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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 $(1)$?

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You don't need to find the roots, you just need to show that they aren't rational. – Git Gud May 19 '13 at 16:00
I'm assuming you mean, e.g. $\sin(10^\circ)$ and not $\sin(10^0)$? – amWhy May 19 '13 at 16:01
Thanks Git Gud but how can i prove that they are not rational – juantheron May 19 '13 at 16:02
to put degree sign on angle use ^\circ between $ sign – iostream007 May 19 '13 at 16:02
up vote 3 down vote accepted

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 $2 \Rightarrow\Leftarrow$.

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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.

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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.

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