在△ABC中,向量m=(cosC/2,sinC/2),向量n=(cosC/2,-sinC/2),且向量m与向量n的夹角为π/3,(1)求C,(2)已知c=7/2 ,S△ABC =3√3/2求a+b的值

m*n=cos^2(C/2)-sin^2(C/2)=cosC=|m|*|n|*cos(π/3)|m|=|n|=cos^2(C/2)+sin^(C/2)=1所以cosC=cos(π/3)=1/2∵C∈(0,π)∴C=π/32）S△ABC=(a*b*sinC)/2=(3/2)√3∴a*b*√3/2*1/2=(3/2)√3∴ab=6根据余弦定理c^2=a^2+...