数学问题，在线等，急

A+B/2是(A+B)/2吧?
是的话应该是
2[sin（A+B)/2）]2+cos2C=1
===>2(cosC)^2+(cosC)^2-(sinC)^2=1
===>4(cosC)^2=2
===>cosC=±√2/2===>C=π/4或3π/4
若c=根号3 ,a+b=3,求a和b的值；
c^2=a^2+b^2-2abcosC=(a+b)^2-2ab(1+cosC)
===>ab=6/(2(1+cosC)
===>a(3-a)=6/(2(1+cosC)
===>a^2-3a+6/(2(1+cosC)
===>a= ,b=

2[sin（A+B)/2）]2+cos2C=1
2(sin(π-C)/2)^2+2(cosC)^2-1=1
2cos(c/2)^2-1+2(cosc)^2=1
2(cosC)^2+cosC-1=0
cosC=-1(舍去)
cosC=1/2
c=60
√(c^2-(a*√3/2)^2)+3a/2=3
3-3a^2/4=(3-3a/2)^2
3-3a^2/4=9+9a^2/4-9a
3a^2-9a+6=0
a^2-3a+2=0
a=1,b=2,
a=2,b=1