Question

How do you sketch one cycle of `y=sec(2x)`?

Answer 1

Graph data: Period : `pi; x in[-pi/2, pi/2]. y in [-pi/2, pi/2]`;
the asymptotes in this single period: `uarr x=+-pi/4 darr`.

Explanation:

`|y|=|sec(2x)|>=1`

Period is`(2pi)/2=pi=3.1416`, nearly,

`y to +-oo`, as `x to (2k+1)pi/2, k = 0, +-1, +-2, +-3, ...`The

graph{(y-1/cos(2x))=0 [-1.57, 1.57, -1.57, 1.57]}