This is an answer showing the limitation of the "computational wish",
taken instead for the lack of a question.
The question is rather a question related to computations in number fields.
(Which are field extensions of finite degree of the base field $\Bbb Q$.)
We are especially dealing with cyclotomic fields.
There is already a notion of a "radical extension" related to extensions of fields,
but this notion also covers $\sqrt[n]{\pm 1}$, and using this notion, the
problem of expressing the cyclotomic units by (such) radicals is already solved.
So we are searching for more explicit formulas.
A natural wish is to use - when possible - only the radicals of some order computed
for a very specific real number. This is too specific, below there will
be some more or less explicit formulas. For instance, we obtain a formula
for
$$
\cos\frac \pi{13} + i\sin\frac \pi{13}
$$
which is not the cheap "$\displaystyle\sqrt[13]{-1}$",
but rather a formula involving some more explicit radicals of smaller degree, of the shape
$C+\sqrt[3]{A+B}+\sqrt[3]{A-B}$, where $A,B,C$ are complex numbers written
explicitly in terms of radicals and algebraic (field) operations.
It is still a task to isolate the real and the imaginary part of
$\sqrt[3]{A+B}$, but well, i will not do it here, hoping that the progress made
is enough.
Same qualitative situation applies to the other cases, however these cases
come with "higher" complications.
The case $p=13$:
I will start with the simplest case, and "compute" the sine and cosine for $\pi/13$. For this, let $u$ be the complex number
$$
u
=
\exp \frac {\pi i}{13}
=
\cos \frac {\pi}{13}
+i
\sin \frac {\pi}{13}
=
\cos \frac {2\pi}{26}
+i
\sin \frac {2\pi}{26}
\ .
$$
Here are some facts about this number.
It is a cyclotomic unit of degree $23$
It satisfies the equation $u^{13}=\exp(\pi i)=-1$, so it is a root of the polynomial of degree $13-1=12=2^2\cdot 3$:
$$
q = X^{12} - X^{11} + X^{10} - X^9 + X^8 - X^7 + X^6 - X^5 + X^4 - X^3 + X^2 - X + 1\ .
$$
The Galois group of the related cyclotomic field is cyclic and abelian,
so we expect a tower of subfields corresponding to the subgroups of $\Bbb Z/12$.
They are joined via edges in a parallel way as the edges of the graph of divisors of $12$, an edge being drawn if there is a divisibility relation.
(The following investigation can be done using either the divisibility path $1|2|4|12$ or the divisibility path $1|3|6|12$ or... My choice will be the one involving third roots as late as possible.)
Let $K=K_{12}=\Bbb Q(u)$ be the field generated by $u$. Then we have a tower of subfields...
$$
\require{AMScd}
\color{darkgreen}{
\begin{CD}
K=K_{12}=\Bbb Q(u)\\
@V V :3 V\\
K_4=\Bbb Q(a) @. \qquad\text{ where }a=\frac14\Big(1\color{brown}{\pm}\sqrt{ 13}\color{blue}{\pm}\sqrt{2(13\color{brown}{\pm} 3\sqrt {13})}\ \Big) \\
@VV :2 V\\
K_2=\Bbb Q(\sqrt{13})\\
@VV :2 V\\
\Bbb Q
\end{CD}}
$$
Being an extension of degree three, $K:K_4$, it is radical, and we can "compute" $u$ in terms of $a$.
The minimal polynomial of $a$ over $\Bbb Q$ is $x^4 - x^3 + 2x^2 + 4x + 3$.
We are not fixing the two signs in the definition of $a$, and compute algebraically $u$ as the root of the polynomial with coefficients in $K_4$:
$$
f = X^3 - aX^2 + \frac 13(a^3 + 2a + 1)X + 1\ .
$$
(Already the idea of applying the formulas of Del Ferro, Tartaglia, Cardano explicitly makes me feel some pain in the finger.) Then $u-\frac 13a$ satisfies
$$
g =
x^3 + \frac 13(a^3 - a^2 + 2a + 3)x + \frac1{27}(a^3 - 3a + 18)\ .
$$
We are searching for two numbers, $s,t$ with $-3st=\frac 13(a^3 - a^2 + 2a + 3)$
and $s^3+t^3=-\frac1{27}(a^3 - 3a + 18)$. We know the sum and product of the numbers $s^3$, and $t^3$, so these two numbers are
$$
\begin{aligned}
s^3, t^3
&=
-\frac1{54}(a^3 - 3a + 18)
\pm
\frac 12\sqrt{\frac 1{27}(a^3 - a + 11)}
\\
&=
\frac 1{108}\Big(-26 + 4\sqrt{13} + i\sqrt{26 + 6\sqrt{13}}\Big)
\ \pm\
\frac 1{36}\sqrt{6(13-3\sqrt{13})}
\\
&=A\pm B
\ .
