Question

Prove that:
$\sin {20}^{\circ }\sin {40}^{\circ }\sin {60}^{\circ }\sin {80}^{\circ }=\frac{3}{16}$

Answer 1

To prove: $\sin {20}^{\circ }\sin {40}^{\circ }\sin {60}^{\circ }\sin {80}^{\circ }=\frac{3}{16}$

L.H.S $=\sin {20}^{\circ }\sin {40}^{\circ }\sin {60}^{\circ }\sin {80}^{\circ }$

Multiplying $\&$ dividing by 2
$=\frac{1}{2}\sin {60}^{\circ }\left[2\sin {20}^{\circ }\sin {40}^{\circ }\right]\sin {80}^{\circ }$
$[\because 2\sin A\sin B=\cos (A-B)-\cos (A+B\left)\right]$
$=\frac{1}{2}\times \frac{\sqrt{3}}{2}[\cos ({20}^{\circ }-{40}^{\circ })-\cos ({20}^{\circ }+{40}^{\circ })]\sin {80}^{\circ }$
$[\because \sin {60}^{\circ }=\frac{\sqrt{3}}{2}]$
$=\frac{\sqrt{3}}{4}[\cos {20}^{\circ }-\cos {60}^{\circ }]\sin {80}^{\circ }$
$\left[\cos \right(-\theta )=\cos \theta ]$
$=\frac{\sqrt{3}}{4}\sin {80}^{\circ }[\cos {20}^{\circ }-\frac{1}{2}]$

$\text{L.H.S}=\frac{\sqrt{3}}{4}\sin {80}^{\circ }[\cos {20}^{\circ }-\frac{1}{2}]$
$=\frac{\sqrt{3}}{4}\sin {80}^{\circ }\cos {20}^{\circ }-\frac{\sqrt{3}}{8}\sin {80}^{\circ }$
$=\frac{\sqrt{3}}{4}\sin ({90}^{\circ }-{10}^{\circ })\cos {20}^{\circ }-\frac{\sqrt{3}}{8}\sin {80}^{\circ }$
$=\frac{\sqrt{3}}{4}\cos {10}^{\circ }\cos {20}^{\circ }-\frac{\sqrt{3}}{8}\sin {80}^{\circ }$
$[\because \sin ({90}^{\circ }-\theta )=\cos \theta ]$
$=\frac{\sqrt{3}}{8}\left[2\cos {10}^{\circ }\cos {20}^{\circ }\right]-\frac{\sqrt{3}}{8}\sin {80}^{\circ }$
$[\because 2\cos A\cos B=\cos (A+B)+\cos (A-B\left)\right]$
$=\frac{\sqrt{3}}{8}\left[\begin{array}{c}\cos ({10}^{\circ }+{20}^{\circ })\\ +\cos ({10}^{\circ }-{20}^{\circ })\end{array}\right]-\frac{\sqrt{3}}{8}\sin {80}^{\circ }$

$\text{L.H.S}=\frac{\sqrt{3}}{8}\left[\begin{array}{c}\cos ({10}^{\circ }+{20}^{\circ })\\ +\cos ({10}^{\circ }-{20}^{\circ })\end{array}\right]-\frac{\sqrt{3}}{8}\sin {80}^{\circ }$
$=\frac{\sqrt{3}}{8}[\cos {30}^{\circ }+\cos (-{10}^{\circ })]-\frac{\sqrt{3}}{8}\sin {80}^{\circ }$
$=\frac{\sqrt{3}}{8}[\cos {30}^{\circ }+\cos \left({10}^{\circ }\right)]-\frac{\sqrt{3}}{8}\sin {80}^{\circ }$
$=\frac{\sqrt{3}}{8}[\cos {30}^{\circ }+\cos ({90}^{\circ }-{80}^{\circ })]-\frac{\sqrt{3}}{8}\sin {80}^{\circ }$
$=\frac{3}{16}+\frac{\sqrt{3}}{8}\sin {80}^{\circ }-\frac{\sqrt{3}}{8}\sin {80}^{\circ }$

$[\because \cos ({90}^{\circ }-\theta )=\sin \theta ]$
$=\frac{3}{16}$
$=\text{R.H.S}$
$\text{Hence, proved.}$