三倍角的函数公式是如何推导出来的?

1个回答

  • 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^2(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

    sin3α=3sinα-4sin^3(α)

    cos3α=4cos^3(α)-3cosα