![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}}](/tpl/images/2667/1808/5480c.png)
![\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}}](/tpl/images/2667/1808/fa3f8.png)
![\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}](/tpl/images/2667/1808/0bb5e.png)
How to solve this half angle identities
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