作业帮化简[sin(2A+B)]/sinA-2cos(A+B)
来源：学生作业帮助网 编辑：作业帮 时间：2024/04/29 05:50:50
![化简[sin(2A+B)]/sinA-2cos(A+B)](/uploads/image/z/1116593-17-3.jpg?t=%E5%8C%96%E7%AE%80%5Bsin%282A%2BB%29%5D%2FsinA-2cos%28A%2BB%29)
化简[sin(2A+B)]/sinA-2cos(A+B)
化简[sin(2A+B)]/sinA-2cos(A+B)
化简[sin(2A+B)]/sinA-2cos(A+B)
[sin(2A+B)]/sinA-2cos(A+B)
=[sin(A+B+A)]/sinA-2cos(A+B)
=[sin(A+B)cosA+cos(A+B)sinA]/sinA-2cos(A+B)
=[sin(A+B)cosA-cos(A+B)sinA]/sinA
=sin(A+B-A)/sinA
=sinB/sinA