Question

$\cos {10}^{\circ }\cos {30}^{\circ }\cos {50}^{\circ }\cos {70}^{\circ }=\frac{k}{16}$. Find the value of $k$.

Answer 1

$\cos {10}^o\cos {30}^o\cos {50}^o\cos {70}^o=\cos ({90}^o-{80}^o)\cos ({90}^o-{60}^o)\cos ({90}^o-{40}^o)\cos ({90}^o-{20}^o)=\sin {80}^o\sin {60}^o\sin {40}^o\sin {20}^o=\sin ({60}^o+{20}^o)\sin \left({20}^o\right)\sin ({60}^o-{20}^o)\sin \left({60}^o\right)$
$=\frac{1\sin {60}^o\sin {60}^o}{4}=\frac{1}{4}{\sin }^2{60}^0=\frac{3}{16}$
$(\because \sin (60+\theta )\sin \theta \sin (60-\theta )=\frac{\sin 3\theta }{4})$