Precalculus (6th Edition)

Published by Pearson
ISBN 10: 013421742X
ISBN 13: 978-0-13421-742-0

Chapter 6 - The Circular Functions and Their Graphs - Chapter 6 Test Prep - Review Exercises - Page 645: 49

Answer

$$s = \frac{\pi }{4}$$

Work Step by Step

$$\eqalign{ & \left[ {0,\frac{\pi }{2}} \right];{\text{ cos }}s = \frac{{\sqrt 2 }}{2} \cr & {\text{From the unit circle we can see that in the interval }}\left[ {0,\frac{\pi }{2}} \right],{\text{ }} \cr & {\text{the arc length }}s = \frac{\pi }{4}{\text{ is associate with the point }}\left( {\frac{{\sqrt 2 }}{2},\frac{{\sqrt 2 }}{2}} \right). \cr & {\text{The first coordinate is }} \cr & {\text{cos }}s = \cos \frac{\pi }{4} = \frac{{\sqrt 2 }}{2} \cr & {\text{then}}\,\,s = \frac{\pi }{4} \cr} $$
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