# Forming a polynomial to show that $\cos\left(\frac{2\pi}{17}\right)$ is algebraic

I know if $x=e^{\frac{2\pi i}{17}}$ then $x^{17}=1$ and $\Re(x)=\cos\left(\frac{2\pi}{17}\right)$.

But how do I form a polynomial which has root $\cos\left(\frac{2\pi}{17}\right)$.

I know you can consider de Moivre's theorem and expand the LHS using binomial theorem but that will take a long time.

• $\cos(\frac{2\pi}{17})$ is a root of $U_{16}$, the Chebyshev polynomial of the second kind. $U_{16}$ has integer coefficients. – Gabriel Romon Mar 11 '16 at 14:27
• Regarding ages, keep in mind that it may well take ages, and probably about the same number of ages, just to write out the answer. – Lee Mosher Mar 11 '16 at 14:27
• $(a+b)^{17}$ would take me 'decades' to expand and then simplify using trig identities. – Tony Mar 11 '16 at 14:31
• The title does not reflect the question. – lhf Mar 11 '16 at 14:56
• I would suggest editing the question title so that it better reflects the actual question in the body. – Wojowu Mar 11 '16 at 15:08

Let $c=\cos\left(\frac{2\pi}{17}\right)$ and $s=\sin\left(\frac{2\pi}{17}\right)$.

Then

$1=\Re(1)=\Re ((c+s\, i)^{17})=c^{17}-136 c^{15} s^2+2380 c^{13} s^4-12376 c^{11} s^6+24310 c^9 s^8-19448 c^7 s^{10}+6188 c^5 s^{12}-680 c^3 s^{14}+17 c s^{16}$

Note that $s$ appears only with even powers. Now replace $s^2=1-c^2$ to get a polynomial in $c$.

• Is there no quicker way? If I had no access to a calculator then computing the binomial coefficients would take a long time. – Tony Mar 11 '16 at 14:46
• @Tony, you could use that $\binom{17}{k+2} = \frac{(k-17) (k-16)}{(k+1) (k+2)} \binom{17}{k}$. – lhf Mar 11 '16 at 14:59
• I have $c (17 - 816 c^2 + 11424 c^4 - 71808 c^6 + 239360 c^8 - 452608 c^{10} + 487424 c^{12} - 278528 c^{14} + 65536 c^{16}$ but this doesn't give the right root. – Tony Mar 11 '16 at 18:39
• @Tony, you probably forgot to subtract $1$. – lhf Mar 11 '16 at 18:47

Probably the best way is to just show that the sum of two algebraic numbers is algebraic. This is not obvious, but if you look at it just right it's much easier than it seems at first.

Regard $\Bbb C$ as a vector space over $\Bbb Q$. Any linear-algebra concepts below refer to $\Bbb Q$-linear subspaces of $\Bbb C$.

Lemma 0. The number $a\in \Bbb C$ is algebraic if and only if the span of $1,a,a^2\dots$ is finite-dimensional.

Proof: Easy exercise. QED.

Lemma 1. Suppose $A,B\subset\Bbb C$ are subspaces, and let $C$ be the span of the $xy$ with $x\in A$, $y\in B$. If $A$ is spanned by $a_1,\dots,a_n$ and $B$ is spanned by $b_1,\dots,b_m$ then $C$ is spanned by $a_jb_k$, $1\le j\le n$, $1\le k\le m$ (so in particular $C$ is finite dimensional).

Proof: Easy exercise. QED.

Theorem. If $a,b\in\Bbb C$ are algebraic then $a+b$ is algebraic.

Proof. Let $A$ be the span of the powers of $a$ and $B$ the span of the powers of $b$. Let $C$ be as in Lemma 1. Now there exist $n$ and $m$ such that $A$ is spanned by $1,a,\dots, a^n$ and $B$ is spanned by $1,b,\dots,b^m$. So Lemma 1 shows that $C$ is finite dimensional.

But every power of $a+b$ lies in $C$. So the span of the powers of $a+b$ is finite dimensional, and hence Lemma 0 shows that $a+b$ is algebraic. QED.

This shows that your number is algebraic since $\cos(t)=(e^{it}+e^{-it})/2$.

Edit. One could use the argument above to find $P$ with $P(a+b)=0$, given $p(a)=0$ and $q(b)=0$. Any power of $A$ can be written explicitly as a linear combination of $1,a,\dots a^n$, and similarly for $b$. So any $a^jb^k$ can be written explicitly as a linear combination of $a^jb^k$ with $0\le j\le n$ and $0\le k\le m$. Hence the same is true of any power of $a+b$. So write down the powers of $a+b$ as such linear combinations, one by one, and check the vectors of coefficients for linear dependence. Eventually a dependence relation appears, and that gives you $P$ with $P(a+b)=0$.

• +1, this answers the question in the title, which is different from the one in the text. – lhf Mar 11 '16 at 14:56
• @lhf It's sufficiently "constructive" that one could use it to get that polynomial explicitly. Not that I'd want to. – David C. Ullrich Mar 11 '16 at 14:58

By adding equalities $$\cos(n+1)x=\cos nx\cos x-\sin nx\sin x\\ \cos(n-1)x=\cos nx\cos x+\sin nx\sin x$$ we get an equality $$\cos(n+1)x+\cos(n-1)x=2\cos nx\cos x$$ If we now define, by induction, Chebyshev polynomials $T_0(y)=1,T_1(y)=y,T_{n+1}(y)=2yT_n(y)-T_{n-1}(y)$ then it follows, by taking $y=\cos x$, that $$T_n(\cos x)=\cos nx$$ It follows that $\cos\frac{2\pi}{17}$ is a root of $T_{17}(x)-\cos 2\pi=T_{17}(x)-1$.

Chebyshev polynomials are a bit tedious to calculate by hand, but thanks to the recurrence relation this can be done in quite a short amount of time. You can draw a Pascal-triangle like table containing their coefficients, which would make it even faster.

The number $x=cos\left(\frac{2\pi}{17}\right)$ is a root of the polynomial $$\sum_{k=0}^{8} \binom{17}{2k+1}x^{2k+1}\cdot i^{16-2k}\cdot (1-x^2)^{8-k}=1$$.

• But how did you get that polynomial? – Tony Mar 11 '16 at 14:39
• Sorry for my lack of explanation. I used binomial coefficients in the expression $\left(cos\left(\frac{2\pi}{17}\right)+isin\left(\frac{2\pi}{17}\right)\right)^{17}=1$, knowing that the sum of all summands with factor an odd power of $i$ is zero. However, since "it would take you decades", I was not sure it was the answer you were looking for – Darío G Mar 11 '16 at 15:37

Consider that

$$\sin{\left ( \frac{9 \pi}{17} \right )} = \sin{\left ( \frac{8 \pi}{17} \right )}$$

or, letting $y = \frac{\pi}{17}$,

$$3 \sin{3 y} - 4 \sin^3{3 y} = 2 \sin{4 y} \cos{4 y}$$

or

$$9 \sin{y} - 120 \sin^3{y} + 432 \sin^5{y} - 576 \sin^7{y} + 256 \sin^9{y} \\= 8 \sin{y} \cos{y} (2 \cos^2{y}-1) [2 (2 \cos^2{y}-1)^2-1]$$

Note that $\sin{y}$ cancels on both sides. Using $\sin^2{y}=1-\cos^2{y}$, we get an $8$th degree polynomial in $\cos{y}$, from which $\cos{2 y} = 2 \cos^2{y}-1$ may be determined.