题目应该要求了ABC为锐角的吧?
否则ABC同时加上720°依然满足等式但A+B+C就变了.
对所给等式进行恒等变形：
cosA+cosB+cosC=1+4sinA/2 * sinB/2 * sinC/2
2cos(A/2+B/2)cos(A/2-B/2)+1-2(sinC/2)^2=1+4sinA/2*sinB/2*sinC/2
cos(A/2+B/2)cos(A/2-B/2)-(sinC/2)^2=2sinA/2*sinB/2*sinC/2
cos(A/2+B/2)cos(A/2-B/2)-(sinC/2)^2=sinC/2*[cos(B/2-A/2)-cos(B/2+A/2)]
(sinC/2)^2+sinC/2*cos(B/2-A/2)-sinC/2*cos(B/2+A/2)-cos(A/2+B/2)cos(A/2-B/2)=0
[sinC/2+cos(B/2-A/2)][sinC/2-cos(B/2+A/2)]=0
前式在我附加的ABC为锐角的情况下显然是不能为0的.
(当然可能是别的条件,总之应该可以说明前面这个不为0)
故只能后式为0
sinC/2=cos(B/2+A/2)
C/2+B/2+A/2=90°
A+B+C=180 °
证毕.