求证：sin2(a+β)-2sinβcosasin(a+β)=sin2a-sin2β．
【正确答案】：左端=sin(a+β)[sin(a+β)-2sinβcosa] =(sinacosβ+cosasinβ)(sinacosβ-cosasinfl) =sin2cos2β-cos2asin2β =sin2a(1-sin2β)-cos2asin2β =sin2a-sin2asin2β-cos2asin2β =sin2a-sin2β(sin2a+cos2a) =sin2a-sin2β =右端．
