$13 \cdot 49$ When $\sin \frac{A}{2}-\sqrt{3} \cos \frac{A}{2}$ has a minimum value, the value of $A$ is
(A) $-180^{\circ}$.
(B) $60^{\circ}$.
(C) $120^{\circ}$.
(D) $0^{\circ}$.
(E) None of these.
(25th American High School Mathematics Examination, 1974)
[Solution] Notice $\sin \frac{A}{2}-\sqrt{3} \cos \frac{A}{2}$
$$
\begin{array}{l}
=2\left(\frac{1}{2} \sin \frac{A}{2}-\frac{\sqrt{3}}{2} \cos \frac{A}{2}\right) \\
=2\left(\cos 60^{\circ} \sin \frac{A}{2}-\sin 60^{\circ} \cos \frac{A}{2}\right) \\
=2 \sin \left(\frac{A}{2}-60^{\circ}\right),
\end{array}
$$

We see that the condition for the last expression to reach its minimum value is $\sin \left(\frac{A}{2}-60^{\circ}\right)=-1$, or $\frac{A}{2}-60^{\circ}=270^{\circ}+k \cdot 360^{\circ}, \quad k=0, \pm 1, \pm 2, \cdots$.
That is, $A=660^{\circ}+k \cdot 720^{\circ}, k=0, \pm 1, \pm 2, \cdots$
None of the options from $(A)$ to $(D)$ satisfy the above equation. Therefore, the answer is $(E)$.