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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?

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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 $\cosh^2y-\sinh^2y=1\implies \cosh y=\sqrt{2x^2+1}$, so that \begin{align*} \sinh 4y&=2\sinh 2y\cosh 2y=4\sinh y\cosh y(1+2\sinh^2y)\\ &=4x\sqrt{2x^2+1}(1+2x^2). \end{align*}

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  • $\begingroup$ excellent, thank you. $\endgroup$ – Brendan Hill Nov 2 '15 at 1:29
  • $\begingroup$ You're welcome. Always glad to help! $\endgroup$ – Bernard Nov 2 '15 at 1:34

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