## Precalculus (6th Edition) Blitzer

The graph does not necessarily have symmetry with respect to the polar axis, the line $\theta =\frac{\pi }{2}$, or the pole. We look for symmetry by making the following substitutions: (a) $\theta \to - \theta$ :$$r=\frac{3\sin 2(-\theta )}{\sin ^3 (-\theta )+\cos ^3 (-\theta )} \quad \Rightarrow \quad r=-\frac{3 \sin 2 \theta}{-\sin ^3 \theta +\cos ^3 \theta }$$Thus, the graph does not necessarily have symmetry with respect to the polar axis. (b) $r \to -r, \quad \theta \to -\theta$ :$$-r =\frac{3 \sin 2 (-\theta )}{\sin ^3 (-\theta )+ \cos ^3(-\theta )} \quad \Rightarrow \quad r=\frac{3\sin 2\theta}{-\sin ^3 \theta +\cos ^3 \theta }$$Thus, the graph does not necessarily have symmetry with respect to the line $\theta=\frac{\pi}{2}$. (c) $r \to -r$ :$$-r =\frac{3\sin 2\theta }{\sin ^3 \theta +\cos ^3 \theta } \quad \Rightarrow \quad r=-\frac{3\sin 2 \theta}{\sin ^3 \theta +\cos ^3 \theta }$$Thus, the graph does not necessarily have symmetry with respect to the pole.