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. 
 A: Hint:
Prosthaphaeresis Formulas: $\cos A+\cos B=2\cos\dfrac{A+B}2\cos\dfrac{A-B}2$
Double angle formula: $1-\cos C=2\sin^2\dfrac C2$
$\dfrac{A+B}2=\dfrac\pi2-\dfrac C2\implies\cos\dfrac{A+B}2=\sin\dfrac C2$
As $0<c<\pi, \sin\dfrac C2\ne0$
So, we have $\cos\dfrac{A-B}2=2\cos\dfrac{A+B}2$
Expand $\cos\dfrac{A\pm B}2$ and divide both sides by $\cos\dfrac A2\cos\dfrac B2$
Finally, we have $\tan\dfrac A2=\sqrt{\dfrac{(s-b)(s-c)}{s(s-a)}}$  where $2s=a+b+c$
A: 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?
A: From : 
$$
\cos(A)+\cos(B)+2\cos(C)=2
\\
\implies \cos(A)+\cos(B)=2-2\cos(C)
\\
\implies \cos(A)+\cos(B)=2[1-\cos(C)]
\\
\\\implies \cos(A)+\cos(B)=4\sin^2\left(\frac{C}{2}\right)
$$
Using Prosthaphaeresis Formulas : 
$$
\cos A+\cos B=2\cos\dfrac{A+B}2\cos\dfrac{A-B}2
$$
And substituting this formula in the first equation, we have :
$$
2\cos\dfrac{A+B}2\cos\dfrac{A-B}2=4\sin^2\left(\frac{C}{2}\right)
$$
Since 
$A+B+C=\pi$
$$
\frac{A+B}{2}=\frac{\pi}{2}-\frac{C}{2}
$$
Taking cosines on both sides :
$$
\cos\left(\frac{A+B}{2}\right)=\cos\left(\frac{\pi}{2}-\frac{C}{2}\right)=\sin\left(\frac{C}{2}\right)
$$
Using this :
$$
2\cos\dfrac{A+B}2\cos\dfrac{A-B}2=4\cos^2\left(\frac{A+B}{2}\right)
\\
\implies \cos\dfrac{A-B}2=2 \cos\dfrac{A+B}2
$$
Multiplying both sides by $2\sin\dfrac{A+B}2$ :
$$
2\sin\dfrac{A+B}2\cos\dfrac{A-B}2=4\sin\dfrac{A+B}2\cos\dfrac{A+B}2
$$
Using Another Prosthaphaeresis Formula :
$$
\sin A+\sin B=2\sin\dfrac{A+B}2\cos\dfrac{A-B}2
$$
Applying the reverse in the obtained equation, we get :
$$
\sin A+\sin B=2\sin\dfrac{A+B}2\cos\dfrac{A-B}2=4\sin\dfrac{A+B}2\cos\dfrac{A+B}2
$$
Using the sine double angle formula $2\sin(\alpha)\cos(\alpha)=\sin(2\alpha)$ :
$$
\sin A+\sin B=2\sin(A+B)=2\sin C
\\
\implies \sin A + \sin B = 2\sin C
$$
Using the sine rule :
$$
a + b = 2c
$$
Hence, the sides of the triangle are in A.P. 
A: 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.
