8
$\begingroup$

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 $\oplus$?

$\endgroup$
2
  • 5
    $\begingroup$ The Gudermannian Function $\text{gd}: \mathbb R \to [-\pi/2,\pi/2]$ may help here. This function is defined by $$\text{gd}(x) = \int^x_{0} \frac{dt}{\cosh(t)}, \,\,\,\,\, x \in \mathbb R.$$ Among other interesting properties, it satisfies $$\tan(\text{gd}(x)) = \sinh(x), \,\,\,\,\, \sec(\text{gd}(x)) = \cosh(x)$$ which shows some explicit link between $\tan / \sec$ and $\sinh / \cosh$. $\endgroup$
    – User8128
    Commented Jul 16, 2018 at 19:37
  • $\begingroup$ Another correspondence: $x = a \cosh t, y = b \sinh t$ and $x = a \sec t, y = b \tan t$ both parameterize identical conics $\endgroup$ Commented Jan 3, 2023 at 7:41

3 Answers 3

8
$\begingroup$

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.


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

$\endgroup$
4
  • 1
    $\begingroup$ Great answer! Incidentally, your $f$ is the inverse of the Gudermannian function which I mentioned in the comments above. $\endgroup$
    – User8128
    Commented Jul 16, 2018 at 20:00
  • $\begingroup$ @User8128: Yes, I noticed. I didn't know it had a name. $\endgroup$ Commented Jul 16, 2018 at 20:35
  • $\begingroup$ Thank you! (By the way, I think you mean $t \in \left ( - \frac {\pi} 2, \frac {\pi} 2 \right )$ otherwise you also get the lower branch.) $\endgroup$ Commented Jul 17, 2018 at 8:53
  • $\begingroup$ @LucaBressan whoops, right. Fixed. $\endgroup$ Commented Jul 17, 2018 at 11:09
1
$\begingroup$

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}

$\endgroup$
2
  • $\begingroup$ Is there a value of $k$ such that $x = a \operatorname{cn}(t, k), y = b \operatorname{sn}(t, k)$ parameterizes a parabola? $\endgroup$ Commented Jan 3, 2023 at 7:44
  • $\begingroup$ Unfortunately, it always defines an ellipse even for $(\pm a\operatorname{sech} z, b\tanh z)$. Same for hyperbola $(\sec, \tan)$ or $(\pm \cosh, \sinh)$ $\endgroup$ Commented Jan 3, 2023 at 7:53
-2
$\begingroup$

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

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

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

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .