(1)求A的大小;
(2)求sin(B+)的值.
解:(1)由m∥n,得2sin2A-1-cosA=0,
即2sin2A+cosA-1=0.
∴cosA=或cosA=-1.
∵A是△ABC的内角,cosA=-1舍去,∴A=.
(2)∵B+c=A,由正弦定理,sinB+sinC=sinA=.
∵B+C=,∴sinB+sin(-B)=. ∴cosB+sinB=,即sin(B+)=.