Question

How to solve this half angle identities

Answer 1

90° ⩽ x ⩽ 180° is another way of saying that the angle "x" is in the II Quadrant, where sine is positive whilst cosine is negative, as you can see in your Unit Circle.  We also know that csc(x) = 2, so

$csc(x)=2\implies \cfrac{1}{sin(x)}=2\implies \boxed{\cfrac{1}{2}=sin(x)}\implies sin^{-1}\left( \cfrac{1}{2} \right)=x \\\\\\ \cfrac{5\pi }{6}=x~\hspace{10em}\boxed{cos\left( \cfrac{5\pi }{6} \right)=-\cfrac{\sqrt{3}}{2}} \\\\[-0.35em] ~\dotfill\\\\ sin\left(\cfrac{x}{2}\right)=\pm \sqrt{\cfrac{1-cos(x)}{2}}$

$\pm \sqrt{\cfrac{1-\left( -\frac{\sqrt{3}}{2} \right)}{2}}\implies \pm \sqrt{\cfrac{1+\left( \frac{\sqrt{3}}{2} \right)}{2}}\implies \pm\sqrt{\cfrac{~~ \frac{2+\sqrt{3}}{2}~~}{2}}\implies \pm\sqrt{\cfrac{2+\sqrt{3}}{4}} \\\\\\ \pm\cfrac{\sqrt{2+\sqrt{3}}}{\sqrt{4}}\implies +\cfrac{\sqrt{2+\sqrt{3}}}{2} \\\\[-0.35em] ~\dotfill\\\\ cos\left(\cfrac{x}{2}\right)=\pm \sqrt{\cfrac{1+cos(x)}{2}}$

$\pm\sqrt{\cfrac{1+\left( -\frac{\sqrt{3}}{2} \right)}{2}}\implies \pm\sqrt{\cfrac{~~\frac{2-\sqrt{3}}{2} ~~}{2}}\implies \pm\sqrt{\cfrac{2-\sqrt{3}}{4}}\implies +\cfrac{\sqrt{2-\sqrt{3}}}{2} \\\\[-0.35em] ~\dotfill\\\\ tan\left(\cfrac{x}{2}\right)=\cfrac{1-cos(x)}{sin(x)} \\\\\\ \cfrac{~~1-\left( -\frac{\sqrt{3}}{2} \right)~~}{\frac{1}{2}}\implies \cfrac{~~1+\frac{\sqrt{3}}{2}~~}{\frac{1}{2}}\implies \cfrac{~~ \frac{2+\sqrt{3}}{2}~~}{\frac{1}{2}}\implies 2+\sqrt{3}$

now, the angle "x/2" is half the arc of the angle "x", and since the angle "x" is in the II Quadrant, we can conclude the half of that arc will be on the I Quadrant, where sine and cosine and tangent are all positive.