求详细过程,非常感谢!
来源:学生作业帮 编辑:百度作业网作业帮 分类:数学作业 时间:2024/07/02 15:09:55
![](http://img.wesiedu.com/upload/5/f4/5f401cc5acbb354e0c651925f7576db9.jpg)
![求详细过程,非常感谢!](/uploads/image/z/5093505-9-5.jpg?t=%E6%B1%82%E8%AF%A6%E7%BB%86%E8%BF%87%E7%A8%8B%EF%BC%8C%E9%9D%9E%E5%B8%B8%E6%84%9F%E8%B0%A2%EF%BC%81)
解题思路: 4cos(A/2)cos(B/2)cos(C/2)-sinA-sinB=sinA(2cos(B/2)^2-1)+sinB(2cos(A/2)^2-1)=sinAcosB+sinBcosA=sin(A+B)=sin(pi-A-B)=sin(A+B)==>sinA+sinB+sinC=4cos(A/2)cos(B/2)cos(C/
解题过程:
4cos(A/2)cos(B/2)cos(C/2)-sinA-sinB=sinA(2cos(B/2)^2-1)+sinB(2cos(A/2)^2-1)
=sinAcosB+sinBcosA=sin(A+B)=sin(pi-A-B)=sin(A+B)
==>
sinA+sinB+sinC=4cos(A/2)cos(B/2)cos(C/
解题过程:
4cos(A/2)cos(B/2)cos(C/2)-sinA-sinB=sinA(2cos(B/2)^2-1)+sinB(2cos(A/2)^2-1)
=sinAcosB+sinBcosA=sin(A+B)=sin(pi-A-B)=sin(A+B)
==>
sinA+sinB+sinC=4cos(A/2)cos(B/2)cos(C/