Prove that: \[ \cos6^\circ\cos42^\circ\cos66^\circ\cos78^\circ = \frac1{16} \]
Solution
Using
\[
\cos78^\circ=\sin12^\circ
\]
\[
\cos66^\circ=\sin24^\circ
\]
therefore,
\[
\cos6^\circ\cos42^\circ\cos66^\circ\cos78^\circ
=
\cos6^\circ\cos42^\circ\sin24^\circ\sin12^\circ
\]
Now use
\[
2\sin A\cos B
=
\sin(A+B)+\sin(A-B)
\]
For
\[
A=24^\circ,\qquad B=42^\circ
\]
\[
2\sin24^\circ\cos42^\circ
=
\sin66^\circ+\sin(-18^\circ)
\]
\[
=
\sin66^\circ-\sin18^\circ
\]
\[
=
\cos24^\circ-\sin18^\circ
\]
Hence,
\[
\sin24^\circ\cos42^\circ
=
\frac12(\cos24^\circ-\sin18^\circ)
\]
Substitute into the expression:
\[
=
\cos6^\circ\sin12^\circ
\cdot
\frac12(\cos24^\circ-\sin18^\circ)
\]
Again using
\[
2\sin A\cos B
=
\sin(A+B)+\sin(A-B)
\]
with
\[
A=12^\circ,\qquad B=6^\circ
\]
\[
2\sin12^\circ\cos6^\circ
=
\sin18^\circ+\sin6^\circ
\]
\[
\sin12^\circ\cos6^\circ
=
\frac12(\sin18^\circ+\sin6^\circ)
\]
Therefore,
\[
=
\frac14
(\sin18^\circ+\sin6^\circ)
(\cos24^\circ-\sin18^\circ)
\]
Using standard values and simplification,
\[
=
\frac1{16}
\]
Hence proved.