Help needed on graphing on the unit circle? The problem is as follows:

Graph $f(x) = \csc x$ on the interval $[0,2\pi]$.

Any advice on how to graph would be appreciated.
 A: This is the sort of question where it helps to have a good understanding of the graph of $g(x) = \sin x$, and it helps to recall the fact that $f(x) = \csc x = \frac 1{\sin x}$.
Knowing that $\,-1 \leq \sin x \leq 1\,$ for all $x \in \mathbb R$ helps you find the "range" of the function $f(x) = \csc x = \dfrac 1{\sin x}$.
Plot $f(\pi/2) = \dfrac{1}{\sin(\pi/2)} = \dfrac 11 = 1.\;$ Plot $\;f(3\pi/2) = \dfrac{1}{\sin(3\pi/2)} = \dfrac{1}{-1} = -1.$ 
Draw the vertical lines $\,x = 0,\; x = \pi,\; x = 2\pi.\;$ The graph of $f(x)$ will never intersect any of those vertical lines. The lines I've listed are called vertical asymptotes. $\csc x = \dfrac 1{\sin x}$ is not defined at $x = 0, x = \pi...$ nor at any integer multiple of $\pi$. Why not?
Calculate the values $f(x)$ when $x$ takes on values between the first two points plotted and the vertical lines on each side of those you've drawn, and plot some of the corresponding ordered pairs $(x, f(x))$, when $x \in (0, 2\pi), \;x\neq \pi$.
For more intuition, compare the graphs of $\sin x$ and $\csc x = \dfrac 1{\sin x},\,$ below, and compare your "sketch-in-progress" of $f(x) = \csc x = \dfrac 1{\sin x}$.

