Question

已知函数 f(x)=sinxcosx-√3sin^2x,①求 f(x)的最小正周期 ②求 f(x)在区间[0，π/2]上的最大值和最小值

详细步骤

Answer 1

f(x)=-√3sin^2x+sinxcosx

=-√3(1-cos2x)/2+sin2x/2

=sin2x/2+√3cos2x/2-√3/2

=sin(2x+π/3)-√3/2
（2）因为X∈[0,π/2],所以2x+π/3∈[π/3,4π/3] 所以sin(2x+π/3)∈[-√3/2,1]

所以f(x)∈[-√3,1-√3/2]

Answer 2

f(x)=1/2sin2x+√3[(1-cos2x)/2]=1/2sin2x+√3/2cos2x-√3/2=sin(2x+π/3)-√3/2
①∴T=π
②∵0≤x≤π/2

∴π/3≤2x+π/3)≤4π/3
∴-√3/2≤sin(2x+π/3)≤1
∴-√3≤f(x)≤1-√3/2