.H.S.

Hence proved.

(iii) $\frac{\cos 4A + \cos 3A + \cos 2A}{\sin 4A + \sin 3A + \sin 2A} = \cot 3A$

Solution:

We have,

L.H.S. = $\frac{\cos 4A + \cos 3A + \cos 2A}{\sin 4A + \sin 3A + \sin 2A}$

= $\frac{\cos 4A + \cos 2A + \cos 3A}{\sin 4A + \sin 2A + \sin 3A}$

= $\frac{2\cos \left( \frac{4A + 2A}{2} \right) \cos \left( \frac{4A - 2A}{2} \right) + \cos 3A}{2\sin \left( \frac{4A + 2A}{2} \right) \cos \left( \frac{4A - 2A}{2} \right) + \sin 3A}$

= $\frac{2\cos 3A \cos A + \cos 3A}{2\sin 3A \cos A + \sin 3A}$

= $\frac{\cos 3A\left[ 2\cos A + 1 \right]}{\sin 3A\left[ 2\cos A + 1 \right]}$

= cot 3A

= R.H.S.

Hence proved.

(iv) $\frac{\sin 3A + \sin 5A + \sin 7A + \sin 9A}{\cos 3A + \cos 5A + \cos 7A + \cos 9A} = \tan 6A$

Solution:

We have,

L.H.S. = $\frac{\sin 3A + \sin 5A + \sin 7A + \sin 9A}{\cos 3A + \cos 5A + \cos 7A + \cos 9A}$

= $\frac{\sin 3A + \sin 9A + \sin 5A + \sin 7A}{\cos 3A + \cos 9A + \cos 5A + \sin 7A}$

= $\frac{2\sin \left( \frac{3A + 9A}{2} \right) \cos \left( \frac{3A - 9A}{2} \right) + 2\sin \left( \frac{5A + 7A}{2} \right) \cos \left( \frac{5A - 7A}{2} \right)}{2\cos \left( \frac{3A + 9A}{2} \right) \cos \left( \frac{3A - 9A}{2} \right) + 2\cos \left( \frac{5A + 7A}{2} \right) \cos \left( \frac{5A - 7A}{2} \right)}$

= $\frac{2\sin 6A \cos \left( - 3A \right) + 2\sin 6A \cos \left( - A \right)}{2\cos 6A \cos \left( - 3A \right) + 2\cos 6A \cos \left( - A \right)}$

= $\frac{2\sin 6A \cos 3A + 2\sin 6A \cos A}{2\cos 6A \cos 3A + 2\cos 6A \cos A}$

= $\frac{2\sin 6A\left[ \cos 3A + \cos A \right]}{2\cos 6A\left[ \cos 3A + \cos A \right]}$

= tan 6A

= R.H.S.

Hence proved.

(v) $\frac{\sin 5A - \sin 7A + \sin 8A - \sin 4A}{\cos 4A + \cos 7A - \cos 5A - \cos 8A} = \cot 6A$

Solution:

We have,

L.H.S. = $\frac{\sin 5A - \sin 7A + \sin 8A - \sin 4A}{\cos 4A + \cos 7A - \cos 5A - \cos 8A}$

= $\frac{\sin 5A - \sin 7A + \sin 8A - \sin 4A}{\cos 4A - \cos 8A + \cos 7A - \sin 5A}$

= $\frac{2\sin \left( \frac{5A - 7A}{2} \right) \cos \left( \frac{5A + 7A}{2} \right) + 2\sin \left( \frac{8A - 4A}{2} \right) \cos \left( \frac{8A + 4A}{2} \right)}{- 2\sin \left( \frac{4A + 8A}{2} \right) \sin \left( \frac{4A - 8A}{2} \right) - 2\sin \left( \frac{7A + 5A}{2} \right) \sin \left( \frac{7A - 5A}{2} \right)}$

= $\frac{2\sin \left( - A \right) \cos 6A + 2\sin 2A \cos 6A}{- 2\sin 6A \sin \left( - 2A \right) - 2\sin 6A \sin A}$

= $\frac{- 2\sin A \cos 6A + 2\sin 2A cos 6A}{2\sin 6A \sin 2A - 2\sin 6A \sin A}$

= $\frac{2\cos 6A\left[ \sin 2A - \sin A \right]}{2\sin 6A\left[ \sin 2A - \sin A \right]}$

= cot 6A

= R.H.S.

