三倍角公式推导
tan3α=sin3α/cos3α
=(sin2αcosα cos2αsinα)/(cos2αcosα-sin2αsinα)
=(2sinαcos^2(α) cos^2(α)sinα-sin^3(α))/(cos^3(α)-cosαsin^2(α)-2sin^2(α)cosα)
上下同除以cos^3(α)，得：
tan3α=(3tanα-tan^3(α))/(1-3tan^2(α))
sin3α=sin(2α α)=sin2αcosα cos2αsinα
=2sinαcos^2(α) (1-2sin^2(α))sinα
=2sinα-2sin^3(α) sinα-2sin^3(α)=3sinα-4sin^3(α)
cos3α=cos(2α α)=cos2αcosα-sin2αsinα
=(2cos^2(α)-1)cosα-2cosαsin^2(α)
=2cos^3(α)-cosα (2cosα-2cos^3(α))
=4cos^3(α)-3cosα
即
sin3α=3sinα-4sin^3(α)
cos3α=4cos^3(α)-3cosα
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