作业帮试证明：cos(4π/n)+cos(8π/n)+...+cos(4(n-1)π/n)+cos(4nπ/n ) = 0
来源：学生作业帮助网 编辑：作业帮 时间：2024/11/30 07:15:47
x){~f3+9X|~6ia遹&yi<[ASV&HZFP]y@Y-Ϧo{0Ɏg ykʋK|9sӎmy
:*u=[83Ol6PnDECBv
.HSVE+N q$c0@̄6"Tjj
fP `IAF@.3 2#dA@?h%
ѧNT0*5$ف
试证明：cos(4π/n)+cos(8π/n)+...+cos(4(n-1)π/n)+cos(4nπ/n ) = 0
试证明：cos(4π/n)+cos(8π/n)+...+cos(4(n-1)π/n)+cos(4nπ/n ) = 0
试证明：cos(4π/n)+cos(8π/n)+...+cos(4(n-1)π/n)+cos(4nπ/n ) = 0
n = 1,2的时候不成立,应该有限制n ≥ 3.
此时sin(2π/n) ≠ 0.
而2sin(2π/n)·(cos(4π/n)+cos(8π/n)+...+cos(4nπ/n))
= 2sin(2π/n)·cos(4π/n)+2sin(2π/n)·cos(8π/n)+...+2sin(2π/n)·cos(4nπ/n)
= (sin(6π/n)-sin(2π/n))+(sin(10π/n)-sin(6π/n))+...+(sin((4n+2)π/n)-sin((4n-2)π/n))
= sin((4n+2)π/n)-sin(2π/n)
= sin(4π+2π/n)-sin(2π/n)
= sin(2π/n)-sin(2π/n)
= 0.
即得cos(4π/n)+cos(8π/n)+...+cos(4nπ/n) = 0.
试证明：cos(4π/n)+cos(8π/n)+...+cos(4(n-1)π/n)+cos(4nπ/n ) = 0
证明：cos( π/2n+1)+cos( 3π/2n+1)+cos( 5π/2n+1)+cos(7 π/2n+1)+...+cos[(2n-1) π/2n+1)]=1/2
证明cos【(4n+1)π/4+x】-cos[(4n-1)π/4-x]=0(n属于z）
cos(π/17)*cos(2π/17)*cos(4π/17)*cos(8π/17)
证明cos(π/(2n+1))cos(2π/(2n+1))cos(3π/(2n+1))……cos(nπ/(2n+1))=1/2^n
证明∫cos^nxsin^nxdx＝(1/2^n)∫＜0,π/2＞cos^nxdx n为整数
化简cos[(4n+1)π/4+x]+cos[(4n-1)π/4+x]
cos(π/32)*cos(π/16)*cos(π/8)=
证明 Σcos((k/n)π)=0;k=0,1,2,...2n-1
为什么cos(n-1)π=cos(n+1)πcos(n-1)π=cos(n-1+2)π=cos(n+1)π 对吗?
cos(2π/7)+cos(4π/7)+cos(6π/7)= -1/2.证明[cos(2π/7)]^2+[cos(4π/7)]^2+[cos(6π/7)]^2=5/4
cos(2π/7)+cos(4π/7)+cos(6π/7)= -1/2.证明[cos(2π/7)]^2+[cos(4π/7)]^2+[cos(6π/7)]^2=5/4
COS(π/17)COS(2π/17)COS(4π/17)COS(8π/17)要过程
化简cos{[(4n+1)π/4]+α}+cos{[(4n-1)π/4]-α},（n∈Z）
请教,化简： cos(4/4n+1 π +a)+cos(4/4n-1 π -a) n E Z
cos(4/4n+1 π +a)+cos(4/4n-1 π -a) n E Z
高二推理证明证明等式2cosπ/4=√2,2cosπ/8=√(2+√2),2cosπ/16=√(2+√(2+√2))…并从中归纳出一般结论
求SINπ/64 COSπ/64 COSπ/32 COSπ/16 COSπ/8