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