已知正弦定理a/sinA = b/sinB = c/sinC = 2R
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已知正弦定理a/sinA = b/sinB = c/sinC = 2R
证明 (a+b+c)/(sinA+sinB+sinC)=(a-b-c)/(sinA-sinB-sinC)=a/sinA = b/sinB = c/sinC = 2R
就是推出它们全部相等.
证明 (a+b+c)/(sinA+sinB+sinC)=(a-b-c)/(sinA-sinB-sinC)=a/sinA = b/sinB = c/sinC = 2R
就是推出它们全部相等.
证明,已知
a/sinA = b/sinB = c/sinC = 2R(1)
a=2RsinA, b=2RsinB,c=2RsinC
(a+b+c)/(sinA+sinB+sinC)=2R(sinA+sinB+sinC)/(sinA+sinB+sinC)=2R(2)
(a-b-c)/(sinA-sinB-sinC)=2R(sinA-sinB-sinC)/(sinA-sinB-sinC)=2R(3)
前2个代入后提取2R就出来了,后面3个是正弦定理已知的
所以由(1)(2)(3)得到
(a+b+c)/(sinA+sinB+sinC)=(a-b-c)/(sinA-sinB-sinC)=a/sinA = b/sinB = c/sinC = 2R
a/sinA = b/sinB = c/sinC = 2R(1)
a=2RsinA, b=2RsinB,c=2RsinC
(a+b+c)/(sinA+sinB+sinC)=2R(sinA+sinB+sinC)/(sinA+sinB+sinC)=2R(2)
(a-b-c)/(sinA-sinB-sinC)=2R(sinA-sinB-sinC)/(sinA-sinB-sinC)=2R(3)
前2个代入后提取2R就出来了,后面3个是正弦定理已知的
所以由(1)(2)(3)得到
(a+b+c)/(sinA+sinB+sinC)=(a-b-c)/(sinA-sinB-sinC)=a/sinA = b/sinB = c/sinC = 2R
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