由S△ABC=Rr(sinA+sinB+sinC)到4Rr·cosA/2 ·cosB/2 ·cosC/2是怎么推导出来的
1个回答

∵在三角形ABC中,

∴A+B+C=180°,得sinA=sin(B+C)

则A/2=90°-(B+C)/2,得cos﹙A/2﹚=sin((B+C)/2)

左边=sin(B+C)+sinB+sinC

则4cos(A/2)cos(B/2)cos(C/2)

=4sin[(B+C)/2]Cos(B/2)Cos(C/2)

=4cos(B/2)cos(C/2)[sinB/2·cos﹙C/2﹚+cosB/2·sin﹙C/2)]

=[4sin(B/2)cos(B/2)(cos²(C/2)]+[4sin(C/2)cos(C/2)(cos²(B/2)

=sinB(cosC+1)+sinC(CosB+1)

=sin(B+C)+sinB+sinC

左边=右边

原式成立