Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 4 - Section 4.2 - Trigonometric Functions: The Unit Circle - Exercise Set - Page 547: 6

Answer

$\frac{\sqrt 3}{2}$

Work Step by Step

($\frac{1}{2}$ , $\frac{\sqrt 3}{2}$) sin $\frac{\pi}{3}$ corresponds to y value So sin $\frac{\pi}{3}$ = $\frac{\sqrt 3}{2}$
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