Question

第六题怎么化简

Answer 1

sin²x-sin²(x-π/6)
=½[1-cos(2x)]-½[1-cos(2x-π/3)]
=½-½cos(2x)-½+½cos(2x-π/3)
=-½cos(2x)+½[cos(2x)cos(π/3)+sin(2x)sin(π/3)]
=-½cos(2x)+½[½cos(2x)+(√3/2)sin(2x)]
=-½cos(2x)+¼cos(2x)+(√3/4)sin(2x)
=½[(√3/2)sin(2x)-½cos(2x)]
=½[sin(2x)cos(π/6)-cos(2x)sin(π/6)]
=½sin(2x-π/6)

Answer 2

f(x)=sin²x-sin²(x-π/6)
= 1/2-1/2cos2x - 1/2 + 1/2cos(2x-π/3)
= -1/2cos2x +1/2(cos2xcosπ/3+sin2xsinπ/3)
= -1/2cos2x +1/2(cos2x*1/2+sin2x*√3/2)
= 1/2(√3/2sin2x-1/2cos2x)
= 1/2(sin2xcosπ/6-cos2xsinπ/6)
= 1/2sin(2x-π/6)