= Solution
Put $z=2\cos(\pi/n)$. The recurrence
$$
S_0(z)=2,\quad S_1(z)=z,\quad S_{k+1}(z)=zS_k(z)-S_{k-1}(z)
$$
defines monic integer <polynomials> and gives $S_k(z)=2\cos(k\pi/n)$. Hence $z$ is an algebraic integer. If $\cos(\pi/n)$ is rational, then the rational algebraic integer $z$ is an integer. Since $-2\leq z<2$, this leaves the values corresponding to
$$
n=1:\cos\pi=-1,\qquad n=2:\cos(\pi/2)=0,\qquad n=3:\cos(\pi/3)=\frac12.
$$
This is the <rational cosine of an integral submultiple of pi> result. For every $n\geq4$, the value lies strictly between $1/2$ and $1$, so it is irrational.
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