在△ABC,sinC+cosC=1-sinC/2,求sinC；若a^2+b^2=4(a+b)-8,求边C的值

(1)原方程变形为2sinC/2cosC/2+cos²C/2-sin²C/2=sin²C/2+cos²C/2-sinC/2
即sinC/2-cosC/2=1/2
(sinC/2-cosC/2)²=1/4=1-2sinC/2cosC/2 得2sinC/2cosC/2=3/4=sinC
即sinC=3/4
(a²-4a+4)+(b²-4b+4)=0 即(a-2)²+(b-2)²=0
得a=b=2
由余弦定理c²=a²+b²-2abcosC=4+4-2X4X√7/4=8-2√7
C=√(8-2√7) =√7-1