$\sin(\frac{2\pi j}{m})^{2}$ is distinct for all $j=1,..., \frac{m-1}{2}$

Question

Let $m$ be an odd positive integer.

I showed that

• $\frac{\sin(mx)}{\sin(x)}$ is a polynomial in $\sin(x)^2$ with degree $\leq \frac{m-1}{2}$ (by induction)

• $\frac{\sin(mx)}{\sin(x)} = 0$ for $x =\frac{2\pi j}{m}$ and $j=1,..., \frac{m-1}{2}$ (obviously).

Now I need to show that $\frac{\sin(mx)}{\sin(x)}$ is a polynomial in $\sin(x)^2$ with degree $= \frac{m-1}{2}$ and that $C$ so that $\frac{\sin(mx)}{\sin(x)} = C \cdot \prod\limits_{j=1}^{(m-1)/2} (\sin(x)^{2} - \sin(\frac{2\pi j}{m})^{2})$ exists.

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$\frac{sin(mx)}{sin(x)}$ has degree $\leq \frac{m-1}{2}$ and if we show that $sin(\frac{2\pi j}{m})^{2}$ is distinct for all $j=1,..., \frac{m-1}{2}$ we should have proved it (?)

I tried to use the unit circle:

If $sin(\frac{2\pi j}{m})^{2}$ shall has to be distinct for all $j$, so has to be $sin(\frac{2\pi j}{m})$. That should make things easier, but I struggle here.

Answer 1

Notice that $4\sin(t/2)^2=2-2\cos(t)$ and go from there.

Answer 2

Use Prove $ \sin(A+B)\sin(A-B)=\sin^2A-\sin^2B $ for $j_1,j_2$

We need $\dfrac{2\pi}m(j_1\pm j_2)=n\pi$ for some integer $n$

$\iff j_1\pm j_2=\dfrac{mn}2$