Construct a right $\triangle{ABC}$ where $\angle{A}=30^\circ$ and $\angle{C}= 90^\circ$. Let the angle bisector of $A$ meet $BC$ and $D$, then $\angle{DAC}=15^\circ$.

Let $AB=2$, then by the property of $30^\circ-60^\circ-90^\circ$ right triangle, $BC$ must equal $1$ and $AC=\sqrt{3}$. Furthermore, assuming $CD=x$, then $BD=1-x$. By the angle bisector theorem, we have $$\frac{BD}{CD}=\frac{AB}{AC}\implies \frac{1-x}{x}=\frac{2}{\sqrt{3}}\implies x=\frac{\sqrt{3}}{2+\sqrt{3}}=2\sqrt{3}-3$$ Applying the Pythagorean theorem on $\triangle{ACD}$ leads $$AD^2=AC^2+CD^2=24-12\sqrt{3}\implies AD=\sqrt{6}\cdot(\sqrt{3}-1)$$ Now, we have $$\sin 15^\circ = \frac{CD}{AD}=\frac{2\sqrt{3}-3}{\sqrt{6}\cdot(\sqrt{3}-1)}=\frac{\sqrt{6}-\sqrt{2}}{4}$$ $$\cos 15^\circ = \frac{AC}{AD}=\frac{\sqrt{3}}{\sqrt{6}\cdot(\sqrt{3}-1)}=\frac{\sqrt{6}+\sqrt{2}}{4}$$