The trigonometic identity $$ \arctan x + \arctan \frac 1 x = \frac\pi 2\quad\text{for }x>0 $$ can be seen to be true by observing that if the lengths of the legs of a right triangle are $x$ and $1$, then the two acute angles of the triangle are the two arctangents above.

Does the identity $$ \arctan x + \arctan \frac{1-x}{1+x} = \frac \pi 4 $$ have a similar geometric justification?

PS: The second identity, like the first, can be established by either of two familiar methods:

  1. Use the usual formula for a sum of two arctangents; or

  2. differentiate the sum with respect to $x$.

PPS: Secondary question: Both of the functions $x\mapsto\dfrac 1 x $ and $x\mapsto\dfrac{1-x}{1+x}$ are involutions. Does that have anything to do with this? Presumably it should mean we should hope for some geometric symmetry, so that $x$ and $\displaystyle\vphantom{\frac\int\int}\frac{1-x}{1+x}$ play symmetrical roles.

PPPS: In a triangle with a $135^\circ$ angle, if the tangent of one of the acute angles is $x$, then the tangent of the other acute angle is $\displaystyle\vphantom{\frac\int\int}\dfrac{1-x}{1+x}$. That follows from the second identity above. If there's a simple way to prove that result by geometry without the identity above, then that should do it.

  • $\begingroup$ @Chappers : Why to you insist on \arctan{x} instead of \arctan x? Enclosing things in braces is sometimes necessary when more than one character is enclosed. For example x^53 and x^{53} yield different results. People do in fact learn MathJax coding from what they see here, and if they see \arctan{x} they sometimes conclude that those braces are actually necessary. ${}\qquad{}$ $\endgroup$ – Michael Hardy Jun 17 '15 at 17:00
  • $\begingroup$ I think it's good practice to always have the curly brackets there: easier to keep track of what's an argument of what. And if people see that, they understand what the curly brackets are for, rather than if one just has a space, which can be misinterpreted as how the syntax works in general. (I also find \frac a b surprisingly hard to read, probably because I've never used it myself.) (What you didn't see was the edit I rejected, which had $\tan^{-1}$s everywhere... I fixed that, and then felt like neatening the whole thing.) $\endgroup$ – Chappers Jun 17 '15 at 17:19
  • $\begingroup$ @Chappers : How could that enable them to understand what curly braces are for? What they're for is keeping things like "53" together in x^{53} so that you see $x^{53}$ rather than $x^53$, which is coded as x^53. ${}\qquad{}$ $\endgroup$ – Michael Hardy Jun 17 '15 at 18:17

Here's an attempt to address @Michael's desire for a solution that treats the two angles symmetrically.

It's based on this preliminary result:

enter image description here

Lemma. If $Q$ is the orthocenter of acute $\triangle ABC$, then $$\tan \angle ABQ = \frac{|\overline{AQ}|}{|\overline{BC}|}$$

(Proof is left as an exercise to the reader. Hint: Note the congruent angles at $A$ and $C$.)

With that, we can construct the following:

enter image description here

$$\tan \alpha = \frac{x\sqrt{2}}{\sqrt{2}} = x \qquad \tan \beta = \frac{1-x}{1+x} \quad\implies\quad \operatorname{atan}x + \operatorname{atan}\frac{1-x}{1+x} = \alpha+\beta = \frac{\pi}{4}$$


enter image description here

In answer to the first question, I have been trying to upload a picture but without success, so I'll have to describe it instead.

  1. Draw an isosceles right-angled triangle OAC with A on the x axis and C on the y axis, and O as the origin.

  2. Choose a point P on the line AC, with Q as the foot of the perpendicular from P onto OA

  3. Let B be the point on OA so that Q is the midpoint of BA

  4. Let OQ be 1 unit long, and PQ be of length $x$, in which case BQ and AQ also have this length

  5. We now have $\tan POQ=x$ and $\tan OCB=\frac{1-x}{1+x}$

  6. It simply remains to show that angle BCP = angle POQ, which is easy enough since triangle CPB is right angled and CP has length $\sqrt2$

