$\underline{\textbf{Direct Proof}}$ Note that $A+B+C=\pi\implies \sin C =\sin(A+B)$, we have \begin{align*} &\sin A + \sin B + \sin C \\ &=\sin A + \sin B + \sin (A+B)\\ &=2\sin\frac{A+B}{2}\cos\frac{A-B}{2} + 2\sin\frac{A+B}{2}\cos\frac{A+B}{2}\\ &=2\sin\frac{A+B}{2}\Big(\cos\frac{A-B}{2}+\cos\frac{A+B}{2}\Big)\\ &=2\sin\Big(\frac{\pi}{2}-\frac{C}{2}\Big)\Big(2\cos\frac{A}{2}\cos\frac{B}{2}\Big)\\ &=4\cos\frac{A}{2}\cos\frac{B}{2}\cos\frac{C}{2} \end{align*} $\underline{\textbf{Indirect Proof}}$ Note that the to be proved identity is a special case of %%HREF%%3851%%. $$\sin\alpha + \sin\beta + \sin\gamma - \sin(\alpha+\beta+\gamma)\nonumber\\ =4\sin\frac{\alpha +\beta}{2}\sin\frac{\beta+\gamma}{2}\sin\frac{\gamma+\alpha}{2} $$ Setting $\alpha=A$, $\beta=B$, $\gamma=C$, and also noting $A+B+C=\pi$: \begin{align*} &\sin A + \sin B + \sin C - \sin(A+B+C) \\ &=4\sin\frac{A +B}{2}\sin\frac{B+C}{2}\sin\frac{C+A}{2}\\ \Leftrightarrow & \sin A + \sin B + \sin C = 4\sin\Big(\frac{\pi}{2}-\frac{C}{2}\Big)\sin\Big(\frac{\pi}{2}-\frac{A}{2}\Big)\sin\Big(\frac{\pi}{2}-\frac{B}{2}\Big)\\ \Leftrightarrow & \sin A + \sin B + \sin C = 4\cos \frac{C}{2}\cos \frac{A}{2} \cos \frac{B}{2}\\ \Leftrightarrow & \sin A + \sin B + \sin C = 4\cos \frac{A}{2}\cos \frac{B}{2} \cos \frac{C}{2} \end{align*}