Chapter 12, Problem 24PT

To determine

To find: Amplitude, period and phase shift of each function.

Answer to Problem 24PT

Amplitude $=undefined$

Period $=\pi$

Phase shift $=\frac{\pi }{2}$

Explanation of Solution

Given information:

  $y=\frac{1}{2}\tan \left(\theta -\frac{\pi }{2}\right)$

Concept:

If equation in the form

  $y=a\tan k\left(x+b\right)$

Then

Amplitude $=undefined$

Period $=\frac{\pi }{\left|k\right|}$

Phase shift $=-b$

Given

  $y=\frac{1}{2}\tan \left(\theta -\frac{\pi }{2}\right)$

On comparing with general equation

We get

  $\begin{array}{l}a=\frac{1}{2}\\ k=1\\ b=-\frac{\pi }{2}\end{array}$

Graph of given equation is

  

Therefore,

Amplitude $=a=undefined$

Period $=\frac{\pi }{\left|k\right|}$

  $\Rightarrow$ Period $=\frac{\pi }{1}$

  $\Rightarrow$ Period $=\pi$

Phase shift $=-b$

  $\Rightarrow$ Phase shift $=-\left(-\frac{\pi }{2}\right)$

  $\Rightarrow$ Phase shift $=\frac{\pi }{2}$

Therefore,

We got

Amplitude $=undefined$

Period $=\pi$

Phase shift $=\frac{\pi }{2}$