Trigonometry (10th Edition)

Published by Pearson
ISBN 10: 0321671775
ISBN 13: 978-0-32167-177-6

Chapter 6 - Inverse Circular Functions and Trigonometric Equations - Section 6.1 Inverse Circular Functions - 6.1 Exercises - Page 258: 37

Answer

arctan($-1$)$=-45^{o}$

Work Step by Step

Inverse Tangent Function $y=\tan^{-1}x$ or $y=$ arctan $x$ means that $x=\tan y$, for $-\displaystyle \frac{\pi}{2} < y < \frac{\pi}{2}$. Or for an angle in degrees, $\theta=\tan^{-1}x$ means $ x=\tan\theta$, for $-90^{o} < \theta < 90^{o}$. --------------- Knowing: $\quad \tan 45^{o}=1$, (and $\tan(-\theta)=-\tan\theta),$ we find $\theta=-45^{o}$ from the interval $-90^{o} < \theta < 90^{o}$ such that $\tan$($-45^{o})=-1.$ So, arctan($-1$)$=-45^{o}$
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