Answer
$\frac{\sqrt {2-\sqrt 3}}{2}=\frac{\sqrt 6-\sqrt 2}{4}$
Work Step by Step
1. $sin15^\circ=sin\frac{30^\circ}{2}=\sqrt {\frac{1-cos30^\circ}{2}}=\sqrt {\frac{1-\sqrt 3/2}{2}}=\frac{\sqrt {2-\sqrt 3}}{2}$
2. $sin15^\circ=sin(45^\circ-30^\circ)=sin45^\circ cos30^\circ-cos45^\circ sin30^\circ=(\frac{\sqrt 2}{2})(\frac{\sqrt 3}{2})-(\frac{\sqrt 2}{2})(\frac{1}{2})=\frac{\sqrt 6-\sqrt 2}{4}$
3. Take the square of each result, we have $(\frac{\sqrt {2-\sqrt 3}}{2})^2=\frac{2-\sqrt 3}{4}$ and $(\frac{\sqrt 6-\sqrt 2}{4})^2=\frac{8-4\sqrt 3}{16}=\frac{2-\sqrt 3}{4}$. Thus the two results are the same.
