$\min$ of expression $\sin \alpha+\sin \beta+\sin \gamma,$satisfying $\alpha+\beta+\gamma = \pi$ 
The $\min$ of expression $\sin \alpha+\sin \beta+\sin \gamma,$ Where $\alpha,\beta,\gamma\in \mathbb{R}$ satisfying $\alpha+\beta+\gamma = \pi$

$\bf{Options ::}$ $(a)\;\; + ve \;\;\;\;\;\;\; (b)\;\; -ve \;\;\;\;\;\; (c)\;\; 0\;\;\;\;\;\; (d)\;\; -3$
$\bf{My\; Try::}$ Putting $\displaystyle \alpha = -\frac{\pi}{2}\;\;\;\;\;\; ,\beta = -\frac{\pi}{2}\;\;\;\;\;\; ,\gamma = 2\pi$
Then we get $\sin\alpha+\sin \beta+\sin \gamma = -ve $
$\bf{Added::}$ Trying using analytical way: Given $\alpha+\beta+\gamma = \pi$
$$\sin \alpha+\sin \beta+\sin \gamma = 2\sin \left(\frac{\alpha+\beta}{2}\right)\cos \left(\frac{\alpha-\beta}{2}\right)+2\sin \frac{\gamma}{2}\cdot \cos \frac{\gamma}{2}$$
So $$ = 2\cos \frac{\gamma}{2}\left[\cos \left(\frac{\alpha-\beta}{2}\right)+\cos \left(\frac{\alpha+\beta}{2}\right)\right] = 4\cos \frac{\alpha}{2}\cos \frac{\beta}{2}\cos \frac{\gamma}{2}$$
Now Using $\displaystyle -1 \leq \cos \frac{\alpha}{2}\;,\cos \frac{\beta}{2}\;,\cos \frac{\gamma}{2}\leq 1$
So we get $$\sin \alpha+\sin \beta+\sin \gamma = 4\cos \frac{\alpha}{2}\cos \frac{\beta}{2}\cos \frac{\gamma}{2}\geq -4$$
But this is not possible bcz $\displaystyle \cos \frac{\alpha}{2}\;,\cos \frac{\beta}{2}\;,\cos \frac{\gamma}{2}\neq -1$ Simultaneously 
I did not understand how can i solve it
Help me , Thanks

With an hopefully obvious notation, $$\sin\alpha+\sin\beta+\sin(\alpha+\beta)=a+b\pm a\sqrt{1-b^2}\pm b\sqrt{1-a^2}.$$
Cancelling the gradient, and rearranging,
$$\begin{cases}\sqrt{1-a^2}\pm\sqrt{1-a^2}\sqrt{1-b^2}\mp ab=0,
\\\sqrt{1-b^2}\mp ab\pm\sqrt{1-a^2}\sqrt{1-b^2}=0.\end{cases}$$
This proves a symmetry, $\sqrt{1-a^2}=\sqrt{1-b^2}$ or $a=\pm b$.
Now,
$$\sqrt{1-a^2}\pm(1-a^2)\mp a^2=0$$ or 
$$c\pm c^2\mp(1-c^2)=0$$
yields the positive solutions
$$c=1,a=0
\\c=\frac12,a=\pm\frac{\sqrt3}2.$$
and the minimum is
$$-\frac{3\sqrt3}2.$$
