$33 \cdot 40$ Given $a_{0}=\sin x, a_{n}=(-1)^{\left[\frac{n}{2}\right]} \sqrt{1-a_{n-1}^{2}}$, where $n \in$ $N$, and $\left[\frac{n}{2}\right]$ is the integer part of $\frac{n}{2}$. Then $a_{1989}$ equals
(A) $\sin x$.
(B) $-\cos x$.
(C) $-\sin x$.
(D) $\cos x$.
(China Sichuan Province High School Mathematics Competition, 1989)
［Solution］From the given, it is not hard to see:
$$
a_{n}=\left\{\begin{array}{ll}
\sin x, & n=4 k \text {; } \\
\cos x, & n=4 k+1 \\
-\sin x, & n=4 k+2 \text {; } \\
-\cos x, & n=4 k+3 \text {; }
\end{array}\right.
$$

Since $1989=497 \cdot 4+1$, then $a_{1989}=\cos x$.
Therefore, the answer is $(D)$.