To solve this question, let’s first address Part (a) and complete the table for the polar equation $r = 2 - 2\cos(\theta)$. For each given value of $\theta$, substitute it into the equation to find the corresponding $r$.

1. Completing the Table

The polar equation is: $r = 2 - 2\cos(\theta)$

We evaluate $r$ at the given values of $\theta$:

1. For $\theta = 0$: $r = 2 - 2\cos(0) = 2 - 2(1) = 0$

2. For $\theta = \frac{\pi}{3}$: $r = 2 - 2\cos\left(\frac{\pi}{3}\right) = 2 - 2\left(\frac{1}{2}\right) = 2 - 1 = 1$

3. For $\theta = \frac{\pi}{2}$: $r = 2 - 2\cos\left(\frac{\pi}{2}\right) = 2 - 2(0) = 2$

4. For $\theta = \frac{2\pi}{3}$: $r = 2 - 2\cos\left(\frac{2\pi}{3}\right) = 2 - 2\left(-\frac{1}{2}\right) = 2 + 1 = 3$

5. For $\theta = \pi$: $r = 2 - 2\cos(\pi) = 2 - 2(-1) = 2 + 2 = 4$

6. For $\theta = \frac{4\pi}{3}$: $r = 2 - 2\cos\left(\frac{4\pi}{3}\right) = 2 - 2\left(-\frac{1}{2}\right) = 2 + 1 = 3$

7. For $\theta = \frac{3\pi}{2}$: $r = 2 - 2\cos\left(\frac{3\pi}{2}\right) = 2 - 2(0) = 2$

8. For $\theta = \frac{5\pi}{3}$: $r = 2 - 2\cos\left(\frac{5\pi}{3}\right) = 2 - 2\left(\frac{1}{2}\right) = 2 - 1 = 1$

9. For $\theta = 2\pi$: $r = 2 - 2\cos(2\pi) = 2 - 2(1) = 0$

The completed table is:

2. Sketching the Graph

In Part (b), use the table values to plot points in the polar graph. For each pair of $(\theta, r)$:

1. Start at the angle $\theta$ and move outward along the radius by $r$.
2. Connect the points to form the graph of $r = 2 - 2\cos(\theta)$.

Let me know if you'd like me to help plot the graph or explain the process further!

Relative Questions:

1. What is the significance of symmetry in polar graphs?
2. How does the graph of $r = 2 - 2\cos(\theta)$ compare to a standard cardioid?
3. What happens if the coefficient of $\cos(\theta)$ changes in the equation?
4. How would the graph look for $r = 2 + 2\cos(\theta)$?
5. Why does $r = 2 - 2\cos(\theta)$ return to $r = 0$ at $2\pi$?

Tip: In polar graphs, pay attention to symmetry and intercepts for faster plotting!