## University Calculus: Early Transcendentals (3rd Edition)

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