Prove that: \( \sin10^\circ \sin50^\circ \sin60^\circ \sin70^\circ = \frac{\sqrt3}{16} \)
Solution:
\[
\sin10^\circ \sin50^\circ \sin60^\circ \sin70^\circ
\]
Using identity,
\[
\sin3\theta = 4\sin\theta \sin(60^\circ+\theta)\sin(60^\circ-\theta)
\]
Putting \( \theta=10^\circ \),
\[
\sin30^\circ
=
4\sin10^\circ \sin50^\circ \sin70^\circ
\]
\[
\frac12
=
4\sin10^\circ \sin50^\circ \sin70^\circ
\]
\[
\sin10^\circ \sin50^\circ \sin70^\circ
=
\frac18
\]
Therefore,
\[
\sin10^\circ \sin50^\circ \sin60^\circ \sin70^\circ
=
\frac18 \times \frac{\sqrt3}{2}
\]
\[
=
\frac{\sqrt3}{16}
\]
\[
\boxed{\sin10^\circ \sin50^\circ \sin60^\circ \sin70^\circ = \frac{\sqrt3}{16}}
\]