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

Answer

$\text{L.H.S.}=(\cos\text{x}-\cos\text{y})^2+(\sin\text{x}-\sin\text{y})^2$$=\Big(-2\sin\frac{\text{x}+\text{y}}{2}\sin\frac{\text{x}-\text{y}}{2}\Big)^2+\Big(2\cos\frac{\text{x}+\text{y}}{2}\sin\frac{\text{x}-\text{y}}{2}\Big)^2$
$\bigg[\because\cos\text{A}-\cos\text{B}=-2\sin\frac{\text{A}+\text{B}}{2}\sin\frac{\text{A}-\text{B}}{2}\text{and} \sin\text{B}=2\cos\frac{\text{A}+\text{B}}{2}\sin\frac{\text{A}-\text{B}}{2}\bigg]$
$=\Big(4\sin^2\frac{\text{x}+\text{y}}{2}\sin^2\frac{\text{x}-\text{y}}{2}\Big)+\Big(4\cos^2\frac{\text{x}+\text{y}}{2}\sin^2\frac{\text{x}-\text{y}}{2}\Big)$
$=4\sin^2\frac{\text{x}-\text{y}}{2}\Big(\sin^2\frac{\text{x}+\text{y}}{2}+\cos^2\frac{\text{x}+\text{y}}{2}\Big)$
$=4\sin^2\frac{\text{x}-\text{y}}{2}=\text{R.H.S.}$

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