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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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www.aydinyayinlari.com.tr 53÷(0/0.&53÷ 5. MODÜL ·/÷7&34÷5&:&)\";*3-*, %m/*m ÖRNEK 16 t (x ) =DPUYGPOLTJZPOV Õ BSBMóOEBCJSFCJS 6ZHVO UBON BSBMóOEB ZFS BMBO Z = G Y GPOLTJZPOV ve örtendir. BõBóEBWFSJMNJõUJS G Y = 2cot ( 3x - 1 ) PMEVôVOBHÌSF G-1 Y JCVMVOV[ y mÕ – Õ O x y 2 Õ Õ Õ Õ y = 2cot(3x - 1) j = cot^ 3x - 1 h 22 2 yy BSDDPUd n = 3x - 1 j arcc cotd n + 1 = 3x 22 y j G-1(x) = arc cotd x n + 1 arc cotd n + 1 2 2 3 =x 3 olur. :VLBSEBWFSJMFODPUYHSBGJóJOEFHËSÐMEÐóÐHJCJ ÖRNEK 17 GPOLTJZPO Õ BSBMóOEBCJSFCJSWFËSUFOEJS %PMBZTZMBCVBSBMLUBUFSTJWBSES 6ZHVOUBONBSBMôOEBZFSBMBOGGPOLTJZPOV G Y =BSDDPU ÕY- 1 ) DPU Y GPOLTJZPOVU Õ Z R ise PMEVôVOBHÌSF G-1 Y JCVMVOV[ t-1 : R Z Õ U-1( x ) = BSDDPUY GPOLTJZP- OVOBLPUBOKBOUGPOLTJZPOVOVOUFSTGPOLTJZPOV y denir. y =BSDDPU ÖY- 1) j = arc cot^ πx - 1 h y =BSDDPUYl x = coty olur. 2 ÖRNEK 14 y y 6ZHVOUBONBSBMôOEB cotd n = πx - 1 j cotd n + 1 = πx arccot ( tanx ) 2 2 JGBEFTJOJOFöJUJOJCVMVOV[ y cot d x n + 1 UBOY= cotd π - x n jBSDDPUd cotd π - x n n cot d n + 1 2 22 j f–1 ^ x h = π olur. 2 π x= π L - x olur. ÖRNEK 18 2 \"öBôEBLJJGBEFMFSJOEFôFSMFSJOJCVMVOV[ ÖRNEK 15 B arccot ( 1 ) C arc cot^ - 3 h arccotx = arctany JGBEFTJOFHÌSFYWFZBSBTOEBCJSCBôOUCVMVOV[ π B BSDDPU= x j cotx = 1 j x = BSDDPUY=B jDPUB= x BSDUBOZ=B jUBOB= y 4 DPUBUBOB=PMEVôVOEBOYZ= 1 olur. C arc cot^ - 3 h = x & cot x = - 3 & x = 5π olur. 6 π 15. YZ arc cotd x n + 1 cotd x n + 1 14. - x 2 2 16. π 18. B ÖC Ö 2 3

TEST - 20 5FST5SJHPOPNFUSJL'POLTJZPOMBS 1. G Y = arcsin c3 - 5xm 5. arcsin 3 + arccosf - 1 p 4 22 GPOLTJZPOVOVOFOHFOJöUBONBSBMôBöBôEB- JöMFNJOJOTPOVDVLBÀUS LJMFSEFOIBOHJTJEJS A) - π B) - 2π $ Õ A) >- 3 , 7 H B) >- 3 , 7 p 3 3 55 55 5π %) 4π E) 3 3 C) f - 1 , 7 H % >- 1 , 7 H 55 55 E) > 1 , 7 H 55 6. arcsin^ cos^ arctan 3 hh 2. arcsinx = arccosy JGBEFTJOJOFöJUJBöBôEBLJMFSEFOIBOHJTJEJS PMEVôVOBHÌSF Y2 + y2UPQMBNLBÀUS A) 3π B) 2π C) π % π E) π 4 3 34 6 B) 7 C) 3 % 5 E) 1 A) 2 4 24 3. 6ZHVOUBONBSBMôOEB sin f arccos 5 + 2. arcsin 2 p sin ( arcsin ( 2x - 1 ) ) = 1 13 2 PMEVôVOBHÌSF YLBÀUS JGBEFTJOJOEFôFSJLBÀUS \" # $ % & A) 12 5 C) 5 % 12 E) 24 13 B) 13 25 25 12 4. 6ZHVOUBONBSBMôOEB 8. 6ZHVOUBONBSBMôOEB f^ x h = 2 sin^ x - 1 h arccos (4x - 2) 3 f (x) = PMEVôVOBHÌSF G-1 Y BöBôEBLJMFSEFOIBOHJ- 6 PMEVôVOBHÌSF G-1 Y BöBôEBLJMFSEFOIBOHJ- TJEJS TJEJS A) arcsin f 3x p + 1 B) 3 arcsin^ x h + 1 2 2 2 + cos 6x 2 - cos 6x 4 + cos 6x A) C) 3 arcsin c x + 1 m % 2 arcsin c x - 1 m B) C) 2 3 4 42 E) arccos f 3 x p + 1 % 4 - cos 6x E) -4 + 6cos6x 2 2 1. D 2. & 3. B 4. A 5. C 6. & C 8. A

5FST5SJHPOPNFUSJL'POLTJZPOMBS TEST - 21 1. BSDUBO=YPMEVôVOBHÌSF 5. arccot ( x + 1 ) + arctan 1 = π tan2 x + cot2 x 22 PMEVôVOBHÌSF YLBÀUS UPQMBNOOEFôFSJLBÀUS A) -1 B) - 1 $ % 1 E) 1 2 2 10 B) 82 $ % 4 9 A) 9 3 E) 3 91 6. arctanx = arccot3x 2. arc cot^ 3x2 - 2x - 4 h = π PMEVôVOB HÌSF Y JO BMBDBô EFôFSMFS UPQMBN LBÀUS 4 EFOLMFNJOJ TBôMBZBO Y EFôFSMFSJ UPQMBN LBÀ- A) - 1 B) - $ % & 1 3 3 US A) - 5 B) -1 C) - 2 % 2 E) 5 3 33 3 sin ( arctanx + arccotx ) JGBEFTJOJOEFôFSJLBÀUS A) π # $ % - π E) -1 2 2 3. a, b, iCJS\"#$ÑÀHFOJOJOEöBÀMBSPMNBLÑ[F- re, arccot ( -2 ) = a + b PMEVôVOBHÌSF DPUiLBÀUS A) - 1 B) - $ % 1 E) 2 8. f^ x h = cotf 3x - 1 p + 1 2 2 2 PMEVôVOBHÌSF G-1 Y BöBôEBLJMFSEFOIBOHJTJ- EJS 2arc cot^ x - 1 h + 1 2arc cot^ x - 1 h - 1 A) B) 33 4. arcsinx = arccosy ve arctanx = arctany 3arc cot^ x - 1 h + 1 3arc cot^ x - 1 h - 1 C) % PMEVôVOBHÌSF Y+ZLBÀPMBCJMJS 22 A) 2 B) 3 $ % & 5 2 arctan^ x - 1 h + 1 E) 3 1. B 2. D 3. & 4. B 51 5. B 6. C B 8. A