Hence proved.

(vi) $\frac{\sin 5A \cos 2A - \sin 6A \cos A}{\sin A \sin 2A - \cos 2A \cos 3A} = \tan A$

Solution:

We have,

L.H.S. = $\frac{\sin 5A \cos 2A - \sin 6A \cos A}{\sin A \sin 2A - \cos 2A \cos 3A}$

Multiplying numerator and denominator by 2, we get

= $\frac{2\sin 5A \cos 2A - 2\sin 6A \cos A}{2\sin A \sin 2A - 2\cos 2A \cos 3A}$

= $\frac{\sin \left( 5A + 2A \right) + \sin \left( 5A - 2A \right) - \sin \left( 6A + A \right) - \sin \left( 6A - A \right)}{\cos \left( A - 2A \right) + \cos \left( A + 2A \right) - \cos \left( 2A + 3A \right) - \cos \left( 2A - 3A \right)}$

= $\frac{\sin 7A + \sin 3A - \sin 7A - \sin 5A}{\cos \left( - A \right) + \cos 3A - \cos 5A - \cos \left( - A \right)}$

= $\frac{\sin 7A + \sin 3A - \sin 7A - \sin 5A}{\cos A + \cos 3A - \cos 5A - cos A}$

= $\frac{\sin 3A - \sin 5A}{\cos 3A - \cos 5A}$

= $\frac{2\sin \left( \frac{3A - 5A}{2} \right) \cos \left( \frac{3A + 5A}{2} \right)}{- 2\cos \left( \frac{3A + 5A}{2} \right) \cos\left( \frac{3A - 5A}{2} \right)}$

= $\frac{\sin \left( - A \right) \cos 4A}{- \cos 4A \cos \left( - A \right)}$

= $\frac{- \sin A \cos 4A}{- \cos 4A\cos A}$

= $\frac{\sin A}{\cos A}$

= tan A

= R.H.S.

Hence proved.

(vii) $\frac{sin11AsinA+sin7Asin3A}{cos11AsinA+cos7Asin3A}=tan8A$

Solution:

We have,

L.H.S. = $\frac{sin11AsinA+sin7Asin3A}{cos11AsinA+cos7Asin3A}$

= $\frac{cos10A-cos12A+cos4A-cos10A}{sin12A-sin10A+sin10A-sin4A}$

= $\frac{cos4A-cos12A}{sin12A-sin4A}$

= $\frac{2sin\frac{16A}{2}sin\frac{8A}{2}}{2cos\frac{16A}{2}sin\frac{8A}{2}}$

= $\frac{2sin8Asin4A}{2cos8Asin4A}$

= tan 8A

= R.H.S.

Hence proved.

(viii) $\frac{\sin 3A \cos 4A - \sin A \cos 2A}{\sin 4A \sin A + \cos 6A \cos A} = \tan 2A$

Solution:

We have,

L.H.S. = $\frac{\sin 3A \cos 4A - \sin A \cos 2A}{\sin 4A sin A + \cos 6A \cos A}$

On multiplying numerator and denominator by 2, we get

= $\frac{2\sin 3A \cos 4A - 2\sin A \cos 2A}{2\sin 4A \sin A + 2\cos 6A \cos A}$

= $\frac{\sin \left( 3A + 4A \right) + \sin \left( 3A - 4A \right) - \sin \left( A + 2A \right) - \sin \left( A - 2A \right)}{\cos \left( 4A - A \right) - \cos \left( 4A + A \right) + \cos \left( 6A + A \right) + \cos \left( 6A - A \right)}$

