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AYT Matematik Ders İşleyiş Modülleri 5. Modül Trigonometri

Published by Nesibe Aydın Eğitim Kurumları, 2019-08-24 01:33:36

Description: AYT Matematik Ders İşleyiş Modülleri 5. Modül Trigonometri

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MODÜL Alt bölümlerin Karma TestlerTrigonometri EDĜOñNODUñQñL©HULU KARMA TEST - 4 TRİGONOMETRİ 1. 0 # x #ÖPMNBLÑ[FSF 4. arccosf 1 p = arc cotf 12 - 4x p cos x + 2 sin π x2 + 1 9 6 =1 ³ Dik Üçgende Trigonometrik Oranlar t 2 cos 5π - sin 3π FöJUMJôJOJTBôMBZBOYEFôFSJLBÀUS 32 A) 3 B) 2 C) 3 D) 7 E) 4 Modülün sonunda PMEVôVOBHÌSF YLBÀUS 22 tüm alt bölümleri L©HUHQNDUPDWHVWOHU ³ Trigonometrik Fonksiyonlar t 11 π π 5π 2π 5π A) B) C) D) E) 6 3 12 3 6 ³ Trigonometrik Fonksiyonların Grafikleri t 35 ³ Ters Trigonometrik Fonksiyonlar t 46 6ñQñIð©LðĜOH\\LĜ 2. C ³ Kosinüs ve Sinüs Teoremleri t 52 5. %JLLPPSEJOBUTJTUFNJOEFWFSJMFOBõBóEBLJHSBGJL y ·/÷7&34÷5&:&)\";*3-*, 5. MODÜL 53÷(0/0.&53÷ x y = a +CTJO DY GPOLTJZPOVOBBJUUJS ³ Toplam - Fark Formülleri t 63 Dy %÷,·¦(&/%&53÷(0/0.&53÷,03\"/-\"3 www.aydinyayinlari.com.tr 2 Formü%mll/e*rmi t ÖRNEK 2 AB O Õ x \\HUDOñU Bir ABC üçgeninde [ BA ] m [ CA ], m ( A%BC ) = a ve ABDC dörtgeninde, [AC] m [CD], [CD] m [DB] –2 ABC dik tan a = 3 olarak veriliyor. ³ İki Kat Açı 74 % I I[ CB ] m[ AB ], m(B%CD) = a, AC = x br, BAC ) 4 üçgeninde m( = a olsun. ·ÀHFOJO[ BC ]LFOBSOOV[VOMVôVCJSJNPMEVôVOB %XE¸O¾PGHNL¸UQHN #VOBHÌSF B+ b +DUPQMBNLBÀUS HÌSF \"#$ÑÀHFOJOJOÀFWSFTJLBÀCJSJNEJS VRUXODUñQ©¸]¾POHULQH A I ICD = y br dir. ³ Trigonometrik Denklemlaer t,PNõV%JL,FOBS 82 \" # $ % & Hipotenüs #VOBHÌSF ZOJOYWFa DJOTJOEFOFöJUJBöB- <HQL1HVLO6RUXODU<(1m1(6m/6258/$5 ôEBLJMFSEFOIBOHJTJEJS ³ Karma Testler t 91 A) x.sina B) x.cosa C) x.tana ³ Yeni Nesil SoruBlar t 101,BSõ%JL,FOBS C D) x.sin2! TrigonomeEtr)i x sin 2a 2 | |A B = c br DNñOOñWDKWDX\\JXODPDVñQGDQ 6. 3. EôFLJMEFLJHJCJCJSCJMBSEPNBTBTOEB9OPLUBTO- | |AC = b br 1. (Ë[IJ[BTWFZFSEÐ[MFNJBSBTNPMBOCJSCBT- 3. LAFUCPMDV¿FNCFSTFWJZFTJ NZÐLTFLMJóJOEFPMBO EBOBUõBIB[SMBOBOCJSPZVODVCFZB[UPQV\"WF | |BC = a br olmak üzere, QPUBZBMJLB¿ZMBBUõZBQZPS O #OPLUBMBSBSBDTOEBCJSZFSF¿BSQUSBSBLLSN[UP- B Q4VWCVSNBxZB¿BMõBDBLUS A 20 0RG¾O¾QJHQHOLQGH\\RUXP \\DSPDDQDOL]HWPHYE ,BSõ%JL,FOBS a ÖRNEK 3 XODĜDELOLUVLQL] O 1m A 5m B 3m EHFHULOHUL¸O©HQNXUJXOX sina = = B F VRUXODUD\\HUYHULOPLĜWLU Hipotenüs 1 b G F a ôFkildeki dikdört- 35° % cosa = ,PNõV%JL,FOBS = c C genler QSJ[NBT x 3,5 m 2m 3m 1,8 m 15 ôFLJMEFLJ0NFSLF[MJçemberde, [EF] m [AD], Hipotenüs b için, C H E | |EF = 15 br % A 7 | |BE = 7 br AD1,F5 B I I'UFóFUOPLUBT m ,BSõ%JL,FOBS a % ( )m= x , AO = 20 cm, 2,7 m tana = = 0NFSLF[MJ¿FNCFSJOZBS¿BQDN m ( BAC ) = A ,PNõV%JL,FOBS c 3 24 I IBC = 4 cm dir. X cota = ,PNõV%JL,FOBS = c | |AB = 24 br ve m(H%GB) = a PMEVôVOBHÌSF UBOa sin 2A = dir. ,BSõ%JL,FOBS a 2 #VOBHÌSF ÀFNCFSJMFCBTLFUCPMDVB#SVBOTBOHEÌBLSFJ UBOYL BÀUS#VOB HÌSF PZVODV \" OPLUBTOEBO LBÀ NFUSF deôFSJLBÀUS I I#VOBHÌSF BC =YLZBBÀUBDZNVE[JBSLMLZBLMBöLLBÀNFUSFEJS V[BôBCFZB[UPQVÀBSQUSSTBLSN[UPQVWVSB- $OW%¸O¾P7HVWOHUL A) 4 5 CC)J2MJS D) 7 E) 3 B) 1 Hipotenüs b Her alt bölümün seca = = = VRQXQGDRE¸O¾POHLOJLOL $( tan 35°=% PMBSB&L BMO[ 33 3 \" # cosa ,PNõV%JL,FOBS c A) 5 # 5 $ % 25 & 25 TEST - 1 Hipotenüs b Dik Üçgende Trigonometrik Oranlar 2. E \"3 .C # $ 94% & 3 4 78 6. A coseca = 1 = = 1. B 4. A 5. A sina ,BSõ%JL,FOBS a 1. A B ÖARBNCEDK ka4re 4. H G ôFLJMEFLJLÐQUF E C ÖRNEK 1 3|DE| = |DC|A ABx Cy dik üçgen m ( D%HB ) = x $\\UñFDPRG¾OVRQXQGD a [ AB ] m [ AC ]F WDPDPñ\\HQLQHVLOVRUXODUGDQ % = a CA % = y ROXĜDQWHVWOHUEXOXQXU m ( AED ) m ( BHF ) 0 < i < π ve sin i = 3 m ( % ) = i [ AH ] m [ BC ] 25 BE3C olEVôVOB göre; UBOi , cosi ve cot iEFôFSMFSJOJCV- | |%AB = 3 birim 4. \"SBMBSOEB NFUSF PMBO JLJ HË[MFNDJ õFLJMEF- LJ HJCJ \" WF # OPLUBMBSOEB EVSNBLUBES #V TSB- MVOVz. ai | |BH = 1 birim 2. - ¿VCVóV N WF N HFOJõMJóJOEFLJ CJSCJSJOF EJL EBÐTUMFSJOEFOHF¿FOCJSV¿Bó\"OPLUBTOEBLJHË[- ZPMMBSEBOPMVõBOLBOBMBõFLJMEFLJHJCJLBOBMEVWBS MFNDJMJLCJSB¿ZMB #OPLUBTOEBLJHË[MFNDJJTF %E C B 1H m ( % ) =Ba JMFiEFSFDFMJLB¿PMVõUVSBDBLõFLJMEFZFSMFõUJSJMNJõ- MJLCJSB¿ZMBHË[MFNMFNFLUFEJS ACB UJS :VLBSEBLJ WFSJMFSF HÌSF UB:OVaLB+SEtBaLnJiWFStJMoFpSF- HÌSF TJOa LBÀUS tan x PSBOLBÀUS 3m :VLBSEBLJWFSJMFSFHÌSF MBNOOTPOVDVLBÀUS cot y A) 3 B) 7 C) 4 D) 9 E) 5 A) 1 B) 1 C) 2 D) 66 WHVWOHU\\HUDOñU 22 E) 2 23 2. H KG ôFLJMEFLJLÐQUF 5. A i 3 44 108 5 x L 1. tan i = , cos i = , cot i = 2 1 4m 4 53 | | | |HK = 3 KG2. 5 3. 4. 3 40° 35° % 3 A B C % H 1000 m DAK ) m( = x 12 #VOBHÌSF -ÀVCVôVOVOV[VOMVôVOVi cinsin- EFOWFSFOJGBEFBöBôEBLJMFSEFOIBOHJTJEJS F y #VOB HÌSF VÀBôO ZFSEFO ZÑLTFLMJôJ ZBLMBöL Ex BC LBÀNFUSFEJS (tan40° = 0,84 ve tan35°= \" TJOi +DPTi # DPTi +TJOi A) 4200 # 4300 $ 4450 11 11 11 AB \"#$Ð¿HFOJOEF [ CH ] m [ AB ] m ( % ) = x $ TFDi +DPTFDi % DPTFDi +TFDi BAC & 4600 11 :VLBSEBLJWFSJMFSFHÌSF TJOYLBÀUS m ( % ) = y UBOZ= tan x = 8 & UBOi +DPUi % 4500 HCB 11 3 B) 4 C) 4 15 A) 5 41 | |BH =CSEJS 5 :VLBSEBLJWFSJMFSFHÌSF | AC | kaç birimEJS 1. C 2. C 101 3. E 4. A 5 6 A) 15 B) 20 C) 25 D) 32 E) 34 D) E) 41 41 3. % E C ABCD kare 6. A \"#$CJSÐ¿HFO F B [ FE ] m [ GE ] |AB| = |AC| x |AF| = 2|DF| % G m ( BAC ) = i |DE| = |EC| cos i = 5 % 13 m ( EGC ) = x m ( A%CB ) = a a AB C :VLBSEBLJWFSJMFSFHÌSF DPUYLBÀUS :VLBSEBLJWFSJMFSFHÌSF UBOa kaçUS A) 2 B) 3 C) 3 D) 4 E) 5 A) 2 B) 3 C) 5 D) 3 E) 4 3 4 2 3 4 3 4 12 2 3 1. D 2. D 3. C 6 4. B 5. E 6. D

ÜNwİVwwE.aRydiSnyaİyTinlEariY.coEm.tHr AZIRLIK ·/÷7&34÷5&:&)\";*3-*, MATEMATİK - 2 5. MODÜL TRİGONOMETRİ ³ Dik Üçgende Trigonometrik Oranlar t 2 ³ Trigonometrik Fonksiyonlar t 11 ³ Trigonometrik Fonksiyonların Grafikleri t 35 ³ Ters Trigonometrik Fonksiyonlar t 46 ³ Kosinüs ve Sinüs Teoremleri t 52 ³ Toplam - Fark Formülleri t 63 ³ İki Kat Açı Formülleri t 74 ³ Trigonometrik Denklemler t 82 ³ Karma Testler t 91 ³ Yeni Nesil Sorular t 101 1

·/÷7&34÷5&:&)\";*3-*, 5. MODÜL 53÷(0/0.&53÷ www.aydinyayinlari.com.tr %÷,·¦(&/%&53÷(0/0.&53÷,03\"/-\"3 %m/*m ÖRNEK 2 ABC dik üçgeninde % = a olsun. Bir ABC üçgeninde [ BA ] m [ CA ], m( % = a ve m ( BAC ) ABC ) A tan a = 3 olarak veriliyor. 4 ,PNõV%JL,FOBS a Hipotenüs ·ÀHFOJO[ BC ]LFOBSOOV[VOMVôVCJSJNPMEVôVOB HÌSF \"#$ÑÀHFOJOJOÀFWSFTJLBÀCJSJNEJS B C B | BC | = 5k =CSWF ,BSõ%JL,FOBS a Ç ( ABC ) =LPMEVôVOEBO 4k | |A B = c br 5k 9 A 3k k = JÀJO 5 | |AC = b br 9 108 Ç ( ABC ) = 12· = br olur. C 55 | |BC = a br olmak üzere, sina = ,BSõ%JL,FOBS = a ÖRNEK 3 Hipotenüs b G F ,PNõV%JL,FOBS c ôFkildeki dikdört- cosa = = a b % C genler QSJ[NBT Hipotenüs 15 15 için, ,BSõ%JL,FOBS a H 25 E | |EF = 15 br tana = = 7 24 7 | |BE = 7 br ,PNõV%JL,FOBS c B A ,PNõV%JL,FOBS c | |AB = 24 br ve % = a PMEVôVOBHÌSF UBOa m ( HGB ) cota = = deôFSJLBÀUS ,BSõ%JL,FOBS a seca = 1 = Hipotenüs = b \"#)ÑÀHFOJOEF - 24 -Ì[FMÑÀHFOJOEFO cosa ,PNõV%JL,FOBS c |HB| =CSCVMVOVS HB 25 5 tan a = = = olur. 1 Hipotenüs b GH 15 3 coseca = = = sina ,BSõ%JL,FOBS a ÖRNEK 1 ÖRNEK 4 ABC dik üçgen 0 < i < π ve sin i = 3 A [ AB ] m [ AC ] 25 ab 3 [ AH ] m [ BC ] olEVôVOB göre; UBOi , cosi ve cot iEFôFSMFSJOJCV- MVOVz. B 1H | |AB = 3 birim a | |BH = 1 birim C % m ( ACB ) = a A 34 :VLBSEBLJWFSJMFSFHÌSF TJOa LBÀUS tan i = , cos i = , 45 i4 cot i = % 3 HAC BÀTOOÌMÀÑTÑOÑbPMBSBLBEMBOESEôN[EB 4k 5k % a + b =PMVS#VEVSVNEB BAC BÀTOOÌMÀÑTÑOÑO B 3k C PMNBTJÀJO % BÀTOOÌMÀÑTÑEFa olur. BAH 1 #)\"EJLÑÀHFOJOEFO sin a = olur. 3 3 44 2 108 5 1 1. tan i = , cos i = , cot i = 2. 3. 4. 4 53 5 3 3

www.aydinyayinlari.com.tr 53÷(0/0.&53÷ 5. MODÜL ·/÷7&34÷5&:&)\";*3-*, ÖRNEK 5 ÖRNEK 7 A FB A ABC üçgeninde k \"#$%LBSF cb | |AB = c br a [ EF ] m [&%] x x–a | |AC = b br Ba E |BE| = |EC| | |BC = a br k5 b m ( B%EF ) = a % k m ( ABC ) = a ve C % = i m ( ACB ) a % 2k C olmak üzere, c.cosa CDPTi UPQMBNOO EFôFSJOJ CVMVOV[ :VLBSEBLJWFSJMFSFHÌSF cosa TJOaWFUBOaEFôFS- MFSJOJCVMVOV[ x x-a PMEVôVOEBO cos a = & cos i = c b % TOO ÌMÀÑTÑOÑ b PMBSBL BEMBOESEôN[EB %$& x ^a-xh DEC c. cos a + b. cos i = c· c + b· b ÑÀHFOJOEF % OJOEFÌMÀÑTÑa olur. = x +B- x =BPMVS CDE |#&| = |&$| = L PMTVO #V EVSVNEB | DC | = 2k olur. %&$ÑÀHFOJOEF1JTBHPSUFPSFNJOEFO|%&| = k 5 ol- ÖRNEK 8 EVôVOEBO cos a = 21 1 , sin a = ve tan a = 55 2 A ABC ikizkenar üçgen PMBSBLCVMVOVS a | AB | = | AC | = 15 br H 15. | BC | = 6 5 br Eb % b m ( HBC ) = a a b [ BH ] m [ AC ] B 35 ÖRNEK 6 D 35 C A :VLBSEBLJWFSJMFSFHÌSF TJOaOOEFôFSJLBÀUS 2k % . ÷LJ[LFOBS ÑÀHFOEF \" LÌöFTJOEFO [BC] LFOBSOB JOEJSJMFO 4k ZÑLTFLMJL [BC] LFOBSO JLJ Fö QBSÀBZB BZSS a BÀTOO ak E UÑNMFSJOJbPMBSBLBEMBOESEôN[EB D%AC OOÌMÀÑTÑEF 35 5 aPMVS%\"$ÑÀHFOJOEFO sin a = = CVMVOVS 15 5 3k 3 3k 30° 6k 60° ÖRNEK 9 B C A 2k % 2k E kB \"#$%EJLEËSUHFO a | | | |\"#$ FõLFOBS Ð¿HFO m(BDC) = a ve 2 \"% = %$ % % m ( DCE ) = a dir. | | | | | |2k 2 EB = AE = \"% :VLBSEBLJWFSJMFSFHÌSF UBOaLBÀUS . a dir. |AD| = 2k ve |DC| =LPMTVO\"#$FöLFOBSÑÀHFOJOEF C [AC]LFOBSOBBJU[#&]ZÑLTFLMJôJÀJ[JMEJôJOEF[AC]öF- :VLBSEBLJWFSJMFSFHÌSF UBOaLBÀUS LJMEFLJHJCJJLJFöQBSÀBZBBZSMS [AB] // [DC]PMEVôVOEBO m ( % ) = % olur. #&$ ÑÀHFOJ – – PMEVôVOEBO CV EVSVNEB DCE m ( CEB ) |&$|= 3k ve |#&| = 3k 3 olur. %&#ÑÀHFOJOEFOUBOa = 3 3 PMBSBLCVMVOVS $#&ÑÀHFOJOEFOUBOa =PMBSBLCVMVOVS 2 11 6. 3 3 3 5 2 5. cos a = , sin a = , tan a = B8. 5 52 5

