Exact values of $\cos(2\pi/7)$ and $\sin(2\pi/7)$
There are various ways of construing and attacking your question.
At the most basic level: it's no problem to write down a cubic polynomial satisfied by $\alpha = \cos(2 \pi/7)$ and hit it with Cardano's cubic formula. For instance, if we put $z = \zeta_7 = e^{2 \pi i/7}$, then $2\alpha = z + \overline{z} = z + \frac{1}{z}$. A little algebra leads to the polynomial $P(t) = t^3 + \frac{1}{2} t^2 - \frac{1}{2}t - \frac{1}{8}$ which is irreducible with $P(\alpha) = 0$. (Note that the noninteger coefficients of $P(t)$ imply that $\alpha$ is not an algebraic integer. In this respect, the quantity $2 \alpha$ is much better behaved, and it is often a good idea to work with $2 \alpha$ instead of $\alpha$.) To see what you get when you apply Cardano's formula, consult the other answers or just google for it: for instance I quickly found this page, among many others (including wikipedia) which does it.
The expression is kind of a mess, which gives you the idea that having these explicit radical expressions for roots of unity (and related quantities like the values of the sine and cosine) may not actually be so useful: if I wanted to compute with $\alpha$ (and it has come up in my work!) I wouldn't get anything out of this formula that I didn't get from $2 \alpha = \zeta_7 + \zeta_7^{-1}$ or the minimal polynomial $P(t)$.
On the other hand, if you know some Galois theory, you know that the Galois group of every cyclotomic polynomial is abelian, so there must exist a radical expression for $\zeta_n$ for any $n \in \mathbb{Z}^+$. (We will usually not be able to get away with only repeatedly extracting square roots; that could only be sufficient when Euler's totient function $\varphi(n)$ is a power of $2$, for instance, so not even when $n = 7$.) From this perspective, applying the cubic formula is a big copout, since there is no analogous formula in degree $d > 4$: the general polynomial of such a degree cannot be solved by radicals...but cyclotomic polynomials can.
So what do you do in general? The answer was known to Gauss, and involves some classical algebra -- resolvents, Gaussian periods, etc. -- that is not very well remembered nowadays. In fact I have never gone through the details myself. But I cast around on the web for a while looking for a nice treatment, and I eventually found this writeup by Paul Garrett. I recommend it to those who want to learn more about this (not so useful, as far as I know, but interesting) classical problem: his notes are consistently excellent, and have the virtue of concision (which I admire especially for lack of ability to produce it myself).
case = {7},
