# Let $m \in \mathbb Z, m>1$, then $\cos(2 \pi/m) \in \mathbb Q$ if and only if $m \in \{1,2,3,4,6\}$. [duplicate]

Possible Duplicate:
When is $\sin(x)$ rational?

Let $m \in \mathbb Z, m\geq1$, then $\cos(2 \pi/m) \in \mathbb Q$ if and only if $m \in \{1,2,3,4,6\}$.

Why is this statement true? Why is $\cos(2 \pi/m)$ always non rational for the integer $m >6$?

Thanks very much.

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## marked as duplicate by Hans Lundmark, Norbert, sdcvvc, Arkamis, rschwiebOct 15 '12 at 18:19

I wonder if there is a Galois theoretic explanation of this? Something to do with cyclotomic extensions of $\mathbb{Q}$? – fretty Oct 13 '12 at 13:47

One approach that may prove helpful:

Lemma: $\theta =\frac{2\pi}{m} \text{ then }\cos \theta \in \mathbb Q \iff \cos \theta \in \{ 0, \pm \frac12, \pm1 \}$

Proof:

Lest $2\cos \theta = \frac ab$ where $a$ and $b$ are co-prime.Then $2\cos 2\theta = {(2\cos \theta)}^2 -2$. $$2\cos 2\theta=\frac {a^2-2b^2}{b^2}$$

Now $\gcd (a^2-2b^2,b^2)=1$.

Proof:Assume $p$ is a prime dividing both Numerator and Denominator. Then, $p|b^2 \text{ hence} p|b$ and $p|(a^2-2b^2) \text { hence }p|a$ giving us contradiction.

So if $b \neq \pm1$, then we get that in $2\cos \theta, 2\cos 2\theta,2\cos 2^2\theta,\ldots$, the denominators get bigger and bigger and $\to \infty$.

Also we have $\cos \theta$ is periodic with period $2\pi$. Hence the sequence noted above can admit at most $m$ different values and then will start repeating contradicting that Denominator tends to infinity.

Hence $b=\pm 1$ and hence our claim is proved.

I suppose it might help as now we know all possible (rational) values of $\cos \theta$ we can check for values corresponding to them.

Thanks.

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