<s>[INST] Prove trigonometry identity for $\cos A+\cos B+\cos C$ I humbly ask for help in the following problem.
If 
\begin{equation}
A+B+C=180
\end{equation}
Then prove
\begin{equation}
\cos A+\cos B+\cos C=1+4\sin(A/2)\sin(B/2)\sin(C/2)
\end{equation}
How would I begin the problem I mean I think $\cos C $ can be $\cos(180-A+B)$. But I am unsure what to do next.
 [/INST] What I might do is start with the right side.  Since I don't remember half-angle formulas, let $a = A/2$, $b=B/2$.  Note that (from the addition formulas for $\cos$) $\sin(x) \sin(y) = (\cos(x-y) - \cos(x+y))/2$, $\cos(x) \cos(y) = (\cos(x+y) + \cos(x-y))/2$, $\sin(90 - x) = \cos(x)$.
$$ \eqalign{1 &{}+ 4 \sin(a) \sin(b) \sin(90 - a - b)\cr
       = & 1 + 2 (\cos(a-b) - \cos(a+b)) \cos(a+b) \cr
       = & 1 + \cos(2a) + \cos(2b) - \cos(2a+2b) - \cos(0)\cr
       = & \cos(2a) + \cos(2b) + \cos(180 - 2a - 2b)\cr} $$
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