In a triangle, if $\tan(A/2)$, $\tan(B/2)$, $\tan(C/2)$ are in arithmetic progression, then so are $\cos A$, $\cos B$, $\cos C$

Question

In a triangle, if $\tan\frac{A}{2}$, $\tan\frac{B}{2}$, $\tan\frac{C}{2}$ are in arithmetic progression, then show that $\cos A$, $\cos B$, $\cos C$ are in arithmetic progression.

$$2\tan\left(\dfrac{B}{2}\right)=\tan\left(\dfrac{A}{2}\right)+\tan\left(\dfrac{C}{2}\right)$$

$$2\sqrt{\dfrac{(s-a)(s-c)}{s(s-b)}}=\sqrt{\dfrac{(s-b)(s-c)}{s(s-a)}}+\sqrt{\dfrac{(s-a)(s-b)}{s(s-c)}}$$

$$2\sqrt{\dfrac{(s-a)(s-c)(s-b)}{s(s-b)^2}}=\sqrt{\dfrac{(s-a)(s-b)(s-c)}{s(s-a)^2}}+\sqrt{\dfrac{(s-a)(s-b)(s-c)}{s(s-c)^2}}$$

$$\dfrac{2}{s-b}=\dfrac{1}{s-a}+\dfrac{1}{s-c}$$

$$\dfrac{2}{s-b}=\dfrac{s-c+s-a}{(s-a)(s-c)}$$ $$\dfrac{2}{s-b}=\dfrac{b}{(s-a)(s-c)}$$

$$2\left(\dfrac{a+b+c}{2}-a\right)\left(\dfrac{a+b+c}{2}-c\right)=b\left(\dfrac{a+b+c}{2}-b\right)$$

$$2\left(\dfrac{b+c-a}{2}\right)\left(\dfrac{a+b-c}{2}\right)=b\left(\dfrac{a+c-b}{2}\right)$$

$$2\left(\dfrac{b+c-a}{2}\right)\left(\dfrac{a+b-c}{2}\right)=b\left(\dfrac{a+c-b}{2}\right)$$

$$b^2-a^2-c^2+2ac=ba+bc-b^2$$ $$2b^2-a^2-c^2+2ac-ba-bc=0\tag{1}$$

$$\cos B=\dfrac{a^2+c^2-b^2}{2ac}$$

$$\cos A=\dfrac{b^2+c^2-a^2}{2bc}$$

$$\cos C=\dfrac{a^2+b^2-c^2}{2ab}$$

$$\cos A+\cos C=\dfrac{ab^2+ac^2-a^3+a^2c+b^2c-c^3}{2abc}$$

$$\cos A+\cos C=\dfrac{ab+bc+\dfrac{ac^2-a^3+a^2c-c^3}{b}}{2ac}$$

$$\cos A+\cos C=\dfrac{ab+bc+\dfrac{ac(a+c)-(a+c)(a^2+c^2-ac)}{b}}{2ac}$$

Using equation $(1)$, $2ac-a^2-c^2=ba+bc-2b^2$

$$\cos A+\cos C=\dfrac{ab+bc+\dfrac{(a+c)(ba+bc-2b^2)}{b}}{2ac}$$

$$\cos A+\cos C=\dfrac{ab+bc+(a+c)(a+c-2b)}{2ac}$$

$$\cos A+\cos C=\dfrac{ab+bc+a^2+c^2+2ac-2ba-2bc}{2ac}$$

$$\cos A+\cos C=\dfrac{a^2+c^2+2ac-ab-bc}{2ac}$$

Using equation $(1)$, $2ac-ba-bc=a^2+c^2-2b^2$

$$\cos A+\cos C=\dfrac{a^2+c^2+a^2+c^2-2b^2}{2ac}$$

$$\cos A+\cos C=\dfrac{2a^2+2c^2-2b^2}{2ac}$$

$$\cos A+\cos C=2\cdot\dfrac{a^2+c^2-b^2}{2ac}$$

$$\cos A+\cos C=2\cos B$$

Is there any nice way to solve this question, mine goes very long. I tried various methods but this was the only way I was able to prove the required result.

Answer 1

Rewrite the equation as $$2\frac{\sin\frac B2}{\cos\frac B2} =\frac{\sin\frac A2}{\cos\frac A2}+\frac{\sin\frac C2}{\cos\frac C2} =\frac{\sin\frac {A+C}2}{\cos\frac A2\cos\frac C2}$$

