## Precalculus (6th Edition) Blitzer

Published by Pearson

# Chapter 6 - Section 6.4 - Graphs of Polar Equations - Exercise Set - Page 754: 26

#### Answer

The graph is symmetric with respect to the line $\theta =\frac{\pi }{2}$. See the graph below:

#### Work Step by Step

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

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