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)...