Why $\frac{1}{2} \arctan(x) = \arctan(x-\sqrt{x^2+1}) + \frac{\pi}{4}$?

Question

Since

$$\dfrac{d}{dx} \left( \dfrac{1}{2} \arctan(x) \right) = \dfrac{d}{dx} \left( \arctan(x-\sqrt{x^2+1}) \right) $$

then the format of their graphs are the same, but separated by a constant, which is $\dfrac{\pi}{4}$. That is,

$$\dfrac{1}{2} \arctan(x) = \arctan(x-\sqrt{x^2+1}) + \dfrac{\pi}{4} $$

My question is: why is this constant and what procedure is done to discover it.

Answer 1

It's even nicer if you use simple trigonometry. Consider the following Figure, for positive $x$ (for negative $x$ see edit, thanks to Steven for pointing this out in comment).

$\hskip1.in$

The triangle $OAB$ is right-angled, and such that $\overline{OA} = 1$, and $\overline{AB} = x$. We have \begin{equation}\angle BOA = 2\alpha = \arctan x\tag{1}\label{eq:0}\end{equation} and $$\overline{OB} = \sqrt{x^2+1}.$$ Extend $BA$ to a segment $BC$ such that $$ BC \cong OB.$$ Then $$\overline{AC} = \sqrt{1+x^2}-x,$$ and so \begin{equation}\beta = \angle AOC = \arctan\left(\sqrt{1+x^2}-x\right).\tag{2}\label{eq:0p}\end{equation} By considering the right-angled triangle $OAB$, you obtain \begin{equation}\angle OBA = \frac{\pi}{2}-2\alpha,\tag{3}\label{eq:1}\end{equation} and, by considering the isosceles triangle $OBC$, you get \begin{equation}\angle OBA = \pi - 2(2\alpha+\beta).\tag{4}\label{eq:2}\end{equation} Equating \eqref{eq:1} and \eqref{eq:2} yields $$\alpha = -\beta + \frac{\pi}{4}.$$ Plugging in \eqref{eq:0} and \eqref{eq:0p} and using the odd symmetry of the $\arctan(\cdot )$ function leads to $$ \frac{1}{2}\cdot \arctan x = \arctan\left(x-\sqrt{1+x^2}\right) + \frac{\pi}{4},$$ the desired result. $\blacksquare$

EDIT The same result applies if $x<0$. Use then the Figure below.

$\hskip1.in$

Now $\overline{AB} = -x$, $\overline{OB} = \overline{BC} = \sqrt{1+x^2}$, and $\overline{CA} = \sqrt{1+x^2}-x$. Then, again using right-angled triangle $OAC$ and isosceles triangle $OBC$, we get $$\angle OCA = \frac{\pi}{2}-\beta = \beta - 2\alpha,$$ where $2\alpha = \angle BOA$, and $\beta = \angle COA$. The equality of OP follows.

Answer 2

To get the constant you plug in an arbitrary number, usually $0$ for simplicity.

Answer 3

I'm going to use $y$ instead of $x$.

Let $P(\theta) = (1, y)$. Then $$\theta = \arctan(y).$$

Then $$\sin \theta = \dfrac{y}{\sqrt{1+y^2}} \qquad \text{and} \qquad \cos \theta = \dfrac{1}{\sqrt{1+y^2}}$$

Then \begin{align} \tan \bigg( \frac 12 \theta - \frac{\pi}{4} \bigg) &= \dfrac{\tan \frac 12 \theta - \tan \frac{\pi}{4}} {1 + \tan(\frac 12 \theta) \tan(\frac{\pi}{4})}\\ &=\dfrac{\left(\dfrac{\sin \theta}{1+\cos \theta}-1 \right)} {\left( 1+\dfrac{\sin \theta}{1+\cos \theta} \right)} \\ &= \frac{\sin \theta - 1 - \cos \theta} {1 + \cos \theta + \sin \theta} \\ &= \frac{\left( \dfrac{y}{\sqrt{1+y^2}} - 1 - \dfrac{1}{\sqrt{1+y^2}} \right)} {\left( 1 + \dfrac{1}{\sqrt{1+y^2}} + \dfrac{y}{\sqrt{1+y^2}} \right)} \\ &= \frac{y - \sqrt{1+y^2} - 1}{\sqrt{1+y^2} + 1 + y} \\ &= \frac{(y - 1) - \sqrt{1+y^2}}{(y + 1) + \sqrt{1+y^2}} \cdot \frac{(y + 1) - \sqrt{1+y^2}}{(y + 1) - \sqrt{1+y^2}} \\ &= \dfrac{2y^2 - 2y\sqrt{1+y^2}}{2y} \\ &= y - \sqrt{1+y^2} \end{align}

Since $\tan \bigg( \dfrac 12 \theta - \dfrac{\pi}{4} \bigg) = y - \sqrt{1+y^2}$, then $$\dfrac 12 \theta - \dfrac{\pi}{4} = \arctan\left(y - \sqrt{1+y^2}\right).$$

In other words

$$\dfrac 12 \arctan(y) = \arctan\left(y - \sqrt{1+y^2}\right) + \dfrac{\pi}{4}.$$

Answer 4

Let $\arctan x-\dfrac\pi2=2y$

Using Principal values, $ -\dfrac\pi2<\dfrac\pi2+2y<\dfrac\pi2\iff-\dfrac\pi2<y<0$

$\arctan x=\dfrac\pi2+2y\implies x=-\cot2y$

$t=x-\sqrt{x^2+1}=-\cot2y-\sqrt{\csc^22y}=-\cot2y+\csc2y$ as $\csc2y=\dfrac1{\sin2y}<0$ for $-\pi<2y<0$

$t=\dfrac{1-\cos2y}{\sin2y}=\tan y\implies\arctan(t)=y$