（2011•江西）在△ABC中，角A，B，C的对边分别是a，b，c，已知sinC+cosC=1-sinC2（1）求sinC的值（2）若a2+b2=4（a+b）-8，求边c的值．

Question

（2011•江西）在△ABC中，角A，B，C的对边分别是a，b，c，已知sinC+cosC=1-sin(C)/(2)
（1）求sinC的值
（2）若 a²+b²=4（a+b）-8，求边c的值．

Answer 1

（1）∵sinC+cosC＝1−sin(C)/(2)
∴2sin(C)/(2)cos(C)/(2)+1−2sin2(C)/(2)＝1−sin(C)/(2)
∴2sin(C)/(2)cos(C)/(2)−2sin2(C)/(2)＝−sin(C)/(2)
∴2sin2(C)/(2)−2sin(C)/(2)cos(C)/(2)＝sin(C)/(2)
∴2sin(C)/(2)(sin (C)/(2)−cos(C)/(2))＝sin(C)/(2)
∴sin(C)/(2)−cos (C)/(2)＝(1)/(2)
∴sin2(C)/(2)−sinC+cos2(C)/(2)＝(1)/(4)
∴sinC＝(3)/(4)
（2）由sin(C)/(2)−cos(C)/(2)＝(1)/(2)＞0得(π)/(4)＜(c)/(2)＜(π)/(2)
即(π)/(2)＜C＜π
∴cosC＝−(√(7))/(4)
∵a²+b²=4（a+b）-8
∴（a-2）²+（b-2）²=0
∴a=2，b=2
由余弦定理得c