How do you graph #y=-2csc(2x)+1#?

Question

2675 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-11-18T02:14:53

Using the Socratic graphic facility, I have inserted a graph.

The period of y is # (2pi)/2=pi#. It is convenient to choose one

period as #x in (-pi/2, pi/2). Note that y has infinite discontinuity at

the ends and the middle x = 0.

A Table for making the graph:

#(x, y): (3.31, -pi/3) (3, -pi/4) (3.1, -pi/6) (oo, 0)#

#(-1.3, pi/6) (-1, pi/4) (-1.3, pi/3)#

The part of the inserted graph that braces the y-axis is the graph for

this period.

ygraph{(y-1) sin (2x)+2=0 [-9.91, 10.09, -3.68, 6.32]}