Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 1 - Review - Concept Check: 13

Answer

(a) The definition is $$\text{for }x\in[-\pi/2,\pi/2]\quad y=\sin{x}\Rightarrow x=\sin^{-1}y.$$ The domain is $[-1,1]$. The range is $[-\pi/2,\pi/2]$. (b) The definition is $$\text{for }x\in[0,\pi]\quad y=\cos{x}\Rightarrow x=\cos^{-1}y.$$ The domain is $[-1,1]$. The range is $[0,\pi]$. (c) The definition is $$\text{for }x\in(-\pi/2,\pi/2)\quad y=\tan{x}\Rightarrow x=\tan^{-1}y.$$ The domain is $(-\infty,\infty)$ The range is $(-\pi/2,\pi/2)$

Work Step by Step

(a) Since $\sin x$ is not one-to-one, we take its restriction on the interval $[-\pi/2,\pi/2]$. Then this restriction is one-to-one. We can define now the inverse sine as the value from $[-\pi/2,\pi/2]$ whose sine is equal to the given value from $[-1,1]$ i.e. $$\text{for }x\in[-\pi/2,\pi/2]\quad y=\sin{x}\Rightarrow x=\sin^{-1}y.$$ The domain of $\sin^{-1}x$ is the range of $\sin x$ when its domain is narrowed down to $[-\pi/2,\pi/2]$ which is $[-1,1]$. The range of $\sin^{-1}$ is the set of all of its possible values which is from its definition $[-\pi/2,\pi/2]$. (b) Since $\cos x$ is not one-to-one we take its restriction on the interval $[0,\pi]$. Then this restriction is one-to-one. We can define now the inverse cosine as the value from $[0,\pi]$ whose cosine is equal to the given value from $[-1,1]$ i.e. $$\text{for }x\in[0,\pi]\quad y=\cos{x}\Rightarrow x=\cos^{-1}y.$$ The domain of $\cos^{-1}x$ is the range of $\cos x$ when its domain is narrowed down to $[0,\pi]$ which is $[-1,1]$. The range of $\cos^{-1}$ is set of all of its possible values which is from its definition $[0,\pi]$. (c) Since $\tan x$ is not one-to-one we take its restriction on the interval $(-\pi/2,\pi/2)$. Then this restriction is one-to-one. We can define now the inverse tangent as the value from $(-\pi/2,\pi/2)$ whose tangent is equal to the given value from $(-\infty,\infty)$ (because the tangent of an angle from $(-\pi/2,\pi/2)$ can be any number from $(-\infty,\infty)$) i.e. $$\text{for }x\in(-\pi/2,\pi/2)\quad y=\tan{x}\Rightarrow x=\tan^{-1}y.$$ The domain of $\tan^{-1}x$ is the range of $\tan x$ when its domain is narrowed down to $(-\pi/2,\pi/2)$ which is $(-\infty,\infty)$. The range of $\tan^{-1}$ is set of all of its possible values which is from its definition $(-\pi/2,\pi/2)$.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.