6. $\sum_{k=0}^{2022} C_{2022}^{k} \cos \frac{(1011-k) \pi}{2}$ The value is $\qquad$ .
$$
\cos \left(\frac{1011 \pi}{2}-\frac{k \pi}{2}\right)=\cos \left(\frac{3 \pi}{2}-\frac{k \pi}{2}\right)=-\sin \frac{k \pi}{2}=\left\{\begin{array}{l}
-1, k=4 m+1, \\
1, k=4 m+3, \\
0, k \text { is even. }
\end{array}\right.
$$

Then the original expression $=\left(C_{2022}^{3}+C_{2022}^{7}+\cdots+C_{2022}^{2019}\right)-\left(C_{2022}^{1}+C_{2022}^{5}+\cdots+C_{2022}^{2021}\right)$. Let $A=C_{n}^{0}+C_{n}^{4}+C_{n}^{8}+\cdots, B=C_{n}^{1}+C_{n}^{5}+C_{n}^{9}+\cdots$, $C=C_{n}^{2}+C_{n}^{6}+C_{n}^{10}+\cdots, D=C_{n}^{3}+C_{n}^{7}+C_{n}^{11}+\cdots$, using the equation $(1+x)^{n}=C_{n}^{0}+C_{n}^{1} x+C_{n}^{2} x^{2}+C_{n}^{3} x^{3}+\cdots$, we get
$$
\left\{\begin{array} { l } 
{ A + B + C + D = 2 ^ { n } , } \\
{ A + B \mathrm { i } - C - D \mathrm { i } = ( 1 + \mathrm { i } ) ^ { n } } \\
{ A - B + C - D = 0 , } \\
{ A - B \mathrm { i } - C + D \mathrm { i } = ( 1 - \mathrm { i } ) ^ { n } }
\end{array} \Rightarrow \left\{\begin{array}{l}
A=\frac{1}{2}\left(2^{n-1}+2^{\frac{n}{2}} \cos \frac{n \pi}{4}\right), \\
B=\frac{1}{2}\left(2^{n-1}+2^{\frac{n}{2}} \sin \frac{n \pi}{4}\right), \\
C=\frac{1}{2}\left(2^{n-1}-2^{\frac{n}{2}} \cos \frac{n \pi}{4}\right), \\
D=\frac{1}{2}\left(2^{n-1}-2^{\frac{n}{2}} \sin \frac{n \pi}{4}\right) .
\end{array}\right.\right.
$$

Taking $n=2022$, so the original expression $=D-B=-2^{1011} \sin \frac{1011 \pi}{2}=2^{1011}$.