RD Sharma Solutions for Class 11 Chapter 8 - Transformation Formulae

RD Sharma Solutions are basically designed for CBSE students in accordance with the latest syllabus prescribed by the CBSE Board. In this chapter, we shall establish two sets of transformation formulae. These two sets of formulae are of fundamental importance and one should have a thorough acquaintance with these formulae. The solutions are designed by BYJU’S experts to boost confidence among students in understanding the concepts covered in this chapter. The practice is an essential task to learn and score well in Mathematics. The PDFs of RD Sharma Class 11 Solutions Maths Chapter 8 Transformation Formulae are provided here. The students are required to go through RD Sharma Solutions thoroughly before the final exams to score well and intensify their problem-solving abilities.

Chapter 8 – Transformation Formulae contains two exercises and the RD Sharma Solutions present on this page provide solutions to the questions present in each exercise. Now, let us have a look at the concepts discussed in this chapter.

  • Formulae to transform the product into sum or difference.
  • Formulae to transform the sum or difference into a product.

Download the Pdf of RD Sharma Solutions for Class 11 Maths Chapter 8 – Transformation Formulae

 

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Also, access exercises of RD Sharma Solutions for Class 11 Maths Chapter 8 – Transformation Formulae

Exercise 8.1 Solutions

Exercise 8.2 Solutions

Access answers to RD Sharma Solutions for Class 11 Maths Chapter 8 – Transformation Formulae

EXERCISE 8.1 PAGE NO: 8.6

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 – B)

2 sin 3x cos x = sin (3x + x) + sin (3x – 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) – sin (A – B)

2 cos 3x sin 2x = sin (3x + 2x) – sin (3x – 2x)

= sin (5x) – sin (x)

= sin 5x – sin x


(iii) 2 sin 4x sin 3x

By using the formula,

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) 2 cos 7x cos 3x

By using the formula,

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

(ii) 2 cos 5π/12 cos π/12 = 1/2

(iii) 2 sin 5π/12 cos π/12 = (√3 + 2)/2

Solution:

(i) 2 sin 5π/12 sin π/12 = 1/2

By using the formula,

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

2 sin 5π/12 sin π/12 = cos (5π/12 – π/12) – cos (5π/12 + π/12)

= cos (4π/12) – cos (6π/12)

= cos (π/3) – cos (π/2)

= cos (180o/3) – cos (180o/2)

= cos 60° – cos 90°

= 1/2 – 0

= 1/2

Hence Proved.

(ii) 2 cos 5π/12 cos π/12 = 1/2

By using the formula,

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

2 cos 5π/12 cos π/12 = cos (5π/12 + π/12) + cos (5π/12 – π/12)

= cos (6π/12) + cos (4π/12)

= cos (π/2) + cos (π/3)

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

= cos 90° + cos 60°

= 0 + 1/2

= 1/2

Hence Proved.

(iii) 2 sin 5π/12 cos π/12 = (√3 + 2)/2

By using the formula,

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

2 sin 5π/12 cos π/12 = sin (5π/12 + π/12) + sin (5π/12 – π/12)

= sin (6π/12) + sin (4π/12)

= sin (π/2) + sin (π/3)

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

= sin 90° + sin 60°

= 1 + √3

= (2 + √3)/2

= (√3 + 2)/2

Hence Proved.

3. show that:

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

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

Solution:

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

By using the formula,

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

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

sin 50o cos 85o = [sin(50o + 85o) + sin (50o – 85o)] / 2

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

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

= [sin (180o – 45o) – sin 35o] / 2

= [sin 45o – sin 35o] / 2

= [(1/√2) – sin 35o] / 2

= [(1 – sin 35o)/2] / 2

= (1 – sin 35o) / 22

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 – B)

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

sin 20o cos 115o = [sin(25o + 115o) + sin (25o – 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 (π/3 + x) cos (π/3 – x) = cos 3x

Solution:

Let us consider LHS:

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

By using the formula,

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

2 cos x (2 cos (π/3+x) cos (π/3 – x)) = 2 cos x (cos (π/3+x + π/3-x) + cos (π/3+x – π/3+ x))

= 2 cos x (cos (2π/3) + cos (2x))

= 2 cos x {cos 120° + cos 2x}

= 2 cos x {cos (180° – 60°) + cos 2x}

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

= 2 cos 2x cos x – 2 cos x cos 60°

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

= cos 3x + cos x – cos x

= cos 3x

= RHS

Hence Proved.

