If we have a polynomial with rational coefficients $P(x)$ which is not reducible over $\mathbb{Q}$, then $\mathbb{Q}[x]/(P(x))$ is a field. In particular, it contains all roots of $P(x)$ and is the smallest field to do so.
As a simple example, consider the polynomial $x^4-10x^2+1$ (which is the minimal polynomial for $\pm\sqrt{2}\pm\sqrt{3}$). Then $\mathbb{Q}[x]/(x^4-10x^2+1)$ is isomorphic to $\mathbb{Q}(\sqrt{2}+\sqrt{3})$, and if we decide that $x$ represents $\sqrt{2}+\sqrt{3}$ then we can represent the other roots in terms of $x$: first $-x$ is $-\sqrt{2}-\sqrt{3}$, then consider $$a(\sqrt{2}+\sqrt{3})^3+b(\sqrt{2}+\sqrt{3})^2+c(\sqrt{2}+\sqrt{3})+d=\sqrt{2}-\sqrt{3}$$ $$a(11\sqrt{2}+9\sqrt{2})+b(5+2\sqrt{6})+c(\sqrt{2}+\sqrt{3})+d=\sqrt{2}-\sqrt{3}$$ $$11a+c=1, 9a+c=-1$$ so $x^3-10x$ is $\sqrt{2}-\sqrt{3}$ and $-x^3+10x$ is $-\sqrt{2}+\sqrt{3}$.
I would be delighted to know if there is a general method to determine the representation of all other roots of an irreducible $P(x)$ in $\mathbb{Q}[x]/(P(x))$ if $x$ represents a root of $P(x)$, but in case this is not possible I will also mention the specific polynomial I am dealing with, where it comes from, and what I've tried.
We have algebraic expressions for the sine and cosine of 30 and 36 degrees, and we have algebraic expressions for the sine and cosine of the difference, bisection, and trisection of angles, thus cos(1) and sin(1) (I'm using degrees here) are algebraic and have minimal polynomials. These are not necessarily the same polynomial, for instance for 30 degrees they are not the same polynomial, but in this case they are the same polynomial. I will not write it down because it is degree 96, but the easiest way to construct it is not the way I originally did with differences and divisions of angles but rather to factor $x^{180}+1$ and consider the largest factor (the fact that the minimal polynomial for sin(1) and cos(1) divide this is not surprising because $e^\frac{i\pi}{180}$ satisfies this equation).
If $x$ represents sin(1), I want to find the representation for cos(1), but computing in this field is impossible by hand to say the least. The most promising approach I have considered was using the $\sin^2 x + \cos^2 x = 1$ relation, so if $x$ represents sin(1) then $\sqrt{1-x^2}$ represents cos(1). Of course, this is a finite degree extension of $\mathbb{Q}$ so it is not closed under square roots, but we know this particular square root is in the field. And if we consider the Taylor expansion of $\sqrt{1-x^2}=1-\sum_{k=1}^\infty{\frac{C_{k-1}}{2\cdot 4^{k-1}}x^{2k}}$, this would become a degree 95 polynomial if we mod out by the degree 96 minimal polynomial for sin(1), but it seems like this would still be extremely hard to compute, especially because the resulting sums do not appear to be rational.
Is it possible to describe the relationship between the roots of $P(x)$ for $\mathbb{Q}[x]/(P(x))$ in general, and if not is it possible to express cos(1) as a degree 95 polynomial with rational coefficients in terms of sin(1) in particular?
I did not find a useful general connection for Galois Conjugates beyond just being roots of some irreducible polynomial, but I did find a connection for trig functions like we see here.
First of all, there is a small error in the original question: the minimal polynomial for sin and cos of 1 degree is actually degree 48, not degree 96. I'll go into how I constructed it in a bit more detail. We have $\mu_{\cos 6^\circ}(x)=256x^8-448x^6+224x^4-32x^2+1$, and we have $\cos 2x=2\cos^2 x - 1$ and $\cos 3x=4\cos^3 x - 3\cos x$. The minimal polynomial for $\cos 6^\circ$ comes from the minimal polynomials for $\cos 36^\circ$ and $\cos 30^\circ$, as given two polynomials $f$ and $g$ we can find a polynomial $h$ so that $f(a)=g(b)=0$ implies $h(a-b)=0$ as follows: let $F$ and $G$ be companion matrices for $f$ and $g$ (ie $f$ and $g$ are their characteristic polynomials), then $F \otimes I - I \otimes G$ has $h$ as its characteristic polynomial, where $\otimes$ is the Kronecker product. Then finding the minimal polynomial for $\cos 6^\circ$ is simply a matter of factoring this characteristic polynomial (originally I thought it did not factor).
