Chapter 12.7, Problem 42HP

To determine

To find: Compare and contrast the graphs of $y=\frac{1}{2}\sin \theta$ and $y=\sin \frac{1}{2}\theta$ .

Answer to Problem 42HP

Both graphs have different in Amplitude and Period.

Explanation of Solution

Given information:

  $\begin{array}{l}y=\frac{1}{2}\sin \theta \\ y=\sin \frac{1}{2}\theta \end{array}$

Concept:

If a function is in the form

  $y=a\sin \left(bx\right)$

Then

Amplitude $=\left|a\right|$

Period $=\frac{2\pi }{\left|b\right|}$

Given

  $y=\frac{1}{2}\sin \theta$

On comaparing with $y=a\sin \left(bx\right)$

We get

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

Amplitude of this function $=\left|a\right|=\frac{1}{2}$

Period $=2\pi$

  $y=\sin \frac{1}{2}\theta$

On comaparing with $y=a\sin \left(bx\right)$

We get

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

Amplitude of this function $=\left|a\right|=1$

Period $=\frac{2\pi }{\frac{1}{2}}=4\pi$

  

Red curve represents the graph of the function $y=\frac{1}{2}\sin \theta$

Green curve represents the graph of the function $y=\sin \frac{1}{2}\theta$

On comparing both curves we can say that

Amplitude and period of both graphs are different.

Amplitude of $y=\frac{1}{2}\sin \theta$ is $\frac{1}{2}$

Period of $y=\frac{1}{2}\sin \theta$ is $2\pi$

Whereas

Amplitude of $y=\sin \frac{1}{2}\theta$ is $1$

Period of $y=\sin \frac{1}{2}\theta$ is $4\pi$

Therefore,

Both graphs have different in Amplitude and Period.