作业帮求证sinθ+sin3θ+...+sin((2n-1)θ)=sin^2(nθ)/sinθ
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求证sinθ+sin3θ+...+sin((2n-1)θ)=sin^2(nθ)/sinθ
求证sinθ+sin3θ+...+sin((2n-1)θ)=sin^2(nθ)/sinθ
求证sinθ+sin3θ+...+sin((2n-1)θ)=sin^2(nθ)/sinθ
需要利用积化和差公式
∵ 2sinθ*[sinθ+sin3θ+...+sin((2n-1)θ)]
=2sinθsinθ+2sinθsin3θ+.+2sinθsin[(2n-1)θ]
=cos0-cos2θ+cos2θ-cos4θ+cos4θ-cos6θ+.+cos(2n-2θ)-cos(2nθ)
=cos0-cos(2nθ)
=1-cos(2nθ)
=2sin²(nθ)
∴ sinθ+sin3θ+...+sin((2n-1)θ)=sin²(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
=...
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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
收起
2sin(2k-1)osino=cos(2k-2)o-cos2ko,
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