n大于等于1,θ不是π的倍数,求证sinθ+sin3θ+...+sin((2n-1)θ)=sin^2(nθ)/sinθ
来源:学生作业帮 编辑:灵鹊做题网作业帮 分类:数学作业 时间:2024/05/02 15:09:10
n大于等于1,θ不是π的倍数,求证sinθ+sin3θ+...+sin((2n-1)θ)=sin^2(nθ)/sinθ
数学归纳法:
(前面略)
假设当N=n时成立,即sinθ+sin(3θ)+...+sin[(2n-1)θ]=sin^2(nθ)/sinθ
则当N=n+1时
sinθ+sin(3θ)+...+sin{[2(n+1)-1]θ}=sin[(2n+1)θ]+sin^2(nθ)/sinθ
={sinθsin[(2n+1)θ]+sin^2(nθ)}/sinθ
={sinθ[sin(2nθ)cosθ+cos(2nθ)sinθ]+[1-cos(2nθ)]/2}/sinθ
={sin(2θ)sin(2nθ)/2+cos(2nθ)[1-cos(2θ)]/2+1/2-cos(2nθ)/2}/sinθ
=[sin(2θ)sin(2nθ)-cos(2θ)cos(2nθ)+1]/(2sinθ)
=[-cos(2nθ+2θ)+1]/(2sinθ)
={-cos[2(n+1)θ]+1}/(2sinθ)
=2sin^2[(n+1)θ]/(2sinθ)
=sin^2[(n+1)θ]/sinθ
所以当N=n时成立,则当N=n+1时也成立,所以原式成立.
(前面略)
假设当N=n时成立,即sinθ+sin(3θ)+...+sin[(2n-1)θ]=sin^2(nθ)/sinθ
则当N=n+1时
sinθ+sin(3θ)+...+sin{[2(n+1)-1]θ}=sin[(2n+1)θ]+sin^2(nθ)/sinθ
={sinθsin[(2n+1)θ]+sin^2(nθ)}/sinθ
={sinθ[sin(2nθ)cosθ+cos(2nθ)sinθ]+[1-cos(2nθ)]/2}/sinθ
={sin(2θ)sin(2nθ)/2+cos(2nθ)[1-cos(2θ)]/2+1/2-cos(2nθ)/2}/sinθ
=[sin(2θ)sin(2nθ)-cos(2θ)cos(2nθ)+1]/(2sinθ)
=[-cos(2nθ+2θ)+1]/(2sinθ)
={-cos[2(n+1)θ]+1}/(2sinθ)
=2sin^2[(n+1)θ]/(2sinθ)
=sin^2[(n+1)θ]/sinθ
所以当N=n时成立,则当N=n+1时也成立,所以原式成立.
求证sinθ/(1+cosθ)+(1+cosθ)/sinθ=2/sinθ
lim(cos^nθ-sin^nθ)/(cos^nθ+sin^nθ)
证明sin(pi/n)*sin(2pi/n)*sin(3pi/n)*…sin((n-1)pi/n)=n/(2^(n-1)
已知 z = cosθ+ i sinθ,求证 Im(z^n + 1/(z^n))=0
证明:(cos3θ+sin3θ)/(cosθ-sinθ) =1+2sin2θ
求证:(1+cosθ+cosθ/2) /(sinθ+sinθ/2)=sinθ/1-cosθ
求证 sinθ-sinφ=2cos[(θ+φ)/2]sin[(θ-φ)/2]
求证(1+sinθ+cosθ)/(1+sinθ-cosθ)+(1-cosθ+sinθ)/(1+cosθ+sinθ)=2/
求证(1-sinθcosθ)除以(cos^2θ-sin^2θ)=(cos^2θ-sin^2θ)除以(1+2sinθcos
求极限lim(1/n)*[(sin(pi/n)+sin(2pi/n)+.+sin(n*pi/n)] n->无穷
limn^2sin^2(θ/n) 当n趋近于无穷的极限怎么算出来的
级数(1/n) × sin(πn/2)的敛散性