# NCERT Solutions for Class 11 Maths Chapter 3 - Trigonometric Functions Exercise 3.3

NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions Exercise 3.3 have been provided here for students to prepare well for the board exam. The steps given in the examples have been followed while providing the NCERT Solutions for Class 11 for the questions present in the exercises. These solutions have been prepared by the subject experts at BYJUâ€™S and are in accordance with the latest CBSE Syllabus 2023-24.

Exercise 3.3 of NCERT Solutions for Class 11 Maths Chapter 3 â€“ Trigonometric Functions are based on Trigonometric Functions of Sum and Difference of Two Angles. Some of these functions are:

1. sin (â€“ x) = â€“ sin x
2. cos (â€“ x) = cos x
3. cos (x + y) = cos x cos y â€“ sin x sin y
4. cos (x â€“ y) = cos x cos y + sin x sin y
5. sin (x + y) = sin x cos y + cos x sin y
6. sin (x â€“ y) = sin x cos y â€“ cos x sin y

## NCERT Solutions for Class 11 Maths Chapter 3 â€“ Trigonometric Functions Exercise 3.3

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

To solve more Trigonometric Function related problems from NCERT Class 11 Maths Solutions, use the links here.

Exercise 3.1 Solutions 7 Questions

Exercise 3.2 Solutions 10 Questions

Exercise 3.4 Solutions 9 Questions

Miscellaneous Exercise On Chapter 3 Solutions 10 Questions

### Access Solutions for Class 11 Maths Chapter 3.3 Exercise

Prove that:

1.

Solution:

2.

Solution:

Here

= 1/2 + 4/4

= 1/2 + 1

= 3/2

= RHS

3.

Solution:

4.

Solution:

5. Find the value of:

(i) sin 75o

(ii) tan 15o

Solution:

(ii) tan 15Â°

It can be written as

= tan (45Â° â€“ 30Â°)

Using formula

Prove the following:

6.

Solution:

7.

Solution:

8.

Solution:

9.

Solution:

Consider

It can be written as

= sin x cos x (tan x + cot x)

So we get

10. sin (nÂ + 1)xÂ sin (nÂ + 2)xÂ + cos (nÂ + 1)xÂ cos (nÂ + 2)xÂ = cosÂ x

Solution:

LHS = sin (nÂ + 1)xÂ sin (nÂ + 2)xÂ + cos (nÂ + 1)xÂ cos (nÂ + 2)x

11.

Solution:

Consider

Using the formula

12. sin2Â 6xÂ â€“ sin2Â 4xÂ = sin 2xÂ sin 10x

Solution:

13. cos2Â 2xÂ â€“ cos2Â 6xÂ = sin 4xÂ sin 8x

Solution:

We get

= [2 cos 4x cos (-2x)] [-2 sin 4x sin (-2x)]

It can be written as

= [2 cos 4xÂ cos 2x] [â€“2 sin 4xÂ (â€“sin 2x)]

So we get

= (2 sin 4xÂ cos 4x) (2 sin 2xÂ cos 2x)

= sin 8x sin 4x

= RHS

14. sin 2xÂ + 2sin 4xÂ + sin 6xÂ = 4cos2Â xÂ sin 4x

Solution:

By further simplification

= 2 sin 4xÂ cos (â€“ 2x) + 2 sin 4x

It can be written as

= 2 sin 4xÂ cos 2xÂ + 2 sin 4x

Taking common terms

= 2 sin 4xÂ (cos 2xÂ + 1)

Using the formula

= 2 sin 4xÂ (2 cos2Â xÂ â€“ 1 + 1)

We get

= 2 sin 4xÂ (2 cos2Â x)

= 4cos2Â xÂ sin 4x

= R.H.S.

15. cot 4xÂ (sin 5xÂ + sin 3x) = cotÂ xÂ (sin 5xÂ â€“ sin 3x)

Solution:

Consider

LHS = cot 4xÂ (sin 5xÂ + sin 3x)

It can be written as

Using the formula

= 2 cos 4x cos x

Hence, LHS = RHS.

16.

Solution:

Consider

Using the formula

17.

Solution:

18.

Solution:

19.

Solution:

20.

Solution:

21.

Solution:

22. cotÂ xÂ cot 2xÂ â€“ cot 2xÂ cot 3xÂ â€“ cot 3xÂ cotÂ xÂ = 1

Solution:

23.

Solution:

Consider

LHS = tan 4x = tan 2(2x)

By using the formula

24. cos 4xÂ = 1 â€“ 8sin2Â xÂ cos2Â x

Solution:

Consider

LHS = cos 4x

We can write it as

= cos 2(2x)

Using the formula cos 2AÂ = 1 â€“ 2 sin2Â A

= 1 â€“ 2 sin2Â 2x

Again by using the formula sin2AÂ = 2sinÂ AÂ cos A

= 1 â€“ 2(2 sinÂ xÂ cosÂ x) 2

So we get

= 1 â€“ 8 sin2xÂ cos2x

= R.H.S.

25. cos 6xÂ = 32 cos6Â xÂ â€“ 48 cos4Â xÂ + 18 cos2Â xÂ â€“ 1

Solution:

Consider

L.H.S. = cos 6x

It can be written as

= cos 3(2x)

Using the formula cos 3AÂ = 4 cos3Â AÂ â€“ 3 cosÂ A

= 4 cos3Â 2xÂ â€“ 3 cosÂ 2x

Again by using formula cos 2xÂ = 2 cos2Â xÂ â€“ 1

= 4 [(2 cos2Â xÂ â€“ 1)3Â â€“ 3 (2 cos2Â xÂ â€“ 1)

By further simplification

= 4 [(2 cos2Â x) 3Â â€“ (1)3Â â€“ 3 (2 cos2Â x) 2Â + 3 (2 cos2Â x)] â€“ 6cos2Â xÂ + 3

We get

= 4 [8cos6xÂ â€“ 1 â€“ 12 cos4xÂ + 6 cos2x] â€“ 6 cos2xÂ + 3

By multiplication

= 32 cos6xÂ â€“ 4 â€“ 48 cos4xÂ + 24 cos2Â xÂ â€“ 6 cos2xÂ + 3

On further calculation

= 32 cos6xÂ â€“ 48 cos4xÂ + 18 cos2xÂ â€“ 1

= R.H.S.