Question

已知向量a=(sinx,根号3cosx),向量b=(cosx,cosx),f(x)=a*b,求f(x)的周期、值域及单调

Answer 1

f(x)=sinxcosx+√3cos²x

=(1/2)sin2x+(√3/2)(cos2x+1)

=(1/2)sin2x+(√3/2)cos2x+√3/2

=sin(2x+π/3)+√3/2

周期 T=2π/2=π,

值域[-1+√3/2,1+√3/2]

令 -π/2+2kπ≤2x+π/3≤π/2+2kπ,

解得 -5π/12+kπ≤x≤π/12+kπ,

即增区间为[-5π/12+kπ,π/12+kπ],k∈Z

同理,得减区间为[π/12+kπ,7π/12+kπ],k∈Z