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锐角三角函数公式sin α=∠α的对边 / 斜边cos α=∠α的邻边 / 斜边tan α=∠α的对边 / ∠α的邻边cot α=∠α的邻边 / ∠α的对边倍角公式Sin2A=2SinA?CosACos2A=CosA^2-SinA^2=1-2SinA^2=2CosA^2-1tan2A=(2tanA)/(1-tanA^2)(注：SinA^2 是sinA的平方 sin2(A) )三倍角公式sin3α=4sinα·sin(π/3+α)sin(π/3-α)cos3α=4cosα·cos(π/3+α)cos(π/3-α)tan3a = tan a · tan(π/3+a)· tan(π/3-a)三倍角公式推导sin3a=sin(2a+a)=sin2acosa+cos2asina辅助角公式Asinα+Bcosα=(A^2+B^2)^(1/2)sin(α+t)，其中sint=B/(A^2+B^2)^(1/2)cost=A/(A^2+B^2)^(1/2)tant=B/AAsinα+Bcosα=(A^2+B^2)^(1/2)cos(α-t)，tant=A/B降幂公式sin^2(α)=(1-cos(2α))/2=versin(2α)/2cos^2(α)=(1+cos(2α))/2=covers(2α)/2tan^2(α)=(1-cos(2α))/(1+cos(2α))推导公式tanα+cotα=2/sin2αtanα-cotα=-2cot2α1+cos2α=2cos^2α1-cos2α=2sin^2α1+sinα=(sinα/2+cosα/2)^2=2sina(1-sin2a)+(1-2sin2a)sina=3sina-4sin3acos3a=cos(2a+a)=cos2acosa-sin2asina=(2cos2a-1)cosa-2(1-sin2a)cosa=4cos3a-3cosasin3a=3sina-4sin3a=4sina(3/4-sin2a)=4sina[(√3/2)2-sin2a]=4sina(sin260°-sin2a)=4sina(sin60°+sina)(sin60°-sina)=4sina*2sin[(60+a)/2]cos[(60°-a)/2]*2sin[(60°-a)/2]cos[(60°-a)/2]=4sinasin(60°+a)sin(60°-a)cos3a=4cos3a-3cosa=4cosa(cos2a-3/4)=4cosa[cos2a-(√3/2)2]=4cosa(cos2a-cos230°)=4cosa(cosa+cos30°)(cosa-cos30°)=4cosa*2cos[(a+30°)/2]cos[(a-30°)/2]*{-2sin[(a+30°)/2]sin[(a-30°)/2]}=-4cosasin(a+30°)sin(a-30°)=-4cosasin[90°-(60°-a)]sin[-90°+(60°+a)]=-4cosacos(60°-a)[-cos(60°+a)]=4cosacos(60°-a)cos(60°+a)上述两式相比可得tan3a=tanatan(60°-a)tan(60°+a)半角公式tan(A/2)=(1-cosA)/sinA=sinA/(1+cosA);cot(A/2)=sinA/(1-cosA)=(1+cosA)/sinA.sin^2(a/2)=(1-cos(a))/2cos^2(a/2)=(1+cos(a))/2tan(a/2)=(1-cos(a))/sin(a)=sin(a)/(1+cos(a))学习方法网[www.xuexifangfa.com]三角和sin(α+β+γ)=sinα·cosβ·cosγ+cosα·sinβ·cosγ+cosα·cosβ·sinγ-sinα·sinβ·sinγcos(α+β+γ)=cosα·cosβ·cosγ-cosα·sinβ·sinγ-sinα·cosβ·sinγ-sinα·sinβ·cosγtan(α+β+γ)=(tanα+tanβ+tanγ-tanα·tanβ·tanγ)/(1-tanα·tanβ-tanβ·tanγ-tanγ·tanα)两角和差cos(α+β)=cosα·cosβ-sinα·sinβcos(α-β)=cosα·cosβ+sinα·sinβsin(α±β)=sinα·cosβ±cosα·sinβtan(α+β)=(tanα+tanβ)/(1-tanα·tanβ)tan(α-β)=(tanα-tanβ)/(1+tanα·tanβ)和差化积sinθ+sinφ = 2 sin[(θ+φ)/2] cos[(θ-φ)/2]sinθ-sinφ = 2 cos[(θ+φ)/2] sin[(θ-φ)/2]cosθ+cosφ = 2 cos[(θ+φ)/2] cos[(θ-φ)/2]cosθ-cosφ = -2 sin[(θ+φ)/2] sin[(θ-φ)/2]tanA+tanB=sin(A+B)/cosAcosB=tan(A+B)(1-tanAtanB)tanA-tanB=sin(A-B)/cosAcosB=tan(A-B)(1+tanAtanB)积化和差sinαsinβ = [cos(α-β)-cos(α+β)] /2cosαcosβ = [cos(α+β)+cos(α-β)]/2sinαcosβ = [sin(α+β)+sin(α-β)]/2cosαsinβ = [sin(α+β)-sin(α-β)]/2诱导公式sin(-α) = -sinαcos(-α) = cosαtan (—a)=-tanαsin(π/2-α) = cosαcos(π/2-α) = sinαsin(π/2+α) = cosαcos(π/2+α) = -sinαsin(π-α) = sinαcos(π-α) = -cosαsin(π+α) = -sinαcos(π+α) = -cosαtanA= sinA/cosAtan(π/2+α)=-cotαtan(π/2-α)=cotαtan(π-α)=-tanαtan(π+α)=tanα诱导公式记背诀窍：奇变偶不变，符号看象限万能公式sinα=2tan(α/2)/[1+tan^(α/2)]cosα=[1-tan^(α/2)]/1+tan^(α/2)]tanα=2tan(α/2)/[1-tan^(α/2)]其它公式(1)(sinα)^2+(cosα)^2=1(2)1+(tanα)^2=(secα)^2(3)1+(cotα)^2=(cscα)^2证明下面两式,只需将一式,左右同除(sinα)^2,第二个除(cosα)^2即可(4)对于任意非直角三角形,总有tanA+tanB+tanC=tanAtanBtanC证:A+B=π-Ctan(A+B)=tan(π-C)(tanA+tanB)/(1-tanAtanB)=(tanπ-tanC)/(1+tanπtanC)整理可得tanA+tanB+tanC=tanAtanBtanC得证同样可以得证,当x+y+z=nπ(n∈Z)时,该关系式也成立由tanA+tanB+tanC=tanAtanBtanC可得出以下结论(5)cotAcotB+cotAcotC+cotBcotC=1(6)cot(A/2)+cot(B/2)+cot(C/2)=cot(A/2)cot(B/2)cot(C/2)(7)(cosA)^2+(cosB)^2+(cosC)^2=1-2cosAcosBcosC(8)(sinA)^2+(sinB)^2+(sinC)^2=2+2cosAcosBcosC(9)sinα+sin(α+2π/n)+sin(α+2π*2/n)+sin(α+2π*3/n)+……+sin[α+2π*(n-1)/n]=0cosα+cos(α+2π/n)+cos(α+2π*2/n)+cos(α+2π*3/n)+……+cos[α+2π*(n-1)/n]=0 以及sin^2(α)+sin^2(α-2π/3)+sin^2(α+2π/3)=3/2tanAtanBtan(A+B)+tanA+tanB-tan(A+B)=0
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