Then, with $A+C = \pi - B$,

$$2\sin\frac B2 \cos\frac A2 \cos\frac C2 = \cos\frac B2\sin\frac {\pi-B}2=\cos^2\frac B2$$

$$2\sin\frac B2(\cos\frac {A+C}2 + \cos\frac {A-C}2) = 1 + \cos B$$

$$2\sin^2\frac B2+ 2\sin\frac B2\cos\frac {A-C}2 = 1 + \cos B$$

$$2\cos\frac {A+C}2\cos\frac {A-C}2 = 2\cos B$$

$$\cos A + \cos C = 2\cos B$$

Answer 2

Writing $A_2$ for $A/2$, etc, and noting that $A_2+B_2+C_2 = \pi/2$, we have

$$\begin{align} \tan A_2-\tan B_2 &= \tan B_2 - \tan C_2 \\[6pt] \frac{\sin A_2 \cos B_2 - \cos A_2 \sin B_2}{\cos A_2 \cos B_2} &= \frac{\sin B_2 \cos C_2 - \cos B_2 \sin C_2}{\cos B_2 \cos C_2} \\[6pt] \sin(A_2-B_2)\cos C_2 &= \sin(B_2-C_2)\cos A_2 \\[6pt] \sin(A_2-B_2)\sin(A_2+B_2) &= \sin(B_2-C_2)\sin(B_2+C_2) \\[6pt] \frac12\left(\cos 2B_2 - \cos 2 A_2\right) &= \frac12\left(\cos 2C_2-\cos 2B_2\right) \\[6pt] \cos B - \cos A &= \cos C - \cos B \end{align}$$

Answer 3

For $A\ne B,A+B+C=\pi$

using http://mathworld.wolfram.com/ProsthaphaeresisFormulas.html

$$f(B,A)=\dfrac{\tan\dfrac B2-\tan\dfrac A2}{\cos B-\cos A}=-\dfrac1{2\cos\dfrac A2\cos\dfrac B2\cos\dfrac C2}$$

By symmetry, $$f(B,A)=f(C,B)$$

Can you take it from here?

Similarly we can establish $$\dfrac{\cot\dfrac B2-\cot\dfrac A2}{\sin B-\sin C}=-\dfrac1{2\sin\dfrac A2\sin\dfrac B2\sin\dfrac C2}$$ so that a similar problem can be proved

Answer 4

Given

\begin{align} \tan\tfrac12A&=u-d ,\quad \tan\tfrac12B=u ,\quad \tan\tfrac12C=u+d ,\quad u,d\in\mathbb{R} \tag{1}\label{1} . \end{align}

We can express $d$ in terms of $u$ using known identity for triangles \begin{align} \tan\tfrac A2\tan\tfrac B2+ \tan\tfrac B2\tan\tfrac C2+ \tan\tfrac C2\tan\tfrac A2&=1 ,\\ (u-d)u+u(u+d)+(u+d)(u-d)=3u^2-d^2 &= 1 ,\\ d&=\sqrt{3u^2-1} \tag{2}\label{2} , \end{align}

and \eqref{1} becomes \begin{align} \tan\tfrac12A&=u-\sqrt{3u^2-1} ,\quad \tan\tfrac12B=u ,\quad \tan\tfrac12C=u+\sqrt{3u^2-1} \tag{3}\label{3} . \end{align}

Since all the tangents must be positive, we have a condition $u\in(\tfrac{\sqrt3}3,\,\tfrac{\sqrt2}2)$, the endpoints correspond to degenerate solutions: one corresponds to the equilateral triangle, $d=0$, and the other corresponds to the degenerate triangle with $\tan\tfrac12A=0$.

We also know that \begin{align} \cos x&=\frac{1-\tan^2\tfrac x2}{1+\tan^2\tfrac x2} , \end{align} so the corresponding three cosines \begin{align} \cos A&= \frac{u-u^3+\sqrt{3u^2-1}}{u(1+u^2)} ,\\ \cos B&= \frac{1-u^2}{1+u^2} ,\\ \cos C&= \frac{u-u^3-\sqrt{3u^2-1}}{u(1+u^2)} \end{align}

are indeed in arithmetic progression, \begin{align} u'-d',\quad &u',\quad u+d' ,\\ u'&=\frac{1-u^2}{1+u^2} ,\qquad d'=-\frac{\sqrt{3u^2-1}}{u(u^2+1)} . \end{align}

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Bonus:

a suitable example of the triangle $ABC$ with given properties has integer side lengths $a=27$, $b=32$ and $c=35$ units, and

\begin{align} u&=\tfrac{4\sqrt{47}}{47} ,\quad d=\tfrac{\sqrt{47}}{47} ,\\ u'&=\tfrac{31}{63} ,\quad d'=-\tfrac{47}{252} ,\\ \alpha&= 2\arctan(\tfrac{3\sqrt{47}}{47})=\arccos(\tfrac{19}{28}) \approx 47.2679^\circ ,\\ \beta&= 2\arctan(\tfrac{4\sqrt{47}}{47})=\arccos(\tfrac{31}{63}) \approx 60.5237^\circ ,\\ \gamma&= 2\arctan(\tfrac{5\sqrt{47}}{47})=\arccos(\tfrac{11}{36}) \approx 72.2084^\circ . \end{align}