Suppose $x\in\Bbb R$ is such that

$$\sin x=\sum_{i=1}^m x_i\sqrt{r_i},\quad \cos x=\sum_{j=1}^n y_j\sqrt{s_j}$$

for some $x_i, r_i, y_j, s_j \in\Bbb Q \ , \ |x_i|=|y_j|=1$. Show that $x=\dfrac{k\pi}{12}$ for some $k\in\Bbb Z$.

  • $\begingroup$ I don't understand what your question is. $\endgroup$
    – pitchounet
    Aug 8, 2013 at 16:27
  • 1
    $\begingroup$ He's saying if $\sin x$ and $\cos x$ can both be represented as the sum of square roots of rationals, then show that $x=\frac{k\pi}{12}$ for some integer $k$. @jibounet $\endgroup$ Aug 8, 2013 at 16:29
  • 1
    $\begingroup$ Would work in what sense? @xavierm02 Did you read the question? He's trying to show for all $x$ for which this is true, that $x=k\pi/12$ for some integer $k$. You've just shown that $x=0$ satisfies this condition, which is sort of obvious. $\endgroup$ Aug 8, 2013 at 16:38
  • $\begingroup$ Ziang, I've edited your question to make it easier to understand. I hope you don't mind. $\endgroup$ Aug 8, 2013 at 16:51
  • $\begingroup$ @RagibZaman,@Thomas Andrews Thanks a lot $\endgroup$
    – ziang chen
    Aug 8, 2013 at 16:57

1 Answer 1


Sketch of a proof. Let $z= \cos x + i\sin x = e^{ix}$, where $x$ is as in in the statement of the problem. Note that $z$ is an algebraic number: $$z = \sum_{j=1}^n y_j\sqrt{s_j} + i \sum_{i=1}^m x_i\sqrt{r_i}.$$

Consider $F={\mathbb Q}[z]$, the splitting field of $z$. Consider the Galois group of $F$ over $\mathbb Q$, $G = \mathrm{Gal}(F/\mathbb{Q}).$ We prove that $G=\mathbb{Z}_2^M$ for some $M$. Indeed, let $p_1,\dots, p_t$ be all prime factors of numbers $r_i$ and $s_j$. Let $E=\mathbb{Q}(\sqrt{-1},\sqrt{p_1},\dots, \sqrt{p_t})$. Since $z = \sum_{j=1}^n y_j\sqrt{s_j} + i \sum_{i=1}^m x_i\sqrt{r_i}$, we have each $\sqrt{s_j} \in E$ and $\sqrt{r_i}\in E$, and thus $z\in E$.

Therefore, $\mathbb{Q}\subset F\subset E$. The Galois group of $E/\mathbb{Q}$ is isomorphic to $\mathbb{Z}_2^{t+1}$. Now, by the fundamental theorem of Galois theory, $G$ is a quotient group of $\mathbb{Z}_2^{t+1}$. Thus $G$ is isomorphic to $\mathbb{Z}_2^M$ for some $M$.

Each $\mathbb Q[\sqrt{s_j}]$ and each $\mathbb Q[\sqrt{r_i}]$ is a subfield of a cyclotomic field (see Square roots of integers and cyclotomic fields). Thus $E$ is a subfield of a cyclotomic field $\mathbb{Q}[\xi]$. (Alternatively, since $G$ is Abelian, by Kronecker–Weber theorem, $F$ is a subfield of a cyclotomic field).

Every complex number of unit norm in $\mathbb{Q}[\xi]$ is a power of $\xi$ (see Elements of absolute value one in cyclotomic fields). Therefore, $z = \xi^u$ for some $u$ and thus $x = (p / q) \cdot 2\pi$, for some coprime integer numbers $p$ and $q$.

We get that $G$ is isomorphic to the group ${\mathbb Z}_q^*$ (the multiplicative group of integers modulo $q$). It is isomorphic to $\mathbb{Z}_2^M$ only if $q$ divides $24$ (for every $q$, $G$ is a direct product of cyclic groups, which can be explicitly described in terms of $q$; all of them are isomorphic to ${\mathbb Z}_2$ if and only if $q$ divides 24; see Wikipedia for details). We have, $$x = \frac{(p\cdot (24/q))\pi}{12}.$$


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .