Co- prefix in hyperbolic functions

Question

I was studying calculus when a question came to my mind.

As written here, in the context of trigonometric functions, the prefix co- stands for complementary, viz. if f is an angular function (mose notably, sine, cosine or tangent), the co-f function maps x into f of the complementary angle of x.

In the context of hyperbolic function there are (hyperbolic) sine, cosine and tangent as well, but at least in the case of tangent, the hyperbolic cotangent is not equal to the hyperbolic tangent of the "complementary angle" (indeed, the argument is not interpreted anymore as an angle). My question (broad, but I hope clear given the previous paragraph) is: given a real number x, is there a real number co(x) which, in the context of hyperbolic functions, plays the same role that the complementary angle plays in goniometric functions (e.g. the hyperbolic cotangent of x is equal to the hyperbolic tangent of co(x), etc.)?

EDIT: Comments suggested that it could be fruitful instead to set the question in the complex field.

Answer 1

Trigonometric functions are periodic, and hence co-functions exist because we can shift the values by a specific parameter (most commonly $\pi/2$) to give the same answer for a specific function.

Hyperbolic functions, on the other hand, are not periodic. They are based on exponential functions that arise from complex numbers, i.e. $$\cosh x = \dfrac {e^x + e^{-x}}{2}$$ and $$\sinh x = \dfrac {e^x - e^{-x}}{2}$$

In addition, $\cosh x$ has no value less than $1$ and is always positive (symmetric about the $y$ axis), while $\sinh x$ can be either positive or negative (symmetric about the $x$ axis.)

If we tried to find the "co-" functions for the hyperbolic functions in $\mathbb {R}$, we'd discover that the values would be very different. By coincidence, $\cosh 0 =1$ by the formula; however $\sinh 1$ has a very different value ($\ln (1+\sqrt{2})$, done by setting $\sinh x=1$ to one and solving for $x$; this results in a quadratic in $e^x$). Also by coincidence, $\sinh 0 = 0$, but $\cosh x = 0$ has no solution (both by solving the equation and noticing that $e^x + e^{-x}$ is never zero and the property that $\cosh x \geq 1$.) Finally, for $\cosh x = \sinh x$, this equation has no solution as $2e^{-x} = 0$ has no solution.

You can also graph $\cosh x$ and $\sinh x$ and notice that the graphs do not intersect, and as both functions increase to $+\infty$, $\cosh x$ is slightly more than $\sinh x$.

Hence there are no hyperbolic "co-" functions in $\mathbb{R}$.

Answer 2

For all $x\in\mathbb{R}$, $\tanh(x)\in(-1,1)$ and $\coth(x)\in(-\infty,-1)\cup(1,\infty).$ Therefore, there is no real function $\operatorname{co}(x)$ such that $\tanh(\operatorname{co}(x))=\coth(x)$.