Prove that:

$\mathrm{LHS}=\cos 6^{\circ} \cos 42^{\circ} \cos 66^{\circ} \cos 78^{\circ}$

$=\frac{1}{4}\left(2 \cos 6^{\circ} \cos 66^{\circ}\right)\left(2 \cos 42^{\circ} \cos 78^{\circ}\right)$

$=\frac{1}{4}\left(\cos 72^{\circ}+\cos 60^{\circ}\right)\left(\cos 120^{\circ}+\cos 36^{\circ}\right) \quad[\because 2 \cos A \cos B=\cos (A+B)+\cos (A-B)]$

$=\frac{1}{4}\left\{\cos \left(90^{\circ}-72^{\circ}\right)+\frac{1}{2}\right\}\left\{-\frac{1}{2}+\frac{\sqrt{5}+1}{4}\right\}$

$=\frac{1}{4}\left(\sin 18^{\circ}+\frac{1}{2}\right)\left(-\frac{1}{2}+\frac{\sqrt{5}+1}{4}\right)$

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

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

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

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

$=\frac{1}{16}=\mathrm{RHS}$

Hence proved.