Value of $\cos(\alpha+\beta)+\cos(\beta+\gamma)+\cos(\gamma+\alpha) =$ 
If $\alpha,\beta,\gamma$ be three distinct real values such that $$\displaystyle \frac{\sin \alpha+\sin \beta+\sin \gamma}{\sin(\alpha+\beta+\gamma)} = \frac{\cos \alpha+\cos \beta+\cos \gamma}{\cos(\alpha+\beta+\gamma)}=2$$
  Then value of $\cos(\alpha+\beta)+\cos(\beta+\gamma)+\cos(\gamma+\alpha) = $

$\bf{My\; Try::}$ Using 
$\sin (\alpha+\beta+\gamma) = \sin \alpha \cos \beta \cos \gamma+\cos \alpha \sin \beta \cos \gamma+\cos \alpha \cos \beta \sin \gamma-\sin \alpha \sin \beta \sin\gamma$
$\cos (\alpha+\beta+\gamma) = \cos \alpha \cos \beta \cos \gamma-\sin \alpha \sin \beta \cos \gamma-\cos \alpha \sin \beta \sin \gamma-\sin \alpha \cos \beta \sin \gamma$
But putting that values in above expression form very lengthy expression,
How can we solve it in less complex way,
Thanks
 A: Let $\ \pmb p_i =(\cos\ \theta_i,\ \sin\ \theta_i)\ $ and $\ \pmb p=\left(\cos \sum_i\theta_i,\ \sin\sum_i\theta_i \right)\ (i=1,2,3).$
Then $ \sum_i\pmb p_i =2\pmb p,$ and so $$ \bigg(\sum_i\ \pmb p_i\bigg)\cdot \pmb p =2.$$
A: Write $\sin(\alpha) $ as $\sin(t-(\beta+\gamma))$ where $t=\alpha+\beta+\gamma$
Now, expand.
We get $$\cos(\beta+\gamma) +\cos(\gamma+\alpha)+cos(\alpha+\beta) -\cot t(\sin(\beta+\gamma) +\sin(\gamma+\alpha)+sin(\alpha+\beta)) = \cos(\beta+\gamma) +\cos(\gamma+\alpha)+cos(\alpha+\beta) +\tan t(\sin(\beta+\gamma) +\sin(\gamma+\alpha)+sin(\alpha+\beta)) = 2 $$
Now, $\tan t$ can't be equal to $-\cot t$.
So, $\sin(\beta+\gamma) +\sin(\gamma+\alpha)+sin(\alpha+\beta)=0$
Substituting back in the above equation, $$\cos(\beta+\gamma) +\cos(\gamma+\alpha)+cos(\alpha+\beta)=2$$
A: The three expressions are symmetric wrt $\alpha, \, \beta, \, \gamma$, which means that,
 if they are valid for a given triple, they shall be valid for any permutation of that triple.
Thus if $(\alpha, \, \beta, \, \gamma)$ is a solution that lies on the plane $\alpha + \beta + \gamma=s$, the permutations will also lie on that plane.
Therefore
$$
\eqalign{
  & \cos \left( {\alpha  + \beta } \right) + \cos \left( {\beta  + \gamma } \right) + \cos \left( {\alpha  + \gamma } \right) =   \cr 
  &  = \cos \left( {s - \alpha } \right) + \cos \left( {s - \beta } \right) + \cos \left( {s - \gamma } \right) =   \cr 
  &  = \cos s\;\left( {\cos \alpha  + \cos \beta  + \cos \gamma } \right) + \sin s\left( {\sin \alpha  + \sin \beta  + \sin \gamma } \right)  \cr 
  &  = \cos ^{\,2} s\;{{\left( {\cos \alpha  + \cos \beta  + \cos \gamma } \right)} \over {\cos s}}
 + \sin ^{\,2} s{{\left( {\sin \alpha  + \sin \beta  + \sin \gamma } \right)} \over {\sin s}} =   \cr 
  &  = 2\left( {\cos ^{\,2} s + \sin ^{\,2} s} \right) = 2 \cr} 
$$
which is independent of $s$, and is therefore the answer.
