Prove that: \( \sin20^\circ \sin40^\circ \sin60^\circ \sin80^\circ = \frac{3}{16} \)
Solution:
\[
\sin20^\circ \sin40^\circ \sin60^\circ \sin80^\circ
\]
Using identity,
\[
\sin3\theta = 4\sin\theta \sin(60^\circ+\theta)\sin(60^\circ-\theta)
\]
Putting \( \theta=20^\circ \),
\[
\sin60^\circ
=
4\sin20^\circ \sin40^\circ \sin80^\circ
\]
\[
\frac{\sqrt3}{2}
=
4\sin20^\circ \sin40^\circ \sin80^\circ
\]
\[
\sin20^\circ \sin40^\circ \sin80^\circ
=
\frac{\sqrt3}{8}
\]
Therefore,
\[
\sin20^\circ \sin40^\circ \sin60^\circ \sin80^\circ
=
\frac{\sqrt3}{8}\times\frac{\sqrt3}{2}
\]
\[
=
\frac{3}{16}
\]
\[
\boxed{\sin20^\circ \sin40^\circ \sin60^\circ \sin80^\circ=\frac{3}{16}}
\]