c sinA=a cosC
c/cosC=a/sinA=c/sinC
所以C=π/4
√3sinA-cos(B+π/4)
=√3sinA-cos(B+C)
=√3sinA+cosA
=2(√3/2sinA+1/2cosA)
=2sin(A+π/6)
因此最大值是2,此时A+π/6=π/2,A=π/3
B=5π/12
c sinA=a cosC
c/cosC=a/sinA=c/sinC
所以C=π/4
√3sinA-cos(B+π/4)
=√3sinA-cos(B+C)
=√3sinA+cosA
=2(√3/2sinA+1/2cosA)
=2sin(A+π/6)
因此最大值是2,此时A+π/6=π/2,A=π/3
B=5π/12