Question

Prove that:

$\text{sin 3x+ sin 2x - sin x = 4 sin x cos}\frac{x}{2}\text{cos}\frac{3x}{2}$

Answer 1

We have L.H.S

= sin 3x + sin 2x - sin x.

= (sin 3x - sin x) + sin 2x

= $\left[\text{2 cos}\left(\frac{3x+x}{2}\right)\text{sin}\left(\frac{3x-x}{2}\right)\right]+\text{2 sin x cos x}[\because \text{sin C- sin D = 2 cos}\frac{C+D}{2}\text{sin}\frac{C-D}{2}]$

= 2 cos 2x sin x + 2 sin x cos x

= 2 sin x [ cos 2x+ cos x]

= 2 sin x $\left[\text{2 cos}\left(\frac{2x+x}{2}\right)\text{cos}\left(\frac{2x-x}{2}\right)\right]$

= 2 sin x $\left[\text{2 cos}\left(\frac{3x}{2}\right)\text{cos}\left(\frac{x}{2}\right)\right]$

= 4 sin x cos $\frac{x}{2}$ cos $\frac{3x}{2}$ = R.H.S.