If, in a triangle, $\cos(A) + \cos(B) + 2\cos(C) = 2$ prove that the sides of the triangle are in AP

Question

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.

Answer 1

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?

Answer 2

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 $2s=a+b+c$

Answer 3

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.

Answer 4

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.