Chapter 14, Problem 73CCSR

Solution

To find: The function that has a period of $4\pi$ and amplitude of 6.

Option D is correct.

Given:

  $\begin{array}{c}y=-6\sin 8\theta \\ y=6\sin 2\theta \\ y=3\sin 6\theta \\ y=-6\sin \frac{1}{2}\theta \end{array}$

Concept used:

The amplitude and period of a function of the form $y=a\sin bx$ are $\left|a\right|$ and $\frac{2\pi }{b}$ respectively.

Calculation:

Consider,

  $y=-6\sin 8\theta$

By comparing with $y=a\sin b\theta$ , it follows that

  $a=-6,b=8$ .

Now the amplitude is,

  $\begin{array}{c}\left|a\right|=\left|-6\right|\\ =6\end{array}$

And the period is,

  $\begin{array}{c}\frac{2\pi }{b}=\frac{2\pi }{8}\\ =\frac{\pi }{4}\end{array}$

Consider,

  $y=6\sin 2\theta$

By comparing with $y=a\sin b\theta$ , it follows that

  $a=6,b=2$ .

Now the amplitude is,

  $\begin{array}{c}\left|a\right|=\left|6\right|\\ =6\end{array}$

And the period is,

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

Consider,

  $y=3\sin 6\theta$

By comparing with $y=a\sin b\theta$ , it follows that

  $a=3,b=6$ .

Now the amplitude is,

  $\begin{array}{c}\left|a\right|=\left|3\right|\\ =3\end{array}$

And the period is,

  $\begin{array}{c}\frac{2\pi }{b}=\frac{2\pi }{6}\\ =\frac{\pi }{3}\end{array}$

Consider,

  $y=-6\sin \frac{1}{2}\theta$

By comparing with $y=a\sin b\theta$ , it follows that

  $a=-6,b=\frac{1}{2}$ .

Now the amplitude is,

  $\begin{array}{c}\left|a\right|=\left|-6\right|\\ =6\end{array}$

And the period is,

  $\begin{array}{c}\frac{2\pi }{b}=\frac{2\pi }{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\\ =4\pi \end{array}$

Hence option D is correct.