TITLE: Can a single trigonometric term be the solutions of a depressed cubic?
QUESTION [0 upvotes]: For example a term like $\cos\big(\frac{\pi}{7}\big)$ (where the argument is a rational multiple of $\pi$), not including terms like $\cos\big(\frac{\pi}{7}\big)+1$

REPLY [3 votes]: Yes, for example $\,\cos\left(\frac{\pi}{9}\right)\,$ is a root of $\,8 x^3 - 6 x - 1\,$, per the triple angle cosine identity:
$$
\frac{1}{2} = \cos \left(3 \cdot \frac{\pi}{9}\right) = 4 \cos^3\left(\frac{\pi}{9}\right) - 3 \cos\left(\frac{\pi}{9}\right)
$$
{"set_name": "stack_exchange", "score": 0, "question_id": 2715062}