Precalculus (6th Edition) Blitzer

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

Chapter 4 - Section 4.7 - Inverse Trigonometric Functions - Exercise Set - Page 626: 53


$\dfrac{\sqrt 2}{2}$

Work Step by Step

Suppose $\theta =\cos^{-1} (\dfrac{\sqrt 2}{2})$ This gives: $\cos \theta=\dfrac{\sqrt 2}{2}$ Since, $ r=\sqrt {x^2+y^2}$ This implies that $ y=\sqrt {r^2-x^2}=\sqrt {2^2-(\sqrt 2)^2}=\sqrt 2$ Therefore, we have $\sin \theta =\sin [\cos^{-1} (\dfrac{\sqrt 2}{2})]=\dfrac{\sqrt 2}{2}$
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