2. Given the curve
$$
(x-20)^{2}+(y-16)^{2}=r^{2} \text { and } y=\sin x
$$

have exactly one common point $P\left(x_{0}, y_{0}\right)$. Then
$$
\frac{1}{2} \sin 2 x_{0}-16 \cos x_{0}+x_{0}=
$$
$\qquad$
2. 20 .

From the given information, there is a common tangent line at point $P\left(x_{0}, y_{0}\right)$.
$$
\begin{array}{l}
\text { Therefore, } \frac{\sin x_{0}-16}{x_{0}-20} \cos x_{0}=-1 \\
\Rightarrow \frac{1}{2} \sin 2 x_{0}-16 \cos x_{0}+x_{0}=20 .
\end{array}
$$