Trigonometry (11th Edition) Clone

Published by Pearson
ISBN 10: 978-0-13-421743-7
ISBN 13: 978-0-13421-743-7

Chapter 6 - Inverse Circular Functions and Trigonometric Equations - Section 6.1 Inverse Circular Functions - 6.1 Exercises - Page 264: 23


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

Work Step by Step

Inverse Cosine Function $y=\cos^{-1}x$ or $y=$ arccos $x$ means that $x=\cos y$, for $ 0 \leq y \leq \pi$. ------------------- In quadrant I, we know $\displaystyle \cos\frac{\pi}{6}=\frac{\sqrt{3}}{2}.$ Also, $\cos(\pi-x)=-\cos x.$ In the interval $0 \leq y \leq \pi, $we find $y=\displaystyle \pi-\frac{\pi}{6} =\displaystyle \frac{5\pi}{6}$ such that $\displaystyle \cos(\frac{5\pi}{6}) =-\displaystyle \frac{\sqrt{3}}{2}$ so $y=$ arccos$(-\displaystyle \frac{\sqrt{3}}{2})=\frac{5\pi}{6}$
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