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设函数f(x)=2x-cosx,{an}是公差为8分之派的等差数列,f(a1)+f(a2)+……+f(a5)=5派,则f...设函数f(x)=2x-cosx,{an}是公差为8分之派的等差数列,f(a1)+f(a2)+……+f(a5)=5派,则
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设函数f(x)=2x-cosx,{an}是公差为8分之派的等差数列,f(a1)+f(a2)+……+f(a5)=5派,则【f...
设函数f(x)=2x-cosx,{an}是公差为8分之派的等差数列,f(a1)+f(a2)+……+f(a5)=5派,则【f(a3)】的平方-a1×a3等于多少?
设函数f(x)=2x-cosx,{an}是公差为8分之派的等差数列,f(a1)+f(a2)+……+f(a5)=5派,则【f(a3)】的平方-a1×a3等于多少?
▼优质解答
答案和解析
设a1=a
an=a+(n-1)*π/8
f(an)=2a+(n-1)*π/4-cos[a+(n-1)*π/8]
f(a1)+f(a2)+...+f(a5)=5(2a+2a+π)/2-cosa-cos(a+π/8)-cos(a+π/4)-cos(a+3π/8)-cos(a+π/2)
=10a+5π/2-[.]
=5π
因为π是无理数,所以a=π/4
a1=a=π/4
a3=a+2*π/8=π/2
[f(a3)]^2-a1*a3
=(π-0)^2-π/4*π/2
=(7π^2)/8
an=a+(n-1)*π/8
f(an)=2a+(n-1)*π/4-cos[a+(n-1)*π/8]
f(a1)+f(a2)+...+f(a5)=5(2a+2a+π)/2-cosa-cos(a+π/8)-cos(a+π/4)-cos(a+3π/8)-cos(a+π/2)
=10a+5π/2-[.]
=5π
因为π是无理数,所以a=π/4
a1=a=π/4
a3=a+2*π/8=π/2
[f(a3)]^2-a1*a3
=(π-0)^2-π/4*π/2
=(7π^2)/8
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