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 5.3 Properties of the Trigonometric Functions - 5.3 Assess Your Understanding - Page 418: 83

Answer

$1$

Work Step by Step

Recall: $\cos{(\theta+360^{\circ})} = \cos{\theta}$ This means that $\cos{400^{\circ}} = \cos{(40^{\circ}+360^{\circ})} = \cos{40^{\circ}}$ Thus, $\cos{400^{\circ}} \cdot \sec{40^{\circ}} = \cos{40^{\circ}}\cdot \sec{40^{\circ}}$ Since $\sec{\theta} = \dfrac{1}{\cos{\theta}}$, then $\sec{40^{\circ}} = \dfrac{1}{\cos{40^{\circ}}}$ Hence, $\cos{40^{\circ}}\cdot \sec{40^{\circ}} = \cos{40^{\circ}}\cdot \dfrac{1}{\cos{40^{\circ}} }= \boxed{1}$
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