[Example 4.4.10] Let $n=1990$, find the value of the following expression:
$$
\frac{1}{2^{n}}\left(1-3 C_{n}^{2}+3^{2} C_{n}^{4}-3^{3} C_{n}^{6}+\cdots+3^{994} C_{n}^{1988}-3^{995} C_{n}^{1990}\right)
$$
From the surface observation, the expression sought is part of a binomial expansion. For easier observation, let $\sqrt{3}=t$, then
$$
\begin{aligned}
A & =\frac{1}{2^{n}}\left(1-3 C_{n}^{2}+3^{2} C_{n}^{4}-3^{3} C_{n}^{6}+\cdots+3^{994} C_{n}^{1998}-3^{995} C_{n}^{1990}\right) \\
& =\frac{1}{2^{n}}\left(1-t^{2} C_{n}^{2}+t^{4} C_{n}^{4}-t^{6} C_{n}^{6}+\cdots+t^{1988} C_{n}^{1988}-t^{1990} C_{n}^{1990}\right)
\end{aligned}
$$

We add the other part of the above expansion to form a complete binomial expansion. For this, let
$$
B=\frac{1}{2^{n}}\left(-t C_{n}^{1}+t^{3} C_{n}^{3}-t^{5} C_{n}^{5}+\cdots+t^{1987} C_{n}^{1987}-t^{1989} C_{n}^{1989}\right) .
$$

At this point, we have
$$
A+B \mathrm{i}=\frac{1}{2^{n}}(-1+\sqrt{3} \mathrm{i})^{n},
$$

and the expression for $A$, which is the real part of the above equation, is
$$
\begin{aligned}
A+B \mathrm{i} & =\frac{1}{2^{n}}(-1+\sqrt{3} \mathrm{i})^{n}=\left(-\frac{1}{2}+\frac{\sqrt{3}}{2} \mathrm{i}\right)^{n} \\
& =\left(\cos \frac{2 \pi}{3}+\mathrm{i} \sin \frac{2 \pi}{3}\right)^{n} \\
& =\cos \frac{2 n \pi}{3}+\mathrm{i} \sin \frac{2 n \pi}{3}
\end{aligned}
$$

Thus,
$$
A=\cos \frac{2 n \pi}{3}
$$

When $n=1990$, $A=-\frac{1}{2}$.