# 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? • Thee is a similar question here: math.stackexchange.com/questions/321343/… , but no answer. – RicardoCruz Jul 24 '13 at 14:14
• It's worth noting that the line tangent to the point with coordinates $(\cosh x, \sinh x)$ on the unit hyperbola has slope $\cosh x/\sinh x = \coth x$, echoing the fact that the line tangent to the unit circle at $(\cos\theta, \sin\theta)$ has slope $-\cot\theta$. – Blue Jul 24 '13 at 14:20
• (Whoops, timed-out.) So, the line through the origin, parallel to that tangent line, meets the vertical line through the appropriate hyperbola vertex, at a point that happens to be $\coth x$ units away from the hyperbola's axis. There's a construction of $\coth$. – Blue Jul 24 '13 at 14:28

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 direct analogues to the circular arrangement.

Edit (7 April, 2018). But, there is this:

Here, the (now-visible) conjugate hyperbola fittingly hosts the cosine as well as cotangent ... though not cosecant, which, like secant, is introduced via an interesting ---if non-obvious--- reciprocal construction. (In an arbitrary hyperbola, the construction relates two segments whose geometric mean is the conjugate radius.) Still, there's a nice balance across the two hyperbolas.

The hyperbolic figure isn't as identity-rich as its circular counterpart, but we see that it covers the essentials. For instance, we have

$$\sinh \cdot \operatorname{csch} = 1 \qquad \cosh \cdot \operatorname{sech} = 1 \qquad \tanh \cdot \coth = 1$$

the last of which is implied by similar triangles; similarity also yields

$$\frac{\tanh}{1} = \frac{\sinh}{\cosh} \qquad\qquad \frac{\coth}{1} = \frac{\cosh}{\sinh}$$

Moreover, since the hyperbola has equation $x^2-y^2=1$, we have these hyperbolic Pythagorean relations $$\cosh^2 - \sinh^2 = 1 \qquad\qquad \coth^2 - \operatorname{csch}^2 = 1$$ while a guest appearance by the unit circle $x^2+y^2=1$ shows $$\operatorname{sech}^2 + \tanh^2 = 1$$

• In the last paragraph, you'd said "and doesn't provide much guidance about the corresponding placement of sech"; on the bright side, at least {sec(x) = 1/cos(x)} sees analytic continuation to {sech(z) = 1/cosh(z)}. mathworld.wolfram.com/HyperbolicSecant.html – Charles Rockafellor Jul 4 '16 at 5:34
• I've been looking into this lately, and I'm wondering if the best way to bring sech and csch into your second diagram is just to measure distances along the arm of the angle using a Minkowski metric instead of a Euclidean metric. Then all the same line segments can be used between the circle diagram and the hyperbola diagram, and you can even say the distance from the origin to the circle/hyperbola is always 1 in both cases. – Matt Dickau Sep 25 '17 at 18:32
• +1 for the new diagram added on April 7th, 2018, almost five years after the initial post. – Lee David Chung Lin Apr 20 '18 at 7:00