Show that $\sin 10^\circ$ is irrational So, this is the problem I am working on. 
Show that $\sin 10^\circ$ is irrational. 
The solution to the problem is $$1/2 = \sin 30^\circ = 3 \sin 10^\circ - 4\sin^3 10^\circ .$$
Let $$x = 2\sin 10^\circ.$$ 
Then we have, $$x^3 - 3x + 1 = 0.$$ 
And, we have to work on this to find out the roots. But, what I don't understand is that why do I have to subtract $4\sin^3 10^\circ$ from $3\sin 10^\circ$. And,  how did they come up with $x^3 - 3x+1 = 0?$ I am confused. Can someone please explain this in details and is there any other way we can do this problem?

mesel's answer about why $2\sin(10^\circ)$ satisfies $x^3-3x+1=0$ is very good.
Let's answer the question about why $x^3-3x+1=0$ implies $x$ is irrational. In this answer, it is shown that if $x^3-3x+1=0$ has a rational root, then that root must be an integer.
Suppose that $|x|\ge2$, then dividing by $x^3$ yields
$$
\begin{align}
1
&=\left|\,\frac3{x^2}-\frac1{x^3}\,\right|\\
&\le\frac34+\frac18\\
&=\frac78
\end{align}
$$
Thus, there can be no solutions for $|x|\ge2$.
Simply checking $\{-1,0,1\}$, we see there are no integer solutions. Therefore, any solution to $x^3-3x+1=0$ is irrational.
