Question

$sin\left(20\right)sin\left(40\right)sin\left(60\right)sin\left(80\right)=\frac{1}{16}$

Answer 1

$sin\left(20\right)sin\left(40\right)sin\left(60\right)sin\left(80\right)=\frac{1}{16}$

$sin\left(20\right)sin\left(40\right)sin\left(60\right)sin\left(80\right)$

substitute $sin\left(60\right)=\frac{\sqrt{3}}{2}$

$\frac{\sqrt{3}}{2}\left[sin\right(20\left)sin\right(40\left)sin\right(80\left)\right]$

$=\left(\frac{\sqrt{3}}{2}\right)sin\left(20\right)\left[sin\right(40\left)sin\right(80\left)\right]$

using the formula $sinAsinB=\left(\frac{1}{2}\right)\left[cos\right(A-B)-cos(A+B\left)\right]=\frac{\sqrt{3}}{2}sin\left(20\right)\left[cos\right(40)+cos(60\left)\right]=\frac{\sqrt{3}}{4}sin\left(20\right)\left[cos\right(40)+cos(60\left)\right]=\frac{\sqrt{3}}{4}sin\left(20\right)\left[cos\right(40)+\frac{1}{2}]=\frac{\sqrt{3}}{4}sin\left(20\right)cos\left(40\right)+\left(\frac{\sqrt{3}}{8}\right)sin\left(20\right)$

use the formula $sinAcosB=\frac{1}{2}\left[sin\right(A+B)+sin(A-B\left)\right]$

$=\left(\frac{\sqrt{3}}{4}\right)\left(\frac{1}{2}\right)\left[sin\right(60)+sin(-20\left)\right]+\left(\frac{\sqrt{3}}{8}\right)sin\left(20\right)=\left(\frac{\sqrt{3}}{8}\right)[\left(\frac{\sqrt{3}}{2}\right)-sin(20\left)\right]+\left(\frac{\sqrt{3}}{8}\right)sin\left(20\right)=\frac{3}{16}-\left(\frac{\sqrt{3}}{8}\right)sin\left(20\right)+\left(\frac{\sqrt{3}}{8}\right)sin\left(20\right)=\frac{3}{16}$