Prove that

Question:

Prove that

$\cos 10^{\circ} \cos 30^{\circ} \cos 50^{\circ} \cos 70^{\circ}=\frac{3}{16}$

Solution:

L.H.S

$=\cos 10^{\circ} \cos 30^{\circ} \cos 50^{\circ} \cos 70^{\circ}$

$=\frac{1}{2}\left(2 \cos 70^{\circ} \cos 10^{\circ}\right) \cos 50^{\circ} \frac{\sqrt{3}}{2}$

$=\frac{\sqrt{3}}{4}\left\{\cos \left(70^{\circ}+10^{\circ}\right)+\cos \left(70^{\circ}-10^{\circ}\right)\right\} \cos 50^{\circ}$

$=\frac{\sqrt{3}}{4}\left\{\cos 80^{\circ} \cos 50^{\circ}+\cos 60^{\circ} \cos 50^{\circ}\right\}$

$\left.=\frac{\sqrt{3}}{4}\left\{\cos 80^{\circ} \cos 50^{\circ}+\frac{1}{2} \cos 50^{\circ}\right\}\right\}$

$=\frac{\sqrt{3}}{8}\left\{2 \cos 80^{\circ} \cos 50^{\circ}+\cos 50^{\circ}\right\}$

$=\frac{\sqrt{3}}{8}\left\{\cos 130^{\circ}-\cos 30^{\circ}+\cos 50^{\circ}\right\}$

$=\frac{\sqrt{3}}{8}\left\{\cos 130^{\circ}-\cos 50^{\circ}+\cos 30^{\circ}\right\}$

$=\frac{\sqrt{3}}{8}\left\{\cos \left(180^{\circ}-50^{\circ}\right)-\cos \left(50^{\circ}\right)+\frac{\sqrt{3}}{2}\right\}$

$=\frac{\sqrt{3}}{8}\left\{\cos 50^{\circ}-\cos 50^{\circ}+\frac{\sqrt{3}}{2}\right\}$

$=\frac{3}{16}$