Let the figure below
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According to same notation of the figure verify if it's possible to construct the point $\displaystyle \zeta=e^{\frac{2\pi i}{13}}$ with straight-edge and compass from $X=\{0,1,\vartheta \}$
I'm not sure if what I did it's right.
First note $\mathbb Q(e^{\frac{2\pi i}{13}})=\mathbb Q(e^{\frac{4\pi i}{13}})$ (they are coprime with $13$). Thus the problem is equivalent to the constructibility of $e^{\frac{4\pi i}{13}}$. Then observe that $\operatorname{Re}(e^{\frac{4\pi i}{13}})=\vartheta$. Thus it's enough to construct $\operatorname{Im}(e^{\frac{4\pi i}{13}})= \sin(\frac{4\pi}{13})$. From the identity:
$$ \sin\left(\frac{4\pi}{13}\right)^2 = 1-\cos^2\left(\frac{4\pi}{13}\right)$$ we deduce that $\sin(\frac{4\pi}{13})^2 \in \mathbb Q (X)$ then $\sin(\frac{4\pi}{13})^2$ is constructible. But since the constructible numbers are closed under squares then $\sin(\frac{4\pi}{13})$ is constructible. Then $e^{\frac{4\pi i}{13}}$ is constructible.