·/÷7&34÷5&:&)\";*3-*, 5. MODÜL 53÷(0/0.&53÷ www.aydinyayinlari.com.tr ,PTJOÑT5FPSFNJ ,04÷/·47&4÷/·45&03&.-&3÷ ÖRNEK 1 %m/*m \"ZõF PLVMEB CVMVOEVóV TSBEB BSLBEBõ \"MJhZF UFMFGP- OVOEBOLPOVNHËOEFSJZPS\"ZõFEBIBTPOSB\"MJhZFLN A V[BLMLUBLJFWJOEFOCJSLPOVNEBIBHËOEFSJZPS ch b OKUL B BmåY H xC :VLBSEBLJ\"#$Ð¿HFOJOEFLFOBSV[VOMVLMBS 5 km x | | | | | |BC = a, AC = b, AB = c; 60° Ali J¿B¿ËM¿ÐMFSJ m (WA) = A m (WB) = B , m (XC) = C olmak üze- 8 km re, [ AH ] m [ BC ] olsun. Ev | | | |HC =YPMEVóVOEBO BH = a - x olur. \"MJhOJOCVMVOEVóVOPLUB PLVMWFFWCJSÐ¿HFOJOLËõFMF- SJPMBDBLõFLJMEFNPEFMMFOEJóJOEF\"MJhOJOCVMVOEVóVLË- \")$EJLÐ¿HFOJOEF1JTBHPSUFPSFNJVZHVMBOE- õFMJLB¿ZBQZPS | | | | | |óOEB AC 2 = AH 2 + HC 2 \"MJhOJOPLVMV[BLMôLNPMEVôVOBHÌSF PLVMJMFFW BSBTLBÀLNPMVS b2 =I2 + x2 jI2 = b2 - x2 olur. (1) x2 = 52 + 82 -DPT \")#EJLÐ¿HFOJOEF1JTBHPSUFPSFNJVZHVMBOE- x2 = 25 + 64 - 80· 1 óOEB 2 |AB|2 = |AH|2 + |BH|2 x2 =- x2 =j x =LN c2 =I2 + ( a - x )2PMEVóVOBHËSF ÖRNEK 2 I2 = c2 -( a - x )2 olur. (2) (1) ve (2)EFOLMFNMFSJCJSMJLUF¿Ë[ÐMEÐóÐOEF A ABC üçgen b2 - x2 = c2 - a2 + 2ax - x2 90°– a | |AB = 6 cm b2 = c2 - a2 + 2ax olur. (3) 6 | |AC = 4 cm AHC dik üçgeninde B 4 cos C = x & x = b. cos C PMVS #V EFóFS (3) % b m ( BAC ) = 90° - a denkleminde yFSJOFZB[MEóOEB sin a = 3 c2 = a2 + b2 - 2ab. cos (XC) elde edilir. C 48 | |:VLBSEBLJWFSJMFSFgöre, BC LBÀDNEJS #FO[FSõFLJMEF | BC |=YPMNBLÑ[FSF a2 = b2 + c2 - 2bc. cos A b2 = a2 + c2 - 2ac. cos B ZB[MBCJMJS DPT - a) =TJOa = 3 j x2 = 62 + 42 - 2.6.4 · 3 48 48 x2 = 36 + 16 - 3 x2 =j x = 52 1. 2.

www.aydinyayinlari.com.tr 53÷(0/0.&53÷ 5. MODÜL ·/÷7&34÷5&:&)\";*3-*, ÖRNEK 3 ÖRNEK 5 A Verilen üçgenlerde B \"#$%LJSJõMFSEËSUHFOJ [ AB ] m [ BC ] A4 | |AB = 4 cm a | |\"% = 5 cm 6 10 | |AB = 6 cm x 6 | |BC = 6 cm ve | |AE = 10 cm | |%$ = 3 cm aE 5 C || ||B x &% = 3 cm 5 EC = 5 cm ve 8 a 180°–a 3 3 D | |BE = x cm dir. C D | |:VLBSEBLJ WFSJMFSF HÌSF DC = Y LBÀ DN PMEVôV- m ( B%AD ) = aPMNBLÑ[FSF DPTaOOEFôFSJOJCVMVOV[ OVCVMVOV[ ,JSJöMFSEÌSUHFOJOEFLBSöMLMBÀMBSUPQMBNEJS 6 4 4 4– c7os4a 4 4 8 84 \"#&ÑÀHFOJOEFO cos a = = x2 = 42 + 52 - 2.4.5.cosa = 62 + 32 - 2.6.3. cos^ 180° - a h 16 + 25 -DPTa = 36 ++ 36 cosa 10 5 41 -DPTa = 45 + 36 cosa -DPTa = 4 j cos a = 4 = - 1 olur. %&$ÑÀHFOJOEFO 32 + 52 - 2.3. 5 · 4 = 2 - 76 19 x + 25 - 24 = x2 5 x2=j x = 10 PMBSBLCVMVOVS ÖRNEK 6 | | | |ABC üçgeninde BC = a br, AC = b br, |AB| = c br ve % m ( ACB ) = a ES¶¿HFnin kenarlaSBSBTOEB af a - b p = ^ c - b h^ c + b h 2 ÖRNEK 4 CBôOUTPMEVôVOBHÌSF, cosaEFôFSJOJCVMVOV[ A 4D B2 - ab = c2 -C2 j c2 =B2 +C2 - ab 2 22 6 a C a c2 =B2+C2 -BCDPTaPMEVôVOEBO x ab 1 54 - = - 2 ab . cos a & = cos a olur. 24 BE | | | | | |[AE] a [#%] = {C}, AC = CE = 4 cm, AB = 6 cm, ÖRNEK 7 | | | | | |BC = 5 cm, $% = 2 cm ve BE = x cm dir. A 6B \"#$%ZBNVL | |:VLBSEBLJ WFSJMFSF HÌSF BC = Y LBÀ DN PMEVôV- a [AB] // [$%] OVCVMVOV[ | |5 57 AB = 6 cm a 180°–a | |BC = 7 cm \"#$ÑÀHFOJOEF D6 1E4 8 | |C $% = 14 cm 62 = 52+ 42 - 2.5.4.cosa | |\"% = 5 cm dir. 36 = 41 -DPTa :VLBSEBLJWFSJMFSFHÌSF m ( D%AB ) = aPMNBLÑ[FSF 1 cosaEFôFSJOJCVMVOV[ -5 = -DPTa j cosa = 8 [AD] // [#&] PMBDBL öFLJMEF CJS [#&] EPôSV QBSÀBT ÀJ[JMJS &$%ÑÀHFOJOEFO #&$ÑÀHFOJOEF 6 4 4 4– c7os4a 4 4 8 x2 = 22 + 42 - 2. 2 . 4 · 1 8 2 = 52 + 82 - 2.5.8. cos^ 180° - a h x2 = 4 + 16 - 2 1 = 25 + 64 +DPTa j - = cos a PMBSBLCVMVOVS x2 = 18 j x = 3 2 PMBSBLCVMVOVS 2 3. 10 4. 3 2 53 1 11 5. - 6. - 19 4 2