\end{aligned}
$$
The above results are easy, when displayed, well, getting them...
Now the three possible values for $\displaystyle v = u-\frac 13a$ are finally the following ones:
$$
\varepsilon^k s + \bar\varepsilon^k t = \varepsilon^k\sqrt[3]{A + B} + \bar\varepsilon^k\sqrt[3]{A - B}\ ,\qquad k=0,1,2\ .
$$
Here $\varepsilon$ is a primitive third root of unity.
And please do not ask now for the explicit form of the real part of these numbers...
(The third roots should be chosen in each line so that the corresponding product $-3st$ has the needed value.)
It is clear, that we obtain formulas using only radicals and algebraic operations for the roots of the polynomials $g$ and $f$ above.
Let us see if the numerical values
of the above expressions have any chance...
a = ( 1 + sqrt(13) + i*sqrt( 2*(13 + 3*sqrt(13)) ) ) / 4
print( f"a = {a} \n ~ {a.n()}" )
print( f"The minimal polynomial of a is {a.minpoly()}\n" )
We introduce an algebraic version of the above a, and related fields.
K4.<a> = NumberField(x^4 - x^3 + 2x^2 + 4x + 3)
R.<X> = PolynomialRing(K4)
L.<u> = K4.extension(X^3 - aX^2 + (1/3a^3 + 2/3a + 1)X + 1)
print( f"u is a root of the polynomial:\n{u.minpoly()}\n" )
print( f"Then we have u^13 = {u**13}" )
var('x');
for ac in a.complex_embeddings():
print(f"a is numerically ~ {ac}")
P = (x^3 -ac*x^2 + (1/3*ac^3 + 2/3*ac + 1)*x + 1)
uc_values = P.roots(ring=CC, multiplicities=False)
for uc in uc_values:
k = round( real( (log(uc)/i/pi*13).n() ), 6 )
print(f" u is numerically {uc} ... angle -> {k:g} pi / 13")
And we obtain first:
a = 1/4*sqrt(13) + 1/4*I*sqrt(6*sqrt(13) + 26) + 1/4
~ 1.15138781886600 + 1.72542218842201*I
The minimal polynomial of a is x^4 - x^3 + 2*x^2 + 4*x + 3
u is a root of the polynomial:
x^3 - ax^2 + (1/3a^3 + 2/3a + 1)x + 1
Then we have u^13 = -1
confirming the fact that all values of $u$ algebraically found by the above receipt are satisfying $u^{13}=-1$. Then we have using numerical approximations for $a$ corresponding roots $u$, written also numerically, and guess the corresponding angle...
a is numerically ~ -0.651387818865997 - 0.522415803456408*I
u is numerically -0.885456025653210 - 0.464723172043769*I ... angle -> -11 pi / 13
u is numerically -0.120536680255323 - 0.992708874098054*I ... angle -> -7 pi / 13
u is numerically 0.354604887042536 + 0.935016242685415*I ... angle -> 5 pi / 13
a is numerically ~ -0.651387818865997 + 0.522415803456408*I
u is numerically -0.885456025653210 + 0.464723172043769*I ... angle -> 11 pi / 13
u is numerically -0.120536680255323 + 0.992708874098054*I ... angle -> 7 pi / 13
u is numerically 0.354604887042536 - 0.935016242685415*I ... angle -> -5 pi / 13
a is numerically ~ 1.15138781886600 - 1.72542218842201*I
u is numerically -0.568064746731156 - 0.822983865893656*I ... angle -> -9 pi / 13
u is numerically 0.748510748171100 - 0.663122658240795*I ... angle -> -3 pi / 13
u is numerically 0.970941817426053 - 0.239315664287557*I ... angle -> -1 pi / 13
a is numerically ~ 1.15138781886600 + 1.72542218842201*I
u is numerically -0.568064746731156 + 0.822983865893656*I ... angle -> 9 pi / 13
u is numerically 0.748510748171100 + 0.663122658240795*I ... angle -> 3 pi / 13
u is numerically 0.970941817426053 + 0.239315664287557*I ... angle -> 1 pi / 13
Now that we also have some formulas involving only radicals (over $\Bbb C$) and algebraic operations, let us implement some of the required angles... The following simple code involves only the mentioned operations and can be easily translated in some other CAS...
