Question

If $\cos \alpha +\cos \beta =0=\sin \alpha +\sin \beta ,$ then $\cos 2\alpha +\cos 2\beta$ is equal to-

• A. $-2\sin (\alpha +\beta )$
• B. $-2\cos (\alpha +\beta )$
• C. $2\sin (\alpha +\beta )$
• D. $2\cos (\alpha +\beta )$

Answer 1

The correct option is B $-2\cos (\alpha +\beta )$

$\cos 2\alpha +\cos 2\beta$

$\cos C+\cos D=\cos \frac{C+D}{2}\cos \frac{C-D}{2}$

$2\cos (\alpha +\beta )+\cos (\alpha -\beta )$

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

$(\cos \alpha +\cos \beta {)}^2-$$(\sin \alpha +\sin \beta {)}^2=0$

${\cos }^2\alpha +{\cos }^2\beta +2\cos \alpha \cos \beta -$${\sin }^2\alpha -{\sin }^2\beta -2\sin \alpha \sin \beta =0$

$({\cos }^2\alpha -{\sin }^2\alpha )+$$({\cos }^2\beta -{\sin }^2\beta )=2(-\sin \alpha \sin \beta +\cos \alpha \cos \beta )$

$({\cos }^2\alpha -{\sin }^2\alpha )+$$({\cos }^2\beta -{\sin }^2\beta )=-2\cos (\alpha +\beta )$

$(\cos 2\alpha +\cos 2\beta )=-2\cos (\alpha +\beta )$