高二數(shù)學(xué)教案：三角函數(shù)兩角和公式(編輯修改稿)
2024-10-11 19:57 本頁(yè)面
【文章內(nèi)容簡(jiǎn)介】 osa(cos178。acos178。30176。)=4cosa(cosa+cos30176。)(cosacos30176。)=4cosa*2cos[(a+30176。)/2]cos[(a30176。)/2]*{2sin[(a+30176。)/2]sin[(a30176。)/2]}=4cosasin(a+30176。)sin(a30176。)=4cosasin[90176。(60176。a)]sin[90176。+(60176。+a)]=4cosacos(60176。a)[cos(60176。+a)]=4cosacos(60176。a)cos(60176。+a)上述兩式相比可得tan3a=tanatan(60176。a)tan(60176。+a)半角公式tan(A/2)=(1cosA)/sinA=sinA/(1+cosA)。cot(A/2)=sinA/(1cosA)=(1+cosA)/^2(a/2)=(1cos(a))/2cos^2(a/2)=(1+cos(a))/2tan(a/2)=(1cos(a))/sin(a)=sin(a)/(1+cos(a))三角和sin(α+β+γ)=sinαcosβcosγ+cosαsinβcosγ+cosαcosβsinγsinαsinβsinγcos(α+β+γ)=cosαcosβcosγcosαsinβsinγsinαcosβsinγsinαsinβcosγtan(α+β+γ)=(tanα+tanβ+tanγtanαtanβtanγ)/(1tanαtanβtanβtanγtanγtanα)兩角和差cos(α+β)=cosαcosβsinαsinβcos(αβ)=cosαcosβ+sinαsinβsin(α177。β)=sinαcosβ177。cosαsinβtan(α+β)=(tanα+tanβ)/(1tanαtanβ)tan(αβ)=(tanαtanβ)/(1+tanαtanβ)和差化積sinθ+sinφ = 2 sin[(θ+φ)/2] cos[(θφ)/2]sinθsinφ = 2 cos[(θ+φ)/2] sin[(θφ)/2]cosθ+cosφ = 2 cos[(θ+φ)/2] cos[(θφ)/2]cosθcosφ =2 sin[(θ+φ)/2] sin[(θφ)/2]tanA+tanB=sin(A+B)/cosAcosB=tan(A+B)(1tanAtanB)tanAtanB=sin(AB)/cosAcosB=tan(AB)(1+tanAtanB)積化和差sinαsinβ = [cos(αβ)cos(α+β)] /2cosαcosβ = [cos(α+β)+cos(αβ)]/2sinαcosβ = [sin(α+β)+sin(αβ)]/2cosαsinβ = [sin(α+β)sin(αβ)]/2誘導(dǎo)公式sin(α)=sinαcos(α)= cosαtan(—a)=tanαsin(π/2α)= cosαcos(π/2α)= sinαsin(π/2+α)= cosαcos(π/2+α)=sinαsin(πα)= sinαcos(πα)=cosαsin(π+α)=sinαcos(π+α)=cosαtanA= sinA/cosAtan（π/2＋α）＝－cotαtan（π/2－α）＝cotαtan（π－α）＝－tanαtan（π＋α）＝tanα誘導(dǎo)公式記背訣竅：奇變偶不變，符號(hào)看象限萬能公式sinα=2tan(α/2）/［1+tan＾(α/2)］cosα=［1tan＾(α/2)］/1+tan＾(α/2)］tanα=2tan(α/2)/［1tan＾(α/2)］其它公式(1)(sinα)^2+(cosα)^2=1(2)1+(tanα)^2=(secα)^2(3)1+(cotα)^2=(cscα)^2證明下面兩式,只需將一式,左右同除(sinα)^2,第二個(gè)除(cosα)^2即可(4)對(duì)于任意非直角三角形,總有tanA+tanB+tanC=tanAtanBtanC證:A+B=πCtan(A+B)=tan(πC)(tanA+tanB)/(1tanAtanB)=(tanπtanC)/(1+tanπtanC)整理可得tanA+tanB+tanC=tanAtanBtanC得證同樣可以得證,當(dāng)x+y+z=nπ(n∈Z)時(shí),該關(guān)系式也成立由tanA+tanB+tanC=tanAtanBtanC可得出以下結(jié)論(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=12cosAcosBcosC(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*(n1)/n]=0cosα+cos(α+2π/n)+cos(α+2π*2/n)+cos(α+2π*3/n)+……+cos[α+2π*(n1)/n]=0 以及sin^2(α)+sin^2(α2π/3)+sin^2(α+2π/3)=3/2tanAtanBtan(A+B)+tanA+tanBtan(A+B)=0第四篇：高中數(shù)學(xué)三角函數(shù)公式兩角和公式sin(A+B)= sinAcosB+cosAsinBsin(AB)= sinAcosBcosAsinBcos(A+B)= cosAcosBsinAsinBcos(AB)= cosAcosB+sinAsinBtan(A+B)=(tanA+tanB)/(1tanAtanB)tan(AB)=(tanAtanB)/(1+tanAtanB)倍角公式tan2A = 2tanA/(1tan^2 A)Sin2A=2SinA?CosACos2A = Cos^2 ASin^2 A=2Cos^2 A—1=1—2sin^2 A 三倍角公式sin
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