Question

test for symmetry and then graph the polar equation.
r = 2 + 3 sinθ
a. is the graph of the polar equation symmetric with respect to the polar axis?
a. yes.
b. the polar equation failed the test for symmetry which means that the graph is not symmetric with respect to the polar axis.
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 = π/2?
a. 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 = π/2.
b. yes.
c. the polar equation failed the test for symmetry which means that the graph is not symmetric with respect to the line θ = π/2.
c. is the graph of the polar equation symmetric with respect to the pole?
a. the polar equation failed the test for symmetry which means that the graph may or may not be symmetric with respect to the pole.
b. the polar equation failed the test for symmetry which means that the graph is not symmetric with respect to the pole.
c. yes.

Explanation:

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

Replace $\theta$ with $-\theta$. The equation becomes $r = 2+3\sin(-\theta)=2 - 3\sin\theta$. Since $2 - 3\sin\theta
eq2 + 3\sin\theta$ (in general, except for specific values of $\theta$), the test for symmetry about the polar axis fails. But a failed test does not guarantee lack of symmetry.

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

Replace $(r,\theta)$ with $(-r,-\theta)$. We get $-r=2+3\sin(-\theta)=2 - 3\sin\theta$, or $r=- 2 + 3\sin\theta$. Also, replacing $\theta$ with $\pi-\theta$, we have $r = 2+3\sin(\pi-\theta)=2 + 3\sin\theta$. The original equation is recovered when we use the second substitution (a valid symmetry test for the line $\theta=\frac{\pi}{2}$ in polar coordinates). But if we consider the first substitution (a common - used test in some textbooks, which is equivalent to a reflection), a failed test (when using the $(-r,-\theta)$ substitution) does not rule out symmetry. However, if we use the $\theta=\pi-\theta$ substitution (a more appropriate test for $\theta=\frac{\pi}{2}$ symmetry in polar coordinates), we can also note that for a general limacon $r=a + b\sin\theta$ ($a,b>0,a
eq b$), it is symmetric about $\theta=\frac{\pi}{2}$.

Step3: Test for symmetry about the pole

Replace $r$ with $-r$. The equation becomes $-r=2+3\sin\theta$, or $r=-2 - 3\sin\theta$. Since $-2 - 3\sin\theta
eq2 + 3\sin\theta$ (in general), the test for symmetry about the pole fails. A failed test for pole - symmetry does not guarantee lack of symmetry.

Answer:

a. 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. B. Yes.
c. A. The polar equation failed the test for symmetry which means that the graph may or may not be symmetric with respect to the pole.