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: 54


$\dfrac{\sqrt 3}{2}$

Work Step by Step

Suppose $\theta =\sin^{-1} (\dfrac{1}{2})$ This gives: $\sin \theta=\dfrac{1}{2}$ Since, $ r=\sqrt {x^2+y^2}$ This implies that $ x=\sqrt {r^2-y^2}=\sqrt {2^2-(1)^2}=\sqrt 3$ Therefore, we have $\cos \theta =\cos [\sin^{-1} (\dfrac{1}{2})]=\dfrac{\sqrt 3}{2}$
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