在△ABC中,若a c=2b,A-C=π/2,则sinB=_____

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a+c=2b == sinA+sinC = 2*sinB== 2*sin[(A+C)/2] *sin[(A-C)/2] = 4sin(B/2)*cos(B/2)== sin[(π-B)/2]*sin(π/4) = 2*sin(B/2)*cos(B/2)== (根号2)/2 = 2*sin(B/2) == sin(B/2) = (根号2)/4== cos(B/2) = (根号14)/4sinB = 2*sin(B/2)*cos(B/2) = (根号7)/4