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Jan 22, 2014 at 20:55 comment added Sh.M1972 All constructable angels can be bisected. If $\theta$ is constructable, then so is $\cos \theta$ hence $[\mathbb{Q}(\cos\theta):\mathbb{Q}]$ is a power of 2. But since $\cos\theta$ is quadratic in terms of $\cos \theta/2$ so the later is also constructable.
Jan 22, 2014 at 19:42 comment added bof Is the same true for bisectable? An angle $\theta$ is bisectable if and only if $\cos(\theta/2)$ is constructible? I thought all angles were bisectable.
Jan 19, 2014 at 19:44 comment added user76758 The above is only addressing when the trisected angle can be built by straightedge and compass starting from nothing but unit length. This is sort of the "wrong" viewpoint, as Quas identifies the correct viewpoint: one seeks trisectability given the angle $\theta$, so it becomes an algebraicity question over $\mathbf{Q}(\cos \theta)$, which in turn looks basically hopeless for a nice answer.
Jan 19, 2014 at 19:41 comment added user76758 Since $\mathbf{Q}(e^{i \alpha})$ has degree 2 over $\mathbf{Q}(\cos(\alpha))$ for $\alpha\not\in \mathbf{Z}\pi$, necessarily $e^{i\theta/3}$ has 2-power degree over $\mathbf{Q}$. But $e^{i z} = e^{(z/\pi)(\pi i)}$ with $\pi i$ a complex log of the algebraic $-1 \ne 0, 1$, so by Gelfond-Schneider it's algebraic iff $z/\pi$ is rational. You seek rational $\alpha$ so that $e^{i \pi \alpha}$ has 2-power degree over $\mathbf{Q}$; the $3\alpha$'s are the $\theta$'s. So the answer is $\theta = a/b$ in reduced form with $b = 2^r \prod p_i$ for distinct Fermat primes $p_i > 3$ (!) (avoiding 6...).
Jan 19, 2014 at 18:41 comment added Manfred Weis thank you for your answer; while it certainly answers the question, I was hoping for a characterization via fractions of $2\pi$ which could eventually be presented in a classroom.
Jan 19, 2014 at 18:28 comment added Sh.M1972 Note that my answer and the answer of Antony Quas are essentially the same.
Jan 19, 2014 at 18:25 history answered Sh.M1972 CC BY-SA 3.0