<s>[INST] 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!
 [/INST] How about a proof with a geometric flavor?
Let $a$, $b$, $c$ be the sides that oppose angles $A$, $B$, $C$, respectively. By the Law of Sines, 
$$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = d$$
where $d$ is the circumdiameter of the triangle. If we conveniently scale the triangle so that $d=1$, then we can say simply that 
$$a = \sin A \qquad b = \sin B \qquad c = \sin C$$
This is a common simplification technique, as it nicely blurs the distinction between edges and angles, giving us things like this wonderfully-symmetric area formula:
$$\mathrm{Area} \; = \frac{1}{2}a b c \qquad \left(\;=\frac{1}{2}ab\sin C =\frac{1}{2}ac\sin B = \frac{1}{2}bc\sin A\right)$$
(When you have this mindset, you can't look at the expression "$\sin A + \sin B + \sin C$" and not think, "That's perimeter!" ... and then you find yourself pursuing proof approaches like this one.) In what follows, I'll continue to use "$a$", "$b$", "$c$", because they're more compact than "$\sin A$", etc, but you should read them as "$\sin A$", etc.
By the Law of Cosines,
$$\cos{C} = \frac{a^2+b^2-c^2}{2ab}$$
By the half-angle formula for cosines,
$$\cos^2\frac{C}{2} = \frac{1+\cos{C}}{2}=\frac{a^2+2ab+b^2-c^2}{4ab}=\frac{(a+b)^2-c^2}{4ab}=\frac{(a+b+c)(a+b-c)}{4ab}$$ 
Likewise for $\cos(A/2)$ and $\cos(B/2)$, so that
$$\cos^2\frac{A}{2} \cos^2\frac{B}{2} \cos^2\frac{C}{2}=\frac{(a+b+c)^2}{4a^2b^2c^2}\cdot \frac{(a+b+c)(-a+b+c)(a-b+c)(a+b-c)}{16}$$
The conveniently-separated second factor just happens to be Heron's formula for the square of the area of the triangle; re-writing the area in wonderfully-symmetric form gives ...
$$\cos^2\frac{A}{2} \cos^2\frac{B}{2} \cos^2\frac{C}{2}=\frac{(a+b+c)^2}{4a^2b^2c^2}\cdot \left(\frac{1}{2}abc\right)^2 = \frac{1}{16}\left(a+b+c\right)^2$$
We can now clear the fraction, expand the symbols "$a$", "$b$", "$c$" as the sines they represent, and take square roots (secure in the knowledge that none of the trig values is negative), so that
$$4 \cos \frac{A}{2} \cos\frac{B}{2} \cos\frac{C}{2} = \sin A + \sin B + \sin C$$
as desired. 
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