SAT1000 - P964

$\begin{aligned} \sum_{k=1}^5 f(a_k) & = 5 \pi \\ \sum_{k=1}^5 \left(2 a_k - \cos a_k \right) & = 5 \pi \\ \sum_{k=1}^5 2 \left(a_1 + \frac \pi 8 (k-1)\right) - \sum_{k=1}^5 \cos a_k & = 5 \pi \\ 10a_1 + \frac \pi 4 \sum_{k=1}^4 k - \sum_{k=1}^5 \cos a_k & = 5 \pi \\ 10a_1 + \frac {5\pi}2 - \sum_{k=1}^5 \cos a_k & = 5 \pi \end{aligned}$

It is obvious that $\displaystyle \sum_{k=1}^5 \cos a_k$ cannot be a multiple of $\pi$ , therefore $\displaystyle \sum_{k=1}^5 \cos a_k = 0$ . Then $10a_1 + \dfrac {5\pi}2 = 5 \pi \implies a_1 = \dfrac \pi 4$ . Then

$\begin{aligned} \sum_{k=1}^5 \cos a_k & = \cos a_1 + \cos \left(a_1 + \frac \pi 8 \right) + \cos \left(a_1 + \frac \pi 4 \right) + \cos \left(a_1 + \frac {3\pi}8 \right) + \cos \left(a_1 + \frac \pi 2 \right) \\ & = \cos \frac \pi 4 + \cos \frac {3\pi}8 + \cos \frac \pi 2 + \cos \frac {5\pi}8 + \cos \frac {3\pi}4 \\ & = \cos \frac \pi 4 + \cos \frac {3\pi}8 + 0 - \cos \frac {3\pi}8 - \cos \frac \pi 4 \\ & = 0 \end{aligned}$

Therefore $a_1 = \dfrac \pi 4$ and

$\begin{aligned} A & = f^2 (a_3) -a_1 a_5 \\ & = f\left( \frac \pi 2\right)^2 - \frac \pi 4 \cdot \frac {3\pi}4 \\ & = \pi^2 - \frac {3\pi^2}{16} \approx 8.019053576 \\ \implies \lfloor 1000 A \rfloor & = \boxed{8019} \end{aligned}$