Functions Modeling Change: A Preparation for Calculus, 5th Edition

Published by Wiley
ISBN 10: 1118583191
ISBN 13: 978-1-11858-319-7

Chapter 8 - Triangle Trigonometry and Polar Coordinates - 8.3 Polar Coordinates - Exercises and Problems for Section 8.3 - Exercises and Problems - Page 347: 22



Work Step by Step

Converting the polar coordinates to Cartesian coordinates, we find: $$x=2\cos \left(\left(\frac{5\pi }{6}\right)\right) =-\sqrt3\\ y=2\sin \left(\left(\frac{5\pi }{6}\right)\right) =1 \\ \left(-\sqrt{3},\:1\right)$$
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