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I know that by Lindemann–Weierstrass theorem(LW) sine and cosine of non-zero algebraic numbers (in radians) produce results that are transcendental.

My question is what are the transcendentals produced? Are there known $\alpha \in \mathbb{Q}$ where $sin(\alpha) = r\pi$, with $r \in \mathbb{Q}$? What about $r \in \mathbb{E}$(constructible)? What about $r \in \mathbb{\overline{Q}}$(algebraic)? What about $\sin(\alpha) = re$? I'm not trying to ask multiple questions, I just want to know if some or all values with algebraic (or especially rational or constructible) input for trig functions produce a certain type or types of transcendental number.

I'm afraid theory is not my strongest suit, as my background is engineering. I'm investigating differential equations involving springs and exploring what kinds of numbers to expect for equations involving simple harmonic motion. Sometimes my angular frequency $\omega$ is rational or constructible. I find the proof of LW on Wikipedia a little dense and hard to follow, so please try to keep your answers approachable to a broader audience if you can.

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  • $\begingroup$ If $\sin(\alpha) = k\pi$ then $\sin(\sin(\alpha)) = 0$ and I personally don't think that happens for rational or algebraic $\alpha$. Probably it's not even EL. Constructible numbers are just algebraics of degree $2^k$. $\endgroup$ Jul 8, 2014 at 16:03
  • $\begingroup$ A promising way of thought : $\sin(\sin(\alpha))$ is the imaginary part of $\exp(i\sin(\alpha))$ which is $\exp(\exp(i\alpha)/2 - \exp(-i\alpha)/2)$ so we are really asking whether $\exp(\exp(\alpha))$ and $\exp(\exp(-\alpha))$ are linearly independent. $\endgroup$ Jul 8, 2014 at 16:11
  • $\begingroup$ You are assuming $k$ is an integer, correct? What if $k$ is rational or algebraic? That would make $\sin(\sin(\alpha))$ an algebraic number (for rational $k$, I'm not sure about algebraic $k$). So is there a way to test if $\exp(\exp(\alpha))$ and $\exp(\exp(-\alpha))$ are linearly independent? $\endgroup$
    – hatch22
    Jul 8, 2014 at 16:19
  • $\begingroup$ Also, multiples of $\pi$ are only one possible set of transcendental numbers. Are there others that $\sin(\alpha)$ might produce? $\endgroup$
    – hatch22
    Jul 8, 2014 at 16:28
  • $\begingroup$ I would like to answer this question but I don't have time to answer it right now so I'm just making this comment in order to find this question later. $\endgroup$ Nov 3, 2023 at 0:13

2 Answers 2

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Consider the following:$$\arccos(x)=\frac{\ln(x\pm\sqrt{x^2-1})}{i}+\frac{\pi}2\pm2\pi n, n=0,1,2,3,\cdots$$Assume C is algebraic. $\pi C=\frac{\ln(x\pm\sqrt{x^2-1})}{i}$$$\arccos(x)=\pi[C+\frac12\pm2n]$$$$x=\cos(\pi[C+\frac12\pm2n])=\cos(\pi)\cos(C+\frac12\pm2n)-\sin(\pi)\sin(C+\frac12\pm2n)=-\cos(C+\frac12\pm2n)$$$$x=-\cos(C+\frac12\pm2n)$$I have assumed C to be algebraic. But for C to be algebraic, the logarithm at the beginning must have had a non-algebraic $x$. I don't know how to do anything beyond this.

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In general one can conjecture that there are no algebraic relations among definable numbers unless such relations can be derived from their definitions.

In particular, Schanuel's conjecture implies that the sine of an algebraic number can't be an algebraic multiple of $\pi$: if $\alpha$ is algebraic, then so is $i\alpha$, and thus $i\alpha$ and $i\pi$ are linearly independent as $i\pi$ is transcendental. This means that Schanuel's conjecture implies that at least two of the four numbers $i\alpha$, $e^{i\alpha}$, $i\pi$ and $e^{i\pi}$ are algebraically independent, and since $e^{i\pi}=-1$ is clearly algebraic, it must be $e^{i\alpha}$ and $i\pi$ that are algebraically independent, meaning that $\sin x = (e^{i\alpha}-e^{-i\alpha})/2i$ is algebraically independent with $\pi$.

By the full Lindemann-Weierstrass theorem the sine of an nonzero real algebraic number can't be an algebraic multiple of $e$: if $\alpha$ is a nonzero real algebraic number, $i\alpha$ is linearly independent with $1$ since one is real and the other is imaginary and $\alpha$ and $i\alpha$ are both algebraic, so Schanuel's conjecture implies that the two numbers $e^{i\alpha}$ and $e$ are algebraically independent, and thus $\sin x = (e^{i\alpha}-e^{-i\alpha})/2i$ is also algrebraically independent with $e$.

To address the comments about EL numbers: A number is an EL number if it is expressible with arithmetic operations and exponential and logarithmic functions. The field of EL numbers is closed under trigonometric functions because if $x$ is an EL number, so is $\sin x = (e^{ix}-e^{-ix})/2i$ (note that $i=e^{\log(-1)/2}$ is an EL number), and there are similar formulas for other trigonometric functions. Similarly, the field of EL numbers is closed under inverse trigonomentric functions because, for example, $\arcsin(x)=i\log(\sqrt{1-x^2}-ix)$. Thus $\sin(\sin(x))$ is an EL number if and only if $x$ is an EL number.

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