w = sqrt(13)
a = ( 1 + w + isqrt( 26 + 6w ) ) / 4
s3 = ( -26 + 4w + isqrt( 26 + 6w ) ) / 108 + 1/36 sqrt( 6(13 - 3w) )
t3 = ( -26 + 4w + isqrt( 26 + 6w ) ) / 108 - 1/36 sqrt( 6(13 - 3w) )
st = -1/9 * (a^3 - a^2 + 2a + 3)
s = s3 ^ (1/3) # extract cubic root
#t = t3 ^ (1/3) # extract cubic root ... but better insure st = st
t = st / s
eps = (-1 + isqrt(3))/2
u1 = a/3 + s + t
u2 = a/3 + eps s + eps^2 * t
u3 = a/3 + eps^2 * s + eps * t
for u in (u1, u2, u3):
k = ZZ(round( real( log(u.n()) / (ipi) 13 ), 6 ))
print(f"u is numerically {u.n()}")
print(f" cos({k} pi / 13) ~ {cos(kpi / 13).n()}")
print(f" sin({k} pi / 13) ~ {sin(kpi / 13).n()}")
And we obtain:
u is numerically 0.970941817426052 + 0.239315664287557*I
cos(1 pi / 13) ~ 0.970941817426052
sin(1 pi / 13) ~ 0.239315664287558
u is numerically -0.568064746731156 + 0.822983865893657*I
cos(9 pi / 13) ~ -0.568064746731156
sin(9 pi / 13) ~ 0.822983865893656
u is numerically 0.748510748171102 + 0.663122658240795*I
cos(3 pi / 13) ~ 0.748510748171101
sin(3 pi / 13) ~ 0.663122658240795
So if we must write a formula, it is:
$$
\color{blue}{
\begin{aligned}
&w := \sqrt{13}\ ,\\
&
\cos\frac\pi{13} +
i\sin\frac\pi{13}
\\
&\qquad=
\frac 1{12}\Big( 1 + w + i\sqrt{26 + 6w} \Big)
\\
&\qquad\qquad
+\left(\ \frac 1{108} ( -26 + 4w + i\sqrt{26 + 6w} )\ +\ \frac 1{36}\sqrt{6(13-3w)}\ \right)^{1/3}
\\
&\qquad\qquad
+\left(\ \frac 1{108} ( -26 + 4w + i\sqrt{26 + 6w} )\ -\ \frac 1{36}\sqrt{6(13-3w)}\ \right)^{1/3}
\ .
\end{aligned}}
$$
Here, the two cubic radicals are taken from complex numbers,
and they have to be inter-coordinated, they have to match in the sense, that their product
- defined up to some third root of unity - should be a known quantity,
$-\frac 19(a^3 - a^2 + 2a + 3)$.
A choice of the first cubic root determines thus the second cubic root, and the value of the above
expression shows the sine and the cosine of $\pi/13$, or of $3\pi/3$, or of $9\pi/13$.
Here is a further possible path to obtain the "simpler" cosine values,
and then by trigonometry the "more complicated" sine values for the angles
with denominator $13$.
We blindly use sage to get the minimal polynomial of the algebraic number
$c=2\cos(\pi/13)$. Then $c$ determines a field of degree six over $\Bbb Q$.
We split this extension again in two parts. The splitting can be either the one corresponding
to $6=2\cdot 3$, or the one corresponding to $3\cdot 2$. (At any rate there are two strategies.)
Depending on strategy (and its meaning, which is still to be defined / anticipated),
we will get expressions involving
- either radicals of expressions involving cubic radicals of "base expressions",
- or cubic radicals of expressions involving radicals of "base expressions",
where the "base expressions" differ from case to case, and may also show radicals.
