Question

If $\cos \alpha +\cos \beta =0=\sin \alpha +\sin \beta$,
then prove that
$\cos 2\alpha +\cos 2\beta =-2\cos (\alpha +\beta )$.

Answer 1

Given:

$\cos \alpha +\cos \beta =0$ and
$\sin \alpha +\sin \beta =0$

Square & subtract given equations, we get
$(\cos \alpha +\cos \beta {)}^2-(\sin \alpha +\sin \beta {)}^2=0$
$\Rightarrow {\cos }^2\alpha +{\cos }^2\beta +2\cos \alpha \cos \beta -{\sin }^2\alpha –{\sin }^2\beta -2\sin \alpha \sin \beta =0$
$\Rightarrow ({\cos }^2\alpha -{\sin }^2\alpha )+({\cos }^2\beta -{\sin }^2\beta )+2(\cos \alpha \cos \beta -\sin \alpha \sin \beta )=0$
$\left[{}_{\cos A\cos B–\sin A\sin B=\cos (A+B)}^{{\cos }^{2}A–{\sin }^{2}A=\cos 2A\text{and}}\right]$
$\Rightarrow \cos 2\alpha +\cos 2\beta +2\cos (\alpha +\beta )=0$
$\Rightarrow \cos 2\alpha +\cos 2\beta =-2\cos (\alpha +\beta )$

Hence, proved.