Can hyperbolic functions be defined in terms of trigonometric functions?

For example, can $\sinh x$ be written as a function of $\sin x$?

Another question, are hyperbolic functions dependent of their trigonometric correspondence in any way?

Yes. For example \begin{align*} \sinh x &= -i \sin(ix) \\ \cosh x &= \cos(ix) \\ \tanh x &= -i \tan(ix) \\ \end{align*}

These identities come from the definitions,

$$\sin x = \frac{e^{xi}-e^{-xi}}{2i} \text{ and } \sinh x = \frac{e^x - e^{-x}}{2}$$ and similar for cosine and tangent.

• From this definition, can we say that a hyperbolic function is a trigonometric function with complex variable and with a complex coefficient for an odd function? Aug 9, 2017 at 15:50
• Might also be worth mentioning the link to Osborne's rule here. Aug 9, 2017 at 21:35
• @Typhon not sure what you mean, $$-i\sin(ix) = \frac 1 i \frac{e^{xi^2}-e^{-xi^2}}{2i} = \frac{e^x-e^{-x}}{2} = \sinh x$$ Aug 10, 2017 at 1:37
• @LeoAuthersh I get what you're saying, but I would not word it like that. Trig functions are already defined for complex variables, i.e. $\cosh z = \cos iz$ for $z\in\mathbb C$. Aug 10, 2017 at 3:42
• @LeoAuthersh no. Simply think of their definitions $\cosh z \equiv 1/2(e^z + e^{-z}),\ \ z\in\mathbb C$ and $\cos z \equiv 1/2(e^{zi}+e^{-zi}),\ z\in\mathbb C$. Both are defined for complex input, but we just have this nice identity that $\cosh z = \cos(iz)$. Aug 11, 2017 at 14:00

Besides the connections between hyperbolic and circular functions which arise from substitutions involving imaginary arguments the functions can also be related using only real arguments via the Gudermannian function defined as $$\text{gd}(x)=\int_0^x\text{sech}\,t\,dt$$ This leads to identities such as $\sinh x = \tan (\text{gd}\,x)$ and $\sin x = \tanh( \text{gd}^{-1}\, x)$.

• Now this is something quite unusual. Aug 10, 2017 at 23:08

This may not be what you mean, but the phrase "as a function of" sometimes has a narrow and precise meaning. Specifically, "$x$ can be written as a function of $y$" means that there is a function $f$ such that $x = f(y)$.

To answer that narrow question, $\sinh x$ cannot be written as a function of $\sin x$ because $\sin$ is periodic along the real axis but $\sinh$ is not.

More specifically:

1. Assume $\forall{x \in R},~ \sinh(x) = f(\sin(x))$
2. Therefore, $\sinh(0) = f(\sin(0))$ and $\sinh(2\pi) = f(\sin(2\pi))$
3. Therefore, $\sinh(0) = f(0) = \sinh(2\pi)$

The last equation is not true.

• Good point about about the need for clear terminology. In particular, the question “can $\sinh x$ be written as a function of $\sin x$?” (answer: no) is very different from “can $\sinh$ be written as a function of $\sin$?”, in fact that's somewhat trivially possible through e.g. $\sinh = \backslash x \to \frac{e^x-e^{-x}}2 + 0\cdot \sin x$. Aug 10, 2017 at 12:28