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: 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 $\sin(\frac{2\pi}{n})$).
A: 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.
A: 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.
A: 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.
