# Is there a usual method for finding the minimal polynomial of trigonometric values?

I've been thinking a bit about finding the minimal polynomials of side lengths of regular $n$-gons inscribed in the unit circle. For example, I recently wanted to find the minimal polynomial of the side length of an inscribed regular nonagon. Using the law of cosines, I was able to find that the side length is $a=\sqrt{2-2\cos(\frac{2\pi}{9})}$.

So $\displaystyle \frac{2-a^2}{2}=\cos \left(\frac{2\pi}{9} \right)$, but since I can't express $\displaystyle\cos \left(\frac{2\pi}{9} \right)$ in terms of rational numbers or their square roots, I'm unsure of how to proceed exactly. Is there a usual method to attack values like this? Possibly for say $\displaystyle \cos \left(\frac{m\pi}{n} \right)$?

Thanks.

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Look up the Chebyshev polynomials: en.wikipedia.org/wiki/Chebyshev_polynomials – Thomas Andrews May 13 '11 at 4:04
@Amy Maybe you're already aware, but if you use displaymode on inline formulas, you can also add \left( or \right) for the parentheses on the cosine expressions, so that the parentheses match the arguments better. – yunone May 13 '11 at 5:04
@yunone: Ah...thanks for the tip. I'm kind of learning as I'm going along. Good to know! – amWhy May 13 '11 at 5:17
@Dom: Take a look here: math.stackexchange.com/questions/45144/… – user9413 Aug 4 '11 at 10:33

EDIT : Following Marty Green's approach, I managed to actually find the minimal polynomial explicitly. I think both of you (Marty and Dom) will appreciate. Write $$2 \cos \left( \frac{2\pi}9 \right) = \zeta_2 + \zeta_7$$ where $\zeta_2$ and $\zeta_7$ are the second and seventh rooth of unity, respectively. Now we try to find the minimal polynomial for $\zeta_2 + \zeta_7$ (if we found it then substituting $\zeta_2 + \zeta_7 = x = 2y$ in the polynomial gives us our wished polynomial). Notice that $$x^9 -1 = \prod_{d \mid n} \Phi_d(x) = (x-1)(x^2+x+1)(x^6 + x^3 + 1)$$ where $\Phi_d(x)$ is the cyclotomic polynomial of the $d^{\text{th}}$ roots of unity. Now since $$\phi_9(x) = x^6 + x^3 + 1 = \prod_{(i,9)=1}(x-\zeta_i)$$ and developping this product to obtain the symmetric polynomials in the $\zeta_i$ as coefficients of $\phi_9(x)$, we obtain the two useful identities : $$\sum_{(i,9)=1} \zeta_i = 0, \quad \underset{i \neq j}{\sum_{(i,9)=(j,9)=1}} \zeta_i \zeta_j = 0.$$ Now we also have the identity $\zeta_6 + \zeta_3 + 1 = 0$ since $\zeta_3$ is a root of $x^2 + x + 1$. Hence we have $$(x-(\zeta_1 + \zeta_8))(x-(\zeta_2 + \zeta_7))(x-(\zeta_4 + \zeta_5)) =$$ $$x^3 - ((\zeta_1 + \zeta_8) + (\zeta_2 + \zeta_7) + (\zeta_4 + \zeta_5))x^2 +$$ $$((\zeta_1 + \zeta_8)(\zeta_2 + \zeta_7) + (\zeta_1 + \zeta_8)(\zeta_4 + \zeta_5) + (\zeta_2 + \zeta_7)(\zeta_4 + \zeta_5))x$$ $$- (\zeta_1 + \zeta_8)(\zeta_2 + \zeta_7)(\zeta_4 + \zeta_5).$$ Since the sum of the roots of $\Phi_9$ is $0$, the quadratic term vanishes. For the linear term, notice that all pairs of roots appear as products of two roots beside the ones that are together in the sum, i.e. $\zeta_1 \zeta_8$, $\zeta_2 \zeta_7$ and $\zeta_4 \zeta_5$. Hence if we call the linear term $\alpha$, $$\alpha = \left( \underset{i \neq j}{\sum_{(i,9)=(j,9)=1}} \zeta_i \zeta_j \right) - \zeta_1\zeta_8 - \zeta_2 \zeta_7 - \zeta_4 \zeta_5 = 0 - \zeta_9 - \zeta_9 - \zeta_9 = -3.$$ Now for the constant in the end, develop it manually to have something like this : $$- (\zeta_1 + \zeta_8)(\zeta_2 + \zeta_7)(\zeta_4 + \zeta_5) =$$ $$- (\zeta_1 \zeta_2 \zeta_4 + \zeta_1 \zeta_2 \zeta_5 + \zeta_1 \zeta_7 \zeta_4 + \zeta_1 \zeta_7 \zeta_5 +$$ $$\zeta_8 \zeta_2 \zeta_4 + \zeta_8 \zeta_2 \zeta_5 + \zeta_8 \zeta_7 \zeta_4 + \zeta_8 \zeta_7 \zeta_5) =$$ $$- (\zeta_7 + \zeta_8 + \zeta_3 + \zeta_4 + \zeta_5 + \zeta_6 + \zeta_1 + \zeta_2) = -(\zeta_3 + \zeta_6) = 1$$ The last line is because the sum of the roots is $0$. Hence a polynomial for which $\zeta_2 + \zeta_7$ is a root is $x^3 - 3x + 1$. Making the variable change to get the polynomial for $\cos(2\pi / 9)$ gives us $8x^3 - 6x + 1 = 0$, which is irreducible (there are many possible answer to this irreducibility question non-rationality of the roots, rational root theorem, etc.) which I leave up to you.

