(cos2B-cos2A)/2 怎样化成 sin(A+B)sin(A-B)要详细点!逆运算：相等1/2（cos2B-cos2A)=1/2[2(cosB)^2-1-2(cosA)^2+1]=(cosB)^2-(cosA)^2sin(A+B)sin(A-B)=(sinAcosB+cosAsinB)(sinAcosB-cosAsinB)=(sinAcosB)^2-(cosAsinB)^2=(cosB)^2[1-(cosA)^2]-(cosA)^2[1-(cosB)^2]=(cosB)^2-(cosBcosA)^2-(cosA)^2+(cosAcosB)^2=(cosB)^2-(cosA)^2综上,化简后 1/2（cos2B-cos2A)=sin(A+B)sin(A-B) 我想知道的是若我们一开始不知道会与sin(A+B)sin(A-B)相等,要怎样划出来?
问题描述：
(cos2B-cos2A)/2 怎样化成 sin(A+B)sin(A-B)要详细点!
逆运算：相等
1/2（cos2B-cos2A)=1/2[2(cosB)^2-1-2(cosA)^2+1]=(cosB)^2-(cosA)^2
sin(A+B)sin(A-B)=(sinAcosB+cosAsinB)(sinAcosB-cosAsinB)
=(sinAcosB)^2-(cosAsinB)^2=(cosB)^2[1-(cosA)^2]-(cosA)^2[1-(cosB)^2]
=(cosB)^2-(cosBcosA)^2-(cosA)^2+(cosAcosB)^2
=(cosB)^2-(cosA)^2
综上,化简后 1/2（cos2B-cos2A)=sin(A+B)sin(A-B)
我想知道的是若我们一开始不知道会与sin(A+B)sin(A-B)相等,要怎样划出来?
答
(cos2b-cos2a)/2
=cos2b/2-cos2a/2
=1/2-sin"b-1/2+sin"a
=sin"a-sin"b
=sin(a+b)sin(a-b)