Question

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

Answer 1

Simplifying the LHS of ${\left(\cos A+\cos B\right)}^2+{\left(\sin A+\sin B\right)}^2=4{\cos }^2\frac{A-B}{2}$.

${\left(\cos A+\cos B\right)}^2+{\left(\sin A+\sin B\right)}^2={\cos }^{2}A+{\cos }^{2}B+2\cos A\cos B+{\sin }^{2}A+{\sin }^{2}B+2\sin A\sin B$

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

$=1+1+2\left(\cos A\cos B+\sin A\sin B\right)$

$=2+2\cos \left(A-B\right)$

$=2\left(1+\cos \left(A-B\right)\right)$

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

$=4{\cos }^2\frac{A-B}{2}$

$=RHS$