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 } \\).
c. is the graph of the polar equation symmetric with respect to the pole?
a. yes.
b. the polar equation failed the test for symmetry which means that the graph may or may not be symmetric with respect to the pole.
c. the polar equation failed the test for symmetry which means that the graph is not symmetric with respect to the pole.

Explanation:

• For symmetry about the polar axis ($\theta = 0$):
• Replace $\theta$ with $-\theta$. The polar equation \(r = 1-\cos\theta\) becomes \(r = 1-\cos(-\theta)\). Since \(\cos(-\theta)=\cos\theta\) (cosine is an even - function), the equation remains \(r = 1 - \cos\theta\). So, the graph is symmetric about the polar axis.
• For symmetry about the line \(\theta=\frac{\pi}{2}\):
• Replace \((r,\theta)\) with \((-r,-\theta)\). The equation becomes \(-r=1-\cos(-\theta)\), or \(r=- 1+\cos\theta\), which is not the same as the original equation \(r = 1-\cos\theta\). Also, replacing \(r\) with \(-r\) gives \(-r = 1-\cos\theta\) or \(r=-1 + \cos\theta\) (not the same as the original). A failed symmetry test does not rule out symmetry (e.g., a circle \(r = 1\) fails the \(\theta=\frac{\pi}{2}\) symmetry test if we use the \((-r,-\theta)\) substitution but is symmetric about \(\theta=\frac{\pi}{2}\)).
• For symmetry about the pole:
• Replace \(r\) with \(-r\). The equation becomes \(-r=1-\cos\theta\) or \(r=-1+\cos\theta\), which is not the same as the original equation \(r = 1-\cos\theta\). A failed symmetry test does not rule out symmetry (e.g., a circle \(r = 1\) fails the pole symmetry test if we use the \(r\to - r\) substitution but is symmetric about the pole).

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 \(\theta=\frac{\pi}{2}\).
c. B. The polar equation failed the test for symmetry which means that the graph may or may not be symmetric with respect to the pole.