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!

Since $C = \pi - A - B$, what needs to be shown is 
$$\sin(A) + \sin(B) + \sin(\pi - A - B) = 4\cos({A \over 2})\cos({B \over 2})\cos({\pi \over 2} - {A + B \over 2})$$
Equivalently, you need that
$$\sin(A) + \sin(B) + \sin( A + B) = 4\cos({A \over 2})\cos({B \over 2})\sin( {A + B \over 2})$$
Using the trig identity $\cos(x)\sin(y) = {1 \over 2}(\sin(x + y) + \sin(y - x))$, this is the same as
$$\sin(A) + \sin(B) + \sin( A + B) = 2\cos({A \over 2})\sin({A \over 2} + B) + 2\cos({A \over 2})\sin({A \over 2}) $$
Using it again on the first term on the right, this becomes
$$\sin(A) + \sin(B) + \sin( A + B) = \sin(A + B) + \sin(B) + 2\cos({A \over 2})\sin({A \over 2}) $$
This collapses into the sine double angle identity
$$\sin(A) = 2\cos({A \over 2})\sin({A \over 2}) $$
Reversing the above steps gives the desired result.