first, set $\zeta^k_7$ as seventh roots of unity, using both Euler and De Moiver Formula, $\zeta^{k}_{7}=\text{e}^{\frac{2ki\pi}{7}}=\cos\frac{2k\pi}{7} \pm i\sin\frac{2k\pi}{7}$ hence $\zeta^{-k}_{7}=\text{e}^{\frac{2ki\pi}{7}}=\cos\frac{2k\pi}{7} \mp i\sin\frac{2k\pi}{7}$ hence $$\zeta^{k}_{7}+\zeta^{-k}_{7}=2\cos\frac{2k\pi}{7}$$ $$\zeta^{k}_{7}-\zeta^{-k}_{7}=2i\sin\frac{2k\pi}{7}$$ $$\zeta^{2k}_{7}-2+\zeta^{-2k}_{7}=-4\sin\frac{2k\pi}{7}$$ $$-4\sin\frac{2k\pi}{7}=2\cos\frac{4k\pi}{7}-2$$ $$2i\sin\frac{2k\pi}{7}=\sqrt{2\cos\frac{4k\pi}{7}-2}$$ and some certain properties of $7^{th}$ roots of unity, $$\zeta^{7}_{7}=1$$ $$\zeta^{k}_{7}=\zeta^{k\text{mod}7}_{7}$$ $$\zeta^{-k}_{7}=\zeta^{7-k}_{7}$$ and sum of all $n^{th}$ (where $n$ is more than $1$) roots of unity is $0$, so we can write $$\zeta^{1}_{7}+\zeta^{2}_{7}+\zeta^{3}_{7}+\zeta^{4}_{7}+\zeta^{5}_{7}+\zeta^{6}_{7}+1=0$$ move 1 $$\zeta^{1}_{7}+\zeta^{2}_{7}+\zeta^{3}_{7}+\zeta^{4}_{7}+\zeta^{5}_{7}+\zeta^{6}_{7}=-1$$ hence $$\zeta^{1}_{7}+\zeta^{2}_{7}+\zeta^{3}_{7}+\zeta^{-3}_{7}+\zeta^{-2}_{7}+\zeta^{-1}_{7}=-1$$ set $$t_{k}=\zeta^{k}_{7}+\zeta^{-k}_{7}=2\cos\frac{2k\pi}{7}$$ this simplifies $$t_1+t_2+t_3=-1$$ multiply $t_1$ with $t_2$ , remember $\zeta^{k}_{7}=\zeta^{k\text{mod}7}_{7}$ $$t_{1}=\left(\zeta^{1}_{7}+\zeta^{6}_{7}\right)$$ $$t_{2}=\left(\zeta^{2}_{7}+\zeta^{5}_{7}\right)$$ $$t_{1} t_{2}=\zeta^{3}_{7}+\zeta^{6}_{7}+\zeta^{1}_{7}+\zeta^{4}_{7}$$ $$t_{1} t_{2}=t_{1}+t_{3}$$ similarly $$t_{1} t_{3}=t_{3}+t_{2}$$ $$t_{2} t_{3}=t_{2}+t_{1}$$ so $$t_{1} t_{2}+t_{1} t_{3}+t_{2} t_{3}=-2$$ then, find $t_{1} t_{2} t_{3}$ $$t_{1} t_{2}=t_{1}+t_{3}$$ since $t_1+t_2+t_3=-1$ $$t_{1} t_{2}=-1-t_{2}$$ $$t_{1} t_{2} t_{3}=\left(-1-t_{2}\right)t_{3}$$ $$t_{1} t_{2} t_{3}=-t_{3}-t_{2}t_{3}$$ $$t_{1} t_{2} t_{3}=-t_{3}-t_{2}-t_{1}$$ $$t_{1} t_{2} t_{3}=1$$ then, $$t_1 + t_2 + t_3 = -1$$ $$t_2 + t_3 = -1 -t_1$$ $$t_1 t_2 t_3 = 1$$ $$t_2 t_3 = \frac {1}{t_1} = t^{-1}_1$$ $$t_1 t_2 + t_1 t_3 + t_2 t_3 +2 = 0$$ $$t_1 ( t_2 + t_3 ) + t_2 t_3 +2 = 0$$ $$t_1 ( -1 -t_1 ) + t^{-1}_1 +2 = 0$$ $$-t^2_1 -t_1 + 2 + t^{-1}_1 = 0$$ $$t^3_1 + t^2_1 -2t_1 -1 = 0$$ this statement also applies for other $t_k$ $$t^3+t^2-2-1=0$$ back to the statement that we've set $$t_{k}=2\cos\frac{2k\pi}{7}$$ the roots of the cubic polynomial $$t^3+t^2-2-1=0$$ have the roots cosine, applying Lagrange Resolvent ( if you want a hard way ) or Tartaglia Formula,
$$2\cos\frac{2\pi}{7}=\frac{-1+\sqrt[3]{\frac{7}{2}\left(1+3\sqrt{-3}\right)}+\sqrt[3]{\frac{7}{2}\left(1+3\sqrt{-3}\right)}}{3}$$ $$2\cos\frac{4\pi}{7}=\frac{-1+\frac{-1-\sqrt{-3}}{2}\sqrt[3]{\frac{7}{2}\left(1+3\sqrt{-3}\right)}+\frac{-1+\sqrt{-3}}{2}\sqrt[3]{\frac{7}{2}\left(1+3\sqrt{-3}\right)}}{3}$$ $$2\cos\frac{6\pi}{7}=\frac{-1+\frac{-1+\sqrt{-3}}{2}\sqrt[3]{\frac{7}{2}\left(1+3\sqrt{-3}\right)}+\frac{-1-\sqrt{-3}}{2}\sqrt[3]{\frac{7}{2}\left(1+3\sqrt{-3}\right)}}{3}$$ and, notice $t_{k}=\zeta^{k}_{7}+\zeta^{-k}_{7}$ $t_{-k}=\zeta^{-k}_{7}+\zeta^{k}_{7}$ so $t_{1}=t_{1} t_{2}=t_{2} t_{3}=t_{3} t_{4}=t_{3} t_{5}=t_{2} t_{6}=t_{1}$ $$\zeta^{2k}_{7}-2+\zeta^{-2k}_{7}=-4\sin\frac{2k\pi}{7}$$ so $$t_{2k}-2=-4\sin\frac{2k\pi}{7}$$
$$\sqrt{t_{2k}-2}=2i\sin\frac{2k\pi}{7}$$
