已知向量AB=(1+tanx,1-tanx),向量AC=(sin(x-π/4),sin(x+π/4)

来源：学生作业帮编辑：作业帮分类：数学作业时间：2024/05/10 09:11:11

已知向量AB=(1+tanx,1-tanx),向量AC=(sin(x-π/4),sin(x+π/4)
(1)求证:角BAC为直角
(2)若x属于[-π/4,π/4],求三角形ABC的边BC的长度的取值范围

1.证明:角BAC为直角,
即,证明:向量AB*向量AC=0,即可,
向量AB*向量AC=(1+tanx)*sin(x-π/4)+(1-tanx)*sin(x+π/4)
=[sin(x-π/4)+sin(x+π/4)]+tanx[sin(x-π/4)-sin(x+π/4)]
=2*sinx*cos(-π/4)+tanx*2cosx*sin(-π/4)
=√2*sinx-√2*sinx
=0,
即,向量AB⊥向量AC,
角BAC为直角 ,得证.
2.|BC|^2=|AB|^2+|AC|^2,
=(1+tanx)^2+(1-tanx)^2+[sin(x-π/4)]^2+[sin(x+π/4)]^2
=2+2tan^2x+[sin(x-π/4)]^2+[sin(x+π/4)]^2.
而,tanx,sinx在区间x属于[-π/4,π/4],都是同增,同减的,
当X=0时,|BC|^2取最小值,
|BC|^2=2+2*(1/2)=3,
|BC|=√3,
当X=-π/4,或π/4时,
|BC|^2=2+2+1=5,
|BC|=√5.
则,三角形ABC的边BC的长度的取值范围是:√3≤|BC|≤√5.

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