How do you graph #f(t)=3sec(2t)# by first sketching the related sine and cosine graphs?

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Answer 1 · 2018-07-17T12:30:20

See the related graphs.

The related cosine graph is for #f ( t ) cos (2 t ) = 3#, with

# 2t ne ( 2k + 1 ) pi/2 rArr t ne ( 2k + 1 ) pi/4, k = 0, +-1,+-2,+-3, ...#.

Graph of # f (t) = 3 sec 2t#,:
graph{ cos ( 2x )- 3/y = 0}

The related sine graph is from # f(t) = 3 cosec( pi/2 - 2t )#

#rArr f(t) sin( pi/2 - 2t ) = 3#. You ought to get the same graph

from this form also.

graph{ sin (pi/2 - 2x) - 3/y = 0}

Graph revealing asymptotes #t = +- pi/2#,

for one period #((2pi)/2) = pi, t in ( - pi/2, pi/ 2 )#:

graph{(cos ( 2x )- 3/y)(x^2-(pi/4)^2) = 0[-1.57 1.57-20 20 ]}