I hope this is clear enough

  • $\begingroup$ For this to be fully valid, $P$ has to be at least as close to $A$ and $P$ is to $C$. If the opposite inequality holds, then $\tan(\angle OCB)=\dfrac{x-1}{x+1}$ $\ne\dfrac{1-x}{1+x}$. ${}\qquad{}$ $\endgroup$ – Michael Hardy Jun 15 '15 at 12:28
  • $\begingroup$ Corollary: $C,P,B,O$ are concyclic. The center of the circle on which all four of these points lie is the midpoint between $B$ and $C$. ${}\qquad{}$ $\endgroup$ – Michael Hardy Jun 15 '15 at 12:30
  • $\begingroup$ Yes the diagram shows when $0 \le \arctan x \le \frac {\pi}{4}$.. When the diagram is drawn as you are suggesting, so that $\frac {\pi}{4} \le\arctan x<\frac {\pi}{2}$, the identity becomes $$\arctan x-\arctan \frac {x-1}{x+1}=\frac {\pi}{4}$$ which is equivalent to the original $\endgroup$ – David Quinn Jun 15 '15 at 12:48
  • 2
    $\begingroup$ @DavidQuinn I've uploaded your beautifully drawn diagram. Feel free to edit your post accordingly. $\endgroup$ – Simon S Jun 15 '15 at 14:37
  • 2
    $\begingroup$ @SimonS: thank you for doing that! $\endgroup$ – David Quinn Jun 15 '15 at 14:52

Edit. Here's a simpler trigonograph that's also purer, in the sense that it avoids auxiliary proportions in favor of relating elements in the identity more directly. It therefore requires less (or perhaps no) explanation.

enter image description here

$$45^\circ = \alpha + \beta = \operatorname{atan}\frac{x}{1} + \operatorname{atan}\frac{1-x}{1+x} \qquad (0 \leq x \leq 1)$$

Interestingly, this diagram has the same basic structure as my Angle-Sum and -Difference trigonographs, as well as the one for $p\sin\theta + q \cos\theta$.

My previous answer:

enter image description here

$$\tan \theta = \frac{x}{1} \qquad \tan \phi = \frac{|\overline{XY}|}{|\overline{XZ}|} = \frac{|\overline{XP}|}{|\overline{XQ}|} = \frac{1-x}{1+x}$$

$$\implies\qquad \operatorname{atan} x + \operatorname{atan}\frac{1-x}{1+x} = \theta + \phi = \frac{\pi}{4}$$


Here's another geometrical picture of what the identity means.

The tangent of an angle can also be interpreted as the slope of a line. That is, take a straight line through the origin of an XY cartesian coordinate system, then its equation in cartesian coordinates is

$$y=\tan{\theta} \; x$$

where $\theta$ is the angle between that line and the X-axis. There is however one line that does not fit in this scheme, but we're going to change that. Obviously, the Y-axis does not fit into that scheme because it's cartesian equation is $x=0$. We'll however extend our possible slopes with $\infty$ so that the Y-axis has slope $\infty$ and therefore $\tan \pi/2$ is defined to be $\infty$.

Now, going back to our cartesian plane, I want you to look at the following linear transformation

$$\mathbb{R}^2\to\mathbb{R}^2:\left(\begin{array}{c}x\\y\end{array}\right)\mapsto\frac{1}{\sqrt{2}}\left(\begin{array}{cc} 1 & 1 \\ 1 & -1\end{array}\right)\left(\begin{array}{c}x\\y\end{array}\right)$$

Since this is a linear transformation, it leaves the origin invariant but more than that it also maps straight lines through the origin to straight lines through the origin. Therefore, we can look to define it in terms of the slopes of these lines. Since the slope of a line is given by $z=y/x$ we have that

$$\mathbb{R}\cup \{\infty\}\to \mathbb{R}\cup \{\infty\} : z\mapsto\frac{1-z}{1+z}$$

This is exactly your transformation from before. Going back to my original linear transformation, we can check that it has eigenvalues $1$ and $-1$ with respective eigenvectors $(1,\sqrt{2}-1)$ and $(1,-\sqrt{2}-1)$. I've written those in such a way that we can immediately read off the slopes as $\sqrt{2}-1$ and $-\sqrt{2}-1$.

