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

solution:

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

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

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

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

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

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

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