1、∵A、B、C是三角形的内角
∴sin(A+B)=sinC
∴√2asin(B+ π/4)=c
√2 sinAsin(B+ π/4)=sinC (根据正弦定理)
√2sinA[(√2/2)sinB+(√2/2)cosB]=sinC
sinA(sinB+cosB)=sin(A+B)
sinAcosB+sinAsinB=sinAcosB+cosAsinB
对照等号两边,得出：sinA=cosA
∴ A=π/4
2、利用积化和差公式,将sinBsinC变形：
sinBsinC
=(-1/2)[cos(B+C)-cos(B-C)]
=(-1/2)[-cosA-cos(B-C)]
=(√2/4)+(1/2)cos(B-C)
∵0