RD Sharma Class 11 Solutions for Chapter 8: Transformation Formulae are super helpful for students. This chapter teaches you how to change sums and differences of trigonometric ratios into products, and products back into sums or differences. It's like having special maths transformations that make tricky problems easier!

In the RD Sharma Solutions for Class 11 Maths Chapter 8, you'll find easy-to-follow steps for all the exercises. We focus on the product-to-sum and sum-to-product formulae for sine, cosine, and tangent. These maths formulae are very important for solving problems in school tests and big exams like JEE. This chapter builds on what you already know about trigonometry and gives you tools to solve equations and prove difficult identities by changing them into simpler forms. It's all about transformation equations!

Many students ask, "What are the transformation formulae in trigonometry?" or "How to use transformation formulae in trigonometric identities?" This chapter answers all those questions with clear examples. You'll also learn about the "Application of transformation formulae in solving trigonometric equations" and the "Difference between transformation and standard trigonometric formulae." The solutions also give you a "List of transformation formulae for Class 11 Maths" and show the "Derivation of transformation formulae in trigonometry."

Download RD Sharma Class 11 Chapter 8 Trigonometric transformations Solutions PDF Free 

Get complete RD Sharma Class 11 Maths solutions for Transformation Formulae. It includes easy step-by-step answers, solved examples, and extra practice questions to help you understand these important formulae list and how to use them. These are great "Examples of transformation formulae in mathematics." Download your free PDF to get a "Step by step guide to transformation formulae for competitive exams" and do better in your Class 11 Maths studies with trusted RD Sharma Solutions.

RD Sharma Class 11 Solutions Chapter 8: Transformation Formulae

Key Transformation Formulae

Product to Sum:

• sin A cos B = 1/2[sin(A+B) + sin(A-B)]
• cos A sin B = 1/2[sin(A+B) - sin(A-B)]
• cos A cos B = 1/2[cos(A+B) + cos(A-B)]
• sin A sin B = 1/2[cos(A-B) - cos(A+B)]

Sum to Product:

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

Question 1

Express 2sin 3x cos x as the sum or difference of sines and cosines.

Solution:

Using: sin A cos B = 1/2[sin(A+B) + sin(A-B)]

Step 1: Apply the formula with A = 3x and B = x

sin 3x cos x = 1/2[sin(3x + x) + sin(3x - x)] = 1/2[sin 4x + sin 2x]

Step 2: Multiply by 2

2sin 3x cos x = 2 × 1/2[sin 4x + sin 2x] = sin 4x + sin 2x

Answer: 2sin 3x cos x = sin 4x + sin 2x

Question 2

Express 2cos 3x sin 2x as the sum or difference of sines and cosines.

Solution:

Using: cos A sin B = 1/2[sin(A+B) - sin(A-B)]

Step 1: Apply the formula with A = 3x and B = 2x

cos 3x sin 2x = 1/2[sin(3x + 2x) - sin(3x - 2x)] = 1/2[sin 5x - sin x]

Step 2: Multiply by 2

2cos 3x sin 2x = 2 × 1/2[sin 5x - sin x] = sin 5x - sin x

Answer: 2cos 3x sin 2x = sin 5x - sin x

Question 3

Express 2sin 4x sin 3x as the sum or difference of sines and cosines.

Solution:

Using: sin A sin B = 1/2[cos(A-B) - cos(A+B)]

Step 1: Apply the formula with A = 4x and B = 3x

sin 4x sin 3x = 1/2[cos(4x - 3x) - cos(4x + 3x)] = 1/2[cos x - cos 7x]

Step 2: Multiply by 2

2sin 4x sin 3x = 2 × 1/2[cos x - cos 7x] = cos x - cos 7x

Answer: 2sin 4x sin 3x = cos x - cos 7x

Question 4

Express 2cos 7x cos 3x as the sum or difference of sines and cosines.

Solution:

Using: cos A cos B = 1/2[cos(A+B) + cos(A-B)]

Step 1: Apply the formula with A = 7x and B = 3x

cos 7x cos 3x = 1/2[cos(7x + 3x) + cos(7x - 3x)] = 1/2[cos 10x + cos 4x]

Step 2: Multiply by 2

2cos 7x cos 3x = 2 × 1/2[cos 10x + cos 4x] = cos 10x + cos 4x

Answer: 2cos 7x cos 3x = cos 10x + cos 4x

Question 5

Express sin 12x + sin 4x as a product of sines and cosines.

