Question

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

Answer 1

See the related graphs.

Explanation:

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 ]}