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

In Class 11 Mathematics, Chapter 8 of RD Sharma's textbook focuses on Transformation formulas which are crucial for simplifying complex trigonometric expressions. This chapter helps students understand how to transform trigonometric functions into more manageable forms using specific formulae. Mastery of these formulae is essential for solving a wide range of problems in trigonometry and calculus.

Transformation Formulas

The Transformation formulae involve the conversion of the trigonometric functions from one form to another. These formulae are used to simplify trigonometric expressions and solve equations involving trigonometric functions. Key formulae include:

• Addition and Subtraction Formulae: For example, sin(A±B) and cos(A±B) which express the sine and cosine of the sum or difference of the two angles.
• Double Angle Formulae: Such as the sin(2A) and cos(2A) which relate functions of double angles to the functions of single angles.
• Half Angle Formulae: For instance, sin²( A/2) and cos²(A/2) express trigonometric functions of the half-angles in terms of functions of the full angles.

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 thattan\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.

Read More:

• Transformation Matrix
• Linear Algebra