∵a,b,c成等差数列,∴2b=a+c,
又∵A-C=90°,A+B+C=180°,
∴C=45°-
,B 2
由正弦定理可得2sinB=sinA+sinC,
∴2sinB=sin(90°+C)+sinC
=cosC+sinC=
sin(C+45°)
2
=
sin(45°-
2
+45°)B 2
=
sin(90°-
2
)=B 2
cos
2
,B 2
∴2sinB=4sin
cosB 2
=B 2
cos
2
,B 2
解得sin
=B 2
,
2
4
∴cosB=1-2sin2
=B 2
3 4
故答案为:
3 4