已知向量m=(sin,1),n=(根号3Acosx,2分之Acos2x)函数f(x)=m·n的最大值为6.求A

问题描述:

已知向量m=(sin,1),n=(根号3Acosx,2分之Acos2x)函数f(x)=m·n的最大值为6.求A

f(x)=m.n=sinx√3Acosx+ (A/2)co2x
= A[√3/2sin2x+(1/2)cos2x]
=Asin(2x+π/6)
∴f(x)max=A=6.
∴A=6.

fx=m.n=sinx√3Acosx+ (A/2)co2x=
f(x)= A[√3/2sin2x+(1/2)cos2x]
∵f(x)max=6.
∴f(x)max=Asin(2x+π/6)=6.即sin(2x+π/6)=1.
∴A=6.