Prove that: \[ \cos78^\circ\cos42^\circ\cos36^\circ=\frac18 \]
Solution
Using
\[
\cos78^\circ=\sin12^\circ
\]
therefore,
\[
\cos78^\circ\cos42^\circ\cos36^\circ
=
\sin12^\circ\cos42^\circ\cos36^\circ
\]
Now use
\[
2\sin A\cos B
=
\sin(A+B)+\sin(A-B)
\]
with
\[
A=12^\circ,\qquad B=42^\circ
\]
Then
\[
2\sin12^\circ\cos42^\circ
=
\sin54^\circ+\sin(-30^\circ)
\]
\[
=
\sin54^\circ-\sin30^\circ
\]
\[
=
\cos36^\circ-\frac12
\]
Hence,
\[
\sin12^\circ\cos42^\circ
=
\frac12\left(\cos36^\circ-\frac12\right)
\]
Therefore,
\[
\cos78^\circ\cos42^\circ\cos36^\circ
=
\frac12\left(\cos36^\circ-\frac12\right)\cos36^\circ
\]
\[
=
\frac12\left(\cos^236^\circ-\frac12\cos36^\circ\right)
\]
Using the standard value
\[
\cos36^\circ=\frac{\sqrt5+1}{4}
\]
Substitute:
\[
=
\frac12\left[
\left(\frac{\sqrt5+1}{4}\right)^2
–
\frac12\cdot\frac{\sqrt5+1}{4}
\right]
\]
\[
=
\frac12\left[
\frac{6+2\sqrt5}{16}
–
\frac{\sqrt5+1}{8}
\right]
\]
\[
=
\frac12\left[
\frac{6+2\sqrt5-2\sqrt5-2}{16}
\right]
\]
\[
=
\frac12\cdot\frac4{16}
\]
\[
=
\frac18
\]
Hence proved.