在三角形ABC中,角A,B,C所对的边分别是a,b,c,已知sinC+cosC=1-sin(C/2),（1)求sinC (2)若a2+b2=4(a+b)-8,求边c

sinC+cosC=1-sin（C/2）sinC=1-cosC-sin（C/2）2sin(C/2)cos(C/2)=2sin²(C/2)-sin（C/2)∵sin(C/2)≠0∴2cos(C/2)=2sin(C/2)-1sin(C/2)-cos(C/2)=1/2[sin(C/2)-cos(C/2)]^2=1/41-sinC=1/4,sinC=3/4(2)∵a^2+b^=...