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: 68


(a) $\cos^{-1}\Big(\frac{1}{2}\Big)=\frac{\pi}{3}$ (b) $\cos^{-1}\Big(\frac{-1}{\sqrt2}\Big)=\frac{3\pi}{4}$ (c) $\cos^{-1}\Big(\frac{\sqrt3}{2}\Big)=\frac{\pi}{6}$

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) $$\cos^{-1}\Big(\frac{1}{2}\Big)=a$$ According to the defintion: $$\cos a=\frac{1}{2}\hspace{1cm}\text{and}\hspace{1cm}a\in[0,\pi]$$ So $a=\frac{\pi}{3}$. In other words, $$\cos^{-1}\Big(\frac{1}{2}\Big)=\frac{\pi}{3}$$ (b) $$\cos^{-1}\Big(\frac{-1}{\sqrt2}\Big)=\cos^{-1}\Big(\frac{-\sqrt2}{2}\Big)=a$$ According to the defintion: $$\cos a=\frac{-\sqrt2}{2}\hspace{1cm}\text{and}\hspace{1cm}a\in[0,\pi]$$ So $a=\frac{3\pi}{4}$. In other words, $$\cos^{-1}\Big(\frac{-1}{\sqrt2}\Big)=\frac{3\pi}{4}$$ (c) $$\cos^{-1}\Big(\frac{\sqrt3}{2}\Big)=a$$ According to the defintion: $$\cos a=\frac{\sqrt3}{2}\hspace{1cm}\text{and}\hspace{1cm}a\in[0,\pi]$$ So $a=\frac{\pi}{6}$. In other words, $$\cos^{-1}\Big(\frac{\sqrt3}{2}\Big)=\frac{\pi}{6}$$
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