Question

How do you sketch one cycle of `y=cscx`?

Answer 1

Like a `sin` graph, but opposite.

Explanation:

`csc(x)=1/sin(x)`

Sketch a `sin` graph, then draw loops down to meet each maximum or minimum point.

graph{csc(x) [-10, 10, -5, 5]}

Notice that at each stationary point on this graph, there would usually be a stationary point for `sin(x)`.

Each time the `sin` graph would approach `0`, the `csc` graph will approach `oo` or `-oo`. Each time the `sin` graph approaches `+-1`, the `csc` graph approaches `+-1`.

These are because

`csc(x)=1/sin(x)`

`sin(x)=0 -> csc(x) = 1/0 = oo`
`sin(x)=1 -> csc(x) = 1/1 = 1`