1.由csinA=acosC 得a/sinA=c/cosC,
∵a/sinA=c/sinC(正弦定理)
∴sinC=cosC
∴∠C=45°
2.∵cos(B+45°)=cos(B+C )
又cos(B+C)=cos(π-A)=-cosA
∴原式=√3sinA-(-cosA)=√3sinA+cosA=2sin(A+30°)
显然∠A=90°时有最大值2,此时B=45°
1.由csinA=acosC 得a/sinA=c/cosC,
∵a/sinA=c/sinC(正弦定理)
∴sinC=cosC
∴∠C=45°
2.∵cos(B+45°)=cos(B+C )
又cos(B+C)=cos(π-A)=-cosA
∴原式=√3sinA-(-cosA)=√3sinA+cosA=2sin(A+30°)
显然∠A=90°时有最大值2,此时B=45°