一道向量与三角函数结合的难题
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.能缩写吗？由已知得a²n= cos ²nπ/7+sin² nπ/7=1，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因为cos nπ/7与sin nπ/7的周期都是14，所以a1+a2+……+a141=10*(a1+a2+……+a14)+ a141=10*(0,0)+a1=a1,(a1+a2+……+a141)b=a1b≤|a1||b|=1∴y=282+2(a1+a2+……+a141)b≤282+2=284.