高中數(shù)學(xué)公式大全

高中數(shù)學(xué)公式大全

高中數(shù)學(xué)公式大全，包括文科必背數(shù)學(xué)公式整理和重點(diǎn)數(shù)學(xué)公式大全。乘法與因式分：a^2-b^2=(a+b)(a-b)a^3+b^3=(a+b)(a^2-ab+b^2)a^3-b^3=(a-b)(a^2+ab+b^2)三角不等式：|a+b|≤|a|+|b||a-b|≤|a|+|b||a|≤b<=>-b≤a≤b|a-b|≥|a|-|b|-|a|≤a≤|a|一元二次方程的解：(-b+√(b^2-4ac))/2a(-b-√(b^2-4ac))/2a根與系數(shù)的關(guān)系：X1+X2=-b/aX1*X2=c/a注：韋達(dá)定理判別式：b^2-4ac=0注：方程有兩個(gè)相等的實(shí)根b^2-4ac>0注：方程有兩個(gè)不等的實(shí)根b^2-4ac<0注：方程沒(méi)有實(shí)根，有共軛復(fù)數(shù)根三角函數(shù)公式：兩角和公式：sin(A+B)=sinAcosB+cosAsinBsin(A-B)=sinAcosB-sinBcosAcos(A+B)=cosAcosB-sinAsinBcos(A-B)=cosAcosB+sinAsinBtan(A+B)=(tanA+tanB)/(1-tanAtanB)tan(A-B)=(tanA-tanB)/(1+tanAtanB)ctg(A+B)=(ctgActgB-1)/(ctgB+ctgA)ctg(A-B)=(ctgActgB+1)/(ctgB-ctgA)倍角公式：tan2A=2tanA/(1-tan^2A)ctg2A=(ctg^2A-1)/2ctgacos2a=cos^2a-sin^2a=2cos^2a-1=1-2sin^2a半角公式：sin(A/2)=√((1-cosA)/2)sin(A/2)=-√((1-cosA)/2)cos(A/2)=√((1+cosA)/2)cos(A/2)=-√((1+cosA)/2)tan(A/2)=√((1-cosA)/((1+cosA))tan(A/2)=-√((1-cosA)/((1+cosA))ctg(A/2)=√((1+cosA)/((1-cosA))ctg(A/2)=-√((1+cosA)/((1-cosA))和差化積：2sinAcosB=sin(A+B)+sin(A-B)2cosAsinB=sin(A+B)-sin(A-B)2cosAcosB=cos(A+B)-sin(A-B)-2sinAsinB=cos(A+B)-cos(A-B)sin(A+B)=2sin((A+B)/2)cos((A-B)/2)cos(A+B)=2cos((A+B)/2)sin((A-B)/2)tan(A+B)=sin(A+B)/(cosAcosB)tan(A-B)=sin(A-B)/(cosAcosB)ctg(A+B)=sin(A+B)/sinAsinBctg(A-B)=sin(A-B)/sinAsinB一些數(shù)列前n項(xiàng)和：1+2+3+4+5+6+7+8+9+...+n=n(n+1)/21+3+5+7+9+11+13+15+...+(2n-1)=n^22+4+6+8+10+12+14+...+2n=n(n+1)1^2+2^2+3^2+4^2+5^2+6^2+...+n^2=n(n+1)(2n+1)/61+2^2+3^2+4^2+5^2+6^2+...+n^2=n(n+1)(n+2)/3正弦定理：a/sinA=b/sinB=c/sinC=2R（其中R表示三角形的外接圓半徑）余弦定理：b^2=a^2+c^2-2accosB（其中角B是邊a和邊c的夾角）圓的標(biāo)準(zhǔn)方程：(x-a)^2+(y-b)^2=r^2（其中(a,b)是圓心坐標(biāo)）圓的一般方程：x^2+y^2+Dx+Ey+F=0（其中D^2+E^2-4F>0）拋物線標(biāo)準(zhǔn)方程：y^2=2px,y^2=-2px,x^2=2py,x^2=-2py直棱柱側(cè)面積：S=c*h，斜棱柱側(cè)面積：S=c'*h正棱錐側(cè)面積：S=1/2c*h'，正棱臺(tái)側(cè)面積：S=1/2(c+c')h'圓臺(tái)側(cè)面積：S=1/2(c+c')l=pi(R+r)l，球的表面積：S=4pi*r^2圓柱側(cè)面積：S=c*h=2pi*h，圓錐側(cè)面積：S=1/2c*l=pi*r*l弧長(zhǎng)公式：l=a*r（其中a是圓心角的弧度數(shù)，r>0）扇形面積公式：s=1/2*l*r錐體體積公式：V=1/3*S*H，圓錐體體積公式：V=1/3*pi*r^2h斜棱柱體積：V=S'L（其中S'是直截面面積，L是側(cè)棱長(zhǎng)）柱體體積公式：V=s*h，圓柱體體積：V=pi*r^2h高中文科數(shù)學(xué)必背公式總結(jié)：公式一：設(shè)α為任意角，終邊相同的角的同一三角函數(shù)的值相等：sin(2kπ+α)=sinα(k∈Z)cos(2kπ+α)=cosα(k∈Z)tan(2kπ+α)=tanα，其中k為整數(shù)。cot(2kπ+α)=cotα，其中k為整數(shù)。對(duì)于任意角α，有π+α的三角函數(shù)值與α的三角函數(shù)值之間的關(guān)系：sin(π+α)=-sinαcos(π+α)=-cosαtan(π+α)=tanαcot(π+α)=cotα任意角α與其相反數(shù)-α的三角函數(shù)值之間的關(guān)系為：sin(-α)=-sinαcos(-α)=cosαtan(-α)=-tanαcot(-α)=-cotα利用公式二和公式三可以得到π-α與α的三角函數(shù)值之間的關(guān)系：sin(π-α)=sinαcos(π-α)=-cosαtan(π-α)=-tanαcot(π-α)=-cotα利用公式一和公式三可以得到2π-α與α的三角函數(shù)值之間的關(guān)系：sin(2π-α)=-sinαcos(2π-α)=cosαtan(2π-α)=-tanαcot(2π-α)=-cotα對(duì)于π/2±α及3π/2±α與α，它們的三角函數(shù)值之間的關(guān)系如下：sin(π/2+α)=cosαcos(π/2+α)=-sinαtan(π/2+α)=-cotαcot(π/2+α)=-tanαsin(π/2-α)=cosαcos(π/2-α)=sinαtan(π/2-α)=cotαcot(π/2-α)=tanαsin(3π/2+α)=-cosαcos(3π/2+α)=sinαtan(3π/2+α)=-cotαcot(3π/2+α)=-tanαsin(3π/2-α)=-cosαcos(3π/2-α)=-sinαtan(3π/2-α)=cotαcot(3π/2-α)=tanα（其中k為整數(shù)）兩角和與差的三角函數(shù)公式如下：sin(α+β)=sinαcosβ+cosαsinβsin(α-β)=sinαcosβ-cosαsinβcos(α+β)=cosαcosβ-sinαsinβcos(α-β)=cosαcosβ+sinαsinβtan(α+β)=(tanα+tanβ)/(1-tanαtanβ)tan(α-β)=(tanα-tanβ)/(1+tanα·tanβ)二倍角的正弦、余弦和正切公式如下：sin2α=2sinαcosαCosinedoubleangleformula:cos2α=cos^2(α)-sin^2(α)=2cos^2(α)-1=1-2sin^2(α)Tangentdoubleangleformula:tan2α=2tanα/[1-tan^2(α)]Half-angleformulasforsine,cosine,andtangent:sin^2(α/2)=(1-cosα)/2cos^2(α/2)=(1+cosα)/2tan^2(α/2)=(1-cosα)/(1+cosα)Anotherformofthetangenthalf-angleformulaistan(α/2)=(1-cosα)/sinα=sinα/(1+cosα)Universalformulasforsine,cosine,andtangent:sinα=2tan(α/2)/[1+tan^2(α/2)]cosα=[1-tan^2(α/2)]/[1+tan^2(α/2)]tanα=2tan(α/2)/[1-tan^2(α/2)]Tripleangleformulasforsine,cosine,andtangent:sin3α=3sinα-4sin^3(α)cos3α=4cos^3(α)-3cosαtan3α=(3tanα-tan^3(α))/(1-3tan^2(α))Methodstoimprovehighschoolmathgrades:1.ActivepreviewingPreviewingistheprocessofactivelyacquiringnewknowledge,whichhelpstostimulatelearninginitiative.Beforethenewknowledgeisexplained,carefullyreadthetextbookandcultivatethehabitofactivelypreviewing,whichisanimportantwaytoacquiremathematicalknowledge.Therefore,payattentiontodevelopingself-learningabilityandlearntoreadbooks.Whenstudyingexampleproblems,understandthecontentoftheexample,theconditionsgiven,whatneedstobesolved,howthebooksolvestheproblem,whyitissolvedthisway,whethertherearenewmethods,andwhatthesolutionstepsare.Grasptheseimportantissues,thinkdeeply,andlearntouseexistingknowledgetoindependentlyexplorenewknowledge.2.ActivethinkingManystudentsonlylistenpassivelyduringclassanddonotactivelythink.Whenfacedwithrealproblems,theydonotknowhowtoapplywhattheyhavelearnedtosolvethem.Themainreasonisthelackofactivethinkingduringclass.Inadditiontofollowingtheteacher'sthinking,weshouldalsothinkmoreaboutwhythingsaredefinedthiswayandwhatthebenefitsareofsolvingproblemsthisway.Activethinkingnotonlyhelpsuslistenmorecarefullyinclassbutalsostimulatesinterestincertainknowledge,whichismoreconducivetolearning.Relyingontheteacher'sguidancetothinkabouttheproblem-solvingprocess;theanswerisnotreallyimportant;themethodis!3.GoodatsummarizingrulesSolvingmathproblemscangenerallyfollowcertainrules.Whensolvingproblems,payattentiontosummarizingtheproblem-solvingrules.Aftersolvingeachpracticeproblem,reviewthefollowingquestions:①Whatisthemostimportantcharacteristicofthisproblem?②Whatbasicknowledgeandbasicshapeswereusedtosolvethisproblem?③Howdidyouobserve,associate,andtransformtoachievethetransformation?4.在解決這道題目時(shí)，我們需要運(yùn)用哪些數(shù)學(xué)思想和方法呢？5.在解決這道題目時(shí)，哪一步是最關(guān)鍵的呢？6.你是否曾經(jīng)遇到過(guò)類似的題目？在解題方法和思路上有什么異同之處？7.你能夠發(fā)現(xiàn)幾種解題方法？哪一種方法最優(yōu)？這種