在△ABC中,求证sin(A+B)/(sinA+sinB)+sin(B+C)/(sinB+sinC)+sin(C+A)/(sinC_数学解答

在△ABC中,求证sin(A+B)/(sinA+sinB)+sin(B+C)/(sinB+sinC)+sin(C+A)/(sinC+sinA)>=3/2

人气：268 ℃　时间：2019-11-04 23:01:58

解答

证明：先用正弦定理.将角度化成边：sin(A+B)/(sinA+sinB)+sin(B+C)/(sinB+sinC)+sin(C+A)/(sinC+sinA)=sinc/(sinA+sinB)+sina/(sinB+sinC)+sinb/(sinC+sinA)=1/(sina/sinc)+(sinb/sinc)+1/(sinb/sina)+(sinc/sina)+1...