·/÷7&34÷5&:&)\";*3-*, 5. MODÜL 53÷(0/0.&53÷ www.aydinyayinlari.com.tr ÖRNEK 8 4JOÑT5FPSFNJ %m/*m A B 1 A CL D 1 G c b ha H BH C a a E1 K 1 F :VLBSEBLJ\"#$Ð¿HFOJOEFLFOBSV[VOMVLMBS | | | | | | | |ôFLJMEFLJLÐQUF EK = KF , #- = -( ve | | | | | |BC = a, AC = b, AB = c olsun. % m ( AKL ) = a ES m (WB) = B , m (WA) = A , m (XC) = C olmak üzere, :VLBSEBLJWFSJMFSFHÌSF DPTaEFôFSJOJCVMVOV[ [ AH ] m [ BC ]PMEVóVOEB ,ÑQÑOCJSLFOBSCSPMTVO AHB dik üçgeninde |\",|2 = |,&|2 + |%&|2 + | AD |2 sin B = ha & ha = c. sin B olur. |\",|2 = 12 + 22 + 22 =j |\",| = 3 c |,-|2 = |,'|2 + |'(|2+ |-(|2 j 12 + 22 + 12= 6 |,-| = 6 Buradan | AL |2 = | AB |2 + | BL |2 j | AL | = 5 Aa A&BC k = a.ha = a.c. sin B LPTJOÑTUFPSFNJOEFO ^ 5 h2 = 2 + ^ 6 h2 - 2.3. 6 . cos a 22 3 5 = 9 + 6 - 6 6 cos a j -= - 6 6 cosa elde edilir. cos a = 10 olur. 66 #FO[FSõFLJMEF & = a.b. sin C ve A ( ABC ) 2 ÖRNEK 9 & = b.c. sin A FõJUMJLMFSJZB[MS A ( ABC ) 6 A 2 a Bu durumda & = a.c. sin B = a.b. sin A A ( ABC ) 7 22 i–a 180°–i b.c. sin A olur. B 5C 2 a.c. sin B = b.c. sin A = a.b. sin a 222 ABC üçgeninde, % = a, % = i - a, FõJUMJLMFSJ a.b.c JGBEFTJOFCËMÐOÐSTF m ( BAC ) m ( ABC ) 2 | | | | | |AC = 7 cm, AB = 6 cm, BC = 5 cm dir. sin A = sin B = sin C FõJUMJóJFMEFFEJMJS abc :VLBSEBLJWFSJMFre göre, cosiEFôFSJOJCVMVOV[ Buradan; 2 = 2 + 2 - 2.5.7. 6 4 4 4– c7os4i 4 48 a = b = c olarak bulunur. cos^ 180° - ih sin A sin B sin C 6 5 7 36 = 25 ++DPTa j -38 =DPTa - 38 - 19 = cos a = 70 35 10 - 19 54 8. 66 35

www.aydinyayinlari.com.tr 53÷(0/0.&53÷ 5. MODÜL ·/÷7&34÷5&:&)\";*3-*, %m/*m ÖRNEK 11 ABC çevrel çemberinin merkezi O ve [#%] çap ôFLJMEF 0 NFSLF[MJ \"#$ Ð¿HFOJOJO ¿FWSFM ¿FNCFSJ WF- olmak üzere, SJMNJõUJS A A bD a 10 cR 180°–2a RO B aC a O B 90°–a WA ile X% BZOZBZHËSEÐLMFSJOEFOËM¿ÐMFSJFõJU- C tir. m (WA) = A ve m (XD) = D olmak üzere, | |AB = 10 cm, % = a ve sin a = 11 sin (XD) = sin (WA) = a ise 2R = a olarak m ( ABO ) dir. 2R sin A 6 bulunur. #FO[FSõFLJMEF YuLBSEBLJ WFSJMFSF HÌSF ÀFWSFM ÀFNCFSJO ÀFWSFTJOJ CVMVOV[ a = b = c = 2R sin A sin B sin C 11 5 sin a = PMEVôVOEBO cos a = olur. PMBSBLJGBEFFEJMJS 66 10 10 = 2R cos a = 2R sin^ 90° - a h 10 = 12 = 2R 5 6 ¦FWSFMÀFNCFSJOÀFWSFTJ=Ö3 PMEVôVOBHÌSF ¦=ÖPMVS ÖRNEK 10 ÖRNEK 12 A #JS \"#$ Ð¿HFOJOEF \" # $ Ð¿HFOJO J¿ B¿MBSOO ËM¿Ð- leridir. 68 sin2 A + sin2 B = sin2 C a b PMEVôVOBHÌSF DPT$EFôFSJOJCVMVOV[ B C | | | |ABC üçgeninde, AB = 6 cm, AC = 8 cm dir. abc k = = = JTF :VLBSEBLJWFSJMFSFHÌSF sin a PSBOOCVMVOVz. sin b sin A sin B sin C 2 22 a b c 2 = = = olur. 222 k sin A sin B sin C 2 A + sin 2 B 2 sin sin C = PMBSBLZB[MS 68 2 2 2 TJOÑTUFPSFNJOFHÌSF = PMEVôVOEBO + b a c sin b sin a sin a 8 4 TJO2A +TJO2B =TJO2C jB2 +C2 = c2 = = olur. sin b 6 3 1JTBHPSUFPSFNJOEFOm (XC) = 90°WFDPT=PMVS 4 55 11. Ö 12. 3

·/÷7&34÷5&:&)\";*3-*, 5. MODÜL 53÷(0/0.&53÷ ÖRNEK 14 www.aydinyayinlari.com.tr 4JOÑT\"MBO'PSNÑMÑ 9 A ABC üçgen %m/*m | |AB = 9 br | |8 A AC = 8 br dir. chb BC :VLBSEBLJ WFSJMFSF HÌSF \" \"#$ FO ÀPL LBÀ CS2 BH C PMVS a 1 A^ ABC h = 1 · a.h 9 A(ABC) = ·9.8. sin A 2 2 sin C = h iseI= b.siO$PMEVóVOEBOCVFõJUMJ- \"MBOOO NBLTJNVN PMNBT sin A OO NBLTJNVN PM- b NBTES sin A FOÀPLPMBDBôOEBO óJ 9 EBZFSJOFZB[Mr. Bu durumda A^ ABC h = 1 ·a.b sin CPMBSBLJGBEFFEJMJS A = 1 ·9.8.1 = 36 2 olur. 2 max 2 br #FO[FSõFLJMEF ÖRNEK 15 A^ ABC h = 1 ·a.c. sin B = 1 ·b.c. sin A 22 GPSNÐMMFSJFMEFFEJMJS 'PSNÐM ZBSENZla alan bulunurken iki kenar A \"#% WF #&$ CJSFS WFCVJLJLFOBSOBSBTOEBLJB¿OOTJOÐTÐOEFO x üçgen olmak üzere, ZBSBSMBOS EA | |4 F BE = 4 br | |B A #% = 5 br | |B 5 D 3 C %$ = 3 br || & & ÖRNEK 13 AE = x br , A^ AEF h = A^ FDC h A :VLBSEBLJWFSJMFSFHÌSF YLBÀCSEJS ABC üçgen | |AC = 4 cm && A^ ABD h = A^ BEC hPMEVôVOEBO 4 | |BC = 6 cm 1 ·^ 4 + x h.5. sin B = 1 ·4.8. sin B 22 60° % + 5x = 32 j 5x = 12 j x = 12 olur. m ( ACB ) = 60° dir. 5 B6 C :VLBSEBLJWFSJMFSFHÌSF A^ & hLBÀDN2EJS ABC 1 13 2 A = .4.6. sin 60° = .4.6. = 6 3 cm olur. 2 22 13. 6 3 56 12 14. 36 15. 5

