大师作文网高中数学三角函数的难题
来源：学生作业帮 编辑：大师作文网作业帮 分类：数学作业 时间：2024/11/11 06:29:38
高中数学三角函数的难题
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?
过程写的详细点,通俗易懂就好,谢谢!
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?
过程写的详细点,通俗易懂就好,谢谢!
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
由于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.
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
由于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.