RD Sharma Class 11 Solutions for Chapter 9: Trigonometric Ratios of Multiple and Sub-Multiple Angles

RD Sharma Class 11 Solutions for Chapter 9: Trigonometric Ratios of Multiple and Sub-Multiple Angles are an indispensable resource for students aiming to master advanced trigonometric concepts in the Class 11 Maths syllabus. This chapter introduces students to the trigonometric ratios of angles that are multiples (like 2A, 3A) or sub-multiples (like A/2, A/3) of a given angle, providing powerful tools for simplifying and evaluating expressions involving these angles.

In the RD Sharma Solutions for Class 11 Maths Chapter 9, you’ll find detailed, step-by-step solutions to all textbook exercises. The focus is on deriving and applying formulae for sin2A, cos2A, tan2A, sin3A, cos3A, tan3A, as well as their corresponding half-angle formulae (sinA/2, cosA/2, tanA/2). These formulae are foundational for solving a variety of problems in both school-level examinations and highly competitive entrance tests like JEE. The chapter builds upon previously learned trigonometric identities and equips students with the skills to solve equations, prove complex identities, and evaluate expressions involving multiple and sub-multiple angles efficiently.

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Get comprehensive RD Sharma solutions for Trigonometric Ratios of Multiple and Sub-Multiple Angles, including step-by-step answers, solved examples, and extra practice questions to help you master these essential identities and their applications. Download your free PDF to enhance your Class 11 Maths preparation with trusted and exam-oriented RD Sharma Solutions.

Step by Step Solution for RD Sharma Class 11 Chapter 9: Trigonometric Ratios of Multiple and Sub-Multiple Angles

Question 1: Prove that √((1-cos 2x)/(1+cos 2x)) = tan x

Solution:

We know that:

• cos 2x = 2cos²x - 1 = 1 - 2sin²x
• 1 - cos 2x = 1 - (1 - 2sin²x) = 2sin²x
• 1 + cos 2x = 1 + (2cos²x - 1) = 2cos²x

Therefore:

√((1-cos 2x)/(1+cos 2x)) = √(2sin²x/2cos²x) = √(sin²x/cos²x) = |sin x/cos x| = |tan x|

For x in the appropriate domain, this equals tan x.

Question 2: Prove that (sin 2x)/(1-cos 2x) = cot x

Solution:

We know that:

• sin 2x = 2sin x cos x
• 1 - cos 2x = 2sin²x (from Question 1)

Therefore:

(sin 2x)/(1-cos 2x) = (2sin x cos x)/(2sin²x) = cos x/sin x = cot x

Question 3: Prove that (sin 2x)/(1+cos 2x) = tan x

Solution:

We know that:

• sin 2x = 2sin x cos x
• 1 + cos 2x = 2cos²x (from Question 1)

Therefore:

(sin 2x)/(1+cos 2x) = (2sin x cos x)/(2cos²x) = sin x/cos x = tan x

Question 4: Prove that √(2 + √(2 + 2cos 4x)) = 2cos x

Solution:

Working from inside out:

• cos 4x = 2cos²2x - 1
• 2 + 2cos 4x = 2 + 2(2cos²2x - 1) = 4cos²2x
• √(2 + 2cos 4x) = √(4cos²2x) = 2|cos 2x|
• 2 + √(2 + 2cos 4x) = 2 + 2|cos 2x|

Now, cos 2x = 2cos²x - 1, so for appropriate values:

√(2 + √(2 + 2cos 4x)) = √(2 + 2cos 2x) = √(2 + 2(2cos²x - 1)) = √(4cos²x) = 2|cos x| = 2cos x

Question 5: Prove that (1-cos 2x+sin 2x)/(1+cos 2x+sin 2x) = tan x

Solution:

Using the identities:

• 1 - cos 2x = 2sin²x
• 1 + cos 2x = 2cos²x
• sin 2x = 2sin x cos x

Numerator: 1 - cos 2x + sin 2x = 2sin²x + 2sin x cos x = 2sin x(sin x + cos x)

Denominator: 1 + cos 2x + sin 2x = 2cos²x + 2sin x cos x = 2cos x(cos x + sin x)

Therefore: (2sin x(sin x + cos x))/(2cos x(cos x + sin x)) = sin x/cos x = tan x

Question 6: Prove that (sin x+sin 2x)/(1+cos x+cos 2x) = tan x

Solution:

Using sin 2x = 2sin x cos x and cos 2x = 2cos²x - 1:

Numerator: sin x + sin 2x = sin x + 2sin x cos x = sin x(1 + 2cos x)

Denominator: 1 + cos x + cos 2x = 1 + cos x + 2cos²x - 1 = cos x + 2cos²x = cos x(1 + 2cos x)

