Question:
Prove that
$(\cos x-\cos y)^{2}+(\sin x-\sin y)^{2}=4 \sin ^{2}\left(\frac{x-y}{2}\right)$
Solution:
$=(\cos x-\cos y)^{2}+(\sin x-\sin y)^{2}$
$=\left(-2 \sin \frac{x+y}{2} \sin \frac{x-y}{2}\right)^{2}+\left(2 \cos \frac{x+y}{2} \sin \frac{x-y}{2}\right)^{2}$
$=4 \sin ^{2}\left(\frac{x-y}{2}\right)\left(\sin ^{2}\left(\frac{x-y}{2}\right)+\cos ^{2}\left(\frac{x-y}{2}\right)\right)$
$=4 \sin ^{2}\left(\frac{x-y}{2}\right)$
Using the formula,
$\cos A-\cos B=-2 \sin \frac{A+B}{2} \sin \frac{A-B}{2}$
$\sin A-\sin B=2 \cos \frac{A+B}{2} \sin \frac{A-B}{2}$
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