在三角形ABC中,角 A,B,C的对边分别为a,b,c.已知A=4分之派,bsin(4分之派+C)-csin(4分之派+B)=a. (1）求证:B-C=2分之派 (2) 若a=根号2.求三角形ABC的面积

1、正弦定理：a/sinA=b/sinB=c/sinC=t
a=tsinA
b=tsinB
c=tsinC
bsin(π/4+C)-csin(π/4+B)=a
所以sinAsin(π/4+C)-sinCsin(π/4+B)=sinA=√2/2
sinA(√2/2sinC+√2/2cosC)-sinC(√2/2sinB+√2/2cosB)
=√2/2(sinBcosC-sinCcosB)
=√2/2sin(B-C)
=√2/2
sin(B-C)=1
A=π/4,所以 B、C∈(0,3π/4)
B-C∈(-3π/4,3π/4)
所以B-C=π/2
2、B-C=π/2,B+C=π-A=3π/4
B=5π/8,C=π/8
b/sinB=c/sinC=a/sinA=2
b²=4sin²B=4sin²5π/8=4·(1-cos5π/4)/2=2+√2
c²=4sin²C=4sin²π/8=4·(1-cosπ/4)/2=2-√2
b²c²=2
bc=√2
S=bcsinA/2=√2·(√2/2)/2=1/2