www.aydinyayinlari.com.tr 53÷(0/0.&53÷ 5. MODÜL ·/÷7&34÷5&:&)\";*3-*, ÖRNEK 16 ÖRNEK 18 A ôFLJMEF41 42WF43 A \"#%CJSÐ¿HFO 4 3 CVMVOEVLMBS CËMHele- 57 | |AB = 5 cm S1 G | |BC = 6 cm B rJO BMBOMBSO HËTUFS- B6 | |AC = 7 cm ve C1 S2 3 mektedir. | |$% = 3 cm dir. | | | |2 C 3D S3 F AB = 2 $% = 4 br 1 | | | |D E BC = EF = 1 br | | | |AG = GF = 3 br :VLBSEBLJWFSJMFSFHÌSF A^ & hOJOLBÀDN2 oldu- #VOBHÌSF S2 PSBOLBÀUS ACD S1 + S3 ôVOVCVMVOV[ 1 \"#$ÑÀHFOJOEFu = 5+6+7 =9 S = ·4.3. sin A = 6. sin A 2 12 11 A^ ABC h = 9.4.2.3 = 6 2 S = ·5.6. sin A - ·4.3. sin A = 9 sin A 22 6 cm olur. 2 1 1 19 A^ ABC h S = ·7.7. sin A - ·5.6. sin A = sin A A^ ACD h = 32 2 2 PMEVôVOEBO 2 #VOBHÌSF S = 9 sin A = 18 A^ ACB h = 3 6 DN2 olur. 2 S +S 19 31 1 3 6 sin A + sin A 2 ÖRNEK 17 ABC üçgen %m/*m ,FOBS V[VOMVLMBS a, b, c ve çevrel A [\"%]B¿PSUBZ A ¿FNCFSJO ZBS¿BQ x y6 | |AC = 6 br c b \"R\" olan ABC üç- A 3A R O \" \"#% =\" \"%$ HFOJOJOBMBO BD C dir. C Ba | |:VLBSEBLJWFSJMFSFHÌSF AB YLBÀCJSJNEJS & ) = a.b.c dir. A ( ABC 4.R 11 ÖRNEK 19 3· ·x.y. sin a = ·y.6. sin a j 3x = 6 j x = 2 olur. ,FOBSV[VOMVLMBSDN DNWFDNPMBOCJSÑÀHF- 22 OJOÀFWSFMÀFNCFSJOJOZBSÀBQOCVMVOV[ %m/*m ,FOBSV[VOMVLMBSWFSJMFOCJSÐ¿HFOJOBMBO A cb 5+6+7 u= =9 Ba C 2 a + b + c = 2u olmak üzere, 5.6.7 Aa & k = u^ u - a h^ u - b h^ u - c h dir. A = 9.4.3.2 = ABC 4R #VGPSNÐMF)FSPO'PSNÑMÑdenir. 6 6 .4R = 5. 6 .7 35 R = olur. 46 18 2 18. 3 6 35 16. 31 46

·/÷7&34÷5&:&)\";*3-*, 5. MODÜL 53÷(0/0.&53÷ www.aydinyayinlari.com.tr %m/*m A a ÖRNEK 22 ABC üçgeninde b rr A | |AB = 5 cm B Or | |AC = 6 cm ve 6 | |BC = 7 cm dir. 5 C c B 7C #JS\"#$Ð¿HFOJOJO0NFSLF[MJJ¿UFóFU¿FNCF- :VLBSEBWFSJMFOMFSFHÌSF JÀUFôFUÀFNCFSJOZBSÀB- QLBÀCJSJNEJS SJOZBS¿BQSWF u = a + b + c olaSBLBMOE- A^ ABC h = u^ u - a h^ u - b h^ u - c h = r.u 2 óOEB & = & + & + & 5+6+7 A ( ABC ) A ( AOC ) A ( AOB ) A ( BOC ) u = = 9 #VEVSVNEB A = 9.4.3.2 = 9.r 2 & r.a r.b r.c 66 26 ABC A = 6 6 = 9r & r = = olur. PMEVóVOEBO A ( ) = + + 93 222 r^ a + b + c h ÖRNEK 23 = = r.u 2 olur. A Yanda verilen üçgende ÖRNEK 20 Yanda verilen üçgende 4 5 IJ¿UFóFU¿FNCFSJONFS- r Ir A I J¿ UFóFU ¿FNCFSJO kezi B r merkezi, | |AB = 4 cm 6 | |AC = 5 cm [I%] m [BC] | |BC = 6 cm dir. | |AB = 5 br ve C | |% = 4 br dir. 5E :VLBSEBWFSJMFOMFSFHÌSF A^ AIB h + A^ AIC h LBÀUS 4 C I A^ BIC h B 4 D I JÀ UFôFU ÀFNCFSJO NFSLF[J PMEVôVOEBO LFOBSMBSB PMBO EJL V[BLMLMBS FöJUUJS #V EVSVNEB TPSVMBO BMBOMBS UBCBOMBSZMBPSBOUMES\" \"IB) =4 \" \"IC) =4WF :VLBSEBLJWFSJMFSFHÌSF \" \"I# ZJCVMVOV[ A(BIC) =4PMVS 4S + 5S 3 = olur. 6S 2 [&I] ve [DI]JÀUFôFUÀFNCFSJOZBSÀBQPMEVôVJÀJO | D | = |& | =CSPMVS ÖRNEK 24 A ( AIB ) = 4.5 2 A õFLJMEFWFSJMFOMFSFHÌ- = 10 br SF \"#$ ÑÀHFOJOJO JÀ- 2 3 UFôFU ÀFNCFSJOJO ZBS- ÀBQ S ÀFWSFM ÀFNCF 5 SJOJO ZBSÀBQ 3 PMEV- C ôVOBHÌSF S3ÀBSQN ÖRNEK 21 B7 LBÀUS ¦FWSFTJ CS PMBO CJS ÑÀHFOJO JÀ UFôFU ÀFNCFSJOJO A^ ABC h = a.b.c = r.uPMEVôVOEBO ZBSÀBQCSPMEVôVOBHÌSF BMBOLBÀCS2EJS 4R A =SVPMEVôVOEBO 3.5.7 = r.d 3 + 5 + 7 n j 105 15 r 14 7 A ==CS2 olur. =j = R.r = 4R 2 4R 2 4 2 olur. 21. 58 26 3 7 22. 23. 24. 3 22

