Answer
Only symmetric with respect to the pole.
See graph.
Work Step by Step
Step 1. We are given the polar equation $r^2=16sin(2\theta)$. To test the symmetry with respect to the polar axis, let $\theta\to -\theta$; we have $r^2=16sin(-2\theta)$ or $r^2=-16sin(2\theta)$. Thus, the equation is not symmetric with respect to the polar axis.
Step 2. To test the symmetry with respect to the line $\theta=\frac{\pi}{2}$, let $r\to -r$ and $\theta\to -\theta$; we have $(-r)^2=16sin(-2\theta)$ or $r^2=-16sin(2\theta)$. Thus, the equation is not symmetric with respect to the line $\theta=\frac{\pi}{2}$.
Step 3. To test the symmetry with respect to the pole, let $r\to -r$; we have $(-r)^2=16sin(2\theta)$ or $r^2=16sin(2\theta)$. Thus, the equation is symmetric with respect to the pole.
Step 4. We can graph the equation as shown in the figure.
