已知向量AB=(1+tanX,1-tanX),向量AC=(sIn(X-45度),sin(X+45度)).(1)求证向量A

1个回答

  • (1)向量AB·向量AC

    =(1+tanx)*sin(x-45)+(1-tanx)*sin(x+45)

    =(1+tanx)(sinx-cosx)/根号2+(1-tanx)(sinx+cosx)/根号2

    =[(1+tanx)(sinx-cosx)+(1-tanx)(sinx+cosx)]/根号2

    =[(sinx+tanxsinx-cosx-sinx)+(sinx+cosx-tanxsinx-sinx)]/根号2

    =0,

    所以向量AB垂直于向量AC;

    (2)因为向量BC=向量AC-向量AB

    =(sin(x-45)-(1+tanx),sin(x+45)-(1-tanx)),

    所以|向量BC|^2

    =[sin(x-45)-(1+tanx)]^2+[sin(x+45)-(1-tanx)]^2

    =[sin(x-45)^2]^2+(1+tanx)^2+[sin(x+45)]^2+(1-tanx)^2

    -2[sin(x-45)(1+tanx)+sin(x+45)(1-tanx)],

    由(1)可得,sin(x-45)(1+tanx)+sin(x+45)(1-tanx)=0,

    而[sin(x+45)^2]+[sin(x-45)]^2

    =(sinx+cosx)^2/2+(sinx-cosx)^2/2

    =(sinx)^2+(cosx)^2

    =1,

    而(1+tanx)^2+(1-tanx)^2

    =2(tanx)^2+2,

    所以|向量BC|^2

    =2(tanx)^2+2+1

    =2(tanx)^2+3,

    因为x属于[-45度,45度],

    所以tanx属于[-1,1],

    所以(tanx)^2属于[0,1],

    所以|向量BC|^2属于[3,5],

    所以|向量BC|属于[根号3,根号5].

    所以|向量BC|的取值范围为[根号3,根号5].