[tex]\bf sin(\theta)=\cfrac{opposite}{hypotenuse}
\qquad
cos(\theta)=\cfrac{adjacent}{hypotenuse}
\quad
% tangent
tan(\theta)=\cfrac{opposite}{adjacent}\\\\
-------------------------------\\\\
sin(\theta )=\cfrac{1}{3}\cfrac{\leftarrow opposite}{\leftarrow hypotenuse}\impliedby
\begin{array}{llll}
\textit{so, let's use the pythagorean}\\
\textit{theorem to get the adjacent}
\end{array}[/tex]
[tex]\bf 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}
\\\\\\
\pm\sqrt{3^2-1^2}=a\implies \pm\sqrt{8}=a\implies \pm2\sqrt{2}=a[/tex]
so hmmm which is it, the +/-? well, we know that tan(θ) < 0, that means the tangent of the angle is negative, well, the tangent is opposite/adjacent, the only way that fraction can be negative, is that if either, no both, just either opposite or adjacent is negative.
well, we know the opposite is +1, so, the adjacent then has to be negative, so is -2√(2) then
[tex]\bf cos(\theta)=\cfrac{adjacent}{hypotenuse}\implies cos(\theta)=\cfrac{-2\sqrt{2}}{3}[/tex]
