Question

Prove that ${\sin }^{2}A={\cos }^2(A-B)+{\cos }^{2}B-2\cos (A-B)\cos A\cos B$.

Answer 1

Taking RHS:

$\cos (A-B)\left[\cos \right(A-B)-2\cos A.\cos B]+{\cos }^{2}B$

$=\cos (A-B)[\cos A.\cos B+\sin A.\sin B-2\cos A.\cos B]+{\cos }^{2}B$

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

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

$=\frac{1}{2}[2{\cos }^{2}B-\cos 2A-\cos 2B]$

$=\frac{1}{2}\left[2{\sin }^{2}A\right]={\sin }^{2}A$