<s>[INST] Prove $\sin(\alpha -\beta)+\sin(\alpha-\gamma)+\sin(\beta-\gamma)=4\cos\frac{\alpha-\beta}{2}\sin\frac{\alpha-\gamma}{2}\cos\frac{\beta-\gamma}{2}$ Here is a problem from Gelfand's Trigonometry:

Let $\alpha, \beta, \gamma$ be any angle, show that $$\sin(\alpha -\beta)+\sin(\alpha-\gamma)+\sin(\beta-\gamma)=4\cos\left(\frac{\alpha-\beta}{2}\right)\sin\left(\frac{\alpha-\gamma}{2}\right)\cos\left(\frac{\beta-\gamma}{2}\right).$$

I have tried to worked through this problem but cannot complete it. If I let $A= \alpha -\beta$, $B=\beta-\gamma$ and $C= \beta-\gamma$, and $A+B+C=\pi$ (now $A$, $B$ and $C$ are angles of a triangle), then I could prove the equality. But without this condition, I am stuck.
Could you show me how to complete this exercise?
 [/INST] Use
\begin{eqnarray*}
\sin A + \sin B = 2 \sin \left( \frac{A+B}{2} \right) \cos \left( \frac{A-B}{2} \right)
\end{eqnarray*}
to give 
\begin{eqnarray*}
\sin (\alpha-\beta) + \sin (\alpha-\gamma) = 2 \sin \left( \frac{2 \alpha-\beta-\gamma}{2} \right) \cos \left( \frac{\beta-\gamma}{2} \right).
\end{eqnarray*}
Now use the double angle formula 
\begin{eqnarray*}
\sin(\beta-\gamma)=2 \sin \left(\frac{\beta-\gamma}{2} \right) \cos \left(\frac{\beta-\gamma}{2}\right).
\end{eqnarray*}
Use the first formula again & the result follows.
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