Let $\epsilon\in (0, \pi/2)$. Suppose that, $x\in\bigcup_{n\in\mathbb{Z}}[n\pi +\epsilon, (n+1)\pi -\epsilon]$. What is an upper bound for the function $$f(x)=\frac{\cos x}{\sin^3x}?$$
It is enough for me to know that it is bounded from above by a constant.
I would say that it is bounded by $1$ using the fact that $-1\le\cos x\le 1$ and $-1\le\sin x\le 1$, but I am not sure about that.
Could someone please help me?
Thank you in advance!
Answer for the question in your comment:Using the fact that $|\sin (n \pi+x)|=|\sin x|$ we see we only need a lower bound for $|\sin x|$ on $[\epsilon, \pi-\epsilon]$. But $\sin x$ is increasing on $[\epsilon, \frac {\pi} 2]$ and decreasing on $[\frac {\pi} 2, \pi-\epsilon]$. Now it is not difficult to see that $|\sin x| \geq |\sin \epsilon]$ on $[\epsilon, \pi-\epsilon]$. Hence, $|\frac {\cos x} {\sin^{3}(x)}| \leq \frac 1{\sin^{3}\epsilon}$.