Question:
Simplify $$2\cos^2(\log(2009)i)+i\sin(\log(4036081)i).$$Note: The logarithms are with respect to base $e.$

Answer:
We have
\begin{align*}
2\cos^2(\log(2009)i)+i\sin(\log(4036081)i) &= 1+\cos(2\log(2009)i)+i\sin(\log(4036081)i) \\
&= 1+\cos(\log(4036081)i)+i\sin(\log(4036081)i) \\
&= 1+e^{i^2\log(4036081)} \\
&= 1+\frac{1}{4036081} \\
&= \boxed{\frac{4036082}{4036081}}.
\end{align*}