Question

If $\cos \alpha +\cos \beta =0=\sin \alpha +\sin \beta$, then $\cos 2\alpha +\cos 2\beta$ $=2\cdot \cos (\pi +\alpha +\beta )$

Answer 1

Given that $\cos \alpha +\cos \beta =0$

$\Rightarrow \cos \beta =-\cos \alpha$ …………..$\left(1\right)$

Given that $\sin \alpha +\sin \beta =0$

$\Rightarrow \sin \beta =-\sin \alpha$ …………$\left(2\right)$

Now,

L.H.S$=\cos 2\alpha +\cos 2\beta$

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

$=({\cos }^{2}2\alpha -{\sin }^{2}2\alpha )+\left[\right(-\cos \alpha {)}^2-(-\sin \alpha {)}^2]$

By substituting based on $\left(1\right)$ & $\left(2\right)$

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

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

$=2(\cos \alpha \cdot \cos \alpha -\sin \alpha \cdot \sin \alpha )$

$=2\left(\cos \alpha \right(-\cos \beta )-\sin \alpha (-\sin \beta )$

$=-2(\cos \alpha \cdot \cos \beta -\sin \alpha \sin \beta )$

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

$=2\cdot \cos (\pi +\alpha +\beta )$

$=$R.H.S.