Show that $\sin 10^\circ$ is irrational

Question

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?

Answer 1

identity: $\sin(3a)=3\sin(a)-4\sin^3(a)$ By using this identity,

$$1/2 = \sin 30^\circ = 3 \sin 10^\circ - 4\sin^3 10^\circ$$

$$1=2\sin 30^\circ = 6 \sin 10^\circ - 8\sin^3 10^\circ$$ Then if you set $x=2\sin(10)$ you will get $$x^3 - 3x+1 = 0.$$

Answer 2

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.