六種三角函數(shù)性質(zhì)(編輯修改稿)
2025-06-12 01:05 本頁(yè)面
【文章內(nèi)容簡(jiǎn)介】 2*secαcscα三倍角公式：sin(3α) = 3sinα4sin^3α = 4sinαsin(60176。+α)sin(60176。α)cos(3α) = 4cos^3α3cosα = 4cosαcos(60176。+α)cos(60176。α)tan(3α) = (3tanαtan^3α)/(13tan^2α) = tanαtan(π/3+α)tan(π/3α)cot(3α)=(cot^3α3cotα)/(3cot^2α1)n倍角公式：sin(nα)=ncos^(n1)αsinαC(n,3)cos^(n3)αsin^3α+C(n,5)cos^(n5)αsin^5α…cos(nα)=cos^nαC(n,2)cos^(n2)αsin^2α+C(n,4)cos^(n4)αsin^4α…半角公式[/url]：sin(α/2)=177?！?(1cosα)/2)cos(α/2)=177。√((1+cosα)/2)tan(α/2)=177。√((1cosα)/(1+cosα))=sinα/(1+cosα)=(1cosα)/sinαcot(α/2)=177。√((1+cosα)/(1cosα))=(1+cosα)/sinα=sinα/(1cosα)sec(α/2)=177?！?(2secα/(secα+1))csc(α/2)=177?！?(2secα/(secα1))輔助角公式：Asinα+Bcosα=√(A^2+B^2)sin(α+φ）（tanφ=B/A）Asinα+Bcosα=√(A^2+B^2)cos(αφ）（tanφ=A/B）萬(wàn)能公式sin(a)= (2tan(a/2))/(1+tan^2(a/2))cos(a)= (1tan^2(a/2))/(1+tan^2(a/2))tan(a)= (2tan(a/2))/(1tan^2(a/2))降冪公式sin^2α=(1cos(2α))/2=versin(2α)/2cos^2α=(1+cos(2α))/2=covers(2α)/2tan^2α=(1cos(2α))/(1+cos(2α))三角和的三角函數(shù)：sin(α+β+γ)=sinαcosβcosγ+cosαsinβcosγ+cosαcosβsinγsinαsinβsinγcos(α+β+γ)=cosαcosβcosγcosαsinβsinγsinαcosβsinγsinαsinβcosγtan(α+β+γ)=(tanα+tanβ+tanγtanαtanβtanγ)/(1tanαtanβtanβ1
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