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: 24


$\displaystyle \frac{2\pi}{3}$

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: $\ \ \displaystyle \cos\frac{\pi}{3}=\frac{1}{2}.$ Also, $\cos(\pi-x)=-\cos x.$ In the interval $0 \leq y \leq \pi, $we find $y=\displaystyle \pi-\frac{\pi}{3} =\displaystyle \frac{2\pi}{3}$ such that $\displaystyle \cos(\frac{2\pi}{3}) =-\displaystyle \frac{1}{2}$ so $y=$ arccos$(-\displaystyle \frac{1}{2})=\frac{2\pi}{3}$
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