Prove that \( \sin20^\circ\sin40^\circ\sin80^\circ=\frac{\sqrt3}{8} \)
Solution
Using identity: \(2\sin A\sin B=\cos(A-B)-\cos(A+B)\)
\[
\begin{aligned}
&\sin20^\circ\sin40^\circ\sin80^\circ\\[4pt]
&=\frac12(\cos20^\circ-\cos60^\circ)\sin80^\circ\\[4pt]
&=\frac12\left(\cos20^\circ-\frac12\right)\sin80^\circ\\[4pt]
&=\frac12\cos20^\circ\sin80^\circ-\frac14\sin80^\circ
\end{aligned}
\]
Using identity: \(2\sin A\cos B=\sin(A+B)+\sin(A-B)\)
\[
\begin{aligned}
&=\frac14(\sin100^\circ+\sin60^\circ)-\frac14\sin80^\circ\\[4pt]
&=\frac14(\cos10^\circ+\frac{\sqrt3}{2})-\frac14\cos10^\circ\\[4pt]
&=\frac{\sqrt3}{8}
\end{aligned}
\]