Show that $\cos\big(\frac{2\pi}{n}\big)$ is an algebraic number

A proof in an entirely different manner:

Consider the matrix $\begin{bmatrix}\cos(\phi)&-\sin(\phi)\\\sin(\phi)&\cos(\phi)\end{bmatrix}$. This is a rotation matrix.

If we take $\phi=\frac{2\pi}{n}$ for some natural number n, we get the following equation: $$\vec{v}\begin{bmatrix}\cos(\phi)&-\sin(\phi)\\\sin(\phi)&\cos(\phi)\end{bmatrix}^n=\vec{v}$$ for all vectors $\vec{v}$. We can now use this identity to generate a polynomial equation in terms of $\cos(\frac{2\pi}{n})$ and $\sin(\frac{2\pi}{n})$. By using $\sin(\frac{2\pi}{n})=\sqrt{1-\cos^2(\frac{2\pi}{n})}$, we can make this into a polynomial in terms of $\cos(\frac{2\pi}{n})$ where all coefficients are integers and all exponents are either integers or fractions with a denominator of 2. By isolating terms and squaring, we can get a polynomial of integer coefficients and integer powers, which means that the variable, $\cos(\frac{2\pi}{n})$, must be algebraic, (as is $\sin(\frac{2\pi}{n})$).

Here’s another approach: suppose $z=x+iy \in \mathbb{C}$ is an algebraic number, $x,y \in \mathbb{R}$. Sums and products of algebraic numbers are algebraic, and any complex conjugate of an algebraic number is algebraic (ask me about why if you don’t know why). So $$x=\frac{z+\overline{z}}{2}$$ is algebraic, just as $$y=\frac{z-\overline{z}}{2i}$$. So we see, real and imaginary parts of algebraic numbers are algebraic! Now $\cos ( \frac{2 \pi}{n})$ is the real part of $$z=e^{\frac{ 2 \pi i}{n}}$$ which satisfies $$z^n-1=0$$ so is algebraic.

This looks good to me. The only note I have is that the argument really uses strong induction rather than weak induction. More specifically, you are assuming the induction hypothesis holds for all $n\leq k$ in order to prove the result for $n=k+1$ (at least you need it for $n=k,k-1$). But this is maybe a minor technical thing; the proof is otherwise good.