·/÷7&34÷5&:&)\";*3-*, 5. MODÜL 53÷(0/0.&53÷ www.aydinyayinlari.com.tr ÖRNEK 10 ÖRNEK 13 A ABC dik üçgen % 4C x 2 [#%] m [ AC ] 8 % E | |\"% = 2 br h % = a a m ( ABD ) b a y B 16 B C . 4A | |:VLBSEBLJWFSJMFSFHÌSF BC OJOa cJOTJOEFOEFôF- \"#$%EJLZBNVL [%$] // [ AB ], [%\"] m [ AB ] SJOJWFSFOJGBEFZJCVMVOV[ | | | |[ AC ] m [%#], %$ = 4 br ve AB = 16 br dir. 2 :VLBSEBLJ WFSJMFSF HÌSF UBOY DPUZ JGBEFTJOJO EF- \"#%ÑÀHFOJOEFO tan a = WF#%$ÑÀHFOJOEFO ôFSJOJCVMVOV[ h h 2h 2 sin a = olur. UBOa TJOa = · = BC h BC BC [AC] // [DF] PMBDBL öFLJMEF [DF] ÀJ[JMJS '%# EJL ÑÀHFOJO- 2 cot a de; ²LMJU UFPSFNJOEFO 4.16 = | DA |2 j | DA | = CS EJS olur. BC = = 2· PMBSBLCVMVOVS tan a. sin a sin a 8 \"%$ ÑÀHFOJOEFO tan x = = 2 WF \"%# ÑÀHFOJOEFO ÖRNEK 11 4 [&\"õFLJMEFLJZBSN¿FNCFSF#OPLUBTOEBUFóFUWF 16 cot y = = 2 PMBSBLCVMVOVS 8 #VEVSVNEBUBOY+ coty = 4 olur. | | | |&% = %$ dir. %m/*m A B x A a 60° a 2x ax 2a E 2a % OC a % 30° m ( BCE ) = x PMEVôVOBHÌSF DPTYJOEFôFSJOFEJS B a3 C 0NFSLF[JOEFO#OPLUBTOBÀJ[JMFOZBSÀBQUFôFUFCVOPL- 30° - 60° -EJLÐ¿HFOJOEFMJLB¿OOLBS- õTOEBLJLFOBSOV[VOMVóVIJQPUFOÐTV[VOMVóVOVO UBEBEJLUJS#VEVSVNEB|&%|= 2 | DO | = 2 | OC | =BPMTVO ZBSTOB FõJUUJS MJL B¿OO LBSõTOEBLJ LFOBSO V[VOMVóVIJQPUFOÐTV[VOMVóVOVOZBSTOO 3 ka- 1 UOBFõJUUJS &#0EJLÑÀHFOJOEFO cos 2x = PMBSBLCVMVOVS 3 ÖRNEK 12 a ÖRNEK 14 3 sin60° + tan30° a JGBEFTJOJOEFôFSJLBÀUS 4. 3 3 53 :VLBSEBWFSJMFOöFLJMFöLBSFEFOPMVöUVôVOBHÌSF += cotaEFôFSJOJCVMVOV[ 23 6 4 cot a = olur. 3 2 cot a 1 4 4 13. 4 53 11. 12. 14. sin a 3 3 6

www.aydinyayinlari.com.tr 53÷(0/0.&53÷ 5. MODÜL ·/÷7&34÷5&:&)\";*3-*, ÖRNEK 15 ÖRNEK 17 \"õBóEB[FNJOFEJLCJSõFLJMEFEVSBOBóB¿FOUFQFTJO- \"öBôEBLJUBCMPZVEPMEVSVOV[ EFLJ$OPLUBTOEBO[FNJOEFLJ\"WF#OPLUBMBSOBJLJIB- MBUJMFCBóMBONõUS \"À 'POLTJZPO 3 C 12 2 Sol 1 sin 2 2 2 4Bó 3 30° 60° 32 1 3 cos 22 Halat 1 Halat 2 1 tan 1 3 60° % 30° cot 3 1 A B [ AC ] m [ BC ] ve m ( % ) = 30° PMEVôVOBHÌSF BôBÀ ÖRNEK 18 ABC Tavan//////////////////////////////////////////// JMFIBMBUBSBTOEBLJBÀOOUBOKBOULBÀUS 3 10 m 45° 10 tan 30° = 52 3 52 %m/*m x=? A 2m 45° zemin a 6[VOMVôVNWFZFSEFOZÑLTFLMJôJNPMBOZVLB- SEBLJTBSLBÀöFLJMEFLJHJCJMJLTBMONZBQUôOEB C TBSLBDOZFSEFOZÑLTFLMJôJLBÀNPMVS a2 x = (12 - 5 2 NEJS 45° a ÖRNEK 19 B #JSJNÀFNCFSÑ[FSJOEFZFSBMBOBöBôEBLJOPLUBMBSO 45° - 45° -EJLÐ¿HFOJOEFMJLB¿MBSOLBS- CFMJSUUJôJNFSLF[BÀMBSCVMVOV[ õMBSOEBLJ LFOBS V[VOMVóVOVO 2 LBU IJQPUFOÐT V[VOMVóVOBFõJUUJS J A f 1 , - 1 p JJ B f - 1 , 1 p 22 22 JJJ C f - 1 , - 1 p JW D f 1 , 1 p 22 22 ÖRNEK 16 y /PLUB .FSLF[\"À 1 sin45°+ cos45° + tan45° - cot45° BD Dd 1 1 n & JGBEFTJOJOEFôFSJOJCVMVOV[ , KD ZBZOHÌSFOMJL –1 45° 45° 22 45° 45° 1 x2 2 QP[JUJGZÌOMÑBÀ + +1-1= 2 CA -1 1 ) yBZO HÌSFO 22 –1 KDB B( , ) 2 2 1MJLQP[JUJGZÌOMÑBÀ -1 -1 ) KDC ZBZO HÌSFO C( , ) MJLQP[JUJGZÌOMÑBÀ 22 Ad 1 , -1 n ) KDA ZBZO HÌSFO 2 2 MJLQP[JUJGZÌOMÑBÀ 3 16. 2 5 18. 12 - 5 2 J JJ JJJ JW 15. 3

TEST - 1 %JL·ÀHFOEF5SJHPOPNFUSJL0SBOMBS 1. A B \"#$%LBSF 4. H G ôFLJMEFLJLÐQUF xy F m ( D%HB ) = x | | | |3 %& = %$ E % m ( B%HF ) = y m ( AED ) = a % m ( BEC ) = i % C ai C AB %E :VLBSEBLJ WFSJMFSF HÌSF UBOa + UBOi UPQ :VLBSEBLJWFSJMFSFHÌSF tan x PSBOLBÀUS MBNOOTPOVDVLBÀUS cot y A) 3 B) 7 $ % 9 E) 5 A) 1 B) 1 $ 66 22 % E) 2 23 2. H KG ôFLJMEFLJLÐQUF 5. A % |HK| = 3|KG| x C m ( D%AK ) = x H 12 Ex F y BC AB ABC üçgeninde, [ CH ] m [ AB ], % = x m ( BAC ) :VLBSEBLJWFSJMFSFHÌSF TJOYLBÀUS % = y, tany = 0,75, tan x = 8 m ( HCB ) , 3 B) 4 C) 4 15 A) 5 41 | |BH = 12 br dir. 5 % 5 6 :VLBSEBLJWFSJMFSFHÌSF | AC | LBÀCJSJNEJS 41 E) 41 \" # $ % & 3. % E C \"#$%LBSF 6. A ABC bir üçgen F [ FE ] m [ GE ] B |AB| = |AC| |AF| = 2|%'| % = i | | | |x m ( BAC ) G %& = EC cos i = 5 % 13 m ( EGC ) = x a % = a C m ( ACB ) AB :VLBSEBLJWFSJMFSFHÌSF DPUYLBÀUS :VLBSEBLJWFSJMFSFgörF UBOaLBÀUS 2 3 C) 3 % 4 5 A) 2 B) 3 C) 5 % 3 E) 4 A) B) 23 E) 3 4 12 2 3 3 4 4 1. D 2. D 3. C 6 4. B 5. & 6. D

%JL·ÀHFOEF5SJHPOPNFUSJL0SBOMBS TEST - 2 1. y 4. % 12 C K A(4, 6) x 15 a x a O B A 27 B %JLLPPSEJOBUTJTUFNJOEF OAB dik üçgen, | |\"#$%ZBNVóVOEB [%$] // [ AB ], %C = 12 br | | | |CB = 15 br, AB = 27 br, m(D%AB) = a , [ OA ] m [ AB ], A ( 4, 6 ) ve % = a ES m ( KOA ) tan a = 3 tür. 4 :VLBSEBLJWFSJMFSFHÌSF UBOaLBÀUS | | :VLBSEBLJWFSJMFSFHÌSF AD = xLBÀCJSJNEJS A) 1 B) 2 C) 4 % & 3 2 3 5 2 \" # $ % & 2. A ABC dik üçgen % 2C m (WB) = 90° 5. x [\"%]B¿PSUBZ H | |#% = 3 br B3 % | |%$ = 5 br A 18 B 5C % m ( ACB ) = a \"#$%EJLZBNVóVOEB [%$] // [ AB ], [ AC ] m [%#], :VLBSEBLJWFSJMFSFHÌSF cosaLBÀUS | | | |AB % = 18 br, %$ = 2 br ve m ( DCA ) = x tir. 3 B) 4 3 :VLBSEBLJWFSJMFSFHÌSF TFDYDPTFDYÀBSQN A) 5 C) LBÀUS 5 7 % 5 13 A) 3 B) 7 C) 10 % 10 E) 10 8 E) 10 10 37 8 3. A 6. A ABC bir üçgen 12 m( % = a BAC ) m ( A%CB ) = i x tan a = 2 B7 H C tan i = 3 BC ABC üçgeninde, [ BA ] m [ CA ], [ AH ] m [ BC ], :VLBSEBLJWFSJMFSFHÌSF BC PSBOLBÀUS | | | |% = 7 br AB m ( ABC ) = x, AC = 12 br ve BH dir. :VLBSEBLJWFSJMFSFHÌSF TJOYLBÀUS 22 2 32 A) B) C) 3 3 2 A) 2 B) 3 C) 4 % 5 E) 7 % 2 6 3 4 56 8 2 E) 3 1. B 2. B 3. B 4. & 5. C 6. A

TEST - 3 %JL·ÀHFOEF5SJHPOPNFUSJL0SBOMBS 1. % ABC üçgen 4. C ôFLJMEFLJ0 2 [%#] m[ BC ] A 16 merkezli çemberde A x |AE| = |EC| B [ AB ] çap 4 | | | |AB = 2 \"% = 4 br | |E 17 O | |AO = 17 br B BC = 6 br | |CB = 16 br m ( B%DE ) = a % = x 6C m ( ABC ) :VLBSEBLJWFSJMFSFHÌSF UBOaLBÀUS :VLBSEBLJWFSJMFSFHÌSF UBOYLBÀUS 15 A) - 8 B) - 15 C) - 15 % 8 E) 1 1 C) 1 % 2 3 A) B) 23 E) 15 17 8 15 8 4 3 4 5. A 2 2. % C \"#$%EJLEËSUHFO BH _ C 30 [%#]LËõFHFO E ABC üçgen, [ AB ] m [ AC ], [AH] m [BC], m % = a | |\"% = 30 br (BCA) x | |AB = 40 br A 40 | | | |B %& = 4 EB | |AB = 2 br m ( E%AB ) = x | | :VLBSEBLJWFSJMFSFHÌSF CH BöBôEBLJMFSEFO IBOHJTJOFFöJUUJS :VLBSEBLJWFSJMFSFHÌSF UBOYLBÀUS A) 2sina - 1 B) 2cot!. cosa A) 3 B) 5 C) 3 % 5 E) 9 C) 2tana . seca % TFDa . cosa 16 16 88 16 E) 2tana . sina 3. % C \"#$%JLJ[LFOBS 6. % C \"#$%LBSF yamuk H H x [%)] m [ CE ] [%$] // [ AB ] x EB = 4 A |\"%| = |CB| AB 5 [ AC ] m [%#] % = x m ( ADH ) B% m ( CAB ) = x AE B :VLBSEBLJWFSJMFSF göre, UBOYLBÀUS :VLBSEBLJWFSJMFSFHÌSF DPUYLBÀUS A) 2 B) 3 $ % 3 E) 4 A) 3 B) 4 $ % 4 E) 5 3 4 2 3 4 5 3 4 1. & 2. A 3. C 8 4. & 5. B 6. B

%JL·ÀHFOEF5SJHPOPNFUSJL0SBOMBS TEST - 4 1. A 4. \"õBóEBLJ EJLEËSUHFO CJ¿JNJOEFLJ CJMBSEP NBTBT- OO\"LËõFTJOEFCVMVOBOUPQõFLJMEFLJZPMVJ[MFZF- SFL % OPLUBTOEB CVMVOBO EFMJóF FO B[ ZPM BMBSBL HJSNJõUJS PB D B %C | | | | | |ABC üçgen, \"% = #% = %$ , m(A%BC) = a, m ( A%CB ) = i ve tana = 2 dir. :VLBSEBLJWFSJMFSFHÌSF cotiLBÀUS AC A) 1 B) 3 $ % & 5PQVO EFMJóF HJSFSLFO J[MFEJóJ ZPMVO EÐõFZ CBOUMB 2 2 | |ZBQUóFOTPOB¿OOUBOKBOU 3 ve AP = 4 cm dir. 4 | | #VOBgöre, AC LBÀDNEJS 2. 1FOBMUOPLUBTOOLBMFZFV[BLMóN LBMFOJOHF- \" # $ % & OJõMJóJN LBMFOJOTBóWFTPMZÐLTFLMJóJJTFNEJS A a 1FOBMUZ LVMMBOBO PZVODV UPQV \" OPLUBTOB 5. %FOHFEFLJõFLMJOTBóWFTPMLFGFMFSJOFTSBTJMFTB- WVSEVôVOBHÌSF öFLJMEFPMVöBOaBÀTOOUBO- KBOULBÀUS 1FOBMUOPLUBTLBMFZJPSUBMZPS dece X ve sadece Y cisimleri konarak terazinin ön- DFTBóWFTPOSBTPMLFGFMFSJOJOZFSFUFNBTFUNFMF- SJTBóMBOZPS9DJTNJLPOVMEVóVOEBEÐ[MFNMFa° lik B¿ :DJTNJLPOVMEVóVOEBEÐ[MFNMFb°MJLB¿PMVõ- NBLUBES XY A) 3 B) 5 C) 6 193 193 193 % 7 E) 8 6m 4m 2m 193 193 3. Mert 10 metrelik bir merdiven ile yerden 6 metre #VOBHÌSF DPTa +TJObUPQMBNLBÀUS (0 < a < π , 0 < b < π ) ZÐLTFLMJLUFLJEÐ[CJSEVWBSB¿LNBLJTUJZPS 22 %VWBSOÑTUÑOFÀLBSLFOLVMMBOEôNFSEJWFOJO 3+4 2 42 2 BZBô JMF ZFS EÑ[MFNJ BSBTOEBLJ BÀOO TJOÑTÑ A) B) C) FOB[LBÀPMNBMES 6 6 6 A) 2 3 C) 4 % 5 E) 7 % 3 + 2 3 5 B) 56 8 66 E) 5 6 1. D 2. C 3. B 4. & 5. A

