University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 1 - Section 1.6 - Inverse Functions and Logarithms - Exercises - Page 50: 69


(a) $\arccos(-1)=\pi$ (b) $\arccos0=\frac{\pi}{2}$

Work Step by Step

*Recall the definition of arccosine: $y=\cos^{-1}x$ is the number in $[0,\pi]$ for which $\cos y=x$ (a) $$\arccos(-1)=a$$ According to the defintion: $$\cos a=-1\hspace{1cm}\text{and}\hspace{1cm}a\in[0,\pi]$$ So $a=\pi$. In other words, $$\arccos(-1)=\pi$$ (b) $$\arccos0=a$$ According to the defintion: $$\cos a=0\hspace{1cm}\text{and}\hspace{1cm}a\in[0,\pi]$$ So $a=\frac{\pi}{2}$. In other words, $$\arccos0=\frac{\pi}{2}$$
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