Question

$\sin {36}^{\circ }\sin {72}^{\circ }.\sin {108}^{\circ }.\sin {144}^{\circ }=\frac{k}{16}.$ Find the value of $k$.

Answer 1

$\sin {36}^o\sin {72}^o\sin {108}^o\sin {144}^o=\sin {36}^o\sin {72}^o\sin ({90}^o+{18}^o)\sin ({90}^o+{34}^o)=\sin {36}^o\sin {72}^o\cos {18}^o\cos {54}^o$
$=\frac{1}{4}\left(2\sin {36}^o\cos {54}^o\right)\left(2\sin {72}^o\cos {18}^o\right)$
$=\frac{1}{4}(\cos ({36}^o-{54}^o)-\cos ({36}^o+{54}^o))(\cos ({72}^o-{18}^o)-\cos ({72}^o+{18}^o))$
$=\frac{1}{4}(\cos {18}^o-\cos {90}^o)(\cos {54}^o-\cos {90}^o)$
$=\frac{1}{4}\cos {18}^o\cos {54}^o=\frac{1}{4}\left(\sqrt{\frac{5}{8}+\frac{\sqrt{5}}{8}}\right)\left(\sqrt{\frac{5}{8}-\frac{\sqrt{5}}{8}}\right)=\frac{5}{16}$