TEST - 5 %JL·ÀHFOEF5SJHPOPNFUSJL0SBOMBS 1. \"õBóEBMJLLBMLõB¿TJMFIBWBMBOBOCJSV¿BL 3. \"õBóEBLJLBSFQJSBNJUJOZBOZÐ[ÐOÐOUBCBOEÐ[MF- HËTUFSJMNJõUJS NJJMFZBQUóB¿UJS B T Uçak C A yer B 30° A #VOBHÌSF VÀBôOZFSEFOGFFUZÑLTFLMJôF EF ÀLBOBLBEBSZBUBZEBBMEôZPMLBÀNEJS CD GFFU, 30,5 cm) | | 1JSBNJEJODJTJNZÑLTFLMJôJ 5& =CSPMEVôV- OBHÌSF ZBOZÑ[ZÑLTFLMJôJLBÀCJSJNEJS A) 610 3 305 C) 610 A) 5 B) 5 3 C) 5 2 B) E) 305 % & 10 2 13 % 305 3 2. #PZVNPMBOBóB¿õFLJMEFLJHJCJHËWEFTJOEFOL- 4. N ZÐLTFLMJLUF V¿BO LVõ N ZÐLTFLMJóJOEFLJ SMZPSWFHËWEFOJOCJSLTNZFSJMFMJLB¿ZBQB- BóBDO ÐTUÐOEFLJ ZVWBTOB HJSNFL J¿JO EJLFZ FL- DBLõFLJMEFZFSFEÐõÐZPS TFOMFMJLB¿ZBQBDBLõFLJMEFJOJõFHF¿JZPS 30 m 45° 30 m 50 m yer 30° #VOBHÌSF CVLVöZVWBTOBHJSNFLJÀJOZBUBZ- Yer EBFOB[LBÀNFUSFZPMBMNöUS #VOB HÌSF HÌWEFOJO BZBLUB EVSBO LTN LBÀ A) 10 B) 20 C) 10 2 NFUSFEJS % 20 2 E) 30 \" # $ % & 1. D 2. A 3. & 4. B

www.aydinyayinlari.com.tr 53÷(0/0.&53÷ 5. MODÜL ·/÷7&34÷5&:&)\";*3-*, :ÌOMÑ\"À 53÷(0/0.&53÷,'0/,4÷:0/-\"3 TANIM CJUJõLFOBS y B B OA (+) yön 1° A x %FSFDFOJO 1 ineEBLJLBdenir. Bu ölçü 1' O CBõMBOH¿LFOBS 60 ,FOBSMBSOEBO CJSJ CBöMBOHÀ EJóFSJ CJUJö ke- biçiminde gösterilir%BLJLBOO 1 ine TBOJ- OBSPMBSBLLBCVMFEJMFOB¿ZBZÌOMÑBÀdenir. 60 :ËOMÐB¿MBSEBCBõMBOH¿LFOBSTBCJU CJUJõLF- OBSIBSFLFUMJEJS ye denir. Bu ölçü 1\" biçiminde gösterilir. #JUJõLFOBSTBBUJOEËONFZËOÐOÐOUFSTZËOÐO- =h=olur. EFIBSFLFUFEFOB¿MBSBQP[JUJGZÌOMÑBÀ denir. #JSB¿OOËM¿ÐTÐBEFSFDFCEBLJLBDTBOJZF :VLBSEBLJ\"0#B¿TQP[JUJGZËOMÐB¿ES ise bu y a° + b' + c\" veya a° b' c\" biçiminde gösterilir. O CBõMBOH¿LFOBS x ÖRNEK 1 A CJUJõLFOBS ²MÀÑTÑhhhPMBOBÀZTBOJZFDJOTJOEFOZB[- (–) yön O[ B ++ 12 = #JUJõ LFOBS TBBUJO EËONF ZËOÐZMF BZO ZËOEF IBSFLFUFEFOB¿MBSBOFHBUJGZÌOMÑBÀ denir.Yu- LBSEBLJ\"0#B¿TOFHBUJGZËOMÐB¿ES \"À²MÀÑ#JSJNMFSJ ÖRNEK 2 7$1,0%m/*m TBOJZFMJLBÀÌMÀÑTÑOÑEFSFDF EBLJLBWFTBOJ- ZFDJOTJOEFOZB[O[ #JS B¿OO ËM¿ÐMNFTJ B¿OO LPMMBS BSBTOEBLJ B¿LMóOCFMJSMFONFTJJMFZBQMS\"¿OOËM¿ÐTÐ- 20180 3600 OÐJGBEFFUNFLJ¿JOderece veya SBEZBO birim- 18000 5° MFSJLVMMBOMS 2180 60 #JSUBN¿FNCFSZBZOOFõQBS¿BZBCËMÐO- 2160 36' NFTJZMFFMEFFEJMFOWFIFSCJSZBZHËSFONFS- LF[B¿OOËM¿ÐTÐOF1 derece denir. Bu ölçü 1° 20'' biçiminde gösterilir. Bu durumda bir çemberin yay ölçüsü 360° olur. hh=hhh 11 1. 2. hhh

·/÷7&34÷5&:&)\";*3-*, 5. MODÜL 53÷(0/0.&53÷ www.aydinyayinlari.com.tr ÖRNEK 3 ÖRNEK 4 m (WA) = 18° 36' 08'' \"öBôEB WFSJMFO BÀ ÌMÀÑMFSJOJO SBEZBO DJOTJOEFO m (WB) = 32° 20' 12'' FöJUMFSJOJCVMVOV[ BÀÌMÀÑMFSJOFHÌSFBöBôEBLJJöMFNMFSJZBQO[ B 60° C 90° c) -120° B m (WA) + m (WB) C m (WB) - m (WA) 60. π π C 90· ππ B = = m (WA) m (WB) 180° 2 c) d) 180 3 2 4 c) - 120. π - 2π = 180° 3 a) 18° 36' 08'' b) 32° 20' 12'' ÖRNEK 5 + 32° 20' 12'' – 18° 36' 08'' 50° 56' 20'' 13° 44' 04'' \"öBôEB WFSJMFO BÀ ÌMÀÑMFSJOJO EFSFDF DJOTJOEFO d) 32° 20' 12''= 8° 5' 3'' FöJUMFSJOJCVMVOV[ 4 c) 18° 36' 08'' = 9° 18' 04'' B π C 2π c) - 7π 2 5 3 4 3BEZBO π 180° 2π 180° B · = 36° C · = 120° %m/*m 5 π 3 π #JS ¿FNCFSEF ZBS¿BQ V[VOMVóVOB FõJU ZBZO 7π 180° = - 315° V[VOMVóVOVHËSFONFSLF[B¿OOËM¿ÐTÐOF c) - · SBEZBOdenir ve bu ölçü 1R ile gösterilir. 4 π A &TBT²MÀÑ r %m/*m r O SBEZBO B Yandaki O merkezli r a çember üzerinde A B Oa WF#OPLUBMBSOCJS- A MFõUJSFO ZBZ HËSFO b NFSLF[B¿OOËM¿Ð- SCJSJNZBZV[VOMVóV SBEZBO sü a olmak üzere ÕSCJSJNZBZV[VOMVóV YSBEZBO çember üzerinde A SYÕS j YÕ OPLUBTOEBO QP[JUJG ZËOEF JMFSMFZFO CJS LJõJ UVSVO BSEOEBO # OPLUBTOEB EVSEVóVOEB CV #JS B¿OO EFSFDF DJOTJOEFO ËM¿ÐTÐ % SBEZBO LJõJOJOLB¿EFSFDFZFSEFóJõUJSEJóJ cinsinden ölçüsü R olmak üzere, b = a + 3GPSNÐMÐJMFCVMVOVS Õ UVSTBZT DR ôFLJMEFHËSÐMEÐóÐHJCJ\"OPLUBTOEBOIBSFLF- te CBõMBZBOLJõJUVSVOBSEOEBO#OPLUBT- %Õ3 OBWBSNõWF¿FNCFSÐ[FSJOEF!B¿TLBEBS ZFSEFóJõJLMJóJZBQNõUS#V!B¿T bB¿T- D = R & D = R elde edilir. OOFTBTËM¿ÐTÐEÐS 360° 2π 180° π #V PSBOU EFSFDF JMF SBEZBO CJrbirJOF EËOÐõ- b = a +FõJUMJóJOEFa, bOOJMF UÐSNFLJ¿JOLVMMBOMS CËMÐNÐOEFOFMEFFEJMFOLBMBOES 3. B '''C hhhD hhhE hhh 12 ππ 2π 4. B C c) - 5. B C D m 32 3

www.aydinyayinlari.com.tr 53÷(0/0.&53÷ 5. MODÜL ·/÷7&34÷5&:&)\";*3-*, TANIM #JSJN¦FNCFS TANIM k ` Z için; y a ` [0°, 360°) olmak üzere, b = a + k.360° ise (0,1) !B¿TOB bBÀTOOFTBTÌMÀÑTÑ denir. (–1,0) (1,0) x i ` [ Õ PMNBLÐ[FSF ËM¿ÐTÐi +LÕPMBO O B¿OOFTBTËM¿ÐTÐiSBEZBOES \"¿MBSOFTBTËM¿ÐMFSJOFHBUJGPMBNB[ ÖRNEK 6 (0,–1) \"öBôEB WFSJMFO BÀMBSO FTBT ÌMÀÑMFSJOJ EFSFDF DJO- .FSLF[J PSJKJO ZBS¿BQ V[VOMVóu 1 birim olan TJOEFOCVMVOV[ çembere CJSJN ÀFNCFS ya da USJHPOPNFUSJL ÀFNCFSdenir. { ( x, y ) : x2 + y2 = 1, x `R , y `R } kümesi birim çemberi belirtir. B 1280° C -790° B =+ C -= -+ ÖRNEK 8 &TBTÌMÀÑ &TBTÌMÀÑ 1 Af , kp 2 OPLUBTCJSJNÀFNCFSÑ[FSJOEFPMEVôVOBHÌSF LOJO BMBDBôEFôFSMFSJCVMVOV[ d 1 2 + 2 = 1 j k2 = 3 n k 24 3 j k=± 2 ÖRNEK 7 \"öBôEB WFSJMFO BÀMBSO FTBT ÌMÀÑMFSJOJ SBEZBO DJO- ÖRNEK 9 TJOEFOCVMVOV[ #JSJNÀFNCFSJOÑ[FSJOEFLJCJSOPLUB B Õ C Õ Af 3, 1 p 22 c) 28π d) - 32π 5 7 olduôVOBHÌSF \"OPLUBTOOUBONMBEôNFSLF[BÀ- OOÌMÀ ÑTÑLBÀEFSFDFEJS B Ö=Ö+Ö &TBTÌMÀÑÖ y C Ö=Ö+ &TBTÌMÀÑ 28π 8π 8π A 3, 1 c) = 2.2π + &TBTÌMÀÑ 22 55 5 10π 32π 10π &TBTÌMÀÑ 11 d) - = - 3.2π + 77 7 30° 2 x O3 B 2 % = 30° EJS m ( BOA ) 6. B C B ÖC D 8π 10π 13 3 d) 8. ± 57 2

·/÷7&34÷5&:&)\";*3-*, 5. MODÜL 53÷(0/0.&53÷ www.aydinyayinlari.com.tr ÖRNEK 10 %m/*m ÜÀLÌöFTJEFCJSJNÀFNCFSÑ[FSJOEFPMBOCJSEJLüçge- 1 - sin2 a = cos2 ! OJOEJLLFOBSMBSOEBOCJSJOJOV[VOMVôV 6 CJSJNPM- 1 - sin2 a = ( 1 - sin!) ( 1 + sin!) 1 - cos2 a = sin2 ! 5 1 - cos2 a = ( 1 - cos!) ( 1 + cos!) EVôVOBHÌSF EJôFSEJLLFOBSV[VOMVôVLBÀCJSJNEJS y d 6 2 AC 2 = 2 ÖRNEK 11 A 5 n+ 2 G x <ÖPMNBLÑ[FSF cos x = 3 6 AC 2 = 64 5 5 25 PMEVôVOBHÌSF TJOYEFôFSMFSJOJCVMVOV[ B1 1 x 8 C AC = 5 cos2x +TJO2x = 1 4JOÑTWF,PTJOÑT'POLTJZPOMBS d 3 2 + 2 = 1 5 n sin x TANIM jTJO2x = 16 25 y 4 1 j sin x = ± P(x,y) = P(cosa TJOa) 5 1 TJOa –1 a1 x ÖRNEK 12 O cosa H #JSJNÀFNCFSEF WFMJLBÀMBSOTJ- OÑTWFLPTJOÑTEFôFSMFSJOJCVMVOV[ –1 y 90° (0, 1) #JSJN¿FNCFSÐ[FSJOEF1 Y Z OPLUBTWFSJMTJO #VOPLUBZPSJKJOMFCJSMFõUJSFO[01OOYFLTFOJ 180° (1, 0) cos -1 1 JMFZBQUóQP[JUJGZËOMÐB¿OOËM¿ÐTÐa olsun. (–1, 0) O 0° x TJO 1 -1 1 OPLUBTOO BQTJTJOF a BÀTOO LPTJOÑTÑ 360° EFOJSWFCVJGBEFcosa ile gösterilir. x = cosa olur. 270° (0, –1) 1OPLUBTOOPSEJOBUOBaBÀTOOTJOÑTÑ de- OJSWFCVJGBEFTJOa ile gösterilir. y =TJOa olur. ÖRNEK 13 Buna göre x eksenine LPTJOÑTFLTFOJ, y ekse- nine TJOÑTFLTFOJ denir. x `3PMNBLÑ[FSF 1OPLUBTCJSJN¿FNCFSÐ[FSJOEFCVMVOEVóVO- 8 + cos2x - 3 EBOBQTJTWFPSEJOBUEFóFSMFSJ-1 den küçük, 1 3 - sin x den büyük olamaz. Buna göre JGBEFTJOJOFOTBEFI»MJOJCVMVOV[ -âTJOaâ -âDPTaâPMVS :VLBSEBLJ CJSJN ¿FNCFSEF 1 OPLUBTOEBO JO- 8 + cos2x -3= 8 + cos 2 - 9 + 3 sin x EJSJMFOEJLNFOJOBZBó)PMNBLÐ[FSF10)EJL x Ð¿HFOJOEF1JTBHPSUFPSFNJVZHVMBOEóOEB 3 - sin x 3 - sin x cos2a +TJO2a = 1 Ë[EFõMJóJFMEFFEJMJS 22 cos x - 1 + 3 sin x - sin x + 3 sin x == 3 - sin x 3 - sin x sin x^ 3 - sin x h = = sin x ^ 3 - sin x h 8 14 4 13. TJOY 11. ± 5 5

www.aydinyayinlari.com.tr 53÷(0/0.&53÷ 5. MODÜL ·/÷7&34÷5&:&)\";*3-*, ÖRNEK 14 ÖRNEK 17 TJOYâ Y`3PMNBLÑ[FSF t `3PMNBLÑ[FSF 1 - sin x - cos2x 2cosx + 3t - 2 = 0 sin x - 1 PMEVôVOBHÌSF UOJOBMBCJMFDFôJUBNTBZEFôFSMFSJ- JGBEFTJOJOFOTBEFI»MJOJCVMVOV[ OJCVMVOV[ 2 2cosx = 2 - 3t 6 44si7n 4x48 2 - 3t 2 sin2x - sin x 1 - cos - sin x = cosx = x 2 sin x - 1 sin x - 1 -1 # cosx # 1 sin x^ sin x - 1 h -1 # 2 - 3t #1 ZJGBEFZJJMFHFOJöMFUFMJN = = sin x ZJGBEeEFOÀLBSBMN 2 sin x - 1 -2 # 2 - 3t # 2 -4 # -3t # Z JGBEFZJ-FCÌMFMJN ÖRNEK 15 4 # t # x `3PMNBLÑ[FSF sin4 x + sin2 x.cos2 x - sin2 x + 1 3 JGBEFTJOJOFOTBEFI»MJOJCVMVOV[ #VEVSVNEBU=WFZBU= 1 olur. 22 2 2 ÖRNEK 18 sin x (1s4in44x 2+ c4o4s4x3) - sin x + 1 a = 3sinx - 4cosy + 2 PMEVôVOBHÌSF BHFSÀFMTBZTOOEFôFSBSBMôOCV- 1 MVOV[ = sJO2x -TJO2x + 1 = 1 - 3 ≤ 3 sin x ≤ 3 ZJGBEFMFSJUBSBGUBSBGBUPQMBZBMN 4 - 4 ≤ - 4 cos y ≤ 4 ÖRNEK 16 -#TJOY- 4cosy # Z JGBEFZFFLMFZFMJN sin x = 2t + 1 -5 #TJOY-4cosy + 2 # 4 0I»MEFBOOEFôFSBSBMô[- ]CVMVOVS PMEVôVOBHÌSF UOJOFOCÑZÑLUBNTBZEFôFSJLBÀ- US -1 #TJOY# 1 -1 # 2t + 1 ≤1 ZJGBEFZJJMFHFOJöMFUFMJN ZJGBEFEFOÀLBSBMN 4 %m/*m -4 # 2t + 1 # 4 y = acosx + bsinx olmak üzere, ZOJOFOCÐZÐLEFóeri = a2 + b2 -5 # 2t # 3 ZJGBEFZJZFCÌMFMJN ZOJOFOLÐ¿ÐLEFóFSJ= - a2 + b2 dir. 53 - GtG 22 #VEVSVNEBUOJOFOCÑZÑLUBNTBZEFôFSJPMVS 14. TJOY 15. 1 16. 1 15 WF 18. <m >

