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 $f\left(a_{1}\right)+f\left(a_{2}\right)+\cdots+f\left(a_{5}\right)=5 \pi$, then $\left[f\left(a_{3}\right)\right]^{2}-a_{1} a_{5}=$
A. 0
B. $\frac{\pi^{2}}{16}$
C. $\frac{\pi^{2}}{8}$
D. $\frac{13 \pi^{2}}{16}$
Solve:
$2\left(a_{1}+a_{2}+\cdots+a_{5}\right)-\left(\cos a_{1}+\cos a_{2}+\cdots+\cos a_{5}\right)=5 \pi$, when $a_{1}+a_{2}+\cdots+a_{5}=\frac{5 \pi}{2}$, $\left(a_{1}, a_{2}, a_{3}, a_{4}, a_{5}\right)=\left(\frac{\pi}{4}, \frac{3 \pi}{8}, \frac{\pi}{2}, \frac{5 \pi}{8}, \frac{3 \pi}{4}\right) \Rightarrow \cos a_{1}+\cos a_{2}+\cdots+\cos a_{5}=0$ satisfies the condition; on the other hand, $f^{\prime}(x)=2+\sin x>0 \Rightarrow f(x)$ is monotonically increasing, i.e., $f\left(a_{1}\right)+f\left(a_{2}\right)+\cdots+f\left(a_{5}\right)=5 \pi$ has a unique solution $\left(a_{1}, a_{2}, a_{3}, a_{4}, a_{5}\right)=\left(\frac{\pi}{4}, \frac{3 \pi}{8}, \frac{\pi}{2}, \frac{5 \pi}{8}, \frac{3 \pi}{4}\right)$. Therefore, $\left[f\left(a_{3}\right)\right]^{2}-a_{1} a_{5}=\pi^{2}-\frac{3 \pi^{2}}{16}=\frac{13 \pi^{2}}{16}$, so the answer is $D$.