Question

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

• A. $p=q=0$
• B. $p+q=1$
• C. $p=q=1$
• D. $p-q=1$

Answer 1

The correct option is A $p=q=0$

Given

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

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

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

Consider

$p^2+q^2$

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

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

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

$=1+1+1+2\left(\cos \right(\alpha -\beta )+\cos (\beta -\gamma )+\cos (\gamma -\alpha \left)\right)$

$({\cos }^2\theta +{\sin }^2\theta =1,\cos (A-B)=\cos A\cos B+\sin A\sin B)$

$=3+2\left(\frac{-3}{2}\right)$

$=⟹3-3=0$