EXERCISE 8.2 PAGE NO: 8.17

1. Express each of the following as the product of sines and cosines:
(i) sin 12x + sin 4x
(ii) sin 5x – sin x
(iii) cos 12x + cos 8x
(iv) cos 12x – cos 4x

(v) sin 2x + cos 4x

Solution:

(i) sin 12x + sin 4x

By using the formula,

sin A + sin B = 2 sin (A+B)/2 cos (A-B)/2

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

= 2 sin 16x/2 cos 8x/2

= 2 sin 8x cos 4x

(ii) sin 5x – sin x

By using the formula,

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

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

= 2 cos 6x/2 sin 4x/2

= 2 cos 3x sin 2x

(iii) cos 12x + cos 8x

By using the formula,

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

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

= 2 cos 20x/2 cos 4x/2

= 2 cos 10x cos 2x

(iv) cos 12x – cos 4x

By using the formula,

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

cos 12x – cos 4x = -2 sin (12x + 4x)/2 sin (12x – 4x)/2

= -2 sin 16x/2 sin 8x/2

= -2 sin 8x sin 4x

(v) sin 2x + cos 4x

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

By using the formula,

sin A + sin B = 2 sin (A+B)/2 cos (A-B)/2

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

= 2 sin (90o – 2x)/2 cos (6x – 90o)/2

= 2 sin (45° – x) cos (3x – 45°)

= 2 sin (45° – x) cos {-(45° – 3x)} (since, {cos (-x) = cos x})

= 2 sin (45° – x) cos (45° – 3x)

= 2 sin (π/4 – x) cos (π/4 – 3x)

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°

Solution:

(i) sin 38° + sin 22° = sin 82°

Let us consider LHS:

sin 38° + sin 22°

By using the formula,

sin A + sin B = 2 sin (A+B)/2 cos (A-B)/2

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

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

= 2 sin 30o cos 8o

= 2 × 1/2 × cos 8o

= cos 8o

= cos (90° – 82°)

= sin 82° (since, {cos (90° – A) = sin A})

= RHS

Hence Proved.

(ii) cos 100° + cos 20° = cos 40°

Let us consider LHS:

cos 100° + cos 20°

By using the formula,

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

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

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

= 2 cos 60o cos 4o

= 2 × 1/2 × cos 40o

= cos 40o

= RHS

Hence Proved.

(iii) sin 50° + sin 10° = cos 20°

Let us consider LHS:

sin 50° + sin 10°

By using the formula,

sin A + sin B = 2 sin (A+B)/2 cos (A-B)/2

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

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

= 2 sin 30o cos 20o

= 2 × 1/2 × cos 20o

= cos 20o

= RHS

Hence Proved.

(iv) sin 23° + sin 37° = cos 7°

Let us consider LHS:

sin 23° + sin 37°

By using the formula,

sin A + sin B = 2 sin (A+B)/2 cos (A-B)/2

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

= 2 sin 60o/2 cos -14o/2

= 2 sin 30o cos -7o

= 2 × 1/2 × cos -7o

= cos 7o (since, {cos (-A) = cos A})

= RHS

Hence Proved.

(v) sin 105° + cos 105° = cos 45°

Let us consider LHS: sin 105° + cos 105°

sin 105° + cos 105° = sin 105o + sin (90o – 105o) [since, {sin (90° – A) = cos A}]

= sin 105o + sin (-15o)

= sin 105o – sin 15o [{sin(-A) = – sin A}]

By using the formula,

Sin A – sin B = 2 cos (A+B)/2 sin (A-B)/2

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

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

= 2 cos 60o sin 45o

= 2 × 1/2 × 1/2

= 1/2

= cos 45o

= RHS

Hence proved.

