Question

已知向量a=(sinx,-1),向量b=(√3cosx ,-½),函数f(x)=(a+b)a-2 (

已知向量a=(sinx,-1),向量b=(√3cosx ,-½),函数f(x)=(a+b)a-2 (1)求函数的最小正周期9
最小正周期T

Answer 1

解：据题，得：
f(x)=(sinx，-1)▪(sinx+√3cosx，-3/2)-2
=sinx(sinx+√3cosx)+3/2-2
=sin²x+√3sinxcosx-½
=1-cos²x+(√3/2)sin2x-½
=-½(1+cos2x)+(√3/2)sin2x+½
=-½-½cos2x+(√3/2)sin2x+½
=(√3/2)sin2x-½cos2x
=sin(2x-π/6)，
∴函数f(x)的最小正周期为：
T=2π/2=π.

Answer 2

解：∵ a=(sinx,-1),b=(√3cosx,-1/2)
∴a+b=(sinx+√3cosx,-3/2)
∴(a+b)a=sin^2x+√3sinxcosx-1/2
∴f(x))=(a+b)a-2=sin(2x-π/6)
周期T=2π/ω=2π/2=π.

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Answer 3

a+b＝sinx+√3cosx -3/2
(a+b)a=(sinx+√3cosx )sinx+0
f(x)=(a+b)a-2 =(sinx+√3cosx )sinx-2
=sin^2x+√3cosx sinx-2
=1/2(1-cos2x )+√3/2 sin2x-2
=√3/2 sin2x-1/2cos2x -3/2
=sin(2x-π/6)-3/2
T=2π/2=π