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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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