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Inspired by a comment by @QC_QAOA on Question 3458920, which mentioned the ratio between the acute angles in a $3:4:5$ triangle, I would like to know if we can prove that this ratio is irrational.

The symmetries of the arguments of the functions suggested that this could be a fun and potentially tractable problem.

We can prove that the ratio is equal to $\displaystyle \frac{\tan^{-1}\frac{3}{4}}{\tan^{-1}\frac{4}{3}}=\frac{\log\left(\frac{7+24i}{25}\right)}{\log\left(\frac{-7+24i}{25}\right)}=0.69395$, with the complex-logarithm definition of $\tan^{-1}x$. So, it could also be expressed as a solution to $$25^{z}\left(7+24i\right)=25\left(-7+24i\right)^{z}$$

I had then tried to use $x^{p/q}=(x^p)^{1/q}$ but as the numbers are complex, it changed their values. The constant, nor its reciprocal, appear in the OEIS and I can't find it elsewhere.

We know that at least one of the angles (and quite likely both) is irrational, as their sum is $\frac\pi2$.

Jam
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    I believe that this ratio is in fact transcendental and that this follows from Lindemann–Weierstrass theorem but I'm not completely sure. – Peter Foreman Dec 01 '19 at 23:39
  • @PeterForeman Could you elaborate please? I think you might be right but I don't see how it works. – Jam Dec 01 '19 at 23:42
  • Well this theorem implies that $\log{(\alpha)}$ is transcendental for all algebraic $\alpha\ne1$ (as stated in this answer for example) and you have written the ratio as a single logarithm albeit with a complex base. – Peter Foreman Dec 01 '19 at 23:45
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    @PeterForeman Thanks, that makes sense. And it helped me find another source corroborating it: (F. M. S. Lima - Some transcendence results Lemma 4) which cites back to Niven. $\frac{\ln \alpha}{\ln \beta}\notin\mathbb{A}$ whenever $\alpha,\beta\in\mathbb{A}/{0}$ and $\frac{\ln \alpha}{\ln \beta}\notin\mathbb{Q}$. – Jam Dec 01 '19 at 23:51
  • (For posterity) On second thoughts regarding your prior comment, Peter, $\ln\alpha\notin\mathbb{A}$ and $\ln\beta\notin\mathbb{A}$ need not necessarily imply $\log_{\beta}\alpha\notin\mathbb{A}$ (take $\log_{2}4$ for example). We need the additional constraint that $\log_{\beta}\alpha\notin\mathbb{Q}$. – Jam Dec 02 '19 at 00:06
  • I didn't think I was implying this. I was just stating an implication of the theorem that seemed similar. – Peter Foreman Dec 02 '19 at 00:10
  • Nice response to that question. A definitive answer then in the negative – QC_QAOA Dec 02 '19 at 11:54

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I believe I've found a proof. Label the two angles, $a=\tan^{-1}\frac34,b=\tan^{-1}\frac43$, and their ratio, $z=\frac{a}{b}$. Then, as $a+b=\frac{\pi}{2}$, we have $z=\frac{\pi}{2b}-1$.

By Niven's theorem, the only values of $0\leq x\leq \frac{\pi}{2}$ that are a rational multiple of $\pi$ and have a rational $\sin$ value are $0,\frac{\pi}{6},\frac{\pi}{2}$. But, as $\sin b = \frac45$, and $b$ is not equal to those values, it cannot be not a rational multiple of $\pi$. Therefore $z$ must be irrational.

Jam
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