$13 \cdot 50$ The function $f(x)=\sin x+\sin 2 x+\sin 3 x$ has
(A) a maximum value of 3.
(B) a maximum value not less than $\frac{3+\sqrt{3}}{2}$.
(C) a maximum value of $\sqrt{3}$.
(D) a minimum value of 1.
(2nd "Hope Cup" National Mathematics Invitational Competition, 1991)
[Solution 1] Since $\sin x \leqslant 1$, and $\sin x, \sin 2 x, \sin 3 x$ are not all 1, hence (A) is not true.
When $x=\frac{\pi}{4}$,
$$
\begin{aligned}
f\left(\frac{\pi}{4}\right) & =\sin \frac{\pi}{4}+\sin \frac{\pi}{2}+\sin \frac{3 \pi}{4}=\frac{\sqrt{2}}{2}+1+\frac{\sqrt{2}}{2} \\
& =1+\sqrt{2}>\sqrt{3}>1,
\end{aligned}
$$

we know that (C) is not true.
When $x=0$, $f(0)=0$, so (D) is not true.
Therefore, the answer is (B).
[Solution 2] When $x=\frac{\pi}{4}$, $f\left(\frac{\pi}{4}\right)=1+\sqrt{2}>\frac{3+\sqrt{2}}{2}$, hence the answer is (B).