Question

Prove:
$\cos {20}^o.\cos {40}^o.\cos {60}^o.\cos {80}^o=\frac{1}{16}$

Answer 1

${\cos 20}^0.{\cos 40}^0.{\cos 60}^0.{\cos 80}^0$

$\Rightarrow \cos \left({20}^0\right)\times \cos ({60}^0-{20}^0)\times \cos ({60}^0+{20}^0)\times \frac{1}{2}$

$\Rightarrow cosA.\cos ({60}^0-A)\cos ({60}^0+A)=\frac{1}{4}\cos 3A$

$=\frac{1}{4}\times \cos (3\times {20}^0)\times \frac{1}{2}$

$\Rightarrow \frac{1}{8}\times \frac{1}{2}=\frac{1}{16}$

Hence, proved.