Class 11 NCERT Solutions- Chapter 3 Trigonometric Function – Exercise 3.3 | Set 2

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

Solution:

Taking LHS in consideration, we get

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

=(sin 5x + sin 3x)

Using the identity,

sin A + sin B = 2 sincos

=(2 sincos)

=(2 sincos)

=(2 sin 4x cos x)

= 2 cos 4x cos x

Now, taking RHS in consideration, we get

= cot x (sin 5x – sin 3x)

=(sin 5x – sin 3x)

Using the identity,

sin A – sin B = 2 cossin

=(2 cossin)

=(2 cossin)

=(2 cos 4x sin x)

= 2 cos 4x cos x

Hence, LHS = RHS

Question 16:

Solution:

Taking LHS in consideration, we get

=

Using the identity,

cos A – cos B = 2 sinsin

sin A – sin B = 2 cossin

=

=

=

=

Hence, LHS = RHS

Question 17:= tan 4x

Solution:

Taking LHS in consideration, we get

=

Using the identity,

sin A + sin B = 2 sincos

cos A + cos B = 2 coscos

=

=

=

=

= tan 4x

Hence, LHS = RHS

Question 18:

Solution:

Taking LHS in consideration, we get

=

Using the identity,

sin A – sin B = 2 cossin

cos A + cos B = 2 coscos

=

=

=

Hence, LHS = RHS

Question 19:= tan 2x

Solution:

Taking LHS in consideration, we get

=

Using the identity,

sin A + sin B = 2 sincos

cos A + cos B = 2 coscos

=

=

=

=

=

= tan 2x

Hence, LHS = RHS

Question 20:= 2 sin x

Solution:

Taking LHS in consideration, we get

=

Using the identity,

sin A – sin B = 2 cossin

cos 2θ = cos² θ – sin² θ

=

=

=

=

= 2 sin (x)

Hence, LHS = RHS

Question 21:= cot 3x

Solution:

Taking LHS in consideration, we get

=

Using the identity,

sin A + sin B = 2 sincos

cos A + cos B = 2 coscos

=

=

=

Taking common, we have

=

=

= cot 3x

Hence, LHS = RHS

Question 22: cot x cot 2x – cot 2x cot 3x – cot 3x cot x = 1

Solution:

Taking LHS in consideration, we get

= cot x cot 2x – cot 2x cot 3x – cot 3x cot x

= cot x cot 2x – cot 3x (cot 2x + cot x)

= cot x cot 2x – cot (2x+x) (cot 2x + cot x)

Using the identity,

cot(A+B) =

= cot x cot 2x –(cot 2x + cot x)

= cot x cot 2x – [cot 2x cot x – 1]

= cot x cot 2x – cot 2x cot x – 1

= 1

Hence, LHS = RHS

Question 23: tan 4x =

Solution:

Taking LHS in consideration, we get

tan 4x = tan 2(2x)

Using the identity,

tan 2θ =

=

Again using the same identity, we get

=

=

=

=

=

=

Hence, LHS = RHS

Question 24: cos 4x = 1 – 8sin²x cos²x

Solution:

Taking LHS in consideration, we get

cos 4x = cos 2 (2x)

Using the identity,

cos 2θ = 1 – 2sin² θ

= 1 – 2sin² (2x)

= 1 – 2(2sin x cos x)² (As, sin 2θ = 2 sin θ cos θ)

= 1 – 2(4sin² x cos² x)

= 1 – 8sin² x cos² x

Hence, LHS = RHS

Question 25: cos 6x = 32 cos⁶x – 48cos⁴x + 18 cos²x – 1

Solution:

Taking LHS in consideration, we get

cos 6x = cos 3 (2x)

Using the identity,

cos 3θ = 4 cos³ θ – 3 cos θ

= 4 cos³ (2x) – 3 cos (2x)

= 4 cos³ (2x) – 3 cos (2x)

Now, Using the identity cos 2θ = 2cos² θ – 1

= 4 (2cos² x – 1)³ – 3 (2cos² x – 1)

Using algebraic identity,

(a-b)³ = a³ + b³ – 3a²b + 3ab²

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

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

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

= 32 cos⁶x – 48 cos⁴x + 18 cos²x – 1

Hence, LHS = RHS