Chapter 12, Problem 47SGR

To determine

To find: Amplitude and period of each function.

Answer to Problem 47SGR

Amplitude $=a=1$

Period $=4\pi$

Explanation of Solution

Given information:

  $y=\cos \frac{1}{2}\theta$

Concept:

If equation in the form

  $y=a\cos \left(bx\right)$

Then

Amplitude $=\left|a\right|$

Period $=\frac{2\pi }{\left|b\right|}$

Given

  $y=\cos \frac{1}{2}\theta$

On comparing with general equation

We get

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

Graph of given equation is

  

X-axis is represented by $\theta$

Therefore,

Amplitude $=a=1$

Period $=\frac{2\pi }{\left|b\right|}$

  $\Rightarrow$ Period $=\frac{2\pi }{\left|\frac{1}{2}\right|}$

  $\Rightarrow$ Period $=2\pi \times 2$

  $\Rightarrow$ Period $=4\pi$

Therefore,

We got

Amplitude $=a=1$

Period $=4\pi$