Now before we look at the connections between the roots of $\mu_{\cos 1^\circ}$, let's think about the roots of $\mu_{\cos 6^\circ}$. Obviously this is an even polynomial of degree 8 so we know we can find radical expressions for all of its roots. By graphing it on Desmos and looking at cos and sin for some integer values, or less illuminatingly looking at Wolfram Alpha, we can find the exact forms of the roots:
$$\pm\frac{1}{8}\left(-\sqrt{10-\sqrt{20}}-\sqrt{3}+\sqrt{15}\right) = \pm\frac{1}{4}\sqrt{7-\sqrt{5}-\sqrt{6(5+\sqrt{5})}} = \pm \sin 12^\circ = \pm \cos 78^\circ$$ $$\pm\frac{1}{8}\left(-\sqrt{10-\sqrt{20}}+\sqrt{3}+\sqrt{15}\right) = \pm\frac{1}{4}\sqrt{7+\sqrt{5}-\sqrt{6(5+\sqrt{5})}} = \pm \sin 24^\circ = \pm \cos 66^\circ$$ $$\pm\frac{1}{8}\left(\sqrt{10-\sqrt{20}}-\sqrt{3}+\sqrt{15}\right) = \pm\frac{1}{4}\sqrt{7-\sqrt{5}+\sqrt{6(5+\sqrt{5})}} = \pm \sin 48^\circ = \pm \cos 42^\circ$$ $$\pm\frac{1}{8}\left(\sqrt{10-\sqrt{20}}+\sqrt{3}+\sqrt{15}\right) = \pm\frac{1}{4}\sqrt{7+\sqrt{5}+\sqrt{6(5+\sqrt{5})}} = \pm \sin 84^\circ = \pm \cos 6^\circ.$$
(The radical expressions wolfram alpha gives are the ones listed second. The ones listed first are from this page on Wikipedia and are nicer since they are de-nested, although the wolfram alpha results more clearly reflect how these solutions are the square roots of the solutions of a degree 4 equation.)
The first thing we might notice about these roots is the positive ones are sin values of angles which double from 12 degrees up to 96 degrees. However, the multiple sin formula involves cosines so this is not necessarily helpful.
However, $42 = 7\cdot 6$, $66 = 11\cdot 6$, and $78 = 13\cdot 6$, so we can use the multiple cos formula:
$$\cos n x = \sum_{k\, \mathrm{even}}{(-1)^\frac{k}{2}\binom{n}{k}\cos^{n-k}x\sin^k x}$$ $$\cos n x = \sum_{k\, \mathrm{even}}{(-1)^\frac{k}{2}\binom{n}{k}\cos^{n-k}x(1-cos^2 x)^\frac{k}{2}}.$$
This gives us a great way to represent $\cos 6z^\circ$ as a degree z polynomial of $\cos 6^\circ$, and therefore if we take $x$ to be $\cos 6^\circ$ we can express all the roots of $\mu_{\cos 6^\circ}$ as polynomials of $x$ of degree 7 or less as follows: $$\pm (4096x^{13} - 13312x^{11} + 16640x^9 - 9984x^7 + 2912x^5 - 364x^3 + 13x)$$ $$\pm (1024x^{11} - 2816x^9 + 2816x^7 - 1232x^5 + 220x^3 - 11x)$$ $$\pm (64x^7 - 112x^5 + 56x^3 - 7x)$$ $$\pm x.$$
(Note that I did not actually reduce the top two modulo $\mu_{\cos 6^\circ}$ so they appear to be degree 13 and 11 but the reduced forms are degree 7. Also, the pattern where only odd terms appear would still hold.)
Let's move on to consider $\cos 1^\circ$ and $\sin 1^\circ$. We know the minimal polynomial for both is $\mu_{\cos 1^\circ}(x) = \mu_{\cos 6^\circ}(2(4x^3-3x)^2-1)$ (this must be irreducible because a divisor of a composition of $f$ and $g$ would lead to a divisor of $f$). Also, we know that $\sin 1^\circ = \cos 89^\circ$ so we have a degree 89 polynomial representation of $\sin 1^\circ$ in terms of $\cos 1^\circ$ by the multiple cos formula. All we need to do to finish up is reduce this degree 89 polynomial modulo $\mu_{\cos 1^\circ}(x)$ to get a degree 47 or less polynomial. I used Sympy to carry out these computations and found that if we take $x$ to be $\cos 1^\circ$ then $\sin 1^\circ$ is $$-1073741824x^{31}+8589934592x^{29}-31071404032x^{27}+67175972864x^{25}-96678707200x^{23}+97626619904x^{21}-71031324672x^{19}+37639421952x^{17}-14510751744x^{15}+4021944320x^{13}-782559232x^{11}+102742016x^9-8542976x^7+404992x^5-9020x^3+60x$$ which we can verify numerically. Once I thought to use the multiple cos formula after looking at the graph for $\mu_{\cos 6^\circ}$, rather than messing around with trying to pry the coefficients out of the quotient field, things became much easier. Here is my Python code:
This is useful for working with sin and cos of rational degree values exactly, since we can represent these trig functions as polynomials in the quotient field for the minimal polynomial of the cos of the smallest degree value we care about (in this case $\cos 1^\circ$). For rational degree values that arise just from dividing 3 degrees into halves or thirds some number of times we could technically do this with a deeply nested radical representation but it would be much messier.