已知M(cosα-sinα,1),N(cosα,sinα),则|MN|的最小值是M是向量M,N是向量N
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MN=(sina,sina-1)
|MN|²=sin²a+(sina-1)²
=2sin²a-2sina+1
=2(sina-1/2)²+1/2
当 sina=1/2
|MN|²有最小值1/2
所以 |MN|有最小值√2/2
======以下答案可供参考======
供参考答案1:
MN=(sina,sina-1)
|MN|²=sin²a+(sina-1)²
=2sin²a-2sina+1
=2(sina-1/2)²+1/2
当 sina=1/2
|MN|²有最小值1/2
所以 |MN|有最小值√2/2
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