·/÷7&34÷5&:&)\";*3-*, 5. MODÜL 53÷(0/0.&53÷ www.aydinyayinlari.com.tr ÖRNEK 19 5BOKBOUWF,PUBOKBOU'POLTJZPOMBS TANIM \"öBôEBLJJGBEFMFSJOFOCÑZÑLWFFOLÑÀÑLEFôFSMF- SJOJCVMVOV[ B 3sinx + 4cosx C 5cosx - 12sinx y x=1 1 c) 2sinx - cosx d) sinx - cosx T(1, UBOa) P TJOa UBOa a1 B NBY 3 2 + 2 = 5 , NJO - 2 + 2 = - 5 x 4 3 4 –1 O cosa H A C NBY 2 + ^ - 12 h2 = 13 , NJO- 2 + ^ - 12 h2 = - 13 UBOKBOU FLTFOJ 5 5 c) NBY 2 + ^ - 1 h2 = 5 , NJO - 2 + ^ - 1 h2 = - 5 –1 2 2 d) NBY 2 + ^ -1 h2 = 2 , NJO - 2 + ^ -1 h2 = - 2 #JSJN ¿FNCFS Ð[FSJOEF 1 OPLUBT WFSJMTJO #V 1 1 OPLUBZ PSJKJOMF CJSMFõUJSFO [01 OO Y FLTFOJ JMF ZBQUóQP[JUJGZËOMÐB¿OOËM¿ÐTÐa olsun. \" OPLUBTOEBCJSJN çembere teóFUPMBO x =EPóSVTVOBUBOKBOUFLTFOJ denir. ÖRNEK 20 \"01 B¿TOO CJUJõ LFOBSOO UBOKBOU FLTFOJOJ LFTUJóJ5OPLUBTOOPSEJOBUOBaBÀTOOUBO- msinx + 4cosx KBOUEFOJSWFCVJGBEFUBOa ile gösterilir. JGBEFTJOJOFOCÑZÑLEFôFSJPME VôVOBHÌSF NOJO BMBCJMFDFôJEFôFSMFSJCVMVO V[ | |TA = tana olur. 2 + 2 = 5 :VLBSEBLJ õFLJMEF 0PH ve OTA benzer üç- genlerdir. m 4 #FO[FSMJLCBóOUMBSZB[MEóOEB N2 + 42 = 52 PH OH N2 =jN= ±3 = olur. TA OA Buradan sin a = cos a & tan a = sin a tan a 1 cos a ÖRNEK 21 elde edilir. A = sinx + acosx ÖRNEK 22 PMNBLÑ[FSF \"OOBMBCJMFDFôJGBSLMUBNTBZEFôF- A = 1 + tan2x SJPMEVôVOBHÌSF BOOBMBDBôFOCÑZÑLWFFOLÑÀÑL FöJUMJôJOJ TBôMBZBO \" HFSÀFM TBZMBSOO EFôFS BSBM- EFôFSMFSJCVMVOV[ ôOCVMVOV[ A =TJOY+BDPTYJTF -ß<UBOY<ß #UBO2x <ß - 2 + 1 # A # 2 + 1 EJS 1 #UBO2x + 1 <ß a a 2 \"OOBMBCJMFDFôJGBSLMEFôFSWBSTBCVUBNTBZMBS 1 # tan x + 1 <ß -3, -2, - PMBCJMJS 1 # A <ß 0I»MEF 22 a +1=3&a =8 j a = ± 2 2 EJS B m C m D 5 , - 5 , d) 2 , - 2 16 22. < ß ±3 21. ± 2 2

www.aydinyayinlari.com.tr 53÷(0/0.&53÷ 5. MODÜL ·/÷7&34÷5&:&)\";*3-*, TANIM ÖRNEK 23 y 0 < x < r PMNBLÑ[FSF 2 LPUBOKBOUFLTFOJ 1 B cota y=1 H a K(cota,1) sin x - cos x = 1 sin x + cos x 2 cosa P PMEVôVOBHÌSF tan x + cot x PSBOLBÀUS TJOa tan x - cot x –1 O a Ax 1 7FSJMFOFöJUMJLUFJÀMFS-EöMBSÀBSQNZBQMSTB –1 TJOY+ cosx =TJOY- 2cosx j 3 cosx =TJOY 3 cos x sin x 1 j cos x = jUBOY= 3 j cot x = olur. cos x 3 #JSJN ¿FNCFS Ð[FSJOEF 1 OPLUBT WFSJMTJO #V 3+ 1 OPLUBZ PSJKJOMF CJSMFõUJSFO [01 OO Y FLTFOJ JMF 0I»MEF tan x + cot x = 3 5 = CVMVOVS ZBQUóQP[JUJGZËOMÐB¿OOËM¿ÐTÐa olsun. tan x - cot x 3 - 1 4 # OPLUBTOda birim¿FNCFSFUFóFUPMBO y =EPóSVTVOBLPUBOKBOUFLTFOJdenir. 3 \"01B¿TOOCJUJõLFOBSOOLPUBOKBOUFLTFOJ- ÖRNEK 24 OJLFTUJóJ,OPLUBTOOBQTJTJOFaBÀTOOLP- UBOKBOUEFOJSWFCVJGBEFcota ile gösterilir. 5BONMPMEVôVBSBMLUB cot x + sin x | |BK = cota olur. 1 + cos x :VLBSEBLJ õFLJMEF 01) WF 0,# CFO[FS Ð¿- JGBEFTJOJOFOTBEFI»MJOJCVMVOV[ HFOMFSEJS#FO[FSMJLCBóOUMBSZB[MEóOEB cos x + sin x PH OH = olur. sin x 1 + cos x KB OB Buradan cos a = sin a & cot a = cos a (1 + cos x) (sin x) cot a 1 sin a 6 4 4 471 4 448 cos x + 2 + sin 2 1 + cos x 1 = cos x x = elde edilir. sin x^ 1 + cos x h sin x^ 1 + cos x h sin x %m/*m ÖRNEK 25 aOOVZHVOUBONBSBMóOEB 5BONMPMEVôVBSBMLUB sin i. cos i ^ 1 - sin i h. tan i JGBEFTJOJOFOTBEFI»MJOJCVMVOV[ tana . cota = 1 olur. /PUDPT2i = 1 -TJO2i = (1 -TJOi) (1 +TJOi) #FO[FSõFLJMEF sin i. cos i sin i . cos i. cos i 2 i ^ 1 - sin i h. sin i == cos cos i sin i .^ 1 - sin i h 1 - sin i tan a = 1 veya cot a = 1 olur. ^ 1 - sin i h^ 1 + sin i h cot a tan a = = 1 +TJOi ^ 1 - sin i h 5 1 25. TJOi 23. 24. 4 sin x

·/÷7&34÷5&:&)\";*3-*, 5. MODÜL 53÷(0/0.&53÷ www.aydinyayinlari.com.tr ÖRNEK 26 4FLBOUWF,PTFLBOU'POLTJZPOMBS 5BONMPMEVôVBSBMLUB TANIM y sin a - sin a M –1 1 1 - cos a 1 + cos a coseca N JGBEFTJOJOFOTBEFI»MJOJCVMVOV[ 1 K sin a - sin a TJOa TJOa 1 - cos a 1 + cos a a1 L x O cosa H (1 + cos a) (1 - cos a) seca sin a + sin a. cos a - sin a + sin a. cos a = ^ 1 - cos a h^ 1 + cos a h 2 sin a. cos a 2 sin a. cos a 2 cos a –1 = = = = 2 cot a 1 - cos2 a 2 sin a a % sin m ( HOK ) = a olmak üzere birim çember üze- SJOEFLJ,OPLUBTOEBO¿J[JMFOUFóFUJOYFLTFOJ- ÖRNEK 27 OJLFTUJóJ-OPLUBTOOBQTJTJOFaBÀTOOTF- LBOUEFOJSWFCVJGBEFseca ile gösterilir. tani + coti = 7 3 | |0- = seca olur. PMEVôVOBHÌSF UBO2 i + cot2 i deôFSJLBÀUS ,OPLUBTOEBO¿J[JMFOUFóFUJOZFLTFOJOJLFTUJóJ .OPLUBTOOPSEJOBUOBaBÀTOOLPTFLBOU ^ tan i + cot i h2 = d 7 2 EFOJSWFCVJGBEF coseca ile gösterilir. 3 n | |OM = coseca olur. UBO2i + 2 = 49 :VLBSEBLJ CJSJN ¿FNCFSEF ,0) JMF -0, Ð¿- 9 21ta4n4i2. c4o4t i3 + cot i % % 1 m ( HKO ) m ( OLK ) UBO2i + 2 + cot2i = 49 genlerinde = olur. Buna gö- 9 SF,0)JMF-0,Ð¿HFOMFSJCFO[FSÐçgenler olur. #FO[FSMJLPSBOZB[MEóOEB UBO2i + cot2i = 49 31 -2= 99 ÖRNEK 28 OH = OK & cos a = 1 1 sec a tan4 i + cot4 i = 3 OK OL PMEVôVOBHÌSF UBOi + cotiUPQMBNOOQP[JUJGEFôF- & sec a = 1 elde edilir. SJLBÀUS cos a UBOi + coti =\"PMTVO \"ZO õFLJMEF ,0/ JMe MOK üçgenlerinin ben- UBOi + coti)2= A2 [FSMJóJOEFO cosec a = 1 elde edilir. UBO2i + 2 + cot2i = A2 sin a UBO2i + cot2i = A2 - 2 UBO2i + cot2i)2 = (A2 - 2)2 %m/*m UBO4 i + 2 + cot4i = (A2 - 2)2 secaOOEFóFSBSBMó -Þ, -1] b [1, +Þ WF UBO4i + cot4i = (A2 - 2)2 - 2 = 3 coseca OOEFóFSBSBMó -Þ, -1] b [1, +Þ PMB- (A2 - 2)2 = 5 A2 - 2 = 5 WFZB\"2 - 2 = - 5 SBLJGBEFFEJMJS A2 = 2 + 5 WFZB\"2 = 2 - 5 <j q A = ± 2 + 5 WF\"OOQP[JUJGEFôFSJ A = 2 + 5 CVMVOVS 26. 2cota 31 28. 2 + 5 18 9

www.aydinyayinlari.com.tr 53÷(0/0.&53÷ 5. MODÜL ·/÷7&34÷5&:&)\";*3-*, ÖRNEK 29 ÖRNEK 32 5BONMPMEVôVBSBMLMBSJÀJOBöBôEBLJJGBEFMFSJOFO 5BONMPMEVôVBSBMLUB TBEFI»MJOJCVMVOV[ 2sec2 x - 5 = 5tanx B 1 + tan2 x C 1 + cot2 x PMEVôVOBHÌSF UBOYEFôFSMFSJOJCVMVOV[ B 1 + 2 = 2 x + 2 = 1 = 2 /PU ÌSOFôJOBNBEEFTJOFCBLO[ sin x cos sin x sec x 2sec2x - 5 =UBOY 2 (1 +UBO2x) - 5 =UBOY 22 2 UBO2x -UBOY- 3 =j UBOY+ 1 UBOY- 3) = cos x cos x cos x 1 C 1 + 2 = sin2x + cos2x 1 2 tan x = - WFZBUBOY=PMBSBLCVMVOVS = = cosec x cos x 2 2 22 sin x sin x sin x ÖRNEK 30 %m/*m sin x + cos x = 3 Ölçüsü a PMBO CJS B¿OO IFSIBOHJ CJS USJHPOPNFU- 4 SJL EFóFSJOJO JõBSFUJOJ CFMJSMFNFL J¿JO CV B¿OO CJ- UJNLFOBSOOCJSJN¿FNCFSJLFTUJóJOPLUBOOLPPSEJ- PMEVôVOB HÌSF TFDY DPTFDY UPQMBNOO EFôFSJOJ OBUMBSOBCBLMS#VOPLUBOOBQTJTJOJOJõBSFUJDPTa CVMVOV[ OO PSEJOBUOOJõBSFUJTJOaOOJõBSFUJEJS y ^ sin x + cos x h2 = d 3 2 4 n cos – cos + TJO + TJO 1 UBOm #ËMHF #ËMHF UBO cot – (–, +) O (+, +) cot + sin2x + 2sin.cosx + cos2x = 9 cos – 16 TJO – #ËMHF #ËMHF x UBO (–, –) (+, –) 9 cot + cos + 1 +TJOYDPTY= TJO – UBO – 16 cot – 9 -7 7 2. sin x. cos x = - 1 = jTJOYDPTY= - olur. 16 16 32 :VLBSEBLJõFLJMEFCJSOPLUBOOLPPSEJOBUMBSOOEP- MBZTZMB CV OPLUBOO LBSõMLM HFMEJóJ NFSLF[ B¿- 11 OO USJHPOPNFUSJL PSBOMBSOO CËMHFMFSF HËSF IBOHJ sec x + cosec x = sin x + cos x JõBSFUMFSJBMBDBóHËTUFSJMNJõUJS 3 cos x + sin x 4 - 24 = = = CVMVOVS sin x. cos x 7 7 - 32 ÖRNEK 31 ÖRNEK 33 5BONMPMEVôVBSBMLUB π < x < πPMNBLÑ[FSF 1 - sin x + cos x 2 cos x 1 - sin x sin x = 3 JGBEFTJOJOFOTBEFI»MJOJCVMVOV[ 5 PMEVôVOB HÌSF UBOY DPUY WF DPTY EFôFSMFSJOJ IF- TBQMBZO[ 6 4 4 471 4 448 π 22 < x < π BSBMô CÌMHFZJ HÌTUFSJS UBO Dot ve cos CV 1 - sin x cos x 1 - 2 sin x + sin x + cos x cos x += 2 1 - sin x cos x^ 1 - sin x h CÌMHFEF-EJS (1 - sin x) (cos x) %PMBZTZMBDPT2x +TJO2x = 1 jDPTY- 4 PMVS#VOBHÌ- 5 2 - 2 sin x 2^ 1 - sin x h 2 34 4 = cos x^ 1 - sin x h = cos x^ 1 - sin x h = cos x = 2 sec x re, tan x = - , cot x = - ve cos x = - olur. 43 5 B TFD2YC DPTFD2x - 24 1 34 4 31. 2secx 32. - ve 3 33. UBOY - DPUY - DPTY - 7 2 43 5

·/÷7&34÷5&:&)\";*3-*, 5. MODÜL 53÷(0/0.&53÷ www.aydinyayinlari.com.tr ÖRNEK 34 %m/*m y \"öBôEBLJJGBEFMFSJOJöBSFUMFSJOJCVMVOV[ –1 1 $ TJOa,cosa) B cos100°, sin210°, tan320°, cot250° D TJOa DPTjCÌMHF - TJOjCÌMHF -) UBOjCÌMHF - DPUjCÌMHF +) 1 cosa TJOa A(cosa TJOa) a 90°–a 1 1x a B O cosa C sin70°, cos170°, tan280°, cot220° –1 %% TJOjCÌMHF + DPTjCÌMHF -) m ( AOB ) = a ve m ( COB ) = 90° - a olsun. UBOjCÌMHF - DPUjCÌMHF +) Buna göre, aOOCJSJNJEFSFDFPMNBLÐ[FSF c) sin300°, cot150°, tan140°, cot190° cos^ 90° - a h = sin a sin^ 90° - a h = cos a tan^ 90° - a h = cot a cot^ 90° - a h = tan a TJOjCÌMHF - DPUjCÌMHF -) UBOjCÌMHF - DPUjCÌMHF +) olur. d) cotf 17π p, cosf - 7π p, sinc - π m, cosf - 45π p aOOCJSJNJSBEZBOPMNBLÐ[FSF 6 36 4 cosc π - a m = sin a sinc π - a m = cos a cotd 17π n jCÌMHF -) 2 2 6 tanc π - a m = cot a cotc π - a m = tan a cosd - 7π n j 4.CÌMHF +) 2 2 3 olur. sind - π n jCÌMHF -) 6 %m/*m cosd - 45π njCÌMHF -) y 4 1 e) sec10°, cosec110° C(–cosa TJOa) A(cosa TJOa) TFDjCÌMHF DPTFDjCÌMHF 1 180°–a 1 TJOa G sin 100° . cos 150° TJOa tan 210° . cot 300° –1 a a 1 x D cosa O cosa B TJOjCÌMHF + DPTjCÌMHF -) UBOjCÌMHF + DPUjCÌMHF -) –1 +. - %% m ( AOB ) = a ve m ( COB ) = 180° - a olsun. =+ Buna göre, aOOCJSJNJEFSFDFPMNBLÐ[FSF +. - cos^ 180° - a h = - cos a sin^ 180° - a h = sin a tan^ 180° - a h = - tan a cot^ 180° - a h = - cot a olur. aOOCJSJNJSBEZBOPMNBLÐ[FSF cos^ π - a h = - cos a sin^ π - a h = sin a tan^ π - a h = - tan a cot^ π - a h = - cot a olur. 34. B m m m C m m D m m m E m m m e) +, + G

