由正弦定理有:a=2RsinA,b=2RsinB,c=2RsinC
所以:(a^2-b^2)/c^2=((sinA)^2-(sinB)^2)/(sinC)^2
=(sinA-sinB)(sinA+sinB)/(sinC)^2
你的题目抄错了,因为sinC=sin(A+B)=sinAcosB+sinBcosA≤sinA+sinB(仅当A=B=0取等号)
因此(sinA+sinB)/sinC>1
所以(sinA-sinB)(sinA+sinB)/(sinC)^2≠(sinA-sinB)/sinC
由正弦定理有:a=2RsinA,b=2RsinB,c=2RsinC
所以:(a^2-b^2)/c^2=((sinA)^2-(sinB)^2)/(sinC)^2
=(sinA-sinB)(sinA+sinB)/(sinC)^2
你的题目抄错了,因为sinC=sin(A+B)=sinAcosB+sinBcosA≤sinA+sinB(仅当A=B=0取等号)
因此(sinA+sinB)/sinC>1
所以(sinA-sinB)(sinA+sinB)/(sinC)^2≠(sinA-sinB)/sinC