When solving a cubic equation, one might have to use trigonometric functions.
In some cases, these trigonometric functions can be expressed in terms of real radicals.
The goal is to find all the angles for which it is possible, for cubic equations with integer coefficients.
CONTEXT:
Let $\text{i}$ be the imaginary unit.
Let $\{a_0,b_0,a_1,b_1,a_2,b_2,a_3\}$ be a set of real integers such that $a_3>0$.
Consider the following cubic equation of unknown $z$:
$a_3z^3+\left(a_2+\text{i}b_2\right)z^2+\left(a_1+\text{i}b_1\right)z+a_0+\text{i}b_0=0$.
Let the real integers:
\begin{equation} \begin{split} c_E&=a_2^2-b_2^2-3a_1a_3\\ c_F&=2a_2b_2-3b_1a_3\\ c_G&=9a_3\left(a_1a_2-b_1b_2-3a_0a_3\right)+2a_2\left(3b_2^2-a_2^2\right)\\ c_H&=9a_3\left(a_1b_2+b_1a_2-3b_0a_3\right)+2b_2\left(b_2^2-3a_2^2\right)\\ c_I&=\frac{4c_E\left(3c_F^2-c_E^2\right)+c_G^2-c_H^2}{27a_3^2}\\ c_J&=\frac{2c_F\left(c_F^2-3c_E^2\right)+c_Gc_H}{27a_3^2}\\ c_K&=c_I^2+4c_J^2\\ c_L&=1+\text{sgn}\left(c_J\right)-\left|\text{sgn}\left(c_J\right)\right|\\ \end{split} \notag \end{equation}
Let $\{r_G,r_H,r_I,r_J,r_K,r_L,r_V,r_W\}$ be a set of real integers such that $r_Gr_H\neq0$.
Let $u\in\{0,1,2\}$.
The useful angles $\alpha_u$ for solving any cubic equation with integer coefficients have the following general expression:
\begin{equation} \alpha_u=\frac{r_G}{3}\text{arccot}\left(\frac{r_J+r_I\sqrt{r_V}+r_W\sqrt{6\left(r_K\sqrt{c_K}+r_L\right)}}{r_Gr_H}\right)+\frac{2u\pi}{3} \notag \end{equation}
There are three cases in which these angles have to be defined:
A) If $c_E=0$ and $c_F=0$ and $c_H\neq0$
\begin{equation} \begin{split} r_G&=-c_L\\ r_H&=\left|c_H\right|\\ r_I&=0\\ r_J&=-c_Gc_L\\ r_K&=0\\ r_L&=0\\ r_V&=0\\ r_W&=0\\ \end{split} \notag \end{equation}
B) If $\left(c_E\neq0\text{ or }c_F\neq0\right)$ and $2c_H^2\neq27a_3^2\left(\sqrt{c_K}-c_I\right)$ and $c_J=0$
\begin{equation} \begin{split} r_G&=\text{sgn}\left(c_H-\frac{3a_3\left(1-\text{sgn}\left(c_I\right)\right)}{2}\sqrt{3\left|c_I\right|}\right)\\ r_H&=\frac{\left(1-\text{sgn}\left(c_I\right)\right)\left(4c_E\left(3c_F^2-c_E^2\right)+c_G^2\right)+\left(1+\text{sgn}\left(c_I\right)\right)c_H^2}{2}\\ r_I&=\frac{3a_3\left(\left(1-\text{sgn}\left(c_I\right)\right)c_G-\left(1+\text{sgn}\left(c_I\right)\right)c_H\right)}{2}\\ r_J&=2c_F\left(3c_E^2-c_F^2\right)\\ r_K&=0\\ r_L&=0\\ r_V&=3\left|c_I\right|\\ r_W&=0\\ \end{split} \notag \end{equation}
C) If $\left(c_E\neq0\text{ or }c_F\neq0\right)$ and $2c_H^2\neq27a_3^2\left(\sqrt{c_K}-c_I\right)$ and $c_J\neq0$
\begin{equation} \begin{split} r_G&=\text{sgn}\left(2c_H-3a_3c_L\sqrt{6\left(\sqrt{c_K}-c_I\right)}\right)\\ r_H&=4\left(c_Ec_H^2\left(3c_F^2-c_E^2\right)-c_F^2\left(c_F^2-3c_E^2\right)^2+c_Fc_Gc_H\left(3c_E^2-c_F^2\right)\right)\\ r_I&=27a_3^2c_F\left(3c_E^2-c_F^2\right)\\ r_J&=c_F\left(3c_E^2-c_F^2\right)\left(4c_E\left(3c_F^2-c_E^2\right)+c_G^2+c_H^2\right)\\ r_K&=c_F^2c_G^2\left(c_F^2-3c_E^2\right)^2-4c_Ec_Fc_Gc_H\left(3c_F^2-c_E^2\right)\left(c_F^2-3c_E^2\right)\\ &+c_H^2\left(4c_E^2\left(3c_F^2-c_E^2\right)^2+c_F^2\left(c_F^2-3c_E^2\right)^2\right)\\ r_L&=\frac{1}{27a_3^2}\left( \begin{split} &-4c_Ec_Fc_Gc_H\left(3c_F^2-c_E^2\right)\left(c_F^2-3c_E^2\right)\left(c_G^2-3c_H^2\right)\\ &+4c_Ec_H^2\left(3c_F^2-c_E^2\right)\left(4c_E^2\left(3c_F^2-c_E^2\right)^2+3c_F^2\left(c_F^2-3c_E^2\right)^2\right)\\ &+c_F^2c_G^2\left(c_F^2-3c_E^2\right)^2\left(4c_E\left(3c_F^2-c_E^2\right)+c_G^2\right)\\ &-8c_Fc_Gc_H\left(c_F^2-3c_E^2\right)\left(2c_E^2\left(3c_F^2-c_E^2\right)^2+c_F^2\left(c_F^2-3c_E^2\right)^2\right)\\ &+2c_G^2c_H^2\left(2c_E^2\left(3c_F^2-c_E^2\right)^2-3c_F^2\left(c_F^2-3c_E^2\right)^2\right)\\ &-c_H^4\left(4c_E^2\left(3c_F^2-c_E^2\right)^2-c_F^2\left(c_F^2-3c_E^2\right)^2\right)\\ \end{split} \right)\\ r_V&=c_K\\ r_W&=3a_3\text{sgn}\left( \begin{split} &\left(c_Fc_G\left(c_F^2-3c_E^2\right)-2c_Ec_H\left(3c_F^2-c_E^2\right)\right)\sqrt{\sqrt{c_K}+c_I}\\ &-c_Fc_Hc_L\left(c_F^2-3c_E^2\right)\sqrt{\sqrt{c_K}-c_I}\\ \end{split} \right)\\ \end{split} \notag \end{equation}
QUESTIONS:
1) Is the set of all angles satisfying the above constraints finite?
