22)
[tex]\bf \textit{let's say that }sin^{-1}\left( \frac{2}{3} \right)=\theta \textit{ this means }sin(\theta )=\cfrac{2}{3}\qquad so
\\\\\\
sin\left[ sin^{-1}\left( \frac{2}{3} \right) \right]\implies sin[\theta ]\implies \cfrac{2}{3}[/tex]
23)
[tex]\bf sec^{-1}\left( \frac{5}{2} \right)=\theta \textit{ this means }sec(\theta )=\cfrac{5}{2}\cfrac{\leftarrow hypotenuse}{\leftarrow adjacent}
\\\\\\
\textit{let's find the \underline{opposite side} then}[/tex]
[tex]\bf \textit{using the pythagorean theorem}\\\\
c^2=a^2+b^2\implies \pm\sqrt{c^2-a^2}=b\qquad
\begin{cases}
c=hypotenuse\\
a=adjacent\\
b=opposite\\
\end{cases}
\\\\\\
\pm\sqrt{5^2-2^2}=b\implies \boxed{\pm \sqrt{21}=b}\\\\
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tan(\theta)=\cfrac{opposite}{adjacent}\qquad \qquad tan(\theta)=\cfrac{\pm\sqrt{21}}{2}[/tex]
again, we dunno what quadrant the angle is at, thus the +/- are both valid.
