Question

fx=sin²x-sin²（x-π/6）化简

Answer 1

f(x)=1/2(1-cos2x)-1/2[1-cos(2x-π/3)]
=1/2-1/2cos2x-1/2+1/2cos(2x-π/3)
=-1/2cos2x+1/2cos2xcosπ/3+1/2sin2xsinπ/3
=-1/2cos2x+1/4cos2x+√3/4sin2x
=-1/4cos2x+√3/4sin2x
=1/2sin(2x-π/6)

Answer 2

引用fanglva的回答：
f(x)=1/2(1-cos2x)-1/2[1-cos(2x-π/3)]
=1/2-1/2cos2x-1/2+1/2cos(2x-π/3)
=-1/2cos2x+1/2cos2xcosπ/3+1/2sin2xsinπ/3
=-1/2cos2x+1/4cos2x+√3/4sin2x
=-1/4cos2x+√3/4sin2x
=1/2sin(2x-π/6)

=-1/2sin(2x-π/6)

Answer 3

sin^2x-sin^2(x-π/6)
=1/2sin(2x-π/6)