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Let $n\ge 4$ postive integer,show that $$\cos{\dfrac{\pi}{n}}\notin\mathbb Q$$

Now I have solve for a case:

Assmue that $$\cos{\dfrac{\pi}{n}}=\dfrac{q}{p},(p,q)=1,p,q\in N^{+}$$ use Chebyshev polynomials? $$T_{n}(\cos{x})=\cos{(nx)}$$ so we have $$T_{n}\left(\dfrac{q}{p}\right)=-1$$ then we have $$2^{n-1}\left(\dfrac{q}{p}\right)^n+a_{n-1}\left(\dfrac{q}{p}\right)^{n-1}+\cdots+a_{0}+1=0$$ so we have $$q|\le a_{0}+1$$ since $|a_{0}|=0,1,-1$,

(1):$a_{0}\neq -1$,so we have $$\dfrac{q}{p}\le\dfrac{2}{3}$$,but $$\cos{\dfrac{\pi}{n}}\ge\cos{\dfrac{\pi}{4}}=\dfrac{\sqrt{2}}{2}>\dfrac{2}{3}$$ contradiction。

(2):But for $a_{0}=-1$, I can't prove it

Kenta S
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math110
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3 Answers3

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Proof 1. Write $\zeta = \exp \left( \frac{ \pi i}{n} \right)$ (a primitive $2n$-th root of unity), so that $\zeta + \zeta^{-1} = 2 \cos \frac{\pi}{n}$ is an algebraic integer, and hence is rational iff it's an integer (e.g. by the rational root theorem). So $\cos \frac{\pi}{n}$ is rational iff it's half an integer, and the only possibilities are $\{ 0, \pm \frac{1}{2}, \pm 1 \}$ which gives $n = 1, 2, 3$.

Proof 2. $\zeta + \zeta^{-1}$ lies in the cyclotomic field $\mathbb{Q}(\zeta)$, which has Galois group $(\mathbb{Z}/2n)^{\times}$ acting via $\zeta \mapsto \zeta^k, k \in (\mathbb{Z}/2n)^{\times}$. It follows that the Galois conjugates of $\zeta + \zeta^{-1}$ are

$$\zeta^k + \zeta^{-k} = 2 \cos \frac{\pi k}{n}, k \in (\mathbb{Z}/2n)^{\times}$$

which are distinct as long as $0 < \frac{\pi k}{n} < \pi$, or equivalently $0 < k < n$, so there are $\frac{\varphi(2n)}{2}$ of them (unless $n = 1$ which is a degenerate case). For $\zeta + \zeta^{-1}$ to be rational this number must be equal to $1$, so $\varphi(2n) = 2$. From an inspection of the prime factorization of $2n$ (which can only contain $2$ and $3$) this only occurs when $2n = 4, 6$ and hence when $n = 2, 3$.

We can also argue that $\mathbb{Q}(\zeta + \zeta^{-1})$ is the fixed field of complex conjugation and so a subfield of index $2$ and order $\frac{\varphi(2n)}{2}$ in $\mathbb{Q}(\zeta)$ (which follows from the Galois correspondence).

Qiaochu Yuan
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The roots of Chebyshev polynomials $U_{n}$ of the second kind are of the form $\cos{\frac{k\pi}{n+1}}$ for $k=1,\dots,n$, so $\cos{\frac{\pi}{n}}$ is a root of $U_{n-1}$.

Assume that $\cos{\frac{\pi}{n}}$ is rational and of the form $p/q$ with $p\in\mathbb N$ and $q\in\mathbb Z$. The leading coefficient of $U_{n-1}$ is $2^{n-1}$ and the constant coefficient is an element of $\{-1,0,1\}$. Therefore $p/q=1/2^m$ for some integer $1\leq m\leq n-1$ by the Rational Root Theorem.

However, for $n\geq 4$ we have $$\cos{\frac{\pi}{n}}\geq\cos{\frac{\pi}{4}}=\frac{1}{2^{1/2}}>\frac{1}{2^m}=\frac{p}{q}$$ for any positive integer $m$.

Randy Marsh
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An elementary proof. From deMoivre's Theorem, if $k\in \Bbb N$ then $\cos kx=\sum_{(0\le 2j\le k)}\binom {k}{2j}(-1)^j(\cos x)^{k-2j}(1-\cos^2 x)^j.$ So if $\cos (\pi/n)\in \Bbb Q$ and $m|n$ then $\cos (\pi/m)\in \Bbb Q.$ So it suffices to show $\cos (\pi/m)\not \in \Bbb Q$ when (i)$\,m=4,$ or (ii) $\,m=6,$ or (iii) $\,m=9,$ or (iv) $\,m \ge 5$ is prime.

For $m=9,$ let $y=2\cos (\pi/9).$ Then $4(y/2)^3-3(y/2)=\cos (\pi/3)=1/2,$ so $y^3-3y-1=0.$ If $y=A/B$ with $A,B\in \Bbb N$ and $\gcd (A,B)=1$ then $A^3-3AB^2-B^3=0 ,$ so we have $$1\equiv 1+A^3-3AB^2-B^3\equiv 1+A+AB+B\equiv (1+A)(1+B) \pmod 2$$ implying $1+A$ and $1+B$ are odd, implying $A$ and $B$ are even, contrary to $\gcd(A,B)=1.$

For prime $m\ge 5,$ if $\cos \pi/m=A/B$ with $A,B\in \Bbb N$ and $\gcd(A,B)=1$ then by deMoivre's Theorem we have $$(\bullet)\quad -B^m=\sum_{j=0}^{(m-1)/2}\binom {m}{2j}A^{m-2j}(B^2-A^2)^j(-1)^j.$$ The RHS of $(\bullet)$ is divisible by $A$ so $A|(-B^m).$ But $\gcd(A,B)=1$ so we must have $A=1.$

$\quad$ Now $B^m\equiv B \mod m,$ as $m$ is prime. And if $m$ is an odd prime and $0< 2j<m$ then $m$ divides $\binom {m}{2j}.$ So from $(\bullet)$ we have $$-B\equiv -B^m\equiv A^m\equiv 1 \pmod m.$$ So $B\ge m-1.$

But now $1/2=\cos (\pi/3)<\cos (\pi/m)=A/B=1/B\le 1/(m-1)\le 1/4,$ a paradox.

DanielWainfleet
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