由cosB=sinA/2sinC,得
2cosB=sinA/sinC=sin(B+C)/sinc=(sinBcosC+cosBsinC)/sinC
=sinBcosC/sinC+cosB
故有cosB=sinBcosC/sinC
即sinCcosB-sinBcosC=sin(C-B)=0
∴C-B=0,即C=B.
代入原式得
2sinBcosB=sinA,即sin2B=sinA,∴2B=A.
于是有A+B+C=2B+B+B=4B=180°
∴B=45°,C=45°,A=90°.
即△ABC是等要直角三角形.