Question 5.4.6: A cosecant function with a transformation Sketch two cycles ...
A cosecant function with a transformation
Sketch two cycles of the graph of y = csc(2x – 2π/3) and determine the period and the range of the function.
Since
\begin{aligned} y &=\csc (2 x-2 \pi / 3) \\ &=\frac{1}{\sin [2(x-\pi / 3)]} \end{aligned}we first graph y = sin[2(x – π/3)]. The period for y = sin[2(x – π/3)] is π with phase shift of π/3. So the fundamental cycle of y = sin x is transformed to occur on the interval [π/3, 4π/3]. Draw at least two cycles of y = sin[2(x – π/3)] with a vertical asymptote (for the cosecant function) at every x-intercept, as shown in Fig. 5.73. Each portion of y = csc[2(x – π/3)] that opens up has a minimum value of 1, and each portion that opens down has a maximum value of -1. The period of the function is p, and the range is (-\infty,-1] \cup[1, \infty).
The calculator graph in Fig. 5.74 supports these conclusions.