,PTJOÑTWF4JOÑT5FPSFNMFSJ TEST - 22 1. A ABC bir üçgen 4. A 120° % = 120° 77 34 m ( BAC ) 3 | |AB = 3 br | |B C AC = 4 br dir. B5 D xC | | :VLBSEBLJWFSJMFSFHÌSF BC LBÀCJSJNEJS | | | |ABC bir üçgen, \"% = 3 cm, #% = 5 cm | AB | = | AC | = 7 cm dir. A) 5 B) 2 7 C) 3 2 | |:VLBSEBLJWFSJMFSF göre, DC =YLBÀDNEJS % 29 E) 37 \" # $ % & 2. A B 8 5 7 D 7 C 3 5. D6 C E \"#$WF$%&CJSFSÐ¿HFO [ AE ] a [#%] = { C } 120° |BC| = |$%| = 7 cm, | AC | = 5 cm, | CE | = 3 cm, 99 | |AB = 8 cm dir. AxB :VLBSEB WFSJMFOMFSF HÌSF \" $&% LBÀ DN2 EJS \" # $ % & 3 ôFLJMEFLJ¿FNCFSEF\"#$%LJSJõMFSEËSUHFOJ | | | | | |\"% = % CB = 9 cm, %$ = 6 cm, m ( DCB ) = 120° dir. A | | :VLBSEBLJWFSJMFSFHÌSF AB =YLBÀDNEJS 3. 2 \" # $ % & E 42 17 B 5 D3 C | | | |ABC bir üçgen, AE = 2 br, EC = 4 2 br 6. #JS\"#$ÑÀHFOJOJOLFOBSMBSBSBTOEB | | | | | |&% = 17 br, #% = 5 br, %$ = 3 br b2 - 2c2 = b - 2 c | |:VLBSEBLJWFSJMFSFHÌSF AB LBÀCJSJNEJS a2 - c2 b A) 2 17 B) 34 34 CBôOUTPMEVôVOBHÌSF \"BÀTOOÌMÀÑTÑLBÀ C) EFSFDFEJS 2 % 3 17 17 \" # $ % & 2 E) 3 1. & 2. C 3. B 4. D 5. D 6. &

TEST - 23 ,PTJOÑTWF4JOÑT5FPSFNMFSJ 1. #JS\"#$ÑÀHFOJOEF 4. A m (WA) = 120° ve a = 4 3 br B F 3 PMEVôVOBHÌSF ÑÀHFOJOÀFWSFMÀFNCFSJOJOZBS ÀBQLBÀCJSJNEJS \" # $ % & C 4 D 2E | |\"$%WF#$&CJSFSüçgen, BC = 3 cm | | | |$% = 4 cm, %& = 2 cm ve \" \"$% = 2.A ( BCE ) dir. | | :VLBSEBLJWFSJMFSFHÌSF AB LBÀDNEJS \" # $ % & 2. ôFLJMEFLJ\"#$Ð¿HFOJOJO¿FWSFM¿FNCFSJOJONFS- A | | | |LF[J0OPLUBT OB = 6 cm ve BC = 8 cm, 5. % m ( BAC ) = a dir. x 30° A 48 BDC O C | | | |ABC bir üçgen, AC = 8 cm, AB = 4 cm 6 B8 | | | |#% = % % %$ , m ( DAC ) = 30° ve m ( BAD ) = x dir. :VLBSEBLJWFSJMFSFHÌSF TJOa LBÀUS :VLBSEBLJWFSJMFSFHÌSF TJOYLBÀUS A) 1 B) 2 C) 3 % 4 5 1 2 C) 3 % 5 E) 1 2 3 45 E) A) B) 46 6 2 3 6. A ABC ikizkenar üçgen 3. \"#$Ð¿HFOJOJO¿FWSFM¿FNCFSJOJOZBS¿BQCJSJN- 75° 6 B |AB| = |AC| dir. % | | | |AB = 4 2 CJSJN AC =CJSJNPMEVôVOBHÌ- m ( ABC ) = 75° SF ÑÀHFOJOFOCÑZÑLJÀBÀTLBÀEFSFDFEJS | |BC = 6 cm dir. \" # $ % & C :VLBSEBLJWFSJMFSFHÌSF \"#$ÑÀHFOJOJOÀFW- SFMÀFNCFSJOJOÀBQLBÀDNEJS \" # $ % & 1. B 2. B 3. D 4. D 5. & 6. C

,PTJOÑTWF4JOÑT5FPSFNMFSJ TEST - 24 1. ,FOBSV[VOMVLMBS 4. ,FOBSV[VOMVLMBSB C DPMBOCJS\"#$ÑÀHFOJO- a = 4 birim, b = 5 birim ve c = 7 birim de, 2sinA - sinB = sinC + sinA PMBO\"#$ÑÀHFOJJÀJOTJO$EFôFSJLBÀUS PMEVôVOBHÌSF a + b + c PSBOLBÀUS A) 3 2 B) 6 C) 2 6 b+c 26 A) 0 B) 1 C) % E) 4 E) % 3 6 5 5. A ABC bir üçgen 2x % = 2x 5 m ( BAC ) 2. A ABC dik üçgen % = x m ( ACB ) 6 [ AB ] m [ BC ] m ( A%BC ) = m ( A%CB ) | |AB = 6 cm | |BC = 8 cm B6 | |x BC = 6 cm | |C AB = 5 cm dir. :VLBSEBLJWFSJMFSFHÌSF \"#$ÑÀHFOJOJOÀFWSFM B 8C ÀFNCFSJOJOZBSÀBQLBÀDNEJS :VLBSEBLJWFSJMFSFHÌSF \"#$ÑÀHFOJOJOÀFWSFM 15 15 C) 25 % 25 25 ÀFNC FSJOJOZBSÀBQOOJÀUFô FUÀFNCFSJOJOZB- A) B) E) SÀBQOBPSBOLBÀUS 4 24 68 A) 2 B) 3 $ % 5 E) 12 5 5 2 5 3. ôFLJMEFLJ\"#$WF'#%Ð¿HFOMFSJOEFUBSBMCËMHF- 6. \"SBMBSOEBLNCVMVOBO\"WF#MJNBOMBSOEBOõF- MFSJOBMBOMBSCJSCJSJOFFõJUUJS LJMEFLJHJCJJLJCPUTSBTZMBLZZMBWFMJLB¿ ZBQBDBLõFLJMEFIBSFLFUFEJZPSMBSWF$OPLUBTOEB A CVMVõVZPSMBS C 7 F 15° 40° LZ 5 A 60 km B E B 10 C x D #VOBHÌSF \"EBOIBSFLFUFEFOCPUZBLMBöLLBÀ LNEBIBGB[MBZPMBMNöUS | | | | | |AF = 7 cm, BF = 5 cm, BC = 10 cm dir. TJO , TJO , TJO , | |YukBSEBLJWFSJMFSFHÌSF CD =YLBÀDNEJS \" # $ % & \" # $ % & 1. & 2. D 3. C 61 4. C 5. & 6. B

