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首页 高考数学公式汇总(Summary of the mathematics formula of c…

高考数学公式汇总(Summary of the mathematics formula of college entrance examination).doc

高考数学公式汇总(Summary of the mathema…

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简介:本文档为《高考数学公式汇总(Summary of the mathematics formula of college entrance examination)doc》,可适用于社会民生领域

高考数学公式汇总(Summaryofthemathematicsformulaofcollegeentranceexamination)高考数学公式汇总(Summaryofthemathematicsformulaofcollegeentranceexamination)AmathematicalformulaParabola:y=ax*bxcYisequaltoaxsquaredplusbxpluscA>,openupA<Whencis,theparabolagoesthroughtheoriginWhenbis,theparaboloidaxisistheYaxisAndthenwehavethepointyisequaltoatimesxplushtimespluskYisequaltoatimesxplushsquaredpluskMinushisthexofthevertexcoordinatesKistheyofthevertexcoordinatesGenerallyusedformaximumandminimumvaluesStandardparabolicequation:y=p^Itsaysthatthefocusoftheparabolaisonthepositivehalfaxisofx,andthefocalpointis(p,),andtheequationofthedirectrixisx=pDuetothefocusoftheparabolacanbeinanyhalfshaft,soatotalofstandardequationy^=p^=ypxx^=pyx^=pyRound:volume=fourthirds(PI)(r^)Area=(PI)(r^)Perimeter=(PI)rThestandardequationofthecircle(xa)(yb)=rnote:(a,b)isthecenterofthecircleThegeneralequationofthecirclexyDxyF=note:DEF>()calculationformulaofellipsecircumferenceTheformulaforthecircumferenceoftheellipseisPIbplustimesaminusbTheellipsecircumferencetheorem:thecircumferenceoftheellipseisequaltothecircumferenceoftheellipsewiththelengthoftheradius(pib)plusfourtimesthedifferencebetweenthelengthoftheellipse(a)andtheshorthalfaxis(b)()calculationformulaofellipticareaEquationofellipticarea:S=PIabTheellipticareatheorem:theellipseareaisequaltoPI(PI)timestheproductofthelengthoftheellipse(a)andtheshorthalfaxis(b)TheellipsecircumferenceandtheareaformuladonotshowtheellipseratioT,butthesetwoformulasarederivedfromtheellipticweekrateTderivationTheconstantisthebody,theformulaisusedEllipticallengthradius*shortradius*PAI*heightTrigonometricfunctions:TwoanglesandformulasSin(AB)=sinAcosBcosAsinBsin=sinAcosBsinBcosA(AB)Cos(AB)=cosAcosBsinAsinBcos(AB)=cosAcosBsinAsinBTan(AB)=(tanAtanB)(tanAtanB)tan(AB)=(tanAtanB)(tanAtanB)Cot(AB)=(cotAcotB)(cotBcotA)cot(AB)=(cotAcotB)(cotBcotA)DoubleAngleformulaTanatanA=(tanA)cotA=cota(cotA)Cosa=cosacosasina==sinaSinealphasin(alphaPIn)sin(alphaPI*n)sin(alphaPI*n)sinalphaPI*(n)n=Cosinealphacos(alphaPIn)cos(alphaPI*n)cos(alphaPI*n)cosalphaPI*(n)n=andSin(alpha)^sin(alphaPI)^sin(alphaPI)^=TanAtanBtan(AB)tanBtanAtan(AB)=universalformula:Thesineofalpha=tan(alpha)(tan^(alpha)Thecosineofalpha=tan^(alpha)tan^(alpha)Tan,tanalpha=(alpha)(tan^(alpha)HalfAngleformula(sin(A)=)(cosA))sin(A)=)((cosA))(cos(A)=)(cosA))cos(A)=)((cosA))Tan(A)=)((cosA)((cosA))tan(A)=(cosA)()((cosA))(cot(A)=)(cosA)((cosA))cot(A)=(cosA)()((cosA))Andbadproductsinacosb=sin(AB)sin(AB)cosasinb=sin(AB)sin(AB)cosacosb=cos(AB)sin(AB)sinasinb=cos(AB)cos(AB)SinAsinB=sin((AB))cos(cosA(AB)cosB=cos((AB))sin(AB)()TanAtanB=sin(AB)cosAcosBtanAtanB=sin(AB)cosAcosBCotAcotBsin(AB)sinAsinBcotAcotBsin(AB)sinAsinBSomeofthetopnentriesn=n(n),(n)=n(n)=n(n)^^^^^^^^n^=n(n)(n)^^^^^^N^=(n(n))^******n(n)=n(n)(n)SinetheoremasinA=bsinB=csinC=Rnote:RstandsfortheouterradiusofthetriangleCosinetheorembisequaltoapluscminusaccosineB,AngleBistheAnglebetweensideaandsidecMultiplicationandthefactorab=(ab)(ab)ab=(ab)(aabb)ab=(abab(ab)Thetriangleinequality,|aplusb|islessthanorequalto|a||||aminusb|islessthanorequalto|a||b|,|,|,|,a|,a|,a|,a|,a|,a|,a|,a|,a|,a|,a|,a|,a|,a|,a|,aBBB|ab|p|||b||a|b|a|orlessorlessThesolutionofthequadraticequationisminusbplusthesquarerootofbminusacoveraminusthesquarerootofbminusacoveraTherelationbetweentherootandthecoefficientisxplusxisequaltominusbaxtimesxisequaltocaThediscriminantba=note:theequationhasthesametwosolidrootsBac>note:theequationhastwodifferentrealrootsBac<note:theequationhasconjugatecomplexrootsFormulaformulaexpressionThestandardequationofthecircle(xa)(yb)=rnote:(a,b)isthecenterofthecircleThegeneralequationofthecirclexyDxyF=note:DEF>Theparabolicstandardequationy=pxy=pxx=pyx=pyThesideareaofthestraightprismisS=c*hobliqueprismsideareaS=c'*hTheareaofthepositiveedgeofthepyramidisS=c*hprimeandthelateralareaS=(cc')h'S=(cc')l=PI(RR),thesurfaceareaoftheball,S=pi*rTheareaofthecylinderisS=c*h=pi*hconicalsideareaS=*c*r*lArclengthformulal=a*raistheradianofthecenterofthecirclerBBBfanareaformulas=*l*rConevolumeformulaV=*S*HconicalvolumeformulaV=*PI*rhTheobliqueprismvolumeV=S'lnotethatS'isthedirectareaandListhesidelengthCylindervolumeformulaV=s*hcylinderGraphgirthsizeformulaThecircumferenceoftherectangle=(lengthwidth)byThecircumferenceofasquareisequaltothelengthofTheareaoftherectangle=lengthxwidthTheareaofthesquareisequaltothelengthofthesideAreaofatriangleWeknowthatthetrianglebottoma,theheighth,thenSisequaltoahoverTrianglewiththreesidesa,b,c,halfperimeterp,S=)p(pa)(b)p(pc)(Helenformula)(p=(abc))And(abc)*(abc)*Solet'ssaythatIhavea,b,andIhavetheAngleC,andthenS=absinCThethreesidesofthetrianglearea,bandc,andtheinnercircleradiusisrSothetriangleareaisequaltoaplusbplusctimesroverThethreesidesofthetrianglearea,bandc,andtheouterradiusisrTriangleareaisequaltoABCrTrianglewiththreesidesa,b,c,S=){c^a^((c^a^b^))^)}("threeobliquequadrature"southernsongdynastychiushao)|ab|DeltaSisequaltotimes|,c,d,,||ef||ab||cd|isthethirdorderdeterminant,thistriangleABCisintheplanerectangularcoordinatesystemA(A,b),b(c,d),c(e,f),hereABC|ef|Districttakebestinanticlockwiseorderstartingfromthetoprightcornertotake,becauseitisgenerallypositive,andtheresultsachievedifnotaccordingtotherules,maybenegative,butitdoesn'tmatter,justtaketheabsolutevalue,willnotaffectthesizeofthetrianglearea!】TheformulafortheareaoftheCentralLineoftheshaoshaotriangle:S=)(MaMbMc)*(MbMcMa)*(McMaMb)*(MaMbMc)AmongthemMa,Mb,McisthemiddlelengthofthetriangleTheareaoftheparallelogramisequaltothebasexheightTheareaoftrapezoid=(bottombottom)xheight=Diameter=radiusxradius=diameterThecircumferenceofacircle=PI(PI)isequaltoPIPIradiustimesTheareaofthecircleisequaltoPItimesradiustimesradiusThesurfaceareaofacuboid=(lengthxwidthlengthxheightwidexheight)xThevolumeofacuboid=lengthxwidthxheightThesurfaceareaofacubeisequaltothelengthoftheedgeoftherectangleThevolumeofacubeisequaltothelengthoftheedgeThesideareaofthecylinder=thecircumferenceofthebottomcircleThesurfaceareaofthecylinder=thesurfaceareaoftheupperandlowerbaseThevolumeofthecylinderisequaltothebaseareatimesheightThevolumeoftheconeisequaltothebaseareaCuboid(cube,cylinder)Volume=floorareaxheightTheplanefigureTheperimeterofthenameisCandtheareaSSquareasidelongC=aS=aRectangleaandblengthC=(ab)S=abTrianglea,b,cminusTheheightofhaShalfofthecircumferenceA,B,CinsideS=(abc)s=ah=absinC=s(sa)(b)s(sc)=asinBsinCsina()OneaftertwohasonlyonelineTheshortestlinesegmentisbetweentwopointsThesupplementaryanglesofthesameAngleorequalAngleareequalTheremaininganglesofthesameAngleorequalAngleareequalIt'salittlebitandthere'sonlyonelinethat'sperpendiculartoagivenlineTheverticalsegmentistheshortestofallsegmentsconnectedtothedotsonastraightlineThesevenparallelaxiomsgothroughtheline,There'sandonlyonelinethat'sparalleltothislineIfbothlinesareparalleltothethird,thetwolinesareparalleltoeachotherCorrespondinganglesareequal,twolinesareparallelTheinterioranglesareequalandthetwolinesareparallelEachofthetwosidesarecomplementary,andthetwolinesareparallelThetwolinesareparallel,correspondinganglesareequalThetwolinesareparallelandtheinterioranglesareequalThetwolinesareparalleltoeachotherandcomplementeachotherThetheoremisthatthesidesofthetrianglearebiggerthanthethirdsideThedifferencebetweenthetwosidesofthetriangleislessthanthethirdthetriangleinsideandtheoremofthreeanglesandequaltoThetwoacuteanglesofarighttrianglearecomplementaryCorollaryanouterAngleofatriangleisequaltothesumoftwoinneranglesthatarenotadjacenttoitCorollarytheoutercornerofthetriangleisgreaterthananyinnerAnglethatisnotadjacenttoitThecorrespondingsidesandcorrespondinganglesofthecongruenttrianglesareequalThetwosidesoftheborderAngleaxiom(sas)havetwotriangularcongruencecorrespondingtotheirAnglecorrespondingtoeachotherTheanglesoftheAngleoftheAngle(asa)havetwoanglescorrespondingtothetwosidesofthetwotrianglesThecorollary(aas)hastwoanglescorrespondingtoonecornerandtwotrianglescongruentTherearethreesidescorrespondingtotheequalpartsofthetriangleThehypotenuse,therightAnglesideaxiom(hl)hasthehypotenuseandarightAnglecorrespondingtothetworighttrianglesTheoremisequaltothedistanceonbothsidesofthisAngleintheAnglebisectorTheoremtothesamepointonbothsidesofanAngle,inthisAnglebisectorThebisectoroftheAngleisthesetofallpointsthatareequaltobothsidesoftheAngleThepropertyoftheisoscelestriangleisequaltothetwobaseanglesoftheisoscelestriangleThebisectedbisectoroftheverticalAngleofanisoscelestriangleisbisectedandperpendiculartothebaseTheverticalbisectoroftheisoscelestriangle,themiddleandthebottomofthebottomedgeoverlapinferenceeveryAngleofanequilateraltriangleareequal,andeachAngleisequaltoIfatrianglehastwoanglesequaltoeachother,thenthesidesofthetwoanglesareequal(equilateraltoequilateral)ThecorollariesthatallthreeanglesareequalareequilateraltrianglesCorollaryhasanAngleisequaltoisoscelestriangleisanequilateraltriangleinarighttriangle,ifanacuteAngleisequaltoitontherightsidetothehypotenuseisequaltohalfThemidlineofthehypotenuseofarighttriangleisequaltohalfofthehypotenuseThepointofthelinesegmentisthesameasthedistancebetweenthetwoendsofthelinesegmentTheinversetheoremisthesamepointasthedistancebetweenthetwoendsofalinesegment,intheverticalbisectorofthelinesegmentTheverticalbisectorofthelinesegmentcanbeseenasacollectionofallpointsequaltothedistancebetweenthetwoendsofalinesegmentTheoremThetwographsofastraightlinesymmetryareallequalTheoremifthetwographsaresymmetricaboutaline,ThenthesymmetryaxisistheperpendicularbisectorofthecorrespondingpointlinetheoremThetwographsaresymmetricaboutacertainline,andiftheircorrespondinglinesegmentorextensionlineintersects,thentheintersectionisontheaxisofsymmetryinversetheoremifthecorrespondingpointofthetwographsisbisectedverticallybythesameline,thenthetwographsaresymmetricaboutthislinethePythagoreantheoremTworightanglesideofarighttriangle,thesquareofthehypotenuseisequaltothesumofthesquaresoftheb,c,namelya^b^=c^TheconversetheoremofPythagoreantheoremIfthethreesidesofthetrianglea,b,cwitha^b^=c^,thenthetriangleisarighttriangletheoremofquadrilateralinsideandisequaltosquareoutside,andisequaltopolygonnedgeanglesandthetheoremofinternalanglesandequalto(n)xinferenceanymultilateralexteriorAngleandequaltoTheparallelogrampropertiestheoremisequaltothediagonaloftheparallelogramTheparallelogrampropertytheoremisthattheoppositesidesoftheparallelogramareequalThecorollaryisequalbetweenparallellinesbetweentwoparallellinesTheparallelogramofaparallelogramisevenlydividedbetweenthediagonallinesoftheparallelogramTheparallelogramoftheparallelogramruledthatthequadrangleofthetwopairsofdiagonalanglesareparallelogramsThefiftysevenparallelogramdeterminesthatthetwopairsofthesamequadrilateralareparallelogramsTheparallelogramoftheparallelogramisaparallelogramTheparallelogramoftheparallelogramisaparallelogramoffourpairsofparallelogramsTherectanglepropertytheoremisthatthefourcornersofarectanglearerightanglesTherectangularpropertytheoremisequaltothediagonaloftherectangleTherectangularrulehasthreeanglesandthequadranglesarerectanglesTheparallelogramofthetwodiagonallinesisrectangularThediamondpropertytheoremisequaltofoursidesofadiamondTherhombicpropertytheoremdiamonddiagonalsareperpendiculartoeachother,andeachdiagonalisdividedintoapairofdiagonalanglesThediamondareaisequaltohalfofthediagonalproduct,whichiss=(acrossb)TherhomboidtheoremisthatthequadrilateralofthefoursidesarethediamondThetwodiagonalparallelogramsperpendiculartoeachotherarediamondshapesThesquarepropertytheoremisthatthefourcornersofasquareareatrightangles,andthefoursidesareequalThetwodiagonalsofasquareoftwosquaresareequal,andareequallydividedbetweeneachother,andeachdiagonalisdividedintoapairofdiagonalsTheoremThetwographsofcentersymmetryarecongruentThetwographsofthecentersymmetry,thesymmetricalpointsoftheline,aredividedequallybetweenthecenterofsymmetryandthecenterofsymmetryTheinversetheoremisthatifthecorrespondingpointsofthetwographsgothroughsomepointandaredividedequallybetweenthem,thenthesetwographsaresymmetricaboutthispointTheisoscelestrapezoidisequaltotwoanglesatthesamebaseThetwodiagonalsoftheisoscelestrapezoidarethesameTheisoscelestrapezoidalruletheoremisthatthetwoequaltrapezoidsonthesamebaseareisoscelestrapezoidTheequaltrapezoidofthethdiagonalisisoscelestrapezoidTheparallelsegmenttheoremofparallellinesisthatifasetofparallellinesisequaltothelinesegmentinastraightline,thelinesegmentsthataretruncatedonotherlinesarealsoequalCorollarythemiddleandthebottomofatrapezoidinastraightlinewillbeequallydividedCorollary:themidpointononesideofthetriangleisparalleltotheotherside,anditwillbeequallydividedThemedianlineofatriangleinatriangleisparalleltothethirdside,andisequaltohalfofitThemedianlineofthetrapezoidinthetrapezoidisparalleltothebaseofthetwo,andisequaltohalfofthebaseandhalfofthesumofthetwoisequalto(ab)Thebasicnatureoftheratioof()isthatifa:b=c:d,thenAD