Question

三角形ABC中sinA/2sinB/2sinC/2=1/8,判断三角形形状

Answer 1

法1（sinA/2)^2+(sinB/2)^2+(sinC/2)^2+2sinA/2sinB/2sinC/2
=sinA/2（（sinA/2)+2sinB/2sinC/2）+(sinB/2)^2+(sinC/2)^2
=sinA/2（（cos（B+C）/2)+2sinB/2sinC/2）+(sinB/2)^2+(sinC/2)^2
=sinA/2*cos（B+C）/2+(1-cosB)/2+(1-cosC)/2(积化和差)
=cos（B+C）/2*cos（B+C）/2+1-cos（B+C）/2*cos（B+C）/2
=1故（sinA/2)^2+(sinB/2)^2+(sinC/2)^2取最小值sinA/2sinB/2sinC/2最大值
（sinA/2)^2+(sinB/2)^2+(sinC/2)^2+2sinA/2sinB/2sinC/2>=3（3次√（sinA/2sinB/2sinC/2）^2）+2sinA/2sinB/2sinC/2