Let's construct $\triangle{ABC}$ where $\angle{A}=36^\circ$ and $\angle{B}=\angle{C}=72^\circ$. Assuming the bisector of $\angle{B}$ meets $AC$ at point $D$. Then, $\angle{ABD}=\angle{DBC}=36^\circ$. It is easy to see that $\triangle{ABC}$, $\triangle{DAB}$ and $\triangle{BCD}$ are all isosceles. Meanwhile, let $H$ be the foot of the altitude drawn from vertex $A$. Because $\triangle{ABC}$ is isosceles and $\angle{A}=36^\circ$, we find $\angle{BAH}=\angle{CAH}=18^\circ$. Then, our objective is to compute $$\sin 18^\circ = \frac{BH}{AB}$$

 For convenience, let's assume $AB=1$ and $BH=HC=x$. Then, because $\triangle{BCD}$ and $\triangle{DAB}$ are both isosceles, it must be true that $$AD=DB=BC=2x$$ \t Accordingly, $$CD = AC-AD=1-2x$$ Meanwhile, we claim that $\triangle{ABC}\sim\triangle{BCD}$ because both of them are $36^\circ-72^\circ-72^\circ$ triangles. \t$$\therefore\quad\frac{AB}{BC}=\frac{BC}{CD}\implies\frac{1}{2x}=\frac{2x}{1-2x}\implies x=\frac{\sqrt{5}-1}{4}$$ \tNow, it follows that $$\sin 18^\circ = \frac{BH}{AB}=\frac{x}{1}=\frac{\sqrt{5}-1}{4}$$