(vi) sin 40° + sin 20° = cos 10°

Let us consider LHS:

sin 40° + sin 20°

By using the formula,

sin A + sin B = 2 sin (A+B)/2 cos (A-B)/2

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

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

= 2 sin 30o cos 10o

= 2 × 1/2 × cos 10o

= cos 10o

= RHS

Hence Proved.

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°

Solution:

(i) cos 55° + cos 65° + cos 175° = 0

Let us consider LHS:

cos 55° + cos 65° + cos 175°

By using the formula,

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

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

= 2 cos 120o/2 cos (-10o)/2 – cos 5o (since, {cos (180° – A) = – cos A})

= 2 cos 60° cos (-5°) – cos 5° (since, {cos (-A) = cos A})

= 2 × 1/2 × cos 5o – cos 5o

= 0

= RHS

Hence Proved.

(ii) sin 50° – sin 70° + sin 10° = 0

Let us consider LHS:

sin 50° – sin 70° + sin 10°

By using the formula,

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

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

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

= 2 cos 60o (- sin 10o) + sin 10o [since,{sin (-A) = -sin (A)}]

= 2 × 1/2 × – sin 10o + sin 10o

= 0

= RHS

Hence proved.

(iii) cos 80° + cos 40° – cos 20° = 0

Let us consider LHS:

cos 80° + cos 40° – cos 20°

By using the formula,

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

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

= 2 cos 120o/2 cos 40o/2 – cos 20o

= 2 cos 60° cos 20o – cos 20°

= 2 × 1/2 × cos 20o – cos 20o

= 0

= RHS

Hence Proved.

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

Let us consider LHS:

cos 20° + cos 100° + cos 140°

By using the formula,

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

cos 20° + cos 100° + cos 140° = 2 cos (20o + 100o)/2 cos (20o – 100o) + cos (180o – 40o)

= 2 cos 120o/2 cos (-80o)/2 – cos 40o (since, {cos (180° – A) = – cos A})

= 2 cos 60° cos (-40°) – cos 40° (since, {cos (-A) = cos A})

= 2 × 1/2 × cos 40o – cos 40o

= 0

= RHS

Hence Proved.

(v) sin 5π/18 – cos 4π/9 = √3 sin π/9

Let us consider LHS:

sin 5π/18 – cos 4π/9 = sin 5π/18 – sin (π/2 – 4π/9) (since, cos A = sin (90o – A))

= sin 5π/18 – sin (9π – 8π)/18

= sin 5π/18 – sin π/18

By using the formula,

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

RD Sharma Solutions for Class 11 Maths Chapter 8 – Transformation Formulae image - 1

= 2 cos (6π/36) sin (4π/36)

= 2 cos π/6 sin π/9

= 2 cos 30o sin π/9

= 2 × √3/2 × sin π/9

= √3 sin π/9

= RHS

Hence proved.

(vi) cos π/12 – sin π/12 = 1/√2

Let us consider LHS:

cos π/12 – sin π/12 = sin (π/2 – π/12) – sin π/12 (since, cos A = sin(90o – A))

= sin (6π – 5π)/12 – sin π/12

= sin 5π/12 – sin π/12

By using the formula,

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

RD Sharma Solutions for Class 11 Maths Chapter 8 – Transformation Formulae image - 2

= 2 cos (6π/24) sin (4π/24)

= 2 cos π/4 sin π/6

= 2 cos 45o sin 30o

= 2 × 1/√2 × 1/2

= 1/√2

= RHS

Hence proved.

(vii) sin 80° – cos 70° = cos 50°

sin 80° = cos 50° + cos 70o

So, now let us consider RHS

cos 50° + cos 70o

By using the formula,

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

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

= 2 cos 120o/2 cos (-20o)/2

= 2 cos 60o cos (-10o)

= 2 × 1/2 × cos 10o (since, cos (-A) = cos A)

= cos 10o

= cos (90° – 80°)

= sin 80° (since, cos (90° – A) = sin A)

= LHS

Hence Proved.

(viii) sin 51° + cos 81° = cos 21°

Let us consider LHS:

sin 51° + cos 81° = sin 51o + sin (90o – 81o)

= sin 51o + sin 9o (since, sin (90° – A) = cos A)

By using the formula,

sin A + sin B = 2 sin (A+B)/2 cos (A-B)/2

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

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

= 2 sin 30o cos 21o

= 2 × 1/2 × cos 21o

= cos 21o

= RHS

Hence proved.