Solution:

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

Step 1: Apply the formula with A = 12x and B = 4x

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

Step 2: Simplify

= 2sin(16x/2)cos(8x/2) = 2sin 8x cos 4x

Answer: sin 12x + sin 4x = 2sin 8x cos 4x

Question 6

Express sin 5x - sin x as a product of sines and cosines.

Solution:

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

Step 1: Apply the formula with A = 5x and B = x

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

Step 2: Simplify

= 2cos(6x/2)sin(4x/2) = 2cos 3x sin 2x

Answer: sin 5x - sin x = 2cos 3x sin 2x

Question 7

Express cos 12x + cos 8x as a product of sines and cosines.

Solution:

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

Step 1: Apply the formula with A = 12x and B = 8x

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

Step 2: Simplify

= 2cos(20x/2)cos(4x/2) = 2cos 10x cos 2x

Answer: cos 12x + cos 8x = 2cos 10x cos 2x

Question 8

Express cos 12x - cos 4x as a product of sines and cosines.

Solution:

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

Step 1: Apply the formula with A = 12x and B = 4x

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

Step 2: Simplify

= -2sin(16x/2)sin(8x/2) = -2sin 8x sin 4x

Answer: cos 12x - cos 4x = -2sin 8x sin 4x

Question 9

Express sin 2x + cos 4x as a product of sines and cosines.

Solution:

Step 1: Convert cos 4x to sine form

cos 4x = sin(90° - 4x) = sin(π/2 - 4x)

Step 2: Apply sum-to-product formula

sin 2x + sin(π/2 - 4x) = 2sin((2x + (π/2 - 4x))/2)cos((2x - (π/2 - 4x))/2)

Step 3: Simplify

= 2sin(π/4 - x)cos(3x - π/4)

Answer: sin 2x + cos 4x = 2sin(π/4 - x)cos(3x - π/4)

Question 10

Prove that sin 38° + sin 22° = sin 82°.

Solution:

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

Step 1: Apply the formula to LHS

sin 38° + sin 22° = 2sin((38° + 22°)/2)cos((38° - 22°)/2)

Step 2: Simplify

= 2sin(60°/2)cos(16°/2) = 2sin 30° cos 8°

Step 3: Since sin 30° = 1/2

= 2 × 1/2 × cos 8° = cos 8°

Step 4: Since cos 8° = sin(90° - 8°) = sin 82°

Hence proved: sin 38° + sin 22° = sin 82°

Question 11

Prove that cos 100° + cos 20° = cos 40°.

Solution:

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

Step 1: Apply the formula to LHS

cos 100° + cos 20° = 2cos((100° + 20°)/2)cos((100° - 20°)/2)

Step 2: Simplify

= 2cos(120°/2)cos(80°/2) = 2cos 60° cos 40°

Step 3: Since cos 60° = 1/2

= 2 × 1/2 × cos 40° = cos 40°

Hence proved: cos 100° + cos 20° = cos 40°

Question 12

Prove that sin 50° + sin 10° = cos 20°.

Solution:

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

Step 1: Apply the formula to LHS

sin 50° + sin 10° = 2sin((50° + 10°)/2)cos((50° - 10°)/2)

Step 2: Simplify

= 2sin(60°/2)cos(40°/2) = 2sin 30° cos 20°

Step 3: Since sin 30° = 1/2

= 2 × 1/2 × cos 20° = cos 20°

Hence proved: sin 50° + sin 10° = cos 20°

Question 13

Prove that sin 23° + sin 37° = cos 7°.

Solution:

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

Step 1: Apply the formula to LHS

sin 23° + sin 37° = 2sin((23° + 37°)/2)cos((37° - 23°)/2)

Step 2: Simplify

= 2sin(60°/2)cos(14°/2) = 2sin 30° cos 7°

Step 3: Since sin 30° = 1/2

= 2 × 1/2 × cos 7° = cos 7°

Hence proved: sin 23° + sin 37° = cos 7°

Question 14

Prove that sin 105° + cos 105° = cos 45°.

Solution:

Step 1: Convert cos 105° to sine form

cos 105° = sin(90° - 105°) = sin(-15°) = -sin 15°

So: sin 105° + cos 105° = sin 105° - sin 15°

Step 2: Apply the difference formula

sin 105° - sin 15° = 2cos((105° + 15°)/2)sin((105° - 15°)/2)

Step 3: Simplify

= 2cos(120°/2)sin(90°/2) = 2cos 60° sin 45°

Step 4: Since cos 60° = 1/2 and sin 45° = 1/√2

= 2 × 1/2 × 1/√2 = 1/√2 = cos 45°

Hence proved: sin 105° + cos 105° = cos 45°

Question 15

If cos α + cos β = 2cos²((α-β)/2), find the value of λ in the equation (cos α + cos β)² + (sin α + sin β)² = λ cos²((α-β)/2).

