If $x=\sinh{\theta}$, is it possible to express $\cosh{n\theta}$ and $\sinh{n\theta}$ in terms of $x$? We know that hyperbolic sine is:
$$\sinh \theta={\frac {e^{\theta}-e^{-\theta}}{2}}$$
and that hyperbolic cosine is
$$\cosh \theta={\frac {e^{\theta}+e^{-\theta}}{2}}$$
Let $n\in\mathbb N$.
If $x=\sinh{\theta}$, is it possible to express $\cosh{n\theta}$ and $\sinh{n\theta}$ in terms of $x$?

$\begin{align}\sinh(\theta)=\frac{e^\theta-e^{-\theta}}{2}=x &\Rightarrow e^{2\theta}-2xe^{\theta}-1 =0\\
&\Rightarrow e^{\theta}= \frac{2x\pm\sqrt{4x^2+4}}{2} = x\pm\sqrt{x^2+1} \\
&\Rightarrow \boxed{e^{\theta} = x+\sqrt{x^2+1}} ~~~\text{ as } e^\theta \ge0,\theta \in \Bbb R\end{align}$
So, $$\sinh(n\theta) = \frac{e^{n\theta}-e^{-n\theta}}{2} = \frac{(x+\sqrt{x^2+1})^n-(x+\sqrt{x^2+1})^{-n}}{2}$$
and
$$\cosh(n\theta) = \frac{e^{n\theta}+e^{-n\theta}}{2} = \frac{(x+\sqrt{x^2+1})^n+(x+\sqrt{x^2+1})^{-n}}{2}$$
