∵sinAcosB+sin²B=sinC∴sinAcosB=sinC-sin²B∵sinC=sinAcosB+cosAsinB∴sinC=sinC-sin²B+cosAsinB∴sin²B=cosAsinB两边同时约去一个sinB∴sinB=cosA∵∠p=π/2，所以sinB=cos(π/2-B)∴cos(π/2-B)=cosA
