Question

If $\cos \left(\alpha -\beta \right)+\cos \left(\beta -\gamma \right)+\cos \left(\gamma -\alpha \right)=-\frac{3}{2}$, Prove
that $\cos \alpha +\cos \beta +\cos \gamma =\sin \alpha +\sin \beta +\sin \gamma =0$

Answer 1

$2\cos \alpha \cos \beta +2\cos \beta \cos \gamma +2\cos \gamma \cos \alpha +2\sin \alpha \sin \beta +2\sin \beta \sin \gamma +2\sin \gamma \sin \alpha +3=0$

$2\cos \alpha \cos \beta +2\cos \beta \cos \gamma +2\cos \gamma \cos \alpha +2\sin \alpha \sin \beta +2\sin \beta \sin \gamma +$

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

${\cos }^2\alpha +{\cos }^2\beta +{\cos }^2\gamma +2\cos \alpha \cos \beta +2\cos \beta \cos \gamma +2\cos \gamma \cos \alpha +{\sin }^2\alpha +{\sin }^2\beta +{\sin }^2\gamma +$

$2\sin \alpha \sin \beta +\sin \beta \sin \gamma +\sin \gamma \sin \alpha$

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

Since a square quantity is always $+ve$ or $zero$

So,

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

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