Chapter 13.1, Problem 62E

To determine

To calculate: The amplitude, period and phase shift for the function $y=8\cos \left(\theta -30°\right)$ .

Answer to Problem 62E

The amplitude is 8, period is $2\pi$ and phase shift is $30°$ .

Explanation of Solution

Given information:

The function $y=8\cos \left(\theta -30°\right)$ .

Formula used:

The function $y=a\cos \left(b\theta -c\right)$ , the amplitude is a , period is $\frac{2\pi }{b}$ and phase shift is $\frac{c}{b}$ .

Calculation:

Consider the function $y=8\cos \left(\theta -30°\right)$ .

Recall that for the function of the form $y=a\cos \left(b\theta -c\right)$ , the amplitude is a , period is $\frac{2\pi }{b}$ and phase shift is $\frac{c}{b}$ .

Compare $y=8\cos \left(\theta -30°\right)$ and $y=a\cos \left(b\theta -c\right)$ .

Here, a is 8, b is 1 and cis $30°$ .

Therefore, amplitude is 8.

Period is,

  $\begin{array}{c}\frac{2\pi }{b}=\frac{2\pi }{1}\\ =2\pi \end{array}$

And phase shift is,

  $\begin{array}{c}\frac{c}{b}=\frac{30°}{1}\\ =30°\end{array}$

Thus, the amplitude is 8, period is $2\pi$ and phase shift is $30°$ .