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

Chapter 3 Trigonometric Functions of Class 11 Maths is categorized under the term – II CBSE Syllabus for 2021-22. 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- 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 term – II prescribed by the CBSE. The NCERT Solutions for Class 11 can help the students in preparing well for the second term examination.

## Download PDF of NCERT Solutions for Class 11 Maths Chapter 3- Trigonometric Functions Miscellaneous Exercise            ### Access other exercise solutions of Class 11 Maths Chapter 3- Trigonometric Functions

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 3+ sin x) sin + (cos 3– cos x) cos = 0

Solution:

Consider

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

By further calculation

= sin 3x sin x + sin2 x + cos 3x cos x – cos2 x

Taking out the common terms

= cos 3x cos x + sin 3x sin x – (cos2 x – sin2 x)

Using the formula

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

= cos (3x – x) – cos 2x

So we get

= cos 2x – cos 2x

= 0

= RHS

3. Solution:

Consider

LHS = (cos x + cos y) 2 + (sin x – sin y) 2

By expanding using formula we get

= cos2 x + cos2 y + 2 cos x cos y + sin2 x + sin2 y – 2 sin x sin y

Grouping the terms

= (cos2 x + sin2 x) + (cos2 y + sin2 y) + 2 (cos x cos y – sin x sin y)

Using the formula cos (A + B) = (cos A cos B – 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 – 1 4. Solution:

LHS = (cos x – cos y) 2 + (sin x – sin y) 2

By expanding using formula

= cos2 x + cos2 y – 2 cos x cos y + sin2 x + sin2 y – 2 sin x sin y

Grouping the terms

= (cos2 x + sin2 x) + (cos2 y + sin2 y) – 2 (cos x cos y + sin x sin y)

Using the formula cos (A – B) = cos A cos B + sin A sin B

= 1 + 1 – 2 [cos (x – y)]

By further calculation

= 2 [1 – cos (x – y)]

From formula cos 2A = 1 – 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.