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:
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!!!
答
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(...