如何证明在三角形中sinA+sinB+sinC=4coc(A/2)*cos(B/2)*cos(C/2)

问题描述:

如何证明在三角形中sinA+sinB+sinC=4coc(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/...