Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 1305071751
ISBN 13: 978-1-30507-175-9

Chapter 5 - Section 5.5 - Inverse Trigonometric Functions and Their Graphs - 5.5 Exercises - Page 444: 1

Answer

(a) $[-\frac{\pi}{2}, \frac{\pi}{2}]$. $sin(y)=x$. $sin^{-1}\frac{1}{2}=\frac{\pi}{6}$, $sin(\frac{\pi}{6})=\frac{1}{2}$. (b) $[0,\pi]$. $cos(y)=x$. $cos^{-1}\frac{1}{2}=\frac{\pi}{3}$, $cos\frac{\pi}{3}=\frac{1}{2}$

Work Step by Step

(a) restrict the domain of sine to the interval $[-\frac{\pi}{2}, \frac{\pi}{2}]$. ... $sin(y)=x$. For example, $sin^{-1}\frac{1}{2}=\frac{\pi}{6}$ because $sin(\frac{\pi}{6})=\frac{1}{2}$. (b) restrict the domain of cosine to the interval $[0,\pi]$. ... $cos(y)=x$. For example, $cos^{-1}\frac{1}{2}=\frac{\pi}{3}$ because $cos\frac{\pi}{3}=\frac{1}{2}$
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