Respuesta :
[tex]\bf sin(\theta )=-\cfrac{1}{2}\cfrac{\leftarrow opposite}{\leftarrow hypotenuse}
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\textit{the hypotenuse is never negative, is just a radius unit, thus}
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sin(\theta )=\cfrac{-1}{2}\cfrac{\leftarrow opposite}{\leftarrow hypotenuse}
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\textit{now, let's find the \underline{adjacent side}}
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\textit{using the pythagorean theorem}\\\\
c^2=a^2+b^2\implies \pm\sqrt{c^2-b^2}=a\qquad
\begin{cases}
c=hypotenuse\\
a=adjacent\\
b=opposite\\
\end{cases}[/tex]
[tex]\bf \\\\\\ \pm\sqrt{(2)^2-(-1)^2}=a\implies \pm\sqrt{3}=a[/tex]
now... we know the angle is NOT in the III quadrant, however the opposite(y) is negative, well the opposite is negative on the III and IV quadrants, thus, the angle must be in the IV quadrant then, and there, the adjacent(x) is positive, thus [tex]\bf \sqrt{3}=a[/tex]
[tex]\bf opposite=-1\qquad \qquad hypotenuse=2\qquad \qquad adjacent=\sqrt{3}\\\\ -------------------------------\\\\ sin(\theta)=\cfrac{opposite}{hypotenuse} \qquad cos(\theta)=\cfrac{adjacent}{hypotenuse} \quad tan(\theta)=\cfrac{opposite}{adjacent} \\\\\\ cot(\theta)=\cfrac{adjacent}{opposite} \qquad csc(\theta)=\cfrac{hypotenuse}{opposite} \quad sec(\theta)=\cfrac{hypotenuse}{adjacent}[/tex]
[tex]\bf \\\\\\ \pm\sqrt{(2)^2-(-1)^2}=a\implies \pm\sqrt{3}=a[/tex]
now... we know the angle is NOT in the III quadrant, however the opposite(y) is negative, well the opposite is negative on the III and IV quadrants, thus, the angle must be in the IV quadrant then, and there, the adjacent(x) is positive, thus [tex]\bf \sqrt{3}=a[/tex]
[tex]\bf opposite=-1\qquad \qquad hypotenuse=2\qquad \qquad adjacent=\sqrt{3}\\\\ -------------------------------\\\\ sin(\theta)=\cfrac{opposite}{hypotenuse} \qquad cos(\theta)=\cfrac{adjacent}{hypotenuse} \quad tan(\theta)=\cfrac{opposite}{adjacent} \\\\\\ cot(\theta)=\cfrac{adjacent}{opposite} \qquad csc(\theta)=\cfrac{hypotenuse}{opposite} \quad sec(\theta)=\cfrac{hypotenuse}{adjacent}[/tex]
