I understand that expressions like $\sinh(\text{arcsinh}(x))$ simplify immediately and expressions like $\sinh(\text{arccosh}(x))$ tend to simplify after some algebra.
However I cannot work out how to simplify something like:
$$\sinh(4\,\text{arcsinh}(x))$$
I get about this far:
$$=\frac{1}{2}[e^{\log\left(x+\sqrt{x^2+1}\right)^4}-e^{\log\left(x+\sqrt{x^2+1}\right)^{-4}}]$$
$$=\frac{1}{2}[\left(x+\sqrt{x^2+1}\right)^4-\left(x+\sqrt{x^2+1}\right)^{-4}]$$
Then the thought of having to expand out the $4$th power of these expressions then find creative ways to simplify seems too much. The answers book tells me this works out to (not sure of the exact constant value $k$, it's mixed with other stuff, maybe $4$?):
$$kx\sqrt{x^2+1}(1+2x^2)$$
How can simplify $\sinh(4\,\text{arcsinh}(x))$ to get this result?
Do some hyperbolic trigonometry. The formulae are similar the the formulae of usual trigonometry, with some sign changes.
Let's denote $y=\operatorname{argsinh} x$. We have $\sinh y=x$ by definition, and $\cosh^2y-\sinh^2y=1\implies \cosh y=\sqrt{x^2+1}$, so that \begin{align*} \sinh 4y&=2\sinh 2y\cosh 2y=4\sinh y\cosh y(1+2\sinh^2y)\\ &=4x\sqrt{x^2+1}(1+2x^2). \end{align*}