Prove that \[ \sin^2\frac{\pi}{8} +\sin^2\frac{3\pi}{8} +\sin^2\frac{5\pi}{8} +\sin^2\frac{7\pi}{8} =2 \]
Proof:
Using the identity
\[
\sin(\pi-\theta)=\sin\theta
\]
therefore,
\[
\sin^2\frac{5\pi}{8}=\sin^2\frac{3\pi}{8}
\]
and
\[
\sin^2\frac{7\pi}{8}=\sin^2\frac{\pi}{8}
\]
Hence,
\[
LHS
=2\sin^2\frac{\pi}{8}
+2\sin^2\frac{3\pi}{8}
\]
\[
=2\left(
\sin^2\frac{\pi}{8}
+\sin^2\frac{3\pi}{8}
\right)
\]
Using
\[
\sin^2A=\frac{1-\cos2A}{2}
\]
we get
\[
LHS
=2\left(
\frac{1-\cos\frac{\pi}{4}}{2}
+
\frac{1-\cos\frac{3\pi}{4}}{2}
\right)
\]
\[
=2\left(
\frac{1-\frac{\sqrt2}{2}}{2}
+
\frac{1+\frac{\sqrt2}{2}}{2}
\right)
\]
\[
=2\left(\frac{2}{2}\right)
\]
\[
=2
\]
Hence proved,
\[
\boxed{
\sin^2\frac{\pi}{8}
+\sin^2\frac{3\pi}{8}
+\sin^2\frac{5\pi}{8}
+\sin^2\frac{7\pi}{8}
=2
}
\]