Show that $\cos\big(\frac{2\pi}{n}\big)$ is an algebraic number

$\bullet~$Problem: Show that $\cos\bigg(\dfrac{2\pi}{n}\bigg)$ is an algebraic number [where $n$ $\in$ $\mathbb{Z} \setminus \{0\}$].

$\bullet~$ My approach:

Let's consider the following polynomial in $\mathbb{Z}[x]$ in recursive terms. \begin{align*} &T_{0}(x) = 1\\ &T_{1}(x) = x\\ &T_{n + 1}(x) = 2x T_{n}(x) - T_{n-1}(x) \end{align*} $\bullet~$ $\textbf{Claim:}$ The polynomial $T_{n}(x)$ for any $n$ $\in$ $\mathbb{N}$ satisfies the following \begin{align*} T_{n}(\cos(\theta)) = \cos(n\theta) \end{align*} $\bullet~$Proof: We'll use induction on $n$ for this proof.

At first, we easily obtain that for $n = 0$ the given is true.

Now for some $n = k$, we assume that \begin{align*} T_{k}(\cos(\theta)) = \cos(k\theta) \end{align*} Therefore we need to prove for $n = (k + 1)$.

Now from the recursion relation of $T_{n}(x)$ we have \begin{align*} T_{k + 1}(\cos(\theta)) & = 2 \cos(\theta)T_{k}(\cos(\theta)) - T_{k -1}(\cos(\theta))\\ & = 2 \cos(\theta) \cos(k\theta) - \cos((k -1)\theta)\\ & = 2 \cos(\theta) \cos(k\theta) - \cos(k\theta) \cos(\theta) - \sin(k\theta)\sin(\theta)\\ & = \cos((k + 1)\theta) \end{align*} Hence by induction hypothesis, we obtain that our claim is true.

Therefore we have \begin{align*} T_{n}\Bigg(\cos\bigg(\frac{2\pi}{n}\bigg)\Bigg) = \cos(2\pi) = 1 \end{align*} Therefore we just need to consider a polynomial $P(x) = T_{n}(x) - 1.~$ As $T_{n}(x) \in \mathbb{Z}[x]$ it implies $P(x) \in \mathbb{Z}[x]$

Therefore we have $\cos\big(\frac{2\pi}{n}\big)$ is an algebraic number.

Please check the solution and point out the glitches.

Can you prove this in a different (like a pretty elementary one (by not using the idea of cyclotomic polynomials or Chebyshev's Polynomials)) way?

$\bullet~$ $\large{\textbf{Edit:}}$

$\blacksquare~$ Alternate Approach:

I have used the expansion of $\cos\bigg( \dfrac{2\pi}{n} \bigg)$. And obviously which comes from de-Moivre's (simple for $n \in \mathbb{Z}$).

Can you please try to give a solution not using these arguments? (de-Moivre's, Cyclotomic Polynomial, $\color{blue}{\text{Chebychev Polynomials}}$, etc etc).

A proof in an entirely different manner:

Consider the matrix $\begin{bmatrix}\cos(\phi)&-\sin(\phi)\\\sin(\phi)&\cos(\phi)\end{bmatrix}$. This is a rotation matrix.

If we take $\phi=\frac{2\pi}{n}$ for some natural number n, we get the following equation: $$\vec{v}\begin{bmatrix}\cos(\phi)&-\sin(\phi)\\\sin(\phi)&\cos(\phi)\end{bmatrix}^n=\vec{v}$$ for all vectors $\vec{v}$. We can now use this identity to generate a polynomial equation in terms of $\cos(\frac{2\pi}{n})$ and $\sin(\frac{2\pi}{n})$. By using $\sin(\frac{2\pi}{n})=\sqrt{1-\cos^2(\frac{2\pi}{n})}$, we can make this into a polynomial in terms of $\cos(\frac{2\pi}{n})$ where all coefficients are integers and all exponents are either integers or fractions with a denominator of 2. By isolating terms and squaring, we can get a polynomial of integer coefficients and integer powers, which means that the variable, $\cos(\frac{2\pi}{n})$, must be algebraic, (as is $\sin(\frac{2\pi}{n})$).

Here’s another approach: suppose $z=x+iy \in \mathbb{C}$ is an algebraic number, $x,y \in \mathbb{R}$. Sums and products of algebraic numbers are algebraic, and any complex conjugate of an algebraic number is algebraic (ask me about why if you don’t know why). So $$x=\frac{z+\overline{z}}{2}$$ is algebraic, just as $$y=\frac{z-\overline{z}}{2i}$$. So we see, real and imaginary parts of algebraic numbers are algebraic! Now $\cos ( \frac{2 \pi}{n})$ is the real part of $$z=e^{\frac{ 2 \pi i}{n}}$$ which satisfies $$z^n-1=0$$ so is algebraic.

This looks good to me. The only note I have is that the argument really uses strong induction rather than weak induction. More specifically, you are assuming the induction hypothesis holds for all $n\leq k$ in order to prove the result for $n=k+1$ (at least you need it for $n=k,k-1$). But this is maybe a minor technical thing; the proof is otherwise good.

Let me give a way of showing that the number $a_n = \cos(2\pi/n)$ is algebraic for every $n \geq 3$ that doesn't use cyclotomic polynomials or roots of unity, and only uses one Chebyshev polynomial (albeit iterated). The polynomial $P(x) = 2x^2 - 1$ has the property that $P(\cos(\theta)) = \cos(2\theta)$. We break into 2 cases to show that $a_n$ is algebraic:

Case 1: $n$ is odd. Since $\gcd(2, n) = 1$, there exists an integer $k$ so that $2^k \equiv 1 \pmod n$. If we recursively define the sequence of polynomials $P_1(x) := P(x)$, $P_j(x) = P(P_{j-1}(x))$, it follows that $a_n$ is a solution to the polynomial equation with integer coefficients $P_k(x) = x$, since $P_k(a_n) = \cos(2^k (2\pi/n)) = \cos(2 \pi/n) = a_n$. Therefore $a_n$ is algebraic.

Case 2: $n = 2^r m$ for $r \geq 1$ and $m$ odd. In this case, $P_r(a_n) = \cos( 2\pi/m ) = a_m$, and we already know $a_m$ is a solution to the polynomial equation $P_k(x) = x$, where $k$ is an integer so that $2^k \equiv 1 \pmod m$. But $a_m = P_r(a_n)$, therefore since $P_k(a_m) = a_m$, we get by substitution that $P_k(P_r(a_n)) = P_r(a_n)$, so $a_n$ is a solution to the polynomial equation with integer coefficients $P_k(P_r(x)) = P_{k+r}(x) = P_r(x)$. Therefore $a_n$ is algebraic.