已知向量m=(sin^2x+(1+cos2x)/2,sinx),n=(1/2cos2x-根号3/2sin2x,2sinx) 函数f(x)=m·n1.求函数f(x)的最小正周期2.若x∈[0,π/2],求函数f(x)的值域
问题描述:
已知向量m=(sin^2x+(1+cos2x)/2,sinx),n=(1/2cos2x-根号3/2sin2x,2sinx) 函数f(x)=m·n
1.求函数f(x)的最小正周期
2.若x∈[0,π/2],求函数f(x)的值域
答
m=(sin²x+cos²x,sinx)=(1,sinx)
(1/2)cos2x-(√3/2)sin2x=cos(2x+π/3),则:
n=(cos(2x+π/3),2sinx)
得:
f(x)=m*n
.=cos(2x+π/3)+2sin²x
.=cos(2x+π/3)-cos2x+1
.=-(1/2)cos2x-(√3/2)sin2x+1
.=-cos(2x-π/3)+1
.=cos(2x+2π/3)+1
最小正周期是2π/2=π;
若x∈[0,π/2],则:2x+2π/3∈[2π/3,5π/2],则:
cos(2x+2π/3)∈[-1,1/2]
则:f(x)∈[0,3/2]