# Thread: Prove that Cos(2Pi/n) is algebraic for n a nonzero integer.

1. ## Prove that Cos(2Pi/n) is algebraic for n a nonzero integer.

I can't figure out what to do. I am supposed to prove that $\cos{\frac{2\pi}{n}$ is algebraic for non-zero integers $n$

This is what I have done so far. I don't know if I'm even on the right track.

The question marks are there because the upper limit will depend on whether $n$ is even or odd.

2. Originally Posted by CropDuster
I can't figure out what to do. I am supposed to prove that $\cos{\frac{2\pi}{n}$ is algebraic for non-zero integers $n$

This is what I have done so far. I don't know if I'm even on the right track.

The question marks are there because the upper limit will depend on whether $n$ is even or odd.

Read this: Re: algebraic order of cyclotomic numbers

Tonio

3. Unfortunately that link didn't really help me. I don't know what a cyclomatic field is, or how I would use it for my proof. Thanks though.

4. let's take a detour. let's prove we can write cos(nx) as a polynomial in u = cos(x). for n = 1, this is trivial. now we assume we can do so for all 0 < k < n.

cos(2x) = 2cos^2(x) - 1 = 2u^2 - 1

cos(3x) = 4cos^3(x) - 3cos(x) = 4u^3 - 3u.

now for n ≥ 3, cos(nx) = cos(x)cos((n-1)x) - cos((n-2)x) = cos(x)[p(cos(x))] - q(cos(x)) = up(u) - q(u), by our induction assumption

(expand the RHS of the trig identity to prove the equality).

since up(u) - q(u) is clearly a polynomial in u, the assertion is shown.

in particular, this means we can write 1 = cos(2π) = f(cos(2π/n)), for some polynomial f.

hence if we set g(x) = f(x) - 1, then g(cos(2π/n)) = f(cos(2π/n)) - 1 = 1 - 1 = 0.

evidently cos(2π/n) is a root of g(x), so cos(2π/n) is algebraic.

(note: although this seems to only show a proof for natural numbers n, it actually shows it for any non-zero integer

since cos(2π/n) = cos(2π/-n), if n is nonzero).

5. I figured out a fairly straight forward solution. Seeing as the text in images doesn't show up as results when searching the forum; I'm going to restate the problem and give my proof:

Prove that for all non-zero integers $n$, $\cos(2\pi/n)$ is an algebraic number.

Proof: From Euler's identity we know that $e^{2\pi i}=1$ and therefore $\left(e^{\frac{2\pi i}{n}}\right)^{n}=1$. And from Euler's Formula we know that

$\left(e^{\frac{2\pi i}{n}}\right)^n=\left(\cos\left(2\pi/n\right)+i\sin\left(2\pi/n\right)\right)^{n}.
$

So if we apply the Binomial theorem to the right, and Euler's identity to the left we have

$1=\sum _{k=0}^n\binom{n}{k}(i)^k\left(\sin\left(2\pi/n\right)\right)^k\left(\cos\left(2\pi/n\right)\right)^{n-k}$

Seperating this sum into two sums: one with terms having only even exponents, and another with terms having only odd exponents:

$\begin{array}{cl}
1&=\sum_{k=0}^p\binom{n}{2 k}(i)^{2 k}\sin ^{2 k}\left(2\pi/n\right)\cos\left(2\pi/n\right)^{n-2 k}\\&\qquad+\sum_{k=0}^q\binom{n}{2 k+1}(i)^{2k+1}\sin^{2 k}\left(2\pi/n\right) \cos\left(2\pi/n\right)^{n-2k-1}\\
\end{array}$

where $p+q=n$. This sum consists of $n+1$ terms. If $n$ is odd, $p=q=\frac{n+1}{2}$. Otherwise $p=\frac{n}{2}+1$ while $q=\frac{n}{2}$. So after some algebra,

$\begin{array}{cl}
1&=\sum_{k=0}^p\binom{n}{2 k}(-1)^{k}\sin ^{2k}\left(2\pi/n\right)\cos\left(2\pi/n\right)^{n-2k}\\
\end{array}$

But the right side of this equation is a complex number of the form $a+ib$ where $a,b\in\mathbb{R}$ and the lefthand side is of the form $1+0i.$ Therefore

$0=\sum_{k=0}^q\binom{n}{2k+1}(-1)^{k}\sin^{2k}\left(2\pi/n\right) \cos\left(2\pi/n\right)^{n-2k-1}.$

Hence,
$\begin{array}{cl}
1&=\sum_{k=0}^p\binom{n}{2 k}(-1)^{k}\sin ^{2k}\left(2\pi/n\right)\cos\left(2\pi/n\right)^{n-2k}\\
&=\sum_{k=0}^p\binom{n}{2 k}(-1)^{k}(\sin^{2}\left(2\pi/n\right))^k\cos\left(2\pi/n\right)^{n-2k}\\
\end{array}$

and applying the trigonometric identity $\sin^2(x)=1-\cos^2(x)$ yields:

$1=\sum_{k=0}^p\binom{n}{2 k}(-1)^{k}(1-\cos^{2}\left(2\pi/n\right))^k\cos\left(2\pi/n\right)^{n-2k}.$

Replacing $\cos\left(2\pi/n\right)$, with a variable $y$ gives

$1=\sum_{k=0}^p\binom{n}{2 k}(-1)^{k}(1-y^{2})^k y^{n-2k}$

And after simplifying:

$0&=y^{n}\sum_{k=0}^p\binom{n}{2k}(1-y^{-2})^k-1$

This is a polynomial with rational coefficients that evaluates to zero at $y=\cos(2\pi/n)$ for all non-zero integers $n$.