2. Since $O P^{2}=1$, point $P$ moves on the unit circle, starting point $P_{1}\left(-\frac{1}{2}, \frac{\sqrt{3}}{2}\right)$,
end point $P_{2}\left(\frac{1}{2}, \frac{\sqrt{3}}{2}\right)$.
The shaded area in the right figure is the required area.
$$
\begin{array}{l}
\because S_{\triangle P_{1} O A}=S_{\triangle P_{2} O A} \text {, } \\
\therefore S_{\triangle P_{1} O B}=S_{\triangle P_{2} B A} \text {. } \\
\end{array}
$$

Therefore, the required area $=S_{\text {quadrilateral } O P_{1} P_{2}}=\frac{1}{2} \cdot 1 \cdot \frac{\pi}{3}=\frac{\pi}{6}$.
2. Let $A(2,0)$ be a fixed point on the plane, and $P\left(\sin \left(2 t-60^{\circ}\right), \cos \right.$ $\left(2 t-60^{\circ}\right)$ ) be a moving point. Then, as $t$ varies from $15^{\circ}$ to $45^{\circ}$, the area of the figure swept by the line segment $A P$ is $\qquad$ .