Chapter 6.8, Problem 47E

To determine

To write: an equation of sine function.

Answer to Problem 47E

  $y=\pm 5\sin \left(\frac{2}{3}\theta +\frac{2\pi }{3}\right)-8$

Explanation of Solution

Given information:

  $\begin{array}{c}\left|A\right|=5\\ \text{Period}=\frac{2\pi }{k}=3\pi \\ \text{Phase shift}=-\frac{c}{k}=-\pi \\ h=-8\end{array}$

Calculation:

The equation of cosine function is as follows:

  $y=A\sin \left(k\theta +c\right)+h$...........(1)

Here $A$ is a coefficient whose absolute value is termed as amplitude, c is a constant, $k=\frac{2\pi }{\text{Period}}$ and h is a vertical shift.

Evaluate the values of $A\text{ and }k$ as shown below.

  $\begin{array}{c}\left|A\right|=5\\ A=\pm 5\\ \\ k=\frac{2\pi }{\text{Period}}\\ =\frac{2\pi }{3\pi }\\ =\frac{2}{3}\end{array}$

Put the value of k in the formula given below and solve c as follows:

  $\begin{array}{c}-\frac{c}{k}=-\pi \\ -\frac{c}{\left(\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.\right)}=-\pi \\ \frac{c}{\left(\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.\right)}=\pi \\ c=\pi \left(\frac{2}{3}\right)\\ c=\frac{2\pi }{3}\end{array}$

Now put the values in equation (1) to get the equation for cosine function.

  $\begin{array}{l}y=\pm 5\sin \left(\frac{2}{3}\theta +\frac{2\pi }{3}\right)+\left(-8\right)\\ y=\pm 5\sin \left(\frac{2}{3}\theta +\frac{2\pi }{3}\right)-8\end{array}$

Thus, the equation is $y=\pm 5\sin \left(\frac{2}{3}\theta +\frac{2\pi }{3}\right)-8$ .