2sinθsin(θ/2)=cos(θ/2)-cos(3θ/2)
2sin2θsin(θ/2)=cos(3θ/2)-cos(5θ/2)
2sin3θsin(θ/2)=cos(5θ/2)-cos(7θ/2)
2sin4θsin(θ/2)=cos(7θ/2)-cos(9θ/2)
……
2sinnθsin(θ/2)=cos(nθ-θ/2)-cos(nθ+θ/2)
相加得到，
2sin(θ/2)[sinθ+sin2θ+……+sinnθ]=cos(θ/2)-cos(nθ+θ/2)
∴ sinθ+sin2θ+……+sinnθ
=[cos(θ/2)-cos(nθ+θ/2)]/[2sin(θ/2)]
最后等于0
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