Precalculus (6th Edition)

Published by Pearson
ISBN 10: 013421742X
ISBN 13: 978-0-13421-742-0

Chapter 7 - Trigonometric Identities and Equations - 7.5 Inverse Circular Functions - 7.5 Exercises - Page 708: 28


$y=\displaystyle \frac{5\pi}{6}$

Work Step by Step

$y=\cot^{-1}x$ Domain: $(-\infty, \infty) $ Range: $(0, \pi)$ ----------- $y$ is the number from $(0, \pi)$ such that $\cot y=-1.$ With $\displaystyle \frac{\pi}{6}$ as reference angle, $\displaystyle \cot(\frac{\pi}{6})=\sqrt{3}$, in quadrant II, $\cot$($\displaystyle \frac{5\pi}{6})=-\sqrt{3}\quad$and$\displaystyle \quad \frac{5\pi}{6}\in(0, \pi)$ so $y=\displaystyle \frac{5\pi}{6}$
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