Chapter 5, Problem 9T

To determine

To evaluate: The period and the graph of the function for $y=\tan \left(2x-\frac{\pi }{2}\right)$.

Answer to Problem 9T

The period of the function $y=\tan \left(2x-\frac{\pi }{2}\right)$ is $\frac{\pi }{2}$.

Explanation of Solution

Given:

The given trigonometric function is,

$y=\tan \left(2x-\frac{\pi }{2}\right)$

Calculation:

The general equation of cosecant function is,

$y=a\tan kx$ (1)

The period of the cosecant function is,

$z=\frac{\pi }{k}$ (2)

The given cosecant function is,

$y=\tan \left(2x-\frac{\pi }{2}\right)$ (3)

Compare equation (1) and (3).

$k=2$

Substitute 2 for k in equation (3),

$z=\frac{\pi }{2}$

Hence, the period of the function $y=\tan \left(2x-\frac{\pi }{2}\right)$ is $\frac{\pi }{2}$.

The interval on which the graphs complete one period is $\left(-\frac{\pi }{2k},\frac{\pi }{2k}\right)$.

Substitute 2 for k in the above interval.

The interval on which the graphs complete one period is,

$\begin{array}{c}z=\left(-\frac{\pi }{2\times 2},\frac{\pi }{2\times 2}\right)\\ z=\left(-\frac{\pi }{4},\frac{\pi }{4}\right)\end{array}$

The graph of the function $y=\tan \left(2x-\frac{\pi }{2}\right)$ is,

Figure (1)

The Figure (1) shows the graph of one complete period of $\left[-\frac{\pi }{4},\frac{\pi }{4}\right]$.

The given function can also be written as,

$\begin{array}{l}y=\tan \left(2x-\frac{\pi }{2}\right)\\ y=\tan 2\left(x-\frac{\pi }{4}\right)\end{array}$

The above equation shows the graph shift to the right by $\frac{\pi }{4}$.