(b-c)*cos²A=b*cos²B-c*cos²C,判断三角形ABC的形状.(b-c)cos^2(π-(B+C))=bcos^2B-ccos^2C(b-c)cos^2(B+C)=bcos^2B-ccos^2C等式左边应该加上负号吧~
来源:学生作业帮助网 编辑:作业帮 时间:2024/06/23 19:11:29
![(b-c)*cos²A=b*cos²B-c*cos²C,判断三角形ABC的形状.(b-c)cos^2(π-(B+C))=bcos^2B-ccos^2C(b-c)cos^2(B+C)=bcos^2B-ccos^2C等式左边应该加上负号吧~](/uploads/image/z/5194249-25-9.jpg?t=%28b-c%29%2Acos%26sup2%3BA%3Db%2Acos%26sup2%3BB-c%2Acos%26sup2%3BC%2C%E5%88%A4%E6%96%AD%E4%B8%89%E8%A7%92%E5%BD%A2ABC%E7%9A%84%E5%BD%A2%E7%8A%B6.%28b-c%29cos%5E2%28%CF%80-%28B%2BC%29%29%3Dbcos%5E2B-ccos%5E2C%28b-c%29cos%5E2%28B%2BC%29%3Dbcos%5E2B-ccos%5E2C%E7%AD%89%E5%BC%8F%E5%B7%A6%E8%BE%B9%E5%BA%94%E8%AF%A5%E5%8A%A0%E4%B8%8A%E8%B4%9F%E5%8F%B7%E5%90%A7%7E)
(b-c)*cos²A=b*cos²B-c*cos²C,判断三角形ABC的形状.(b-c)cos^2(π-(B+C))=bcos^2B-ccos^2C(b-c)cos^2(B+C)=bcos^2B-ccos^2C等式左边应该加上负号吧~
(b-c)*cos²A=b*cos²B-c*cos²C,判断三角形ABC的形状.
(b-c)cos^2(π-(B+C))=bcos^2B-ccos^2C
(b-c)cos^2(B+C)=bcos^2B-ccos^2C
等式左边应该加上负号吧~
(b-c)*cos²A=b*cos²B-c*cos²C,判断三角形ABC的形状.(b-c)cos^2(π-(B+C))=bcos^2B-ccos^2C(b-c)cos^2(B+C)=bcos^2B-ccos^2C等式左边应该加上负号吧~
(b-c)cos^2A=bcos^2B-ccos^2C
(b-c)cos^2(π-(B+C))=bcos^2B-ccos^2C
(b-c)cos^2(B+C)=bcos^2B-ccos^2C
b(cos^2(B+C)-cos^2B)=c(cos^2(B+C)-cos^C)
cos^2(B+C)-cos^2B
=(cos(B+C)+cosB)(cos(B+C)-cosB)
=2cos(2B+C)/2cosC/2*(-sin(2B+C)/2sinC/2)
=-[2sin(2B+C)/2cos(2B+C)/2]*[2sinC/2cosC/2]
=-sin(2B+C)sinC
同样
cos^2(B+C)-cos^C=-sin(B+2C)sinB
所以
bsin(2B+C)sinC=csin(B+2C)sinB
b/sinB*sin(2B+C)=c/sinC*sin(B+2C)
因为:b/sinB=c/sinC
所以,sin(2B+C)=sin(B+2C)
2B+C=B+2C,或,2B+C+(B+2C)=π
B=C,或,B+C=π/3
所以,△abc是等腰三角形,或,A=120度的三角形