sin2A-sin2B=2sin(A-B)cos(A+B)是怎么出来的

网友回答

2A=(A＋B)＋(A－B),则：
sin2A=sin[(A＋B)＋(A－B)]=sin(A＋B)cos(A－B)＋cos(A＋B)sin(A－B) ------------(1)
2B=(A＋B)－(A－B),则：
sin2B=sin[(A＋B)－(A－B)]=sin(A＋B)cos(A－B)－cos(A＋B)sin(A－B) ------------(2)
(1)＋(2),得：
sin2A＋sin2B=2sin(A＋B)cos(A－B)
======以下答案可供参考======
供参考答案1:
sin2A-sin2B=2sinAcosA-2sinBcosB=2(sinAcosA-sinBcosB)
=2[sinAcosA(sin^2B+cos^2B)-sinBcosB(sin^2A+cos^2A)]
=2[(sinAcosB-sinBcosA)cosAcosB-(sinAcosB-sinBcosA)sinAsinB]
=2(sinAcosB-sinBcosA)(cosAcosB-sinAsinB)
=2sin(A-B)cos(A+B)
供参考答案2:
=sinAcosA+cosA*sinA-（sinB*cosB+cosB*sinB）
=2sinAcosA-2sinB*cosB
=2(sinAcosA-sinBcosB)
=2sin(A-B)cos(A+B)