2) If yes, what are the values of all these angles?
3) For all these values, express $\cos\left(\alpha_u\right)$ and $\sin\left(\alpha_u\right)$ in terms of real radicals if possible.
Thanks in advance for your help.
ALREADY DONE:
1) As far as I know, the only angles allowing to handle the $\text{arccot}$ function are such that $\text{arccot}\left(\dfrac{r_J+r_I\sqrt{r_V}+r_W\sqrt{6\left(r_K\sqrt{c_K}+r_L\right)}}{r_Gr_H}\right)=\dfrac{N\pi}{2^kF}$, with $k$ being a non-negative integer, $N$ and $\left(2^kF\right)$ being coprimes, and $F$ being either $1$ or a Fermat prime or a product of distinct Fermat primes.
As Trebor mentioned in the comments of his/her answer below, we have to limit ourselves to the $5$ known Fermat primes, that is the $32$ known values of $F$.
However, this might not be a problem.
Indeed, in the expression of the trigonometric functions of such angles, both the number of radicals and their nesting depths seem to increase as either $k$ or $F$ increases.
(Note that I haven't found a way to denest $\sqrt{6\left(r_K\sqrt{c_K}+r_L\right)}$ in (C) since $\left(r_L^2-r_K^2c_K\right)$ actually is the negative of a perfect square).
Here the $\text{arccot}$ function contains at most 2 square roots, and at most 1 is nested with a single square root.
Thus this might result in a finite set of angles, although I haven't been able to prove it yet.
Any idea?
2) I tested it for several values of $k$ and $F$, for now only focusing on the number of radicals, their nesting depths, and the fact that $\left(r_L^2-r_K^2c_K\right)$ is the negative of a perfect square in (C):
- For $F=1$, it works up to $k=3$, it doesn't for $k=4$, and I haven't yet found the expressions for $\cot\left(\frac{N\pi}{32}\right)$ or $\tan\left(\frac{N\pi}{32}\right)$.
- For $F=3$, it works up to $k=2$, it doesn't for $k=3$, and I haven't yet found the expressions for $\cot\left(\frac{N\pi}{48}\right)$ or $\tan\left(\frac{N\pi}{48}\right)$.
- For $F=5$, it doesn't work up to $k=2$, and I haven't yet found the expressions for $\cot\left(\frac{N\pi}{40}\right)$ or $\tan\left(\frac{N\pi}{40}\right)$.
- For $F=15$, it doesn't work up to $k=2$, and I haven't yet found the expressions for $\cot\left(\frac{N\pi}{120}\right)$ or $\tan\left(\frac{N\pi}{120}\right)$.
- For $F\geq17$, I haven't yet found the expressions for $\cot\left(\frac{N\pi}{2^kF}\right)$ or $\tan\left(\frac{N\pi}{2^kF}\right)$.
If anyone knows any canonical denested expression for the $\cot$ or $\tan$ functions of the angles I'm missing, please answer.
3) It will be possible to express $\cos\left(\alpha_u\right)$ and $\sin\left(\alpha_u\right)$ for the angles found at (2) if, after reduction of $\dfrac{r_GN+2^{k+1}u}{2^k3F}$, $\left(r_GN+2^{k+1}u\right)$ and $\left(2^k3F\right)$ are coprimes and $\left(3F\right)$ is either $1$ or a Fermat prime or a product of distinct Fermat primes. For the angles already found at (2), it works if $F=1$ and it doesn't if $F=3$. More generally, I guess that it will work if $F$ isn't a multiple of $3$, and it won't if $F$ is a multiple of $3$. Is that correct?
If you allow quadratic roots only, then see here for a list of values. If higher radicals such as cubic roots (inevitably involving complex numbers, see Casus Irreducibilis), then you can replace the 'Fermat prime' with 'Pierpont prime' and so on.