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 BYJU’S and are in accordance with the latest CBSE Syllabus 2023-24.

Exercise 3.3 of NCERT Solutions for Class 11 Maths Chapter 3 – Trigonometric Functions are 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

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

Download PDF

carouselExampleControls112

Previous





Next

Access other exercise solutions of Class 11 Maths Chapter 3 – 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 75^o

(ii) tan 15^o

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 4x sin 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 4x (–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 – 8sin² x cos² x

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 x – 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 x – 1

= 4 [(2 cos² x – 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 cos⁶x – 48 cos⁴x + 18 cos²x – 1

= R.H.S.