在三角形ABC中求证sinA+sinB+sinC=4cos(A/2)COS(B/2)COS(C/2)证到这步然后怎么证:原式=4sin(B/2)COS(B/2)COS^2(C/2)+4sin(C/2)COS(C/2)COS^2(B/2)=? 从这步开始, 复制的别来步骤不要跳!!!

问题描述:

在三角形ABC中求证sinA+sinB+sinC=4cos(A/2)COS(B/2)COS(C/2)证到这步然后怎么证:
原式=4sin(B/2)COS(B/2)COS^2(C/2)+4sin(C/2)COS(C/2)COS^2(B/2)=? 从这步开始, 复制的别来
步骤不要跳!!!

4cos(A/2)COS(B/2)COS(C/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+SinC
∵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)
sinA+sinB+sinC=Sin(B+C)+SinB+SinC

2sin[(A+B)/2]cos[(A-B)/2]+2sin(C/2)cos(C/2)=2sin(π/2-C/2)cos[(A-B)/2]+2sin(C/2)cos(C/2)=2cos(C/2){cos[(A-B)/2]+sin(C/2)}=2cos(C/2){cos[(A-B)/2]+cos[(A+B)/2]}=4cos(C/2)cos(A/2)cos(B/2)

sinA+sinB+sinC=2sin[(A+B)/2]cos[(A-B)/2]+2sin(C/2)cos(C/2)=2sin(π/2-C/2)cos[(A-B)/2]+2sin(C/2)cos(C/2)=2cos(C/2){cos[(A-B)/2]+sin(C/2)}=2cos(C/2){cos[(A-B)/2]+cos[(A+B)/2]}=4cos(C/2)cos(...