Question

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

Answer 1

Given:${(\cos x+\cos y)}^2+{(\sin x-\sin y)}^2$

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

Hence proved.