因为cos(-3π/2)+cos(-π/2)+cosπ/2+cos3π/2+cos5π/2=0
2(-3π/2-π/2+π/2+3π/2+5π/2)=5π
所以a1=-3π/2
a2=-π/2
a3=π/2
a4=3π/2
a5=5π/2
设函数f(x)=2x-cosx,{An}是公差为π的等差数列,f(a1)+f(a2)+…f(a5)=5π,则[f(a)]^2-a1*a3=
设函数f(x)=2x-cosx,{An}是公差为π的等差数列,f(a1)+f(a2)+…f(a5)=5π,则[f(a)]^2-a1*a3=
数学人气：459 ℃时间：2020-01-26 02:08:23
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