Proving $\sin 20^\circ \, \sin 40^\circ \, \sin 60^\circ \, \sin 80^\circ =\frac{3}{16}$ The problem is to prove that
$$\sin 20^\circ \, \sin 40^\circ \, \sin 60^\circ \, \sin 80^\circ =\frac{3}{16}$$
All my attempts were to get them in $\sin (2A)$ form after eliminating $\sin 60^\circ$ in both sides. Unfortunately, all these attempts were futile.
Any hints are welcomed.

Denote $Q = \sin 20 \sin 40 \sin 60 \sin 80$.  Observe first that
\begin{align}
\sin 60 & = \sin (20+40) \\
        & = \sin 20 \cos 40 + \cos 20 \sin 40 \\
        & = \sin 20 (2\cos^2 20 - 1) + \cos 20 (2 \sin 20 \cos 20) \\
        & = (4\cos^2 20 - 1) \sin 20
\end{align}
We can now write
\begin{align}
Q & = \sin 20 \sin (60-20) \sin 60 \sin (60+20) \\
  & = \sin 20 (\sin 60 \cos 20 - \cos 60 \sin 20)
      \sin 60 (\sin 60 \cos 20 + \cos 60 \sin 20) \\
  & = \sin 20 \sin 60 (\sin^2 60 \cos^2 20 - \cos^2 60 \sin^2 20) \\
  & = \sin 20 \sin 60 \frac{3\cos^2 20 - \sin^2 20}{4} \\
  & = \sin 20 \sin 60 \frac{4\cos^2 20 - 1}{4} \\
  & = \sin 60 \frac{(4\cos^2 20 - 1)\sin 20}{4} \\
  & = \sin 60 \frac{\sin 60}{4} \\
  & = \frac{\sin^2 60}{4} = \frac{3}{16}
\end{align}
