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

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.

• No \dfrac in titles please.
– Did
Feb 5 '19 at 6:52
• @Did im curious, why? Feb 5 '19 at 6:53
• @JohnGlenn Practical: This takes some unnecessary vertical space on the page of questions. (Esthetical: The result is ugly. Logical: \dfrac is made for d- -isplayed equations and titles have none.)
– Did
Feb 5 '19 at 6:56

## 4 Answers

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 $$$$\angle BOA = 2\alpha = \arctan x\tag{1}\label{eq:0}$$$$ and $$\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,$$ and so $$$$\beta = \angle AOC = \arctan\left(\sqrt{1+x^2}-x\right).\tag{2}\label{eq:0p}$$$$ By considering the right-angled triangle $$OAB$$, you obtain $$$$\angle OBA = \frac{\pi}{2}-2\alpha,\tag{3}\label{eq:1}$$$$ and, by considering the isosceles triangle $$OBC$$, you get $$$$\angle OBA = \pi - 2(2\alpha+\beta).\tag{4}\label{eq:2}$$$$ Equating \eqref{eq:1} and \eqref{eq:2} yields $$\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.

• What about if $x < 0$? Feb 5 '19 at 7:57
• @stevengregory I'd use again symmetry of $\tan(\cdot)$. Should I edit the answer? I thought it was clear enough. Well, thinking about that, I have to modify the triangle, also. Thanks!
– dfnu
Feb 5 '19 at 7:59
• @stevengregory, now it should more complete.
– dfnu
Feb 5 '19 at 8:43

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

• The question is asking why $\frac{1}{2} \arctan(x) - \arctan(x-\sqrt{x^2+1})$ is a constant, not just what the constant's value is. Feb 5 '19 at 14:14
• @MichaelSeifert Yeah, the formulation in the OP was not really clear to me because the reason for this expression to be constant is already in there. In any case, taking derivatives easily shows that it must be constant. Feb 5 '19 at 14:38

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

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

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

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

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

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

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