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I'm looking for an analytical solution to the equation $$\frac{\sin n\theta}{\sin m\theta}=a \quad (n,m\in\mathbb Z)$$ where the constant $a$ is real and can be both positive and negative.

The solution is needed for the unknown $\theta$; the rest of the parameters are known.

One approach I took was to write the sine functions as $\sim (e^{i x} - e^{-ix})$ which resulted in a polynomial equation of degrees set by $n$ and $m$, with no apparent analytical solution.

Is there an alternative route that can be taken, even by using special functions?

ACB
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tourmentedSoul
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    I don't see what did you mean by analytical solution? Do you want to solve 'n' and 'm' in terms of a ? – RAHUL Jan 29 '22 at 07:47
  • I want to solve for $\theta$, given the rest of the parameters. – tourmentedSoul Jan 29 '22 at 07:50
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    You can write, using the Gauss hypergeometric function, $$ \frac{{\sin (n\theta )}}{{\sin (m\theta )}} = \frac{n}{m}{}_2F_1! \left( {\frac{{1 + n/m}}{2},\frac{{1 - n/m}}{2};\frac{3}{2};\sin ^2 (m\theta )} \right). $$ Now you may use series reversion to write ${\sin ^2 (m\theta )}$ as a power series in $a$. This can then be easily solved for $\theta$. – Gary Jan 29 '22 at 07:59
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    @Gary How does one get this identity? Is it something standard? What surprises me is that the LHS might have a pole for $\sin(m \theta)=0$, but the RHS doesn't or am I missing something? – Severin Schraven Jan 29 '22 at 08:14
  • @SeverinSchraven You are correct, the identity is true only if $|m\theta|<\pi/2$. – Gary Jan 29 '22 at 08:18
  • @Gary Sorry, I know nothing about the hypergeometric function except that is defined as a power series around the origin in the last coordinate. So if that identity holds and I sent $\theta$ to $\pi/(2m)$, then the RHS will converge to the constant term of the power series expansion of the hypergeometric function, but the LHS will blow up (for the correct choice of $n,m$). Maybe we need a bit more distance to the bad point? Or where am I going wrong? – Severin Schraven Jan 29 '22 at 08:27
  • @SeverinSchraven If $\theta=\pi/(2m)$ then $\sin(m\theta)=1$ so there is no blowup on the LHS. On the other hand there is a branch point of the hypergeometric function at $1$, so you cannot go beyond this disk of validity. – Gary Jan 29 '22 at 08:32
  • @Gary Right. I should go to bed. Thank you for your patience – Severin Schraven Jan 29 '22 at 08:35
  • Thanks much @Gary, @Severin! I'll need some time to digest this :) – tourmentedSoul Jan 29 '22 at 08:36
  • @Gary, could you clarify the second step you suggested? Basically, if n, m, and a are known, doesn't the equation turn effectively into solving for z in: $_2 F_1(a,b,c;z) = d$, where all but $z$ is known? If so, is there an analytical way to solve for $z$? I couldn't find an inverse hypergeometric function that could do that. And if not, are you suggesting that the function form $\sin^2(m\theta)$ is important for finding $\theta$? – tourmentedSoul Jan 29 '22 at 09:05
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    You use the power series of ${}_2 F_1$ and the inversion. – Gary Jan 29 '22 at 09:20
  • That's what I thought I was supposed to do. Thanks again!! – tourmentedSoul Jan 29 '22 at 09:22
  • @Gary I have attempted without success to connect the hypergeometric formula I gave in my solution for $\sin(n \theta)$ to your formula upward for the quotient $\sin(n \theta)/\sin(m\theta)$. Do you know a way to do it ? Said otherwise, what are the general hypergeometric formulas that would give the result ? Thanks in advance. – Jean Marie Feb 01 '22 at 21:03
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    Substitute $x=\ln(w)$ to get a quadrinomial equation and use this method for a series solution – Тyma Gaidash Nov 29 '23 at 20:17

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You said you have attempted to convert your issue into a polynomial equation. But do you know that there already exists "on the shelf" Chebyshev polynomials of the second kind $U_n$ :

$$\displaystyle U_{n-1}(\cos \theta )=\frac {\sin n \,\theta}{\sin \theta }$$

giving

$$U_{n-1}(x)=a U_{m-1}(x)$$

which is a $\max(n,m)-1$ polynomial equation in $x:=\cos \theta$, keeping the roots whose absolute values are at most $1$.

Therefore, your question has essentially a negative answer: looking for an explicit analytical formula is elusive, because in general, above degree 4, most polynomial equations do not have "formulas" for their roots...

Remark 1: connection with hypergeometric function $ {}_{2}F_{1}$ (you find it in the Wikipedia article):

$$U_{n}(x)=(n+1)\ {}_{2}F_{1}\left(-n,n+2;{\tfrac {3}{2}};{\tfrac {1}{2}}(1-x)\right)$$

Remark 2: A completely different approach would be to write your equation under a form involving the cardinal sine function $\operatorname{sinc}$ i.e.,

$$ \frac{\operatorname{sinc}(n \theta)}{\operatorname{sinc}(m \theta)}=\frac{n}{m} a$$

Jean Marie
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