Chapter 13.6, Problem 23PPSE

Solution

To sketch: The graph of the function $y=\tan \left(2\theta \right)$ in the interval from $0\text{ to }2\pi$ .

Given:

The tangent function is $y=\tan \left(2\theta \right)$ .

Concept used:

The tangent curve of the form $y=a\tan \left(b\theta \right)$ where $a\ne 0$ and $b>0$ has period $\frac{\pi }{b}$ , the number of cycles in the interval from $-\frac{\pi }{2b}\text{ to }\frac{\pi }{2b}$ is $1$ and two vertical asymptotes $-\frac{\pi }{2b},\frac{\pi }{2b}$ at each end of the cycle.

Graph:

With the help of concept stated above, evaluate the period $\left(T\right)$ , one cycle in the interval from $-\frac{\pi }{2b}\text{ to }\frac{\pi }{2b}$ and two vertical asymptotes of the function $y=\tan \left(\theta \right)$ as:

  $\begin{array}{c}\text{Period}\left(T\right)=\frac{\pi }{b}\\ =\frac{\pi }{\left(2\right)}\mathrm{....}\left(\because b=2\right)\\ =\frac{\pi }{2}\end{array}$

  $\begin{array}{c}\text{One cycle of the interval}=-\frac{\pi }{2b}\text{ to }\frac{\pi }{2b}\\ =-\frac{\pi }{2\left(2\right)}\text{ to }\frac{\pi }{2\left(2\right)}\mathrm{....}\left(\because b=2\right)\\ =-\frac{\pi }{4}\text{ to }\frac{\pi }{4}\end{array}$

  $\begin{array}{c}\text{Two vertical asymptotes}\left(\theta \right)=-\frac{\pi }{2b},\frac{\pi }{2b}\\ =-\frac{\pi }{2\cdot \left(2\right)},\frac{\pi }{2\cdot \left(2\right)}\mathrm{....}\left(\because b=2\right)\\ =-\frac{\pi }{4},\frac{\pi }{4}\end{array}$

So, three consecutive vertical asymptotes occur at $-\frac{\pi }{4},\frac{\pi }{4},\frac{3\pi }{4},\frac{5\pi }{4},\frac{7\pi }{4}$ among which $\frac{\pi }{4},\frac{3\pi }{4},\frac{5\pi }{4},\frac{7\pi }{4}$ lie in the interval $\left[0,2\pi \right]$ as shown in graph.

Now divide the period into fourths and locate three points between the period as shown below in graph and draw vertical asymptotes $\frac{\pi }{4},\frac{3\pi }{4},\frac{5\pi }{4},\frac{7\pi }{4}$ with dotted lines.