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. 

Inspired by Dr. Sonnhard's answer.
$$cos A + cos B + 2cos C = 2$$
=> $$cos A + cos B = 2 (1 - cos C)$$
=> $$(b²+c²-a²)/2bc + (c²+a²-b²)/2ca = 2[1 -
(a²+b²-c²)/2ab]$$
=> $$a(b²+c²-a²) + b(c²+a²-b²) = 2c[2ab - (a²
+b²-c²)]$$
=>$$ ab² + ac² -a³ + bc² + ba² - b³ = 2c[2ab -
(a²+b²-c²)]$$
=> $$ab² + ba² + ac² + bc² - a³ - b³ = 2c[2ab -
(a²+b²-c²)]$$
=> $$ab(a+b) + c²(a+b) - (a+b)(a²-ab+b²) = 2c
[2ab - (a²+b²-c²)]$$
=>$$ (a+b)(ab+c² - a²+ab-b²) = 2c[2ab - (a²
+b²-c²)]$$
=> $$(a+b)[2ab - (a²+b²-c²)] = 2c[2ab - (a²
+b²-c²)]$$
=>$$ a+b = 2c$$
=> sides of the triangle are in A.P.