This indicates that our transformation is a reflection about the line with slope $\sqrt{2}-1$. This slope corresponds to an angle of $\pi/8$. Now, if I reflect an arbitrary line, the image of the line and the line both make an angle with the X-axis, the sum of which is $\pi/4$. You have to be careful in defining the angles for this to work, as the "orientation" of the line you reflect matters in defining the sign of the angle. Or you could work $\mod\pi$, which is a symmetry of the $\tan$ function anyway.

The transformation

$$\mathbb{C}\cup \{\infty\}\to \mathbb{C}\cup \{\infty\} : z\mapsto\frac{1-z}{1+z}$$

is also known as a Möbius transformation when it is working in the complex plane (or rather the Riemann sphere). But before that, they were also refered to as linear transformations, which at first may seem odd, because the prescription of the transformation is anything but linear. However, because of the connection I explicitly built up here with the group of linear transformations, it is more clear why the name is fitting. More precisely it is known that the group of Möbius transformations is related to the projective special linear group $PSL(2,\mathbb{C})$.

If I have more time, I'll try to look further into the geometric meaning of your identity in the context of the Riemann sphere.

  • 1
    $\begingroup$ Think of $\mathbb C$ as a horizontal plane and the circle $|z|=1$ as the equator, and put the north pole one unit above the plane, directly above $0$. Project points in $\mathbb C$ onto the sphere stereographically onto that sphere with the center of projection at the north pole. Then $\vphantom{\displaystyle\frac\int\int}z\mapsto\dfrac{1-z}{1+z}$ rotates that sphere $180^\circ$, and one of the two poles of that rotation -- its fixed points -- is at latitude $45^\circ$ north and longitude the same as the positive real axis. ${}\qquad{}$ $\endgroup$ – Michael Hardy Jun 15 '15 at 15:55
  • $\begingroup$ Well, you beat me to it. :D $\endgroup$ – Raskolnikov Jun 16 '15 at 7:29
  • $\begingroup$ Now I'm momentarily wondering whether I should have said $45^\circ$ _south_${}\,\ldots\qquad{}$ $\endgroup$ – Michael Hardy Jun 16 '15 at 17:01

Yes. Taking tan on both sides, $$ \frac{x + \frac{1-x}{1+x}}{1- x \frac{1-x}{1+x} } = 1 $$ $$ \rightarrow \tan ^{-1} 1 = \pi/4 $$

EDIT1: This is not a geometry based answer, but an analysis based, also am not too sure.

If $$F(x) + F(1/x)= 2 c , $$

then, $$ F(x) = F(1/x) = c. \tag{1} $$

Proof by contradiction is that if it is not so,the Taylor expansion of F(x) would lead to singularities of F(1/x) for real variable expansion in 1/x.

So given that

$$ \tan^{-1} x + \tan^{-1} (1/x) = \pi/2 \tag{2} $$

Adding/subtracting same particular angle that may be included in geometric construction $ (+\beta - \beta) $ ( This had been done by Blue and David Quinn in a geometric proof)

$$ \tan^{-1} x + \tan^{-1}\frac {1-x} {1+x} - \tan^{-1} \frac {1-x} {1+x} + \tan^{-1} (1/x) = \pi/2 $$

$$ (\tan^{-1} x + \tan^{-1}\frac {1-x} {1+x} ) + ( \tan^{-1} (1/x) + \tan^{-1} \frac{1-1/x} {1+1/x}) = \pi/2 \tag{3} $$

which by virtue of (1) gives

$$ (\tan^{-1} x + \tan^{-1} \frac{1-x} {1+x} ) = \pi/4. \tag{4} $$

  • $\begingroup$ How does that answer the question? The question is whether there is a geometric justification similar to the one given for the first identity. ${}\qquad{}$ $\endgroup$ – Michael Hardy Jun 14 '15 at 19:18
  • $\begingroup$ Another way of showing this involves differentiating this sum of arctangents with respect to $x$ and seeing that the derivative is $0$. ${}\qquad{}$ $\endgroup$ – Michael Hardy Jun 14 '15 at 19:33
  • $\begingroup$ You'll notice I changed $tan$ to $\tan$, coded as \tan, in your answer. That not only prevents italicization but also results in proper spacing in expressions like $a\tan b$. ${}\qquad{}$ $\endgroup$ – Michael Hardy Jun 14 '15 at 19:34
  • $\begingroup$ the OP is looking for a geometrical argument $\endgroup$ – David Quinn Jun 15 '15 at 13:07

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