Question

$\sin {20}^{\circ }\sin {40}^{\circ }\sin {60}^{\circ }\sin {80}^{\circ }=\frac{k}{16}$. Find the value of $k$.

Answer 1

$\sin {20}^o\sin {40}^o\sin {60}^o\sin {80}^o$
$=\sin ({60}^o-{20}^o)\sin {20}^o\sin ({60}^o+{20}^o)\sin {60}^o$

$=\frac{1}{4}\sin (3\times {20}^o)\sin {60}^o=\frac{1}{4}{\sin }^2{60}^o=\frac{3}{16}$$(\because \sin (60-\theta )\sin (60+\theta )\sin \theta =\frac{1}{4}\sin 3\theta )$

$k=3$