<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] Your observation that $C=180^\circ-(A+B)$ is a good one. Recall also the following trigonometric identities: $$\sin(x\pm y)=\sin x\cos y\pm\cos x\sin y$$ $$\cos(x\pm y)=\cos x\cos y\mp\sin x\sin y$$ By LHS, I'll denote the expression on the left-hand side of the desired identity; by RHS, the expression on the right-hand side.
Now, the RHS is messier, so we'll start there. Let's first apply your observation to $\sin(C/2)$, along with the angle difference and sum formulas for sine, and the angle sum formula for cosine, to see that 
$\begin{eqnarray*}
\sin(C/2) & = & \sin\bigl(90^\circ-(A+B)/2\bigr)\\
& = & \sin 90^\circ\cos\bigl((A+B)/2\bigr)-\cos 90^\circ\sin\bigl((A+B)/2\bigr)\\
& = & \cos(A/2+B/2)\\
& = & \cos(A/2)\cos(B/2)-\sin(A/2)\sin(B/2),
\end{eqnarray*}$
since $\sin 90^\circ=1$ and $\cos 90^\circ=0$. Thus, we see that
$$\mathrm{RHS} = 1+4\sin(A/2)\cos(A/2)\sin(B/2)\cos(B/2)-4\sin^2(A/2)\sin^2(B/2)\tag{1}$$
Recall also the double angle formulas for sine (a special case of angle sum with $x=y$): $$\sin(2x)=2\sin x\cos x.$$
Also, the Pythogorean identity and the angle sum formula for cosine (with $x=y$) gives us the following double angle formula for cosine: $$\cos(2x)=\cos^2x-\sin^2x=1-2\sin^2x,$$ from which we derive the identity $$2\sin^2x=1-\cos(2x).$$ Applying this identity, along with the double angle and angle sum formulas for sine, to $(1)$ gives us
$\begin{eqnarray*}
\mathrm{RHS} & = & 1+\bigl(2\sin(A/2)\cos(A/2)\bigr)\bigl(2\sin(B/2)\cos(B/2)\bigr)-\bigl(2\sin^2(A/2)\bigr)(2\sin^2(B/2)\bigr)\\
& = & 1+\sin A\sin B-(1-\cos A)(1-\cos B)\\
& = & \cos A + \cos B - \cos A\cos B+\sin A\sin B\\
& = & \cos A + \cos B - \cos(A+B).
\end{eqnarray*}$
At this point, we can apply your observation again, along with the angle difference formula for cosine, to see that
$\begin{eqnarray*}
\mathrm{LHS} & = & \cos A + \cos B + \cos 180^\circ\cos(A+B)-\sin 180^\circ\sin(A+B)\\
& = & \cos A + \cos B - \cos(A+B),
\end{eqnarray*}$
since $\cos 180^\circ=-1$ and $\sin 180^\circ=0$. Thus, LHS = RHS, as desired.
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