cos$\mathrm{cos}(A\pi )$ where $A$ is an irrational algebraic number

ntaraxq
2022-07-06
Answered

cos$\mathrm{cos}(A\pi )$ where $A$ is an irrational algebraic number

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Pranav Greer

Answered 2022-07-07
Author has **13** answers

Let $A$ be algebraic irrational. We claim that $\mathrm{cos}(A\pi )$ transcendental. Assume not: $\mathrm{cos}(A\pi )$ is algebraic. Then $\mathrm{sin}(A\pi )=\pm \sqrt{1-{\mathrm{cos}}^{2}(A\pi )}$ is algebraic. Then ${e}^{iA\pi}=\mathrm{cos}(A\pi )+i\mathrm{sin}(A\pi )$ is algebraic. Also ${e}^{iA\pi}\ne 0$ and ${e}^{iA\pi}\ne 1$. Now $2/A$ is algebraic irrational. Note

$(}{e}^{iA\pi}{{\textstyle )}}^{(2/A)}={e}^{2\pi i}=1$

is algebraic. This contradicts Gelfond-Schneider.

$(}{e}^{iA\pi}{{\textstyle )}}^{(2/A)}={e}^{2\pi i}=1$

is algebraic. This contradicts Gelfond-Schneider.

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