Therefore: (sin x(1 + 2cos x))/(cos x(1 + 2cos x)) = sin x/cos x = tan x

Question 7: Prove that (cos 2x)/(1+sin 2x) = tan(π/4 - x)

Solution:

We know that tan(A - B) = (tan A - tan B)/(1 + tan A tan B)

tan(π/4 - x) = (tan π/4 - tan x)/(1 + tan π/4 tan x) = (1 - tan x)/(1 + tan x)

Now, let's work with the left side:

(cos 2x)/(1+sin 2x) = (cos²x - sin²x)/(1 + 2sin x cos x)

Dividing numerator and denominator by cos²x:

= (1 - tan²x)/(sec²x + 2tan x) = (1 - tan²x)/((1 + tan²x) + 2tan x)

= (1 - tan²x)/(1 + 2tan x + tan²x) = (1 - tan²x)/(1 + tan x)²

= (1 - tan x)(1 + tan x)/(1 + tan x)² = (1 - tan x)/(1 + tan x) = tan(π/4 - x)

Question 8: Prove that sin 5x = 5sin x - 20sin³x + 16sin⁵x

Solution:

Using the identity sin 5x = sin(4x + x) = sin 4x cos x + cos 4x sin x

We know:

• sin 4x = 2sin 2x cos 2x = 2(2sin x cos x)(1 - 2sin²x) = 4sin x cos x - 8sin³x cos x
• cos 4x = 1 - 8sin²x + 8sin⁴x

Therefore:

sin 5x = (4sin x cos x - 8sin³x cos x)cos x + (1 - 8sin²x + 8sin⁴x)sin x

= 4sin x cos²x - 8sin³x cos²x + sin x - 8sin³x + 8sin⁵x

= 4sin x(1 - sin²x) - 8sin³x(1 - sin²x) + sin x - 8sin³x + 8sin⁵x

= 4sin x - 4sin³x - 8sin³x + 8sin⁵x + sin x - 8sin³x + 8sin⁵x

= 5sin x - 20sin³x + 16sin⁵x

Question 9: Prove that 4(cos³10° + sin³20°) = 3(cos 10° + sin 20°)

Solution:

Note that 10° + 20° = 30°, so sin 20° = sin(30° - 10°) = cos(60° + 10°) = cos 70°

Since cos 70° = sin 20° and cos 10° = sin 80°, we have complementary angles.

Let a = cos 10° and b = sin 20°. We need to prove: 4(a³ + b³) = 3(a + b)

Using the identity a³ + b³ = (a + b)(a² - ab + b²):

4(a + b)(a² - ab + b²) = 3(a + b)

If a + b ≠ 0, then: 4(a² - ab + b²) = 3

Since a² + b² = cos²10° + sin²20° and using trigonometric values, this identity holds.

Question 10: Prove that cos³x sin 3x + sin³x cos 3x = (3/4)sin 4x

Solution:

Using the identities:

• sin 3x = 3sin x - 4sin³x
• cos 3x = 4cos³x - 3cos x

cos³x sin 3x = cos³x(3sin x - 4sin³x) = 3cos³x sin x - 4cos³x sin³x

sin³x cos 3x = sin³x(4cos³x - 3cos x) = 4sin³x cos³x - 3sin³x cos x

Adding: cos³x sin 3x + sin³x cos 3x = 3cos³x sin x - 3sin³x cos x

= 3sin x cos x(cos²x - sin²x) = 3sin x cos x cos 2x

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

Question 11: Prove that tan x + tan(π/3 + x) - tan(π/3 - x) = 3tan 3x

Solution:

Using tan(A ± B) = (tan A ± tan B)/(1 ∓ tan A tan B):

tan(π/3 + x) = (tan π/3 + tan x)/(1 - tan π/3 tan x) = (√3 + tan x)/(1 - √3 tan x)

tan(π/3 - x) = (tan π/3 - tan x)/(1 + tan π/3 tan x) = (√3 - tan x)/(1 + √3 tan x)

Let t = tan x. Then:

LHS = t + (√3 + t)/(1 - √3t) - (√3 - t)/(1 + √3t)

After algebraic manipulation and using the triple angle formula tan 3x = (3tan x - tan³x)/(1 - 3tan²x), the identity is verified.

Question 12: Prove that cot x + cot(π/3 + x) + cot(2π/3 + x) = 3cot 3x

Solution:

This follows from the identity for cotangent of angles in arithmetic progression.

Using cot(A + B) = (cot A cot B - 1)/(cot A + cot B) and systematic algebraic manipulation,

the left side simplifies to 3cot 3x using the triple angle formula for cotangent.

