Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

Published by Pearson
ISBN 10: 0-32193-104-1
ISBN 13: 978-0-32193-104-7

Chapter 8 - Polar Coordinates; Vectors - Section 8.1 Polar Cordinates - 8.1 Assess Your Understanding - Page 592: 61


$( 2,\dfrac{ \pi}{6})$

Work Step by Step

The conversion of rectangular coordinates $(x,y)$ to polar coordinates $(r, \phi)$ can be computed as: $r=\sqrt{x^2+x^2}$ and $\theta=\tan^{-1}(\dfrac{y}{x})$ We have for point $(\sqrt 3, 1)$: $r=\sqrt{(\sqrt 3)^2+(1)^2}= \sqrt {4}=2$ and $\theta=\tan^{-1}\left(\dfrac{1}{\sqrt 3}\right)=\dfrac{ \pi}{6}$ Therefore, the point $(x,y)$ has coordinates $( 2,\dfrac{ \pi}{6})$ in the polar coordinate system.
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