Question

test for symmetry and then graph the polar equation.
r = 1 - cosθ
a. is the graph of the polar equation symmetric with respect to the polar axis?
a. the polar equation failed the test for symmetry which means that the graph is not symmetric with respect to the polar axis.
b. yes.
c. the polar equation failed the test for symmetry which means that the graph may or may not be symmetric with respect to the polar axis.
b. is the graph of the polar equation symmetric with respect to the line θ = \\( \frac{\pi}{2} \\)?
a. yes.
b. the polar equation failed the test for symmetry which means that the graph is not symmetric with respect to the line θ = \\( \frac{\pi}{2} \\).
c. the polar equation failed the test for symmetry which means that the graph may or may not be symmetric with respect to the line θ = \\( \frac{\pi}{2} \\).

Explanation:

Step1: Test for symmetry about the polar axis ($\theta = 0$)

Replace $\theta$ with $-\theta$.
For the polar equation $r = 1-\cos\theta$, when we substitute $\theta$ with $-\theta$, we get $r = 1-\cos(-\theta)$.
Since $\cos(-\theta)=\cos\theta$ (cosine is an even - function), the equation becomes $r = 1-\cos\theta$, which is the same as the original equation. So, the graph is symmetric about the polar axis.

Step2: Test for symmetry about the line $\theta=\frac{\pi}{2}$

Replace $(r,\theta)$ with $(-r,-\theta)$.
The original equation is $r = 1-\cos\theta$. After substitution, we have $-r=1 - \cos(-\theta)$. Since $\cos(-\theta)=\cos\theta$, the equation becomes $-r = 1-\cos\theta$ or $r=-1 + \cos\theta$, which is not the same as the original equation $r = 1-\cos\theta$.
Another way is to use the formula: replace $r$ with $-r$ and $\theta$ with $\pi-\theta$.
Original equation $r = 1-\cos\theta$. After substitution $-r=1-\cos(\pi - \theta)$. Since $\cos(\pi-\theta)=-\cos\theta$, we have $-r=1+\cos\theta$ or $r=-1-\cos\theta$, which is not the same as the original equation. So, we cannot be sure of symmetry about the line $\theta=\frac{\pi}{2}$ from the test.

Answer:

a. B. Yes.
b. C. The polar equation failed the test for symmetry which means that the graph may or may not be symmetric with respect to the line $\theta=\frac{\pi}{2}$.