Question

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

Answer 1

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

Hence proved.