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Apart from the cases where $\theta$ is a rational multiple of $\pi$, if $\theta$ is a transcendental number, is $\tan(\theta)$ necessarily transcendental?

  • If yes, would it be possible, in some cases, to express both numbers via algebraic expressions, in the sense of using a finite number of basic algebraic operations plus exponentiation ($+,-,\times,\div,\wedge$) to express the number? For example, $2^{\sqrt{2}}$ is transcendental.
  • If no, is there an easy counterexample?

The main motivation behind my question is to understand the nature of the relation between an angle and its tangent.

Edit: Slight correction regarding the cases involving $\pi$ that I'm not interested in.

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    $\begingroup$ What do you mean by "trivial cases involving $\pi$"? Can you please clarify your notion of "trivial"? $\endgroup$ – Xander Henderson Sep 5 at 16:33
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    $\begingroup$ Let $\theta$ be any number such that $\tan(\theta)=\sqrt{2}$. By Lindemann-Weierstrass theorem this $\theta$ cannot be algebraic, and one can prove that $\theta$ is not a rational multiple of $\pi$ (fun fact, it was actually a problem at the Polish Mathematical Olympiad three years ago). Using Gelfond-Schneider theorem, you can then conclude $\theta$ is not an algebraic multiple of $\pi$. Hope that is enough to say $\theta$ is not a "trivial case involving $\pi$". $\endgroup$ – Wojowu Sep 5 at 16:34
  • $\begingroup$ Related: math.stackexchange.com/questions/112938/… $\endgroup$ – Xander Henderson Sep 5 at 16:35
  • $\begingroup$ I mean all rational multiples of $\pi$. $\endgroup$ – sam wolfe Sep 5 at 16:37
  • $\begingroup$ @samwolfe If that is what you mean, please edit your question to explain. Also related: math.stackexchange.com/questions/2308528/… $\endgroup$ – Xander Henderson Sep 5 at 16:38
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The answer is no, there are transcendental numbers $\theta$ which aren't rational multiples of $\pi$ such that $\tan(\theta)$ is algebraic. Indeed, since $\tan$ is onto $\mathbb R$, there exists a value of $\theta$ for which $\tan(\theta)=\sqrt{2}$. On one hand, by Lindemann-Weierstrass theorem, $\theta$ has to be transcendental, for otherwise $e^{i\theta}$ would be transcendental and the equality $$-\frac{i}{2}\frac{e^{i\theta}-e^{-i\theta}}{e^{i\theta}+e^{-i\theta}}=\tan(\theta)=\sqrt{2}$$ would be impossible. So now we want to argue that $\theta$ is not a rational multiple of $\pi$. Let me only outline the solution.

Suppose $\theta$ was a rational multiple of $\pi$. Then, because $\tan$ is $\pi$-periodic, the sequence $\tan(2^n\theta)$ for $n=0,1,2,\dots$ would take only finitely many values (since it only depends on $n$ modulo the denominator of $\theta/\pi$). However, by repeatedly using the tangent of double angle formula, we can derive $\tan(2^n\theta)=\frac{p_n}{q_n}\sqrt{2}$, where the sequences $p_n,q_n$ are defined by $p_0=q_0=1,p_{n+1}=2p_nq_n,q_{n+1}=q_n^2-2p_n^2$ and we can easily check that $p_n,q_n$ are nonzero and relatively prime. But then the values of $\tan(2^n\theta)$ are all clearly pairwise distinct, since $p_n$ is strictly increasing, which is a contradiction.

Just for fun, we can now conclude that $\theta$ is not an algebraic multiple of $\pi$ either, for if $\theta=\alpha\pi$, then $\alpha$ would be irrational and we would deduce that $e^{i\theta}=(e^{i\pi})^\alpha=(-1)^\alpha$ is transcendental by the Gelfond-Schneider theorem.

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