3. Let the function $f(x)=2 x-\cos x,\left\{a_{n}\right\}$ be an arithmetic sequence with a common difference of $\frac{\pi}{8}$, and $\sum_{k=1}^{5} f\left(a_{k}\right)=5 \pi$. Then $f^{2}\left(a_{3}\right)-a_{1} a_{2}=(\quad)$.
(A) 0
(B) $\frac{\pi^{2}}{16}$
(C) $\frac{\pi^{2}}{8}$
(D) $\frac{13 \pi^{2}}{16}$
3. D.

Since $\left\{a_{n}\right\}$ is an arithmetic sequence with a common difference of $\frac{\pi}{8}$, and
$$
\sum_{k=1}^{5} f\left(a_{k}\right)=\sum_{k=1}^{5}\left(2 a_{k}-\cos a_{k}\right)=5 \pi \text {, }
$$

it follows that $10 a_{3}-\sum_{k=-2}^{2} \cos \left(a_{3}+\frac{k \pi}{8}\right)=5 \pi$.
Let $g(x)$
$$
=10 x-\left(2 \cos \frac{\pi}{4}+2 \cos \frac{\pi}{8}+1\right) \cos x-5 \pi \text {. }
$$

Then $g^{\prime}(x)$
$$
=10+\left(2 \cos \frac{\pi}{4}+2 \cos \frac{\pi}{8}+1\right) \sin x>0 \text {, }
$$

which means $g(x)$ is a monotonically increasing function on $\mathbf{R}$, with a unique zero at $x=\frac{\pi}{2}$.
Thus, $a_{3}=\frac{\pi}{2}$.
$$
\begin{array}{l}
\text { Therefore, } f^{2}\left(a_{3}\right)-a_{1} a_{2} \\
=\left(2 \times \frac{\pi}{5}-0\right)^{2}-\left(\frac{\pi}{2}-\frac{\pi}{4}\right)\left(\frac{\pi}{2}+\frac{\pi}{4}\right) \\
=\frac{13 \pi^{2}}{16} .
\end{array}
$$