Question

已知a(sinx,-cosx),b(cosx,根号3cosx)f(x)=ab+(根号3/2),求f(x)最小正周期,当0

Answer 1

f(x)=ab+√3/2

=sinxcosx-√3cos²x+√3/2

=1/2sin2x-√3/2(2cos²x-1)

=1/2sin2x-√3/2cos2x

=sin(2x-π/3)

f(x)最小正周期T=2π/2=π

∵0≤x≤π/2

∴0≤2x≤π

∴-π/3≤2x-π/3≤2π/3

∴-√3/2≤sin(2x-π/3)≤1

∴f(x)的值域为[-√3/2,1]