Chapter 10.1, Problem 10E

Solution

To graph: The function $y=\frac{2}{3}\sin x$ .

Given information: The given function is $y=\frac{2}{3}\sin x$ .

Formula used: The general equation of a sine function is $y=a\sin bx$ . The formula to find the amplitude and period of the sine function is given below.

  $\begin{array}{l}\text{Amplitude}=\left|a\right|\\ \text{Period}=\frac{2\pi }{\left|b\right|}\end{array}$

Here, $a$ and $b$ are nonzero real numbers.

Graph:

Compare the given equation $y=\frac{2}{3}\sin x$ with the general equation $y=a\sin bx$ to get $a=\frac{2}{3}$ and $b=1$ .

Calculate the amplitude of the given equation.

  $\left|a\right|=\frac{2}{3}$

Calculate the period of the given equation.

  $\frac{2\pi }{\left|1\right|}=2\pi$

One intercept of the given equation is $\left(0,0\right)$ .

Calculate the other two intercepts of the given equation.

  $\begin{array}{l}\left(\frac{1}{2}\cdot 2\pi ,0\right)=\left(\pi ,0\right)\\ \left(\frac{2\pi }{1},0\right)=\left(2\pi ,0\right)\end{array}$

Calculate the maximum of the equation.

  $\begin{array}{c}\left(\frac{1}{4}\cdot \frac{2\pi }{b},a\right)=\left(\frac{1}{4}\cdot \frac{2\pi }{1},\frac{2}{3}\right)\\ =\left(\frac{\pi }{2},\frac{2}{3}\right)\end{array}$

Calculate the minimum of the equation.

  $\begin{array}{c}\left(\frac{3}{4}\cdot \frac{2\pi }{b},-a\right)=\left(\frac{3}{4}\cdot \frac{2\pi }{1},-\frac{2}{3}\right)\\ =\left(\frac{3\pi }{2},-\frac{2}{3}\right)\end{array}$

Draw the graph of the equation $y=\frac{2}{3}\sin x$ as shown below.

  

Figure (1)

Interpretation: From the graph it can be observed that the domain of the given equation is the set of all real numbers.