Question

The value of $\sin 20^{\circ }\cdot \sin 40^{\circ }\cdot \sin 60^{\circ }\cdot \sin 80^{\circ }$ is

• A. $\frac{-3}{16}$
• B. $\frac{5}{16}$
• C. $\frac{3}{16}$
• D. $\frac{1}{16}$

Answer 1

Answer: (C)

$\sin 20^{\circ }\sin 40^{\circ }\sin 60^{\circ }\sin 80^{\circ }$
$=\frac{\sqrt{3}}{2}\sin 20^{\circ }\sin \left(60^{\circ }-20^{\circ }\right)\sin \left(60^{\circ }+20^{\circ }\right)$
$\left(\because \sin 60^{\circ }=\frac{\sqrt{3}}{2}\right)$
$=\frac{\sqrt{3}}{2}\sin 20^{\circ }\left[{\sin }^{2}60^{\circ }-{\sin }^{2}20^{\circ }\right]$
$\left[\because \sin (A+B)\cdot \sin (A-B)={\sin }^{2}A-{\sin }^{2}B\right]$
$=\frac{\sqrt{3}}{2}\sin 20^{\circ }\left[\frac{3}{4}-{\sin }^{2}20^{\circ }\right]$
$=\frac{\sqrt{3}}{2}\times \frac{1}{4}\left[3\sin 20^{\circ }-4{\sin }^{3}20^{\circ }\right]$
$=\frac{\sqrt{3}}{2}\times \frac{1}{4}\left(\sin 60^{\circ }\right)\left(\because \sin 3A=3\sin A-4{\sin }^{3}A\right)$
$=\frac{\sqrt{3}}{2}\times \frac{1}{4}\times \frac{\sqrt{3}}{2}=\frac{3}{16}(\because \sin )$