已知α∈(π/2,π)且sin0.5α+cos0.5α=根号6/2.一,求cosα的值（已求出为负的根号3/2）二,si

（一）sin(a/2)+cos(a/2)=√6/2.两边平方得:sin²(a/2)+2sin(a/2)cos(a/2)+cos²(a/2)=3/2.===>1+sina=3/2.===>sina=1/2.又sin²a+cos²a=1.===>cos²a=3/4.===>cosa=±√3/2.∵a∈(π/2,π).∴cosa=-√3/2.(二）∵π/2＜b＜π,π/2＜a＜π.sin(a-b)=-3/5.∴π/2＜a＜b＜π.且cos(a-b)=4/5.∴cosb=cos[a+(b-a)]=cosacos(b-a)-sinasin(b-a)=cosacos(a-b)+sinasin(a-b)=-(3+4√3)/10.即cosb=-(3+4√3)/10.