TITLE: How can I prove that $\sin (10^\circ), \sin(1^\circ), \sin(2^\circ), \sin(3^\circ), \tan(10^\circ)$ are irrational
QUESTION [5 upvotes]: Prove that $\sin (10^\circ)$, $\sin(1^\circ)$, $\sin(2^\circ)$, $\sin(3^\circ)$, and $\tan(10^\circ)$ are irrational.

My Attempt:
Let $x = 10^\circ$. Then
$$
\begin{align}
x &= 10^\circ\\
3x &= 30^\circ\\
\sin (3x) &= \sin (30^\circ)\\
3\sin (10^\circ)-4\sin^3(0^\circ) &= \frac{1}{2}
\end{align}
$$
Now let $y = \sin (10^\circ)$. Then
$$
\begin{align}
3y-4y^3 &= \frac{1}{2}\\
6y-8y^3 &= 1\\
\tag18y^3-6y+1 &= 0
\end{align}
$$
How can I calculate the roots of $(1)$?

REPLY [3 votes]: Here is a different approach for $\sin(1^\circ)=\sin\left(\frac{2\pi}{360}\right)$ (the other cases are similar).
The complex numbers $\zeta_{1,2}=e^{\pm\frac{2\pi}{360}i}$ are algebraic integers (are roots of $x^{360}-1$) and therefore $\zeta_1+\zeta_2=2\sin\left(\frac{2\pi}{360}\right)$ is algebraic integer. If $\sin\left(\frac{2\pi}{360}\right)$ is rational then $2\sin\left(\frac{2\pi}{360}\right)$ is rational and therefore integer (the only rational algebraic integers are the integers). 
So $2\sin(1^\circ)=-2,-1,0,1$ or $2 \Rightarrow\Leftarrow$.