If, in a triangle, $\cos(A) + \cos(B) + 2\cos(C) = 2$ prove that the sides of the triangle are in AP By using the formula : 
$$
\cos(A)+\cos(B)+\cos(C) = 1 + 4 \sin\left(\frac{A}{2}\right)\sin\left(\frac{B}{2}\right) \sin\left(\frac{C}{2}\right)
$$
I've managed to simplify it to : 
$$
2\sin\left(\frac{A}{2}\right)\sin\left(\frac{B}{2}\right)=\sin\left(\frac{C}{2}\right)$$
But I have no idea how to proceed. 

using $$\cos(\alpha)=\frac{b^2+c^2-a^2}{2bc}$$ and so on and plugging these equations in your equation and factorizing we get
$$-1/2\,{\frac { \left( c+a-b \right)  \left( -c+a-b \right)  \left( -2
\,c+a+b \right) }{bca}}
=0$$
can you proceed?
