# Geometric construction of hyperbolic trigonometric functions

If we have a circle we can geometrically construct the trigonometric functions as shown. The functions all derive from sin and cos. If we say that the circle is a conic section and imagine it on the cone we can draw a hyperbola perpendicular to it. I believe that the hyperbolic trigonometric functions can be plotted geometrically as well but I cannot find and representation of it. What do the hyperbolic trigonometric functions actually tell you about a hyperbola?

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If you consider this alternative rendering of the circular trig diagram …

… then there’s this hyperbolic analogue:

In each case, the point where the inclined ray meets the curve determines sine and cosine, and the points where it meets the vertical and horizontal tangent lines determine tangent and cotangent. (In the hyperbolic case, the “horizontal tangent line” is actually tangent to the (invisible) conjugate hyperbola.)

What about the secant and cosecant segments? In the circular case, these are the portions of the inclined ray that form hypotenuses of the $1$-and-$\tan$ and $1$-and-$\cot$ right triangles. In the hyperbolic case … well … there don’t seem to be obvious analogues to the circular case. (That’s not to say that there aren’t lots of ways to introduce segments whose lengths happen to be $\operatorname{sech}$ and $\operatorname{csch}$.) Importantly, the relations

$$1 – \tanh^2 = \operatorname{sech}^2 \qquad\qquad \coth^2 – 1 = \operatorname{csch}^2$$

cast the ostensible “$1$” segment into different roles —hypotenuse or leg— in the corresponding right triangles. This is unlike the circular case, where the figures illustrating “co-identities” match (apart from swapping the roles of trig functions and co-functions). I haven’t (yet?) found a particularly-satisfying way to handle these.

Edit. In comments following @MvG’s answer to a similar question, I describe a way to bring $\operatorname{sech}$ into the diagram. Here’s another way to approach the same result: project the endpoint of the $\operatorname{coth}$ segment vertically onto the hyperbola itself.

The $y$ coordinate of that point —the length of the segment from that point to the horizontal axis— happens to be $\operatorname{csch}$, and the Pythagorean relation

$$\operatorname{coth}^2 – \operatorname{csch}^2 = 1$$

simply recognizes that the point is on the hyperbola $x^2 – y^2 = 1$. Unfortunately, the relation

$$\operatorname{csch} = \frac{1}{\sinh}$$

isn’t at all clear visually. (@MvG’s $\operatorname{csch}$ segment makes the reciprocation relation clear, but not the Pythagorean relation.) By interesting coincidence, the line joining the origin to the point where the (extended-if-necessary) $\sinh$ segment meets $y=1$ happens to pass through the new point on the hyperbola, and thus provides a compatible construction of the $\operatorname{csch}$ segment. So, we’ve (re)discovered a method of using a hyperbola to construct reciprocal segments. Perhaps if that method were more popularly known, I might consider this placement of $\operatorname{csch}$ to be fairly “natural”. As it is, the placement still seems somewhat contrived (and doesn’t provide much guidance about the corresponding placement of $\operatorname{sech}$).