Along the lines of my lines of my previous question about irrational angles "$45^\circ$ Rubik's Cube: proving $\arccos ( \frac{\sqrt{2}}{2} - \frac{1}{4} )$ is an irrational angle?", I was working on a puzzle and I ran into an interesting question about an irrational angle.
Take a puzzle made out of a triangular bipyramid that turns on its vertices:
The piece in the center must be a circle because it gets rotated by an irrational amount after some combination of moves. Here is the path a point takes after repeatedly turning the purple axis followed by the green axis 90 degrees:
The rotation matrix for this operation is $$\left( \begin{array}{ccc} \frac{1}{8} & \frac{3}{4} & \frac{3}{8}\sqrt{3} \\ \frac{-3}{4} & \frac{1}{2} & \frac{-1}{4}\sqrt{3} \\ \frac{-3}{8}\sqrt{3} & \frac{-1}{4}\sqrt{3} & \frac{5}{8} \end{array} \right)$$
From there you can find the axis that points are rotating around is $[0, \sqrt{3}/2, -1]$ And the angle $\theta$ they're rotating through is $\arccos{\frac{1}{8}}$.
$\arccos{\frac{1}{8}}$ is approximately $82.819244218541^{\circ}$ and it's an irrational angle because the only rational values that correspond to rational angles for $\cos{\theta}$ are $\{-1, \frac{-1}{2}, 0, \frac{1}{2}, 1\}$.
So what do we know about $\theta = \arccos{\frac{1}{8}}$? It's irrational angle but is it an algebraic angle? In general if $\cos \theta$ is rational does that tell us anything (beyond rationality) about $\theta$?
The Gelfond-Schneider theorem says if $a\notin \{0,1\}$ is algebraic and $b$ is algebraic and irrational, then $a^b$ is transcendental. Apply this to $a = e^{i\pi/180}$, and you find that if $b$ is an irrational algebraic, $\omega = \cos(b^\circ) + i \sin(b^\circ)$ is transcendental. Then $\cos(b^\circ) = (\omega + \omega^{-1})/2$ and $\sin(b^\circ) = (\omega - \omega^{-1})/(2i)$ are also transcendental.