Answer
$\frac{\sqrt {2-\sqrt 3}}{2}$,
$\frac{\sqrt 6-\sqrt 2}{4}$
Work Step by Step
Step 1. $sin15^\circ=sin(\frac{30^\circ}{2})=\sqrt {\frac{1-cos(30^\circ)}{2}}=\sqrt {\frac{1-\frac{\sqrt 3}{2}}{2}}=\frac{\sqrt {2-\sqrt 3}}{2}$
Step 2. $sin15^\circ=sin(45-30)^\circ=sin(45^\circ)cos(30^\circ)-cos(45^\circ)sin(30^\circ)=(\frac{\sqrt 2}{2})(\frac{\sqrt 3}{2})-(\frac{\sqrt 2}{2})(\frac{1}{2})=\frac{\sqrt 6-\sqrt 2}{4}$
Step 3. To show the results from step 1 and 2 are equal, take the square for both to get $(\frac{\sqrt {2-\sqrt 3}}{2})^2=\frac{2-\sqrt 3}{4}$ and $(\frac{\sqrt 6-\sqrt 2}{4})^2=\frac{6+2-4\sqrt 3}{16}=\frac{2-\sqrt 3}{4}$
