在△ABC中,角ABC对边分别为a,b,c,4sin2(A+B)/2-cos2C=7/2,5 c=根号7,求△ABC的面积最大值

4{sin[(A+B)/2]}^2-cos2C=7/2,
2[1-cos(A+B)]-cos2C=7/2,
2+2cosC-2(cosC)^2+1=7/2,
2(cosC)^2-2cosC+1/2=0,
cosC=1/2,C=60°,
5 c=√7,c=√7/5,
由正弦定理,a=csinA/sinC,b=csinB/sinC,
∴△ABC的面积=(1/2)absinC=(c^2/2)sinAsinB/sinC=[7/(25√3)]sinAsinB
=[7/(50√3)][cos(A-B)-cos(A+B)]=[7/(50√3)][cos(A-B)+1/2],
所求最大值=7√3/100.