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.

$\cos A+\cos B+\cos C$
$=2\cos\frac{A+B}{2}\cos\frac{A-B}{2}+1-2\sin^2\frac{C}{2}$ as $\cos2x=1-2\sin^2x$
Now $\cos\frac{A+B}{2}=\cos\frac{180^\circ - C}{2}=\cos(90^\circ-\frac{C}{2})=\sin\frac{C}{2}$
So, $\cos A+\cos B+\cos C$ becomes  $2\sin\frac{C}{2}\cos\frac{A-B}{2}+1-2\sin^2\frac{C}{2}$
$=1+2\sin\frac{C}{2}(\cos\frac{A-B}{2}-\sin\frac{C}{2})$
$=1+2\sin\frac{C}{2}(\cos\frac{A-B}{2}-\cos\frac{A+B}{2})$ replacing $\sin\frac{C}{2}$ with $\cos\frac{A+B}{2}$
$=1+2\sin\frac{C}{2}(2\sin\frac{A}{2}\sin\frac{B}{2})$  applying $\cos(x-y)-\cos(x+y)= 2 \sin x \sin y$
$=1+4\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}$
