Prove that: \[ \cos36^\circ\cos42^\circ\cos60^\circ\cos78^\circ = \frac1{16} \]
Solution
Since
\[
\cos60^\circ=\frac12
\]
therefore,
\[
\cos36^\circ\cos42^\circ\cos60^\circ\cos78^\circ
=
\frac12\cos36^\circ\cos42^\circ\cos78^\circ
\]
Also,
\[
\cos78^\circ=\sin12^\circ
\]
Hence,
\[
=
\frac12\cos36^\circ\cos42^\circ\sin12^\circ
\]
Using the identity
\[
2\sin A\cos B
=
\sin(A+B)+\sin(A-B)
\]
with
\[
A=12^\circ,\qquad B=42^\circ
\]
\[
2\sin12^\circ\cos42^\circ
=
\sin54^\circ+\sin(-30^\circ)
\]
\[
=
\sin54^\circ-\sin30^\circ
\]
\[
=
\cos36^\circ-\frac12
\]
Therefore,
\[
\sin12^\circ\cos42^\circ
=
\frac12\left(\cos36^\circ-\frac12\right)
\]
Substitute this value:
\[
=
\frac12\cos36^\circ
\cdot
\frac12\left(\cos36^\circ-\frac12\right)
\]
\[
=
\frac14
\left(
\cos^236^\circ-\frac12\cos36^\circ
\right)
\]
Using the standard value
\[
\cos36^\circ=\frac{\sqrt5+1}{4}
\]
Therefore,
\[
\cos^236^\circ
=
\left(\frac{\sqrt5+1}{4}\right)^2
=
\frac{3+\sqrt5}{8}
\]
Hence,
\[
=
\frac14
\left(
\frac{3+\sqrt5}{8}
–
\frac{\sqrt5+1}{8}
\right)
\]
\[
=
\frac14\cdot\frac2{8}
\]
\[
=
\frac1{16}
\]
Hence proved.