The picture would be
$$
\require{AMScd}
\color{darkgreen}{
\begin{CD}
K=\Bbb Q(c) @. \qquad\text{ where $c$ is a root of }x^6 - x^5 - 5x^4 + 4x^3 + 6x^2 - 3x - 1\in\Bbb Q[x]\\
@V V :2 V\\
K_3=\Bbb Q(g) @. \qquad\text{ where $g$ is a root of }x^3 - x^2 - 4x - 1\in\Bbb Q[x]\\
@VV :3 V\\
\Bbb Q
\end{CD}}
$$
and the polynomials are extracted from...
p = ( 2*cos(pi/13) ).minpoly()
print(f"c := 2 cos( pi/13 ) is a root of the polynomial:\n {p} .")
K.<c> = NumberField( p )
L, morphism, _ = K.subfields( degree=3, name='g' )[0]
print(L)
We obtain:
c := 2 cos( pi/13 ) is a root of the polynomial:
x^6 - x^5 - 5*x^4 + 4*x^3 + 6*x^2 - 3*x - 1 .
Number Field in g0 with defining polynomial x^3 - x^2 - 4*x - 1
For the above picture, we have to solve the
equation of degree three for $g$, using the formula of Del Ferro, Tartaglia, Cardano,
then an equation of degree two with coefficients depending on $g$.
There is also the "other way", the picture would be:
$$
\require{AMScd}
\color{darkgreen}{
\begin{CD}
K=\Bbb Q(c) @. \qquad\text{ where $c$ is a root of }
X^3-\frac 12(h+1)X^2 -X +\frac 12(h+3)\in K_2[x]\\
@V V :3 V\\
K_2=\Bbb Q(h) @. \qquad\text{ where $h=\sqrt {13}$}\\
@VV :3 V\\
\Bbb Q
\end{CD}}
$$
To obtain this picture, we first ask sage for the subfield(s) of degree two,
obtain only $K_2=\Bbb Q(h)=\Bbb Q(\sqrt{13})$, then use it, and split the now known minimal
polynomial of $c$ over $K_2$:
L.<h> = QuadraticField(13)
RL.<X> = PolynomialRing(L)
factor( X^6 - X^5 - 5X^4 + 4X^3 + 6X^2 - 3X - 1 )
and we obtain the factorization:
(X^3 + (-1/2*h - 1/2)*X^2 - X + 1/2*h + 3/2) * (X^3 + (1/2*h - 1/2)*X^2 - X - 1/2*h + 3/2)
We took one of the factors above.
(Note that the free coefficients $\frac 12(h\pm 3)$ are units in the ring of integers
of $K_2$.)
And the same formula of Del Ferro - Tartaglia - Cardano
gives an "explicit formula" involving only (more or less explicit) radicals
for $c=2\cos(\pi/13)$.
It is always a pleasure to work in a cyclotomic field.
This concludes the case $p=13$.
The case $p=11$:
We will adapt one of the above strategies only for this case.
Let $u$ be the complex number
$$
u
=
\exp \frac {\pi i}{11}
=
\cos \frac {\pi}{11}
+i
\sin \frac {\pi}{11}
=
\cos \frac {2\pi}{22}
+i
\sin \frac {2\pi}{22}
\ .
$$
It satisfies the equation $u^{11}=\exp(\pi i)=-1$,
so it is a root of the polynomial of degree $10=2\cdot 5$:
$$
q = X^{10} - X^9 + X^8 - X^7 + X^6 - X^5 + X^4 - X^3 + X^2 - X + 1\ .
$$
It is a cyclotomic unit of degree $22$.
The Galois group of the related cyclotomic field is cyclic and abelian,
so we expect a tower of subfields corresponding to the subgroups of $\Bbb Z/10$.
We have in a picture for $K=K_{10}:=\Bbb Q(u)$, the field generated by $u$,
the following
tower of subfields...
$$
\require{AMScd}
\color{darkgreen}{
\begin{CD}
K=K_{10}=\Bbb Q(u)\\
@V V :2 V\\
K_5=\Bbb Q(c) @. \qquad\text{ where $c$ is a root of }x^5 - x^4 - 4x^3 + 3x^2 + 3x - 1\in\Bbb Q[x]\\
@VV :5 V\\
\Bbb Q
\end{CD}}
$$
(Here, i want "first" the equation of degree five.)
The variable $c$, one of its complex embeddings, is a posteriori $2\cos(\pi/11)$.
We are again not allowing ourselves to use the expression $\sqrt[11]{-1}$ as a "solution",
instead, more explicit radical expressions are searched.
Being a Galois extension of degree five, $K_5:\Bbb Q$,
we have informations about the possible structure of its roots.
