Question

在线急求，高中数学问题，望高手解答，需过程，急急急急急

已知R是实数集,x属于R,平面向量a=(1,sinx的平方-cosx的平方),平面向量b=(cos(2x-派/3),1),函数f(x)=向量a*向量b.求f(x)的最小正周期.求设函数F(x)=(f(x))的平方+f(x),求F(x)的值域

Answer 1

f(x)=[sin(x)]^2-[cos(x)]^2+cos(2x-pi/3)
=-cos(2x)+cos(2x-pi/3)
=sin(2x-pi/6)
f(x)的最小正周期T为pi
F(x)=[f(x)]^2+f(x) 而-1<=f(x)<=1
F(x)=[f(x)+1/2]^2-1/4
-1/2时最小 1时最大
所以-1/4<=F(x)<=1

Answer 2

f(x)=cos(2x-π/3)+[sin(x)]^2-[cos(x)]^2=cos(2x-π/3）-cos(2x)=√3/2sin2x+1/2cos2x-cos2x=
√3/2sin2x-1/2cos2x=sin(2x-π/6)
周π
令f(x)=t∈〔-1,1〕
f(x)=(t+1/2)^2-1/4
-1/2时最小1时大
值〔-1/4,2〕

Answer 3

f(x)=[sin(x)]^2-[cos(x)]^2+cos(2x-pi/3)
=-cos(2x)+cos(2x-pi/3)
=sin(2x-pi/6)
f(x)的最小正周期为pi
F(x)=[f(x)]^2+f(x) 而-1<=f(x)<=1
所以-1/4<=F(x)<=1

Answer 4

1、a=(1,sinx的平方-cosx的平方） b=(cos(2x-π/3,1) =>f(x)=1xcos(2x-π/3)+1x(sinx的平方-cosx的平方）=cos(2x-π/3)-cos2x=sinπ/3sin2x-1/2cos2x=sin(2x-π/6)=>f(x)的最小正周期为2π/w=2π/2=π