5
$\begingroup$

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 $(1)$?

$\endgroup$
  • 1
    $\begingroup$ You don't need to find the roots, you just need to show that they aren't rational. $\endgroup$ – Git Gud May 19 '13 at 16:00
  • 1
    $\begingroup$ I'm assuming you mean, e.g. $\sin(10^\circ)$ and not $\sin(10^0)$? $\endgroup$ – Namaste May 19 '13 at 16:01
  • 1
    $\begingroup$ Thanks Git Gud but how can i prove that they are not rational $\endgroup$ – juantheron May 19 '13 at 16:02
  • 1
    $\begingroup$ to put degree sign on angle use ^\circ between $ sign $\endgroup$ – iostream007 May 19 '13 at 16:02
  • 2
    $\begingroup$ Related : math.stackexchange.com/questions/87756/when-is-sinx-rational $\endgroup$ – lab bhattacharjee May 19 '13 at 16:04
3
$\begingroup$

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 $2 \Rightarrow\Leftarrow$.

$\endgroup$
9
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.