$ \arctan \sqrt 2 $ is not a rational multiple of $\pi $

Asked Mar 27 '19 at 08:30

Active Mar 27 '19 at 08:30

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How to show that $ \arctan \sqrt 2 \not =q\pi$ for $q\in \mathbb Q $ ?

asked Mar 27 '19 at 08:30

1

See https://math.stackexchange.com/questions/579031/arctan-of-a-square-root-as-a-rational-multiple-of-pi – Robert Z Mar 27 '19 at 08:36
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Also related: https://math.stackexchange.com/questions/79861/arctan2-a-rational-multiple-of-pi – detnvvp Mar 27 '19 at 08:37

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