• 三角恒等式


    两角和与差

    cos(α+β)=cosα·cosβ-sinα·sinβ

    cos(α-β)=cosα·cosβ+sinα·sinβ

    sin(α±β)=sinα·cosβ±cosα·sinβ

    tan(α+β)=(tanα+tanβ)/(1-tanα·tanβ)

    tan(α-β)=(tanα-tanβ)/(1+tanα·tanβ)

    和差化积

    sinα+sinβ=2sin[(α+β)/2]cos[(α-β)/2]

    sinα-sinβ=2cos[(α+β)/2]sin[(α-β)/2]

    cosα+cosβ=2cos[(α+β)/2]cos[(α-β)/2]

    cosα-cosβ=-2sin[(α+β)/2]sin[(α-β)/2]

    积化和差

    sinα·cosβ=(1/2)[sin(α+β)+sin(α-β)]

    cosα·sinβ=(1/2)[sin(α+β)-sin(α-β)]

    cosα·cosβ=(1/2)[cos(α+β)+cos(α-β)]

    sinα·sinβ=-(1/2)[cos(α+β)-cos(α-β)]

    倍角公式

    sin(2α)=2sinα·cosα=2/(tanα+cotα)

    cos(2α)=(cosα)^2-(sinα)^2=2(cosα)^2-1=1-2(sinα)^2

    tan(2α)=2tanα/[1-(tanα)^2]

    cot(2α)=(cot²α-1)/(2cotα)

    sec(2α)=sec²α/(1-tan²α)

    csc(2α)=1/2*secα·cscα

    三倍角公式

    sin(3α) = 3sinα-4sin³α = 4sinα·sin(60°+α)sin(60°-α)

    cos(3α) = 4cos³α-3cosα = 4cosα·cos(60°+α)cos(60°-α)

    tan(3α) = (3tanα-tan³α)/(1-3tan²α) = tanαtan(π/3+α)tan(π/3-α)

    cot(3α)=(cot³α-3cotα)/(3cotα-1)

    n倍角公式

    根据欧拉公式(cosθ+isinθ)^n=cosnθ+isinnθ

    将左边用二项式定理展开分别整理实部和虚部可以得到下面两组公式

    sin(nα)=ncos^(n-1)α·sinα-C(n,3)cos^(n-3)α·sin^3α+C(n,5)cos^(n-5)α·sin^5α-…

    cos(nα)=cos^nα-C(n,2)cos^(n-2)α·sin^2α+C(n,4)cos^(n-4)α·sin^4α

    半角公式

    sin(α/2)=±√[(1-cosα)/2]

    cos(α/2)=±√[(1+cosα)/2]

    tan(α/2)=±√[(1-cosα)/(1+cosα)]=sinα/(1+cosα)=(1-cosα)/sinα=cscα-cotα

    cot(α/2)=±√[(1+cosα)/(1-cosα)]=(1+cosα)/sinα=sinα/(1-coα)=cscα+cotα

    sec(α/2)=±√[(2secα/(secα+1)]

    csc(α/2)=±√[(2secα/(secα-1)]

    辅助角公式

    Asinα+Bcosα=√(A^2+B^2)sin[α+arctan(B/A)]

    Asinα-Bcosα=√(A^2+B^2)cos[α-arctan(A/B)]

    万能公式

    sin(a)=[2tan(a/2)]/[1+tan^2(a/2)]

    cos(a)=[1-tan^2(a/2)]/[1+tan^2(a/2)]

    tan(a)=[2tan(a/2)]/[1-tan^2(a/2)]

    降幂公式

    sin^2α=[1-cos(2α)]/2

    cos^2α=[1+cos(2α)]/2=1-sin(2α)/2

    tan^2α=[1-cos(2α)]/[1+cos(2α)]

    三角和

    sin(α+β+γ)=sinα·cosβ·cosγ+cosα·sinβ·cosγ+cosα·cosβ·sinγ-sinα·sinβ·sinγ

    cos(α+β+γ)=cosα·cosβ·cosγ-cosα·sinβ·sinγ-sinα·cosβ·sinγ-sinα·sinβ·cosγ

    tan(α+β+γ)=(tanα+tanβ+tanγ-tanα·tanβ·tanγ)÷(1-tanα·tanβ-tanβ·tanγ-tanγ·tanα)

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  • 原文地址:https://www.cnblogs.com/swtool/p/3840242.html
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