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

Solution:

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

Let us consider LHS:

cos (3π/4 + x) – cos (3π/4 – x)

By using the formula,

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 (since, (π-A) = sin A)

= -2 × 1/√2 × sin x

= -√2 sin x

= RHS

Hence proved.

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

Let us consider LHS:

cos (π/4 + x) + cos (π/4 – x)

By using the formula,

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

= RHS

Hence proved.

5. Prove that:

(i) sin 65o + cos 65o = √2 cos 20o

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

Solution:

(i) sin 65o + cos 65o = √2 cos 20o

Let us consider LHS:

sin 65o + cos 65o = sin 65o + sin (90o – 65o)

= sin 65o + sin 25o

By using the formula,

sin A + sin B = 2 sin (A+B)/2 cos (A-B)/2

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

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

= 2 sin 45o cos 20o

= 2 × 1/√2 × cos 20o

= √2 cos 20o

= RHS

Hence proved.

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

Let us consider LHS:

sin 47o + cos 77o = sin 47o + sin (90o – 77o)

= sin 47o + sin 13o

By using the formula,

sin A + sin B = 2 sin (A+B)/2 cos (A-B)/2

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

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

= 2 sin 30o cos 17o

= 2 × 1/2 × cos 17o

= cos 17o

= RHS

Hence proved.

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

Solution:

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

Let us consider LHS:

cos 3A + cos 5A + cos 7A + cos 15A

So now,

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

By using the formula,

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)

Again by using the formula,

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

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

= RHS

Hence proved.

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

Let us consider LHS:

cos A + cos 3A + cos 5A + cos 7A

So now,

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

By using the formula,

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)

Again by using the formula,

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

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

= RHS

Hence proved.

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

Let us consider LHS:

sin A + sin 2A + sin 4A + sin 5A

So now,

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

By using the formula,

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)

Again by using the formula,

sin A + sin B = 2 sin (A+B)/2 cos (A-B)/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

= RHS

Hence proved.

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

Let us consider LHS:

sin 3A + sin 2A – sin A

So now,

(sin 3A – sin A) + sin 2A

By using the formula,

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

We know that, sin 2A = 2 sin A cos A

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

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

Again by using the formula,

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

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

= RHS

Hence proved.

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

Let us consider LHS:

cos 20o cos 100o + cos 100o cos 140o – cos 140o cos 200o =

We shall multiply and divide by 2 we get,

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

We know that 2 cos A cos B = cos (A+B) + cos (A-B)

So,

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

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

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

We know, cos (180o + A) = – cos A, cos (90o + A) = – sin A, cos (360o – A) = cos A

So,

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

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

Again by using the formula,

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

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

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

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

= 1/2 [- 1/2 + 2×1/2×cos 20o – cos 20o – 1]

= 1/2 [-1/2 + cos 20o – cos 20o – 1]

= 1/2 [-1/2 -1]

= 1/2 [-3/2]

= -3/4

= RHS

Hence proved.

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

Let us consider LHS:

sin x/2 sin 7x/2 + sin 3x/2 sin 11x/2 =

We shall multiply and divide by 2 we get,

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

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

So,

= 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]

Again by using the formula,

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

= -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

= RHS

Hence proved.

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

Let us consider LHS:

cos x cos x/2 – cos 3x cos 9x/2 =

We shall multiply and divide by 2 we get,

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

We know that 2 cos A cos B = cos (A+B) + cos (A-B)

So,

= 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]

Again by using the formula,

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

= RHS

Hence proved.

