Answer :
Let's solve the problem step-by-step:
### Step 1: Populate the Table of Values
We need to fill in the table with values of [tex]\( r \)[/tex] for different values of [tex]\( \theta \)[/tex] using the polar equation [tex]\( r = \frac{2}{\sin \theta} \)[/tex].
Using the given values of [tex]\(\theta\)[/tex]:
1. For [tex]\(\theta = 0\)[/tex]:
- [tex]\(\sin(0) = 0 \)[/tex]
- [tex]\( r = \frac{2}{0} = \infty \)[/tex] (undefined)
2. For [tex]\(\theta = \frac{\pi}{6}\)[/tex]:
- [tex]\(\sin\left(\frac{\pi}{6}\right) = \frac{1}{2} \)[/tex]
- [tex]\( r = \frac{2}{\frac{1}{2}} = 4 \)[/tex]
3. For [tex]\(\theta = \frac{\pi}{4}\)[/tex]:
- [tex]\(\sin\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2} \)[/tex]
- [tex]\( r = \frac{2}{\frac{\sqrt{2}}{2}} = 2\sqrt{2} \approx 2.828 \)[/tex]
4. For [tex]\(\theta = \frac{\pi}{3}\)[/tex]:
- [tex]\(\sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2} \)[/tex]
- [tex]\( r = \frac{2}{\frac{\sqrt{3}}{2}} = \frac{4}{\sqrt{3}} \approx 2.309 \)[/tex]
5. For [tex]\(\theta = \frac{\pi}{2}\)[/tex]:
- [tex]\(\sin\left(\frac{\pi}{2}\right) = 1 \)[/tex]
- [tex]\( r = \frac{2}{1} = 2 \)[/tex]
6. For [tex]\(\theta = \frac{2\pi}{3}\)[/tex]:
- [tex]\(\sin\left(\frac{2\pi}{3}\right) = \frac{\sqrt{3}}{2} \)[/tex]
- [tex]\( r = \frac{2}{\frac{\sqrt{3}}{2}} = \frac{4}{\sqrt{3}} \approx 2.309 \)[/tex]
7. For [tex]\(\theta = \frac{3\pi}{4}\)[/tex]:
- [tex]\(\sin\left(\frac{3\pi}{4}\right) = \frac{\sqrt{2}}{2} \)[/tex]
- [tex]\( r = \frac{2}{\frac{\sqrt{2}}{2}} = 2\sqrt{2} \approx 2.828 \)[/tex]
8. For [tex]\(\theta = \frac{5\pi}{6}\)[/tex]:
- [tex]\(\sin\left(\frac{5\pi}{6}\right) = \frac{1}{2} \)[/tex]
- [tex]\( r = \frac{2}{\frac{1}{2}} = 4 \)[/tex]
9. For [tex]\(\theta = \pi\)[/tex]:
- [tex]\(\sin(\pi) = 0 \)[/tex]
- [tex]\( r = \frac{2}{0} = \infty \)[/tex] (undefined)
10. For [tex]\(\theta = \frac{3\pi}{2}\)[/tex]:
- [tex]\(\sin\left(\frac{3\pi}{2}\right) = -1 \)[/tex]
- [tex]\( r = \frac{2}{-1} = -2 \)[/tex]
11. For [tex]\(\theta = 2\pi\)[/tex]:
- [tex]\(\sin(2\pi) = 0 \)[/tex]
- [tex]\( r = \frac{2}{0} = \infty \)[/tex] (undefined)
Now, we populate the table:
[tex]\[ \begin{tabular}{|c|c|} \hline $\theta$ & $r=2 / \sin \theta$ \\ \hline 0 & \(\infty\) \\ \hline $\pi / 6$ & 4 \\ \hline $\pi / 4$ & 2.828 \\ \hline $\pi / 3$ & 2.309 \\ \hline $\pi / 2$ & 2 \\ \hline $2\pi / 3$ & 2.309 \\ \hline $3\pi / 4$ & 2.828 \\ \hline $5\pi / 6$ & 4 \\ \hline $\pi$ & \(\infty\) \\ \hline $3\pi / 2$ & -2 \\ \hline $2\pi$ & \(\infty\) \\ \hline \end{tabular} \][/tex]
### Step 2: Type of Graph
By observing the values, we notice that as [tex]\(\theta\)[/tex] approaches 0, [tex]\(\pi\)[/tex], or [tex]\(2\pi\)[/tex], [tex]\( r \)[/tex] tends to infinity.
