first off, let's recall the symmetry trig identity that cos(-θ) = cos(θ).
now, the hypotenuse is just a radius unit, and therefore never negative, so in that cosine equation, the negative must be the numerator.
sin(θ) > 0, is another way to say, the sine is positive, and since the hypotenuse is always positive, then the opposite side in the angle is also positive.
[tex]\bf cos(-\theta )=-\cfrac{\sqrt{2}}{5}\implies \stackrel{\textit{symmetry identity}}{cos(\theta )=-\cfrac{\sqrt{2}}{5}}\implies cos(\theta )=\cfrac{\stackrel{adjacent}{-\sqrt{2}}}{\stackrel{hypotenuse}{5}} \\\\\\ \textit{let's find the \underline{opposite side}} \\\\\\ \stackrel{\textit{using the pythagorean theorem}}{c^2=a^2+b^2\implies \sqrt{c^2-a^2}=b \qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases}}[/tex]
[tex]\bf \pm\sqrt{5^2-(-\sqrt{2})^2}=b\implies \pm\sqrt{25-2}=b \\\\\\ \pm\sqrt{23}=b\implies \stackrel{\textit{recall that }sin(\theta )>0}{+\sqrt{23}=b} \\\\\\ sin(\theta )=\cfrac{\stackrel{opposite}{\sqrt{23}}}{\stackrel{hypotenuse}{5}}[/tex]
