Question

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

Answer 1

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

`\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√(c^2-b^2)=a \qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases} \\\\\\ \pm√(5^2-(-1)^2)=a\implies \pm√(25-1)=a\implies \pm√(24)=a\implies \stackrel{\textit{III Quadrant}}{-√(24)=a} \\\\[-0.35em] ~\dotfill`

`\bf cos(x) = \cfrac{\stackrel{adjacent}{-√(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-(√(24))/(5)}{2}}\implies cos\left(\cfrac{x}{2}\right)=\pm \sqrt{\cfrac{5-√(24)}{10}} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ~\hfill cos\left(\cfrac{x}{2}\right)\approx \pm 0.10051~\hfill`