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I am looking for a formula for $\sin(n\theta)$ in terms of $\sin(\theta)$, where $n \in \Bbb R$. The multiple-angle formula only seems to work with integers, because of the $\prod$-function. Is there a generalization?

user110391
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    Probably $n\in\Bbb Q$ would give multiple/fractional angle type formulas. How would you expand $\sin(\sqrt 2\theta)$, which is transcendental, for example? – Тyma Gaidash Jul 26 '23 at 01:24
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    Why not just use de Moivre's theorem? – Chris Lewis Jul 26 '23 at 02:27
  • @ChrisLewis How would it be helpful? – user110391 Jul 26 '23 at 02:59
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    $\sin{n\theta}=\Im \left(\sqrt{1-\sin^2 {\theta}}+i \sin{\theta}\right)^n$, although you have to be careful about the sign of the square root (so you'd get a piecewise defined function). – Chris Lewis Jul 26 '23 at 03:28
  • I realise this is the obvious answer, but I'm wondering why it wouldn't be an answer to your question as it's worded. As @TymaGaidash says, you're going to run into difficulties with irrational values of $n$ with any other formulation. – Chris Lewis Jul 26 '23 at 03:34
  • A even less helpful answer would be $\sin\left(n,\sin^{-1}(\sin \theta)\right)$ – Henry Jul 26 '23 at 08:10
  • @ChrisLewis What is the function $\displaystyle \mathfrak J(x)$ called? I had no idea about this, so it very well might be the obvious answer, I am just not knowledgeable about this :) – user110391 Jul 26 '23 at 21:31
  • It just means the imaginary part. You can avoid it altogether if you instead write $$\sin{n\theta}=\frac{1}{2i}\left(\left(\sqrt{1-\sin^2 {\theta}}+i \sin{\theta}\right)^n-\left(\sqrt{1-\sin^2 {\theta}}-i \sin{\theta}\right)^n\right)$$ but I'm not sure how useful this is for your particular question – Chris Lewis Jul 26 '23 at 22:02
  • @ChrisLewis Well, I am a bit confused as to how the ratio of the opposite to the hypotenuse could be a complex number. Is there a way to convert it into a real number? – user110391 Aug 04 '23 at 22:10
  • The imaginary part of a complex number is real (for instance, the imaginary part of $2+3i$ is $3$). Have you come across $e^{i\theta}=\cos{\theta}+i\sin{\theta}$ before? If not, the references I made to de Moivre's theorem won't make much sense. – Chris Lewis Aug 04 '23 at 22:46
  • @ChrisLewis I have come across that equation yes, though I don't remember the proof. I thought the imaginary part of $2 + 3i$ would be $3i$, but I see now its just the multiple of $i$. However, the equation for $\sin n\theta$ you showed has multiple $i$'s, so taking the imaginary part would mean factoring a little, in order to get one factor of $i$, and then just removing it? – user110391 Aug 04 '23 at 22:49
  • @ChrisLewis Or is there perhaps some cancelling out between the $i$ in the denominator and the $i$'s in the numerator? – user110391 Aug 04 '23 at 22:55
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    The $i$ cancels out, as you say. It's similar to the formula $\sin{\theta}=\frac{1}{2i}\left(e^{i\theta}-e^{-i\theta}\right)$ – Chris Lewis Aug 04 '23 at 23:48
  • @ChrisLewis Okay, thanks :) If you make this into an answer (linking to a proof of it, or proving it yourself), I will accept it. Feel free to add any other formulas. I don't know yet if I will be able to use this formula to help me with my current problem, so any other formulas are welcome as well. – user110391 Aug 04 '23 at 23:52
  • @ChrisLewis I just read De Moivre's formula doesn't hold for integer $n$'s. If your formula is derived from De Moivre's formula, how is it able to hold for non-integers? – user110391 Aug 05 '23 at 03:37
  • I think you missed my earlier point that I didn't think this was a useful answer!! (I think, by the way, there are extensions to de Moivre you can use for non integer powers, but it gets messy). You mention you have a current problem you're working on, can you share that? – Chris Lewis Aug 05 '23 at 09:01
  • @ChrisLewis Since it doesn't work for non-integers, it isn't a useless answer; instead, it isn't an answer at all to my question. The problem I was working on was wanting to take the expression $c\arcsin(a) + \arcsin(b)$ and turn it into $c\arcsin(a) + c\arcsin(f_{1/c}(b))$, where $f_z(x)$ is a function that takes $\sin(x)$ to $\sin(zx)$. – user110391 Aug 06 '23 at 09:18
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    Read https://ccrma.stanford.edu/~jos/st/De_Moivre_s_Theorem.html#:~:text=can%20be%20any%20real%20number%2C%20not%20just%20an%20integer. – NadiKeUssPar Aug 06 '23 at 09:33

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