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Question Number 97657 by Vishal Sharma last updated on 09/Jun/20

$$\mathrm{If}\cos \alpha +\cos \beta =0=\sin \alpha +\sin \beta ,$$$$\mathrm{then}\cos 2\alpha +\cos 2\beta =$$

Commented by bemath last updated on 09/Jun/20

$$\left(1\right)\cos \alpha +\cos \beta =0$$$$2\cos \left(\frac{\alpha +\beta }{2}\right)\cos \left(\frac{\alpha -\beta }{2}\right)=0$$$$\mathrm{take}\frac{\alpha +\beta }{2}=\frac{\pi }{2}\Rightarrow \alpha +\beta =\pi$$$$\left(2\right)\cos {}^2\alpha +\cos {}^2\beta +2\cos \alpha \cos \beta =0$$$$\sin {}^2\alpha +\sin {}^2\beta +2\sin \alpha \sin \beta =0$$$$\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}+$$$$2+2\cos (\alpha -\beta )=0$$$$\Rightarrow \cos (\alpha -\beta )=-1$$$$\left(3\right)\cos 2\alpha +\cos 2\beta =$$$$2\cos (\alpha +\beta )\cos (\alpha -\beta )=$$$$2\times \cos \pi \times (-1)=2$$

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