The following sage code finds the above minimal polynomial for $c=2\cos\frac\pi{11}$:
sage: p = (2*cos(pi/11)).minpoly()
sage: p
x^5 - x^4 - 4*x^3 + 3*x^2 + 3*x - 1
sage: p.discriminant().factor()
11^4
It turns out that the roots of the polynomial $x^5 - x^4 - 4x^3 + 3x^2 + 3x - 1$
have the structural description:
$$
\begin{aligned}
c_1 &= &&\frac 15(1 + s + t + u + v)
&&\color{gray}{=\frac 15(1 + \sqrt[5]S + \sqrt[5]T + \sqrt[5]U + \sqrt[5]V)}\ ,\\
%
c_2 &=&&\frac 15(1 + \zeta s + \zeta^2 t + \zeta^3 u + \zeta^4 v)
&&\color{gray}{=\frac 15(1 + \zeta \sqrt[5]S + \zeta^2 \sqrt[5]T + \zeta^3 \sqrt[5]U + \zeta^4 \sqrt[5]V)}\ ,\\
%
c_3 &=&&\frac 15(1 + \zeta^2 s + \zeta^4 t + \zeta u + \zeta^3 v)
&&\color{gray}{=\frac 15(1 + \zeta^2 \sqrt[5]S + \zeta^4 \sqrt[5]T + \zeta \sqrt[5]U + \zeta^3 \sqrt[5]V)}\ ,\\
%
c_4 &=&&\frac 15(1 + \zeta^3 s + \zeta t + \zeta^4 u + \zeta^2 v)
&&\color{gray}{=\frac 15(1 + \zeta^3 \sqrt[5]S + \zeta \sqrt[5]T + \zeta^4 \sqrt[5]U + \zeta^2 \sqrt[5]V)}\ ,\\
%
c_5 &=&&\frac 15(1 + \zeta^4 s + \zeta^3 t + \zeta^2 u + \zeta v)
&&\color{gray}{=\frac 15(1 + \zeta^4 \sqrt[5]S + \zeta^3 \sqrt[5]T + \zeta^2 \sqrt[5]U + \zeta \sqrt[5]V)}\ ,
\end{aligned}
$$
or in a compact, structural writing:
$$
\begin{bmatrix}
c_1\\c_2\\c_3\\c_4\\c_5
\end{bmatrix}
=
\frac 15
\underbrace{
\begin{bmatrix}
1 & 1 & 1 & 1 & 1\\
1 & \zeta^{1\cdot 1}& \zeta^{1\cdot 2}& \zeta^{1\cdot 3}& \zeta^{1\cdot 4}\\
1 & \zeta^{2\cdot 1}& \zeta^{2\cdot 2}& \zeta^{2\cdot 3}& \zeta^{2\cdot 4}\\
1 & \zeta^{3\cdot 1}& \zeta^{3\cdot 2}& \zeta^{3\cdot 3}& \zeta^{3\cdot 4}\\
1 & \zeta^{4\cdot 1}& \zeta^{4\cdot 2}& \zeta^{4\cdot 3}& \zeta^{4\cdot 4}\\
\end{bmatrix}}_{=:A(\zeta)}
\begin{bmatrix}
1\\s\\t\\u\\v
\end{bmatrix}
$$
The inverse of the matrix $A(\zeta)$ is up to a factor a matrix of the same shape,
in fact, if we formally (Galois) substitute $\zeta^{-1}$ instead of $\zeta$ in $A(\zeta)$, and compute
$A(\zeta)A(\zeta^{-1})$, then we obtain five times the unit matrix, so we have direct access to $s,t,u,v$
in the form:
$$
\underbrace{
\begin{bmatrix}
1 & 1 & 1 & 1 & 1\\
1 & \zeta^{-1\cdot 1}& \zeta^{-1\cdot 2}& \zeta^{-1\cdot 3}& \zeta^{-1\cdot 4}\\
1 & \zeta^{-2\cdot 1}& \zeta^{-2\cdot 2}& \zeta^{-2\cdot 3}& \zeta^{-2\cdot 4}\\
1 & \zeta^{-3\cdot 1}& \zeta^{-3\cdot 2}& \zeta^{-3\cdot 3}& \zeta^{-3\cdot 4}\\
1 & \zeta^{-4\cdot 1}& \zeta^{-4\cdot 2}& \zeta^{-4\cdot 3}& \zeta^{-4\cdot 4}\\
\end{bmatrix}}_{=A(\zeta^{-1})}
\begin{bmatrix}
c_1\\c_2\\c_3\\c_4\\c_5
\end{bmatrix}
=
\begin{bmatrix}
1\\s\\t\\u\\v
\end{bmatrix}
$$
So far, we have cheaply complicated the story. Why do we introduce
$s,t,u,v$ in a complicated manner, when the conjugated cos-values are that simple?
