= Solution
Let $A$ be the top vertex, $O$ the center, and $B,C$ the two other points on the <circle>. The diameter bisects the angle at $A$, so the two right <triangles> obtained by drawing the diameter to the opposite point show that $B,C$ have coordinates $(\mp\sin\theta,-\cos\theta)$ when $O=(0,0)$ and $A=(0,1)$. The chord $BC$ cuts off a lower <circular segment> of central angle $2\theta$. The <triangle> $ABC$ has area $\sin\theta(1+\cos\theta)$, while the segment has area $\theta-\sin\theta\cos\theta$, by subtracting the central triangle from its <circular sector>. Adding gives
$$
\boxed{\text{shaded area}=\theta+\sin\theta,}
$$
with $\theta$ measured in <radians>. This is the unit-radius case of the <area enclosed by two chords and a circular arc>.
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