Chapter 6.7, Problem 38E

To determine

To write: an equation for the function.

Answer to Problem 38E

  $y=\sec \left(2\theta +\frac{\pi }{2}\right)-10$

Explanation of Solution

Given information:

  $\begin{array}{c}\text{Function}=\text{ secant}\\ \text{Period}=\frac{2\pi }{k}=\pi \\ \text{Phase shift}=-\frac{c}{k}=-\frac{\pi }{4}\\ h=-10\end{array}$

Calculation:

The equation of cosine function is as follows:

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

Here c is a constant, $k=\frac{2\pi }{\text{Period}}$ and h is a vertical shift.

Evaluate the value of $k$ as shown below.

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

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

  $\begin{array}{c}-\frac{c}{k}=-\frac{\text{π}}{4}\\ -\frac{c}{2}=-\frac{\text{π}}{4}\\ c=-\frac{\text{π}\left(2\right)}{4\left(-1\right)}\\ =\frac{\pi }{2}\end{array}$

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

  $\begin{array}{l}y=\sec \left(2\theta +\frac{\pi }{2}\right)+\left(-10\right)\\ y=\sec \left(2\theta +\frac{\pi }{2}\right)-10\end{array}$

Thus, the equation is $y=\sec \left(2\theta +\frac{\pi }{2}\right)-10$ .