在三角形ABC中，若tanB=cos(C-B)/sinA+sin(C-B)则该三角形是？

显示全部楼层 · 2009-7-23 22:05:04

tanB=cos(C－B)/〔sinA+sin(C－B)〕＝cos(C－B)/〔sin（B+C)+sin(C－B)〕 ===>tanB=(cosBcosC+sinBsinC)/(2sinCcosB) ===>2sinBsinC=cosBcosC+sinBsinC ===>cosBcosC-sinBsinC=0 ===>cos(B+C)=0===>B+C=90o∴△ABC是直角三角形...

千问 · 2009-7-23 22:05:04

交叉相乘得sinBsinA+sinBsin(C-B)=cosBcos(C-B) 拆开得sinBsinA+sinCsinBcosB-cosC(sinB)^2=(cosB)^2*cosC+sinCsinBcosB 两边合并同类项cosC((sinB)^2+(cosB)^2)=sinAsinB 即cosC=sinAsinB 又因为cosC=-cos...