www.aydinyayinlari.com.tr 53÷(0/0.&53÷ .0%·- ·/÷7&34÷5&:&)\";*3-*, %m/*m %m/*m y y 1 $ mTJOa,cosa) 1 TJOa D 1 1cosaA(cosa TJOa) 270°–a A(cosa TJOa) a90°+a TJOa 1 cosa –1 TJOa–1 a 1 x a1 x O cosa B a O cosa B 1 TJOa D $ mTJOa,–cosa) –1 –1 %% %% m ( AOB ) = a ve m ( COB ) = 90° + a m ( AOB ) = m ( COD ) = aPMTVO Buna göre, aOOCJSJNJEFSFDFPMNBLÐ[FSF Buna göre, aOOCJSJNJEFSFDFPMNBLÐ[FSF cos^ 90° + a h = - sin a sin^ 90° + a h = cos a cos^ 270° - a h = - sin a sin^ 270° - a h = - cos a tan^ 90° + a h = - cot a cot^ 90° + a h = - tan a tan^ 270° - a h = cot a cot^ 270° - a h = tan a PMVS aOOCJSJNJSBEZBOPMNBLÐ[FSF PMVS aOOCJSJNJSBEZBOPMNBLÐ[FSF cosc π + a m = - sin a sinc π + a m = cos a 2 2 3π 3π cosf - a p = - sin a sinf - a p = - cos a tanc π + a m = - cot a cotc π + a m = - tan a 2 2 2 2 PMVS 3π 3π tanf - a p = cot a cotf - a p = tan a 2 2 PMVS %m/*m %m/*m y y 1 1 180°+a 1 A(cosa TJOa) A(cosa TJOa) 1 TJOa –1 D cosa TJOa a TJOa TJOa a1 x –1 aa cosa 1 x O cosa B B O 1 360°–a 1 C(–cosa mTJOa) C(cosa mTJOa) –1 –1 %% % = % = a PMTVO m ( AOB ) = m ( BOC ) = aPMTVO m ( AOB ) m ( DOC ) Buna göre, aOOCJSJNJEFSFDFPMNBLÐ[FSF Buna göre, aOOCJSJNJEFSFDFPMNBLÐ[FSF cos^ 360° - a h = cos a sin^ 360° - a h = - sin a tan^ 360° - a h = - tan a cot^ 360° - a h = - cot a cos^ 180° + a h = - cos a sin^ 180° + a h = - sin a tan^ 180° + a h = tan a cot^ 180° + a h = cot a PMVS aOOCJSJNJSBEZBOPMNBLÐ[FSF PMVS aOOCJSJNJSBEZBOPMNBLÐ[FSF cos^ π + a h = - cos a sin^ π + a h = - sin a cos^ 2π - a h = cosa sin^ 2π - a h = - sin a tan^ π + a h = tan a cot^ π + a h = cot a tan^ 2π - a h = - tan a cot^ 2π - a h = - cot a PMVS PMVS 21

·/÷7&34÷5&:&)\";*3-*, 5. MODÜL 53÷(0/0.&53÷ www.aydinyayinlari.com.tr %m/*m ÖRNEK 35 y \"öBôEBLJ JGBEFMFSJO EPôSV ZB EB ZBOMö PMEVLMBSO 1 CVMVOV[ A(cosa TJOa) B sin150° = sin30° C cos210° = -cos60° 270°+acosa 1 TJOa –1 a1 x c) tan330° = -cot30° d) cot120° = -tan30° O a cosa B e) sinf 3π - i p = - cos i G tan ( i -Õ = coti 1 2 D TJOa $ TJOa,–cosa) –1 B TJO - = +TJO % m( % = % ) = a olsun. C DPT + = -DPT : AOB ) m ( DOC Buna göre, aOOCJSJNJEFSFDFPMNBLÐ[FSF D UBO + = -DPU : cos^ 270° + a h = sin a sin^ 270° + a h = - cos a E DPU + = -UBO % tan^ 270° + a h = - cot a cot^ 270° + a h = - tan a e) sind 3π - i n = -cosi (D) 2 olur. G UBO i -Ö =UBOi (Y) aOOCJSJNJSBEZBOPMNBLÐ[FSF cosf 3π + a p = sin a sinf 3π + a p = - cos a ÖRNEK 36 2 2 \"öBôEBLJJöMFNMFSJOTBZTBMTPOVÀMBSOCVMVOV[ tanf 3π + a p = - cot a cotf 3π + a p = - tan a B cos120° + tan300° - sin210° + cot120° 2 2 C cos300° - sin150° + tan240° + cot150° olur. B -DPT-UBO+TJO-DPU 1 1 3 -4 3 - - 3+ - = 2 23 3 %m/*m C DPT-TJO+UBO-DPU 11 y - + 3- 3=0 1 22 A(cosa TJOa) –1 360°–a acosa 1TJOa TJOa x ÖRNEK 37 O π < i < 3π ve sin i = - 1 –a B 23 C(cosa mTJOa) PMEVôVOBHÌSF UBOi + cot) LBÀUS –1 A CÌMHFEF UBOKBOU WF LPUBO- %% i KBOUQP[JUJGPMEVôVOEBO m ( AOB ) = m ( BOC ) = a olsun. Buna göre, 22 3 UBOi + coti = 1 +2 2 cos ( -!) = cosa sin ( -!) = -sina 22 tan ( -!) = -tana cot ( -!) = -cota B1 92 = C4 olur. 22 35. B %C :D :E %F %G : 36. B -4 3 C 92 34

www.aydinyayinlari.com.tr 53÷(0/0.&53÷ 5. MODÜL ·/÷7&34÷5&:&)\";*3-*, ÖRNEK 38 ÖRNEK 41 #JSJNLBSFMFSEFOPMVöBO ZBOEBLJ öFLJMEF UBOY 3r YÖPMEVôVOBHÌSF AA EFôFSJOJCVMVOV[ 2 180°–x x sin2x + cos2x - tan x . cos x 180°–x JöMFNJOJOTPOVDVOFEJS BB CC |TJOY| + | cosx | - |UBOY|. cosx jCÌMHFEF -+ - TJOY< cosx > UBOY<ES = -TJOY+ cosx +UBOYDPTY \"#$ÑÀHFOJOEFO 4 sin x 4 = -TJOY+ cosx + cos x · cos x UBO - x ) = -UBOY= jUBOY= - 3 3 = -TJOY+ cosx +TJOY= cosx ÖRNEK 42 A 10 B ÖRNEK 39 x 5 12 12 16i =ÖPMEVôVOBHÌSF sin 6i. sin 3i x 180°–x 13 cos 5i. cos 2i 10 JGBEFTJOJOFöJUJOJCVMVOV[ D E 23 C π | |\"#$%ZBNVóVOEB [ AB ] // [%$], AB = 10 br, 8i = PMEVôVOEBO TJOi = cos2i ve | | | | | |BC = 5 br, \"% = 12 br, %$ = 23 br, m(D%AB) = x 2 tir. TJOi = cos5i olur. :VLBSEBLJWFSJMFSFHÌSF UBOYEFôFSJOJCVMVOV[ sin 6i. sin 3i =1 [ AD ] // [#&]PMBDBLöFLJMEF[#&]OÀJ[FMJN cos 5i. cos 2i 5 Bu dVSVNEBUBO - x ) = -UBOY= 12 UBOY= -5 olur. 12 ÖRNEK 40 ÖRNEK 43 12i =ÖPMEVôVOBHÌSF a + b = 60° cos 3i. tan 7i. sin 5i PMEVôVOBHÌSF TJO a + 4b JGBEFTJOJOFOTBEFIB- sin 7i. cos 9i. tan 5i MJOJCVMVOV[ JGBEFTJOJOFöJUJOJCVMVOV[ 3a + 3b = =TJO a + 3b + b ) =TJO + b ) 12i =ÖPMEVôVOEBO DPTi = -cos3i = -TJOb UBOi = -UBOi TJOi =TJOiES#VEVSVNEBDFWBQPMVS 23 4 5 43. mTJOb 41. - 42. - 38. cosx 1 1 3 12

·/÷7&34÷5&:&)\";*3-*, 5. MODÜL 53÷(0/0.&53÷ www.aydinyayinlari.com.tr ÖRNEK 44 5SJHPOPNFUSJL 'POLTJZPOMBSO \"À %FôFSMFSJOF (ÌSF4SBMBONBT #JS\"#$ÑÀHFOJJÀJO sin^ A + B h + sin C %m/*m cos^ A + B h - cos C 7FSJMFO B¿MBS CJSJN ¿FNCFSJO IBOHJ CËMHFTJO- JGBEFTJOJOFOTBEFI»MJOJCVMVOV[ EFPMVSTBPMTVOCVB¿MBSOUSJHPOPNFUSJLEFóF- SJEBSB¿DJOTJOEFOJGBEFFEJMJS A + B + C =j A + B =- C 7FSJMFOB¿MBSOUSJHPOPNFUSJLEFóFSMFSJBSBTO- sin^ 180° - C h + sin C da büyüklük-LÐ¿ÐLMÐL JMJõLJTJOJO CFMJSMFOFCJM- NFTJJ¿JOUSJHPOPNFUSJLEFóFSMFSEJLFZFLTFOMF- cos^ 180° - C h - cos C SF TJOÐT UBOKBOU WFZB ZBUBZ FLTFOMFSF LPTJ- OÐT LPUBOKBOU UBõOS sin C + sin C 2 sin C = = = - tan C CËMHFEFLJB¿MBSOCÐZÐLMÐLMFSJBSUUL¿BTJOÐT - cos C - cos C - 2 cos C EFóFSMFSJBSUBS LPTJOÐTEFóFSMFSJB[BMS ÖRNEK 45 CËMHFEFLJCJSB¿OOUBOKBOUEFóFSJTJOÐTEF- óFSJOEFOEBJNBCÐZÐLUÐS #JS\"#$ÑÀHFOJJÀJO ÖRNEK 47 tanf A + B p - cot C 22 \"öBôEBWFSJMFO JGBEFMFSJCÑZÑLUFOLÑÀÑôFTSBMBZ- O[ sinf A + B p + cos C B sin40°, cos40°, cot40° 22 C sin20°, cos50°, tan60°, cot10° JGBEFTJOJOFOTBEFI»MJOJCVMVOV[ B TJO TJO UBOWFUBO>TJO>TJO DPU>DPT>TJOEJS A + B + C = A+B+C = 90°JTF C TJO TJO UBO UBOWF 2 UBO>UBO>TJO>TJOPMEVôVOEBO DPU>UBO>DPT>TJOEJS A+B C = 90° - olur. 22 tand 90° - C n - cot C CC 22 cot - cot 2 2 sind 90° - C n + cos C = =0 22 CC cos + cos 22 ÖRNEK 46 ÖRNEK 48 TJO=YPMNBLÑzere, π <a<i<π cos2 50° + sin 130° . cos 220° 2 tan 240° JLFOBöBôEBLJMFSEFOIBOHJMFSJEBJNBEPôSVEVS JGBEFTJOJOEFôFSJOJCVMVOV[ I. cosi > sina II. tani < tana III. cosi > cos! IV. sini > sina sin240° + cos 40° .^ - cos 40° h #JSJNÀFNCFSÑ[FSJOEFBÀMBSJöBSFUMFOFDFLZBQMBCJMF- DFôJHJCJEFôFSWFSFSFLEFZBQMBCJMJS tan 60° i =WFa =PMTVO DPT>TJO j -DPT>TJO : sin240° - cos240° sin 2 40° - a 1 - 2 k UBO<UBO j -UBO< -UBO jUBO>UBO : sin 40° DPT>DPT j -DPT> -DPT == jDPT<DPT : TJO>TJO j TJO>TJO : 33 2 sin240° - 1 2 == 2x - 1 33 44. mUBO$ 45. 46. ^ 2 - 1 h / 3 24 B DPUDPTTJO48. Y, Y, Y, Y 2x C DPUUBODPTTJO

5SJHPOPNFUSJL'POLTJZPOMBS TEST - 6 1. Ölçüsü 236425PMBOBÀOOÌMÀÑTÑBöBôEBLJ- 5. MJLBÀOOFTBTÌMÀÑTÑLBÀEFSFDFEJS MFSEFOIBOHJTJOFFöJUUJS \" # $ % & A) 65° 4124 B) 65° 4225 C) 65° 3924 % 25 E) 65° 4024 2. A = 36° 4527 ve B = 43° 37 55 PMEVôVOBHÌSF \"+#UPQMBNBöBôEBLJMFSEFO 6. -MJLBÀOOFTBTÌMÀÑTÑLBÀEFSFDFEJS IBOHJTJOFFöJUUJS \" # $ % & A) 117° 348 B) 117° 849 C) 117° 1319 % 29 E) 116° 4329 3. ²MÀÑTÑ PMBOBÀOOÌMÀÑTÑBöBôEBLJMFS- MJLBÀOOFTBTÌMÀÑTÑLBÀSBEZBOES EFOIBOHJTJOFFöJUUJS A) π B) 2π C) 4π % 8π E) 10π 9 9 A) 2° 4323 B) 2° 4423 C) 2° 44 43 99 9 % 43 E) 2° 4523 4. #JS\"#$ÑÀHFOJOEF 8. Ölçüsü 31Õ SBEZBO PMBO BÀOO FTBT ÌMÀÑTÑ m (WA) = 63° 39 47 , m (WB) = 54° 48 51 4 LBÀSBEZBOES PMEVôVOBHÌSF m ( XC )BöBôEBLJMFSEFOIBOHJTJ- EJS A) 3π B) 5π C) 7π % 9π E) 5π 4 4 4 88 A) 60° 32 24 B) 60° 31 24 C) 61° 32 22 % 22 E) 62° 31 22 1. D 2. B 3. B 4. D 25 5. & 6. D & 8. C

TEST - 7 5SJHPOPNFUSJL'POLTJZPOMBS 1. Ölçüsü - 46π SBEZBOPMBOBÀOOFTBTÌMÀÑTÑ 5. 2 sin x + 3t = 2 9 5 LBÀSBEZBOES FöJUMJôJOEFUOJOBMBCJMFDFôJUBNTBZEFôFSMFSJ A) 11π B) 8π C) 7π % 5π E) 5π UPQMBNLBÀUS 99 9 98 \" # $ % & 2. \"öBôEBLJMFSEFO IBOHJTJOJO FTBT ÌMÀÑTÑ 6. sin243° , cos317° , tan124° , cot191° EJS JGBEFMFSJOJO JöBSFUMFSJ TSBTZMB BöBôEBLJMFSEFO IBOHJTJEJS A) 4200° B) -3400° 21π C) 6 A) -, -, -, + B) -, +, -, - C) -, +, -, + % - 13π E) -300° % -, +, +, + E) +, +, -, + 3 3. A ABC bir üçgen x <\"%>B¿PSUBZ cos257° , sin312° , tan227° , cot158° BD m ( % ) = 7π BAC 15 TBZMBSOO JöBSFUMFSJ TSBTZMB BöBôEBLJMFSEFO IBOHJTJEJS % = 2730' m ( ABC ) C A) -, -, -, - B) -, +, +, - C) -, -, +, + % % -, -, +, - E) +, - + - :VLBSEBLJWFSJMFSFHÌSF m ( ADC ) = x LBÀEFSF- DFEJS \" # $ % & 4. I. 3780° = 180° 8. 11π 13π sin , cos 75 II. -2520° = 120° 21π , cot f 162π p tan III. - 49π = 5π 57 33 TBZMBSOO JöBSFUMFSJ TSBTZMB BöBôEBLJMFSEFO IV. 127π = 3π IBOHJTJEJS 77 A) -, -, -, - B) -, -, +, - C) -, +, -, - :VLBSEBLJFöJUMJLMFSEFOLBÀUBOFTJEPôSVEVS \" # $ % & % -, -, +, + E) +, -, -, - 1. B 2. D 3. A 4. B 26 5. A 6. C D 8. D

5SJHPOPNFUSJL'POLTJZPOMBS TEST - 8 1. 7 5. cosec x - sin x Afx, - p 4 1 + cos x JGBEFTJOJOFöJUJBöBôEBLJMFSE FOIBOHJTJEJS OPLUBTCJSJNÀFNCFSÑ[FSJOEFPMEVô VOBHÌSF Y JO BMBCJMFDFôJ OFHBUJG EFôFS BöBôEBLJMFSEFO A) tanx B) cotx C) secx & TJOY IBOHJTJEJS A) - 1 B) - 1 C) - 3 % -1 E) - 5 % DPTFDY 4 2 4 4 2. A = -3 + 2sin ( x - 4 ) (1 + sin x) . (tan x - sec x) PMEVôVOBHÌSF \"OOBMBCJMFDFôJLBÀGBSLMUBN 6. TBZEFôFSJWBSES sin x JGBEFTJOJOFöJUJBöBôE BLJMFSEFOIBOHJTJEJS A) -tanx B) tanx C) -cotx \" # $ % & % DPUY & -cosec 3. N`;PMNBLÑ[FSF 1 + tan2x + cosec2x 3sinx + 2m - 8 = 0 cosec2x PMEVôVOBHÌSF NOJOBMBCJMFDFôJLBÀGBSLMEF- JGBEFTJOJOFöJUJBöBôE BLJleSEFOIBOHJTJEJS ôFSWBSES A) tan2 x B) cot2 x C) sec2 x % DPTFD2 x E) sin2 x \" # $ % & 4. 2 sin2 270° + 3 cos180° - tan360° 8. tan x - cot x JGBEFTJOJOFöJUJBöBôEBLJMFSEFOIBOHJTJEJS cosec x - 1 A) -5 B) -3 C) - % & JGBEFTJOJOFöJUJBöBôEBLJMFSd FOIBOHJTJEJS A) -secx B) secx C) -cosecx % DPTFDY & -tanx 1. C 2. D 3. B 4. C 5. B 6. C C 8. A

