Chapter 9.4, Problem 22E

(a)

To determine

To find: The equation of the form $y=4\cos \pi x$ where $a>0$ and $b>0$ such that the graph has the amplitude and period.

Answer to Problem 22E

The required equation is $y=\sin \frac{2\pi }{5}x$ .

Explanation of Solution

Given:

The given amplitude is 1 and the period is 5.

Calculation:

Consider thegeneral form is,

  $y=4\cos \pi x$

Consider the formula to obtain the period is,

  $\begin{array}{l}\frac{2\pi }{b}=5\\ b=\frac{2\pi }{5}\end{array}$

For $b$ as the subject the equation is of the form,

  $y=a\sin bx$

Then,

  $y=\sin \frac{2\pi }{5}x$

(b)

To determine

To find: The equation of the form $y=4\cos \pi x$ where $a>0$ and $b>0$ such that the graph has the amplitude and period.

Answer to Problem 22E

The required equation is $y=10\sin \frac{\pi }{2}x$ .

Explanation of Solution

Given:

The given amplitude is 10 and the period is 4.

Calculation:

Consider the general form is,

  $y=4\cos \pi x$

Consider the formula to obtain the period is,

  $\begin{array}{l}\frac{2\pi }{b}=4\\ b=\frac{\pi }{2}\end{array}$

For $b$ as the subject the equation is of the form,

  $y=a\sin bx$

Then,

  $y=10\sin \frac{\pi }{2}x$

(c)

To determine

To find: The equation of the form $y=4\cos \pi x$ where $a>0$ and $b>0$ such that the graph has the amplitude and period.

Answer to Problem 22E

The required equation is $y=2\sin x$ .

Explanation of Solution

Given:

The given amplitude is 2 and the period is $2\pi$ .

Calculation:

Consider the general form is,

  $y=4\cos \pi x$

Consider the formula to obtain the period is,

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

For $b$ as the subject the equation is of the form,

  $y=a\sin bx$

Then,

  $y=2\sin x$

(d)

To determine

To find: The equation of the form $y=4\cos \pi x$ where $a>0$ and $b>0$ such that the graph has the amplitude and period.

Answer to Problem 22E

The required equation is $y=\frac{1}{2}\sin \frac{2}{3}x$ .

Explanation of Solution

Given:

The given amplitude is $\frac{1}{2}$ and the period is $3\pi$ .

Calculation:

Consider the general form is,

  $y=4\cos \pi x$

Consider the formula to obtain the period is,

  $\begin{array}{l}\frac{2\pi }{b}=3\pi \\ b=\frac{2}{3}\end{array}$

For $b$ as the subject the equation is of the form,

  $y=a\sin bx$

Then,

  $y=\frac{1}{2}\sin \frac{2}{3}x$