Graph of the function $\frac{\sin 5\theta}{\sin^5 \theta}$. I was trying to  visualize a the graph of the function $\frac{\sin 5\theta}{\sin^5 \theta}$. After considering where the function is increasing and where it is decreasing  , I can get the idea of it's graph. Can anyone please check it if it is correct or not.
 A: I will provide a method  to graph this function over $[0,2\pi]$. The passages you have to follow are:


*

*Let $f(\theta)=\frac{\sin(5\theta)}{\sin^5(\theta)}$.

*Calculate the domain of $f(\theta)$. So, you have to impose $\sin^5(\theta)\neq0\leftrightarrow\theta\neq0 + k\pi, k\in Z$. These are the asymptotes.

*Compute the $\Im(f(\theta))$. This is very simple: $\Im(f(\theta))=[-4,+\infty]$.

*Calculate the intersections with $x=0$. In this case $f(0)$ is not defined. 

*Compute the intersections with $y=0$. You have $f(\theta)=0\leftrightarrow\theta=\frac{k}{5}\pi,k\in Z$.

*Calculate when $f'(\theta)=0$ where $f'(\theta)$ is the derivate of $f(\theta)$. You have that: $$f'(\theta)=5\csc^5(\theta)(\cos(5\theta)-\sin(5\theta)\cot(\theta))=0$$ 
The solutions are: $$\theta_1=\frac{\pi}{4}+k\pi\: \vee \:\theta_2=\frac{3\pi}{4}+k\pi\: \vee \: \theta_3=\frac{\pi}{2}+k\pi,k \in Z.$$
At $\theta_1=\frac{\pi}{4}+k\pi$ and at $\theta_2=\frac{3\pi}{4}+k\pi$ you have a minimum. 
At $\theta_3=\frac{\pi}{2}+k\pi$ a relative maximum. Also: $f(\theta_1)=f(\theta_2)=-4$ and $f(\theta_3)=1$.

*When $\theta$ approaches $0,\pi$ from both sides you have $f(\theta)\rightarrow+\infty   $.

*Graph the function $f(\theta)$ in $[0,2\pi]$:

The two plots are exactly the same.
