Answer
$\frac{8}{7}$
Work Step by Step
Let $sin^{-1}\frac{1}{4}=y$, we have $sin(y)=\frac{1}{4},$ and
$sec(2sin^{-1}\frac{1}{4})=sec(2y)
=\frac{1}{cos(2y)}=\frac{1}{1-2sin^2y}=\frac{1}{1-2/16}=\frac{8}{7}$
Precalculus: Mathematics for Calculus, 7th Edition
by Stewart, James; Redlin, Lothar; Watson, Saleem
Published by
Brooks Cole
ISBN 10:
1305071751
ISBN 13:
978-1-30507-175-9
Chapter 7 - Section 7.3 - Double-Angle, Half-Angle, and Product-Sum Formulas - 7.3 Exercises - Page 561: 49
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