A sector with acute central angle $\theta$ is cut from a circle of radius 6. The radius of the circle circumscribed about the sector is 
$\textbf{(A)}\ 3\cos\theta \qquad \textbf{(B)}\ 3\sec\theta \qquad \textbf{(C)}\ 3 \cos \frac12 \theta \qquad \textbf{(D)}\ 3 \sec \frac12 \theta \qquad \textbf{(E)}\ 3$
Let $O$ be the center of the circle and $A,B$ be two points on the circle such that $\angle AOB = \theta$.  If the circle circumscribes the sector, then the circle must circumscribe $\triangle AOB$.

Draw the perpendicular bisectors of $OA$ and $OB$ and mark the intersection as point $C$, and draw a line from $C$ to $O$.  By HL Congruency and CPCTC, $\angle AOC = \angle BOC = \theta /2$.
Let $R$ be the circumradius of the triangle.  Using the definition of cosine for right triangles,
\[\cos (\theta /2) = \frac{3}{R}\]
\[R = \frac{3}{\cos (\theta /2)}\]
\[R = 3 \sec (\theta /2)\]
Answer choices A, C, and E are smaller, so they are eliminated.  However, as $\theta$ aproaches $90^\circ$, the value $3\sec\theta$ would approach infinity while $3\sec \tfrac12 \theta$ would approach $\tfrac{3\sqrt{2}}{2}$.  A super large circle would definitely not be a circumcircle if $\theta$ is close to $90^\circ$, so we can confirm that the answer is $\boxed{\textbf{(D)}\ 3 \sec \frac12 \theta}$.