My attempt: Given $P:=\{0, 1, \cos\theta\}$, clearly we can construct the point $\sin\theta$ on the complex plane since $\sin\theta=\sqrt{1-\cos^2\theta}$, meaning $\sin\theta\in\mathbb{Q}(P)^{py}$. Now, I want to construct the point $e^{\theta i}$, because then I can trisect the angle $\theta$ by the assumption and get $e^{\frac{\theta i}{3}}$, then I can draw the perpendicular from $e^{\frac{\theta i}{3}}$ to the x-axis, which would intersect the x-axis at $\cos\frac{\theta}{3}$, done. But how exactly can I construct the point $e^{\theta i}$, or is it not constructible?
Any hint would be greatly appreciated.
Draw a circle of of radius $1$ with center $O$, as well as a diameter. Mark a point $X$ where the diameter meets the circle. (Let's picture the diameter as horizontal so that $OX$ forms the $x$-axis in the usual way.) Mark point $C$ at distance $\cos \theta$ along this axis, and construct perpendicular line there. Mark point $P$ where this perpendicular meets the circle (in either direction, it doesn't matter). Evidently, angle $\measuredangle XOP$ measures $\theta$.
Now, since you have an angle trisection tool/technique, you can find the point $Q$ on the circle such that $\measuredangle XOQ$ measures $\tfrac{\theta}{3}$. Now drop line through $Q$, perpendicular to $OX$, meeting the axis at point $R$. The length of $OR$ is $\cos \frac{\theta}{3}$, as desired.
Notice, that you need to be able to go back and forth between an angle and the cosine of the angle, both of which require a unit circle, an axis, and construction of perpendiculars.
You can play around with this setup here.