Prove that: \[ \sin^2\frac{2\pi}{5} – \sin^2\frac{\pi}{10} = \frac{\sqrt5-1}{8} \]
Solution
Using the identity
\[
\sin^2A-\sin^2B
=
\sin(A+B)\sin(A-B)
\]
Let
\[
A=\frac{2\pi}{5}, \qquad B=\frac{\pi}{10}
\]
Then
\[
A+B
=
\frac{2\pi}{5}+\frac{\pi}{10}
=
\frac{5\pi}{10}
=
\frac{\pi}{2}
\]
\[
A-B
=
\frac{2\pi}{5}-\frac{\pi}{10}
=
\frac{3\pi}{10}
\]
Therefore,
\[
\sin^2\frac{2\pi}{5}
–
\sin^2\frac{\pi}{10}
=
\sin\frac{\pi}{2}\sin\frac{3\pi}{10}
\]
\[
=
1\cdot\sin\frac{3\pi}{10}
\]
\[
=
\sin54^\circ
\]
Now use the known value
\[
\sin54^\circ
=
\cos36^\circ
=
\frac{\sqrt5+1}{4}
\]
Thus,
\[
\sin^2\frac{2\pi}{5}
–
\sin^2\frac{\pi}{10}
=
\frac{\sqrt5+1}{4}
\]
Hence proved.