Question

Prove that ${(\cos A+\cos B)}^2+{(\sin A-\sin B)}^2=4{\cos }^2\left(\frac{A+B}{2}\right)$

Answer 1

$LHS={\cos }^{2}A+{\cos }^{2}B+2\cos A\cos B+{\sin }^{2}A+{\sin }^{2}B-2\sin A\sin B$

$({\cos }^{2}A+{\sin }^{2}A)+({\cos }^{2}B+{\sin }^{2}B)+2\cos A\cos B-2\sin A\sin B$

$2+2\cos A\cos B-2\sin A\sin B$

$2+\cos (A+B)+\cos (A-B)-\cos (A-B)+\cos (A+B)$

$2+2\cos (A+B)$ $(\therefore \cos x=2{\cos }^2\frac{x}{2}-1)$

$2+2[2{\cos }^2\frac{(A+B)}{2}-1]$

$2+4{\cos }^2\frac{(A+B)}{2}-2=4{\cos }^2\frac{(A+B)}{2}=R.H.S$

Hence proved.