Prove that: \[ \sin\frac{\pi}{5} \sin\frac{2\pi}{5} \sin\frac{3\pi}{5} \sin\frac{4\pi}{5} = \frac{5}{16} \]
Solution
Using
\[
\sin(\pi-\theta)=\sin\theta
\]
we get
\[
\sin\frac{3\pi}{5}
=
\sin\frac{2\pi}{5}
\]
\[
\sin\frac{4\pi}{5}
=
\sin\frac{\pi}{5}
\]
Therefore,
\[
\sin\frac{\pi}{5}
\sin\frac{2\pi}{5}
\sin\frac{3\pi}{5}
\sin\frac{4\pi}{5}
=
\sin^2\frac{\pi}{5}
\sin^2\frac{2\pi}{5}
\]
Now use the identity
\[
2\sin A\sin B
=
\cos(A-B)-\cos(A+B)
\]
with
\[
A=\frac{\pi}{5},\qquad B=\frac{2\pi}{5}
\]
\[
2\sin\frac{\pi}{5}\sin\frac{2\pi}{5}
=
\cos\frac{\pi}{5}
–
\cos\frac{3\pi}{5}
\]
Since
\[
\cos\frac{3\pi}{5}
=
-\cos\frac{2\pi}{5}
\]
therefore,
\[
2\sin\frac{\pi}{5}\sin\frac{2\pi}{5}
=
\cos\frac{\pi}{5}
+
\cos\frac{2\pi}{5}
\]
Using the standard values
\[
\cos36^\circ=\frac{\sqrt5+1}{4}
\]
\[
\cos72^\circ=\frac{\sqrt5-1}{4}
\]
Hence,
\[
2\sin\frac{\pi}{5}\sin\frac{2\pi}{5}
=
\frac{\sqrt5+1}{4}
+
\frac{\sqrt5-1}{4}
\]
\[
=
\frac{2\sqrt5}{4}
=
\frac{\sqrt5}{2}
\]
\[
\sin\frac{\pi}{5}\sin\frac{2\pi}{5}
=
\frac{\sqrt5}{4}
\]
Squaring both sides,
\[
\sin^2\frac{\pi}{5}
\sin^2\frac{2\pi}{5}
=
\frac{5}{16}
\]
Therefore,
\[
\sin\frac{\pi}{5}
\sin\frac{2\pi}{5}
\sin\frac{3\pi}{5}
\sin\frac{4\pi}{5}
=
\frac{5}{16}
\]
Hence proved.