Here is the reason.
Let us consider $S=s^5$, $T=t^5$, $U=u^5$, $V=v^5$.
Then $S,T,U,V$ are the roots of the polynomial of degree four:
$$
(X^4 - 979X^3 + 467181X^2 - 157668929X + 25937424601)=$$
$$(X^4 - 11\cdot 89 X^3 + 3^3\cdot11^3\cdot13 X^2 - 11^6\cdot 89 X + 11^{10})\ .
$$
(All the above depends on a good order of the cosine related values $c_1,c_2,c_3,c_4,c_5$.)
Degree four is good news when formulas involving radicals are needed.
And we have an explicit formula for these roots, simpler than the general one - for Galois reasons,
since there is no degree $3$ field extension involved.
They come in two pairs of conjugated complex numbers:
$$
S,T,U,V \qquad=\qquad \frac {11}4(89 \pm 25\sqrt 5)\qquad\boxed{\pm}\qquad i\frac {55}4\sqrt{2(205\mp 89\sqrt 5)}\ ,
$$
where the signs $\pm$ and $\mp$ correspond in their choice (if the first one is plus, the second one is minus,
and conversely), and the sign $\boxed{\pm}$ is chosen independently.
The four numbers $S,T,U,V$ have each the absolute value $11^{5/2}$.
Now we extract the fifth root from each of the above numbers,
and obtain the four numbers $s,t,u,v$, each one has the absolute value $11^{1/2}=\sqrt{11}$.
It turns out that $stuv=11^2$.
So we have some constraints, after choices and a possible renotation we even have / can take $t=\bar s=11/s$, $v=\bar u=11/u$.
Choices of $s,u$ with $su\ne 11$ determine the situation. We will try all possiblities, but only in some cases we will have a match.
Here is the computer check, there are $5^2=25$ cases computed numerically, and $5$ of them lead to wanted cosine values:
a = sqrt(5)
S = 11/4(89 + 25a) + i55/4sqrt(2(205 - 89a))
U = 11/4(89 - 25a) + i55/4sqrt(2(205 + 89a))
s = S.n()^(1/5)
t = s.conjugate()
u = U.n()^(1/5)
v = u.conjugate()
units5 = [ exp(2pii*k/5).n() for k in [0..4] ]
for z5, w5 in cartesian_product([ units5, units5 ]):
s1, t1, u1, v1 = sz5, t/z5, uw5, v/w5
c = ( 1 + s1 + t1 + u1 + v1 ).real() / 10. # potential cos value
if c < -1 or c > 1:
continue
angle_denominator = arccos(c) / pi.n() * 11
k = round(angle_denominator, 0)
if abs(angle_denominator - k) < 1e-5:
print(f'Radical expression = {c:18.15f} :: cos({k}pi/11) = {cos(k*pi/11).n():18.15f}...')
This delivers:
Radical expression = 0.142314838273285 :: cos(5pi/11) = 0.142314838273285...
Radical expression = 0.654860733945285 :: cos(3pi/11) = 0.654860733945285...
Radical expression = -0.415415013001886 :: cos(7pi/11) = -0.415415013001886...
Radical expression = -0.841253532831181 :: cos(9pi/11) = -0.841253532831181...
Radical expression = 0.959492973614497 :: cos(1pi/11) = 0.959492973614497...
This concludes the case $p=11$.
The case $p=19$:
Let $u$ be the complex number
$$
u
=
\exp \frac {\pi i}{19}
=
\cos \frac {\pi}{19}
+i
\sin \frac {\pi}{19}
\ .
$$
It satisfies the equation $u^{19}=\exp(\pi i)=-1$,
so it is a root of the polynomial of degree $18=2\cdot 3 ²$:
$$
q = X^{18} - X^{17} + \dots + X^4 - X^3 + X^2 - X + 1\ .
$$
The Galois group of the related cyclotomic field is cyclic and abelian,
so we expect a tower of subfields corresponding to the subgroups of $\Bbb Z/18$.
