sin(α+β)cosα-[sin(2α+β)+sinβ]/2
=sin(α+β)cosα-[sin(α+β+α)+sin（α+β-α）]/2
=sin(α+β)cosα-[sin（α+β）cosα+cos（α+β）sinα+sin（α+β）cosα-cos（α+β）sinα]/2
=sin(α+β)cosα-sin(α+β)cosα
=0