一道向量与三角函数结合的难题已知向量an=(cos nπ/7,sin nπ/7)(n∈N+),ㄧbㄧ=1,则函数y=ㄧa1+bㄧ²+ㄧa2+bㄧ²+ㄧa3+bㄧ²+···+ㄧa141+bㄧ²的最大值为多少?知道的人才快说下啦,急死了,
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![一道向量与三角函数结合的难题已知向量an=(cos nπ/7,sin nπ/7)(n∈N+),ㄧbㄧ=1,则函数y=ㄧa1+bㄧ²+ㄧa2+bㄧ²+ㄧa3+bㄧ²+···+ㄧa141+bㄧ²的最大值为多少?知道的人才快说下啦,急死了,](/uploads/image/z/5161357-37-7.jpg?t=%E4%B8%80%E9%81%93%E5%90%91%E9%87%8F%E4%B8%8E%E4%B8%89%E8%A7%92%E5%87%BD%E6%95%B0%E7%BB%93%E5%90%88%E7%9A%84%E9%9A%BE%E9%A2%98%E5%B7%B2%E7%9F%A5%E5%90%91%E9%87%8Fan%3D%28cos+n%CF%80%2F7%2Csin+n%CF%80%2F7%EF%BC%89%EF%BC%88n%E2%88%88N%2B%EF%BC%89%2C%E3%84%A7b%E3%84%A7%3D1%2C%E5%88%99%E5%87%BD%E6%95%B0y%3D%E3%84%A7a1%2Bb%E3%84%A7%26%23178%3B%2B%E3%84%A7a2%2Bb%E3%84%A7%26%23178%3B%2B%E3%84%A7a3%2Bb%E3%84%A7%26%23178%3B%2B%C2%B7%C2%B7%C2%B7%2B%E3%84%A7a141%2Bb%E3%84%A7%26%23178%3B%E7%9A%84%E6%9C%80%E5%A4%A7%E5%80%BC%E4%B8%BA%E5%A4%9A%E5%B0%91%3F%E7%9F%A5%E9%81%93%E7%9A%84%E4%BA%BA%E6%89%8D%E5%BF%AB%E8%AF%B4%E4%B8%8B%E5%95%A6%2C%E6%80%A5%E6%AD%BB%E4%BA%86%2C)
一道向量与三角函数结合的难题已知向量an=(cos nπ/7,sin nπ/7)(n∈N+),ㄧbㄧ=1,则函数y=ㄧa1+bㄧ²+ㄧa2+bㄧ²+ㄧa3+bㄧ²+···+ㄧa141+bㄧ²的最大值为多少?知道的人才快说下啦,急死了,
一道向量与三角函数结合的难题
已知向量an=(cos nπ/7,sin nπ/7)(n∈N+),ㄧbㄧ=1,则函数y=ㄧa1+bㄧ²+ㄧa2+bㄧ²+ㄧa3+bㄧ²+···+ㄧa141+bㄧ²的最大值为多少?
知道的人才快说下啦,急死了,
一道向量与三角函数结合的难题已知向量an=(cos nπ/7,sin nπ/7)(n∈N+),ㄧbㄧ=1,则函数y=ㄧa1+bㄧ²+ㄧa2+bㄧ²+ㄧa3+bㄧ²+···+ㄧa141+bㄧ²的最大值为多少?知道的人才快说下啦,急死了,
an=(cos nπ/7,sin nπ/7)
则a²n= cos ²nπ/7+sin² nπ/7=1,
ㄧbㄧ=1,
则b²=1.
y=ㄧa1+bㄧ²+ㄧa2+bㄧ²+ㄧa3+bㄧ²+•••+ㄧa141+bㄧ²
=a²1+2a1b+ b²+ a²2+2a2b+ b²+•••+ a²141+2a141b+ b²
=1+2a1b+1+1+2a2b+1+•••1+2a141b+1
=282+2(a1+a2+……+a141)b
cosπ/7+ cos2π/7+ cos2π/7+ cos2π/7+……+ cos7π/7+
cos8π/7+ cos9π/7+ cos10π/7+ cos11π/7+……+ cos14π/7
= cosπ/7+ cos2π/7+ cos2π/7+ cos2π/7+……+ cos7π/7
-cosπ/7-cos2π/7- cos3π/7- cos4π/7+……- cos7π/7
=0,
sinπ/7+ sin2π/7+ sin2π/7+ sin2π/7+……+ sin7π/7+
sin8π/7+ sin9π/7+ sin10π/7+ sin11π/7+……+ sin14π/7
= sinπ/7+ sin2π/7+ sin2π/7+ sin2π/7+……+ sin7π/7
-sinπ/7-sin2π/7- sin3π/7- sin4π/7+……- sin7π/7
=0,
因为cos nπ/7与sin nπ/7的周期都是14,
所以a1+a2+……+a141=a1
(a1+a2+……+a141)b=a1b≤|a1||b|=1
∴y=282+2(a1+a2+……+a141)b
≤282+2=284.
已知向量a=(cos3θ/2,sin3θ/2),b=(cosθ/2,-sinθ/2), θ属于[1、a·b/(|a b|)=cos(3θ/2 θ/2)÷{2cos3θ/2cosθ/2-2sin3