Question

If
$sin\alpha +sin\beta +sin\gamma =0=cos\alpha +cos\beta +cos\gamma$
then value of
$cos(\alpha -\beta )+cos(\beta -\gamma )+cos(\gamma -\alpha )is$

Answer 1

We have given that
$sin\alpha +sin\beta +sin\gamma =0=cos\alpha +cos\beta +cos\gamma$
So
$sin\alpha +sin\beta =-sin\gamma$
and
$cos\alpha +cos\beta =-cos\gamma$

$(sin\alpha +sin\beta {)}^2+(cos\alpha +cos\beta {)}^2=(-sin\gamma {)}^2+(-cos\gamma {)}^2\Rightarrow 1+1+2cos(\alpha -\beta )=1$
$\because \cos \alpha .cos\beta +\sin \alpha .\sin \beta =\cos (\alpha -\beta )$
$\Rightarrow cos(\alpha -\beta )=\frac{-1}{2}Similarlycos(\beta -\gamma )=\frac{-1}{2}andcos(\gamma -\alpha )=\frac{-1}{2}$

$\therefore$ Thus the value of expression is $\frac{-3}{2}$