Proof of trig identity using t-formulae

Question

Show that $$2\arctan x = \arccos\frac{1-x^2}{1+x^2}$$

if $ x > 0$, and $$2\arctan x = -\arccos\frac{1-x^2}{1+x^2}$$ if $x <0.$

I have been able to equate the expressions by letting $x=\tan(y/2)$, but I do not know how to show the different signs of the equation for $x>0$ and $x<0$.

Answer 1

The simplest way to go is probably to calculate $\;\cos(2\arctan x)$.

Set $\theta=\arctan x$ and use the duplication formula: \begin{align} \cos 2\theta&=\frac{1-\tan^2\theta}{1+\tan^2\theta}=\frac{1-x^2}{1+x^2},\\ \text{so }\qquad 2\arctan x&\equiv \pm\arccos\frac{1-x^2}{1+x^2}\mod 2\pi. \end{align} Now either $\;0\le \theta<\frac\pi2,\;$ so $\;0\le 2\theta<\pi$, which is (included in, strictly speaking) the range of arccos. In this case we conclude that $$ 2\arctan x=\arccos\frac{1-x^2}{1+x^2},.$$ or $\;-\frac\pi 2<\theta<0$, so $\;-\pi<2\theta<0$, and for the same reason, we conclude in this case that $$ 2\arctan x=-\arccos\frac{1-x^2}{1+x^2}.$$

Answer 2

Let $\arctan x=y\implies-\dfrac\pi2<x<\dfrac\pi2,\tan y=x$

$$\arccos\dfrac{1-x^2}{1+x^2}=\arccos(\cos2y)=\begin{cases}2y &\mbox{if } 0\le 2y\le\pi\\ -2y & \mbox{if } -\pi\le2y<0 \end{cases}$$

See Principal values