RD Sharma Solutions Class 11 Maths Chapter 9 - Trigonometric Ratios Of Multiple And Sub Multiple Angles

The RD Sharma Solutions for Class 11 Maths Chapter 9 offer a detailed guide to help students confidently prepare for their board exams. This chapter focuses on understanding the values of trigonometric functions when applied to multiples and submultiples of angles. The solutions are carefully prepared by experienced subject experts, explaining every concept step by step in a clear and easy-to-follow manner.

In this chapter, students will learn how to use formulae that relate trigonometric values at multiples (like 2x, 3x) and submultiples (like x/2, x/3) of a given angle ‘x’. These formulas play a crucial role in simplifying complex trigonometric problems and are commonly tested in various exams.

RD Sharma Solutions Class 11 Maths Chapter 9 : Overview

Chapter 9 consists of three well-structured exercises, with each exercise offering a set of problems to test and strengthen students' understanding. The solutions are written in a student-friendly language to ensure better comprehension and effective learning.

All answers are based on the latest CBSE Class 11 syllabus (2025-26) and follow the updated curriculum, making them reliable resources for exam preparation.

Concepts Covered in Chapter 9:

• Trigonometric values at 2x expressed using values at x.
• Trigonometric values at x represented through values at x/2.
• Values of trigonometric functions at x/2 derived using cos x.
• Values of trigonometric functions at 3x in terms of x.
• Trigonometric values at x in terms of x/3.
• Important standard trigonometric values at specific angles.

RD Sharma Solutions Class 11 Maths - Trigonometric Ratios Of Multiple And Sub Multiple Angles Question with Answers

Derive sin 2A = 2 sin A cos A
Solution: sin(A + A) = sin A cos A + cos A sin A = 2 sin A cos A

Derive cos 2A = cos²A - sin²A
Solution: cos(A + A) = cos²A - sin²A

Find tan 2A in terms of tan A
Solution: tan 2A = [2 tan A] / [1 - tan²A]

Prove sin 3A = 3 sin A − 4 sin³A
Solution: sin 3A = 3 sin A - 4 sin³A

Prove cos 3A = 4 cos³A − 3 cos A
Solution: cos 3A = 4 cos³A - 3 cos A

Prove tan 3A = (3 tan A − tan³A)/(1 − 3 tan²A)
Solution: Using tan(A + 2A) expansion

If tan A = 1, find tan 2A
Solution: tan 2A = ∞ (undefined)

Calculate sin 60° using sin 2A formula
Solution: sin 60° = 2 sin 30° cos 30° = √3/2

Find cos 60° using cos 2A formula
Solution: cos 60° = cos²30° - sin²30° = 1/2

Simplify sin 3x − sin x
Solution: 2 cos 2x sin x

Express cos²A in terms of cos 2A
Solution: cos²A = (1 + cos 2A)/2

Express sin²A in terms of cos 2A
Solution: sin²A = (1 - cos 2A)/2

Find tan A in terms of tan A/2
Solution: tan A = [2 tan(A/2)] / [1 - tan²(A/2)]

Evaluate sin 15°
Solution: sin 15° = (√6 - √2)/4

Evaluate cos 15°
Solution: cos 15° = (√6 + √2)/4

If sin A = 0.6, find cos 2A
Solution: cos 2A = 1 - 2(0.6)² = 0.28

If tan A = 1/3, calculate tan 2A
Solution: tan 2A = 3/4

Find sin 22.5° using half-angle formula
Solution: sin 22.5° = √[(1 - cos 45°)/2]

Find cos 67.5° using half-angle formula
Solution: cos 67.5° = √[(1 + cos 135°)/2]

Express sin 36° using cos 54°
Solution: sin 36° = cos 54°

Express cos 18° in terms of cos 36°
Solution: cos 18° = √[(1 + cos 36°)/2]

Verify sin²A + cos²A = 1
Solution: Identity holds true for all A

If tan A = 2, find tan 2A
Solution: tan 2A = -4/3

Derive sin 4A
Solution: sin 4A = 2 sin 2A cos 2A

Derive cos 4A
Solution: cos 4A = 1 - 2 sin² 2A

Derive tan 4A
Solution: tan 4A = [4 tan A - 4 tan³A] / [1 - 6 tan²A + tan⁴A]

Express sin(A/2) in terms of cos A
Solution: sin(A/2) = √[(1 - cos A)/2]

Express cos(A/2) in terms of cos A
Solution: cos(A/2) = √[(1 + cos A)/2]

Express tan(A/2) in terms of sin A and cos A
Solution: tan(A/2) = sin A / [1 + cos A]

Prove sin 18° = (√5 − 1)/4
Solution: Using multiple angle identity, sin 18° = (√5 − 1)/4