The value of sin 12° sin 24° sin 48° sin 84°, is

We have,

$\sin {12}^{\circ }\sin {24}^{\circ }\sin {48}^{\circ }\sin {84}^{\circ }$

$\begin{array}{l}=\frac{1}{4}\left(2\sin {12}^{\circ }\sin {48}^{\circ }\right)\left(2\sin {24}^{\circ }\sin {84}^{\circ }\right)\\ =\frac{1}{2}\left(\cos {36}^{\circ }-\cos {60}^{\circ }\right)\left(\cos {60}^{\circ }-\cos {108}^{\circ }\right)\\ =\frac{1}{4}\left(\cos {36}^{\circ }-\frac{1}{2}\right)\left(\frac{1}{2}+\sin {18}^{\circ }\right)\\ =\frac{1}{4}\left\{\frac{1}{4}(\sqrt{5}+1)-\frac{1}{2}\right\}\left\{\frac{1}{2}+\frac{1}{4}(\sqrt{5}-1)\right\}=\frac{1}{16}\end{array}$

and, $\cos {20}^{\circ }\cos {40}^{\circ }\cos {60}^{\circ }\cos 80$

$\begin{array}{l}=\frac{1}{2}\left[\cos \left({60}^{\circ }-{20}^{\circ }\right)\cos {20}^{\circ }\cos \left({60}^{\circ }+{20}^{\circ }\right)\right]\\ =\frac{1}{2}\left\{\frac{1}{4}\cos 3\left({20}^{\circ }\right)\right\}=\frac{1}{8}\cos {60}^{\circ }=\frac{1}{2}\times \frac{1}{8}=\frac{1}{16}\end{array}$