cos x/2, given sin x = 5, with 180° < x < 270°

we know that the angle  is 80° < x < 270°, or namely on the III Quadrant, where cosine is negative and sine is negative as well.let's take a peek, sin(x) = -(1/5), now, the hypotenuse is always just a radius distance and thus is never negative, thus[tex]\bf sin(x)=\cfrac{\stackrel{opposite}{-1}}{\stackrel{hypotenuse}{5}}\qquad \impliedby \textit{let's find the \underline{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{5^2-(-1)^2}=a\implies \pm\sqrt{25-1}=a\implies \pm\sqrt{24}=a\implies \stackrel{\textit{III Quadrant}}{-\sqrt{24}=a} \\\\[-0.35em] ~\dotfill[/tex][tex]\bf cos(x) = \cfrac{\stackrel{adjacent}{-\sqrt{24}}}{\stackrel{hypotenuse}{5}}~\hfill cos\left(\cfrac{x}{2}\right)=\pm \sqrt{\cfrac{1+cos(x)}{2}} \\\\\\ cos\left(\cfrac{x}{2}\right)=\pm \sqrt{\cfrac{1-\frac{\sqrt{24}}{5}}{2}}\implies cos\left(\cfrac{x}{2}\right)=\pm \sqrt{\cfrac{5-\sqrt{24}}{10}} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ~\hfill cos\left(\cfrac{x}{2}\right)\approx \pm 0.10051~\hfill[/tex]