求高中三角函数所有课本要求的公式 ~~~~

复习用的
2025-02-20 13:48:28
推荐回答(2个)
回答1:

三角函数内容规律

  三角函数看似很多,很复杂,但只要掌握了三角函数的本质及内部规律就会发现三角函数各个公式之间有强大的联系。而掌握三角函数的内部规律及本质也是学好三角函数的关键所在.

  1、三角函数本质:

  三角函数的本质来源于定义,如下图:

  根据右图,有

  sinθ=y/ R; cosθ=x/R; tanθ=y/x; cotθ=x/y。

  深刻理解了这一点,下面所有的三角公式都可以从这里出发推导出来,比如以推导

  sin(A+B) = sinAcosB+cosAsinB 为例:

  推导:

  首先画单位圆交X轴于C,D,在单位圆上有任意A,B点。角AOD为α,BOD为β,旋转AOB使OB与OD重合,形成新A'OD。

  A(cosα,sinα),B(cosβ,sinβ),A'(cos(α-β),sin(α-β))

  OA'=OA=OB=OD=1,D(1,0)

  ∴[cos(α-β)-1]^2+[sin(α-β)]^2=(cosα-cosβ)^2+(sinα-sinβ)^2

  和差化积及积化和差用还原法结合上面公式可推出(换(a+b)/2与(a-b)/2)

  [1] 

  两角和公式

  sin(A+B) = sinAcosB+cosAsinB 

  sin(A-B) = sinAcosB-cosAsinB � 

  cos(A+B) = cosAcosB-sinAsinB 

  cos(A-B) = cosAcosB+sinAsinB 

  tan(A+B) = (tanA+tanB)/(1-tanAtanB) 

  tan(A-B) = (tanA-tanB)/(1+tanAtanB) 

  cot(A+B) = (cotAcotB-1)/(cotB+cotA) � 

  cot(A-B) = (cotAcotB+1)/(cotB-cotA)

[编辑本段]倍角公式

  Sin2A=2SinA•CosA

  Cos2A=CosA^2-SinA^2=1-2SinA^2=2CosA^2-1

  tan2A=2tanA/(1-tanA^2)

  (注:SinA^2 是sinA的平方 sin2(A) )

[编辑本段]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)

[编辑本段]三倍角公式推导

  sin3a

  =sin(2a+a)

  =sin2acosa+cos2asina

  =2sina(1-sin²a)+(1-2sin²a)sina

  =3sina-4sin³a

  cos3a

  =cos(2a+a)

  =cos2acosa-sin2asina

  =(2cos²a-1)cosa-2(1-sin²a)cosa

  =4cos³a-3cosa

  sin3a=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)

[编辑本段]半角公式

  tan(A/2)=(1-cosA)/sinA=sinA/(1+cosA);

  cot(A/2)=sinA/(1-cosA)=(1+cosA)/sinA.

[编辑本段]和差化积

  sinθ+sinφ = 2sin[(θ+φ)/2]cos[(θ-φ)/2] 

  sinθ-sinφ = 2cos[(θ+φ)/2]sin[(θ-φ)/2] 

  cosθ+cosφ = 2cos[(θ+φ)/2]cos[(θ-φ)/2] 

  cosθ-cosφ = -2sin[(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)

[编辑本段]积化和差

  sinαsinβ = -1/2*[cos(α+β)-cos(α-β)] 

  cosαcosβ = 1/2*[cos(α+β)+cos(α-β)] 

  sinαcosβ = 1/2*[sin(α+β)+sin(α-β)]

  cosαsinβ = 1/2*[sin(α+β)-sin(α-β)]

[编辑本段]诱导公式

  sin(-α) = -sinα

  cos(-α) = cosα

  sin(π/2-α) = cosα 

  cos(π/2-α) = sinα

  sin(π/2+α) = cosα 

  cos(π/2+α) = -sinα

  sin(π-α) = sinα

  cos(π-α) = -cosα

  sin(π+α) = -sinα 

  cos(π+α) = -cosα 

  tanA= sinA/cosA

  tan(π/2+α)=-cotα 

  tan(π/2-α)=cotα 

  tan(π-α)=-tanα 

  tan(π+α)=tanα

[编辑本段]万能公式

  

[编辑本段]其它公式

  (sinα)^2+(cosα)^2=1

  1+(tanα)^2=(secα)^2

  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

  对于任意非直角三角形,总有

  tanA+tanB+tanC=tanAtanBtanC

  证:

  A+B=π-C

  tan(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

[编辑本段]其他非重点三角函数

  csc(a) = 1/sin(a) 

  sec(a) = 1/cos(a)

  

[编辑本段]双曲函数

  sinh(a) = [e^a-e^(-a)]/2 

  cosh(a) = [e^a+e^(-a)]/2 

  tg h(a) = sin h(a)/cos h(a)

  公式一: 

  设α为任意角,终边相同的角的同一三角函数的值相等: 

  sin(2kπ+α)= sinα 

  cos(2kπ+α)= cosα 

  tan(kπ+α)= tanα 

  cot(kπ+α)= cotα 

  公式二: 

  设α为任意角,π+α的三角函数值与α的三角函数值之间的关系: 

  sin(π+α)= -sinα 

  cos(π+α)= -cosα 

  tan(π+α)= tanα 

  cot(π+α)= cotα 

  公式三: 

  任意角α与 -α的三角函数值之间的关系: 

  sin(-α)= -sinα 

  cos(-α)= cosα 

  tan(-α)= -tanα 

  cot(-α)= -cotα 

  公式四: 

  利用公式二和公式三可以得到π-α与α的三角函数值之间的关系: 

  sin(π-α)= sinα 

  cos(π-α)= -cosα 

  tan(π-α)= -tanα 

  cot(π-α)= -cotα 

  公式五: 

  利用公式-和公式三可以得到2π-α与α的三角函数值之间的关系: 

  sin(2π-α)= -sinα 

  cos(2π-α)= cosα 

  tan(2π-α)= -tanα 

  cot(2π-α)= -cotα 

  公式六: 

  π/2±α及3π/2±α与α的三角函数值之间的关系: 

  sin(π/2+α)= cosα 

  cos(π/2+α)= -sinα

  tan(π/2+α)= -cotα 

  cot(π/2+α)= -tanα 

  sin(π/2-α)= cosα 

  cos(π/2-α)= sinα 

  tan(π/2-α)= cotα 

  cot(π/2-α)= tanα 

  sin(3π/2+α)= -cosα 

  cos(3π/2+α)= sinα 

  tan(3π/2+α)= -cotα 

  cot(3π/2+α)= -tanα 

  sin(3π/2-α)= -cosα 

  cos(3π/2-α)= -sinα 

  tan(3π/2-α)= cotα 

  cot(3π/2-α)= tanα 

  (以上k∈Z) 

A·sin(ωt+θ)+ B·sin(ωt+φ) =

  √{(A^2 +B^2 +2ABcos(θ-φ)} • sin{ ωt + arcsin[ (A•sinθ+B•sinφ) / √{A^2 +B^2; +2ABcos(θ-φ)} }

  √表示根号,包括{……}中的内容

回答2:

其实买一本好的参考书是很有必要的,我今年用 《五年高考三年模拟》,里面要考要点都有了,我数学也得了135,所以推荐给你

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