P.S. I will never do that again, it's so painful... but it was fun. XD Hope you liked it.

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Good calculation. – quanta May 15 '11 at 21:12
Wow, amazing. Thanks for taking the time and effort to do that. – Dom May 21 '11 at 6:51
No problem ; I suggest you look more closely to the other answers also! They were inspiring for this answer, mosty Marty's. Galois Theory is always so much fun... =) – Patrick Da Silva May 21 '11 at 8:30

Usually it's a pain to do so, for instance for $\displaystyle \cos(\frac {2\pi}9)$, you use the triple angle cosine formula : $$\cos 3\theta = 4\cos^3(\theta) - 3 \cos (\theta).$$ since substituting $\displaystyle \theta = \frac{2\pi}9$, you can evaluate $\displaystyle \cos \left( \frac{2\pi}3 \right) = -1/2$, which gives you rational coefficients in $\displaystyle \cos \left(\frac{2\pi}9\right)$. I wouldn't say there is a "general method" because I would never do that in general, but a trick to compute the minimal polynomial of $\displaystyle \cos \left(\frac{m\pi}n \right)$ would be to use trigonometric identities to express $\cos n\theta$ in a polynomial that is in terms of $\cos \theta$, or at least $\cos k\theta$ where $k$ divides $n$ so that when you substitute $\theta =$ your angle, you obtain a remarquable angle (like in the $\displaystyle \frac{2\pi}9$ case, we chose $3$ instead of $9$) . Such a polynomial exists, and they are detailed in the following link :

http://en.wikipedia.org/wiki/List_of_trigonometric_identities#Multiple-angle_formulae

Look for the multiple angle formulas, Tchebyshev and spread polynomials. They give you formulas for the sine and cosine.

For instance, would you be computing the minimal polynomial of $\displaystyle \cos \left(\frac{7\pi}{16} \right)$ for any reason, since the quarters are remarquable angles, I'd go for the $4^{\text{th}}$ Tchebyshev polynomial : $T_4(x) = 8x^4 - 8x^2 + 1$, so that $$\cos(4 \theta) = 8\cos^4 \theta - 8 \cos^2 \theta + 1,$$ hence $\displaystyle \cos \left(\frac{7\pi}4 \right) = 1/\sqrt{2}$, so that $$(8x^4 - 8x^2 + 1)^2 - 1/2 = 0$$ would be a polynomial of which $\displaystyle \cos \left(\frac{7\pi}{16} \right)$ is a root. Expanding it and multiplying by $2$ to get rid of fractions will give you $$128x^8 - 256x^6 + 160x^4 - 32x^2 + 1 = 0$$

