Question

Evaluate: $\sin {20}^{\circ }\sin {40}^{\circ }\sin {80}^{\circ }=\frac{\sqrt{3}}{8}$

Answer 1

Starting with L.H.S,

$sin\left(20\right)sin\left(40\right)sin\left(80\right)$

$=\frac{1}{2}\left[cos\right(20)-cos(60\left)\right]sin\left(80\right)$

$=\frac{1}{2}\left[cos\right(20\left)sin\right(80)-cos(60\left)sin\right(80\left)\right]$

$=\frac{1}{2}\left[\frac{1}{2}\right(sin\left(100\right)+sin\left(60\right))-\frac{1}{2}(sin\left(140\right)+sin\left(20\right)\left)\right]$

$=\frac{1}{4}\left[sin\right(80)+sin(60)-sin(40)-sin(20\left)\right]$

$=\frac{1}{4}sin\left(60\right)$

$\{since,sin(80)-sin(40)-sin(20)=sin(60+20)-sin(60-20)-sin(20)=2cos(60\left)sin\right(20)-sin(20)=0\}$

$=\frac{\sqrt{3}}{8}$ which is R.H.S