Question

Prove that:
$(cosx-cosy{)}^2+(sinx-siny{)}^2=4si{n}^2\frac{x-y}{2}$

Answer 1

$LHS=(cosx-cosy{)}^2+(sinx-siny{)}^2$

$={\cos }^{2}x+{\cos }^{2}y-2\cos x\cos y+{\sin }^{2}x+{\sin }^{2}y-2\sin x\sin y$

$=({\cos }^{2}x+{\sin }^{2}x)+({\cos }^{2}y+{\sin }^{2}y)-2(\cos x\cos y+\sin x\sin y)=1+1-2\cos (x-y)$

$=2(1-\cos (x-y))=4si{n}^2\frac{x-y}{2}=RHS$.............$(\because$ $1-cos2x=2si{n}^{2}x)$

Hence proved