Example 3 A sector-shaped plate, with a central angle of $120^{\circ}$ and a radius of 1, is to be trimmed to form a rectangle of maximum area (the axis of symmetry of the sector is perpendicular to one side of the rectangle). What is the area of the part that will be cut off?
(1958, Wuhan City High School Mathematics Competition)
Solution: As shown in Figure 3, let $OAB$ be a sector,
with $\angle AOB = 120^\circ$,
$OA = 1$, and $CDEF$ be the inscribed rectangle.
Let $CD = x$, $OD = y$, and draw a line through $O$ parallel to $ED$ intersecting the extension of $CD$ at $K$. Then
$$
DK = \frac{y}{2}, \quad KO = \frac{\sqrt{3}}{2} y,
$$

In $\triangle OCD$,
$$
\begin{array}{l}
\because OC^2 = CD^2 + OD^2 + CD \cdot OD, \\
\therefore 1 = x^2 + y^2 + xy .
\end{array}
$$

Solving for $y$ gives $y = \frac{-x + \sqrt{4 - 3x^2}}{2}$.
Thus, $S = \frac{\sqrt{3}}{2} \left(-x^2 + \sqrt{4x^2 - 3x^4}\right)$. Simplifying, we get
$$
3x^4 + (\sqrt{3}S - 3)x^2 + S^2 = 0.
$$

Since $x^2 > 0$, we have $(\sqrt{3}S - 3)^2 - 12S^2 \geq 0$.
Solving this, we get $0 < S \leq \frac{\sqrt{3}}{3}$.
Thus, the maximum area of the rectangle is $\frac{\sqrt{3}}{3}$, at which point $x = \frac{\sqrt{3}}{3}$. Substituting into $S = \sqrt{3}xy$, we get $y = \frac{\sqrt{3}}{3} = x$. Therefore, $\angle COD = \angle OCD = 30^\circ$.