Answer
The graph is symmetric with respect to the pole.
Work Step by Step
We look for symmetry by making the following substitutions:
(a) $\theta \to - \theta$ :$$r^2=16\sin 2(- \theta ) \quad \Rightarrow \quad r^2=-16\sin 2 \theta$$Thus, the graph does not necessarily have symmetry with respect to the polar axis.
(b) $r \to -r, \quad \theta \to -\theta$ :$$(-r)^2 =16\sin 2 (-\theta ) \quad \Rightarrow \quad r^2=-16 \sin 2 \theta$$Thus, the graph does not necessarily have symmetry with respect to the line $\theta=\frac{\pi}{2}$.
(c) $r \to -r$ :$$(-r)^2 =16\sin 2 \theta \quad \Rightarrow \quad r^2=16\sin 2 \theta$$Thus, the graph is symmetric with respect to the pole.
