Chapter 5.4, Problem 39E

To determine

To find:

The functionof period and graph of the function

Answer to Problem 39E

The functions tangent have period $\frac{\pi }{2}$

Explanation of Solution

Given:

The function is

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

Concept used:

The functions tangent and cotangent have period $\pi$

  $\begin{array}{l}\tan \left(x+\pi \right)=\tan x\\ \cot \left(x+\pi \right)=\cot x\end{array}$

The functions cosecant and secant have period $2\pi$

Calculation:

The function is

  $y=\tan 2\left(x+\frac{\pi }{2}\right)\mathrm{...................}\left(1\right)$

Since the tangent and cotangent function have period $\pi$

The function

  $\begin{array}{l}y=a\tan kxand\\ y=a\cot kx\left(k>0\right)\end{array}$

Complete one period as $kx$ varies from $0to\pi$

That is for $0\le kx\le \pi$

Solving this inequality

  $0\le x\le \frac{\pi }{k}$

So each period is $\frac{\pi }{k}$

The functions cosec have period

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

period is $\frac{\pi }{k}=\frac{\pi }{2}$

Draw the graph of the function

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

Draw the table

Test one point in each of the region formed by the graph

If the point satisfies the function then shade the entire region to denote that every point in the region satisfies the function

$x-axis$	$-\pi$	$-\frac{\pi }{2}$	$\frac{\pi }{2}$	$\pi$	$-\frac{3\pi }{2}$	$\frac{3\pi }{2}$
$y-axis$	$0$	$0$	$0$	$0$	$0$	$0$