सिद्ध कीजिए कि: $(\cos x-\cos y)^{2}+(\sin x-\sin y)^{2}=4 \sin ^{2} \frac{x-y}{2}$

(A) $L.H.S. = (\cos x-\cos y)^{2}+(\sin x-\sin y)^{2}$
$= (\cos^{2} x + \cos^{2} y - 2 \cos x \cos y) + (\sin^{2} x + \sin^{2} y - 2 \sin x \sin y)$
$= (\cos^{2} x + \sin^{2} x) + (\cos^{2} y + \sin^{2} y) - 2(\cos x \cos y + \sin x \sin y)$
$= 1 + 1 - 2 \cos(x - y)$
$= 2 - 2 \cos(x - y)$
$= 2(1 - \cos(x - y))$
$= 2 \times 2 \sin^{2} \left(\frac{x - y}{2}\right)$
$= 4 \sin^{2} \left(\frac{x - y}{2}\right) = R.H.S.$