作业帮求证：1/2(cos2B-cos2A)=sin(A+B)sin(A-B)(一定要从1/2(cos2B-cos2A）推到sin(A+B)sin(A-B)不可以像这样：/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-(
来源：学生作业帮助网 编辑：作业帮 时间：2024/04/29 18:13:26
![求证：1/2(cos2B-cos2A)=sin(A+B)sin(A-B)(一定要从1/2(cos2B-cos2A）推到sin(A+B)sin(A-B)不可以像这样：/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-(](/uploads/image/z/2510266-58-6.jpg?t=%E6%B1%82%E8%AF%81%EF%BC%9A1%2F2%28cos2B-cos2A%29%3Dsin%28A%2BB%29sin%28A-B%29%28%E4%B8%80%E5%AE%9A%E8%A6%81%E4%BB%8E1%2F2%28cos2B-cos2A%EF%BC%89%E6%8E%A8%E5%88%B0sin%28A%2BB%29sin%28A-B%29%E4%B8%8D%E5%8F%AF%E4%BB%A5%E5%83%8F%E8%BF%99%E6%A0%B7%EF%BC%9A%2F2%EF%BC%88cos2B-cos2A%29%3D1%2F2%5B2%28cosB%29%5E2-1-2%28cosA%29%5E2%2B1%5D%3D%28cosB%29%5E2-%28cosA%29%5E2sin%28A%2BB%29sin%28A-B%29%3D%28sinAcosB%2BcosAsinB%29%28sinAcosB-cosAsinB%29%3D%28sinAcosB%29%5E2-%28)
求证：1/2(cos2B-cos2A)=sin(A+B)sin(A-B)(一定要从1/2(cos2B-cos2A）推到sin(A+B)sin(A-B)不可以像这样：/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-(
求证：1/2(cos2B-cos2A)=sin(A+B)sin(A-B)
(一定要从1/2(cos2B-cos2A）推到sin(A+B)sin(A-B)
不可以像这样：
/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) ）
求证：1/2(cos2B-cos2A)=sin(A+B)sin(A-B)(一定要从1/2(cos2B-cos2A）推到sin(A+B)sin(A-B)不可以像这样：/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-(
左边=1/2{cos[(A+B)-(A-B)]-cos[(A+B)+(A-B)]}
=1/2{[cos(A+B)cos(A-B)+sin(A+B)sin(A-B)]-[cos(A+B)cos(A-B)-sin(A+B)sin(A-B)]}
=1/2[cos(A+B)cos(A-B)+sin(A+B)sin(A-B)-cos(A+B)cos(A-B)+sin(A+B)sin(A-B)]
=1/2[2sin(A+B)sin(A-B)]
=sin(A+B)sin(A-B)