Question

cosA+cosB+cosC=1+4sinA/2 * sinB/2 * sinC/2求证A+B+C=180

cosA+cosB+cosC=1+4sinA/2 * sinB/2 * sinC/2求证A+B+C=180

Answer 1

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。

Answer 2

cosA+cosB=1-cosC+4sinA/2sinB/2sinC/2
2cos(A+B/2)cos(A-B/2)=2(sinC/2)^2+4sinA/2sinB/2sinC/2
cos(A+B/2)cos(A-B/2)=(sinC/2)^2+sinC/2*(cos(A-B/2)-cos(A+B/2))=(sinC/2)^2+sinC/2*cos(A-B/2)-sinC/2cos(A+B/2)
∴cos(A+B/2)(cos(A-B/2)+sinC/2)=sinC/2*(sinC/2+cos(A-B/2))
即 cos(A+B/2)=sinC/2或者 cos(A-B/2)+sinC/2=0
这里需要有一些条件才能说明A+B+C=180°，比如0 °<A,B,C<=180°
如果有上述条件，那么 cos（A-B/2)+sinC/2>0
那么有cos(A+B/2)=cos(90°-C/2)
(A+B)/2=90°-C/2
A+B+C=180°