Prove that:

$\mathrm{LHS}=\cos 36^{\circ} \cos 42^{\circ} \cos 60^{\circ} \cos 78^{\circ}$

$=\frac{1}{2} \cos 36^{\circ} \cos 60^{\circ}\left(2 \cos 42^{\circ} \cos 78^{\circ}\right)$

$[2 \cos A \cos B=\cos (A+B)+\cos (A-B)]$

$=\frac{1}{2}\left(\frac{\sqrt{5}+1}{4}\right) \times \frac{1}{2}\left(\cos 120^{\circ}+\cos 36^{\circ}\right)$

$=\left(\frac{\sqrt{5}+1}{16}\right)\left(-\frac{1}{2}+\frac{\sqrt{5}+1}{4}\right)$

$=\frac{(\sqrt{5}+1)(\sqrt{5}-1)}{64}$

$=\frac{5-1}{64}$

$=\frac{1}{16}$

= RHS
Hence proved.