We recall that

`csc x=1/(sin x)`

So we will have asymptotes where `sin x` has value zero, that is:

x = ..., -3π, -2π, -π, 0, π, 2π, 3π, 4π, ...

We draw the graph of y = sin x first:

π−π-2π1-1xyOpen image in a new page

Graph of `y=sin x`.

Next, we consider the reciprocals of all the y-values in the above graph (similar to what we did with the y = sec x table we created above).

`x` `y` `= sin x` `csc x` `= 1/(sin x)`
0.01 0.01 100
0.5 0.48 2.09
`pi/2` 1 1
2 0.91 1.10
3 0.14 7.09
3.1 0.04 24.05

I chose values close to `0` and `pi`, and some values in between. The pattern will be similar for the region from `pi` to `2pi` except it will be on the negative side of the axis.

We continue on both sides and realise the pattern will repeat. Now for the graph of y = csc x:

π−π-2π246810-2-4-6-8-10xyOpen image in a new page

Graph of y = csc x.