University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 3 - Section 3.9 - Inverse Trigonometric Functions - Exercises - Page 181: 46

Answer

$\csc^{-1}{\frac{1}{2}}$ is undefined and $\csc^{-1}{{2}}$ is defined

Work Step by Step

a) To determine if $\csc^{-1}{\frac{1}{2}}$ is defined, we note the following: $\csc{x}={\frac{1}{2}}$ ${\frac{1}{\sin{x}}}={\frac{1}{2}}$ ${\sin{x}}=2$ But that value is outside the range of sin, thus $\csc^{-1}{\frac{1}{2}}$ is undefined. b) To determine if $\csc^{-1}{{2}}$ is defined, we note the following: let $\csc^{-1}{{2}}=x$ $\csc{x}={{2}}$ ${{\sin{x}}}={\frac{1}{2}}$ Which is inside the range of sin, thus $\csc^{-1}{{2}}$ is defined.
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