Chapter 7.3, Problem 54E

To determine

To find: the amplitude, period, and phase shift for the function.

Answer to Problem 54E

  $\begin{array}{c}\text{Amplitude}=8\end{array}$

  $\begin{array}{c}\text{Period}=2\pi \end{array}$

  $\begin{array}{c}\text{Phase shift}=\frac{\pi }{12}\end{array}$

Explanation of Solution

Given information:

  $y=8\cos \left(\theta -30°\right)$

Calculation:

The equation of cosine function is as follows:

  $y=A\cos \left(k\theta +c\right)$    … (1)

The cosine function can be rewritten after changing degree into radian measure as follows:

  $y=8\cos \left(\theta -\frac{\pi }{6}\right)$    … (2)

Compare equation (1) and (2) to get the constants $A,k\text{ and }c$ as follows:

  $A=8,k=1\text{ and }c=-\frac{\pi }{6}$

The amplitude can be evaluated as follows:

  $\begin{array}{c}\text{Amplitude}=\left|A\right|\end{array}$

  $\begin{array}{c}=\left|8\right|\left[\text{Put the value of }A\right]\end{array}$

  $\begin{array}{c}=8\end{array}$

The period of the function is evaluated as follows:

  $\begin{array}{c}\text{Period}=\frac{2\pi }{k}\end{array}$

  $\begin{array}{c}=\frac{2\pi }{1}\left[\text{Put the value of }k\right]\end{array}$

  $\begin{array}{c}=2\pi \end{array}$

The phase shift is evaluated as follows:

  $\begin{array}{c}\text{Phase shift}=\frac{-c}{k}\end{array}$

  $\begin{array}{c}=-\frac{\left(\raisebox{1ex}{$-\pi $}\!\left/ \!\raisebox{-1ex}{$6$}\right.\right)}{2}\left[\text{Put the value of }c\text{ and }k\right]\end{array}$

  $\begin{array}{c}=\frac{\pi }{12}\end{array}$

Thus the amplitude, period and phase shift of the given function is 8, $2\pi$ and $\frac{\pi }{6}$ respectively.