Question

Prove geometrically that:
$\cos (A+B)=\cos A\cos B-\sin A\sin B$

Answer 1

$\cos (a+b)=AC-BC=AC-FD$

$=\cos \left(a\right)\cos \left(b\right)-\sin \left(a\right)\sin \left(b\right)$

$\sin (a+b)=BF+FE=CA+FE$

$=\sin \left(a\right)\cos \left(b\right)+\cos \left(a\right)\sin \left(b\right)$