7. Prove that:

RD Sharma Solutions for Class 11 Maths Chapter 8 – Transformation Formulae image - 3

RD Sharma Solutions for Class 11 Maths Chapter 8 – Transformation Formulae image - 4

Solution:

RD Sharma Solutions for Class 11 Maths Chapter 8 – Transformation Formulae image - 5

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RD Sharma Solutions for Class 11 Maths Chapter 8 – Transformation Formulae image - 7

RD Sharma Solutions for Class 11 Maths Chapter 8 – Transformation Formulae image - 8

RD Sharma Solutions for Class 11 Maths Chapter 8 – Transformation Formulae image - 9

RD Sharma Solutions for Class 11 Maths Chapter 8 – Transformation Formulae image - 10

RD Sharma Solutions for Class 11 Maths Chapter 8 – Transformation Formulae image - 11

8. Prove that:

RD Sharma Solutions for Class 11 Maths Chapter 8 – Transformation Formulae image - 12

RD Sharma Solutions for Class 11 Maths Chapter 8 – Transformation Formulae image - 13

RD Sharma Solutions for Class 11 Maths Chapter 8 – Transformation Formulae image - 14

Solution:

RD Sharma Solutions for Class 11 Maths Chapter 8 – Transformation Formulae image - 15
RD Sharma Solutions for Class 11 Maths Chapter 8 – Transformation Formulae image - 16

RD Sharma Solutions for Class 11 Maths Chapter 8 – Transformation Formulae image - 17

RD Sharma Solutions for Class 11 Maths Chapter 8 – Transformation Formulae image - 18

RD Sharma Solutions for Class 11 Maths Chapter 8 – Transformation Formulae image - 19
RD Sharma Solutions for Class 11 Maths Chapter 8 – Transformation Formulae image - 20

RD Sharma Solutions for Class 11 Maths Chapter 8 – Transformation Formulae image - 21
RD Sharma Solutions for Class 11 Maths Chapter 8 – Transformation Formulae image - 22

RD Sharma Solutions for Class 11 Maths Chapter 8 – Transformation Formulae image - 23

RD Sharma Solutions for Class 11 Maths Chapter 8 – Transformation Formulae image - 24

= cot 6A

= RHS

Hence proved.

RD Sharma Solutions for Class 11 Maths Chapter 8 – Transformation Formulae image - 25

RD Sharma Solutions for Class 11 Maths Chapter 8 – Transformation Formulae image - 26

RD Sharma Solutions for Class 11 Maths Chapter 8 – Transformation Formulae image - 27

RD Sharma Solutions for Class 11 Maths Chapter 8 – Transformation Formulae image - 28

By using the formulas,

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

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

RD Sharma Solutions for Class 11 Maths Chapter 8 – Transformation Formulae image - 29

RD Sharma Solutions for Class 11 Maths Chapter 8 – Transformation Formulae image - 30

RD Sharma Solutions for Class 11 Maths Chapter 8 – Transformation Formulae image - 31

RD Sharma Solutions for Class 11 Maths Chapter 8 – Transformation Formulae image - 32

RD Sharma Solutions for Class 11 Maths Chapter 8 – Transformation Formulae image - 33

= RHS

Hence proved.

RD Sharma Solutions for Class 11 Maths Chapter 8 – Transformation Formulae image - 34

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RD Sharma Solutions for Class 11 Maths Chapter 8 – Transformation Formulae image - 36

RD Sharma Solutions for Class 11 Maths Chapter 8 – Transformation Formulae image - 37

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

Solution:

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

Let us consider LHS:

sin α + sin β + sin γ – sin (α + β + γ)

By using the formulas,

Sin A + sin B = 2 sin (A+B)/2 cos (A-B)/2

Sin A – sin B = 2 cos (A+B)/2 sin (A-B)/2

RD Sharma Solutions for Class 11 Maths Chapter 8 – Transformation Formulae image - 38

RD Sharma Solutions for Class 11 Maths Chapter 8 – Transformation Formulae image - 39

= RHS

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

Let us consider LHS:

cos (A + B + C) + cos (A – B + C) + cos (A + B – C) + cos (-A + B + C)

so,

cos (A + B + C) + cos (A – B + C) + cos (A + B – C) + cos (-A + B + C) =

={cos (A + B + C) + cos (A – B + C)} + {cos (A + B – C) + cos (-A + B + C)}

By using the formula,

Cos A + cos B = 2 cos (A+B)/2 cos (A-B)/2

RD Sharma Solutions for Class 11 Maths Chapter 8 – Transformation Formulae image - 40

= 4 cos A cos B cos C

= RHS

Hence proved.

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