take a peek at your Unit Circle, let's notice that in Quadrant IV the cosine is positive and the sine is negative, or namely the adjacent side is positive and the opposite side is negative, so hmmm
[tex]cos(\theta )=\cfrac{\stackrel{adjacent}{4}}{\underset{hypotenuse}{41}}\qquad \impliedby \textit{let's find the \underline{opposite side}} \\\\\\ \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]\pm\sqrt{41^2-4^2}=b\implies \pm\sqrt{1665}=b\implies \pm 3\sqrt{185}=b\implies \stackrel{IV~Quadrant}{-3\sqrt{185}=b} \\\\[-0.35em] ~\dotfill\\\\ ~\hfill sin(\theta )=\cfrac{\stackrel{opposite}{-3\sqrt{185}}}{\underset{hypotenuse}{41}}~\hfill[/tex]
