Period of $\frac{\sin(Ny)}{\sin y}$ with $N$ odd?

Question

The function

$$f(y) = \displaystyle \frac{\sin(Ny)}{\sin y}$$

is periodic with period $2 \pi$ in general. But tracing the graphic of that function for $N$ odd it seems that for $0 \leq x < \pi$ it is the same as $\pi \leq x < 2 \pi$.

So can we state that the period of $f(y)$ with $N$ odd is just $\pi$?

I found only that if $N = 2k + 1$, $k \in \mathbb{Z}$, and $y' = y + \pi$, then

$$\sin(N y') = \sin[2k(y + \pi) + y + \pi] = \sin [(2k + 1) \pi + (2k + 1)y]$$

Is there an analytical method to prove that periodicity?

Answer 1

Note that

$$\sin (x+\pi) = -\sin x$$

for all $x$. Hence

$$\sin (x + k\pi) = (-1)^k\sin x.$$

So we have

$$\sin (N(y+\pi)) = \sin (Ny + N\pi) = (-1)^N\sin (Ny).$$

Thus, whenever $N$ and $M$ have the same parity,

$$\frac{\sin (Ny)}{\sin (My)}$$

has period $\pi$.

Answer 2

Writing sine in exponential form we have $$f(x+\pi) = \frac{e^{iN(x+\pi)}-e^{-iN(x+\pi)}}{e^{i(x+\pi)}-e^{-i(x+\pi)}}$$ Now for $N$ odd, we have $e^{-iN\pi} = e^{iN\pi} = e^{i\pi}=e^{-i\pi}=-1$, so $$f(x+\pi) = \frac{-e^{iNx}+e^{-iNx}}{-e^{ix}+e^{-ix}} = f(x)$$