若f(x)=sin(∏x/3),则f(1)+f(2)+f(3)+...+f(2003)=
来源:百度知道 编辑:UC知道 时间:2024/09/20 02:31:23
当x=6k-5,(k∈N※,下同).
则f(6k-5)
=sin(2kπ-5π/3)
=-sin(5π/3)
=√3/2.
f(6k-4)
=sin(2kπ-4π/3)
=√3/2
f(6k-3)
=sin(2kπ-3π/3)
=0
f(6k-2)
=sin(2kπ-2π/3)
=-√3/2
f(6k-1)
=sin(2kπ-π/3)
=-√3/2
f(6k)
=sin2kπ
=0.
∴f(6k-5)+f(6k-4)+f(6k-3)+f(6k-2)+f(6k-1)+f(6k)=0.
(可见,用x除以6的余数分类,1至2004的整数可分成344组,每组的函数值的代数和为0).
[f(1)+f(2)+f(3)+…+f(6)]+[f(7)+…f(12)]+...+[f(1999)+…f(2003)+f(2004)]-f(2004)(共344个方括号)
=0×344-0
=0.
若f(x)满足2f(-sin x)+3f(sin x)=4sin xcos x(x大于等于负的二分之派,小于等于二分之派).求f(x).
F(X)=6SIN
f(x)=sin|k*pi*x|
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