Question

Prove the following :

cos 20° cos 40° cos 60° cos 80° = ${\displaystyle \frac{1}{16}}$

Answer 1

L.H.S. = cos 20°· cos 40° · cos 60° · cos 80°

${\displaystyle =\frac{1}{2}\times (2 {\cos 20}^{\circ }\cdot {\cos 40}^{\circ })\times \frac{1}{2}\times {\cos 80}^{\circ } ...[2\cos \text{A}\cdot \cos \text{B}=\cos \left(\text{A + B}\right)+\cos (\text{A}\cdot \text{B})]}$

${\displaystyle =\frac{1}{2}\left[{\cos 60}^{\circ }+\cos (-{20}^{\circ })\right]\times \frac{1}{2}\times {\cos 80}^{\circ }}$

${\displaystyle =\frac{1}{2}({\cos 60}^{\circ }+ {\cos 20}^{\circ })\times \frac{1}{2}\times {\cos 80}^{\circ }}$

${\displaystyle =\frac{1}{2}(\frac{1}{2}+{\cos 20}^{\circ })\times \frac{1}{2}{\cos 80}^{\circ }}$

${\displaystyle =\frac{1}{8}{\cos 80}^{\circ }+\frac{1}{4}{\cos 20}^{\circ }\cdot {\cos 80}^{\circ }}$

${\displaystyle =\frac{1}{8}{\cos 80}^{\circ }+\frac{1}{4}\times \frac{1}{2}(2{\cos 20}^{\circ }\cdot {\cos 80}^{\circ }) ...[2\cos \text{A}+\cos \text{B}=\cos \left(\text{A + B}\right)+\cos \left(\text{A - B}\right)]}$

${\displaystyle =\frac{1}{8}{\cos 80}^{\circ }+\frac{1}{8}\left[{\cos 100}^{\circ }+{\cos (-60)}^{\circ }\right]}$

${\displaystyle =\frac{1}{8}{\cos 80}^{\circ }+\frac{1}{8}[\cos ({180}^{\circ }-{80}^{\circ })+{\cos 60}^{\circ }]}$

${\displaystyle =\frac{1}{8}{\cos 80}^{\circ }+\frac{1}{8}[-{\cos 80}^{\circ }+\frac{1}{2}]}$

${\displaystyle =\frac{1}{8}{\cos 80}^{\circ }-\frac{1}{8}{\cos 80}^{\circ }+\frac{1}{16}}$

${\displaystyle =\cancel{\frac{1}{8}{\cos 80}^{\circ }}-\cancel{\frac{1}{8}{\cos 80}^{\circ }}+\frac{1}{16}}$

${\displaystyle =\frac{1}{16}}$

= R.H.S.