TEST - 9 5SJHPOPNFUSJL'POLTJZPOMBS 1. tan4 x + 2sec2 x - sec4 x 5. 1 - sin2 x - sin x JGBEFTJOJOFöJUJBöBôEBL JMFSEFOIBOHJTJEJS sin3 x. cot x (1 - cos2 x) . tan x JGBEFTJOJOFöJUJBöBôEBLJMFSEFOIBOHJTJEJS A) -1 B) 0 C) 1 % TJOY & DPTY A) –1 B) 0 C) 1 % TFDY & DPTFDY 2. 2 cos x - 3 sin x = 2 6. sin2 x 4 cos x + 3 sin x sin2 x + cos x - 1 PMEVôVOB HÌSF DPUY BöBôEBLJMFSEFO IBOHJTJ- JGBEFTJOJOFöJUJBöBôEBLJMFSE FOIBOHJTJEJS EJS A) - 3 B) - 2 C) 2 % 3 A) secx B) cosecx C) 1 + secx 2 E) 1 33 2 % + cosecx E) 1 - cosecx 3. sin2 x - 1 UBOY- cotx = 3 PMEVôVOBHÌSF 1 - sin2 x (1 + cos2 x) 2 tan2 x + 1 tan2x JGBEFTJOJOFöJUJBöBôEBLJMFSJOIBOHJTJEJS JGBEFTJOJOFöJUJBöBôEBLJMFSEFOIBOHJTJEJS A) -tan2 x B) -cot2 x C) -sec2 x A) 1 5 C) 9 % 13 E) 17 4 B) 4 44 % -cosec2 x E) cot2 x 4 4. sin2 x - (sin x - cos x) . (sin x + cos x) 8. secx + cosx = 3 1 - tan2 x PMEVôVOB HÌSF TFDY - DPTY JGBEFTJOJO FöJUJ JGBEFTJOJOFöJUJBöBôEBLJMFSEFOIBOHJTJEJS BöBôEBLJMFSEFOIBOHJTJPMBCJMJS A) 5 B) 3 C) 5 % 3 E) 7 A) -1 B) 0 C) 1 28 5. B 6. C & 8. C % UBOY & DPUY 1. C 2. A 3. C 4. C

5SJHPOPNFUSJL'POLTJZPOMBS TEST - 10 (1 - cot x) . (1 + cot x) . sin2x 5. sinx - cosx = 2 1. 3 PMEVôVOBHÌSF UBOY+DPUYUPQMBNLBÀUS sin x - cos x JGBEFTJOJOFöJUJBöBôEBLJMFSEFOIBOHJTJEJS A) 13 B) 18 $ % 21 E) 23 5 5 55 A) sinx + cosx B) sinx - cosx C) -1 % DPTY & DPTY- sinx 2. 5sinx = 7 - 5cosx 6. sin x - 2 cos x = 3 PMEVôVOBHÌSF TJOYDPTYJGBE FTJOJOFöJUJBöB- sin x + cos x 4 ôEBLJMFSEFOIBOHJTJEJS PMEVôVOBHÌSF UBOYLBÀUS A) 24 18 C) 12 % 9 8 \" # $ % & 25 B) E) 25 25 25 25 3. sec x + cosec x - cos x 0 < x < 3π , cos x = 3 tan x + cot x 25 JGBEFTJOJOFöJUJBöBôEBLJMFSEFOIBOHJTJEJS PMEVôVOBHÌSF TJOYLBÀUS A) cosx B) -cosx C) sinx A) - 3 B) - 4 C) - 3 % 4 4 5 5 43 E) 5 % -sinx E) 0 4. sin2 x. cot x. cosec x. cos x 8. π < x < 3π , tan x = 1 22 3 sec x. cos2 x PMEVôVOBHÌSF cPTYTJOYÀBSQNLBÀUS JGBEFTJOJOFöJUJBöBôEBLJMFSEFOIBOHJTJEJS A) cosx B) cos2 x C) sin2 x A) - 1 B) - 3 C) 1 10 10 10 % 3 % TJOY E) 1 10 E) 1 1. A 2. C 3. C 4. A 5. B 6. A & 8. D

TEST - 11 5SJHPOPNFUSJL'POLTJZPOMBS 1. TJO=BPMEVôVOBHÌSF 5. bCJSEBSBÀPMNBLÑ[FSF cos 250° . sin 340° 3 sin (b - 5π ) - sin (- b) - 2 cosc π + b m sin 160° . sin 200° 2 JöMFNJOJOTPOVDVBöBôEBLJMFSEFOIBOHJTJEJS JGBEFTJOJOFöJUJBöBôEBLJMFSEFOIBOHJTJEJS % - 1 E) 1 A) 0 B) -4sinb C) 4sinb 2 A) -1 B) -B $ B % TJOb E) 6sinb 2. \"öBôEBLJMFSEFO IBOHJTJOJO EFôFSJ FO CÑZÑL- 6. \"öBôEBLJMFSEFOIBOHJTJDPTZFFöJUEFôJM UÑS EJS A) sin20° B) cos40° C) sin60° A) sin200° B) cos110° C) sin340° % DPT & TJO % -cos70° E) cos290° 3. YCJSEBSBÀPMNBLÑ[FSF I. -tan40° II. -cot50° IV. cot310° cot f x - 17π p III. tan320° VI. -cot40° 2 V. cot130° cos f x - 7π p + cos f 19π - x p :VLBSEBLJMFSEFOLBÀUBOFTJUBOZFFöJUUJS 22 JGBEFTJOJOFöJUJBöBôEBLJMFSEFOIBOHJTJEJS A) secx B) 2secx sec x \" # $ % & C) % DPTFDY 2 E) cosec x 2 4. iCJSEBSBÀPMNBLÑ[FSF sin 340° + cot (- 240° ) + cos 290° sin (11π - i) - cosf 5π - i p + tanf 7π + i p 8. 22 tan 30° JGBEFTJOJOFöJUJBöBôEBLJMFSEFOIBOHJTJEJS JGBEFTJOJOFöJUJBöBôEBLJMFSEFOIBOHJTJEJS A) - 3 3 C) -1 3 B) 3 A) -coti B) coti C) -tani % E) 3 % UBOi E) -seci 1. A 2. & 3. C 4. A 5. A 6. & & 8. C

5SJHPOPNFUSJL'POLTJZPOMBS TEST - 12 tan (- 45° ) 5. x = sin27° , y = cos73° 1. z = tan46° , k = cot200° sin 150° - cos 120° PMEVôVOBHÌSF BöBôEBLJTSBMBNBMBSEBOIBO- JöMFNJOJOTPOVDVBöBôEBLJMFSEFOIBOHJTJEJS HJTJEPôSVEVS A) -2 B) - $ % & A) y < x < k < z B) y < x < z < k C) x < y < k <[ % Y< y < z < k E) y < z < x < k 2. cot105° . cot120° . cot135° . cot150° . cot165° 6. a = cos109° , b = cos123° ÀBSQNOOTPOVDVBöBôEBLJMFSEFOIBOHJTJEJS c = cos305° , d = cos216° A) - 3 B) 3 C) - % & -2 PMEVôVOB HÌSF BöBôEBLJ TSBMBNBMBSEBO IBO- 88 HJTJEPôSVEVS A) a < b < d < c B) c < a < b < d C) a < d < b <D % E< b < a < c E) d < b < c < a 3. tan 20° . tan 21° . tan 22° . . . tan 70° 3π < a < b < 2πPMNBLÑ[FSF cot 10° . cot 11° . cot 12° . . . cot 80° 2 JöMFNJOJOTPO VDVBöBôEBLJMFSEFOIBOHJTJEJS 2 3 I. cosa > cosb II. sina > sinb A) C) B) 1 III. tana < tanb IV. cota < cotb 2 3 % 3 E) 2 V. coseca > cosecb FöJUTJ[MJLMFSJOEFOLBÀUBOFTJEPôSVEVS \" # $ % & 4. cot ( 570° ) + tan ( -660° ) 8. 3sin ( x - 20°) + 4cos ( y - 10°) JöMFNJOJOTPOVDVBöBôEBLJMFSEFOIBOHJTJEJS C) - 2 3 JGBEFTJOJOBMBCJMFD FôJFOCÑZÑLEFôFSLBÀUS 3 A) - 2 3 B) 0 E) 2 3 \" # $ % & 51 % 1. B 2. C 3. B 4. & 31 5. B 6. D B 8. D

TEST - 13 5SJHPOPNFUSJL'POLTJZPOMBS A 4. A ABC bir dik üçgen 1. [ #%] m [AC] D | |D AC = 10 birim 8C E % = a a m ( BAC ) B2 F % = b m ( ACD ) \"NFSLF[MJ¿FZSFLEBJSF 'OPLUBTOEB[ BC ] ye te- BC | | | |óFU [ AB ] m [ AC ], BF = 2 br, CF = 8 br cota + cotb = 5 % | |:VLBSEBLJWFSJMFSFHÌSF BD LBÀCJSJNEJS m ( ABC ) = a ES \" # $ % & :VLBSEBLJWFSJMFSFHÌSF DPUaLBÀUS 5. A ABC ikizkenar üçgen A) 1 B) 2 C) 1 % 1 E) 1 248 D B | | | |AB = AC = 10 cm [ AB ] m [$%] % = a m ( DCB ) 2. A F B \"#$%EJLEËSUHFO | |a BC = 12 cm dir. C 1 [ EF ] m [ FC ] Ea |AE| = |FB| m ( D%EF ) = a :VLBSEBLJWFSJMFSFHÌSF DPTaLBÀUS D | |C EF = 1 birim A) 2 B) 3 C) 4 % 3 E) 1 3 4 55 | |:VLBSEBLJWFSJMFSFHÌSF BC BöBôEBLJMFSEFO 6. ôFLJMEFCJSLËõFTJCJSJN¿FNCFSÐ[FSJOEFPMBO\"#0 IBOHJTJOFFöJUUJS EJLÐ¿HFOJWFSJMNJõUJS y A) sina B) cosa C) -sina 1 % DPTa E) tana A 3. \"õBóEBLJõFLJMË[EFõLBSFEFOPMVõNVõUVS C a x –1 B O [AB], x eksenine dik ve m ( A%OB ) = aES ab BC PSBOBöBôEBLJMFSEFOIBO- #VOBHÌSF AB HJTJOFFöJUUJS #VOBHÌSF DPUaUBObLBÀUS A) coseca - cota B) seca - tana A) -4 B) -2 C) - % C) tana - cota % TJOa - cosa & E) seca - coseca 1. C 2. C 3. B 32 4. A 5. C 6. A

5SJHPOPNFUSJL'POLTJZPOMBS TEST - 14 1. y 4. Y=ÖPMEVôVOBHÌSF C cot 4x + sin2 3x + sin2 2x a tan x JGBEFTJOJOEFôFSJLBÀUS E B A) -2 B) - $ % & x O DA ôFLJMEF[ AC ], O merkezli birim çembere B nokta- TOEBUFóFU [OB ] m [AC ] ve m ( % ) = a ES ACD #VOB HÌSF \"0$ ÑÀHFOJOJO BMBO BöBôEBLJMFS- EFOIBOHJTJOFFöJUUJS A) sin a - 1 B) cos a + sin a 2 2 C) tan a % cot a + tan a 2 2 E) sin a - cos a 5. TJO=BPMEVôVOBHÌSF 2 2 cos 280° + sin 350° sec 80° JGBEFTJOJO B DJOTJOEFO EFôFSJ BöBôEBLJMFSEFO IBOHJTJEJS A) a2 B) 1 $ B % 1 E) 1 a2 a 2. tanx - cotx = 2 PMEVôVOBHÌSF UBO3 x - cot3 YJGBEFTJOJOEFôF- SJLBÀUS \" # $ % & 3. x + y = π PMEVôVOBHÌSF 6. a ve iEBSBÀPMNBLÑ[FSF 4 tana < tani tan ( 4x + 5y ) PMEVôVOB HÌSF BöBôEBLJMFSEFO IBOHJTJ EBJNB EPôSVEVS JGBEFTJOJOFöJUJBöBôEBLJMFSEFOIBOHJTJEJS A) tan ( 90° + x ) B) tan ( 90° + y ) A) sina > sini B) cosa < cosi C) -UBOZ % UBOY C) cota > coti % TJOa > cosi E) tany E) cota < tani 1. D 2. A 3. & 33 4. & 5. A 6. C

TEST - 15 5SJHPOPNFUSJL'POLTJZPOMBS 1. k `/+PMNBLÑ[FSF 4. YHFOJöBÀPMNBLÑ[FSF 175π = 2π .k + , 1 - cos x . 1 + cos x 9 1 + sin x 1 - sin x JGBEFTJOJOFöJUJBöBôEBLJMFSEFOIBOHJTJEJS PMEVôVOBHÌSF LùJGBEFTJOJOFOLÑÀÑLQP[J- UJGEFôFSJLBÀUS A) sinx B) cotx C) -tanx \" Õ # Õ $ Õ % Õ & Õ % -secx E) -cotx 2. \"öBôEBLJMFSEFOIBOHJTJCJSJNÀFNCFSÑ[FSJO- 5. YEBSBÀPMNBLÑ[FSF EFZFSBMBOCJSOPLUBES 1 # cosecx # 2 LPöVMVOV TBôMBZBO Y BÀMBSOO EFSFDF PMBSBL A) f - 5 , 12 p B) ( 1, -1 ) 13 13 LBÀUBOFUBNTBZEFôFSJWBSES % f 3 , 1 p \" # $ % & 44 C) ( tan15°, cot75° ) E) ( sin30°, cos60° ) 3. sec x - cosec x 6. \"#$% B 1 + tan x 1 + cot x A paralelkenar JGBEFTJOJOFöJUJBöBôEBLJMFSEFOIBOHJTJEJS x m (XC) = y D m (XD) = x y tan x = 5 8 C A) 0 B) 1 C) 1 :VLBSEBLJWFSJMFSFHÌSF DPUZLBÀUS % UBOY 2 & DPUY A) 3 B) 5 C) - 3 % - 8 E) 1 8 8 85 1. B 2. A 3. A 34 4. C 5. B 6. D

www.aydinyayinlari.com.tr 53÷(0/0.&53÷ 5. MODÜL ·/÷7&34÷5&:&)\";*3-*, 53÷(0/0.&53÷,'0/,4÷:0/-\"3*/(3\"'÷,-&3÷ 1FSJZPUWF1FSJZPEJL'POLTJZPO ÖRNEK 4 %m/*m G\"Z B ve rx ` A için, G Y+ 7) =G Y PMBSBLUBONMBOZPS G Y GPOLTJZPOVOVO UBON LÐNFTJOEFLJ IFS Y G = -WFG = 2 PMEVôVOBHÌSF FMFNBOJ¿JOG Y =G Y+5 FõJUMJóJOJTBóMBZBO G +G - UPQMBNOOEFôFSJOJCVMVOV[ T ` R+WBSTBGGPOLTJZPOVOBQFSJZPEJLGPOLTJ- ZPO FOLÐ¿ÐL5TBZTOBCVGPOLTJZPOVOQF- G Y+ =G Y PMEVôVOEBOGGPOLTJZPOVOVOQFSJZPEV SJZPEV denir. EJS#VEVSVNEB G =G = ... =G = -1 ve 1FSJZPU 5 BZOEFóFSMFSJOUFLSBSFUUJóJFOLÐ- G =G -5 ) = ... =G -33) =PMBSBLCVMVOVS ¿ÐLBSBMLUS G +G -33 ) = 2 ( -1 ) + 2 =PMVS ÖRNEK 1 x `;PMNBLÑ[FSF G Y = \"x in 3 ile bölümünden kalan\" öFLMJOEFUBONMB- OBOGPOLTJZPOVOVOQFSJZPEVOVCVMVOV[ G Y =G Y+5 PMBDBLöFLMEFLJFOLÑÀÑL5` R+EFôF- ÖRNEK 5 SJUÑS G = G = 1 \"õBóEBWFSJMFOEËONFEPMBCOZFSEFOZÐLTFLMJóJNWF G = G = 2 ZBS¿BQNEJS G = G = PMEVôVOEBOGPOLTJZPOVOQFSJZPEVUÑS A ÖRNEK 2 10 m R+ Z3PMNBLÑ[FSF 2m G Y = x2 + 2 GPOLTJZPOVOVOQFSJZPEJLPMVQPMNBEôOCVMVOV[ 'POLTJZPOUBONMPMEVôVBSBMLUBEBJNBBSUBOPMEVôV JÀJOQFSJZPEJLEFôJMEJS ÖRNEK 3 #VEÌONFEPMBQCJSUVSVOVTBOJZFEFUBNBNMBE- ôOB HÌSF \" OPLUBTOO ZFSEFO ZÑLTFLMJôJOJO N GGPOLTJZPOVOVOQFSJZPEVPMEVôVOBHÌSF PMEVôV IFSIBOHJ JLJ BO JÀJO CV TÑSFMFSJO GBSL FO B[ g ( x ) =G Y+ 1 ) - 3 LBÀTBOJZFEJS GPOLTJZPOVOVOQFSJZPEVOVCVMVOV[ \" OPLUBTOO ZFSEFO ZÑLTFLMJôJOJONPMEV- Tg =HGPOLTJZPOVOVOQFSJZPEV A A ôVBO ZFSEÑ[MFNJOFQB- 12 m 12 m SBMFMPMBOÀBQOJLJVDVO- EB PMEVôV [BNBOES #V TG =GGPOLTJZPOVOVOQFSJZPEVPMNBLÑ[FSF EVSVN \" OPLUBTOO ZB- T SNUVSBUUô[BNBOHFS- Tg = 3 ÀFLMFöJS 0 ZÑ[EFO TÑSF G = GBSLFOB[TOPMVS 2 2 1. 3 2. 1FSJZPEJLEFôJMEJS 3 35 4. 5. 3. 2

