Solution:
Clearly $a_{1}<1$, or else $1 \leq a_{1} \leq a_{2} \leq a_{3} \leq \ldots$
We can therefore write $a_{1}=\cos \theta$ for some $0<\theta<90^{\circ}$.
Note that $\cos \frac{\theta}{2}=\sqrt{\frac{1+\cos \theta}{2}}$, and $\cos 15^{\circ}=\frac{\sqrt{6}+\sqrt{2}}{4}$.
Hence, the possibilities for $a_{1}$ are $\cos 15^{\circ}, \cos 30^{\circ}$, and $\cos 60^{\circ}$, which are $\frac{\sqrt{2}+\sqrt{6}}{2}, \frac{\sqrt{3}}{2}$, and $\frac{1}{2}$.