已知函数f(x)满足:f(1)=1/4,4f(x)f(y)=f(x+y)+f(x-y) (x ,y∈R),则f(2010)=?已知函数f(x)满足:f(1)=1/4,4f(x)f(y)=f(x+y)+f(x-y) (x ,y∈R),则f(2010)=?解法:f(1)=1/4,4f(x)f(y)=f(x+y)+f(x-y) (x ,y∈R),构造符

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已知函数f(x)满足:f(1)=1/4,4f(x)f(y)=f(x+y)+f(x-y) (x ,y∈R),则f(2010)=?已知函数f(x)满足:f(1)=1/4,4f(x)f(y)=f(x+y)+f(x-y) (x ,y∈R),则f(2010)=?解法:f(1)=1/4,4f(x)f(y)=f(x+y)+f(x-y) (x ,y∈R),构造符

已知函数f(x)满足:f(1)=1/4,4f(x)f(y)=f(x+y)+f(x-y) (x ,y∈R),则f(2010)=?已知函数f(x)满足:f(1)=1/4,4f(x)f(y)=f(x+y)+f(x-y) (x ,y∈R),则f(2010)=?解法:f(1)=1/4,4f(x)f(y)=f(x+y)+f(x-y) (x ,y∈R),构造符
已知函数f(x)满足:f(1)=1/4,4f(x)f(y)=f(x+y)+f(x-y) (x ,y∈R),则f(2010)=?
已知函数f(x)满足:f(1)=1/4,4f(x)f(y)=f(x+y)+f(x-y) (x ,y∈R),则f(2010)=?
解法:f(1)=1/4,4f(x)f(y)=f(x+y)+f(x-y) (x ,y∈R),构造符合题意的函数f(x)=1/2cos(派/3*x) 所以f(2010)=cos(派/3*2010)=1/2

已知函数f(x)满足:f(1)=1/4,4f(x)f(y)=f(x+y)+f(x-y) (x ,y∈R),则f(2010)=?已知函数f(x)满足:f(1)=1/4,4f(x)f(y)=f(x+y)+f(x-y) (x ,y∈R),则f(2010)=?解法:f(1)=1/4,4f(x)f(y)=f(x+y)+f(x-y) (x ,y∈R),构造符
根据cosxcosy=1/2[cos(x+y)+cos(x-y)]
2cosxcosy=cos(x+y)+cos(x-y)
令f(x)=AcosBx
4f(x)f(y)=f(x+y)(x-y)=>4AcosxAcosy=A[cos(Bx+By)+cos(Bx-By)]
4Acosxcosy=cos(Bx+By)+cos(Bx-By)
4A=2
A=1/2
即f(x)=1/2cosBx
f(1)=1/2cosBx=1/4
所以B可以是π/3
所以可令f(x)=1/2cos(πx/3)

构造特殊函数f(x)=(1/2)*cos(πx/3),当x=1时,符合题意,也可解决这个问题。(这个函数不是太容易想到)
填空题可以,其他题型用其他方法好一些

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