Question

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

Answer 1

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

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

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

$=2(1+\cos (x+y))$

$=4co{s}^2\frac{x+y}{2}=RHS$...........$(\because$ $1+cos2x=2co{s}^{2}x)$

Hence proved