sin 3θ=3sin θ–4sin³ θ

⇒4 sin³θ=3sinθ–sin 3θ

Now,

Substituting equation (i) in above LHS, we get

We know,

(as sin θ =sin (180°–θ))

Similarly,

(as –sin θ =sin (180°+θ))

Substituting the equation (iii) and (iv) in equation (ii), we get

We know,

Substituting this in the above equation, we get

Hence proved