We have, L.H.S.= (sin3x + sinx) sinx + (cos3x – cosx) cosx = $\left(2\sin\left(\frac{3x+x}{2}\right)\cos\left(\frac{3x-x}{2}\right)\right)$ sin x + $\left(-2\sin\left(\frac{3x+x}{2}\right)\sin\left(\frac{3x-x}{2}\right)\right)$ cos x Since sin A + sin B = 2 sin $\left(\frac{A+B}{2}\right)$ cos $\left(\frac{A-B}{2}\right)$ and cos A - cos B = - 2 sin $\left(\frac{A+B}{2}\right)$ cos $\left(\frac{A-B}{2}\right)$ = [(2 sin2x cosx) sinx] – [(2 sin2x sinx) cosx] = [2 sin2x sinx cosx – 2 sin2x·sinx cosx] = 0 = R.H.S.