How do I write a sum of cosines as a product of sines? I am trying to prove that 
$$\cos A+\cos B+\cos C=4\sin\frac A2\sin\frac B2\sin\frac C2$$ for ABC is a triangle.
I tried up to the stage of
$$-2\sin^2 C+2\cos\frac{180-C}2 \cos\frac{A+B}2$$ 
but how do I proceed from here?
 A: $$\cos A+\cos B+\cos C$$
$$=2\cos\frac{A+B}2\cos\frac{A-B}2+1-2\sin^2\frac C2$$
$$=2\sin\frac C2\cos\frac{A-B}2+1-2\sin^2\frac C2\text{ as } \cos\frac{A+B}2=\cos\left(\frac{\pi-C}2\right)=\sin \frac C2$$
$$=1+2\sin\frac C2\left(\cos\frac{A-B}2-\cos\frac{A+B}2\right)\text{ as }\sin \frac C2=\cos\frac{A+B}2$$
$$=1+2\sin\frac C22\sin\frac A2\sin\frac B2$$
$$=1+4\sin\frac A2\sin\frac B2\sin\frac C2$$
A: Recall the identities
$$
\begin{align}
2\sin(x)\sin(y)&=\cos(x-y)-\cos(x+y)\tag{1}\\
2\sin(x)\cos(y)&=\sin(x-y)+\sin(x+y)\tag{2}
\end{align}
$$

First use $(1)$ on $2\sin\left(\frac{A}{2}\right)\sin\left(\frac{B}{2}\right)$, then $(2)$ on the results. Finally, use $A+B+C=\pi$.
$$
\begin{align}
&4\sin\left(\frac{A}{2}\right)\sin\left(\frac{B}{2}\right)\sin\left(\frac{C}{2}\right)\\
&=\color{#C00000}{2\sin\left(\frac{C}{2}\right)\cos\left(\frac{A-B}{2}\right)}
\color{#00A000}{-2\sin\left(\frac{C}{2}\right)\cos\left(\frac{A+B}{2}\right)}\\
&=\color{#C00000}{\sin\left(\frac{C+B-A}{2}\right)+\sin\left(\frac{C+A-B}{2}\right)}
\color{#00A000}{+\sin\left(\frac{A+B-C}{2}\right)-\sin\left(\frac{A+B+C}{2}\right)}\\[6pt]
&=\cos(A)+\cos(B)+\cos(C)-1
\end{align}
$$
A: PROOF: There is something wrong in the question
TO PROVE :$\cos A+\cos B+\cos C -1=4\sin\frac A2\sin\frac B2\sin\frac C2$
PROOF: You can proceed from the right hand side too
$4\sin\frac A2\sin\frac B2\sin\frac C2 = 2\sin\frac A2\sin\frac B2 2\sin\frac C2$
NOW, $2\sin\frac A2\sin\frac B2 = cos\frac{A-B}2 - \cos\frac{A+B}2 = cos\frac{A-B}2 - \sin\frac C2$
The expression becomes,
$2\sin\frac C2\cos\frac{A-B}2 - 2\sin^2\frac C2 = \sin\frac{A-B+C}2 - \sin\frac{A-B-C}2 + cosC - 1$   = $cosB + cosA+ cosC- 1$
