Question

$\cos {60}^{\circ }\cos {36}^{\circ }\cos {42}^{\circ }\cos {78}^{\circ }=\frac{1}{16}$

Answer 1

L.H.S. = $\frac{1}{2}.\frac{(\sqrt{5}+1)}{4}.\frac{1}{2}$ (2 \cos ${42}^{\circ }$ \cos ${78}^{\circ }$)
$=\frac{1}{16}(\sqrt{5}+1)$ (cos ${120}^{\circ }$ + cos ${36}^{\circ }$)
$=\frac{1}{16}(\sqrt{5}+1)(-\frac{1}{2}+\frac{\sqrt{5}+1}{4})$
$=\frac{1}{16}(\sqrt{5}+1).\frac{\sqrt{5}-1}{4}=\frac{1}{16}.\frac{5-1}{4}=\frac{1}{16}.$