 # 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 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 BYJU’S and are in accordance with the NCERT syllabus and guidelines.

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

1. sin (– x) = – sin x
2. cos (– x) = cos x
3. cos (x + y) = cos x cos y – sin x sin y
4. cos (x – y) = cos x cos y + sin x sin y
5. sin (x + y) = sin x cos y + cos x sin y
6. sin (x – y) = sin x cos y – cos x sin y

## Download PDF of NCERT Solutions for Class 11 Maths Chapter 3- Trigonometric Functions Exercise 3.3                  ### 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 3.4 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 15°

It can be written as

= tan (45° – 30°)

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 (n + 1)x sin (n + 2)x + cos (n + 1)x cos (n + 2)x = cos x

Solution:

LHS = sin (n + 1)x sin (n + 2)x + cos (n + 1)x cos (n + 2)x 11. Solution:

Consider Using the formula 12. sin2 6x – sin2 4x = sin 2x sin 10x

Solution:  13. cos2 2x – cos2 6x = sin 4sin 8x

Solution: We get

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

It can be written as

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

So we get

= (2 sin 4x cos 4x) (2 sin 2x cos 2x)

= sin 8x sin 4x

= RHS

14. sin 2x + 2sin 4x + sin 6x = 4cos2 x sin 4x

Solution: By further simplification

= 2 sin 4x cos (– 2x) + 2 sin 4x

It can be written as

= 2 sin 4x cos 2x + 2 sin 4x

Taking common terms

= 2 sin 4x (cos 2x + 1)

Using the formula

= 2 sin 4x (2 cos2 x – 1 + 1)

We get

= 2 sin 4x (2 cos2 x)

= 4cos2 x sin 4x

= R.H.S.

15. cot 4x (sin 5x + sin 3x) = cot x (sin 5x – sin 3x)

Solution:

Consider

LHS = cot 4x (sin 5x + 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. cot x cot 2x – cot 2x cot 3x – cot 3x cot x = 1

Solution: 23. Solution:

Consider

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

By using the formula  24. cos 4x = 1 – 8sincosx

Solution:

Consider

LHS = cos 4x

We can write it as

= cos 2(2x)

Using the formula cos 2A = 1 – 2 sin2 A

= 1 – 2 sin2 2x

Again by using the formula sin2A = 2sin A cos A

= 1 – 2(2 sin x cos x) 2

So we get

= 1 – 8 sin2x cos2x

= R.H.S.

25. cos 6x = 32 cos6 x – 48 cos4 x + 18 cos2 – 1

Solution:

Consider

L.H.S. = cos 6x

It can be written as

= cos 3(2x)

Using the formula cos 3A = 4 cos3 A – 3 cos A

= 4 cos3 2x – 3 cos 2x

Again by using formula cos 2x = 2 cos2 – 1

= 4 [(2 cos2 – 1)3 – 3 (2 cos2 x – 1)

By further simplification

= 4 [(2 cos2 x) 3 – (1)3 – 3 (2 cos2 x) 2 + 3 (2 cos2 x)] – 6cos2 x + 3

We get

= 4 [8cos6x – 1 – 12 cos4x + 6 cos2x] – 6 cos2x + 3

By multiplication

= 32 cos6x – 4 – 48 cos4x + 24 cos2 x – 6 cos2x + 3

On further calculation

= 32 cos6– 48 cos4x + 18 cos2x – 1

= R.H.S.