= $\frac{\sin 7A + \sin \left( - A \right) - \sin 3A - \sin \left( - A \right)}{\cos 3A - \cos 5A + \cos 7A + \cos 5A}$

= $\frac{\sin 7A - \sin A - \sin 3A + \sin A}{\cos 3A - \cos 5A + \cos 7A + \cos 5A}$

= $\frac{\sin 7A - \sin 3A}{\cos 3A + \cos 7A}$

= $\frac{2\sin \left( \frac{7A - 3A}{2} \right) \cos \left( \frac{7A + 3A}{2} \right)}{2\cos \left( \frac{3A + 7A}{2} \right) \cos \left( \frac{3A - 7A}{2} \right)}$

= $\frac{\sin 2A \cos 5A}{\cos 5A \cos \left( - 2A \right)}$

= $\frac{\sin 2A \cos 5A}{\cos 5A \cos 2A}$

= tan 2A

= R.H.S.

Hence proved.

(ix) $\frac{\sin A \sin 2A + \sin 3A \sin 6A}{\sin A \cos 2A + \sin 3A \cos 6A} = \tan 5A$

Solution:

We have,

L.H.S. = $\frac{\sin A \sin 2A + \sin 3A \sin 6A}{\sin A \cos 2A + \sin 3A \cos 6A}$

On multiplying numerator and denominator by 2, we get

= $\frac{2\sin A \sin 2A + 2\sin 3A \sin 6A}{2\sin A \cos 2A + 2\sin 3A \cos 6A}$

= $\frac{\cos \left( A - 2A \right) - \cos \left( A + 2A \right) + \cos \left( 3A - 6A \right) - \cos \left( 3A + 6A \right)}{\sin \left( A + 2A \right) + \sin \left( A - 2A \right) + \sin \left( 3A + 6A \right) + \sin \left( 3A - 6A \right)}$

= $\frac{\cos\left( - A \right) - \cos 3A + \cos \left( - 3A \right) - \cos 9A}{\sin 3A \sin\left( - A \right) + \sin 9A + \sin \left( - 3A \right)}$

= $\frac{\cos A - \cos 3A + \cos 3A - \cos 9A}{\sin 3A - \sin A + \sin 9A - \sin 3A}$

= $\frac{\cos A - \cos 9A}{\sin 9A - \sin A}$

= $\frac{- 2\sin \left( \frac{A + 9A}{2} \right) \sin \left( \frac{A - 9A}{2} \right)}{2\cos \left( \frac{A + 9A}{2} \right) \sin \left( \frac{9A - A}{2} \right)}$

= $\frac{\sin5A\cos4A}{\sin 5A \cos \left( - 4A \right)}$

= tan 5A

= R.H.S.

Hence proved.

(x) $\frac{\sin A + 2 \sin 3A + \sin 5A}{\sin 3A + 2 \sin 5A + \sin 7A} = \frac{\sin 3A}{\sin 5A}$

Solution:

We have,

L.H.S. = $\frac{\sin A + 2 \sin 3A + \sin 5A}{\sin 3A + 2 \sin 5A + \sin 7A}$

= $\frac{\sin 5A \sin A + 2\sin 3A}{\sin 7A \sin 3A + 2\sin 5A }$

= $\frac{2\sin \left( \frac{5A +A}{2} \right) \cos \left( \frac{5A - A}{2}\right) + 2\sin \left(3A \right)}{2\sin \left(\frac{ 7A + 3A}{2} \right) \cos \left(\frac{7A - 3A}{2} \right) + 2\sin \left( 5A \right) }$

= $\frac{2\sin3A\cos2A+2\sin3A}{2\sin5A\cos2A + 2\sin5A}$

= $\frac{2\sin3A(\cos2A +1)}{2\sin5A(\cos2A +1)}$

= sin3A/sin5A

= R.H.S.

Hence proved.

