高一數(shù)學(xué)三角函數(shù)公式匯總(參考版)
2025-04-14 03:44本頁(yè)面
【正文】 a)tan(60176。a)cos(60176。a)[cos(60176。+(60176。(60176。)sin(a30176。)/2]sin[(a30176。)/2]cos[(a30176。)(cosacos30176。a) cos3a=4cosa3cosa =4cosa(cosa3/4) =4cosa[cosa(√3/2)] =4cosa(cosacos30176。a)/2] =4sinasin(60176。a)/2]*2sin[(60176。+sina)(sin60176?！ an3α=sin3α/cos3α =(sin2αcosα+cos2αsinα)/(cos2αcosαsin2αsinα) =(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))/(13tan^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 首先,我們知道sin(a+b)=sina*cosb+cosa*sinb,sin(ab)=sina*cosbcosa*sinb 我們把兩式相加就得到sin(a+b)+
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