题目内容
A.sin2A B.cos2A C.sin2B D.cos2B
解析:原式=cos2A+cos(A+B)[cos(A+B)-2cosAcosB]
=cos2A+cos(A+B)[cosAcosB-sinAsinB-2cosAcosB]
=cos2A-cos(A+B)·cos(A-B)
=cos2A-cos2Acos2B+sin2Asin2B
=cos2A-[cos2A(1-sin2B)-(1-cos2A)sin2B]
=cos2A-[cos2A-sin2B]
=sin2B
答案:C