(xi) $\frac{sin(θ+Ø)-2sinθ+sin(θ-Ø)}{cos(θ+Ø)-2cosθ+cos(θ-Ø)}=tanθ$

Solution:

We have,

L.H.S. = $\frac{sin(θ+Ø)-2sinθ+sin(θ-Ø)}{cos(θ+Ø)-2cosθ+cos(θ-Ø)}$

= $\frac{sin(θ+Ø)+sin(θ-Ø)-2sinθ}{cos(θ+Ø)+cos(θ-Ø)-2cosθ}$

= $\frac{2sin\frac{(θ+Ø+θ-Ø)}{2}cos\frac{(θ+Ø-θ+Ø)}{2}-2sinθ}{2cos\frac{(θ+Ø+θ-Ø)}{2}cos\frac{(θ+Ø-θ+Ø)}{2}-2cosθ}$

= $\frac{2sin\frac{2θ}{2}cos\frac{2Ø}{2}-2sinθ}{2cos\frac{2θ}{2}cos\frac{2Ø}{2}-2cosθ}$

= $\frac{2sinθcosØ-2sinθ}{2cosθcosØ-2cosθ}$

= $\frac{2sinθ(cosØ-1)}{2cosθ(cosØ-1)}$

= $\frac{sinθ}{cosθ}$

= 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)

= $2sin\frac{α+β}{2}cos\frac{α-β}{2}+2cos(\frac{γ+α+β+γ}{2})sin(\frac{γ-α-β-γ}{2})$

= $2sin\frac{α+β}{2}cos\frac{α-β}{2}+2cos(\frac{α+β+2γ}{2})sin(\frac{-(α+β)}{2})$

= $2sin\frac{α+β}{2}cos\frac{α-β}{2}-2cos(\frac{α+β+2γ}{2})sin(\frac{α+β}{2})$

= $2sin\frac{α+β}{2}(cos\frac{α-β}{2}-cos\frac{α+β+2γ}{2})$

= $2sin\frac{α+β}{2}(-2sin\frac{\frac{α-β+α+β+2γ}{2}}{2}sin\frac{\frac{α-β-α-β-2γ}{2}}{2})$

= $2sin\frac{α+β}{2}(2sin\frac{α+γ}{2}sin\frac{β+γ}{2})$

= 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)

= $2cos\frac{A+B+C+A-B+C}{2}cos\frac{A+B+C-A+B-C}{2}+2cos\frac{A+B-C-A+B+C}{2}cos\frac{A+B-C+A-B-C}{2}$

= $2cos\frac{2A+2C}{2}cos\frac{2B}{2}+2cos\frac{2B}{2}cos\frac{2A-2C}{2}$

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

= 2 cos B [cos (A + C) + cos (A − C)]

= 2 cos B $\left[2cos\frac{A+C+A-C}{2}cos\frac{A+C-A+C}{2}\right]$

= 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 $tan\frac{A+B}{2}=\frac{1}{2}$ .

Solution:

We have,

cos A + cos B = 1/2

sin A + sin B = 1/4

=> $\frac{sinA+sinB}{cosA+cosB}=\frac{\frac{1}{4}}{\frac{1}{2}}$

=> $\frac{sinA+sinB}{cosA+cosB}=\frac{1}{2}$

=> $\frac{2sin\frac{A+B}{2}cos\frac{A-B}{2}}{2cos\frac{A+B}{2}cos\frac{A-B}{2}}=\frac{1}{2}$

=> $\frac{sin\frac{A+B}{2}}{cos\frac{A+B}{2}}=\frac{1}{2}$

=> $tan\frac{A+B}{2}=\frac{1}{2}$

Hence proved.

