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?
 A: 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.
A: 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 $\sin x = \tanh( \text{gd}^{-1}\, x)$.
A: 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:


*

*Assume $\forall{x \in R},~ \sinh(x) = f(\sin(x))$

*Therefore, $\sinh(0) = f(\sin(0))$ and $\sinh(2\pi) = f(\sin(2\pi))$

*Therefore, $\sinh(0) = f(0) = \sinh(2\pi)$


The last equation is not true.
