在△ABC中 向量m=(2sinB,2-cos2B)n=(2sin²(π/4+B/2),-1)且m⊥n ①求角

(1)
m.n=0
(2sinB,2-cos2B).(2sin²(π/4+B/2),-1)=0
4sinBsin²(π/4+B/2) -2+cos2B=0
2sinB(1-cos2(π/4+B/2))-2+cos2B=0
2sinB(1+sinB)-2+cos2B=0
2sinB+2(sinB)^2-2 +(1-2(sinB)^2)=0
2sinB-1=0
sinB=1/2
B=π/6
(2)
y=sinA+cosC
= sinA + cos(π-B-A)
= sinA + cos(5π/6-A)
y'= cosA +sin(5π/6-A)=0
cosA+(1/2)cosA+(√3/2)sinA=0
(√3/2)sinA = -(3/2)cosA
tanA =-√3/3
A= 5π/6(max) or 11π/6(min)
maxy =y(5π/6) = sin(5π/6) = 1/2
miny = y(11π/6) = sin(11π/6) + cos(-5π/6) = -1/2+√3/2
y值域=[-1/2+√3/2,1/2]