## Trigonometry (11th Edition) Clone

Published by Pearson

# Chapter 8 - Test - Page 412: 11

#### Answer

$r = 1 - cos~\theta$ $\theta$ in $[0,2\pi)$ This graph is a cardioid. We can see the graph below:

#### Work Step by Step

$r = 1 - cos~\theta$ $\theta$ in $[0,2\pi)$ When $\theta = 0$: $r = 1 - cos 0= 0$ When $\theta = \frac{\pi}{4}$: $r = 1- cos~\frac{\pi}{4} = 1 - \frac{\sqrt{2}}{2} = 0.29$ When $\theta = \frac{\pi}{2}$: $r = 1 - cos~\frac{\pi}{2} = 1$ When $\theta = \frac{3\pi}{4}$: $r = 1 - cos~\frac{3\pi}{4} = 1 + \frac{\sqrt{2}}{2} = 1.71$ When $\theta = \pi$: $r = 1 - cos~\pi = 2$ When $\theta = \frac{5\pi}{4}$: $r = 1 - cos~\frac{5\pi}{4} = 1+\frac{\sqrt{2}}{2} = 1.71$ When $\theta = \frac{3\pi}{2}$: $r = 1 - cos~\frac{3\pi}{2} = 1$ When $\theta = \frac{7\pi}{4}$: $r = 1 - cos~\frac{7\pi}{4} = 1 - \frac{\sqrt{2}}{2} = 0.29$ This graph is a cardioid. We can see the graph below:

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