21. [10] Simplify
$$
2 \cos ^{2}(\ln (2009) i)+i \sin (\ln (4036081) i) .
$$
Answer: $\frac{4036082}{4036081}$ $\square$ Solution:
We have
$$
\begin{aligned}
2 \cos ^{2}(\ln (2009) i)+i \sin (\ln (4036081) i) & =1+\cos (2 \ln (2009) i)+i \sin (\ln (4036081) i) \\
& =1+\cos (\ln (4036081) i)+i \sin (\ln (4036081) i) \\
& =1+e^{i^{2} \ln (4036081)} \\
& =1+\frac{1}{4036081} \\
& =\frac{4036082}{4036081}
\end{aligned}
$$
as desired.