Respuesta :
we know that π/2 < θ < π, which is another way of saying that θ is in the II Quadrant, so half that angle will most likely be located on the I Quadrant, where cosine as well as sine are both positive.
Now, let's keep in mind that θ itself is in the II Quadrant, where the cosine is negative whilst the sine is positive.
[tex]\textit{Half-Angle Identities} \\\\ sin\left(\cfrac{\theta}{2}\right)=\pm \sqrt{\cfrac{1-cos(\theta)}{2}} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ sin(\theta )=\cfrac{\stackrel{opposite}{7}}{\underset{hypotenuse}{25}}\qquad \textit{let's now 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}[/tex]
[tex]\pm\sqrt{25^2-7^2}=a\implies \pm\sqrt{576}=a\implies \pm 24=a\implies \stackrel{II~Quadrant}{-24=a} \\\\[-0.35em] ~\dotfill\\\\ cos(\theta )=\cfrac{\stackrel{adjacent}{-24}}{\underset{hypotenuse}{25}} \\\\\\ sin\left(\cfrac{\theta}{2}\right)\implies \pm \sqrt{\cfrac{1-\left( -\frac{24}{25} \right)}{2}}\implies \pm \sqrt{\cfrac{1 +\frac{24}{25} }{2}}[/tex]
[tex]\pm\sqrt{\cfrac{~~\frac{49}{25} ~~}{2}}\implies \pm\sqrt{\cfrac{49}{50}}\implies \stackrel{I~Quadrant}{+\sqrt{\cfrac{49}{50}}}\implies \cfrac{\sqrt{49}}{\sqrt{50}}\implies \cfrac{7}{5\sqrt{2}} \\\\\\ \stackrel{\textit{rationalizing the denominator}}{\cfrac{7}{5\sqrt{2}}\cdot \cfrac{\sqrt{2}}{\sqrt{2}}\implies \cfrac{7\sqrt{2}}{5\sqrt{2^2}}}\implies \cfrac{7\sqrt{2}}{10}[/tex]
