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: 55

Answer

$\dfrac{4}{\sqrt {15}}$ or, $\dfrac{4\sqrt {15}}{15}$

Work Step by Step

Suppose $\theta =\sin^{-1} (-\dfrac{1}{4})$ This gives: $\sin \theta=-\dfrac{1}{4}$ Since, $ r=\sqrt {x^2+y^2}$ This implies that $ x=\sqrt {r^2-y^2}=\sqrt {42^2-(1)^2}=\sqrt 5$ Therefore, we have $\sec \theta =\sec [\sin^{-1} (-\dfrac{1}{4})]=\dfrac{4}{\sqrt {15}}=\dfrac{4\sqrt {15}}{15}$
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