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In particular, there are two (sometimes four) kinds of Chebyshev polynomials. The first kind $T_n(x)$ satisfies $T_n(\cos\,\theta)=\cos\,n\theta$, while the second kind $U_n(x)$ satisfies $U_n(\cos\,\theta)=\frac{\sin((n+1)\theta)}{\sin\,\theta}$. – J. M. May 13 '11 at 4:18
@J.M. The link I proposed details those well enough, and clearly $U_n$ doesn't make clever identities for this context, but I must admit this is a cool thing to know. Thanks J.M. – Patrick Da Silva May 13 '11 at 4:34
It's a good idea to keep things self-contained if possible, especially considering that you're linking to a Wikipedia entry. ;) – J. M. May 13 '11 at 4:36
Actually that should be $128 x^8 - 256 x^6 + 160 x^4 - 32 x^2 + 1$. Note that these polynomials obtained from Chebyshev polynomials are not necessarily irreducible over the rationals (although this one is). – Robert Israel May 13 '11 at 4:41
Typo. Thanks Robert, I'll edit it. – Patrick Da Silva May 14 '11 at 4:46

I think I can point you in the direction of a very general algebraic way to construct things like the minimal polynomial for cos(2π/9 ). Start with the equation x^9 - 1 = 0, factor out the trivial factor of (x-1) and you get x^8 + x^7 .... + 1 = 0, which has eight roots. Then factor out x^2 + x + 1 to get rid of the cube roots of 1, and you have six roots left. Because of their relation as ninth roots of unity, they can be enumerated as a, a^2, a^4, a^8, a^16 (=a^7) and a^32 (=a^5). The next term in the series is a^64 which is just equal to a. We can call these roots r1,r2,r4,r5,r7, and r8. It is then easy to choose sums of these roots to be the cosines of 40 degrees, 80 degrees, and 160 degrees:

r1 + r8

r2 + r7

r4 + r5

If we call these a, b, and c, then the problem becomes one of constructing the symmetric polynomial:

abc

ab + bc + ca

a + b + c

The information available is the values for the symmetric polynomials in r1,r2,r4,r5,r7 and r8. (which you get directly from the coefficient of the six-degree equation for the r's). In addition you have the known relations between the r's, that is eg. r1r2 = r3.

I don't have the strength to carry it all through right now but I've done stuff like this in the past and its actually kind of fun and somehow it works, if I'm not mistaken.

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This looks like a nice way to do it, but very particular and deserves detailing. I'd love to work it out myself tomorrow or to see you've worked it out in some way, I think it's worth a shot. I'm wondering if it has some generality that would make finding the minimal polynomials real quick. – Patrick Da Silva May 13 '11 at 5:38
Years ago I worked this out for the fifth and seventh roots of unity, but never the ninth. I remember reducing the expressions for the symmetric polynomials to be a very picturesque exercise, but it's not especially quick. My guess is that it works for all roots of unity in terms of deriving the minimal polynomial, but solving it for an algebraic expression is of course another matter. – Marty Green May 13 '11 at 11:19
Thanks, Patrick. I'm amazed and honored that someone actually took my advice on how to solve a problem. Yes, it's a little painful but oddly satisfying, isn't it? (I wanted to add this comment to your posted answer but I couldn't find the COMMENT field so I posted it here.) – Marty Green May 14 '11 at 13:21
In fact Gauss used your method to express the 17th roots of unity and others. – quanta May 15 '11 at 21:13
@quanta : Oh, quite cool =) I think I'll try it someday! (The 17th root thing). @Marty Green : Well well, clever advices make clever answers, the thanks goes back at you! It's weird that you couldn't find the comment field, it's at the same spot where you commented here, but below my answer instead of yours. =P – Patrick Da Silva May 16 '11 at 3:38

The Chebyshev polynomials of the second kind have the property that:

$$U_n(\cos\theta) = \frac {\sin (n+1)\theta}{\sin \theta}$$

So $U_n(x)$ has roots equal to $\displaystyle \cos\left( {\frac{\pi k}{n+1}}\right)$.

You have to eliminate the roots of $U_8(x)$ that are roots of smaller polynomials.

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See formulae 22.16.4 and 22.16.5 here. – J. M. May 13 '11 at 4:34