∵sinC＝(sinA+sinB)/(cosA+cosB)
∴sinC- (sinA+sinB)/(cosA+cosB) ＝0
∴sinC- 2sin[(A+B)/2]cos[(A-B)/2]/ 2cos[(A+B)/2]cos[(A-B)/2]＝0
∴sinC- sin[(A+B)/2] / cos[(A+B)/2]＝0
∴2sin(C/2)cos(C/2)- cos(C/2) / sin(C/2)＝0,
又∵cos(C/2)≠0
∴2sin(C/2) - 1 / sin(C/2)＝0
∴2sin^2 (C/2) - 1＝0
∴2sin^2 (C/2)＝1
∵sin(C/2)＞0
∴sin(C/2)＝√2/2,
则C/2＝π/4
∴C＝π/2,
∴△ABC是以C为直角顶点的直角三角形.
∵2sinAsinC
=-[cos(A+c)-cos(A-c)]
=-[-cosB-cos(A-c)]
=cosB+cos(A-c)
∴2sinAsinC-cosB=1
可以化成:
cosB+cos(A-c)-cosB=1
即:cos(A-c)=1
即A=C
△ABC为等腰三角形
A+B+C=180
A/2+B/2+C/2=90
(sinA/2)^2+2(sinB/2)^2+(sinC/2)^2=1
(sinA/2)^2+(sinB/2)^2+(sinC/2)^2
=1-(sinB/2)^2
=(cosB/2)^2
=[sin(A/2+C/2)]^2
(sinA/2)^2+(sinB/2)^2
=[sin(A/2+C/2)]^2-(sinC/2)^2
=[sin(A/2+C/2)-sinC/2][sin(A/2+C/2)+sinC/2]
=2cos(A/4+C/2)sin(A/4)*2sin(A/4+C/2)cos(A/4)
=sin(A/2+C)sinA/2
所以再把左边的(sinA/2)^2移到右边,得到:
(sinB/2)^2=sin(A/2+C)sinA/2-(sinA/2)^2
=sinA/2*[sin(A/2+C)-sinA/2]
=sin A/2*2cos(A/2+C)sinC/2
=2sinA/2sinB/2sinC/2
所以左右约去sinB/2得到:
SinB/2=2sinA/2sinC/2
又因为:
sinB/2=cos(A/2+C/2)=cosA/2cosC/2-sinA/2sinC/2
2sinA/2sinC/2=cosA/2cosC/2-sinA/2sinC/2
也就是:
3 sinA/2sinC/2= cosA/2cosC/2
左右都除以cosA/2cosC/2就得到:
tanA/2tanC/2=1/3
所以就得到结果:
tanA/2tanC/2=1/3
cosA+cosB+cosC
=1-2(1-2(sinA/2)^2+1-2(sinB/2)^2+1-2(sinC/2)^2
=3-2[(sinA/2)^2+(sinB/2)^2+(sinC/2)^2]
=1+4sinA/2sinB/2sinC/2
3-2[(sinA/2)^2+(sinB/2)^2+(sinC/2)^2=1+4sinA/2sinB/2sinC/2
(sinA/2)^2+(sinB/2)^2+(sinC/2)^2+2sinA/2sinB/2sinC/2-1=0
(sinC/2)^2+2sinA/2sinB/2sinC/2+(sinA/2)^2+(sinB/2)^2-1=0
因为C为锐角,所以C/2也为为锐角
即sinC/2>0
根据求根公式：
sinC/2=[-2sinA/2sinB/2+√{(2sinA/2sinB/2)^2-4[(sinA/2)^2+(sinB/2)^2-1]}]/2
=[-2sinA/2sinB/2+2√{(sinA/2sinB/2)^2-[(sinA/2)^2+(sinB/2)^2-1]}/2
=-sinA/2sinB/2+√{(sinA/2sinB/2)^2-[(sinA/2)^2+(sinB/2)^2-1]}
=-sinA/2sinB/2+√[1-(sinA/2)^2(1-sinB/2)^2]
=-sinA/2sinB/2+√[(cosA/2)^2(cosB/2)^2]
=-sinA/2sinB/2+cosA/2cosB/2
=cosA/2cosB/2-sinA/2sinB/2
=cos(A/2+B/2)
=sin(π/2- A/2-B/2)
所以A+B+C=π
即A+B+C=180°