Analogies between $(\tan, \sec)$ and $(\sinh, \cosh)$

Question

The pair of functions $(\tan, \sec)$ shares some interesting properties with the pair $(\sinh, \cosh)$.

First of all, they satisfy the same quadratic equation, namely $$\sec^2 x - \tan^2 x = 1 \qquad \cosh^2 x - \sinh^2 x = 1$$ for any $x$ in the respective domains.

Moreover, $\tan$ and $\sinh$ are both odd functions, while $\sec$ and $\cosh$ are both even functions.

Now, suppose that we define a binary operation $\oplus$ on some subset of real numbers such that $$\tan (x \oplus y) = \tan x \sec y + \sec x \tan y$$ whenever $x \oplus y$ is defined. Then one can prove that $$\sec(x \oplus y) = \sec x \sec y + \tan x \tan y$$ and these two formulas look exactly like the addition formulas for the hyperbolic functions. (For the subtraction formulas it is enough to let $x \ominus y = x \oplus (-y)$ whenever it is defined.)

There's more: one can also prove that an analogue to De Moivre's formula holds, i.e., $$(\sec x + \tan x)^n = \sec (\mathring n x) + \tan (\mathring n x)$$ where $\mathring n x$ denotes $x \oplus x \oplus \dotsb \oplus x$ with $n$ addends. Finally, if we define an analogue of the derivative with this new operation by letting $$\mathring D f(x) = \lim_{h \to 0} \frac {f(x \oplus h) - f(x)} h$$ then we obtain $$\mathring D \tan x = \sec x \qquad \mathring D \sec x = \tan x$$ similarly to what happens with the hyperbolic functions.

My questions are:

Is there a way to make this correspondence precise so that one can give a simple unique explanation for all of these analogies (and possibly others that might hold)?

How can we interpret the operation $\oplus$?

Answer 1

I think everything follows from your first equation. Since the signs of tan, sec go like the signs of sinh, cosh, this equation tells us that the parametric graphs $$ t\in(-\pi/2,\pi/2) \mapsto(\tan t, \sec t) \qquad \qquad t\in\mathbb R \mapsto (\sinh t, \cosh t) $$ consist of the same points in a different parameterization (in fact it's the upper branch of a hyperbola).

So if we define $f(x) = \sinh^{-1}(\tan t)$, then we have $$ \sinh \circ f = \tan \qquad\qquad \cosh \circ f = \sec $$ It's just a particular transformation of the horizontal axis that make the functions into each other.

This means that we have $x\oplus y = f^{-1}(f(x)+f(y))$; in other words $\oplus$ is just ordinary addition transfered through this bijection.

And this also means that your $\mathring D$ is just ordinary differentiation transfered through the bijection too.

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The same $f$ will turn also $\sin$ and $\cos$ into $\tanh$ and $\operatorname{sech}$, for more correspondences.

Answer 2

Trigonometric functions and hyperbolic functions can be regarded as limiting cases for Jacobi elliptic functions:

\begin{array}{|c|c|c|}\hline & k\to 0 & k\to 1 \\ \hline \operatorname{sn} (z,k) & \sin z & \tanh z \\ \operatorname{cn} (z,k) & \cos z & \operatorname{sech} z \\ \operatorname{nc} (z,k) & \sec z & \cosh z \\ \operatorname{sc} (z,k) & \tan z & \sinh z \\ \operatorname{ns} (z,k) & \csc z & \coth z \\ \operatorname{cs} (z,k) & \cot z & \operatorname{csch} z \\ \hline \end{array}

Answer 3

$$\cos(z)=\frac{e^{iz}+e^{-iz}}{2}=$$ $$\cosh(iz)$$

by the same,

$$\sin(z)=\frac{e^{iz}-e^{-iz}}{2i}=$$ $$-i\sinh(iz)$$

To find hyperbolic formulas from trogonometric ones, you just replace

$$\cos \text{ by } \cosh$$ and $$\sin \text{ by } -i\sinh$$

For example $$1-\cos(x)=2\sin^2(\frac{x}{2})$$ becomes

$$1-\cosh(x)=-2\sinh^2(\frac{x}{2})$$