Precalculus (10th Edition)

Published by Pearson
ISBN 10: 0-32197-907-9
ISBN 13: 978-0-32197-907-0

Chapter 6 - Trigonometric Functions - 6.3 Properties of the Trigonometric Functions - 6.3 Assess Your Understanding - Page 395: 83



Work Step by Step

I know that $\cos$ has a period of $360^\circ$, hence $\cos{\theta}=\cos{(\theta-360^\circ)}.$ Hence: $\cos{400^\circ}\sec{40^\circ}=\cos{(400^\circ-360^\circ)}\sec{40^\circ}=\cos{40^\circ}\sec{40^\circ}$. We also know that $\cos{\theta}\sec{\theta}=1$. Thus: $\cos{40^\circ}\sec{40^\circ}=1$.
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