Ivan Shishkin, Rye (1878)

Problems/Number theoryReviewed

Valeurs rationnelles de tan(rπ)\tan⁡(r\pi)

by Uettechat·
37
Difficulty scaleÉchelle de difficulté

This score reflects both the level of the required concepts and the difficulty of the solution.Ce score tient compte à la fois du niveau des notions nécessaires et de la difficulté de la résolution.

  1. 110First steps / middle schoolPremiers pas / collège
  2. 1125Beginner / high schoolDébutant / lycée
  3. 2650Intermediate / undergraduateIntermédiaire / licence
  4. 5170Advanced / graduateAvancé / master
  5. 7190Expert / specializedExpert / spécialisé
  6. 91100Research levelNiveau recherche
These levels are approximate guides.Ces niveaux sont des repères approximatifs.
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Déterminer les rQr \in \mathbb{Q} tels que tan(rπ)Q\tan(r\pi) \in \mathbb{Q}.

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Solution by Uettechat

Discussions0 useful votes

On admet que les rQr \in \mathbb{Q} tels que cos(rπ)Q\cos(r\pi) \in \mathbb{Q} est l’ensemble 13Z12Z\frac{1}{3}\mathbb{Z} \cup \frac{1}{2}\mathbb{Z} (voir problèmes liés).
On a
cos(2rπ)=2cos2(rπ)1=21+tan2(rπ)1=1tan2(rπ)1+tan2(rπ)\cos(2r\pi)=2\cos^{2}(r\pi)-1=\frac{2}{1+\tan^{2}(r\pi)}-1=\frac{1-\tan^{2}(r\pi)}{1+\tan^{2}(r\pi)}Si tan(rπ)Q\tan(r\pi) \in \mathbb{Q}, alors 1tan2(rπ)1+tan2(rπ)=cos(2rπ)Q\frac{1-\tan^{2}(r\pi)}{1+\tan^{2}(r\pi)} = \cos(2r\pi) \in \mathbb{Q}, donc r16Z14Zr \in \frac{1}{6}\mathbb{Z} \cup \frac{1}{4}\mathbb{Z}, donc tan(rπ){1,0,1}Q\tan(r\pi) \in \{-1,0,1\} \subset \mathbb{Q}
Donc l’ensemble des solutions est
r16Z14Zr \in \frac{1}{6}\mathbb{Z} \cup \frac{1}{4}\mathbb{Z}

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