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I've learned in school that all the trigonometric functions can be constructed geometrically in terms of a unit circle:

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Can the hyperbolic functions be constructed geometrically as well? I know that $\sinh$ and $\cosh$ can be constructed based on the area between a ray through the origin and the unit hyperbola, but what about $\tanh$, $\mathrm{csch}$, $\mathrm{sech}$ and $\coth$?

enter image description here

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2 Answers 2

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The easy arrangement around the circle, as you showed it, makes use of the fact that every radius of the circle is a unit length, so triangles with that unit length as one edge and a right angle at one corner are easy to find. Furthermore, right angles can be used to transfer the angle argument $\theta$ between these triangles.

In the hyperbola situation, things are a bit more complicated. The point $(1,0)$ is easily obtained as the location of one vertex, which has distance $1$ from the origin. Using e.g. a line parallel to one asymptote, you can also obtain a point $(0,1)$. The area cannot easly be transferred, so one has to make do with the single occurence of that area, and find everything else in relation to it.

Using these two unit points together with the coordinate axes and various right angles, you can find all four hyperbolic functions you mentioned like this:

Illustration

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    $\begingroup$ +1. My answer to a similar question intentionally left out $\operatorname{sech}$ and $\operatorname{csch}$. Where you've put these values is pretty good ---the reciprocal relations with $\sinh$ and $\cosh$ are clear--- but I'm holding out for a configuration that also neatly demonstrates the corresponding Pythagorean relations. $\endgroup$
    – Blue
    Jul 30, 2013 at 22:39
  • $\begingroup$ @Blue: The distance between the origin and the point $(\operatorname{csch},1)$ is $\operatorname{coth}$, so there you have the equation $\operatorname{csch}^2+1^2=\operatorname{coth}^2$. And if you form a right triangle with $(0,0)$ and $(1,0)$ as endpoints of its hypothenuse, and with one leg pointing in the $(1,\sinh)$ direction (along the upper of the two cyan lines), then the lengths of the leg will be $\operatorname{sech}$ and the other leg will be $\operatorname{tanh}$, together prooving $\operatorname{sech}^2+\operatorname{tanh}^2=1$. Marking these lengths in an image is tricky, though. $\endgroup$
    – MvG
    Jul 30, 2013 at 23:06
  • $\begingroup$ Certainly, the relations are (and must be) implicitly there. As I mention, there are lots of ways to introduce segments whose lengths happen to be whatever we need them to be. But, eg, the only way we know that the distance from origin to $(\operatorname{csch}, 1)$ is $\operatorname{coth}$ is by already knowing the Pythagorean identity; there's no clear visual tie between that hypotenuse and the segment your figure defines to have length $\operatorname{coth}$. I want a figure that makes everything as obvious as in the circular trig diagram (which, btw, inspired my company logo). $\endgroup$
    – Blue
    Jul 30, 2013 at 23:26
  • $\begingroup$ For dis-satisfaction from other side: Consider the point $P(u,v)$ where the line joining the origin to $(\cosh, 1)$ meets the hyperbola in your figure. By virtue of $P$ lying on the curve, the coordinates satisfy the hyp-Pythagorean relation: $u^2-v^2=1$. Terrific! One can check we must in fact have $u = \operatorname{coth}$ and $v = \operatorname{csch}$ ... but this time we lack a clear visual connection to the proportionality relations $\operatorname{coth} = \cosh/\sinh$ and $\operatorname{csch} = 1/\sinh$. Somewhere is an ideal diagram that shows both Pythagorean-ness and proportionality. $\endgroup$
    – Blue
    Jul 30, 2013 at 23:57
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Unit hyperbola with all hyperbolic-functions lengths visible Since you wanted all Pythagorean relationships to be visible, I had to made right-triangles near the origin. This results in sech being drawn vertically and horizontally, unfortunately. Otherwise, co-functions are horizontal lines, and the rest are vertical lines. The triangles on the line y=tanh(A)*x all have horrible hypotenuses, unavoidably. I did not draw the line y=coth(A)*x, which would've clouded the picture even further.

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  • $\begingroup$ One thing I forgot to add is that all the right-triangles feature the Gudermannian and its complement. $\endgroup$
    – Robert Marshall Murphy
    Mar 29, 2018 at 16:18

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