One can do this problem using Chebyshef polynomials and some computation. But here's a proof using the brutal fork of abstract algebra The primitive $n$-th roots of unity have algebraic degree $\varphi(n)$ (showing this requires us to prove the irreducibility if the cyclotomic polynomials, but I will skip it now). Let $k= \frac mn$ with $m,n$ coprime integers and let $\zeta= e^{\frac{2\pi m i}{n}}$ (a primitive $n$-th root of unity). Since $c:=2\cdot \cos \left( \frac{m}{n} \right) = \zeta + \zeta^{-1}$, we get that $\zeta$ is a root of $x^2-2c x +1=0$. Especially, the dimension $[\mathbb{Q}[\zeta]:\mathbb{Q}[c]]$ is $1$ or $2$. If $n>2$, then $\zeta$ is nonreal, but $c$ is, giving that the degree is in fact $2$. Thus the algebraic degree of $c$ is $\frac{\varphi(n)}2$ if $n \geq 3$, thus $c$ is rational iff $n=1,2,3,4,6$.

So, what is your final answer to the question? For those denominators you have an algebraic number, but the question is when it is rational... or did I misunderstand you?

Read the last line again At the beginning, I showed that it's algebraic and then computed it's algebraic degree.

In $ c: = 2\cdot \cos \left( \frac {m}{n} \right)$ you are probably missing a $ \pi$ there, but I still don't get it, you say $ \cos\left(\frac {1}{6}\pi\right)=\frac{\sqrt{3}}{2}$ is rational?

Sorry, let me kill the little errors: The primitive $ n$-th roots of unity have algebraic degree $ \varphi(n)$ (showing this requires us to prove the irreducibility if the cyclotomic polynomials, but I will skip it now). Let $ k = \frac {2m}n$ with $ m,n$ coprime integers and let $ \zeta = e^{\frac {2\pi m i}{n}}$ (a primitive $ n$-th root of unity). Since $ c: = 2\cdot \cos \left( \frac {2m}{n} \pi \right) = \zeta + \zeta^{ - 1}$, we get that $ \zeta$ is a root of $ x^2 - 2c x + 1 = 0$. Especially, the dimension $ [\mathbb{Q}[\zeta]: \mathbb{Q}[c]]$ is $ 1$ or $ 2$. If $ n > 2$, then $ \zeta$ is nonreal, but $ c$ is real, giving that the degree is in fact $ 2$. Thus the algebraic degree of $ c$ is $ \frac {\varphi(n)}2$ if $ n \geq 3$, thus $ c$ is rational iff $ n = 1,2,3,4,6$.

Now I get it, thanks!

Another proof: Every $ \zeta = e^{\frac {2\pi m i}{n}}$ is an algebraic integer (it is a root of $ x^n-1$). But then $ c: =2\cdot \cos \left( \frac {2m}{n} \pi \right) = \zeta + \zeta^{ - 1}$ is an algebraic integer, too. Thus $ c$ is rational iff it is an integer (as a rational is an algebraic integer iff it is an integer). Especially if $ c$ is rational,then $ c \in \{ -2 ,-1 , 0 , 1 , 2 \}$ as $ -1\leq \cos(x) \leq 1$ for all $ x$. Now checking these cases we get the same as before.