let's recall that on the IV Quadrant the x/cosine is positive and the y/sine is negative, and of course the hypotenuse is just a radius unit and therefore never negative.
[tex]\bf sin(\theta )=\cfrac{\stackrel{opposite}{-7}}{\stackrel{hypotenuse}{15}}\impliedby \textit{let's find the \underline{adjacent side}} \\\\\\ \textit{using the pythagorean theorem} \\\\ c^2=a^2+b^2\implies \sqrt{c^2-b^2}=a \qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases}[/tex]
[tex]\bf \pm\sqrt{15^2-(-7)^2}=a\implies \pm\sqrt{176}=a\implies \stackrel{\textit{IV Quadrant}}{+\sqrt{176}=a} \\\\[-0.35em] ~\dotfill\\\\ ~\hfill cos(\theta )=\cfrac{\stackrel{adjacent}{\sqrt{176}}}{\stackrel{hypotenuse}{15}}~\hfill[/tex]
