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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.

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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 $$ x = 2\sinh t = 2\sinh\left( \frac{\sinh^{-1}\left(-\frac{b}{2}\right)}{5} \right)\tag2 $$

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    $\begingroup$ Thanks! It turns out the general quintic can then be solved in terms of the hyperbolic functions. See my addendum to your answer. $\endgroup$ – Tito Piezas III Nov 27 '15 at 7:19
  • $\begingroup$ By the way, do you know the expressions for all five roots of $(1)$? (The value $(2)$ is just one of the roots.) $\endgroup$ – Tito Piezas III Nov 28 '15 at 7:14
  • $\begingroup$ @TitoPiezasIII. Sorry for the late response. I was about to say adding $2\pi i n$ on the numerator, then I found that you have already done that. :-) $\endgroup$ – hbp Nov 30 '15 at 5:17
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(This is an addendum to hbp's answer.)

I'm glad I raised a bounty for this question because, thanks to hbp, we can show that the general quintic can be solved in terms of trigonometric or hyperbolic functions, plus some special functions.

The general quintic can be reduced to the one-parameter Brioschi form,

$$w^5-10cw^3+45c^2w-c^2=0\tag1$$

with the five solutions (see also this post),

$$w_n=\pm\sqrt{\frac{-c\,(x^2+4)(x^2-2x-4)^2}{b-11}}\tag2$$

and for $n=0,1,2,3,4$,

$$x_n=-2\,i\sin\Bigg(\tfrac{i\log\Big(\tfrac{b+\sqrt{b^2+4}}{2}\Big)\,-\,2\pi\, n}{5} \Bigg)=2\sinh\Bigg(\tfrac{\sinh^{-1}\Big(\tfrac{b}{2}\Big)\,+\,2\pi\,i\, n}{5}\Bigg) \tag3$$

$$b=\frac{v(v-5)^2}{(v-1)^2}+11$$

$$v=\left(\frac{\vartheta_2(0,p)}{\vartheta_2(0,p^5)}\right)^2$$

$$p=e^{\pi i \tau}=\exp(\pi i \tau)$$

$$\tau = i\frac{K(k')}{K(k)}= i\,\frac{\text{EllipticK[1-m]}}{\text{EllipticK[m]}}\tag4$$

$$m =\tfrac{1}{2}\left(1\pm\sqrt{1-4u}\right)\tag5$$

and $u$ is a root of the cubic,

$$\frac{256(1-u)^3}{u^2}=\frac{1728c-1}{c}$$

The solution also uses the Jacobi theta function $\vartheta_j(0,p)$ (where $j=3$ or $4$ will work as well), the complete elliptic integral of the first kind $K(k)$ and elliptic parameter $m=k^2$ (with $\tau$ also given in Mathematica syntax above).

Note 1: Since $(2)$ and $(5)$ uses square roots, then the proper sign has to be chosen.

Note 2: At first, I was missing a neat form for $(3)$ so, after hbp's answer, I'm glad I posted this question.

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    $\begingroup$ Wow, that's cool! Glad to offer some little help to a mighty mind :-) $\endgroup$ – hbp Nov 27 '15 at 18:16
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    $\begingroup$ You claim that a general quintic can be solved in terms of trigonometric/hyperbolic functions. But your further explanation includes theta function and eliptic integrals. $\endgroup$ – Anixx Nov 28 '15 at 3:37
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    $\begingroup$ @Anixx: I didn't say only. trigonometric functions. For its argument, it uses other special functions. But it's not easy to find any function in the first place that can solve the general quintic. $\endgroup$ – Tito Piezas III Nov 28 '15 at 4:12
  • $\begingroup$ @TitoPiezasIII If I may ask, what field of math do you work in? I am trying to figure out which field gives you such vast experience with algebraic equations and identities. $\endgroup$ – Ovi Mar 21 '17 at 6:40

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