Question

cos³x sin 3x + sin³x cos 3x =3/4 sin 4x

Answer 1

In the attachment I have answered this problem.

I have applied Cos3A and Sin3A formulae to remove the cubes occurred in the expression.

The final answer comes from the
formula

Sin(A+B) = SinA CosB + CosA SinB

I hope this answer helps you

Answer 2

Hello.

Question is correct, but in the Question, you have not clearly mentioned that we need to prove the identity.

Given Trigonometric Equation ⇒
Cos³ x + Sin 3x + Sin³x + 3 Cos x = 3/4 × Sin 4x

Proof ⇒ 

From L.H.S.
Cos³ x . Sin 3x + Sin³x .  Cos3x 

Now, We know,
Cos 3 θ = 4 Cos³θ - 3 Cos θ.
∴ 4 Cos³θ  = Cos 3 θ +  3 Cos θ.
∴ Cos³θ  = ( Cos 3 θ +  3 Cos θ)/4

∴ Cos³ x . Sin 3x =  (Cos 3 x +  3 Cos x)/4 . Sin 3x
    
Also,  
Sin 3θ = 3 Sin θ - 4 Sin³θ
∴ 4 Sin³θ = 3 Sin θ - Sin 3θ
∴  Sin³θ = (3 Sin θ - Sin 3θ)/4

∴  Sin³x . Cos 3x  = (3 Sin x - Sin 3x)/4 . Cos 3x


∴ Cos³ x . Sin 3x + Sin³x . 3 Cos x  =  (Cos 3 x +  3 Cos x)/4 . Sin 3x + (3 Sin x - Sin 3x)/4 . Cos 3x
= 1/4[(Cos 3 x +  3 Cos x)Sin 3x + ( Sin 3x - Sin 3x) . Cos3x
= 1/4[Sin 3x.Cos 3x + Sin 3x. 3Cos x + 3Sin x.Cos 3 x  - Sin 3x  Cos 3x].

= 3/4[Sin 3x Cos x + Cos 3x Sinx]
(∵  SinA CosB + CosA SinB = Sin(A+B) )

= 3/4[Sin 4x] 
= R.H.S.

Hence, Proved.


Hope it helps.