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

显示全部楼层 · 2009-8-15 21:40:53

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/2)(SinB/2·CosC/2+CosB/2·SiNC/2) =4Sin(B/2)Cos(B/2)(Cos(C/2))^2+4Sin(C/2)Cos(C/2)(Cos(B/2))^2 =SinB(CosC+1)+SinC(CosB+1) =Sin(B+C)+SinB...