Question

If $\cos (\beta -\gamma )+\cos (\gamma -\alpha )+\cos (\alpha -\beta )=-\frac{3}{2}$, then

• A. $\sum \cos \alpha =0$
• B. $\sum \sin \alpha =0$
• C. $\sum \cos \alpha \sin \alpha =0$
• D. $\sum (\cos \alpha +\sin \alpha )=0$

Answer 1

Answer: (A), (B), (D)

$\sum \cos \alpha =0$
$\sum \sin \alpha =0$
$\sum (\cos \alpha +\sin \alpha )=0$

Given $\cos (\beta -\gamma )+\cos (\gamma -\alpha )+\cos (\alpha -\beta )=-\frac{3}{2}$

$⟹\cos \beta \cos \gamma +\sin \beta \sin \gamma +\cos \gamma \cos \alpha +\sin \gamma \sin a+\cos \alpha \cos \beta +\sin a\sin \beta =-\frac{3}{2}$

$⟹2(\cos \beta \cos \gamma +\sin \beta \sin \gamma +\cos \gamma \cos \alpha +\sin \gamma \sin a+\cos \alpha \cos \beta +\sin a\sin \beta )+1+1+2=0$

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

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

$⟹\sum \cos \alpha =0$ and $\sum \sin \alpha =0$

$⟹\sum (\cos \alpha +\sin \alpha )=0$

Hence, options 'A', 'B' and 'D' are correct.