Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 1 - Functions - 1.4 Trigonometric Functions and Their Inverse - 1.4 Exercises - Page 48: 77

Answer

$\cos(\sec^{-1}(x)) =\frac{1}{x}$

Work Step by Step

We are trying to find $\cos(\sec^{-1}(x))$. Let's construct a right triangle to help us solve this problem. Since we have a $\sec^{-1}(x)$, we will create a right triangle with a hypotenuse of length $x$, a leg with length $1$, and the other leg with a length of $\sqrt{1-x^2}$. Let $\theta$ be the angle between the side with length $1$ and hypotenuse. Thus, $\sec(\theta) = x$, so $\theta = \sec^{-1}(x)$. This is part of the expression we are trying to find, so we can substitute it in: $\cos(\sec^{-1}(x)) = \cos(\theta) = \frac{1}{x}$.
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