Proving $\sin\left(\frac{2\pi}{p}\right)$ is irrational where $p\geqslant3$ is prime

Question

Now I have proven that $\cos\left(\frac{2\pi}{p}\right)$ is irrational by taking $\left(x^2-2x\cos\left(\frac{2\pi}{p}\right) +1\right)$ as the minimal polynomial dividing $\frac{(x^p-1)}{(x-1)}\cdots 1$ and then using Eisentiens irreducibility criteria to show that $(1)$ is irreducible in $\Bbb Q[x]$ .So I wanted to ask can someone help me to find a similar polynomial where $\cos\left(\frac{2\pi}{p}\right)$is replaced by $\sin\left(\frac{2\pi}{p}\right)$.

Answer 1

It would be easy if you know the facts that

• the sum and product of two algebraic integers are again algebraic integers;
• $\Bbb Q$ is integrally closed, i.e. the only algebraic integers in $\Bbb Q$ are the integers, $\Bbb Z$.

Using these two facts, we write $2\sin(\frac{2\pi}p) = -i(e^{\frac{2\pi i}p} - e^{-\frac{2 \pi i}p})$, which shows that it is an algebraic integer.

If it is also rational, then it must be an integer. But $2\sin(\frac{2\pi}p)$ lies between $-2$ and $2$, leaving only $5$ possible values.