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About the transcendence of trigonometric functions I know that:

1) if $x$ is an algebraic number $\ne 0$ than $\cos x$ is transcendental.

2) if $p=\dfrac{m}{2^n}$ with $m,n \in \mathbb{Z}$ than $\cos (p\pi)$ is a constructible number (using bisection formula).

After the comments and the answer it's also proved that:

3) For any rational number $q$, $\cos (q\pi)$ is algebraic ? (since $e^{iq \pi}$ is a root of unity, hence algebraic)

So the open questions are:

4) if $a$ is an algebraic number than $\cos (a \pi)$ is algebraic or transcendental?

5) there exists a transcendental number $\alpha$ such that $\alpha \ne r \pi , \forall r \in \mathbb{Q}$ and if $p=\dfrac{m}{2^n}$ with $m,n > \in \mathbb{Z}$ than $\cos (p\alpha)$ is a constructible number?

Someone know if there is some answer to this related questions?

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    $\begingroup$ Part 3 is easy. If $q$ is rational, then $z=e^{\pi i q}$ is a root of unity, hence algebraic. You get the cosine as $\dfrac12(z+\dfrac1z)$ as a sum of two algebraic numbers. $\endgroup$ – Jyrki Lahtonen Apr 21 '15 at 10:13
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    $\begingroup$ There is, of course, an exception to 1), namely, $x=0$. $\endgroup$ – Gerry Myerson Apr 21 '15 at 10:13
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    $\begingroup$ There are many constructible numbers (between $-1$ and $1$) that are not of the form $\cos(r\pi)$ with $r$ rational. Choose one such. It will be $\cos \alpha$ for some $\alpha$. Then $\cos p\alpha$ will be constructible for $p=m/2^n$. $\endgroup$ – Gerry Myerson Apr 21 '15 at 10:18
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    $\begingroup$ I was thinking along the same lines as Gerry, but I got stuck in justifying why, say, $\alpha=\arccos(1/\sqrt7)$ would be transcendental. It does follow immediately from item 1, so ... :-) Only the problem of proving that it isn't a rational multiple of $\pi$ remains. But that may be a tractable one. $\endgroup$ – Jyrki Lahtonen Apr 21 '15 at 10:23
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    $\begingroup$ Algebraic conjugates = zeros of the same polynomial with rational coefficients. For example let $u$ be one of the complex numbers on the unit circle such that its real part is $\sqrt3-\sqrt2$. One of its conjugates (probably, I didn't check everything) has real part $\sqrt3+\sqrt2$, and hence cannot be on the unit circle. $\endgroup$ – Jyrki Lahtonen Apr 21 '15 at 15:10
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(1) to show that $\cos(\frac {m}{n} \pi)$ is algebraic it is enough to show that $\cos(\frac{1}{n} 2\pi)$ is algebraic. So it comes down to showing that the real part of an $n$-th root of unity is algebraic.

(3) there is a transcendental $a$ not equal to any rational multiple of $\pi$. Just take square root of $(\pi -1)$, lest $\pi$ would be algebraic.

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  • $\begingroup$ Your statement (1) answers my question 3), but your (3) does not fit my question 5). And you have some idea about my 4)? $\endgroup$ – Emilio Novati Apr 21 '15 at 14:06

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