Prove that

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

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

$=4 \sin ^{2}\left(\frac{x-y}{2}\right)\left(\sin ^{2}\left(\frac{x-y}{2}\right)+\cos ^{2}\left(\frac{x-y}{2}\right)\right)$

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

Using the formula,

$\cos A-\cos B=-2 \sin \frac{A+B}{2} \sin \frac{A-B}{2}$

$\sin A-\sin B=2 \cos \frac{A+B}{2} \sin \frac{A-B}{2}$