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?

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,
$2\sin\frac C2\cos\frac{A-B}2 - 2\sin^2\frac C2 = \sin\frac{A-B+C}2 - \sin\frac{A-B-C}2 + cosC - 1$   = $cosB + cosA+ cosC- 1$
