We have, L.H.S.= sin2x + 2 sin 4x + sin 6x = sin2x + sin 6x + 2 sin 4x = 2 sin $\frac{2x+6x}{2}$ cos $\frac{2x-6x}{2}$ + 2 sin 4x Since sin A + sin B = 2 sin $\frac{A+B}{2}$ . cos $\frac{A-B}{2}$ = 2sin4x cos(–2x) + 2 sin4x = (2sin4x)(1 + cos (2x)) [Since cos (–θ) = cos θ] = 2 sin 4x (2 $\cos^2$ x) [Since cos2x = 2 $\cos^2$ x – 1] = 4 cos2x sin4x = R.H.S.