Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I recently came across this curious trigonometric sum:


which has a neat proof here: How to prove that: $\tan(3\pi/11) + 4\sin(2\pi/11) = \sqrt{11}$

For what values of $k$ does the following general identity have integer solutions for $a,b,x,y$?


share|cite|improve this question
For the proof of equation $\tan{\frac{3\pi}{11}}+4\sin{\frac{2\pi}{11}}=\sqrt{11}$, please see…. –  pipi Feb 27 '13 at 9:34
@pipi thank you. –  Vincent Tjeng Feb 27 '13 at 9:37
There's a proof using complex numbers, too ! –  Inceptio Feb 27 '13 at 13:22
I think the first proof in the linked question is by complex numbers - is that what you're referring to? –  Vincent Tjeng Feb 27 '13 at 14:03

1 Answer 1

Here are a few thoughts :

Let $\zeta_n = e^{2i\pi/n}$ and $K_n = \Bbb Q(\zeta_n)$ the cyclotomic extension.
It is well-known that this extension is Galois and its Galois group is $\{\sigma_k : \zeta_n \mapsto \zeta_n^k, k \in (\Bbb Z/n\Bbb Z)^*\}$.
$K_n$ contains $\cos(2\pi/n), i\sin(2\pi/n)$, and $i\tan(\pi/n)$.
Hence, $i(a\tan(x\pi/n)+b\sin(y\pi/n)) \in K_n \cap i \Bbb R$.

So first, it would be good to know when $\sqrt{-n} \in K_n$.

If $n$ is even, then $K_n$ never contains $\sqrt {\pm n}$ (you would need $K_{4n}$ for that).
If $n \equiv 1 \pmod 4$, then $K_n$ contains $\sqrt n$ and not $i$, so it doesn't contain $\sqrt{-n}$, and we can also stop.

If $n \equiv 3 \pmod 4$, then $K_n$ does contain $\sqrt {-n}$. Let $\chi_n : (\Bbb Z/n\Bbb Z)^* \to \{\pm 1\}$ be the character satisfying $\sigma_k(\sqrt{-n}) = \chi_n(k) \sqrt{-n}$, and $H_n = \ker \chi_n$.

We have $\Bbb Q(\sqrt{-n}) = K_n^{H_n}$, and the numbers of the form $a\sqrt{-n} \in K_n^{H_n}$ are the numbers $x \in K$ satisfying $\sigma_k(x) = \chi_n(k)x$. Since $\sigma_k(i\sin(x\pi/n)) = i\sin(kx\pi/n)$ and similarly with $\tan$, we want to find integers $a,b,x,y$ such that forall $k$ prime with $n$, $a\tan(kx\pi/n)+b\sin(2ky\pi/n) = \chi_n(k)(a\tan(x\pi/n)+b\sin(2y\pi/n))$.

It is enough to find one with $x=1$ and to include only a set of generators of $(\Bbb Z/n\Bbb Z)^*$ for $k$. After a bit of rewriting, we have to find an integer $y$ such that forall $k$ in a generating set, $\frac{\chi_n(k)\tan(\pi/n)-\tan(k\pi/n)}{\chi_n(k)\sin(2y\pi/n)-\sin(2ky\pi/n)} = -b/a \in \Bbb Q$

Forgetting that this rational has to be independant from $k$, checking that this is a rational is the same as checking that it is invariant by the $\sigma_k$, which means we should check that forall $k,k'$ in a generating set,

$(\chi_n(k)\tan(\pi/n) - \tan(k\pi/n))(\chi_n(k)\sin(2k'y\pi/n)-\sin(2kk'y\pi/n)) - (\chi_n(k)\tan(k'\pi/n) - \tan(k'k\pi/n))(\chi_n(k)\sin(2y\pi/n)-\sin(2ky\pi/n)) = 0$

For example with $n=11$, we can pick $k=k'=2$, we have $\chi_{11}(2) = -1$, and indeed $y=4$ or $y=7$ works.

For $n=7$, $3$ is a primitive root, and $y= 1$ works, which gives $-b/a = 4$, and finally we get $4\sin(2\pi/7) - \tan(\pi/7) = \sqrt 7$.

For $n=19$, $2$ is again a primitive root, and computing this expression for $y= 1 \ldots 18$, we don't find any zero, which shows that there is no such formula (neither for $n$ prime less than $700$, and probably never again)

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.