Answer
$\dfrac{\sqrt{2+\sqrt{3}}}{2}$
Work Step by Step
Double -angle Theorem can be defined as: $\sin \dfrac{\theta}{2}=\pm \sqrt {\dfrac{1- \cos \theta}{2}}$
$\sin 105 ^{\circ}=\sin \dfrac{210 ^{\circ}}{2}=\pm \sqrt {\dfrac{1- \cos 210 ^{\circ}}{2}}$
Since $\sin$ is positive in the second quadrant:
Thus, $ \sqrt {\dfrac{1- \cos 210 ^{\circ}}{2}}= \sqrt {\dfrac{1- \cos (180 ^{\circ}+30 ^{\circ})}{2}}= \sqrt {\dfrac{1+ \cos 30 ^{\circ}}{2}}$
and $\sqrt {\dfrac{1+ \cos 30 ^{\circ}}{2}}=\sqrt {\dfrac{1+ (\sqrt 3/2)}{2}}=\dfrac{\sqrt{2+\sqrt{3}}}{2}$
