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](/uploads/image/z/17824255-7-5.jpg?t=%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)
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
∵ 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
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