[tex]\bf sin(\theta)=\cfrac{opposite}{hypotenuse}
\qquad
cos(\theta)=\cfrac{adjacent}{hypotenuse}
\quad
% tangent
tan(\theta)=\cfrac{opposite}{adjacent}\\\\
-------------------------------\\\\[/tex]
[tex]\bf sin(x)=\cfrac{\sqrt{2}}{2}\cfrac{\leftarrow opposite}{\leftarrow hypotenuse}\impliedby \textit{now, let's find the adjacent side}
\\\\\\
\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}
\\\\\\
\pm\sqrt{(2)^2-(\sqrt{2})^2}=a\implies \pm\sqrt{4-2}=a\implies \pm\sqrt{2}=a[/tex]
now, the adjacent side, or the x-coordinate for the angle, can be negative and positive, is positive in the I quadrant and negative in the II quadrant, with a positive y-coordinate... so it can be either, the +/- one.
[tex]\bf cos(x)=\cfrac{\pm\sqrt{2}}{2}\qquad tan(x)=\cfrac{\frac{\sqrt{2}}{2}}{\pm\frac{\sqrt{2}}{2}}\implies tan(x)=\pm 1[/tex]
