CAN PLAY!已知向量AB=(1-tanx,1+tanx),AC=(sin(x-π/4),sin(x+π/4))(1)求证AB⊥AC(2)若x∈(-π/4,π/4),求│ BC │的取值范围.
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![CAN PLAY!已知向量AB=(1-tanx,1+tanx),AC=(sin(x-π/4),sin(x+π/4))(1)求证AB⊥AC(2)若x∈(-π/4,π/4),求│ BC │的取值范围.](/uploads/image/z/10605555-27-5.jpg?t=CAN+PLAY%21%E5%B7%B2%E7%9F%A5%E5%90%91%E9%87%8FAB%3D%281-tanx%2C1%2Btanx%29%2CAC%3D%28sin%28x-%CF%80%2F4%29%2Csin%28x%2B%CF%80%2F4%29%29%281%29%E6%B1%82%E8%AF%81AB%E2%8A%A5AC%282%29%E8%8B%A5x%E2%88%88%28-%CF%80%2F4%2C%CF%80%2F4%29%2C%E6%B1%82%E2%94%82+BC+%E2%94%82%E7%9A%84%E5%8F%96%E5%80%BC%E8%8C%83%E5%9B%B4.)
CAN PLAY!已知向量AB=(1-tanx,1+tanx),AC=(sin(x-π/4),sin(x+π/4))(1)求证AB⊥AC(2)若x∈(-π/4,π/4),求│ BC │的取值范围.
CAN PLAY!
已知向量AB=(1-tanx,1+tanx),AC=(sin(x-π/4),sin(x+π/4))
(1)求证AB⊥AC
(2)若x∈(-π/4,π/4),求│ BC │的取值范围.
CAN PLAY!已知向量AB=(1-tanx,1+tanx),AC=(sin(x-π/4),sin(x+π/4))(1)求证AB⊥AC(2)若x∈(-π/4,π/4),求│ BC │的取值范围.
向量AB+向量AC=0,就可以证明AB⊥AC
(1)证明:
1-tanx=(cosx-sinx)/cosx
1+tanx=(cosx+sinx)/cosx
sin(x-π/4)=√2/2(sinx-cosx)
sin(x+π/4)=√2/2(sinx+cosx)
若AB⊥AC 则向量AB·向量AC=0
(cosx-sinx)/cosx·√2/2(sinx-cosx)+(cosx+sinx)/co...
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(1)证明:
1-tanx=(cosx-sinx)/cosx
1+tanx=(cosx+sinx)/cosx
sin(x-π/4)=√2/2(sinx-cosx)
sin(x+π/4)=√2/2(sinx+cosx)
若AB⊥AC 则向量AB·向量AC=0
(cosx-sinx)/cosx·√2/2(sinx-cosx)+(cosx+sinx)/cosx·√2/2(sinx+cosx)
=√2/2cosx·(-(cosx-sinx)^2+(cosx+sinx)^2
≠0
怎么了?是不是倒过来了?
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若证ABAC垂直,即证其向量点乘得零