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)...
求证sinθ+sin3θ+...+sin((2n-1)θ)=sin^2(nθ)/sinθ
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