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For reasons I'm not going to get too into, I'm interested in a number of the form $p+q \cos(\tfrac \pi{18})$, where $p^2,q^2$ are rational. A question that occurred to me is whether or not such a representation is unique, namely could there exist some different $p',q'$ such that $p'+q'\cos(\tfrac \pi{18})$ is equal to the same value.

If this were to occur, then we could write $$\cos \left(\frac \pi{18} \right)=\frac{p-p^{\prime}}{q^{\prime}-q}=\frac{\left(p-p^{\prime} \right) \left(q+q^{\prime} \right)}{q^{{\prime}^{2}}-q^2}$$.

I suspect that this is not the case, as any closed form of $\cos(\tfrac \pi {18})$ I can find online is an absolute mess. That being said, I'm not totally sure how to prove it cannot be done.

I'm quite rusty with abstract algebra, so everything I am about to say might be totally wrong, but a potential approach that occurred to me is to consider field extensions. Namely, the minimal polynomial of $\cos(\tfrac \pi {18})$ over $\mathbb Q$ has degree $6$, while $\frac{(p-p')(q+q')}{q'^2-q^2}$ can be obtained by adding (at most) four square roots to $\mathbb Q$, which has degree some factor of $16$. In particular, one side has a factor of $3$, while the other does not, which leads me to believe we have some sort of contradiction, but I'm not solid enough on field extensions to suss out the details.

Does this general idea work? If so, could someone possibly help me fill in the missing details on how to finish; if not, are there any other approaches that could work?

Bumblebee
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Connor Gordon
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    Yes that works, because extension degrees are multiplicative. – Steve D Jun 20 '23 at 05:40
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    Your number $\cos \frac{\pi}{18}$ is not constructible – orangeskid Jun 20 '23 at 05:44
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    As a note: $$ \cos\left(\frac{\pi}{18}\right) = \frac{1}{2} , \sqrt{ 2 + \sqrt[3]{\frac{1 + i , \sqrt{3}}{2}} + \sqrt[3]{ \frac{1 - i , \sqrt{3}}{2} } }. $$ Since this involves cube roots in a radical it would less likely be able to be placed into a form or a ratio of squares. – Leucippus Jun 20 '23 at 05:59
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    There was this theorem about what cosines of the form $\frac{\pi}{n}$ could be expressed in closed form, depending on $\varphi(n)$, Euler phi function. I can't remember the name though, and have been looking for it for ages. – Zima Jun 20 '23 at 06:29
  • Related: Is it possible to express $\sin \frac{\pi}{9}$ in terms of radicals? – dxiv Jun 20 '23 at 06:47
  • I have the tendency to see constructible trigonometric numbers as being derived from regular polygons of size $n$. This works for $n$ being $3$, $4$, $5$, $6$, $8$, $12$, ..., which gives the possibility to calculate the trigonometric numbers of $60°$, $90°$, $108°$, $120°$, .... In top of that you can calculate the half of a number and $45°$ and $180°$ are well-known. This all leds to the smallest natural number to be calculated being $3$, as $3$ can be written as $(30+45)-(180-108)$, so all multiples of $3$ can be calculated, but $\pi/18=10$ is not a multiple of $3$. – Dominique Jun 20 '23 at 06:51
  • @Zima I think you're referencing the Gauss-Wantzel theorem: https://en.m.wikipedia.org/wiki/Constructible_polygon The link with Euler's totient function is that $\varphi(n)$ is the degree of the minimal polynomial of $\cos(2\pi/n)$ over $\mathbb{Q}$ I believe. – Bruno B Jun 20 '23 at 08:01
  • @BrunoB yes, thank you – Zima Jun 20 '23 at 08:06
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    @Zima: There was this theorem about what cosines of the form $\frac{\pi}{n}$ could be expressed in closed form --- Others have commented about the possibility of expressing $\frac{\pi}{n}$ using arithmetic operations and arithmetic combinations involving square roots (using only rational numbers and numbers previously constructed). For the more general situation involving arithmetic operations and arithmetic combinations of positive integer roots (without using complex numbers), see this 26 September 2005 sci.math post. – Dave L. Renfro Jun 20 '23 at 11:03

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