在三角形ABC中asinA+bsinB=csinC,试用余弦定理证明△ABC为直角三角形

2个回答

  • asinA+bsinB=csinC,

    a=2R*sinA,b=2R*sinB,c=2R*sinC,即有

    (sinA)^2+(sinB)^2=(sinC)^2,

    (sinA)^2=(sinC)^2-(sinB)^2=(sinC+sinB)*(sinC-sinB),

    而,sinC+sinB=2sin[(C+B)/2]*cos[(C-B)/2],

    sinC-sinB=2cos[(C+B)/2]*sin[(C-B)/2],

    (sinA)^2=[2*sin(A/2)*cos(A/2)]^2,

    A+B+C=180,A/2=[180-(C+B)]/2=90-(C+B)/2,

    即有,sin(A/2)=sin[90-(C+B)/2]=cos[(C+B)/2],

    cos(A/2)=cos[(90-(C+B)/2]=sin[(C+B)/2],

    可得到,

    2cos(A/2)*sin(A/2)=2cos[(C-B)/2]*sin[(C-B)/2],

    sinA=sin(C-B),

    sinA-sin(C-B)=0,

    2cos[(A+C-B)/2]*sin[(A+B-C)/2]=0,

    cos[(A+C-B)/2]=0,或sin[(A+B-C)/2]=0,

    A+C=180-B,或A+B=180-C,

    cos[(A+C-B)/2]=cos(90-B)=sinB=0(不合,舍去).

    当sin[(A+B-C)/2]=0,时,

    sin(90-C)=cosC=0,即有,

    cosC=(a^2+b^2-c^2)/2ab=0,

    a^2+b^2=c^2,

    C=90度,ABC为直角三角形.