Use an appropriate Half-Angle Formula to find the exact value of $\sin 105^\circ$ Is this correct? $$\sin 105^\circ = \frac{\sqrt{2+\sqrt{3}}}{4}$$
Because $105^\circ$ is in the second quadrant the square root after the equal sign is $+$. However, because $\cos 210^\circ$ is negative, it value is negative  $-\frac{\sqrt{3}}{2}$. 
Is my thought process correct? 
Also, can I factor out $\frac{1}{4}$? 

This is an equivalent way to get the same answer as above:
$\displaystyle\sin 105^{\circ}=\sqrt{\frac{1-\cos210^{\circ}}{2}}
=\sqrt{\frac{1-(-\frac{\sqrt{3}}{2})}{2}}=\sqrt{\frac{2+\sqrt{3}}{4}}=\frac{\sqrt{2+\sqrt{3}}}{2}$.

Notice that using the addition formula for the sine gives a different-looking answer:
$\sin 105^{\circ}=\sin(60^{\circ}+45^{\circ})=\sin60^{\circ}\cos45^{\circ}+\cos60^{\circ}\sin45^{\circ}=\frac{\sqrt{3}}{2}\cdot\frac{\sqrt{2}}{2}+\frac{1}{2}\cdot\frac{\sqrt{2}}{2}=\frac{\sqrt{6}+\sqrt{2}}{4}$.