TEST - 25 ,PTJOÑTWF4JOÑT5FPSFNMFSJ 1. ,FOBSV[VOMVLMBSB C DPMBOCJSÐ¿HFOJO¿FWSFTJ 4. A \"%$Ð¿HFO 36 cm dir. 5 [#%] m [%$] m (XA) = A , m (XB) = B ve m (XC) = CPMNBLÑ[F- B re, | |AB = 5 cm x | |15 2sinC = sinA + sinB D BC = 15 cm PMEVôVOB HÌSF D LFOBSOO V[VOMVôV LBÀ DN EJS | |%$ = 12 cm \" # $ % & 12 C dir. | | :VLBSEBLJWFSJMFSFHÌre, AD =YLBÀDNEJS A) 2 B) 3 C) 13 % 34 E) 4 10 2. ,FOBSV[VOMVLMBSB C DPMBOCJSÑÀHFOEF 5. #JS\"#$Ð¿HFOJOJO¿FWSFM¿FNCFSJOJO¿FWSFTJÕCS A ( ABC ) = 9 3 cm2 ve dir. ( a + b - c ) ( a + b + c ) - ab = 0 PMEVôVOBHÌSF, BCÀBSQNLBÀUS m % = a PMNBLÑ[FSF ( BAC ) \" # $ % & sin a = 5 6 PMEVôVOBHÌSF [BC]LFOBSOOV[VOMVôVLBÀCS EJS \" # $ % & 3. \"õBóEBLJ LPPSEJOBU EÐ[MFNJOEF 0 NFSLF[MJ CS 6. B ZBS¿BQM¿FZSFL¿FNCFSWFSJMNJõUJS A1 2 Z C A 4 B 3 D a X | |\"#$%LJSJõMFSEËSUHFni, m( D%AC ) = a, AB = 1 br, O C | | | | | |BC = 2 br, $% = 3 br ve \"% = 4 br dir. | |BC = 8 sin a ve % = a ES :VLBSEBLJWFSJMFSFHÌSF DPTa LBÀUS m ( BOC ) :VLBSEBLJWFSJMFSFHÌSF,TJOa . cosaEFôFSJLBÀ- A) 1 B) 1 C) 1 % 1 E) 1 US 2 3 45 6 A) 0 B) 1 C) 1 % 1 E) 1 2 345 1. B 2. A 3. A 62 4. & 5. A 6. D

www.aydinyayinlari.com.tr 53÷(0/0.&53÷ 5. MODÜL ·/÷7&34÷5&:&)\";*3-*, 501-\".'\"3,'03.·--&3÷ %m/*m ÖRNEK 4 sin ( a + b ) = sina.cosb + cosa.sinb sin a + cos b = 1 WFcos a + sin b = 1 :VLBSEBLJGPSNÐMEFCZFSJOF -C ZB[MEóO- 34 EB PMEVôVOBHÌSF Tin ( a +C JGBEFTJOJOFöJUJOJCVMVOV[ sin ( a - b ) = sina.cos ( -b ) + cosa.sin ( -b ) 1 = ^ sin a + cos b h2 = 2 + 2 sin a. cos b + 2 = sina.cosb - cosa.sinb PMVS 9 sin a cos b )BUSMBUNB cos ( -b ) = cosb 1 = ^ cos a + sin b h2 = 2 + 2 cos a. sin b + 2 sin ( -b ) = -TJOCEJS 16 cos a sin b ÖRNEK 1 1 J sin a. cos b N cos70°. sin50° + sin40°. cos20° + K O JGBEFTJOJOFOTBEFI»MJOJCVMVOV[ 9 1 2 2+ c4o4s42a3 2KK O 2 2+ c4o4s42b3 16 = + K + O + sin40° = cos50°ve cos20° =TJOPMEVôVOEBO 1s4in44a 1 L b. cos O 1s4in44b4 1 cos70°.sin50° + cos50°.sin70° P sin a 3 = sin(50° + 70°) = sin120° = PMVS 25 = 2 + 2 sin^ a + b h j sin^ a + b h = - 263 PMVS 144 288 2 ÖRNEK 5 π < x < 3π ve 3π < y < 2πPMNBLÑ[FSF 22 tan x = 4 ve cos y = 12 3 13 PMEVôVOBHÌSF TJO Y+ y ) ifadesiniOFöJUJOJCVMVOV[ ÖRNEK 2 4 -4 -3 CÌMHFEF tan x = ve sin x = cos x = cosx . sin ( x - 30 ) + sinx . cos ( 30 - x ) JGBEFTJOJOFöJUJOJCVMVOV[ 3 55 DPTYGPOLTJZPOVÀJGUGPOLTJZPOPMEVôVOEBO CÌMHFEF cos y = 12 - 5 cos ( 30° - x ) = cos ( x - PMVS ise sin y = UÑS = cosx.sin ( x - 30° ) + sinx.cos ( x - 30° ) 13 13 = sin ( x + x - 30° ) = sin ( 2x - PMVS sin(x + y) = sinx.cosy + siny.cosx - 4 12 - 5 - 3 - 33 PMVS = ·+ · = 5 13 13 5 65 ÖRNEK 3 ÖRNEK 6 sin^ x + y h = 1 vFcos^ x - y h = 1 sin 10° + 3 cos 10° 23 cos 20° PMEVôVOBHÌSF TJOYJOFöJUJOJCVMVOV[ ifadesJOJOFöJUJOJCVMVOV[ Not: 3 = tan 60°FõJUMJóJOJLVMMBOO[ sin 10° + sin 60° . cos 10° sin 10° + tan 60° . cos 10° = cos 60° cos 20° cos 20° cos^ x + y h = 1 ve sin^ x - y h = 2 2 tür. sin 10° . cos 60° + sin 60° . cos 10° 23 = sin2x = sin ( x + y ) . cos ( x - y ) + sin ( x - y ).cos ( x + y ) cos 60° . cos 20° sin 70° 1 = = = 2 PMVS cos 60° . cos 20° 1 1 1 2 2 1 2 2+1 ·+ ·= 23 3 2 2 32 3 2. sin(2x – 30°) 2 2+1 63 - 263 - 33 6. 2 1. 3. 4. 5. 2 32 288 65

