sin3α=[e^(i3α)-e^(-i3α)]/(2i)=｛[e^(iα)]^3-[e^(-iα)]^3｝/(2i)=[e^(iα)-e^(-iα)][e^(i2α)＋e^(-i2α)＋1]/(2i)=[2isinα][2cos(2α)＋1]/(2i)=sinα｛2[1-2(sinα)^2]＋1｝=sinα[3-4(sinα)^2]=4sinα[(√3/2)^2-(sinα)^2]=4sinα[(sinπ/3)^2-(sinα)^2]，即sin3α=4sinα[sin(π/3)-sinα][sin(π/3)＋sinα]=4sinα[2cos(π/6＋α/2)sin(π/6-α/2)][2sin(π/6＋α/2)cos(π/6-α/2)]=4sinα[2sin(π/6＋α/2)cos(π/6＋α/2)][2sin(π/6-α/2)cos(π/6-α/2)]，即sin3α=4sinαsin(π/3＋α)sin(π/3-α)。
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