【正文】 (2α))/2=covers(2α)/2tan^2(α)=(1cos(2α))/(1+cos(2α))推導(dǎo)公式tanα+cotα=2/sin2αtanαcotα=2cot2α1+cos2α=2cos^2α1cos2α=2sin^2α1+sinα=(sinα/2+cosα/2)^2=2sina(1sin178。a)/2]*2sin[(60176。)/2]sin[(a30176。cosγ+cosαtanγ)/(1tanα tan(π/3a)半角公式sin(A/2)= √{(1cosA)/2}cos(A/2)= √{(1+cosA)/2}tan(A/2)= √{(1cosA)/(1+cosA)}tan(A/2)=(1cosA)/sinA=sinA/(1+cosA)和差化積sin(a)+sin(b)= 2sin[(a+b)/2]cos[(ab)/2]sin(a)sin(b)= 2cos[(a+b)/2]sin[(ab)/2]cos(a)+cos(b)= 2cos[(a+b)/2]cos[(ab)/2]cos(a)cos(b)=2sin[(a+b)/2]sin[(ab)/2] tanA+tanB=sin(A+B)/cosAcosB 積化和差sin(a)sin(b)=1/2*[cos(a+b)cos(ab)]cos(a)cos(b)= 1/2*[cos(a+b)+cos(ab)]sin(a)cos(b)= 1/2*[sin(a+b)+sin(ab)] cos(a)sin(b)= 1/2*[sin(a+b)sin(ab)] 誘導(dǎo)公式sin(a)=sin(a)cos(a)= cos(a)sin(π/2a)= cos(a)cos(π/2a)= sin(a)sin(π/2+a)= cos(a)cos(π/2+a)=sin(a)sin(πa)= sin(a)cos(πa)=cos(a)sin(π+a)=sin(a)cos(π+a)=cos(a)tanA = sinA/cosA 萬(wàn)能公式sin(a)= [2tan(a/2)] / {1+[tan(a/2)]^2}cos(a)= {1[tan(a/2)]^2} / {1+[tan(a/2)]^2} tan(a)= [2tan(a/2)]/{1[tan(a/2)]^2}其它公式a計(jì)算證明角先行，注意結(jié)構(gòu)函數(shù)名，保持基本量不變，繁難向著簡(jiǎn)易變。(3)1+(cotα)178。(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)cos178。條件等式的證明，方程思想指路明。cos(a)= [√(a^2+b^2)]*sin(a+c)[其中，tan(c)=b/a]atanγtanγcosγ+cosα)sin(a30176。a)/2]=4sinasin(60176。a)sina成都家教濟(jì)南家教=3sina4sin179。a)cosa=4cos179。a3cosa=4cosa(cos178。a)]sin[90176。sinβsinβcos(αβ)=cosα1sin(a)= [sin(a/2)cos(a/2)]^2。山西鐵路工程建設(shè)監(jiān)理有限公司劉榮申第四篇：高中數(shù)學(xué)反三角函數(shù)的公式小結(jié)高中數(shù)學(xué)反三角函數(shù)的公式小結(jié)反三角函數(shù)主要是三個(gè)：y＝arcsin(x)，定義域[1,1]，值域[π/2,π/2]圖象用紅色線條；y=arccos(x)，定義域[1,1]，值域[0,π