Because the left side is a sum of several terms and the right side is a product, let's natural to apply the sum-product transformation. \begin{align*} & \sin\alpha + \sin\beta + \sin\gamma - \sin(\alpha+\beta+\gamma)\\ =& 2\sin\frac{\alpha+\beta}{2}\cos\frac{\alpha-\beta}{2} + 2\sin\frac{\gamma-(\alpha+\beta+\gamma)}{2}\cos\frac{\gamma+(\alpha+\beta+\gamma)}{2}\\ =&2\sin\frac{\alpha+\beta}{2}\Big(\cos\frac{\alpha-\beta}{2}-\cos\frac{2\gamma+\alpha+\beta}{2}\Big)\\ =&2\sin\frac{\alpha+\beta}{2}\cdot 2\sin\frac{(\alpha-\beta)+(2\gamma +\alpha+\beta)}{4}\sin\frac{(\alpha-\beta)-(2\gamma +\alpha+\beta)}{4}\\ =&4\sin\frac{\alpha+\beta}{2}\sin\frac{\beta+\gamma}{2}\sin\frac{\gamma+\alpha}{2} \end{align*} Quiz: what if $\alpha +\beta +\gamma=\pi$?