·/÷7&34÷5&:&)\";*3-*, 5. MODÜL 53÷(0/0.&53÷ www.aydinyayinlari.com.tr %m/*m ÖRNEK 10 cos (a + b) = cosa.cosb - sina.sinb cos^ a + b h = 1 WFcos^ a - b h = 1 :VLBSEBLJGPSNÐMEFCZFSJOF-CZB[MEóO- 23 EB PMEVôVOB HÌSF DPTBDPTC JGBEFTJOJO FöJUJOJ CVMV- cos (a - b) = cosa.cos (-b) - sina.sin (-b) OV[ = cosa.cosb + sina.sinb 1 PMVS DPTBDPTC-TJOBTJOC= ÖRNEK 7 2 1 cos105° DPTBDPTC+TJOBTJOC= JGBEFTJOJOEFôFSJOJCVMVOV[ 3 11 5 DPTBDPTC= + = 23 6 5 cPTBDPTC= PMVS 12 cos ( 60° + 45° ) = cos60°.cos45° - sin60°.sin45° 12 32 2- 6 PMVS =· - · = 22 2 2 4 ÖRNEK 11 ÖRNEK 8 sin a. sin b = 1 WFcos a. cos b = 3 cos ( 50° - a ).cos ( 10° + a ) - sin ( 50° - a).sin ( a + 10° ) 38 JGBEFTJOJOEFôFSJOJCVMVOV[ PMEVôVOBHÌSF cos2 ( a +C - cos2 ( a -C JGBEFTJ- = cos(50° - a + a + 10°) OJOEFôFSJOJCVMVOV[ 1 31 1 = cos 60° = PMVS cos(a +C =DPTBDPTC-TJOBTJOC= - = 2 8 3 24 3 1 17 cos(a -C = cPTBDPTC+TJOBTJOC= + = 8 3 24 d 1 2 17 2 - 288 1 = - PMVS n -d n= 24 24 576 2 ÖRNEK 9 ÖRNEK 12 #JS\"#$ÑÀHFOJOEF m ( XA ) = A, m ( XC ) = C ve cos x = - sin x m ( XB ) = BPMNBLÑ[FSF sin y cos y cos A = 3 WFcos B = 12 | |PMEVôVOBHÌSF x - y ifadesiniOBMBDBôFOLÑÀÑL 5 13 QP[JUJGEFôFSJCVMVOV[ PMEVôVOBHÌSF cos COJOEFôFSJOJCVMVOV[ cosx.cosy = -sinx.siny A + B +$= $= 180° - ( A + B) cosx.cosy + sinx.siny = 0 cos ( x - y ) = 0 DPT$= cos ( 180°- ( A + B) = -cos ( A + B) π = - ( cosA.cosA- sinA.sinA ) x-y = = -d 3 · 12 - 4 · 5 n = -d 36 - 20 n = - 16 PMVS 2 5 13 5 13 65 65 2- 6 1 16 64 5 1 π 7. 8. 9. - 10. 11. - 12. 4 2 65 12 2 2

www.aydinyayinlari.com.tr 53÷(0/0.&53÷ 5. MODÜL ·/÷7&34÷5&:&)\";*3-*, %m/*m ÖRNEK 15 tan^ a + b h = tan a + tan b 1 + tan 65° . tan 5° 1 - tan a. tan b tan 65° - tan 5° :VLBSEBLJGPSNÐMEFCZFSJOF-CZB[MEón- JGBEFTJOJOEFôFSJOJCVMVOV[ EB 11 3 tan^ a - b h = tan a + tan^ - b h = tan a - tan b = = cot 60° = PMVS 1 - tan a. tan^ - b h 1 + tan a. tan b tan 65° - tan 5° tan 60° 3 PMVS 1 + tan 65° . tan 5° )BUSMBUNBUBO -a ) = -UBOBES ÖRNEK 16 ,PUBOKBOUGPOLTJZPOVOVOUPQMBNWFGBSLGPSNÐ- MÐCVMVOVSLFO Y Z ` c 0 , π mPMNBLÑ[FSF tan^ a + b h = 1 WF 2 cot^ a + b h tan x = 1 WFDPUZ= 3 tan^ a - b h = 1 2 PMNBLÑ[FSF UBO Y+Z JGBEFTJOJOEFôFSJOJCVMVOV[ cot^ a - b h cot y = 3 & tan y = 1 Ë[EFõMJLMFSJOEFOGBZEBMBOMS 3 1+1 tan x + tan y 23 tan ( x + y ) = = = 1PMVS 1 - tan x. tan y 11 1- · 23 ÖRNEK 13 ÖRNEK 17 UBOWFDPU JGBEFMFSJOEFôFSMFSJOJCVMVOV[ 1 - tan x 1 + tan x tan 45° + tan 30° JGBEFTJOJOFöJUJOJCVMVOV[ tan ( 45°+ 30° ) = 1 =UBOPMEVôVOEBO 1 - tan 45° . tan 30° tan 45° - tan x = tan^ 45° - x hPMVS 3 3+ 3 1 + tan 45° . tan x 1+ 3 3+ 3 3 == = 3 3- 3 3- 3 1 - 1· 3 3 1 3- 3 tan 75° = PMEVôVOEBO cot 75° = PMVS cot 75° 3+ 3 ÖRNEK 14 ÖRNEK 18 tan 20° + tan 25° tan2 20° - tan2 10° tan 20° . tan 25° - 1 1 - tan2 20° . tan2 10° JGBEFTJOJOEFôFSJOJCVMVOV[ JGBEFTJOJOFöJUJOJCVMVOV[ tan 20° + tan 25° ^ tan 20° - tan 10° h ^ tan 20° + tan 10° h - = - tan 45° = -PMVS = 1 - tan 20° . tan 25° ^ 1 + tan 20° . tan 10° h ^ 1 - tan 20° . tan 10° h = tan ( 20° - 10° ) . tan ( 20° + 10° ) = ( tan10°.tan30°) = 3 tan 10°PMVS 3 3+ 3 3- 3 14. –1 65 3 3 13. , 15. 16. 1 17. tan(45° – x) 18. tan 10° 33 3- 3 3+ 3

