Question

已知函数F(x)=(sinx-cosx)sinx,x∈R,求F(x)的最小正周期

已知函数F(x)=(sinx-cosx)sinx,x∈R,求F(x)的最小正周期。（麻烦写出详细的解答步骤;用笔书写,然后再用摄像头\手机\照相机拍成照片传上来也可以.)

Answer 1

(sinx-cosx)sinx
=(sinx)^2-sinxcosx
=[-1+2(sinx)^2]/2+1/2-(sin2x)/2
=(-1/2)(cos2x+sin2x)+1/2
=(-√2/2)sin(2x+φ)+1/2
=(-√2/2)sin[2(x+φ/2)]+1/2
∴最小正周期是(2π)/2=π

Answer 2

f（x）=(sinx)^2-sinx*cosx
=(1-cos2x)/2-sin2x/2
=1/2-(sin2x+cos2x)/2
=1/2-根号2sin(2x+π/4)/2
所以T=2π/2=π

Answer 3

sinx-cosx)sinx
=(sinx)^2-sinxcosx
=[-1+2(sinx)^2]/2+1/2-(sin2x)/2
=(-1/2)(cos2x+sin2x)+1/2
=(-√2/2)sin(2x+φ)+1/2
=(-√2/2)sin[2(x+φ/2)]+1/2