[tex]\bf sin(\theta)=\cfrac{opposite}{hypotenuse}
\qquad \qquad
% cosine
cos(\theta)=\cfrac{adjacent}{hypotenuse}\\\\
-------------------------------\\\\
cos(\theta)=\cfrac{\sqrt{3}}{3}\cfrac{\leftarrow adjacent=a}{\leftarrow hypotenuse=c}
\\\\\\
\textit{using the pythagorean theorem}\implies c^2=a^2+b^2\implies \pm\sqrt{c^2-a^2}=b
\\\\\\
\pm\sqrt{3^2-(\sqrt{3})^2}=b\implies \pm\sqrt{6}=b[/tex]
now the root gives us the +/-, which is it? well, we know the sine is negative, so the "b" value must be negative, so it has to be -√6 = b
thus [tex]\bf sin(\theta)=\cfrac{opposite}{hypotenuse}\implies sin(\theta)=-\cfrac{\sqrt{6}}{3}[/tex]
