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What is the value of n $\in \mathbb{Z} $ for which the function $\displaystyle f(x) = \frac{\sin nx} { \sin \biggl( \frac{x}{n} \biggr) } \text { has } 4\pi $ as period?

Also could it be possible to solve this if we need $x\pi$ as period ?I am interested in learning the general approach for this particular type of the problem.

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    @ Bryan Yocks : Done!2010-11-30

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With $n=2$, $\sin(2x)$ has period $\pi$ and $\sin(x/2)$ has period $4\pi$ so their ratio must have a period of $4\pi$ since the latter period is an integral multiple of the former.

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    @Debanjan: And did you plug in and check? From where I am sitting, if $k$ is$a$common integer multiple of the periods of both $f$ and $g$, then $f/g$ and $fg$ are both periodic with period $k$. But you claim it does not.2010-12-01
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You want $\frac{\sin n(x + 4\pi)}{\sin \frac{x + 4\pi}{n}} = \frac{\sin nx}{\sin \frac{x}{n}}.$

This is equivalent with $\sin \frac{x + 4\pi}{n} = \sin \frac{x}{n}.$

Therefore $\frac{x}{n} = \frac{x + 4\pi}{n} + 2k\pi$ or $\frac{x}{n} = \pi - \frac{x + 4\pi}{n} + 2k\pi$ for some $k \in \mathbb{Z}$. In the first case $x = x + 4 \pi + 2k \pi n$ and thus $n = \frac{4}{2k}$. In the second case $x = \pi n - x - 4\pi + 2k\pi n$ and thus $n = \frac{2x + 4\pi}{2k\pi + \pi}$, which is impossible since this should hold for every $x \in \mathbb{R}$.

Thus $n = \pm 1$ or $\pm 2$.