We expect that
$$
c = 2\cos\frac \pi{19}
$$
satisfies an equation of degree $9=18/2$. This is obtained via...
sage: c = 2*cos(pi/19)
sage: c.minpoly()
x^9 - x^8 - 8*x^7 + 7*x^6 + 21*x^5 - 15*x^4 - 20*x^3 + 10*x^2 + 5*x - 1
sage: K.<c> = NumberField( (2cos(pi/19)).minpoly() )
sage: L, _, _ = K.subfields(degree=3)[0]
sage: L
Number Field in c0 with defining polynomial x^3 - x^2 - 6x + 7
And that we can split the situation, two extensions of degree three are needed "only".
We have in a picture for $K=K_9:=\Bbb Q(c)$, the field generated by $c$,
the following
tower of subfields...
$$
\require{AMScd}
\color{darkgreen}{
\begin{CD}
K=K_9=\Bbb Q(c)\\
@VV :3 V\\
K_3=\Bbb Q(g) @. \qquad\text{ where $g$ is a root of }x^3 - x^2 - 6x + 7\in\Bbb Q[x]\\
@VV :3 V\\
\Bbb Q
\end{CD}}
$$
And from here we have to get the possible values for $g$,
then solve the corresponding equation(s) of degree three with coefficients in $K_3$
to obtain $c$.
These last equations can be obtained by factorizing the minimal polynomial of $c$
over $K_3$:
sage: P = ( 2*cos(pi/19) ).minpoly()
sage: Q = x^3 - x^2 - 6*x + 7
sage: K.<g> = NumberField(Q)
sage: RK.<X> = PolynomialRing(K)
sage: factor(P(X))
(X^3 - g *X^2 + ( g^2 - 5)*X - g^2 + 6)
* (X^3 + (-g^2 + 4)*X^2 + (-g^2 - g + 4)*X + g^2 + g - 3)
* (X^3 + ( g^2 + g - 5)*X^2 + ( g - 1)*X - g + 2)
The factorization was manually adjusted.
This concludes the case $p=19$, one could say more, but...
The case $p=23$:
Let $u$ be the complex number
$$
u
=
\exp \frac {\pi i}{23}
=
\cos \frac {\pi}{23}
+i
\sin \frac {\pi}{23}
=
\cos \frac {2\pi}{46}
+i
\sin \frac {2\pi}{46}
\ .
$$
It is a cyclotomic unit of order $46$,
and satisfies the equation $u^{23}=\exp(\pi i)=-1$,
so it is a root of the polynomial of degree $23-1=22=2\cdot 11$:
$$
q = X^{22} - X^{21} + \dots + X^4 - X^3 + X^2 - X + 1\ .
$$
The Galois group of the related cyclotomic field is cyclic and abelian,
so we expect a tower of subfields corresponding to the subgroups of $\Bbb Z/22$.
We expect that
$$
c = 2\cos\frac \pi{23} = u + \frac 1u = u^1 + u^{22}
$$
satisfies an equation of degree $22/2=11$. This is obtained as above in sage,
$c$ and its conjugates are roots of the polynomial:
$$
P = x^{11} - x^{10} - 10 x^{9} + 9 x^{8} + 36 x^{7} - 28 x^{6} - 56 x^{5} + 35 x^{4} + 35 x^{3} - 15 x^{2} - 6 x + 1
\ .
$$
The roots of the above polynomials are
$$
c_k = 2\cos\frac {k\pi}{23} = u^k + u^{23-k}\ ,
$$
where $k$ takes the half of the values between $1$ and $22$,
we can restrict for instance to the odd ones. (We have $c_1=c_{22}$, $c_2=c_{21}$, ...)
We expect again for Galois theoretical reasons that solutions are of the shape
$$
\frac 1{11}
\Big(\ 1 + s_1 + s_2 + \dots s_{10} \ \Big)
=
\frac 1{11}
\Big(\ 1 + S_1^{1/11} + S_2^{1/11} + \dots S_{10}^{1/11} \ \Big)
\ ,
$$
with suitable, inter-coordinated, matching choices for the many radicals $\displaystyle s_k=\sqrt[11]{S_k}$
and we have an equation of degree $10$ over $\Bbb Q$ for the numbers
$S_1,S_2,\dots,S_{10}$ to solve only.
This time we really need Galois theory.
The number $c_1=c$ generates a field of degree $11$, a prime degree, over $\Bbb Q$.
