trigonometry - Prove $sin(alpha -beta)+sin(alpha-gamma)+sin(beta-gamma)=4cosfrac{alpha-beta}{2}sinfrac{alpha-gamma}{2}cosfrac{beta-gamma}{2}$

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?

Answer

$$
\begin{align}
\color{#C00}{\sin(x)+\sin(y)}+\color{#090}{\sin(x+y)}
&=\color{#C00}{2\sin\left(\frac{x+y}2\right)\cos\left(\frac{x-y}2\right)}+\color{#090}{2\sin\left(\frac{x+y}2\right)\cos\left(\frac{x+y}2\right)}\\
&=2\sin\left(\frac{x+y}2\right)\left[\cos\left(\frac{x-y}2\right)+\cos\left(\frac{x+y}2\right)\right]\\
%&=2\sin\left(\frac{x+y}2\right)\,\color{#00F}{2\cos\left(\frac x2\right)\cos\left(\frac y2\right)}\\
%&=4\sin\left(\frac{x+y}2\right)\cos\left(\frac x2\right)\cos\left(\frac y2\right)
\end{align}
$$
Finish off by using the formula for the cosine of a sum/difference, then set $x=\alpha-\beta$ and $y=\beta-\gamma$.