Prove that: $(\cos x+\cos y)^{2}+(\sin x-\sin y)^{2}=4 \cos ^{2} \frac{x+y}{2}$

solution:

L.H.S. $=(\cos x+\cos y)^{2}+(\sin x-\sin y)^{2}$

$=\cos ^{2} x+\cos ^{2} y+2 \cos x \cos y+\sin ^{2} x+\sin ^{2} y-2 \sin x \sin y$

$=\left(\cos ^{2} x+\sin ^{2} x\right)+\left(\cos ^{2} y+\sin ^{2} y\right)+2(\cos x \cos y-\sin x \sin y)$

$=1+1+2 \cos (x+y) \quad[\cos (A+B)=(\cos A \cos B-\sin A \sin B)]$

$=2+2 \cos (x+y)$

$=2[1+\cos (x+y)]$

$=2\left[1+2 \cos ^{2}\left(\frac{x+y}{2}\right)-1\right] \quad\left[\cos 2 A=2 \cos ^{2} A-1\right]$

$=4 \cos ^{2}\left(\frac{x+y}{2}\right)=$ R.H.S.