Its conjugates are $c_k$, $k=1,3,5,7,9,11,13,15,17,19$,
and there is a Galois substitution $\tau$ over $\Bbb Q$
permuting them.
One can take for $\tau$ the map determined by $u\to u^g$, where $g\in(\Bbb Z/23)^\times$
has order $11$, i will use $g=3$ below.
Since in such cases it is useful to have also
the roots of unity of degree $11$, we will work
with the following extension of fields:
$$
L=\Bbb Q(c,\zeta)
\text{ over }
K=\Bbb Q(\zeta)
\ .
$$
Here,
we adjoin the root of unity of degree $11$,
$\zeta=\zeta_{11}$.
The extension of $\tau$ as a Galois substitution of $L$,
trivially action on $\zeta$ will be denoted also by $\tau$.
Then $\zeta\in K$ has norm one:
$$
N_{L/K}(\zeta)
=\prod_{k}\tau^k(\zeta)
=\prod_{k}\zeta
=1\ .
$$
Hilbert 90 insures the existence of some $b\in L$ with
$$
\zeta=\frac{b}{\tau(b)}\ .
$$
In fact, this existence is constructive, so let us consider
(the Lagrange rezolvent)
$$
b =
c
+ \zeta\tau(c)
+ \zeta^2\tau^2(c)
+ \dots
+ \zeta^{10}\tau^{10}(c)
\ .
$$
Then
$$
\frac{b}{\tau(b)}
=
\frac{c
+ \zeta\tau(c)
+ \zeta^2\tau^2(c)
+ \dots
+ \zeta^{10}\tau^{10}(c)
}{
\tau(c)
+ \zeta\tau^2(c)
+ \zeta^2\tau^3(c)
+ \dots
+ \zeta^{10}\tau^{11}(c)
}
=\zeta
\ .
$$
Then $b^{11}$ is fixed by $\tau$, because of
$$
\frac{b^{11}}{\tau(b^{11})} =
\frac{b^{11}}{\tau(b)^{11}} =
\left(\frac{b}{\tau(b)}\right)^{11} =
\zeta^{11}
=1\ .
$$
So it is an element of $K$, and we can proceed by building the $11$.th root of
this element.
Computations in sage deliver this polynomial:
p, q = 23, 11
F.<Z> = CyclotomicField( 2*p*q )
u, z = Z^q, Z^(2*p)
g = 3
b = sum([ (u^(g^j) + 1/u^(g^j)) * z^j for j in range(q) ])
mp = (b^q).minpoly()
print(f'g = {g} :: obtained minimal polynomial of degree {mp.degree()}\n{mp}')
We obtain:
g = 3 :: obtained minimal polynomial of degree 10
x^10 + 52918009*x^9 + 1085397499666778*x^8
+ 29708537267066813398222*x^7
+ 359857016915639527522755372296*x^6
- 16252543266366411222903696822416826813*x^5
+ 342875277171018431617211044718085541208366392*x^4
+ 26970789640316423479491732431077295067790114290557038*x^3
+ 938874200966107161710606291340323350904334419758675374826774*x^2
+ 43614237139618047406820949477941119609397648573092468104280673454169*x
+ 785291652424037263548733517675617858730558099314688805476989869947753298407
(The result was broken manually.)
The free coefficient is $23^{55}$.
The coefficients in $x^9$ and $x^1$ are $23 \cdot 53 \cdot 43411$ and $23^{45} \cdot 53 \cdot 43411$.
The coefficients in $x^8$ and $x^2$ are
$23^3 \cdot 2\cdot 20693 \cdot 2155519$
and
$23^{36} \cdot 2\cdot 20693 \cdot 2155519$.
Similar coincidences apply for the further coefficients.
From the point of view of the theory of radical extensions we are already done here.
This polynomial splits in two conjugated factors if we adjoin $\sqrt{-11}$,
and then we have to get "explicitly" the roots of a polynomial of degree five...
(The above polynomial splits in linear factors over $\Bbb Q(\zeta_{11})$,
but the roots are not "easy", and the knowledge of an explicit expression for $\zeta_{11}$
in terms of radicals is useless.)
Things may be computed "explicitly" further, but this is not the best place to do it,
and even if the job is done till the final end,
i do not see a direct purpose for the obtained expressions... So let me stop here,
i hope the ideas, and the computational framework can be identified and applied,
when a "purpose" is present.
