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 $(1)$?
Note: All that you want to show is that it's not rational.
Hint: Apply the rational root theorem. What possible rational roots are there? Now check if those are indeed roots.
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 $2 \Rightarrow\Leftarrow$.
One can use a feature of complex numbers, and the span of a finite set.
Consider the set of cyclotomic numbers, ie $C(n) = \operatorname{cis}(2\pi/n)^m$, where $\operatorname{cis}(x)=\cos(x)+i\sin(x)$. Such a set is closed to multiplication. The 'Z-span' of the set is the set of values $\sum(a_m \operatorname{cis}(2\pi/n)$, over n, is also closed to multiplication.
We now begin with the observation that a span of a finite set, closed to multiplication, can not include the fractions. This is proved by showing that if a rational number, not an integer, is in the set, so must all of its powers. (ie if $1/2$ is constructible by steps at multiples of $N°$, (eg a random walk of unit-size steps at exact degrees), so must all values of $1/2^a$).
Since this means that that the intersection of the cyclotomic numbers $\mathbb{C}_n$ and the rationals $\mathbb{F}$ can not include any fractions, and thus must give $\mathbb{Z}$.
The double-cosine of the half-angles, are given by $1-\operatorname{cis}(2\pi/n)$, and therefore we see that the only rational numbers that can occur in the sines and cosines, is $1/2$. The chord, and the supplement-chords are entirely free of rationals, and further, no product of such numbers can be rational.