·/÷7&34÷5&:&)\";*3-*, .0%·- 53÷(0/0.&53÷ www.aydinyayinlari.com.tr 4JOÑTWF,PTJOÑT'POLTJZPOMBSOO1FSJZPUMBS %m/*m Y π Õ 3π Õ 5π Õ 7π Õ 2222 TJOY 1 -1 1 -1 :VLBSEBLJUBCMPEBTJOÐTGPOLTJZPOVJ¿JO[ Õ], [Õ Õ] BSBMLMBSOEB 0, 1, 0, -1, 0EFóFSMFSJUFLSBSFUNFL- UFEJS#VSBEBOTJOY=TJO Y+Õ =TJO Y+Õ ==TJO Y+LÕ PMEVóVHËSÐMÐSrY` R, k ` Z+PMNBL Ð[FSFTJO Y+LÕ =TJOYPMEVóVOEBOTJOÐTGPOLTJZPOVOVOQFSJZPEVFOLÐ¿ÐLL` Z+J¿JO5=ÕPMVS Y π Õ 3π Õ 5π Õ 7π Õ DPTY 1 2222 -1 1 -1 1 :VLBSEBLJUBCMPEBLPTJOÐTGPOLTJZPOVJ¿JO[ Õ], [Õ Õ] BSBMLMBSOEB1, 0, -1, 0, 1EFóFSMFSJOJOUFLSBS- MBOEóHËSÐMNFLUFEJS#VSBEBOrY` R, k ` Z+J¿JODPT Y+LÕ =DPTYPMEVóVOEBOLPTJOÐTGPOLTJZPOVOVO QFSJZPEVFOLÐ¿ÐLL` Z+J¿JOT =ÖPMVS %m/*m ÖRNEK 6 y \"öBôEB WFSJMFO GPOLTJZPOMBSO QFSJZPUMBSO CVMV- 1 OV[ a) G Y =TJO Y+ P b) f^ x h = 3 cosf 3x + 1 p Õa 1 5 x c) f^ x h = 2 sin2f x - 1 p –1 a O 7 E f^ x h = 1 - 3 cos3f 2x + 1 p –1 4 :VLBSEBLJ CJSJN ¿FNCFSEF HËSÐMEÐóÐ HJCJ a F G Y =TJOY+DPTY B¿TOBÕWFÕOJOLBUMBSFLMFOEJóJOEFTJOÐT WFLPTJOÐTEFóFSMFSJEFóJõNF[ f) G Y =TJO TJOa =TJO a +LÕ 2π 2π 2π 10π a) T = = b) = DPTa =DPT a +LÕ 33 33 %m/*m 5 L B C`3WFLáPMNBLÐ[FSF π 2π G Y =LTJOO BY+C c) = 7π E = 4π H Y =LDPTO BY+C GPOLTJZPOMBSOEB 1 2 OUFLEPóBMTBZJTF T = 2π 7 4 a O¿JGUEPóBMTBZJTF T = π Oá F TJO2 x +DPT2 x = PMEVôVOEBOQFSJZPEVZPLUVS a G G Y TBCJUGPOLJTZPOPMEVôVJÀJOQFSJZPEVZPLUVS PMBSBLCVMVOVS 2π 10π 6. a) b) D ÖE ÖF QFSJZPEVZPLUVS 36 33 G QFSJZPEVZPLUVS

www.aydinyayinlari.com.tr 53÷(0/0.&53÷ .0%·- ·/÷7&34÷5&:&)\";*3-*, 5BOKBOUWF,PUBOKBOU'POLTJZPOMBSOO1FSJZPUMBS %m/*m 0 π Õ 3π Õ 5π Õ 7π Õ ... ... x 22 22 tanx 0 5BONT[ 0 5BONT[ 0 5BONT[ 0 5BONT[ 0 :VLBSEBLJUBCMPEBUBOKBOUGPOLTJZPOVJ¿JO[ Õ], [Õ Õ], [Õ Õ] BSBMLMBSOEB UBONT[ õFLMJOEFUFL- SBSFEFOTPOV¿MBSFMEFFEJMJS#VSBEBOUBOY= ( x +Õ = ... = tan ( x +LÕ PMEVóVHËSÐMÐSr x ` R, k ` Z+J¿JO tan ( x +LÕ =UBOYPMEVóVOEBOUBOKBOUGPOLTJZPOVOVOQFSJZPEVFOLÐ¿ÐLL` Z+J¿JO5=ÕPMVS x 0 π Õ 3π Õ 5π Õ 7π Õ ... 22 22 DPUY 0 5BONT[ 0 5BONT[ 0 5BONT[ 0 5BONT[ 0 ... :VLBSEBLJUBCMPEBLPUBOKBOUGPOLTJZPOVJ¿JO[ Õ], [Õ Õ], [Õ Õ] BSBMLMBSOEB UBONT[ UBONT[ õFLMJOEFUFLSBSFEFOTPOV¿MBSFMEFFEJMJS#VSBEBODPUBOY=DPU Y+Õ =DPU Y+Õ = ... =DPU Y+LÕ PM- EVóVHËSÐMÐSr x ` R, k ` Z+J¿JOUBO Y+LÕ =UBOYPMEVóVOEBOUBOKBOUGPOLTJZPOVOVOQFSJZPEVFOLÐ¿ÐL k ` Z+J¿JO5=ÕPMVS ÖRNEK 7 %m/*m \"öBôEB WFSJMFO GPOLTJZPOMBSO QFSJZPUMBSO CVMV- #JSEFO GB[MB USJHPOPNFUSJL GPOLTJZPOVOVO UPQ- OV[ MBNOOWFZBGBSLOOQFSJZPEVCVGPOLTJZPOMB- SOQFSJZPUMBSOO0,&,hJOFFõJUUJS a) G Y = tan ( 2x - b) H Y DPU ÕY- I Y =G Y ÷H Y JTF c) I Y = tan2 ÕY- d) U Y =DPU3 ( 3x -Õ TI Y =0,&, 5G Y , TH Y PMBSBLCVMVOVS e) h^ x h = 2 tan2f πx - 3 p - 1 ÖRNEK 8 4 G Y =DPT2 c x + 40° m + tan3 ( x - f) s^ x h = 3 cot3f 2x - 1 p + 5 2 5 GPOLTJZPOVOVOFTBTQFSJZPEVOVCVMVOV[ ππ π DPT2 d x + 40° nOFTBTQFSJZPEV T = π = 2π a) = b) = 1 2 11 22 π 2 π ππ tan3 ^ x - 10° hVOFTBTQFSJZPEV c) = 1 d) = π π 33 T = = π JTF π π 5π 21 e) = 4 f) = Tf(x) =0,&, 51, T2) =0,&, Ö Ö =ÖPMVS. π 22 4 5 7. a) π π 5π 37 8. Ö b) 1 c) 1 d) e) 4 f) 2 32

·/÷7&34÷5&:&)\";*3-*, .0%·- 53÷(0/0.&53÷ www.aydinyayinlari.com.tr 4JOÑT'POLTJZPOVOVO(SBGJôJ %m/*m x -Õ - 3π -Õ - π 0 π Õ 3π Õ 22 0 22 sinx 0 1 0 -1 1 0 -1 0 5BCMPEBPMVõUVSVMBO Y TJOY OPLUBMBSBOBMJUJLEÐ[MFNEFJõBSFUMFOFSFLBSEõLOPLUBMBSBVZHVOCJSõFLJMEFCJS- MFõUJSJMEJóJOEFTJOYGPOLTJZPOVOVOHSBGJóJBõBóEBLJHJCJ¿J[JMJS y 1 Õ mÕ – Õ OÕ Õ 2x 2 2 Õ mÕ –1 :VLBSEBLJHSBGJóJOPSJKJOFHËSFTJNFUSJLPMEVóVHËSÐMNFLUFEJS0SJKJOFHËSFTJNFUSJLGPOLTJZPOMBSUFLGPOLTJZPO- EVSGGPOLTJZPOVOVOUBONLÐNFTJOEFLJIFSYEFóFSJJ¿JO G -x ) = -G Y JTFGUFLUJS G -x ) =G Y JTFG¿JGUUJS #VSBEBUBONHFSFóJTJO -x ) = -TJOYPMEVóVOEBOTJOYGPOLTJZPOVUFLGPOLTJZPOEVS ÖRNEK 9 ÖRNEK 10 \"öBôEB WFSJMFO GPOLTJZPOMBSO HSBGJLMFSJOJ JODFMFZJ- \"öBôEB WFSJMFO GPOLTJZPOMBSO HSBGJLMFSJOJ JODFMFZJ- OJ[ OJ[ B y = sin ( 2x ) B y = -sin ( 3x ) C y = sin ( 2x ) + 1 C y = -sin ( 3x ) + 1 D y = 2sin ( sin2x ) + 1 D y = -2sin ( sin3x ) + 1 ¦Ì[ÑN ¦Ì[ÑN y y y = 2sin(2x) + 1 y = –2sin(3x) +1 3 3 2 1 2 1 –3 –2 –1 1 2 3 x –1 4 –4 –3 –2 –1 O 12 34 x –2 y = sin2x –1 y = sin(2x) + 1 y = –sin(3x) –3 –2 y = –sin(3x) +1 38

www.aydinyayinlari.com.tr 53÷(0/0.&53÷ .0%·- ·/÷7&34÷5&:&)\";*3-*, ,PTJOÑT'POLTJZPOVOVO(SBGJôJ %m/*m r x ` R DPT Y+LÕ =DPTYPMEVóVOEBODPTYGPOLTJZPOVOVOQFSJZPEVÕPMVS#VGPOLTJZPOVOHSBGJóJ[ Õ] ¿J[JMJS(SBGJL[-Õ -Õ] [-Õ ] [ Õ] [Õ Õ]BSBMLMBSOEBUFLSBSFEFS[ Õ]CJSLB¿EFóFSTF¿JMEJ- óJOEFBõBóEBLJUBCMPPMVõVS x -Õ - 3π -Õ - π 0 π Õ 3π Õ 22 1 22 cosx 1 0 -1 0 0 -1 0 1 5BCMPEBPMVõUVSVMBO Y DPTY OPLUBMBSBOBMJUJLEÐ[MFNEFJõBSFUMFOFSFLBSEõLOPLUBMBSVZHVOCJSõFLJMEFCJS- MFõUJSJMEJóJOEFDPTYGPOLTJZPOVOVOHSBGJóJBõBóEBLJHJCJ¿J[JMJS y 1 mÕ Õx Õ Õ mÕ – Õ –Õ O Õ 2 2 2 2 –1 :VLBSEBLJHSBGJóJOZFLTFOJOFHËSFTJNFUSJLPMEVóVHËSÐMNFLUFEJS ,PTJOÐTGPOLTJZPOVOVOHSBGJóJZFLTFOJOFHËSFTJNFUSJLPMEVóVOEBOLPTJOÐTGPOLTJZPOV¿JGUGPOLTJZPOEVS%P- MBZTZMBDPT -x ) =DPTYPMVS ÖRNEK 11 ÖRNEK 12 \"öBôEB WFSJMFO GPOLTJZPOMBSO HSBGJLMFSJOJ JODFMFZJ- \"öBôEB WFSJMFO GPOLTJZPOMBSO HSBGJLMFSJOJ JODFMFZJ- OJ[ OJ[ B y = cos ( 3x ) B y = -cos ( 2x ) C y = cos ( 3x ) - 1 C y = -cos ( 2x ) - 1 D y = 2cos ( 3x ) - 1 D y = -2cos ( 2x ) - 1 ¦Ì[ÑN ¦Ì[ÑN y y y = –2cos(2x)–1 3 1 y = cos3x 2 O x 1 –4 –3 –2 –1 1 23 –1 –4 –3 –2 –1 O 1 2 3 4 x –2 –1 y = cos(2x)+1 y = –cos2x –2 y = 2cos(3x)–1 y = cos(3x)–2 39

·/÷7&34÷5&:&)\";*3-*, 5. MODÜL 53÷(0/0.&53÷ www.aydinyayinlari.com.tr 5BOKBOU'POLTJZPOVOVO(SBGJôJ %m/*m r x ` R, tan ( x +LÕ =UBOYPMEVóVOEBOUBOYGPOLTJZPOVOVOQFSJZPEVÕPMVS#VZÐ[EFOUBOKBOUGPOLTJZPOV- OVOHSBGJóJ;- π , 0 m , f π , 3π p gibi arBMLMBSEBUFLSBSFEFS 2 22 x 0 π Õ 3π Õ 5π Õ 7π Õ ... ... 22 22 tanx 0 5BONT[ 0 5BONT[ 0 5BONT[ 0 5BONT[ 0 y –Õ Õ Õ 2 x mÕ OÕ Õ 22 (SBGJLUFOUBOYGPOLTJZPOVOVO L+ 1 ) . π ( k `; B¿ËM¿ÐMFSJJ¿JOUBONT[PMEVóVOBEJLLBUFEJOJ[ 2 UBOYGPOLTJZPOVPSJKJOFHËSFTJNFUSJLPMEVóVOEBOUFLGPOLTJZPOEur. UBOYGPOLTJZPOVOVOQFSJZPEVÕEJS#VQFSJZPEBLBSõMLHFMFOBSBMLc - π , π m dir. 22 ÖRNEK 13 ÖRNEK 14 \"öBôEB WFSJMFO GPOLTJZPOMBSO HSBGJLMFSJOJ JODFMFZJ- \"öBôEB WFSJMFO GPOLTJZPOMBSO HSBGJLMFSJOJ JODFMFZJ- OJ[ OJ[ B y = tan ( 2x ) B y = -tan ( 2x ) C y = 3tan ( 2x ) C y = -3tan ( 2x ) ¦Ì[ÑN ¦Ì[ÑN y –3 –2 y –3 –2 3 2 3 2 1 1 2 –1 O 1 3 x –1 2 –2 x –1 O 1 3 –3 –1 –2 –3 y = –tan2x y = –3tan(2x) y = tan2x y = 3tan(2x)

www.aydinyayinlari.com.tr 53÷(0/0.&53÷ 5. MODÜL ·/÷7&34÷5&:&)\";*3-*, ,PUBOKBOU'POLTJZPOVOVO(SBüôJ %m/*m r x ` R, cot ( x +LÕ =DPUYPMEVóVOEBOLPUBOKBOUGPOLTJZPOVOVOQFSJZPEVÕPMVS#VZÐ[EFOLPUBOKBOUGPOL- TJZPOVOVOHSBGJóJ( -Õ Õ HJCJBSBMLMBSEBUFLSBSFEFS x 0 π Õ 3π Õ 5π Õ 7π Õ ... ... 22 22 cotx 0 5BONT[ 0 5BONT[ 0 5BONT[ 0 5BONT[ 0 y mÕ – Õ O Õ Õ Õ Õ x 2 2 2 (SBGJLUFUBOYGPOLTJZPOVOVO L+ 1 ) ( k `; B¿ËM¿ÐMFSJDPUYGPOLTJZPOVOVOUBONT[PMEVóVOBEJLLBUFEJOJ[ DPUYGPOLTJZPOVPSJKJOFHËSFTJNFUSJLPMEVóVOEBOUFLGPOLTJZPOEur. DPUYGPOLTJZPOVOVOQFSJZPEVÕEJS#VQFSJZPEBLBSõMLHFMFOBSBML Õ EJS ÖRNEK 15 ÖRNEK 16 \"öBôEB WFSJMFO GPOLTJZPOMBSO HSBGJLMFSJOJ JODFMFZJ- \"öBôEB WFSJMFO GPOLTJZPOMBSO HSBGJLMFSJOJ JODFMFZJ- OJ[ OJ[ B y = cot ( 2x ) B y = -cot ( 2x ) C y = 3cot ( 2x ) C y = -3cot ( 2x ) ¦Ì[ÑN ¦Ì[ÑN y y 3 3 2 2 1 1 –2 1 –3 –1 2 –3 –1 O 2 3 x x –1 –2 O 1 3 –2 –1 –3 –2 –3 y = 3cot(2x) y = cot2x y = –3cot(2x) y = –cot2x 41

