(i) Using the expansions of \(\cos(3x-x)\) and \(\cos(3x+x)\), prove that \(\frac{1}{2}(\cos 2x - \cos 4x) \equiv \sin 3x \sin x\).

(ii) Hence show that \(\int_{\frac{1}{6}\pi}^{\frac{1}{3}\pi} \sin 3x \sin x \, dx = \frac{1}{8}\sqrt{3}\).