Question

Prove $sin{20}^{0}sin{40}^{0}sin{60}^{0}sin{60}^0=\frac{3}{16}$

Answer 1

$=sin20sin40\times \frac{\sqrt{3}}{2}\times sin80$
$=\frac{\sqrt{3}}{2}\times \frac{2}{2}sin20sin40sin80$
$=\frac{\sqrt{3}}{4}\times \left(2sin20sin40\right)sin80$
$=\frac{\sqrt{3}}{4}\left[cos\right(20-40)-cos(20+40\left)\right]sin80$
$=\frac{\sqrt{3}}{4}[cos20-cos60]sin80$
$=\frac{\sqrt{3}}{4}[cos20-\frac{1}{2}]sin80$
$=\frac{\sqrt{3}}{4}[cos20sin80-\frac{sin80}{2}]$
$=\frac{\sqrt{3}}{4}[\frac{2sin80cos20}{2}-\frac{sin80}{2}]$
$=\frac{\sqrt{3}}{4}[\frac{sin(80+20)+sin(80-20)}{2}-\frac{sin80}{2}]$
$=\frac{\sqrt{3}}{4}[\frac{sin100+sin60}{2}-\frac{sin80}{2}]$
$=\frac{\sqrt{3}}{4}[\frac{sin100-sin80}{2}+\frac{\sqrt{3}}{4}]$
$=\frac{\sqrt{3}}{4}[\frac{2cos\left(\frac{100+80}{2}\right)sin\left(\frac{100-80}{2}\right)}{2}+\frac{\sqrt{3}}{4}]$
$=\frac{\sqrt{3}}{4}[\frac{cos90sin10}{2}+\frac{sqrt3}{4}]$
$=\frac{\sqrt{3}}{4}[\frac{0\times sin10}{2}+\frac{\sqrt{3}}{4}]$
$=\frac{\sqrt{3}}{4}\times \frac{\sqrt{3}}{4}$
$=\frac{3}{16}$