Exact values of $\cos(2\pi/7)$ and $\sin(2\pi/7)$ What are the exact values of $\cos(2\pi/7)$ and $\sin(2\pi/7)$ and how do I work it out?
I know that $\cos(2\pi/7)$ and $\sin(2\pi/7)$ are the real and imaginary parts of $e^{2\pi i/7}$ but I am not sure if that helps me...    
 A: 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).
A: $\cos2\pi/7$ 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. 
A: There is a little problem if you want to express $\cos(2\pi/7)$ in terms of radicals. As Gerry Myerson wrote, it is a root of a degree $3$ polynomial ($e^{2\pi i /7}$ is a root of $x^6+x^5+\dots+1=0$, $\cos(2\pi/7)=y=(x+x^{-1})/2$, which gives $8 y^3+4 y^2-4 y-1 = 0$). That polynomial ($8 y^3+4 y^2-4 y-1$) is irreducible (over $\mathbb{Q}$) and has real roots (namely $\cos(2\pi/7),\cos(4\pi/7),\cos(6\pi/7)$). There is a famous theorem (casus irreducibilis) saying that the roots of the polynomial cannot be expressed using real radicals. So you will need complex numbers in your formula for $\cos(2\pi/7)$. 
(Cardano's formula gives
$$y= -1/6+7^{2/3}/(3\times 2^{2/3} (1+3 i \sqrt{3})^{1/3})+ (7/2\times (1+3 i \sqrt{3}))^{1/3}/6 $$
- computed by Wolfram Alpha) 
A: Are you sure there is an "exact" value? Well, it depends on what you mean by exact. My point is that I don't think heptagons are constructible with ruler and compass, which means, if I remember correctly, that the sine and cosine cannot be expressed as sum of fractions and square roots of fractions.
I know $\cos\dfrac{2\pi}{17}$ is a known value, maybe that's what you meant.
A: I arrived at an equation (third degree polynomial below):
$$\binom71 x^3 - \binom73 x^2 + \binom75 x - \binom77 = 0\;.$$
The zeroes are  $ x_1= \cot (\pi/7)^2,  x_2= \cot (2\pi/7)^2$ and $x_3=\cot(3\pi/7)^2$.
I hope this will help. I arrived at an answer to this
problem after solving the above cubic polynomial. This took
some time, but the answer was not very messy.
A: I recently figured a different way of getting to the cubic polynomial. You can see that the three roots $A$, $B$, and $C$ of the cubic must correspond to
$$\begin{align*}
w^1 + w^6  &= A\\
w^2 + w^5  &= B\\
w^4 + w^3  &= C
\end{align*}$$
where $w$ is the seventh root of unity...no, to just "the" seventh root of unity, but any non-trivial seventh root of unity. You can also see that that the group of automorphisms of the splitting field is the cyclic group on three elements. How? because if you adjoin any one of $A$, $B$, or $C$ to the rationals, you get a group of order three which contains all of $A$, $B$, and $C$. In particular it is easy to see that:
$$\begin{align*}
A^2 - 2 &= B\\
B^2 - 2 &= C\\
C^2 - 2 &= A
\end{align*}$$
If we substitute for $A$ recursively in these three equations, we get the eight-degree equation:
$$A^8 - 8A^6 + 20A^4 - 16A^2 + 2 = A$$
Why eighth degree? Because we have perhaps forgotten that the recursive relation is also satisfied if we allow $A^ -2 = A$. So we use synthetic division to divide out these "trivial" solutions and we then get the unlikely sixth-degree equation:
$$A^6 + A^5 - 5A^4 - 3A^3 + 7A^2 + A - 1 = 0$$
This sixth degree equation factors into two third degree equations. One of those third degree equations contains the three cosines of interest as its solution.  The other has a different set of cosine solutions, namely $2\cos(2\pi m/9)$ where $m = 1$, $2$, or $4$.  If, for instance, we select $A = 2\cos(2\pi/9)$, then using the double angle formula for cosine and the above definitions we get $B = 2\cos(4\pi/9)$, then $C = 2\cos(8\pi/9)$, then $A = 2\cos(16\pi/9) = 2\cos(2\pi/9)$ because $$\frac{16\pi}{9} + \frac{2\pi}9 = 2\pi.$$
