In the proof that the poles of a Chebyshev filter lie on an ellipse, there is the following transformation, for the $s$ values correspondant to the poles. From
(1) $$s_{pm} = j \cos \left[ \displaystyle \frac{1}{n} \arccos \left( \frac{j}{\epsilon} \right) + \frac{m \pi}{n} \right]$$
to
(2) $$s_{pm} = \sinh \left[ \displaystyle \frac{1}{n} \mathrm{arsinh} \left( \frac{1}{\epsilon} \right) \right] \sin (\theta_m) + j \cosh \left[ \frac{1}{n} \mathrm{arsinh} \left( \frac{1}{\epsilon} \right) \right] \cos (\theta_m)$$
where $s$ is the complex variable of the Laplace domain; $m = 1, 2, \ldots, n$ (with $n$ integer) and
$$\theta_m = \displaystyle \frac{\pi}{2} \frac{2m - 1}{n}$$
$\epsilon$ is a real number and $j$ is the imaginary unit.
There are so many expressions for the complex $\arccos$. For example I found that, if
$$w = \arccos (z) = \arccos \displaystyle \frac{j}{\epsilon}$$
with $z$ purely imaginary like in (1), then
$$w = \displaystyle \frac{\pi}{2} + j \mathrm{arsinh} \left( - \displaystyle \frac{1}{\epsilon} \right)$$
or
$$w = \displaystyle \frac{3\pi}{2} + j \mathrm{arsinh} \left( \displaystyle \frac{1}{\epsilon} \right)$$
but I don't know how to use these results to obtain the expression (2) from (1).
In particular, I can't figure out how the terms $\sin (\theta_m)$ and $\cos (\theta_m)$ arise in (2) from $m \pi / n$ in (1).
Which relations should be used and how? Any hint?