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

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Answer 1 · 2016-12-30T12:17:02

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

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

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]}

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