Show that $\dfrac{1}{2}\left(\cos 4x + \cos 2x\right) = \cos 3x \cos x$
$\cos 4x = \cos \left(3x + x\right)$ and $\cos 2x = \left(3x - x\right)$ $\cos \left(3x + x\right) = \cos 3x \cos x - \sin 3x \sin x$ $\cos \left(3x - x\right) = \cos 3x \cos x + \sin 3x \sin x$ $ \begin{aligned} \therefore \dfrac{1}{2}\left(\cos 4x + \cos 2x\right) & = \dfrac{1}{2} \left[\cos \left(3x + x\right) + \cos \left(3x - x\right)\right] \\ & = \dfrac{1}{2} \left(\cos 3x \cos x - \sin 3x \sin x + \cos 3x \cos x + \sin 3x \sin x\right) \\ & = \dfrac{1}{2} \times 2 \cos 3x \cos x \\ & = \cos 3x \cos x = \text{RHS} \end{aligned} $