Prove that:

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.

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