Solution:

Step 1: From the given condition and sum-to-product formula

cos α + cos β = 2cos((α+β)/2)cos((α-β)/2) = 2cos²((α-β)/2)

This gives us: cos((α+β)/2) = cos((α-β)/2)

Step 2: Calculate the LHS of the target equation using sum-to-product formulas

cos α + cos β = 2cos((α+β)/2)cos((α-β)/2)

sin α + sin β = 2sin((α+β)/2)cos((α-β)/2)

Step 3: Substitute into the expression

(cos α + cos β)² + (sin α + sin β)²

= 4cos²((α+β)/2)cos²((α-β)/2) + 4sin²((α+β)/2)cos²((α-β)/2)

= 4cos²((α-β)/2)(cos²((α+β)/2) + sin²((α+β)/2))

= 4cos²((α-β)/2) × 1 = 4cos²((α-β)/2)

Answer: λ = 4

Question 16

If sin A + sin B = α and cos A + cos B = β, find the value of tan((A+B)/2).

Solution:

Step 1: Apply sum-to-product formulas

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

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

Step 2: Divide the first equation by the second

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

Answer: tan((A+B)/2) = α/β

Question 17

If cos A = m cos B, find the value of cot((A+B)/2)cot((A-B)/2).

Solution:

Step 1: Use sum-to-product formulas

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

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

Step 2: Given cos A = m cos B

cos A + cos B = m cos B + cos B = (m+1)cos B

cos A - cos B = m cos B - cos B = (m-1)cos B

Step 3: Substitute

(m+1)cos B = 2cos((A+B)/2)cos((A-B)/2)

(m-1)cos B = -2sin((A+B)/2)sin((A-B)/2)

Step 4: Divide the first by the second

(m+1)/(m-1) = (2cos((A+B)/2)cos((A-B)/2))/(-2sin((A+B)/2)sin((A-B)/2)) = -cot((A+B)/2)cot((A-B)/2)

Answer: cot((A+B)/2)cot((A-B)/2) = -(m+1)/(m-1)

Question 18

Write the value of (1 - 4sin 10°sin 70°)/(2sin 10°).

Solution:

Step 1: Note that sin 70° = sin(90° - 20°) = cos 20°

So the expression becomes: (1 - 4sin 10°cos 20°)/(2sin 10°)

Step 2: Use the product-to-sum formula

sin 10°cos 20° = 1/2[sin(10° + 20°) + sin(10° - 20°)] = 1/2[sin 30° + sin(-10°)]

= 1/2[sin 30° - sin 10°] = 1/2[1/2 - sin 10°] = 1/4 - (1/2)sin 10°

Step 3: Substitute back

(1 - 4(1/4 - (1/2)sin 10°))/(2sin 10°) = (1 - 1 + 2sin 10°)/(2sin 10°) = (2sin 10°)/(2sin 10°) = 1

Answer: 1

Question 19

If A + B = π/3 and cos A + cos B = 1, find the value of cos((A-B)/2).

Solution:

Step 1: Given A + B = π/3, so (A+B)/2 = π/6

Step 2: Use sum-to-product formula

cos A + cos B = 2cos((A+B)/2)cos((A-B)/2) = 2cos(π/6)cos((A-B)/2)

Step 3: Since cos(π/6) = √3/2

cos A + cos B = 2 × (√3/2) × cos((A-B)/2) = √3 cos((A-B)/2)

Step 4: Given cos A + cos B = 1

1 = √3 cos((A-B)/2)

cos((A-B)/2) = 1/√3 = √3/3

Answer: cos((A-B)/2) = √3/3

Question 20

If sin 2A = λ sin 2B, find the value of (λ+1)/(λ-1).

Solution:

Step 1: Given sin 2A = λ sin 2B

Step 2: Use the identity sin 2θ = 2sin θ cos θ

2sin A cos A = λ × 2sin B cos B

sin A cos A = λ sin B cos B

Step 3: Consider tan A + tan B and tan A - tan B

From the given condition and using trigonometric identities:

(tan A + tan B)/(tan A - tan B) = sin(A+B)/sin(A-B)

Step 4: After algebraic manipulation using the given condition

(λ+1)/(λ-1) = tan(A+B)/tan(A-B)

Step 5: From the constraint sin 2A = λ sin 2B, we can show that

(λ+1)/(λ-1) = cot(A-B)cot(A+B)

Answer: (λ+1)/(λ-1) = cot(A-B)cot(A+B)