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All of these are algebraic integers:

$$\frac1{1+\cos\tfrac\tau4}=\frac1{1+0}=1$$ $$\frac1{1+\cos\tfrac\tau8}=\frac1{1+\tfrac12\sqrt2}=2-\sqrt2$$ $$\frac1{1+\cos\tfrac\tau{12}}=\frac1{1+\tfrac12\sqrt3}=4-2\sqrt3$$ $$\frac1{1+\cos\tfrac\tau{16}}=\frac1{1+\tfrac12\sqrt{2+\sqrt2}}=(2+\sqrt2)\Big(2-\sqrt{2+\sqrt2}\Big)$$ $$\frac1{1+\cos\tfrac\tau{20}}=\frac1{1+\tfrac12\sqrt{2+\varphi}}=2(1+\varphi)\Big(2-\sqrt{2+\varphi}\Big)$$ $$\bigg(\varphi=\frac{1+\sqrt5}2\bigg)$$ $$\frac1{1+\cos\tfrac\tau{24}}=\frac1{1+\tfrac12\sqrt{2+\sqrt3}}=2(2+\sqrt3)\Big(2-\sqrt{2+\sqrt3}\Big)$$

And I checked a few more values of $1/(1+\cos\tfrac\tau{4n})$ using WolframAlpha to calculate the minimal polynomials. Does this hold for all $n$?

I know that $2\cos\tfrac\tau n$ is always an algebraic integer, being a sum of roots of unity. Also, here's a similar question: For what integer values of n is $\tan (\pi /n)$ an algebraic integer?

It would be sufficient to prove that $1/(1-\cos\tfrac\tau{2n})$ is an algebraic integer, because

$$\frac1{1+\cos\tfrac\tau{4n}}=\frac{1-\cos\tfrac\tau{4n}}{\sin^2\tfrac\tau{4n}}=\frac{2-2\cos\tfrac\tau{4n}}{1-\cos\tfrac\tau{2n}}.$$

Hence the title. (Note $\pi=\tau/2$. See https://tauday.com/tau-manifesto )

mr_e_man
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1 Answers1

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Claim. $\frac{1}{1 + \cos \frac{\pi}{2n}}$ is an algebraic integer for all $n \ge 1$.

Proof. Write $2 \cos \frac{\pi}{2n} = e^{\frac{i\pi}{2n}} + e^{\frac{-i\pi}{2n}}$. Then $$\frac{1}{1 + \cos \frac{\pi}{2n}} = \frac{2}{2 + e^{\frac{i\pi}{2n}} + e^{\frac{-i\pi}{2n}}} = \frac{2}{\left(e^{\frac{i\pi}{4n}} + e^{\frac{-i\pi}{4n}}\right)^2} = \frac{2 e^{\frac{i\pi}{2n}}}{\left(e^{\frac{i\pi}{2n}} + 1\right)^2}$$

Since $e^{\frac{i\pi}{2n}}$ is a root of unity, it suffices to show $$\frac{2}{\left(e^{\frac{i\pi}{2n}} + 1\right)^2}$$ is an algebraic integer.

Let $u = e^{\frac{i\pi}{2n}}$, which is an algebraic integer (root of unity). We want to show $$\frac{2}{\left(u + 1\right)^2}$$ is an algebraic integer. Note that $u^{2n} = -1$, and $$\frac{2}{(u+1)^2} = \frac{2(u-1)^2}{(u^2-1)^2} = \frac{2(u-1)^2(u^{2(n-1)} + u^{2(n-2)} + \cdots + 1)^2}{(u^{2n}-1)^2} = \frac{(u-1)^2(u^{2(n-1)} + u^{2(n-2)} + \cdots + 1)^2}{2}$$

Note that $(u-1)^2 = u^2 - 1 + 2(-u + 1)$. Hence $$ \begin{align*} \frac{2}{\left(u + 1\right)^2} = \, & \frac{(u-1)^2(u^{2(n-1)} + u^{2(n-2)} + \cdots + 1)^2}{2} \\ = \, & \frac{(u^2 - 1 + 2(-u + 1))(u^{2(n-1)} + u^{2(n-2)} + \cdots + 1)^2}{2} \\ = \, & \frac{(u^2 - 1)(u^{2(n-1)} + u^{2(n-2)} + \cdots + 1)^2}{2} + (-u + 1) (u^{2(n-1)} + u^{2(n-2)} + \cdots + 1)^2 \\ = \, & \frac{(u^{2n} - 1)(u^{2(n-1)} + u^{2(n-2)} + \cdots + 1)}{2} + (-u + 1) (u^{2(n-1)} + u^{2(n-2)} + \cdots + 1)^2 \\ = \, & -(u^{2(n-1)} + u^{2(n-2)} + \cdots + 1) + (-u + 1) (u^{2(n-1)} + u^{2(n-2)} + \cdots + 1)^2 \end{align*} $$ is sum/product of roots of unity, and is hence an algebraic integer.

dummy
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  • I'll add that $1/(1-\cos\tfrac\tau{2n})=1/(1+\cos(n-1)\tfrac\tau{2n})$ is an algebraic conjugate of $1/(1+\cos\tfrac\tau{2n})$ when $n$ is even (because $n-1$ is coprime to $n$, and to $2n$), so it is integral because its conjugate is integral. It remains to answer the title question when $n$ is odd. – mr_e_man Oct 06 '25 at 02:12
  • updated - i thought all your $\tau$ in your post is $\pi$. – dummy Oct 06 '25 at 03:48
  • Looks good. That resolves $1/(1+\cos)$, but I'm still wondering about $1/(1-\cos)$. – mr_e_man Oct 06 '25 at 04:24
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    The answer is still yes for odd $n > 1$. Using the same first step in the answer, you are asking whether $\frac{2}{(u-1)^2}$ is an algebraic integer if $u$ is a primitive $2n$-th root of unity for $n$ odd. The key fact here is $u-1$ is in fact a unit in algebraic integer when $n > 1$ odd; for example, by https://math.stackexchange.com/questions/49956/value-of-cyclotomic-polynomial-evaluated-at-1. – dummy Oct 06 '25 at 04:42