Question

已知函数f(x)=cosx·cos(x-θ)-1/2cosθ,其中x∈R,0<θ<π,已知当x=π/3时,f（x）取得最大值.

⑴求θ的值
⑵设g(x)=2f(3/2x),求函数g(x)在[0，π/3]上的最小值

Answer 1

(1)
cos(2x-θ)=cos[x+(x-θ)]=cosxcos(x-θ)-sinxsin(x-θ)
cosθ = cos[x-(x-θ)]=cosxcos(x-θ)+sinxsin(x-θ)
两式相加得：cos(2x-θ)+cosθ=2cosxcos(x-θ)
cosxcos(x-θ)=(1/2)[cos(2x-θ)+cosθ]
f(x)=(1/2)[cos(2x-θ)+cosθ] - (1/2)cosθ=(1/2)cos(2x-θ)
f(x)(MAX)=1/2
f(π/3)=(1/2)[cos(2π/3-θ)=1/2
所以θ=2π/3
(2)
g(x)=2*[(1/2)cos(3x-2π/3)]=cos(3x-2π/3)
∵0≤x≤π/3
∴-2π/3≤3x-2π/3≤π/3
∴g(x)的最小值为cos(-2π/3) = - 1/2