Trigonometry (11th Edition) Clone

Published by Pearson
ISBN 10: 978-0-13-421743-7
ISBN 13: 978-0-13421-743-7

Chapter 8 - Complex Numbers, Polar Equations, and Parametric Equations - Section 8.5 Polar Equations and Graphs - 8.5 Exercises - Page 396: 79


The two points of intersection are: $(2, \frac{\pi}{6})$ and $(2, \frac{5\pi}{6})$

Work Step by Step

$r = 4~sin~\theta$ $r = 1+2~sin~\theta$ To find the points of intersection, we can equate the expressions for $r$: $4~sin~\theta = 1+2~sin~\theta$ $2~sin~\theta = 1$ $sin~\theta = \frac{1}{2}$ $\theta = \frac{\pi}{6}, \frac{5\pi}{6}$ We can find $r$: $r = 4~sin~\theta$ $r = (4)~(\frac{1}{2})$ $r = 2$ The two points of intersection are: $(2, \frac{\pi}{6})$ and $(2, \frac{5\pi}{6})$
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