Prove that: $(\cos x-\cos y)^{2}+(\sin x-\sin y)^{2}=4 \sin ^{2} \frac{x-y}{2}$
Prove that: $(\cos x-\cos y)^{2}+(\sin x-\sin y)^{2}=4 \sin ^{2} \frac{x-y}{2}$
$L.H.S.$ $(\cos x-\cos y)^{2}+(\sin x-\sin y)^{2}$
$=\cos ^{2} x+\cos ^{2} y-2 \cos x \cos y+\sin ^{2} x+\sin ^{2} y-2 \sin x \sin y$
$=\left(\cos ^{2} x+\sin ^{2} x\right)+\left(\cos ^{2} y+\sin ^{2} y\right)-2[\cos x \cos y+\sin x \sin y]$
$=1+1-2[\cos (x-y)]$
$[\cos (A-B)=\cos A \cos B+\sin A \sin B]$
$=2[1-\cos (x-y)]$
$=2\left[1-\left\{1-2 \sin ^{2}\left(\frac{x-y}{2}\right)\right\}\right] \quad\left[\cos 2 A=1-2 \sin ^{2} A\right]$
$=4 \sin ^{2}\left(\frac{x-y}{2}\right)= R . H.S.$
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- [JEE MAIN 2021]