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I was asked to find a solution to $$\frac{\sin^2(nx)}{n^2\sin^2(x)}=2^{-1/2}$$ where $n$ is a fixed integer greater than 1.

Numerically, there's a solution just above 1/n so I decided to find this one (rather than one of the other roots). I hoped to find an obvious pattern that I could use to find some sort of closed-form solution.

It turns out that the power series of the root jumps out, numerically. Since x is only used inside a sine, the power series has only odd terms. The coefficients look like

1.0019063576966069771401037205331682344703361047262779848987578751140847/n +

0.466422280109844953629560577622677335783739044717/n^3 +

0.3247444758314010835749/n^5 + ...

which are annoyingly close to 1, 1/2, 1/3, ... but are clearly distinct.

Any idea how to solve this? This is, I'm sure, well beyond anything the original question-asker cares about, but now my interest is piqued.

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Let me nit-pick a little bit: what you wrote is not a trig identity but a trigonometric equation. – Mariano Suárez-Alvarez Oct 19 '10 at 5:08
The power series of what, exactly? – Hans Lundmark Oct 19 '10 at 6:37
@Hans: The power series of the (specified) root of the equation. – Charles Oct 20 '10 at 13:58
Ah, thanks. Now that it's written as a series in $n$ it's clear. – Hans Lundmark Oct 20 '10 at 14:17
My mistake -- I was writing it on paper with different variables and I mistranslated. – Charles Oct 20 '10 at 18:59
up vote 2 down vote accepted

Just the $x=\arccos\;y$ substitution suggested by Mariano suffices; what you will then have will be the equation $\left(\frac{U_{n-1}(y)}{n}\right)^2=\frac1{\sqrt{2}}$ where $U_n(y)$ is the Chebyshev polynomial of the second kind.

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Let $x=\arccos y$, replace the $\sin^2$ in the numerator by $1-\cos^2$, and use the trigonometric definition of Chebyshev's polynomials to transform the equation into a polynomial equation on $y$. Then solve that using any of the usual ways to solve it (numerically!)

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