Inspired by this question
The methodology used in the answers to that question can be used to prove that $\cos(A\pi)$ is an algebraic number for all rational numbers $A$. This got me thinking: if A is an irrational algebraic, is $\cos(A\pi)$ algebraic?
Some playing around on Wolfram Alpha told me that for $A=\sqrt{2}, \sqrt{3}, \sqrt{5},\sqrt[\leftroot{2}\uproot{-1}3]{2}$, and $\sqrt[\leftroot{2}\uproot{-1}3]{5}$ are all transcendental numbers, but no reasoning for this is provided. Looking at Wikipedia, numbers of this type are not listed as known transcendental numbers.
This leaves me with two questions:
- Can anyone provide a source that $\cos(\sqrt{2}\pi)$, for example, is transcendental?
- Are there any irrational algebraics for which $\cos(A\pi)$ is algebraic? This is equivalent to the question of if for all algebraic numbers X between -1 and 1, is $\frac{\cos^{-1}(X)}{\pi}$ either a rational number or a transcendental number, or can it also be an irrational algebraic number?
Best Answer
Let $A$ be algebraic irrational. We claim that $\cos(A\pi)$ transcendental. Assume not: $\cos(A\pi)$ is algebraic. Then $\sin(A\pi) = \pm\sqrt{1-\cos^2(A\pi)}$ is algebraic. Then $e^{iA\pi}=\cos(A\pi)+i\sin(A\pi)$ is algebraic. Also $e^{iA\pi} \ne 0$ and $e^{iA\pi} \ne 1$. Now $2/A$ is algebraic irrational. Note $$ \big(e^{iA\pi}\big)^{(2/A)} = e^{2\pi i} = 1 $$ is algebraic. This contradicts Gelfond-Schneider.