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

ncert sol class 11 maths ch 3 ex 3 01
ncert sol class 11 maths ch 3 ex 3 02
ncert sol class 11 maths ch 3 ex 3 03
ncert sol class 11 maths ch 3 ex 3 04
ncert sol class 11 maths ch 3 ex 3 05
ncert sol class 11 maths ch 3 ex 3 06
ncert sol class 11 maths ch 3 ex 3 07
ncert sol class 11 maths ch 3 ex 3 08
ncert sol class 11 maths ch 3 ex 3 09
ncert sol class 11 maths ch 3 ex 3 10
ncert sol class 11 maths ch 3 ex 3 11
ncert sol class 11 maths ch 3 ex 3 12
ncert sol class 11 maths ch 3 ex 3 13
ncert sol class 11 maths ch 3 ex 3 14
ncert sol class 11 maths ch 3 ex 3 15
ncert sol class 11 maths ch 3 ex 3 16
ncert sol class 11 maths ch 3 ex 3 17
ncert sol class 11 maths ch 3 ex 3 18

 

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.

NCERT Solutions for Class 11 Chapter 3 Ex 3.3 Image 1

Solution:

NCERT Solutions for Class 11 Chapter 3 Ex 3.3 Image 2

2.

NCERT Solutions for Class 11 Chapter 3 Ex 3.3 Image 3

Solution:

NCERT Solutions for Class 11 Chapter 3 Ex 3.3 Image 4

Here

= 1/2 + 4/4

= 1/2 + 1

= 3/2

= RHS

3.

NCERT Solutions for Class 11 Chapter 3 Ex 3.3 Image 5

Solution:

NCERT Solutions for Class 11 Chapter 3 Ex 3.3 Image 6

4.

NCERT Solutions for Class 11 Chapter 3 Ex 3.3 Image 7

Solution:

NCERT Solutions for Class 11 Chapter 3 Ex 3.3 Image 8

NCERT Solutions for Class 11 Chapter 3 Ex 3.3 Image 9

5. Find the value of:

(i) sin 75o

(ii) tan 15o

Solution:

NCERT Solutions for Class 11 Chapter 3 Ex 3.3 Image 10

(ii) tan 15°

It can be written as

= tan (45° – 30°)

Using formula

NCERT Solutions for Class 11 Chapter 3 Ex 3.3 Image 11

NCERT Solutions for Class 11 Chapter 3 Ex 3.3 Image 12

Prove the following:

6.

NCERT Solutions for Class 11 Chapter 3 Ex 3.3 Image 13

Solution:

NCERT Solutions for Class 11 Chapter 3 Ex 3.3 Image 14

NCERT Solutions for Class 11 Chapter 3 Ex 3.3 Image 15

7.

NCERT Solutions for Class 11 Chapter 3 Ex 3.3 Image 16

Solution:

NCERT Solutions for Class 11 Chapter 3 Ex 3.3 Image 17

NCERT Solutions for Class 11 Chapter 3 Ex 3.3 Image 18

8.

NCERT Solutions for Class 11 Chapter 3 Ex 3.3 Image 19

Solution:

NCERT Solutions for Class 11 Chapter 3 Ex 3.3 Image 20

9.

NCERT Solutions for Class 11 Chapter 3 Ex 3.3 Image 21

Solution:

Consider

NCERT Solutions for Class 11 Chapter 3 Ex 3.3 Image 22

It can be written as

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

So we get

NCERT Solutions for Class 11 Chapter 3 Ex 3.3 Image 23

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

NCERT Solutions for Class 11 Chapter 3 Ex 3.3 Image 24

11.

NCERT Solutions for Class 11 Chapter 3 Ex 3.3 Image 25

Solution:

Consider

NCERT Solutions for Class 11 Chapter 3 Ex 3.3 Image 26

Using the formula

NCERT Solutions for Class 11 Chapter 3 Ex 3.3 Image 27

12. sin2 6x – sin2 4x = sin 2x sin 10x

Solution:

NCERT Solutions for Class 11 Chapter 3 Ex 3.3 Image 28

NCERT Solutions for Class 11 Chapter 3 Ex 3.3 Image 29

13. cos2 2x – cos2 6x = sin 4sin 8x

Solution:

NCERT Solutions for Class 11 Chapter 3 Ex 3.3 Image 30

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:

NCERT Solutions for Class 11 Chapter 3 Ex 3.3 Image 31

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

NCERT Solutions for Class 11 Chapter 3 Ex 3.3 Image 32

Using the formula

NCERT Solutions for Class 11 Chapter 3 Ex 3.3 Image 33

= 2 cos 4x cos x

Hence, LHS = RHS.

16.

NCERT Solutions for Class 11 Chapter 3 Ex 3.3 Image 34

Solution:

Consider

NCERT Solutions for Class 11 Chapter 3 Ex 3.3 Image 35

Using the formula

NCERT Solutions for Class 11 Chapter 3 Ex 3.3 Image 36

17.

NCERT Solutions for Class 11 Chapter 3 Ex 3.3 Image 37

Solution:

NCERT Solutions for Class 11 Chapter 3 Ex 3.3 Image 28

NCERT Solutions for Class 11 Chapter 3 Ex 3.3 Image 39

18.

NCERT Solutions for Class 11 Chapter 3 Ex 3.3 Image 40

Solution:

NCERT Solutions for Class 11 Chapter 3 Ex 3.3 Image 41

19.

NCERT Solutions for Class 11 Chapter 3 Ex 3.3 Image 42

Solution:

NCERT Solutions for Class 11 Chapter 3 Ex 3.3 Image 43

20.

NCERT Solutions for Class 11 Chapter 3 Ex 3.3 Image 44

Solution:

NCERT Solutions for Class 11 Chapter 3 Ex 3.3 Image 45

NCERT Solutions for Class 11 Chapter 3 Ex 3.3 Image 46

21.

NCERT Solutions for Class 11 Chapter 3 Ex 3.3 Image 47

Solution:

NCERT Solutions for Class 11 Chapter 3 Ex 3.3 Image 48

NCERT Solutions for Class 11 Chapter 3 Ex 3.3 Image 49

22. cot x cot 2x – cot 2x cot 3x – cot 3x cot x = 1

Solution:

NCERT Solutions for Class 11 Chapter 3 Ex 3.3 Image 50

23.

NCERT Solutions for Class 11 Chapter 3 Ex 3.3 Image 51

Solution:

Consider

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

By using the formula

NCERT Solutions for Class 11 Chapter 3 Ex 3.3 Image 52

NCERT Solutions for Class 11 Chapter 3 Ex 3.3 Image 53

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.

Leave a Comment

Your email address will not be published. Required fields are marked *