已知向量a=(cosa,1+sina)向量b=(1+cosa,sina)2.设向量c=(-cosa,-2)求(向量a+向量c)X向量b的范围
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![已知向量a=(cosa,1+sina)向量b=(1+cosa,sina)2.设向量c=(-cosa,-2)求(向量a+向量c)X向量b的范围](/uploads/image/z/3768302-38-2.jpg?t=%E5%B7%B2%E7%9F%A5%E5%90%91%E9%87%8Fa%3D%EF%BC%88cosa%2C1%2Bsina%29%E5%90%91%E9%87%8Fb%3D%EF%BC%881%2Bcosa%2Csina%292.%E8%AE%BE%E5%90%91%E9%87%8Fc%3D%EF%BC%88-cosa%2C-2%29%E6%B1%82%EF%BC%88%E5%90%91%E9%87%8Fa%2B%E5%90%91%E9%87%8Fc%EF%BC%89X%E5%90%91%E9%87%8Fb%E7%9A%84%E8%8C%83%E5%9B%B4)
已知向量a=(cosa,1+sina)向量b=(1+cosa,sina)2.设向量c=(-cosa,-2)求(向量a+向量c)X向量b的范围
已知向量a=(cosa,1+sina)向量b=(1+cosa,sina)
2.设向量c=(-cosa,-2)求(向量a+向量c)X向量b的范围
已知向量a=(cosa,1+sina)向量b=(1+cosa,sina)2.设向量c=(-cosa,-2)求(向量a+向量c)X向量b的范围
(a+c)·b
=(0,sina-1)·(1+cosa,sina)
=0+sin²a-sina
=sin²a-sina
令t=sina,t∈[-1,1],则
(a+c)·b
=t²-t
=(t-1/2)²-1/4
∴t-1/2∈[-3/2,1/2]
(t-1/2)²∈[0,9/4]
y∈[-1/4,2]