The trigonometric solution to the solvable DeMoivre quintic? Using the relations for the Rogers-Ramanujan cfrac described in this post,
$$\frac{1}{r}-r = x$$
$$\frac{1}{r^5}-r^5 = y$$
and eliminating $r$ yields,
$$x^5+5x^3+5x = y$$
This is the case $a=1$ of the solvable DeMoivre quintic,
$$x^5+5ax^3+5a^2x+b = 0\tag1$$
In general, it has the solution,
$$x = \left(\tfrac{-b+\sqrt{b^2+4a^5}}{2}\right)^{1/5}-a\left(\tfrac{-b+\sqrt{b^2+4a^5}}{2}\right)^{-1/5}\tag2$$
When $a=-1$, it can also be solved as,
$$x = 2\cos\Bigg(\tfrac{\arccos\Big(\tfrac{-b}{2}\Big)}{5}\Bigg) = 2\cos\Bigg(\tfrac{2\arctan\Big(\tfrac{\sqrt{2+b}}{\sqrt{2-b}}\Big)}{5}\Bigg)\tag3$$

Q: When $\color{red}{a=1}$, is there a neat trigonometric solution similar to $(3)$?

P.S. A google search revealed that it is may indeed be possible.  

Thanks to Tito for the nice question. Here is a solution in terms of hyperbolic sine, which may not be what you want.
$$
\sinh(5t) = 5 \sinh t+ 20 \sinh^3 t + 16 \sinh^5 t.
$$
With $x = 2\sinh t$ and $b = -2\sinh 5t$, we have
$$
x^5+ 5 x^3 + 5 x + b = 0\tag1,
$$
which is the $a = 1$ case. So the solution is
$$
x = 2\sinh t = 2\sinh\left( \frac{\sinh^{-1}\left(-\frac{b}{2}\right)}{5} \right)\tag2
$$
