Prove that: \[ \sin6^\circ\sin42^\circ\sin66^\circ\sin78^\circ = \frac1{16} \]
Solution
Using
\[
\sin78^\circ=\cos12^\circ
\]
\[
\sin66^\circ=\cos24^\circ
\]
Therefore,
\[
\sin6^\circ\sin42^\circ\sin66^\circ\sin78^\circ
=
\sin6^\circ\sin42^\circ\cos24^\circ\cos12^\circ
\]
Now use the identity
\[
2\sin A\cos B
=
\sin(A+B)+\sin(A-B)
\]
For
\[
A=42^\circ,\qquad B=24^\circ
\]
\[
2\sin42^\circ\cos24^\circ
=
\sin66^\circ+\sin18^\circ
\]
\[
=
\cos24^\circ+\sin18^\circ
\]
Hence,
\[
\sin42^\circ\cos24^\circ
=
\frac12(\cos24^\circ+\sin18^\circ)
\]
Substitute into the expression:
\[
=
\sin6^\circ\cos12^\circ
\cdot
\frac12(\cos24^\circ+\sin18^\circ)
\]
Again using
\[
2\sin A\cos B
=
\sin(A+B)+\sin(A-B)
\]
with
\[
A=6^\circ,\qquad B=12^\circ
\]
\[
2\sin6^\circ\cos12^\circ
=
\sin18^\circ-\sin6^\circ
\]
\[
\sin6^\circ\cos12^\circ
=
\frac12(\sin18^\circ-\sin6^\circ)
\]
Therefore,
\[
=
\frac14
(\sin18^\circ-\sin6^\circ)
(\cos24^\circ+\sin18^\circ)
\]
Using standard trigonometric values and simplification,
\[
=
\frac1{16}
\]
Hence proved.