Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

Published by Pearson
ISBN 10: 0-32193-104-1
ISBN 13: 978-0-32193-104-7

Chapter 5 - Trigonometric Functions - Section 5.2 Trigonometric Functions: Unit Circle Approach - 5.2 Assess Your Understanding - Page 402: 29



Work Step by Step

Recall: $ \sec \pi =\dfrac{1}{\cos \pi}$ We need to compute $\sec (- \pi)$. The trigonometric function $\cos$ and its inverse trigonometric function $\sec x$ has a period of $2\pi$. Also, because cosine and secant is even so we can write $\cos (-x)=\cos (x)$ and $\sec (- \pi) = \sec \pi$ Thus, we can simplify as follows: $\sec (- \pi) = \sec \pi =\dfrac{1}{\cos \pi}\\=\dfrac{1}{-1} \\ =-1$
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