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 - Chapter Review - Review Exercises - Page 458: 10



Work Step by Step

We know that $\cos$ has a period of $ 2 \pi$ or $360^\circ$, so we can write $\cos{\theta}=\cos{(\theta-2\pi)}$ This implies that $\cos{540^\circ}=\cos{(540^\circ-360^\circ)}=\cos{180^\circ}$ Since the trigonometric function $\tan$ is an odd function, we can write $f(-\theta)=-f(\theta) \implies \tan{-45^\circ}=-\tan{45^\circ}$. Therefore, $\cos{540^\circ}-\tan{-45^\circ}=\cos{180^\circ}+\tan{45^\circ} \\=-1+1\\=0$
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