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

Answer

To prove: $(\cos x-\cos y)^{2}+(\sin x-\sin y)^{2}=4 \sin ^{2} \frac{x-y}{2}$
We know that:
$\cos x-\cos y=-2 \sin \left(\frac{x+y}{2}\right) \sin \left(\frac{x-y}{2}\right)$
$\sin x-\sin y=2 \cos \left(\frac{x+y}{2}\right) \sin \left(\frac{x-y}{2}\right)$
Using these formulae in L.H.S. we get,
L.H.S $=(\cos x-\cos y)^2+(\sin x-\sin y)^2$
$=\left(-2 \sin \left(\frac{\mathrm{x}+\mathrm{y}}{2}\right) \sin \left(\frac{\mathrm{x}-\mathrm{y}}{2}\right)\right)^{2}+\left(2 \cos \left(\frac{\mathrm{x}+\mathrm{y}}{2}\right) \sin \left(\frac{\mathrm{x}-\mathrm{y}}{2}\right)\right)^{2}$
$=4 \sin ^{2}\left(\frac{\mathrm{x}+\mathrm{y}}{2}\right) \sin ^{2}\left(\frac{\mathrm{x}-\mathrm{y}}{2}\right)+4 \cos ^{2}\left(\frac{\mathrm{x}+\mathrm{y}}{2}\right) \sin ^{2}\left(\frac{\mathrm{x}-\mathrm{y}}{2}\right)$
$=4 \sin ^{2}\left(\frac{\mathrm{x}-\mathrm{y}}{2}\right)\left(\sin ^{2}\left(\frac{\mathrm{x}+\mathrm{y}}{2}\right)+\cos ^{2}\left(\frac{\mathrm{x}+\mathrm{y}}{2}\right)\right)$
= 4 $\sin ^{2}\left(\frac{\mathrm{x}-\mathrm{y}}{2}\right)$
= R.H.S
Hence, L.H.S = R. H. S

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