We have, L.H.S.= sin 3x + sin 2x – sinx = sin 3x – sin x + sin 2x = 2 cos $\left(\frac{3x+x}{2}\right)\sin\left(\frac{3x-x}{2}\right)$ + sin 2x Since sin A - sin B = 2 cos $\left(\frac{A+B}{2}\right)\sin\left(\frac{A-B}{2}\right)$ = 2cos(2x) sinx + 2 sinx cosx [Since sin 2x = 2 sinx cosx] = 2 sinx [cos 2x + cosx] = 2 sin x $\left[2\cos\left(\frac{2x+x}{2}\right)\cos\left(\frac{2x-x}{2}\right)\right]$ Since cos A + cos B = 2 cos $\left(\frac{A+B}{2}\right)$ cos $\left(\frac{A-B}{2}\right)$ = 2 sin x 2 cos $\frac{3x}{2} \cos \frac{x}{2}$ = 4 sin x cos $\frac{3x}{2}$ cos $\frac{x}{2}$ = R.H.S.