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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?

J. W. Tanner
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Leo Authersh
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3 Answers3

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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.

Dando18
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    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? – Leo Authersh Aug 09 '17 at 15:50
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    Might also be worth mentioning the link to [Osborne's rule](https://math.stackexchange.com/q/138842/72968) here. – Silverfish Aug 09 '17 at 21:35
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    @Typhon not sure what you mean, $$ -i\sin(ix) = \frac 1 i \frac{e^{xi^2}-e^{-xi^2}}{2i} = \frac{e^x-e^{-x}}{2} = \sinh x $$ – Dando18 Aug 10 '17 at 01:37
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    @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 $z\in\mathbb C$. – Dando18 Aug 10 '17 at 03:42
  • @Dando18 Are all the trigonometric functions with complex variable hyperbolic functions? Or are hyperbolic functions just one of many classifications of trigonometric function with complex variable? – Leo Authersh Aug 11 '17 at 12:59
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    @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 $\cosh z = \cos(iz)$. – Dando18 Aug 11 '17 at 14:00
  • Good. So, it means that both circular functions and hyperbolic functions can be defined in real function as well as complex function. But coshz = Cos(iz) is just another spectacular equation. Thank you for the explanation. – Leo Authersh Aug 11 '17 at 16:47
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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 $\sin x = \tanh( \text{gd}^{-1}\, x)$.

sharding4
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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.

user64742
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Owen
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  • 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. $\sinh = \backslash x \to \frac{e^x-e^{-x}}2 + 0\cdot \sin x$. – leftaroundabout Aug 10 '17 at 12:28