初三数学复习：三倍角公式推导过程
<p> tan3α=sin3α/cos3α</p><p>=(sin2αcosα+cos2αsinα)/(cos2αcosα-sin2αsinα)</p><p>=(2sinαcos^2(α)+cos^2(α)sinα-sin^3(α))/(cos^3(α)-cosαsin^2(α)-2sin^2(α)cosα)</p><p>上下同除以cos^3(α)，得：</p><p>tan3α=(3tanα-tan^3(α))/(1-3tan^2(α))</p><p>sin3α=sin(2α+α)=sin2αcosα+cos2αsinα</p><p>=2sinαcos^2(α)+(1-2sin^2(α))sinα</p><p>=2sinα-2sin^3(α)+sinα-2sin^3(α)</p><p>=3sinα-4sin^3(α)</p><p>cos3α=cos(2α+α)=cos2αcosα-sin2αsinα</p><p>=(2cos^2(α)-1)cosα-2cosαsin^2(α)</p><p>=2cos^3(α)-cosα+(2cosα-2cos^3(α))</p><p>=4cos^3(α)-3cosα</p><p>即</p><p>sin3α=3sinα-4sin^3(α)</p><p>cos3α=4cos^3(α)-3cosα</p>
页:
[1]