Prove that \[ (\cos\alpha+\cos\beta)^2+(\sin\alpha+\sin\beta)^2 =4\cos^2\left(\frac{\alpha-\beta}{2}\right) \]
Proof:
\[
LHS=(\cos\alpha+\cos\beta)^2+(\sin\alpha+\sin\beta)^2
\]
Expanding both squares:
\[
=\cos^2\alpha+\cos^2\beta+2\cos\alpha\cos\beta
\]
\[
+\sin^2\alpha+\sin^2\beta+2\sin\alpha\sin\beta
\]
Grouping terms:
\[
=(\cos^2\alpha+\sin^2\alpha)
+(\cos^2\beta+\sin^2\beta)
\]
\[
+2(\cos\alpha\cos\beta+\sin\alpha\sin\beta)
\]
Using
\[
\cos^2\theta+\sin^2\theta=1
\]
and
\[
\cos(\alpha-\beta)
=\cos\alpha\cos\beta+\sin\alpha\sin\beta
\]
we get
\[
LHS=1+1+2\cos(\alpha-\beta)
\]
\[
=2+2\cos(\alpha-\beta)
\]
Using the identity
\[
1+\cos\theta=2\cos^2\frac{\theta}{2}
\]
therefore,
\[
2+2\cos(\alpha-\beta)
=2\left(1+\cos(\alpha-\beta)\right)
\]
\[
=2\left(2\cos^2\frac{\alpha-\beta}{2}\right)
\]
\[
=4\cos^2\left(\frac{\alpha-\beta}{2}\right)
\]
Hence proved,
\[
\boxed{
(\cos\alpha+\cos\beta)^2+(\sin\alpha+\sin\beta)^2
=4\cos^2\left(\frac{\alpha-\beta}{2}\right)
}
\]