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Last Updated : 19 Jul, 2021

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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

(ii) sin 5θ – sin θ

(iii) cos 12θ + cos 8θ

(iv) cos 12θ – cos 4θ

(v) sin 2θ + cos 4θ

Question 2. Prove that :

(i) sin 38° + sin 22° = sin 82°

(ii) cos 100° + cos 20° = cos 40°

(iii) sin 50° + sin 10° = cos 20°

(iv) sin 23° + sin 37° = cos 7°

(v) sin 105° + cos 105° = cos 45°

(vi) sin 40° + sin 20° = cos 10°

Question 3. Prove that:

(i) cos 55° + cos 65° + cos 175° = 0

(ii) sin 50° – sin 70° + sin 10° = 0

(iii) cos 80° + cos 40° – cos 20° = 0

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

(v) sin 5π/18 – cos 4π/9 = √3 sin π/9

(vi) cos π/12 – sin π/12 = 1/√2

(vii) sin 80° – cos 70° = cos 50°

(viii) sin 51° + cos 81° = cos 21°

Question 4. Prove that:

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

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

Question 5. Prove that:

(i) sin 65o + cos 65o = √2 cos 20o

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

Question 6. Prove that:

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

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

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

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

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

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

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

Question 7. Prove that:

(i)

(ii)

(iii)

(iv)

(iv)

Question 8. Prove that:

(i)

(ii)

(iii)

(iv)

(v)

(vi)

(vii)

(viii)

(ix)

(x)

(xi)

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

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

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

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Data Structures and Algorithms - Self Paced

CBSE Class 12 Computer Science

GATE CS & IT 2024

(i) sin 65^o + cos 65^o = √2 cos 20^o

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

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

(i) $\frac{\sin A + \sin 3A}{\cos A - \cos 3A} = \cot A$

(ii) $\frac{\sin 9A - \sin 7A}{\cos 7A - \cos 9A} = \cot 8A$

(iii) $\frac{sinA-sinB}{cosA+cosB}=tan\frac{A-B}{2}$

(iv) $\frac{sinA+sinB}{sinA-sinB}=tan(\frac{A+B}{2})cot(\frac{A-B}{2})$

(iv) $\frac{cosA+cosB}{cosB-cosA}=cot(\frac{A+B}{2})cot(\frac{A-B}{2})$

(i) $\frac{\sin A + \sin 3A + \sin 5A}{\cos A + \cos 3A + \cos 5A} = \tan 3A$

(ii) $\frac{cos3A+2cos5A+cos7A}{cosA+2cos3A+cos5A}=\frac{cos5A}{cos3A}$

(iii) $\frac{\cos 4A + \cos 3A + \cos 2A}{\sin 4A + \sin 3A + \sin 2A} = \cot 3A$

(iv) $\frac{\sin 3A + \sin 5A + \sin 7A + \sin 9A}{\cos 3A + \cos 5A + \cos 7A + \cos 9A} = \tan 6A$

(v) $\frac{\sin 5A - \sin 7A + \sin 8A - \sin 4A}{\cos 4A + \cos 7A - \cos 5A - \cos 8A} = \cot 6A$

(vi) $\frac{\sin 5A \cos 2A - \sin 6A \cos A}{\sin A \sin 2A - \cos 2A \cos 3A} = \tan A$

(vii) $\frac{sin11AsinA+sin7Asin3A}{cos11AsinA+cos7Asin3A}=tan8A$

(viii) $\frac{\sin 3A \cos 4A - \sin A \cos 2A}{\sin 4A \sin A + \cos 6A \cos A} = \tan 2A$

(ix) $\frac{\sin A \sin 2A + \sin 3A \sin 6A}{\sin A \cos 2A + \sin 3A \cos 6A} = \tan 5A$

(x) $\frac{\sin A + 2 \sin 3A + \sin 5A}{\sin 3A + 2 \sin 5A + \sin 7A} = \frac{\sin 3A}{\sin 5A}$

(xi) $\frac{sin(θ+Ø)-2sinθ+sin(θ-Ø)}{cos(θ+Ø)-2cosθ+cos(θ-Ø)}=tanθ$

Question 10. If cos A + cos B = 1/2 and sin A + sin B = 1/4, prove that $tan\frac{A+B}{2}=\frac{1}{2}$ .