When graphing these points in polar coordinates, you will see that for most [tex]\(\theta\)[/tex] values, the radius [tex]\(r\)[/tex] is positive and finite, and yet for [tex]\(\theta = 0\)[/tex], [tex]\(\pi\)[/tex], and [tex]\(2\pi\)[/tex], it becomes infinite. As [tex]\(\theta\)[/tex] moves from 0 to [tex]\(2\pi\)[/tex], the values stabilize and recreate a straight horizontal line.
### Step 3: Converting Polar Equation to Rectangular Equation
To justify the type of graph, we convert the polar equation [tex]\( r = \frac{2}{\sin \theta} \)[/tex] to its rectangular form.
Given:
[tex]\[ r = \frac{2}{\sin \theta} \][/tex]
Multiplying both sides by [tex]\(\sin \theta\)[/tex]:
[tex]\[ r \sin \theta = 2 \][/tex]
Using the conversions for polar to rectangular coordinates:
[tex]\[ x = r \cos \theta \][/tex]
[tex]\[ y = r \sin \theta \][/tex]
We substitute [tex]\( r \sin \theta \)[/tex] with [tex]\( y \)[/tex]:
[tex]\[ y = 2 \][/tex]
This is a linear equation of a horizontal line at [tex]\( y = 2 \)[/tex] in the rectangular coordinate system. Therefore, the graph of the polar equation [tex]\( r = \frac{2}{\sin \theta} \)[/tex] corresponds to the line [tex]\(y = 2\)[/tex] in the rectangular coordinate system.
### Step 1: Populate the Table of Values
We need to fill in the table with values of [tex]\( r \)[/tex] for different values of [tex]\( \theta \)[/tex] using the polar equation [tex]\( r = \frac{2}{\sin \theta} \)[/tex].
Using the given values of [tex]\(\theta\)[/tex]:
1. For [tex]\(\theta = 0\)[/tex]:
- [tex]\(\sin(0) = 0 \)[/tex]
- [tex]\( r = \frac{2}{0} = \infty \)[/tex] (undefined)
2. For [tex]\(\theta = \frac{\pi}{6}\)[/tex]:
- [tex]\(\sin\left(\frac{\pi}{6}\right) = \frac{1}{2} \)[/tex]
- [tex]\( r = \frac{2}{\frac{1}{2}} = 4 \)[/tex]
3. For [tex]\(\theta = \frac{\pi}{4}\)[/tex]:
- [tex]\(\sin\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2} \)[/tex]
- [tex]\( r = \frac{2}{\frac{\sqrt{2}}{2}} = 2\sqrt{2} \approx 2.828 \)[/tex]
4. For [tex]\(\theta = \frac{\pi}{3}\)[/tex]:
- [tex]\(\sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2} \)[/tex]
- [tex]\( r = \frac{2}{\frac{\sqrt{3}}{2}} = \frac{4}{\sqrt{3}} \approx 2.309 \)[/tex]
5. For [tex]\(\theta = \frac{\pi}{2}\)[/tex]:
- [tex]\(\sin\left(\frac{\pi}{2}\right) = 1 \)[/tex]
- [tex]\( r = \frac{2}{1} = 2 \)[/tex]
6. For [tex]\(\theta = \frac{2\pi}{3}\)[/tex]:
- [tex]\(\sin\left(\frac{2\pi}{3}\right) = \frac{\sqrt{3}}{2} \)[/tex]
- [tex]\( r = \frac{2}{\frac{\sqrt{3}}{2}} = \frac{4}{\sqrt{3}} \approx 2.309 \)[/tex]
7. For [tex]\(\theta = \frac{3\pi}{4}\)[/tex]:
