Question:
Prove that:
$\left(\cos \alpha+\cos \beta^{2}\right)+(\sin \alpha+\sin \beta)^{2}=4 \cos ^{2}\left(\frac{\alpha-\beta}{2}\right)$
Solution:
LHS $=(\cos \alpha+\cos \beta)^{2}+(\sin \alpha+\sin \beta)^{2}$
$=\cos ^{2} \alpha+\cos ^{2} \beta+2 \cos \alpha \cos \beta+\sin ^{2} \alpha+\sin ^{2} \beta+2 \sin \alpha \sin \beta$
$=\left(\cos ^{2} \alpha+\sin ^{2} \alpha\right)+\left(\cos ^{2} \beta+\sin ^{2} \beta\right)+2(\cos \alpha \cos \beta+\sin \alpha \sin \beta)$
$=1+1+2 \cos (\alpha-\beta)$
$=2\{1+\cos (\alpha-\beta)\}$
$=2\left\{2 \cos ^{2}\left(\frac{\alpha-\beta}{2}\right)\right\}$
$=4 \cos ^{2}\left(\frac{\alpha-\beta}{2}\right)=\mathrm{RHS}$
Hence proved.