11.在△ABC中,求证sinA+sinB+sinC=4cos(A/2)cos(B/2)cos(C/2)
来源:学生作业帮助网 编辑:作业帮 时间:2024/07/04 03:28:22
![11.在△ABC中,求证sinA+sinB+sinC=4cos(A/2)cos(B/2)cos(C/2)](/uploads/image/z/3286642-58-2.jpg?t=11.%E5%9C%A8%E2%96%B3ABC%E4%B8%AD%2C%E6%B1%82%E8%AF%81sinA%2BsinB%2BsinC%3D4cos%28A%2F2%29cos%28B%2F2%29cos%28C%2F2%29)
11.在△ABC中,求证sinA+sinB+sinC=4cos(A/2)cos(B/2)cos(C/2)
11.在△ABC中,求证sinA+sinB+sinC=4cos(A/2)cos(B/2)cos(C/2)
11.在△ABC中,求证sinA+sinB+sinC=4cos(A/2)cos(B/2)cos(C/2)
证明:
∵在三角形ABC中,
∴A+B+C=180度,得SINA=SIN(B+C)
则A/2=90度-(B+C)/2,得COSA/2=SIN((B+C)/2)
左边=Sin(B+C)+SinB+SinC
则4Cos(A/2)Cos(B/2)Cos(C/2)
=4Sin((B+C)/2)Cos(B/2)Cos(C/2)
=4Cos(B/2)Cos(C/2)(SinB/2·CosC/2+CosB/2·SiNC/2)
=4Sin(B/2)Cos(B/2)(Cos(C/2))^2+4Sin(C/2)Cos(C/2)(Cos(B/2))^2
=SinB(CosC+1)+SinC(CosB+1)
=Sin(B+C)+SinB+SinC
左边=右边
原式成立!