[tex]\bf \textit{let's say, the angle is }\theta \textit{ so then }cos^{-1}\left( \frac{2}{3} \right)=\theta
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\textit{this means }cos(\theta )=\cfrac{2}{3}\cfrac{\leftarrow adjacent}{\leftarrow hypotenuse}\impliedby \textit{now, let's find the \underline{opposite}}
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\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}[/tex]
[tex]\bf \pm\sqrt{3^2-2^2}=b\implies \pm\sqrt{9-4}=b\implies\boxed{ \pm\sqrt{5}=b}\\\\
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cos^{-1}\left( \frac{2}{3} \right)=\theta \implies sin\left[ cos^{-1}\left( \frac{2}{3} \right) \right]\implies sin(\theta )
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sin(\theta )=\cfrac{\pm\sqrt{5}}{3}\cfrac{\leftarrow opposite}{\leftarrow hypotenuse}[/tex]
it doesn't say the angle is in a certain quadrant, thus the +/- versions of it are both valid.
