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.

Solution 1:

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 $2s=a+b+c$

Solution 2:

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?