# NCERT Solutions for Class 11 Maths Chapter 3 - Trigonometric Functions Miscellaneous Exercise

Chapter 3, Trigonometric Functions of Class 11 Maths, is categorised under the CBSE Syllabus for 2023-24. The miscellaneous exercise contains questions covering the entire topics present in the chapter. The last exercise of NCERT Solutions for Class 11 Maths Chapter 3 â€“ Trigonometric Functions is based on the following topics:

1. Angles
2. Trigonometric Functions
3. Trigonometric Functions of Sum and Difference of Two Angles
4. Trigonometric Equations

The solutions provided here are based on the latest guidelines of the syllabus prescribed by the CBSE. The NCERT Solutions for Class 11 can help the students in preparing well for the board examination.

## NCERT Solutions for Class 11 Maths Chapter 3 â€“ Trigonometric Functions Miscellaneous Exercise

### Access other exercise solutions of Class 11 Maths Chapter 3 â€“ Trigonometric Functions

The NCERT Class 11 Maths Solutions for Chapter 3 can be referred to in PDF format by checking the links below.

Exercise 3.1 Solutions 7 Questions

Exercise 3.2 Solutions 10 Questions

Exercise 3.3 Solutions 25 Questions

Exercise 3.4 Solutions 9 Questions

#### Access Solutions for Class 11 Maths Chapter 3 Miscellaneous Exercise

Prove that:

1.

Solution:

We get

= 0

= RHS

2. (sin 3xÂ + sinÂ x) sinÂ xÂ + (cos 3xÂ â€“ cosÂ x) cosÂ xÂ = 0

Solution:

Consider

LHS = (sin 3x + sin x) sin x + (cos 3x â€“ cos x) cos x

By further calculation,

= sin 3x sin x + sin2 x + cos 3x cos x â€“ cos2 x

Taking out the common terms,

= cos 3x cos x + sin 3x sin x â€“ (cos2 x â€“ sin2 x)

Using the formula

cos (A â€“ B) = cos A cos B + sin A sin B

= cos (3x â€“ x) â€“ cos 2x

So, we get

= cos 2x â€“ cos 2x

= 0

= RHS

3.

Solution:

Consider

LHS = (cos x + cos y) 2 + (sin x â€“ sin y) 2

By expanding using the formula, we get

= cos2 x + cos2 y + 2 cos x cos y + sin2 x + sin2 y â€“ 2 sin x sin y

Grouping the terms,

= (cos2 x + sin2 x) + (cos2 y + sin2 y) + 2 (cos x cos y â€“ sin x sin y)

Using the formula cos (A + B) = (cos A cos B â€“ sin A sin B)

= 1 + 1 + 2 cos (x + y)

By further calculation,

= 2 + 2 cos (x + y)

Taking 2 as common

= 2 [1 + cos (x + y)]

From the formula cos 2A = 2 cos2 A â€“ 1

4.

Solution:

LHS = (cos x â€“ cos y) 2 + (sin x â€“ sin y) 2

By expanding using the formula,

= cos2 x + cos2 y â€“ 2 cos x cos y + sin2 x + sin2 y â€“ 2 sin x sin y

Grouping the terms,

= (cos2 x + sin2 x) + (cos2 y + sin2 y) â€“ 2 (cos x cos y + sin x sin y)

Using the formula cos (A â€“ B) = cos A cos B + sin A sin B

= 1 + 1 â€“ 2 [cos (x â€“ y)]

By further calculation,

= 2 [1 â€“ cos (x â€“ y)]

From formula cos 2A = 1 â€“ 2 sin2 A

5. sin x + sin 3x + sin 5x + sin 7x = 4 cos x cos 2x sin 4x

Solution:

6.

Solution:

7.

Solution:

8. Find sin x/2, cos x/2 and tan x/2 in each of the following.

Solution:

cos x = -3/5

From the formula,

9. cos x = -1/3, x in quadrant III

Solution:

10. sin x = 1/4, x in quadrant II

Solution:

Class 11 Maths NCERT supplementary or miscellaneous exercise solutions PDFs are provided here.