- [tex]\(\sin\left(\frac{3\pi}{4}\right) = \frac{\sqrt{2}}{2} \)[/tex]
- [tex]\( r = \frac{2}{\frac{\sqrt{2}}{2}} = 2\sqrt{2} \approx 2.828 \)[/tex]
8. For [tex]\(\theta = \frac{5\pi}{6}\)[/tex]:
- [tex]\(\sin\left(\frac{5\pi}{6}\right) = \frac{1}{2} \)[/tex]
- [tex]\( r = \frac{2}{\frac{1}{2}} = 4 \)[/tex]
9. For [tex]\(\theta = \pi\)[/tex]:
- [tex]\(\sin(\pi) = 0 \)[/tex]
- [tex]\( r = \frac{2}{0} = \infty \)[/tex] (undefined)
10. For [tex]\(\theta = \frac{3\pi}{2}\)[/tex]:
- [tex]\(\sin\left(\frac{3\pi}{2}\right) = -1 \)[/tex]
- [tex]\( r = \frac{2}{-1} = -2 \)[/tex]
11. For [tex]\(\theta = 2\pi\)[/tex]:
- [tex]\(\sin(2\pi) = 0 \)[/tex]
- [tex]\( r = \frac{2}{0} = \infty \)[/tex] (undefined)
Now, we populate the table:
[tex]\[ \begin{tabular}{|c|c|} \hline $\theta$ & $r=2 / \sin \theta$ \\ \hline 0 & \(\infty\) \\ \hline $\pi / 6$ & 4 \\ \hline $\pi / 4$ & 2.828 \\ \hline $\pi / 3$ & 2.309 \\ \hline $\pi / 2$ & 2 \\ \hline $2\pi / 3$ & 2.309 \\ \hline $3\pi / 4$ & 2.828 \\ \hline $5\pi / 6$ & 4 \\ \hline $\pi$ & \(\infty\) \\ \hline $3\pi / 2$ & -2 \\ \hline $2\pi$ & \(\infty\) \\ \hline \end{tabular} \][/tex]
### Step 2: Type of Graph
By observing the values, we notice that as [tex]\(\theta\)[/tex] approaches 0, [tex]\(\pi\)[/tex], or [tex]\(2\pi\)[/tex], [tex]\( r \)[/tex] tends to infinity.
When graphing these points in polar coordinates, you will see that for most [tex]\(\theta\)[/tex] values, the radius [tex]\(r\)[/tex] is positive and finite, and yet for [tex]\(\theta = 0\)[/tex], [tex]\(\pi\)[/tex], and [tex]\(2\pi\)[/tex], it becomes infinite. As [tex]\(\theta\)[/tex] moves from 0 to [tex]\(2\pi\)[/tex], the values stabilize and recreate a straight horizontal line.
### Step 3: Converting Polar Equation to Rectangular Equation
To justify the type of graph, we convert the polar equation [tex]\( r = \frac{2}{\sin \theta} \)[/tex] to its rectangular form.
Given:
[tex]\[ r = \frac{2}{\sin \theta} \][/tex]
Multiplying both sides by [tex]\(\sin \theta\)[/tex]:
[tex]\[ r \sin \theta = 2 \][/tex]
Using the conversions for polar to rectangular coordinates:
[tex]\[ x = r \cos \theta \][/tex]
[tex]\[ y = r \sin \theta \][/tex]
We substitute [tex]\( r \sin \theta \)[/tex] with [tex]\( y \)[/tex]:
[tex]\[ y = 2 \][/tex]
This is a linear equation of a horizontal line at [tex]\( y = 2 \)[/tex] in the rectangular coordinate system. Therefore, the graph of the polar equation [tex]\( r = \frac{2}{\sin \theta} \)[/tex] corresponds to the line [tex]\(y = 2\)[/tex] in the rectangular coordinate system.
