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
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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
![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](/uploads/image/z/17824256-8-6.jpg?t=n%E5%A4%A7%E4%BA%8E%E7%AD%89%E4%BA%8E1%2C%CE%B8%E4%B8%8D%E6%98%AF%CF%80%E7%9A%84%E5%80%8D%E6%95%B0%2C%E6%B1%82%E8%AF%81sin%CE%B8%2Bsin3%CE%B8%2B...%2Bsin%28%282n-1%29%CE%B8%29%3Dsin%5E2%28n%CE%B8%29%2Fsin%CE%B8)
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.
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