Tangent and transcendental numbers

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.

• What do you mean by "trivial cases involving $\pi$"? Can you please clarify your notion of "trivial"? – Xander Henderson Sep 5 at 16:33
• 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 $\pi$". – Wojowu Sep 5 at 16:34
• – Xander Henderson Sep 5 at 16:35
• I mean all rational multiples of $\pi$. – sam wolfe Sep 5 at 16:37
• @samwolfe If that is what you mean, please edit your question to explain. Also related: math.stackexchange.com/questions/2308528/… – Xander Henderson Sep 5 at 16:38

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.