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 5 - Trigonometric Functions - Section 5.2 Trigonometric Functions: Unit Circle Approach - 5.2 Assess Your Understanding - Page 402: 10

Answer

$\ P=\left(\frac{1}{2}, \frac{\sqrt{3}}{2}\right)$

Work Step by Step

$\theta=\frac{\pi}{3}$ The point on the Unit circle $=(\cos \theta, \sin \theta)$ $=\left(\cos \frac{\pi}{3}, \sin \frac{\pi}{3}\right)$ $\therefore P=\left(\frac{1}{2}, \frac{\sqrt{3}}{2}\right)$
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