[tex]\bf sin(\theta)=\cfrac{opposite}{hypotenuse}
\qquad \qquad
% cosine
cos(\theta)=\cfrac{adjacent}{hypotenuse}\\\\
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cos(\theta)=\cfrac{4}{7}\cfrac{\leftarrow adjacent=a}{\leftarrow hypotenuse=c}
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\textit{using the pythagorean theorem to get the opposite side or "b"}
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c^2=a^2+b^2\implies \pm\sqrt{c^2-a^2}=b\implies \pm\sqrt{7^2-4^2}=b
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\pm \sqrt{33}[/tex]
now, the square root gives us both +/- versions, which is it? well, we know the angle θ is in the IV quadrant, well, the sine is "b" is negative there, thus is the negative version then
thus [tex]\bf sin(\theta)=\cfrac{opposite}{hypotenuse}\implies sin(\theta)=\cfrac{-\sqrt{33}}{7}[/tex]
