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I was just curious on wolframalpha and typed in : ${\rm arcsin}(\frac {\pi}{3})$ of course, it's a complex number. It was able to show me the number and i couldn't help but notice it had an exact value for it's magnitude in polar form. It being : $\frac 12\sqrt{\pi^2+ 4 \rm log^2(\frac 13(\pi + \sqrt{\pi^2-9}))}$

I think it looks insane, how would you find/prove this kind of result ?

Antony Theo.
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pebao
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    Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking. – Community Feb 17 '25 at 15:10
  • Arsin makes me think of some deadly poison. Aren't you referring to $\arcsin$? – Dominique Feb 17 '25 at 15:22
  • @Dominique the use of "ar" instead of "arc" for inverse trigonometric and hyperbolic functions has gained traction. "Ar" means area, and both inverse trig and inverse hyperbolic functions can be interpretes in terms of sector area; whereas "arc" for length of an arc doesn't work for the hyperbolic case unless you use a Lorentzian pseudo-metric. – Oscar Lanzi Feb 17 '25 at 15:27
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    @Community The question is very clear. I wonder why your comment was upvoted. – Anne Bauval Feb 17 '25 at 15:28
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    This question is similar to: The logarithmic form of $\text{arcsin}(x)$ and its implications. If you believe it’s different, please [edit] the question, make it clear how it’s different and/or how the answers on that question are not helpful for your problem. – Mark S. Feb 17 '25 at 15:35
  • @AnneBauval (I don't think the Community bot receives pings.) – Mark S. Feb 17 '25 at 15:35
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    @MarkS. Neither do I. It was just a way to share with other users my disagreement with the bot's comment. – Anne Bauval Feb 17 '25 at 15:39

1 Answers1

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The Logarithmic form for the inverse sine of a complex number $z$ is $$\arcsin(z)=i\log\left(\sqrt{1-z^2}-iz\right)\stackrel{z:=\pi/3}{\longrightarrow}\arcsin(\pi/3)=i\log\left(\sqrt{1-\left(\frac\pi3\right)^2}-i\frac\pi3\right)$$ And the result follows.

Anne Bauval
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Antony Theo.
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