Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 1 - Functions - Review Exercises - Page 53: 38

Answer

$\dfrac{2\pi}{3}$

Work Step by Step

We are given the expression: $\cos^{-1} \left(-\dfrac{1}{2}\right)$ $y=\cos^{-1} x$ is the value of $y$ so that $\cos y=x$, where $0\leq y\leq\pi$. So $y=\cos^{-1} \left(-\dfrac{1}{2}\right)$ is the value of $y$ so that $\cos y=-\dfrac{1}{2}$, where $0\leq y\leq\pi$ From the unit circle we have: $x=\cos \left(\dfrac{2\pi}{3}\right)$ Therefore the value of $y$ for which $x=-\dfrac{1}{2}$ in the interval $\left[0,\pi\right]$ is $\dfrac{2\pi}{3}$, so we got: $\cos^{-1} \left(-\dfrac{1}{2}\right)=\dfrac{2\pi}{3}$
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