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.
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