3. The general term of the sequence $\left\{a_{n}\right\}$ is $a_{n}=\sin \frac{n}{3} \pi$, then the value of $a_{1}+a_{2}+\cdots+a_{2007}+a_{2008}$ is
A. 0
B. $\frac{\sqrt{3}}{2}$
C. $-\frac{\sqrt{3}}{2}$
D. $\sqrt{3}$
3. B From the general term being in trigonometric form, we know that the sequence $\left\{a_{n}\right\}$ is a function with the smallest positive period of $\frac{2 \pi}{\frac{\pi}{3}}=6$,

Since $a_{1}+a_{2}+a_{3}+a_{1}+a_{5}+a_{6}=\sin \frac{\pi}{3}+\sin \frac{2 \pi}{3}+\sin \pi+\sin \frac{4 \pi}{3}+\sin \frac{5 \pi}{3}+\sin 2 \pi=0, 2008=334 \times 6+4$,
Therefore, $a_{1}+a_{2}+\cdots+a_{2007}+a_{2008}=334\left(a_{1}+a_{2}+a_{3}+a_{1}+a_{5}+a_{6}\right)+a_{1}+a_{2}+a_{3}+a_{4}=\frac{\sqrt{3}}{2}+\frac{\sqrt{3}}{2}+0-\frac{\sqrt{3}}{2}$ $=\frac{\sqrt{3}}{2}$. Hence, the answer is B.