=BC=BC,thena:b=c:dIfabisequaltocd,then(ab)b=(cd)d()thegeometricpropertiesifab=cd=It'smovern(bdnnot),so(acm)(bdn)=abThelinesegmentoftheparallellineistheproportionofthethreeparallellines,andthecorrespondinglinesegmentsareproportionalcorollariesparalleltoonesideofthetriangle(orextensioncordsonbothsides),thecorrespondinglinesegmentsareproportionalIfalineisproportionaltothelinesegmentonbothsidesofthetriangle(orextensioncordonbothsides),thenthislineisparalleltothethirdsideofthetriangleparalleltothesideofthetriangle,andthelinethatintersectstheothertwosides,thethreesidesofthetriangleareproportionaltothethreesidesoftheoriginaltriangleThetheoremisparalleltothelineofthetriangleandintersectsbothsides(orextensioncordsonbothsides),andthetriangleissimilartotheoriginaltriangleAsimilartriangledeterminesthatthetwoanglescorrespondtoeachother,andthetwotrianglesaresimilar(asa)ThetworighttrianglesofthehypotenusearesimilartotheoriginaltrianglesThetwosidesofthetwotrianglesaresimilar(sas)judgmenttheoremthreesidescorrespondingtoproportional,twotriangularsimilarity(SSS)ThetheoremisthatifthehypotenuseofarighttriangleandarightAngleareproportionaltothehypotenuseofanotherrighttriangleandarightAngle,thenthesetworighttrianglesaresimilarTheratioofasimilartriangletoahigherthanthecorrespondingangularbisectorisequaltotheratioofthecorrespondingangularbisectorTheratioofasimilartriangleisequaltotheratioofthecircumferenceofatriangleTheratioofthesimilartriangleareaisequaltothesquareofthesimilarityratioThesineofanyacuteAngleisequaltothecosineofitsremainingAngle,cosineofanyacuteAngleIt'sthesineofitsremainingAngleThetangentofanyacuteAngleisequaltotheremainingtangentofitsremainingAngle,andtheremainingtangentofanyacuteAngleisequaltothetangentofitsremainingAngleThecircleisthesetoffixedpointsTheinsideofthendcirclecanberegardedasasetofpointswithacenterdistancelessthanaradiusTheoutsideoftherdcirclecanbeviewedasasetofpointswithadistancegreaterthantheradiusTheradiusofthecircleisequaltothatofthecircleThedistancefromthefixedpointtothefixedpointisthetrajectoryofthespecifiedlengthAcirclewitharadiusThedistancebetweenthetwoendpointsoftheknownlinesegmentistheperpendicularbisectorofthelinesegmentThedistancebetweenthetwosidesofthetwosidesoftheknownAngleisthesamethingastheangularbisectorThetrajectoryofthedistancebetweenthetwoparallellinesofthetwoparallellinesisastraightlinethatisparalleltotheparallellinestheoremsdonotspecifyacircleonthethreepointsonthesamelineTheverticalpaththeoremisperpendiculartothediameterofthestringandbisectsthisstringandbisectsthetwopiecesofthestringcorollaryisthediameterofthestring(notthediameterofthediameter)perpendiculartothestringandthetwoarcsthatthestringisrightTheverticalbisectingofthestringpassesthroughthecenterandbisectsthetwoarcofthestringThediameterofanarcthatthestringisoppositeisbisected,andtheotherarcofthestringisdividedequallyThetwoparallelstringsofthetwocirclesareequableThecircleisthecenterofsymmetryinthecenterofthecenterofthecenterInthesamecircleorequalcircle,thesamearcisequaltotheAngleofthecircle,thestringisequal,thestringofthestringisequaltothecenterofthestringThecorollariesarethatinacircleorequalcircle,iftwoconcentricangles,twoarcs,twostrings,ortwostringshaveanequalnumberofstrings,therestofthegroupscorrespondtoeachotherThecircleAngleofthearcisequaltohalfoftheAngleofthecenterofthecircleTheequivalentofthecircleAngleofanarcorequalarcThesamearcisequaltothecircleorequalcircleThecircleAngleofthesemicircle(ordiameter)oftheinfernoistherightAngleTheinscribedAngleofthestringisdiameterSoifyousay,well,ifthemiddlelineononesideofthetriangleisequaltohalfofthisside,thenthistrianglerighthereisarighttriangleTheinsideofthecircleofiscomplementarytothediagonalofthequadrangle,andanyouterAngleisequaltoitsinnerdiagonalLinelandcircleointersectd<rLinelandcircled=rThelinelandthecircleareseparatefromd>rThedecisiontheoremofthetangentofisthetangentofthecircletotheoutsideoftheradiusandthelineperpendiculartotheradiusThepropertyofthetangentofthetangentlineisperpendiculartotheradiusofthetangentpointcorollaryastraightlinethatiscenteredandperpendiculartothetangentlinemustbecutThelinethatfollowsthetangentlineandisperpendiculartothetangentlinewillpassthroughthecenterofthecircleThelengthofthetangentlineisthetwotangentsthatdrawthecirclefromtheoutsideofthecircleThetangentsofthetangentsareequaltotheAnglebetweenthetwotangentsThetwogroupsoftheoutertangentquadrilateralareoppositeandequalThestringAngleofthestringAngleisequaltotheAngleofthearcthatitisholdingIfthearcoftwostringsisequal,thenthetwoanglesareequalThetwointersectingstringsinthecircleoftheintersectionstringtheoremarethesameasthetwolinesoftheintersectionThecorollaryifthestringintersectsthediameter,halfofthestringisthediameterofthestringTheratioofthetwosegmentsThecutlinetheoremisthetangentandsecantofacirclefromoutsidethecircleThetangentlineisthepointtocutTheratioofthetwolinesofalinetoacircleThecorollaryisthetwolinesthatdrawthecirclefromoutsidethecircle,whichisthesameasthelengthofthetwolinesoftheintersectionofeachlineandthecircleIftwocirclesaretangenttoeachother,thenthetangentmustbeonthelineIt'stwocirclesawayfromd>rplusrtworoundandd=rrThetwocirclesintersectrr<rr(r>r)Thetworoundsoftangentd=rr(rBBBr)areincludedinthetwocirclesThetheoremisthatthetwocirclesofthetwocirclesareperpendiculartothecommonstringofthetwocirclesTheruleofdividesthecircleinton(nisgreaterthanorequalto):ThepolygonsobtainedfromthepointsinturnaretheinsideofthecircleThetangentofacircleisthetangentofacircle,theintersectionofadjacenttangentsisthevertexpolygonistheoutertangentofthecircleThetheoremisthatanypositivepolygonhasanoutercircleandaninnercircle,andthesetwocirclesareconcentriccirclesnsideofeachinteriorAngleisequalto(n)xnTheradiusofthenedgeshapeandtheedgedistanceofthepositivenaredividedintoncongruentrightangledtrianglesAreasn=PNRNprepresentstheperimeterofnedgesSo,thesquarerootofthistriangleisthesquarerootofadividedbyaifyouhavekinthevicinityofavertex,youknow,becauseofthesumoftheseangles,sok*(n)ton=(n),(k)=Calculationformulaofarclength:l=nPIrsectorareaformula:ssector=nPIr=lrIn,thetangentlineisequaltod(rr)ThetwobottomlegsofanisoscelestriangleThetopAnglebisectoroftheisoscelestriangle,themiddlelineatthebottomedgeandtheheightofthebottomedgeoverlapiftwoanglesofatriangleareequal,theoppositesidesofthetwoanglesareequalThetrianglethatisequaltothethreesidesiscalledanequilateraltriangle

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高考数学公式汇总(Summary of the mathematics formula of college entrance examination)

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