设函数f(x)=2x-cos4x,{An}是公差为π/8的等差数列f(a1)+f(a2)+…f(a8）=11π,则f[(a2)]^2-a1a5=

Question

设函数f(x)=2x-cos4x,{An}是公差为π/8的等差数列f(a1)+f(a2)+…f(a8）=11π,则f[(a2)]^2-a1a5=

Answer 1

f[(a2)]^2应该是[f(a2)]^2吧,否则算不出来结果.
数列{an}是公差为π/8的等差数列,
且f(a1)+f(a2)+……+f(a8)=11π
2a1-cos4a1+2a2-cos4a2+2a3-cos4a3+2a4-cos4a4+2a5cos4a5+.2a8cos4a8=11π
∴2(a1+a2+……+a8)-(cos4a1+cos4a2+……+cos4a8)=11π
∴(cos4a1+cosa2+……+cos4a8)=0
即2(a1+a2+……+a8)=2×4(a1+a8)=11π,
8(a1+a8)=11π,a1+a8=11π/8
2a1+7π/8=11π/8,a1=π/4,a2=3π/8,a5=3π/4
∴[f(a2)]^2-a1a5
=(2a2-cos4a2)^2-a1a5
=(2*3π/8-cos3π/2)^2-π/4*3π/4
=9π^2/16-3π^2/16
=6π^2/16
=3π^2/8