During my work, I encounter the function like $\frac{\sin(n \omega)}{n \sin \omega}$. I'm puzzled by this function and knew nothing about this function before. There are several problems to be solved.

Given integer $n>1$, how to find a simple function to bound $|\frac{\sin(n \omega)}{n \sin \omega}|$ as tight as possible for $0 \le \omega \le \pi/2$. That is to say, we need to find a polynomial $g(n,\omega)$ such that $|\frac{\sin(n \omega)}{n \sin \omega}| \le g(n,\omega)$ and $g(n,\omega)$ is as close to $|\frac{\sin(n \omega)}{n \sin \omega}|$ as possible. To the simpleness, I think polynomials with lower order meet the requirements.

My idea is to bound the absolute value individually. For $0 \le \omega \le \frac{\pi}{2n}$, we bound $|\frac{\sin(n \omega)}{n \sin \omega}|$ by Taylor expansion at $\omega = 0$. However, when $\frac{\pi}{2n} \le \omega \le \frac{\pi}{2}$, I don't know how to find a appropriate polynomial to bound $|\frac{\sin(n \omega)}{n \sin \omega}|$.

For $\frac{\pi}{2n} \le \omega \le \frac{\pi}{2}$, the envelope of $|\frac{\sin(n \omega)}{n \sin \omega}|$ is $\frac{1}{n \sin \omega}$, i.e., $|\frac{\sin(n \omega)}{n \sin \omega}| \le \frac{1}{n \sin \omega}$. One of my ideas is to find a simple polynomial to bound $\frac{1}{n \sin \omega}$ as tight as possible. But how to find?

Thanks for any help!

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