Question

Prove that: $\sin 6^{\circ }\sin {42}^{\circ }\sin {66}^{\circ }\sin {78}^{\circ }=\frac{1}{16}$.

Answer 1

L.H.S. = $\frac{1}{2}$[(2 sin ${}^{\circ }$ sin ${66}^{\circ }$)(2 sin ${42}^{\circ }$ sin ${78}^{\circ }$)]
$=\frac{1}{4}$ [(cos ${60}^{\circ }$ - cos ${72}^{\circ }$)(cos ${36}^{\circ }$ -cos ${120}^{\circ }$)]
$=\frac{1}{4}\left[(\frac{1}{2}-\frac{\sqrt{5}-1}{4})\left(\frac{\sqrt{5}+1}{4}\frac{1}{2}\right)\right]$
$=\frac{1}{4}\frac{3-\sqrt{5}}{4}.\frac{3+\sqrt{5}}{4}=\frac{1}{64}.(9-5)=\frac{1}{16}.$