$\bullet~$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).

  • 2
    $\begingroup$ Looks good to me. $\endgroup$
    – Zarrax
    Jul 30, 2020 at 20:45
  • 9
    $\begingroup$ Looks really nice. I wonder if you have considered an argument based on the fact that $\zeta := e^{2\pi i / n}$ is algebraic (since $\zeta^n - 1 = 0$), and then $\alpha := \cos(\tfrac {2\pi }n)$ is algebraic (since $\alpha = \tfrac 1 2 (\zeta + \zeta^{-1})$)? $\endgroup$
    – Kenny Wong
    Jul 30, 2020 at 21:07
  • 2
    $\begingroup$ @KennyWong I just wanted to use the argument of Chebyshev polynomials. In elementary contest problems, we used your argument and we used $$ x + \frac{1}{x} < 2 $$ has no solution on $\mathbb{R}$. Hence $x = r \cdot e^{i{2\pi} / {n}}$ and so on... $\endgroup$ Jul 30, 2020 at 21:14
  • $\begingroup$ I'm thinking of another way. The number equals the real part of $e^i(2\pi/n)$, which is the solution of $z^n=1$, so $e^i(2\pi/n)$ is algebraic. Is there a way to prove $\cos(2\pi/n)$ is algebraic from that? $\endgroup$ Jul 30, 2020 at 23:49
  • $\begingroup$ @KennyWong @Zarrax In the question title, I noticed a successfully compiled \textbf{algebraic number} which is absurd in MathJax! \textbf{} doesn't work in MathJax. What happened here? $\endgroup$ Aug 4, 2020 at 8:11

4 Answers 4


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

  • 2
    $\begingroup$ Alternatively, we can use the fact that for any natural number A, $\cos(A\phi)$ can be written as a polynomial of degree A with integer coefficients of variable $\cos(\phi)$. We can therefore say that $1=\cos(n\frac{2\pi}{n})=\sum_{k=0}^{n}B_k\cos^k(\frac{2\pi}{n})$ where $B_k$ are integers $\endgroup$
    – Moko19
    Aug 4, 2020 at 11:00
  • $\begingroup$ How do you know the polynomial you get is nonzero? $\endgroup$
    – M. Van
    Aug 9, 2020 at 10:44
  • $\begingroup$ @M.Van: The polynomial you get from the rotation is the same as the polynomial you get from using the expansion of $\cos(n\frac{2\pi}{n})$ as discussed in the comment. In this polynomial, $B_n$ is always nonzero, which means that the polynomial as a whole is nonzero $\endgroup$
    – Moko19
    Aug 9, 2020 at 11:05

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.

  • 1
    $\begingroup$ Is it really relevant to distinguish strong and weak induction here? At some point, one realizes they are the same thing... $\endgroup$
    – Pedro
    Aug 9, 2020 at 12:14
  • $\begingroup$ @Pedro Tamaroff I think it is a small technical detail as I say, but as written the weak induction hypothesis is technically not sufficient. When the OP originally posted the question this was all the feedback I could give (the rest of their proof was very good); now it seems they've completely updated the question to asking for other proofs. $\endgroup$
    – Dave
    Aug 9, 2020 at 17:49

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.

  • $\begingroup$ If you see my approach, it's also based on Chebyshev Polynomials. I wanted another type of solution. Nevertheless, Thank you $\endgroup$ Aug 8, 2020 at 21:45
  • $\begingroup$ @RalphClausen So what properties of $\cos$ do you want to use? You can prove it using the double angle formula, which is pretty basic. If that's not allowed, you have to allow something, as otherwise you have nothing to work with. The original definition? $\endgroup$ Aug 8, 2020 at 21:59
  • $\begingroup$ @Ralph Sorry! I used a different approach with the Chebyshev polynomials than you did, I thought that would be of interest. $\endgroup$ Aug 8, 2020 at 23:27
  • $\begingroup$ @RiversMcForge yeah, that's why I upvoted your answer too. Thanks! $\endgroup$ Aug 9, 2020 at 7:09

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