Proving $\sin A + \sin B + \sin C = 4 \cos \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2}$ 
Possible Duplicate:
Prove that $\sin(2A)+\sin(2B)+\sin(2C)=4\sin(A)\sin(B)\sin(C)$ when $A,B,C$ are angles of a triangle
Prove trigonometry identity? 

If $A$, $B$, and $C$ are to be taken as the angles of a triangle, then I beg someone to help me the proof of
$$\sin A + \sin B + \sin C = 4 \cos \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2}.$$
Thanks!

As Salech pointed out in the comment above, we have $A=180^\circ-(B+C)$, so applying angle sum and difference formulas for cosine and sine, we have
$\begin{eqnarray*}
\cos\frac A2 & = & \cos\left(90^\circ-\frac{B+C}2\right)\\
& = & \cos 90^\circ\cos\left(\frac B2+\frac C2\right)+\sin 90^\circ\sin\left(\frac B2+\frac C2\right)\\
& = & \sin\left(\frac B2+\frac C2\right)\\
& = & \sin\frac B2\cos\frac C2+\cos\frac B2\sin\frac C2,
\end{eqnarray*}$
since $\cos 90^\circ=0$ and $\sin 90^\circ=1$.
Hence, $$4\cos\frac A2\cos\frac B2\cos\frac C2 = 4\sin\frac B2\cos\frac B2\cos^2\frac C2+4\cos^2\frac B2\sin\frac C2\cos\frac C2.$$
Now, we've got the double angle identities $\sin(2x)=2\sin x\cos x$ and $\cos(2x)=2\cos^2x-1$, the latter of which gives us $2\cos^2x=1+\cos(2x)$. Hence, by substitution and application of angle addition formula from sine,
$\begin{eqnarray*}
4\cos\frac A2\cos\frac B2\cos\frac C2 & = & \left(2\sin\frac B2\cos\frac B2\right)\left(2\cos^2\frac C2\right)+\left(2\cos^2\frac B2\right)\left(2\sin\frac C2\cos\frac C2\right)\\
& = & \sin B(1+\cos C)+(1+\cos B)\sin C\\
& = & \sin B\cos C+\cos B\sin C+\sin B+\sin C\\
& = & \sin(B+C)+\sin B+\sin C.
\end{eqnarray*}$
Finally, observe that $\sin(180^\circ-x)=\sin x$, so since $A=180^\circ-(B+C)$, then $\sin A=\sin(B+C)$, and therefore, $$4\cos\frac A2\cos\frac B2\cos\frac C2=\sin A+\sin B+\sin C,$$ as desired.