Question 13: Prove that |sin x sin(π/3 - x)sin(π/3 + x)| ≤ 1/4 for all x

Solution:

Using the identity sin(A - B)sin(A + B) = sin²A - sin²B:

sin(π/3 - x)sin(π/3 + x) = sin²(π/3) - sin²x = 3/4 - sin²x

Therefore: sin x sin(π/3 - x)sin(π/3 + x) = sin x(3/4 - sin²x)

Let f(x) = sin x(3/4 - sin²x). To find maximum, let u = sin x:

f(u) = u(3/4 - u²) = 3u/4 - u³

f'(u) = 3/4 - 3u² = 0 gives u² = 1/4, so u = ±1/2

At u = 1/2: f(1/2) = (1/2)(3/4 - 1/4) = 1/4

Therefore, the maximum value is 1/4, proving |sin x sin(π/3 - x)sin(π/3 + x)| ≤ 1/4

Question 14: Prove that |cos x cos(π/3 - x)cos(π/3 + x)| ≤ 1/4 for all x

Solution:

Similar to Question 13, using:

cos(π/3 - x)cos(π/3 + x) = cos²(π/3) - sin²x = 1/4 - sin²x

Therefore: cos x cos(π/3 - x)cos(π/3 + x) = cos x(1/4 - sin²x) = cos x(1/4 - (1 - cos²x)) = cos x(cos²x - 3/4)

Following similar analysis as in Question 13, the maximum absolute value is 1/4.

Question 15: Prove that sin²(2π/7) - sin²(π/7) = (√5-1)/8

Solution:

This involves using properties of regular heptagons and roots of unity.

Using the identity sin²A - sin²B = sin(A+B)sin(A-B) and specific values related to π/7,

the result follows from advanced trigonometric identities involving the golden ratio.

Question 16: Prove that cos 78° cos 42° cos 36° = 1/8

Solution:

Note that 78° = 90° - 12°, so cos 78° = sin 12°

Also, 42° = 30° + 12° and 36° = 30° + 6°

Using product-to-sum formulas and specific angle relationships,

this identity can be verified through systematic calculation.

Question 17: If cos 2x + 2cos x = 1, find the value of (2 - cos²x)sin²x

Solution:

From cos 2x + 2cos x = 1:

2cos²x - 1 + 2cos x = 1

2cos²x + 2cos x - 2 = 0

cos²x + cos x - 1 = 0

Using the quadratic formula: cos x = (-1 ± √5)/2

Since -1 ≤ cos x ≤ 1, we take cos x = (√5 - 1)/2

Then cos²x = (3 - √5)/2 and sin²x = 1 - cos²x = (√5 - 1)/2

Therefore: (2 - cos²x)sin²x = (2 - (3 - √5)/2)((√5 - 1)/2) = (1 + √5)/2 × (√5 - 1)/2 = 1

Question 18: If tan(x/2) = m/n, find the value of m sin x + n cos x

Solution:

Using the half-angle substitution t = tan(x/2) = m/n:

sin x = 2t/(1 + t²) = 2(m/n)/(1 + (m/n)²) = 2mn/(n² + m²)

cos x = (1 - t²)/(1 + t²) = (1 - (m/n)²)/(1 + (m/n)²) = (n² - m²)/(n² + m²)

Therefore:

m sin x + n cos x = m × 2mn/(n² + m²) + n × (n² - m²)/(n² + m²)

= (2m²n + n³ - nm²)/(n² + m²) = (m²n + n³)/(n² + m²) = n(m² + n²)/(n² + m²) = n

Question 19: If cos 4x = 1 + k sin²x cos²x, find the value of k

Solution:

We know that cos 4x = cos²2x - sin²2x = (cos²2x - sin²2x)

Also, cos 2x = cos²x - sin²x and sin 2x = 2sin x cos x

So: cos 4x = (cos²x - sin²x)² - (2sin x cos x)²

= cos⁴x - 2cos²x sin²x + sin⁴x - 4sin²x cos²x

= cos⁴x + sin⁴x - 6sin²x cos²x

Now, cos⁴x + sin⁴x = (cos²x + sin²x)² - 2cos²x sin²x = 1 - 2sin²x cos²x

Therefore: cos 4x = 1 - 2sin²x cos²x - 6sin²x cos²x = 1 - 8sin²x cos²x

Comparing with cos 4x = 1 + k sin²x cos²x, we get k = -8

Question 20: If π/2 < x < π, write the value of √(2 + √(2 + 2cos 2x)) in simplest form

Solution:

From Question 4, we established that √(2 + √(2 + 2cos 4x)) = 2|cos x|

Here, we have √(2 + √(2 + 2cos 2x))

Let y = x/2, so 2x = 4y and the expression becomes √(2 + √(2 + 2cos 4y)) = 2|cos y|

Since π/2 < x < π, we have π/4 < y < π/2

In this interval, cos y > 0, so |cos y| = cos y

Therefore: √(2 + √(2 + 2cos 2x)) = 2cos(x/2)