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
来源:学生作业帮 编辑:神马作文网作业帮 分类:数学作业 时间:2024/09/25 16:37:32
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)=?
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
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)
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)
已知函数f(x)满足:f(1)=14,4f(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
已知函数f (x )满足f(1)=1/4,4f(x)f(y)=f(x+y)+f(x-y)(x,y属于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属于R),求f(2010)
已知函数f(x)满足f(1)=1/4,f(x)+f(y)=4f(x+y/2)*f(x-y/2)则f(-2011)=?
已知函数f(x)满足:f(1)=1/4,4f(x)f(y)=f(x+y)+f(x-y)(x,y属于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属于R),则f(2010
已知函数f(x)满足f(x+y)+f(x-y)=2f(x)f(y)(x∈R,y∈R),且f(0)≠1.
已知函数f(x)满足f(2)=1/2,2f(x)f(y)=f(x+y)+f(x-y),则f(2012)=?
函数的性质及应用.已知函数f(x)满足f(2)=1//2 ,2f(x)f(y)=f(x+y)+f(x-y)(x,y∈R)
已知函数y=f(x)满足f(x)=2f(1x
已知函数f(x)满足:f(1)=1/4,4f(x)f(y)=f(x+y)+f(x-y).则f(2010)=?(x,y属于