TEST - 16 5SJHPOPNFUSJL'POLTJZPOMBSO(SBGJLMFSJ 1. G Y = 2cos3x + 1 5. G Y = 2cos3x -\"PMNBLÑ[FSF GPOLTJZPOVOVO FTBT QFSJZPEV BöBôEBLJMFSEFO f^ x h =- 4 IBOHJTJEJS 7 π π C) 2π % Õ & Õ EFOLMFNJOJO CJS LÌLÑ Y1 JTF ff x1 + 4π p LBÀ- A) B) 3 US 3 2 3 A) - 16 B) - 3 C) - 8 21 7 7 % - 8 E) - 4 21 7 2. 6ZHVOUBONBSBMôOEB G Y = tan5x . cos5x - 1 GPOLTJZPOVOVO FTBT QFSJZPEV BöBôEBLJMFSEFO 6. cos2x = cos2x -TJO2YPMNBLÑ[FSF IBOHJTJEJS G Y = 3cos2x - sin2x - 1 $ π \" Õ # Õ 5 GPOLTJZPOVOVOFTBTQFSJZPEVLBÀUS % 2π E) - 2π A) π B) π $ Õ % 3π & Õ 5 5 4 2 2 3. f^ x h = 2 + tanc x + 1 m G B = cot ( x + 3 ) + tan ( 2a - 1 ) 3 GPOLTJZPOVOVOFTBTQFSJZPEVLBÀUS GPOLTJZPOVOVO QFSJZPEV BöBôEBLJMFSEFO IBO- A) 3π # Õ $ Õ HJTJPMBCJMJS 2 \" Õ # Õ $ Õ % Õ & Õ % π 2 E) Yoktur. 4. 6ZHVOUBONBSBMôOEB 8. 6ZHVOUBONBSBMôOEBWFSJMFOGGPOLTJZPOVOEB G Y = tanx.cotx + 1 k ≠ 1 PMNBLÑ[FSF 2 GPOLTJZPOVOVOQFSJZPEVJMFJMHJMJBöBôEBLJCJM- f^ k h = 2k - 1 HJMFSEFOIBOHJTJEPôSVEVS tan3^ πk - 1 h \" :PLUVS # Õ $ Õ % π 3π GPOLTJZPOVOVOFTBTQFSJZPEVLBÀUS 2 E) \" Õ # 1 $ Õ2 % & -Õ 2 π 1. C 2. D 3. D 4. A 42 5. & 6. C D 8. D

5SJHPOPNFUSJL'POLTJZPOMBSO(SBGJLMFSJ TEST - 17 1. y 3. y 3 1 Õ Õ 1 O 2 4 Õx O Õ 4 Õ 4 –1 –1 Õ x 12 Õ –3 :VLBSEBLJ öFLJMEF WFSJMFO HSBGJL BöBôEBLJ :VLBSEBLJHSBGJLZ=BTJO CY +DGPOLTJZPOV- GPOLTJZPOMBSEBOIBOHJTJOFBJUPMBCJMJS OBBJUJTFB+C+DLBÀUS A) 1 + 2sinx B) 2 + sinx C) 1 +TJOY % + 2sin2x \" # $ % -3 E) -5 E) 2 + sin2x 2. y 4. G Y TJOY a GPOLTJZPOVOVOHSBGJôJBöBôEBLJMFSEFOIBOHJTJ- EJS A) y B) y 5 5 1 3 2 Õ O Õ 4 bx x 4 x Õ Õ Õ ÕÕ Õ ÕÕ Õ –1 12 6 3 12 6 3 2 –2 –1 C) y D) y 2 6 Õ :VLBSEBWFSJMFOHSBGJL 4 2 Õ –1 Õx x 3 4 G Y = 1 + 2sinx ÕÕ Õ 12 6 3 GPOLTJZPOVOBBJUJTFB+CLBÀUS –1 A) 2 + 7π B) 2 + 3π C) 3 + 6π E) y 6 2 5 5 % 5 + 3π E) 3c 1 + π m 2 x 22 2 Õ 4 ÕÕ Õ 12 6 3 –1 1. D 2. & 43 3. B 4. &

TEST - 18 5SJHPOPNFUSJL'POLTJZPOMBSO(SBGJLMFSJ 1. y 3. y 1 1 Õ Õ Õ ÕÕ O 2 Õ2 x O 6 32 Õ x Õ 3 –1 –1 –3 –3 :VLBSEBLJ öFLJMEF WFSJMFO HSBGJL BöBôEBLJ :VLBSEBLJHSBGJLZ=BDPT CY +DGPOLTJZPOV- GPOLTJZPOMBSEBOIBOHJTJOFBJUPMBCJMJS OBBJUJTFB+C+DLBÀUS A) cos2x - 1 B) 2cosx - 1 \" # $ % & C) 2cos2x - % DPTY- 2 E) cos2x - 2 4. G Y = -2 + 4cos2x 2. y GPOLTJZPOVOVOHSBGJôJBöBôEBLJMFSEFOIBOHJTJ- EJS 2 A) y Õ B) y b 2 4 4 Õ x Õ Õx 2 2 x Õ aÕ Õ –6 –2 12 O Õ 42 C) y D) y 4 2 :VLBSEBWFSJMFOHSBGJLZ= cos2x +GPOLTJZPOVOB Õ Õx Õ Õx aittir. 2 2 –4 –6 #VOBHÌSF B+CLBÀUS A) 1 + 3π B) 1 + 2π C) 3 + 4π E) y Õx 4 23 23 Õ % 3 + 3π E) 3 + 3π 2 44 24 –2 –6 1. B 2. A 44 3. C 4. D

5SJHPOPNFUSJL'POLTJZPOMBSO(SBGJLMFSJ 3. TEST - 19 1. y y Õ Õ Õ 1 mÕ O Õ x 44 4x O :VLBSEBWFSJMFOHSBGJLBöBôEBLJGPOLTJZPOMBS- :VLBSEBWFSJMFOHSBGJLZ= tan ( ax ) +CGPOLTJZPOV- na aittir. EBOIBOHJTJOFBJUPMBCJMJS A) y = tanx B) y = tan x + π #VOB HÌSF B C TSBM JLJMJTJ BöBôEBLJMFSEFO 4 IBOHJTJEJS C) y = tanc x + π m % y = cot x + π 1 B) f - 1 , 1 p 4 4 A) f , 1 p 2 E) y = cotc x + π m 2 4 C) f 1 , - 1 p % f - 1 , - 1 p 2 2 1 E) f 1 , p 2 2. y 4. y 2 O Õ Õx OÕ Õ x 2 6 Õ 3 3 Õ 2 :VLBSEBWFSJMFOHSBGJLZ=DPUBYGPOLTJZPOVOBBJU- :VLBSEBLJ GPOLTJZPO HSBGJôJ BöBôEBLJMFSEFO tir. IBOHJTJOFBJUUJS #VOBHÌSF BLBÀUS A) cot ( 2x + 3 ) B) cot ( 3x ) + 2 A) -2 B) - $ % & C) tan ( 2x + % UBO Y + 3 E) tan x + 3 1. C 2. & 45 3. A 4. B

·/÷7&34÷5&:&)\";*3-*, 5. MODÜL 53÷(0/0.&53÷ www.aydinyayinlari.com.tr 5&3453÷(0/0.&53÷,'0/,4÷:0/-\"3 %m/*m ÖRNEK 1 G\"Z#UBONMCJSGGPOLTJZPOVOUFSTJOJOGPOL- \"öBôEBLJJGBEFMFSJOEFôFSMFSJOJCVMVOV[ TJZPOPMBCJMNFTJJ¿JOCVGPOLTJZPOVOCJSFCJSWF ËSUFOPMNBTHFSFLJS B arcsin 1 C arcsin ( -1 ) 2 TJOY DPTY WF UBOY GPOLTJZPOMBSOO HSBGJLMFSJ 3 d) arcsin 0 ZBUBZEPóSVUFTUJOFUBCJUVUVMEVóVOEBCVGPOL- c) arcsinf - p TJZPOMBSO CJSF CJS PMNBELMBS HËSÐMÐS %PMBZ- TZMBNFWDVUUBONLÐNFMFSJOEFCVGPOLTJZPO- 2 MBSOUFSTGPOLTJZPOMBSZPLUVS 1 1π #VGPOLTJZPOMBSOUBONLÐNFMFSJOJOCJSFCJSWF B arcsin = x j = sin x & x = ËSUFO PMBO BMU LÐNFMFSJOEFO CJSJ UBON LÐNFTJ PMBSBL TF¿JMEJóJOEF GPOLTJZPOMBSO CV LÐNFEF 2 26 UFSTGPOLTJZPOMBSIFTBQMBOBCJMJS C BSDTJO-1 = x π j - 1 = sin x & x = - 2 c) arcsinf - 3 p=x & - 3π = sin x & x = - 22 3 E BSDTJOi = x j i =TJOYj x = G Y TJOY'POLTJZPOVOVO5FSTJ ÖRNEK 2 %m/*m cosf arcsin 5 p 13 TJOY GPOLTJZPOVOVO UBON LÐNFTJ ;- π , π E 22 EFôFSJOJCVMVOV[ PMBSBLBMOEónda bVBSBMLUBGPOLTJZPOCJSFCJS A 55 ve örten olur. x arcsin = x & sin x = 12 13 13 B 13 \"#$ÑÀHFOJOEFO 5 y cos x = 12 olur. 1 13 –Õ sinx C 2 x mÕ O Õ Õ 2 –1 ôFLJMEF WFSJMFO TJOY HSBGJóJOEF HËSÐMEÐóÐ HJCJ ÖRNEK 3 GPOLTJZPO;- π , π E bire bir ve örtendir. 22 \"öBôEBLJJGBEFMFSJOEFôFSMFSJOJCVMVOV[ TJOYGPOLTJZonu f : ;- π , π E \" 6- 1,1@, 22 B) sinf arcsin 1 p C 5π p 2 arcsinf sin G Y =TJOYPMBSBLUBONMBOEóOEB 3 f–1 : 6- 1, 1@ \" ;- π , π E G-1( x ) = arcsinx 22 G–1PG Y =GPG-1(x) =YPMEVôVOEBO GPOLTJZPOVOB TJOÑT GPOLTJZPOVOVO UFST B sin= arcsind 1 nG = 1 olur. GPOLTJZPOV denir. 22 olur. y = arcsinx l x = siny olur. C arc sin= sin 5π G = 5π 33 π ππ 12 1 5π 46 1. B C - c) - E 2. 3. B C 6 23 13 2 3

www.aydinyayinlari.com.tr 53÷(0/0.&53÷ 5. MODÜL ·/÷7&34÷5&:&)\";*3-*, ÖRNEK 4 H Y DPTY'POLTJZPOVOVO5FSTJ tanf arcsin 1 + arcsinf - 1 p p %m/*m 2 DPTYGPOLTJZPOVOVOUBONLÐNFTJ[ Õ] olarak JGBEFTJOJOEFôFSJOJCVMVOV[ BMOEóOEBCVBSBMLUBGPOLTJZPOCJSFCJSWFËS- tendir. y π 1 BSDTJO= x jTJOY= 1 j x = cosx 2 Õ BSDTJOd - 1 n = x jTJOY= - 1 π 2 Õ j x =- Õ 2x 2 26 –Õ O tand π - π n = cot π = 3 2 26 6 –1 ôFLJMEFWFSJMFODPTYHSBGJóJOEFHËSÐMEÐóÐHJCJ GPOLTJZPO[ Õ]BSBMóOEBCJSFCJSWFËSUFOEJS ÖRNEK 5 DPTYGPOLTJZPOVH[ Õ] Z [-1, 1], arcsinf 3 - t p g ( x) =DPTYPMBSBLUBONMBOEóOEB 2 g-1: [-1, 1] Z [ Õ], g-1 ( x ) =BSDDPTYGPOL- JGBEFTJSFFMTBZJTFUOJOBMBCJMFDFôJFOHFOJöEFôFS BSBMôOCVMVOV[ TJZPOVOB DPTY GPOLTJZPOVOVO UFST GPOLTJZPOV denir. y = arccosx l x = cosy olur. ÖRNEK 7 \"SDTJOY GPOLTJZPOVOVO UBON LÑNFTJ BZO [BNBOEB \"öBôEBLJJGBEFMFSJOEFôFSMFSJOJCVMVOV[ TJOYGPOLTJZPOVOVOHÌSÑOUÑLÑNFTJPMVQ[-1, 1]BSB- B arccos 1 C arccosf - 1 p 2 2 MôOEBZFSBMS c) arccos0 d) arccos1 -1 # 3-t #1 2 -2 # 3 - t # 2 -5 # -t # -1 1#t#5 1 1 π B arccos = x j cos = j x = 2 2 3 2π C arccosd - 1 n = x j cos x = - 1 j x = 22 3 π D BSDDPT= x j cosx = j x = 2 ÖRNEK 6 E BSDDPT= x j cosx = 1 j x = G Y = 2sinx - 1 PMEVôVOBHÌSF G-1 Y JCVMVOV[ ÖRNEK 8 6ZHVOUBONBSBMôOEBUBO BSDDPTY JGBEFTJOJOFöJ- UJOJCVMVOV[ y =TJOY- 1 j y + 1 =TJOY y+1 y+1 A BSDDPTY=BjDPTB= x = sin x j arcsinf p=x a \"#$ÑÀHFOJOEFO 22 x1 G-1(x) =BSDTJOd x + 1 n 1 - x2 2 tan a = x olur. B 1 – x2 C 4. 3 5. [1, 5] 6. ZBSDTJOd x + 1 n π 2π π 1 - x2 2 B C c) E 8. x 332

·/÷7&34÷5&:&)\";*3-*, 5. MODÜL 53÷(0/0.&53÷ www.aydinyayinlari.com.tr I Y UBOY'POLTJZPOVOVO5FSTJ ÖRNEK 10 %m/*m arctanf tan 7π pJGBEFTJOJOEFôFSJOJCVMVOV[ 4 I Y =UBOYGPOLTJZPOVc - π , π mBSBMóOEB 22 GPG-1(x) =G-1PG Y =YPMEVôVOEBO arctand tan 7π n = 7π PMBSBLCVMVOVS bire bir ve örtendir. y 44 ÖRNEK 11 –Õ O Õ Õ Õ x tan^ a + b h = tan a + tan b PMEVôVOBHÌSF 2 22 1 - tan a. tan b arctan 1 + arctan 1 JGBEFTJOJOEFôFSJOJCVMVOV[ 32 :VLBSEBWFSJMFOUBOYHSBGJóJOEFHËSÐMEÐóÐHJCJ 11 arctan = a & tan a = GPOksiyon c - π , π m aralóOEBCJSFCJSWFËS- 33 tendir. 22 11 arctan = b & tan b = 22 #V BSBMLUB UBOY GPOLTJZPOVOVO UFST GPOLTJZP- tan^ a + b h = tan a + tan b PMEVôVOEBO OVWBSESUBOYGPOLTJZPOVIc - π , π m Z R, 1 - tan a. tan b 22 11 I( x ) = tanYPMBSBLUBONMBOEóOda + I-1 : R Z c - π , π m I-1( x ) =BSDUBOYGPOL- tan^ a + b h = 3 2π = 1 jB+C= olur. 22 11 4 siyonuna tanYGPOLTJZPOVOVOUFSTGPOLTJZPOV 1- · denir. 32 y =BSDUBOYl x =UBOZolur. ÖRNEK 12 sinf arctanf 1 p + 3π pJGBEFTJOJOEFôFSJOJCVMVOV[ 22 ÖRNEK 9 A 11 arctan = a & tan a = \"öBôEBLJJGBEFMFSJOEFôFSMFSJOJCVMVOV[ a 22 5 sind 3π 2 + a n = - cos a = - 2 25 B arctan1 C arctan^ - 3 h B1 olur. c) arctan^ 3 h C d) arctanf - 3 ÖRNEK 13 p 3 6ZHVOUBONBSBMôOEB arctanx = 2a -Õ B BSDUBO= x jUBOY= 1 π C BSDUBO - 3 = x j tan x = - 3 j x= PMEVôVOBHÌSF YJOEFôFSBSBMôOCVMVOV[ D BSDUBO 3 = x j tan 3 = x 4 BSDUBOY= 2a -Ö π x =UBO a -Ö j x = -UBO Ö- 2a) x =UBOa j x ` (-ß, ß) olur. j x =- 3 π j x= 3 d) arctanf - 3 p = x j tan x = - 3π j x =- 3 36 π ππ π 48 7π π2 13. mß ß B C - c) d) - 11. 12. - 4 45 4 33 6

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AYT Matematik Ders İşleyiş Modülleri 5. Modül Trigonometri

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AYT Matematik Ders İşleyiş Modülleri 5. Modül Trigonometri

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