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

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

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

?Chapter – 3?

✍ Trigonometric Functions✍

?Exercise 3.3?

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

Solution:Solution:

13. cos ² 2x – cos ² 6 x = sin 4 x sin 8 x

Solution:Solution:

We get
= [2 cos 4 x cos (-2x)] [-2 sin 4 x sin (-2x)]
It can be written as
= [2 cos 4 x cos 2x] [–2 sin 4 x (–sin 2x)]
So we get
= (2 sin 4 x cos 4 x) (2 sin 2x cos 2x)
= sin 8 x sin 4 x
= RHS

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

Solution:Solution:

By further simplification
= 2 sin 4 x cos (– 2x) + 2 sin 4 x
It can be written as
= 2 sin 4 x cos 2x + 2 sin 4 x
Taking common terms
= 2 sin 4 x (cos ² x + 1)
Using the formula
= 2 sin 4 x (2 cos ² x – 1 + 1)
We get
= 2 sin 4 x (2 cos ² x)
= 4 cos ² x sin 4 x
= R.H.S.

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

Solution: Solution: Consider

LHS = cot 4 x (sin 5 x + sin 3 x)

It can be written as

Using the formula

= 2 cos 4 x cos x

Hence, LHS = RHS.

16.

Solution: Solution: Consider

Using the formula

17.

Solution:Solution:

18.

Solution:Solution:

19.

Solution:Solution:

20.

Solution:Solution:

21.

Solution:

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

Solution:

23.

Solution: Consider

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

By using the formula

24. cos 4 x = 1 – 8 sin ² x cos ² x

Solution:

Consider
LHS = cos 4 x
We can write it as
= cos 2(2x)
Using the formula cos 2 A = 1 – 2 sin² A
= 1 – 2 sin ² 2x
Again by using the formula sin 2 A = 2 sin A cos A
= 1 – 2(2 sin x cos x) ²
So we get
= 1 – 8 sin ²x cos ²x
= R.H.S.

25. cos 6 x = 32 cos ⁶ x – 48 cos ⁴ x + 18 cos ² 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.