1.设向量a=(1,0),b=(cosθ,sinθ),其中0≤θ≤π,则|a-b|的最大值是_____.2.2.已知n∈Z,在下列三角函数中,与sin数值相同的是( )①sin(nπ+);②cos(2nπ+);③sin(2nπ+);④cos[(2n+1)π-];⑤sin[(2n+1)π-].A.①② B.①
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![1.设向量a=(1,0),b=(cosθ,sinθ),其中0≤θ≤π,则|a-b|的最大值是_____.2.2.已知n∈Z,在下列三角函数中,与sin数值相同的是( )①sin(nπ+);②cos(2nπ+);③sin(2nπ+);④cos[(2n+1)π-];⑤sin[(2n+1)π-].A.①② B.①](/uploads/image/z/7636370-50-0.jpg?t=1.%E8%AE%BE%E5%90%91%E9%87%8Fa%3D%281%2C0%29%2Cb%3D%28cos%CE%B8%2Csin%CE%B8%29%2C%E5%85%B6%E4%B8%AD0%E2%89%A4%CE%B8%E2%89%A4%CF%80%2C%E5%88%99%7Ca-b%7C%E7%9A%84%E6%9C%80%E5%A4%A7%E5%80%BC%E6%98%AF_____.2.2.%E5%B7%B2%E7%9F%A5n%E2%88%88Z%2C%E5%9C%A8%E4%B8%8B%E5%88%97%E4%B8%89%E8%A7%92%E5%87%BD%E6%95%B0%E4%B8%AD%2C%E4%B8%8Esin%E6%95%B0%E5%80%BC%E7%9B%B8%E5%90%8C%E7%9A%84%E6%98%AF%28+%29%E2%91%A0sin%28n%CF%80%2B%29%3B%E2%91%A1cos%282n%CF%80%2B%29%3B%E2%91%A2sin%282n%CF%80%2B%29%3B%E2%91%A3cos%EF%BC%BB%282n%2B1%29%CF%80-%EF%BC%BD%3B%E2%91%A4sin%EF%BC%BB%282n%2B1%29%CF%80-%EF%BC%BD.A.%E2%91%A0%E2%91%A1+B.%E2%91%A0)
1.设向量a=(1,0),b=(cosθ,sinθ),其中0≤θ≤π,则|a-b|的最大值是_____.2.2.已知n∈Z,在下列三角函数中,与sin数值相同的是( )①sin(nπ+);②cos(2nπ+);③sin(2nπ+);④cos[(2n+1)π-];⑤sin[(2n+1)π-].A.①② B.①
1.设向量a=(1,0),b=(cosθ,sinθ),其中0≤θ≤π,则|a-b|的最大值是_____.
2.
2.已知n∈Z,在下列三角函数中,与sin数值相同的是( )
①sin(nπ+);②cos(2nπ+);③sin(2nπ+);④cos[(2n+1)π-];
⑤sin[(2n+1)π-].
A.①② B.①③④ C.②③⑤ D.①③⑤
1.设向量a=(1,0),b=(cosθ,sinθ),其中0≤θ≤π,则|a-b|的最大值是_____.2.2.已知n∈Z,在下列三角函数中,与sin数值相同的是( )①sin(nπ+);②cos(2nπ+);③sin(2nπ+);④cos[(2n+1)π-];⑤sin[(2n+1)π-].A.①② B.①
此题第2问没有正确答案:
1
|a-b|^2=(a-b) dot (a-b)=|a|^2+|b|^2-2(a dot b)=1+1-2cosθ=2-2cosθ
当cosθ=-1时,|a-b|取得最大值:2
2
当n=2k时,sin(nπ+θ)=sin(2kπ+θ)=sinθ
当n=2k+1时,sin(nπ+θ)=sin(2kπ+π+θ)=sin(π+θ)=-sinθ
cos(2nπ+θ)=cosθ
sin(2nπ+θ)=sinθ
cos((2n+1)π-θ)=cos(π-θ)=-cosθ
sin((2n+1)π-θ)=sin(π-θ)=sinθ
所以,只有③⑤满足条件,无正确答案.
1
a-b=(1-cosθ,-sinθ)
|a-b|=√[(1-cosθ)²+sin²θ]=√(2-2cosθ)≤2
2
???这道题实在不懂 请在追问或补问中提出
1.a-b=(1-cosθ,-sinθ)
|a-b|=√[(1-cosθ)²+sin²θ]=√(2-2cosθ)≤√4=2
2.①sin(nπ+a)=-sina或sina
②cos(2nπ+a)=cosa
③sin(2nπ+a)=siba
④cos[(2n+1)π-a]=cos[π-a]=-cosa
⑤sin[(2n+1)π-a]=sin[π-a]=sina.
所以,选D
最大值是2