$$\sqrt{t_{2}-2}=2i\sin\frac{2\pi}{7}$$ $$\sqrt{t_{3}-2}=2i\sin\frac{4\pi}{7}$$ $$\sqrt{t_{1}-2}=2i\sin\frac{6\pi}{7}$$ substitute each $t_{k}$ $$\sqrt{t_{2}-2}=\sqrt{\frac{-1+\frac{-1-\sqrt{-3}}{2}\sqrt[3]{\frac{7}{2}\left(1+3\sqrt{-3}\right)}+\frac{-1+\sqrt{-3}}{2}\sqrt[3]{\frac{7}{2}\left(1+3\sqrt{-3}\right)}}{3}-2}$$ $$\sqrt{t_{2}-2}=\sqrt{\frac{-7+\frac{-1-\sqrt{-3}}{2}\sqrt[3]{\frac{7}{2}\left(1+3\sqrt{-3}\right)}+\frac{-1+\sqrt{-3}}{2}\sqrt[3]{\frac{7}{2}\left(1+3\sqrt{-3}\right)}}{3}}$$ $$\sqrt{t_{2}-2}=\sqrt{\frac{3\left(-7+\frac{-1-\sqrt{-3}}{2}\sqrt[3]{\frac{7}{2}\left(1+3\sqrt{-3}\right)}+\frac{-1+\sqrt{-3}}{2}\sqrt[3]{\frac{7}{2}\left(1+3\sqrt{-3}\right)}\right)}{9}}$$ $$\sqrt{t_{2}-2}=\frac{\sqrt{3\left(-7+\frac{-1-\sqrt{-3}}{2}\sqrt[3]{\frac{7}{2}\left(1+3\sqrt{-3}\right)}+\frac{-1+\sqrt{-3}}{2}\sqrt[3]{\frac{7}{2}\left(1+3\sqrt{-3}\right)}\right)}}{3}$$ $$2i\sin\frac{2\pi}{7}=\frac{\sqrt{3\left(-7+\frac{-1-\sqrt{-3}}{2}\sqrt[3]{\frac{7}{2}\left(1+3\sqrt{-3}\right)}+\frac{-1+\sqrt{-3}}{2}\sqrt[3]{\frac{7}{2}\left(1+3\sqrt{-3}\right)}\right)}}{3}$$ so $$2\sin\frac{2\pi}{7}=\frac{\sqrt{-3\left(-7+\frac{-1-\sqrt{-3}}{2}\sqrt[3]{\frac{7}{2}\left(1+3\sqrt{-3}\right)}+\frac{-1+\sqrt{-3}}{2}\sqrt[3]{\frac{7}{2}\left(1+3\sqrt{-3}\right)}\right)}}{3}$$ $$2\sin\frac{4\pi}{7}=\frac{\sqrt{-3\left(-7+\frac{-1+\sqrt{-3}}{2}\sqrt[3]{\frac{7}{2}\left(1+3\sqrt{-3}\right)}+\frac{-1-\sqrt{-3}}{2}\sqrt[3]{\frac{7}{2}\left(1+3\sqrt{-3}\right)}\right)}}{3}$$ $$2\sin\frac{6\pi}{7}=\frac{\sqrt{-3\left(-7+\sqrt[3]{\frac{7}{2}\left(1+3\sqrt{-3}\right)}+\sqrt[3]{\frac{7}{2}\left(1+3\sqrt{-3}\right)}\right)}}{3}$$ now, we are done with
$$\cos\frac{2\pi}{7}=\frac{-1+\sqrt[3]{\frac{7}{2}\left(1+3\sqrt{-3}\right)}+\sqrt[3]{\frac{7}{2}\left(1+3\sqrt{-3}\right)}}{6}$$ $$\cos\frac{4\pi}{7}=\frac{-1+\frac{-1-\sqrt{-3}}{2}\sqrt[3]{\frac{7}{2}\left(1+3\sqrt{-3}\right)}+\frac{-1+\sqrt{-3}}{2}\sqrt[3]{\frac{7}{2}\left(1+3\sqrt{-3}\right)}}{6}$$ $$\cos\frac{6\pi}{7}=\frac{-1+\frac{-1+\sqrt{-3}}{2}\sqrt[3]{\frac{7}{2}\left(1+3\sqrt{-3}\right)}+\frac{-1-\sqrt{-3}}{2}\sqrt[3]{\frac{7}{2}\left(1+3\sqrt{-3}\right)}}{6}$$ $$\sin\frac{2\pi}{7}=\frac{\sqrt{-3\left(-7+\frac{-1-\sqrt{-3}}{2}\sqrt[3]{\frac{7}{2}\left(1+3\sqrt{-3}\right)}+\frac{-1+\sqrt{-3}}{2}\sqrt[3]{\frac{7}{2}\left(1+3\sqrt{-3}\right)}\right)}}{6}$$ $$\sin\frac{4\pi}{7}=\frac{\sqrt{-3\left(-7+\frac{-1+\sqrt{-3}}{2}\sqrt[3]{\frac{7}{2}\left(1+3\sqrt{-3}\right)}+\frac{-1-\sqrt{-3}}{2}\sqrt[3]{\frac{7}{2}\left(1+3\sqrt{-3}\right)}\right)}}{6}$$ $$\sin\frac{6\pi}{7}=\frac{\sqrt{-3\left(-7+\sqrt[3]{\frac{7}{2}\left(1+3\sqrt{-3}\right)}+\sqrt[3]{\frac{7}{2}\left(1+3\sqrt{-3}\right)}\right)}}{6}$$
$\cos2\pi/7$ is a root of a cubic equation with integer coefficients. You can find that cubic by using $\cos\theta=(1/2)(e^{i\theta}+e^{-i\theta})$, computing the square and the cube, and looking for linear relations, bearing in mind that the $7$ $7$th roots of unity add up to zero. Then you can use Cardano's formula to solve the cubic. I don't know if I recommend actually doing all this - I'm sure you get a mess, although the discriminant will be a perfect square, so you'll get some simplification there.