If \( (\cos\alpha+\cos\beta)^2+(\sin\alpha+\sin\beta)^2=\lambda\cos^2\frac{\alpha-\beta}{2} \), find \( \lambda \)
Solution:
\[
(\cos\alpha+\cos\beta)^2+(\sin\alpha+\sin\beta)^2
\]
Expanding,
\[
=\cos^2\alpha+\cos^2\beta+2\cos\alpha\cos\beta
\]
\[
+\sin^2\alpha+\sin^2\beta+2\sin\alpha\sin\beta
\]
Using,
\[
\sin^2\theta+\cos^2\theta=1
\]
\[
=1+1+2(\cos\alpha\cos\beta+\sin\alpha\sin\beta)
\]
Using identity,
\[
\cos(\alpha-\beta)
=
\cos\alpha\cos\beta+\sin\alpha\sin\beta
\]
\[
=2+2\cos(\alpha-\beta)
\]
Using,
\[
1+\cos\theta=2\cos^2\frac{\theta}{2}
\]
\[
=2\left[1+\cos(\alpha-\beta)\right]
\]
\[
=2\left[2\cos^2\frac{\alpha-\beta}{2}\right]
\]
\[
=4\cos^2\frac{\alpha-\beta}{2}
\]
Comparing with
\[
\lambda\cos^2\frac{\alpha-\beta}{2}
\]
\[
\lambda=4
\]
\[
\boxed{4}
\]