We have, L.H.S.= sinx + sin3x + sin5x + sin7x = (sinx + sin 7x) + (sin 3x + sin 5x) = 2 sin $\left(\frac{x+7x}{2}\right)$ cos $\left(\frac{x-7x}{2}\right)$ + 2 sin $\left(\frac{3x+5x}{2}\right)$ cos $\left(\frac{3x-5x}{2}\right)$ = 2 sin (4x) cos (–3x) + 2 sin 4x cos (–x) = 2 sin 4x cos 3x + 2 sin 4x cos x [Since cos (–θ) = cos θ] = 2 sin 4x(cos 3x + cosx) = 2 sin 4x $\left(2\cos\left(\frac{3x+x}{2}\right)\cos\left(\frac{3x-x}{2}\right)\right)$ = 2 sin 4x(2cos 2x cosx) = 4 cos x cos 2x sin 4x = R.H.S.