Example 9 Let $n=1990$. Then
$$
\frac{1}{2^{n}}\left(1-3 \mathrm{C}_{n}^{2}+3^{2} \mathrm{C}_{n}^{4}-\cdots+3^{99} \mathrm{C}_{n}^{108}-3^{90} \mathrm{C}_{n}^{900}\right) \text {. }
$$
(1990, National High School Mathematics Competition)
Solution: Original expression
$$
\begin{aligned}
= & \left(\frac{1}{2}\right)^{1990}-C_{1990}^{2}\left(\frac{1}{2}\right)^{1988}\left(\frac{\sqrt{3}}{2}\right)^{2}+\cdots+ \\
& C_{1990}^{1998}\left(\frac{1}{2}\right)^{2}\left(\frac{\sqrt{3}}{2}\right)^{1988}-C_{1990}^{1990}\left(\frac{\sqrt{3}}{2}\right)^{1990} .
\end{aligned}
$$

Therefore, the original expression is the real part of $\left(\frac{1}{2}+\frac{\sqrt{3}}{2} \mathrm{i}\right)^{1990}$. And
$$
\begin{array}{l}
\left(\frac{1}{2}+\frac{\sqrt{3}}{2} i\right)^{1990}=\left(\cos \frac{\pi}{3}+i \sin \frac{\pi}{3}\right)^{1990} \\
=\cos \frac{1990}{3} \pi+i \sin \frac{1990}{3} \pi
\end{array}
$$

Thus, the original expression $=\cos \frac{1990}{3} \pi=-\frac{1}{2}$.