Question

化简：sin(α+β)cosα-1/2[sin(2α+β)-sinβ]

Answer 1

原式=（sinαcosβ+cosαsinβ）cosα-1/2[sin2αcosβ+cos2αsinβ-sinβ]
=sinαcosβcosα+cos²αsinβ-1/2[sin2αcosβ+(cos2α-1)sinβ]
=1/2sin2αcosβ+cos²αsinβ-1/2sin2αcosβ-1/2(1-2sin²α-1)sinβ
=cos²αsinβ-1/2(-2sin²α)sinβ
=cos²αsinβ+sin²αsinβ
=(cos²α+sin²α)sinβ
=sinβ

Answer 2

公式：sin(α+β)=sin α×cosβ+sinβ×cos α;sin(α-β)=sin α×cosβ-sinβ×cos α
sin(α+β）cosα-1/2[sin（2α+β）-sinβ]
=sin(α+β）cosα-1/2[sin(α+β）cosα+cos(α+β)sinα-sinβ]
=1/2[sin(α+β）cosα-cos(α+β)sinα+sinβ]
=1/2[sinβ+sinβ]
=sinβ