·/÷7&34÷5&:&)\";*3-*, 5. MODÜL 53÷(0/0.&53÷ www.aydinyayinlari.com.tr ÖRNEK 19 4E B ÖRNEK 21 ôFLJM Ë[EFõ LBSFMFS- xa y 3 EFOPMVõNVõUVS A 3 A xy m( % = a a ABC ) D4 7 3C B PMEVôVOBHÌSF tana EFôFSJOJ CVMV- | | | |ôFLJMEF\"#$%EJLEËSUHFO \"& =DN #$ =DN C OV[ | |%$ =DNWFm ( % ) = aES BEC :VLBSEBLJWFSJMFSFHÌSF UBOaEFôFSJOJCVMVOV[ a = x +ZPMEVôVOEBOUBOa = tan ( x +Z PMVS tan x + tan y 19 +2 a = x +ZPMEVôVOEBOtana = tan ( x + y ) = 1 - tan x. tan y tan x + tan y 4 49 4 7 tan ( x + y ) = 1 - tan x. tan y = == +1 1 12 1 - ·2 33 42 = = =-7 1 - 4 ·1 - 1 33 ÖRNEK 22 ôFLJMEFLJ0NFSLF[MJZBSN¿FNCFSJOZBS¿BQDNEJS C D E 3 2 a 42 ÖRNEK 20 x 3 O y A 3B A \"#$EJLÐ¿HFO [#$]¿FNCFSF#OPLUBTOEBUFóFU m ( A%EB ) = a a | |\"# =DN | | | |\"% =DN #$ =DNEJS 3 | |#% =DN :VLBSEBLJWFSJMFSFHÌSF cotaEFôFSJOJCVMVOV[ | |%$ =DNWF ib % a + x + y = 180° j a = 180° - (x + y) D 2 C m ( DAC ) = a ES B4 1 cota = cot (180°- (x + y ) ) = -cot ( x + y ) = - tan^ x + y h :VLBSEBLJWFSJMFSFHÌSF DotaEFôFSJOJCVMVOV[. 31 1 tan x = = ve tan y = 62 22 a = i - b DPUa = cot ( a - b ) 1- 1 · 1 31 1 - tan x. tan y 2 22 tan i = ve tan b = dir. =- =- 42 cot^ i - b h = 1 1 + tan i. tan b tan x + tan y 1+ 1 = tan^ i - b h tan i - tan b 2 22 31 3 1 4 2-1 1+ . 1+ 1- 42 8 11 4 11 4 2 4 2 4 2-1 3-1 = = ·= = - = - = - PMVS 42 1 81 2 2+1 2+1 2 2+2 4 22 22 19. –7 11 66 9 4 2-1 20. 21. 22. 2 2 2+2 2

www.aydinyayinlari.com.tr 53÷(0/0.&53÷ 5. MODÜL ·/÷7&34÷5&:&)\";*3-*, ÖRNEK 23 ÖRNEK 25 :FSEÐ[MFNJOFEJLPMBOEJSFóFJLJJQõFLJMEFLJHJCJ\"WF$ A G 2B \"#$%LBSF OPLUBMBSOBLB[LMB¿BLMNõUS a [(&] a [#%] = {F} B 6 F6 | | | |#( = %& =DN ab a6 | |\"% =DNWF 5 % 2 m ( BFG ) = a A 3 D2 C xy 2C 2 D2 E 2 | |BD = 5 m AD = 3 m %$ N m ( A%BD) = a :VLBSEBLJWFSJMFSFHÌSF UBOaEFôFSJOJCVMVOV[ 22 % = b ES m ( DBC ) :VLBSEBLJWFSJMFSFHÌSF tan ( a + b LBÀUS a = y - x j tana = tan ( y -Y PMVS tanx = 1 ve tany =PMEVôVOEBO tan y - tan x 3-1 2 1 tan ( y - x )= = = = PMVS tan a + tan b 1 + tan y. tan x 1 + 3.1 4 2 tan ( a + b ) = 1 - tan a. tan b 34 7 + 55 5 7 25 35 = = =· = 3 4 13 5 13 13 1- · 5 5 25 ÖRNEK 24 ÖRNEK 26 ôFLJMEFLJ\"WF#OPLUBMBSOEBOBZOBOO$WF%LËõFMF- 6ZHVOUBONBSBMôOEB SJOF CBLBO JLJ LJõJOJO ZFS EÐ[MFNJZMF ZBQUó B¿MBS 9 WF tan f arcsin 3 + arccos 5 p :EJS 5 13 JGBEFTJOFZFFöJUUJS D C a 6 447a 448 6 4 447b 4 4 8 xy Yer 35 AB düzlemi tan (arcsin + arccos ) :VLBSEBLJWFSJMFSFHÌSF UBOaLBÀUS 5 13 3 33 arcsin = a & sin a = & tan a = PMVS 5 54 5 5 12 arccos = b & cos b = & tan b = 13 13 5 #VEVSVNEB a + x + y = 180° j a = 180°- (x + y) 3 + 12 tana = tan ( 180° - (x + y)) = -tan ( x + y ) tan^ a + b h = tan a + tan b = 4 5 63 =- 4 7 1– tan a. tan b 3 12 3 1- · 16 =- tan x + tan y 1+ 1 45 3 - - =- 3 =7 1 - tan x. tan y 4 1 - 1· 3 35 24. 7 67 1 - 63 23. 25. 26. 2 16 13

·/÷7&34÷5&:&)\";*3-*, 5. MODÜL 53÷(0/0.&53÷ www.aydinyayinlari.com.tr ÖRNEK 27 ÖRNEK 29 A :BOEBWFSJMFOõFLJMEF \"õBóEBLJõFLJMEFGVUCPMDV\"OPLUBTOEBO[#%]EVWBS- 1 [\"#] m[#$] OBEPóSVUPQBWVSEVóVOEBUPQEPóSVTBMPMBSBL$OPLUB- [ FC ] a [\"%] = {&} TOBHJEJZPS5FLSBSBZOOPLUBEBOUPQBWVSEVóVOEBUPQ F CVTFGFSEPóSVTBMPMBSBL#OPLUBTOB¿BSQZPS E | |\"' =DN B ax x 5 7 | |z y 20 15 C B 3 D 3 C #' =DN y | | | |#% =DN m % a ES = %$ ( DEF ) = #VOBHÌSF UBOaEFôFSJOJCVMVOV[ a9 x =[-Z UBOY= tan ( 180° - a) = -tana, A 12 D 5 | | | | | |\"% =N \"$ =N \"# =NWF UBO[= 2 ve tany = PMEVôVOEBO 6 5 7 m( % = a ES 2- BAC ) tan x = tan^ z - y h = 6 6 73 7 :VLBSEBLJWFSJMFSFHÌSF tanaEFôFSJOJCVMVOV[ = =·= 5 8 6 8 16 1 + 2· 63 7 a = y - x j tana = tan ( y - x ) tan a = - tan x = - PMVS 16 43 tany = j tanx = 34 ÖRNEK 28 43 7 - ôFLJMEFCJSIFMJLPQUFSJOGBSOEBO¿LBOõLMBSOBZEOMBU- 34 12 7 UóCËMHFWFSJMNJõUJS tan a = = = 43 2 A 1+ · 24 45° a 34 5 45° y ÖRNEK 30 D5 B xC #JS\"#$ÑÀHFOJOEF VZHVOBÀEFôFSMFSJJÀJO UBO\"+UBO#+UBO$=UBO\"UBO#UBO$ | |'BSO ZFSEFO ZÐLTFLMJóJ \"% = LN m ( B%AD) = 45° FöJUMJôJOJOEPôSVPMEVôVOVHÌTUFSJOJ[ % tan a = 3 EJS tanA + tanB =UBO\"UBO#UBO$-UBO$ m ( BAC ) = aWF 5 tanA + tanB =UBO$ UBO\"UBO#- 1 ) | |:VLBSEBLJWFSJMFSFHÌSF B$ GBSOBZEOMBUUôCÌM- tan A + tan B = tan C HFOJOV[VOMVôVOVCVMVOV[ tan A. tan B - 1 a = 45°-ZPMEVôVOEBOUBOa = tan(45° -Z PMVS -tan (A + B) =UBO$ \"+ B +$=PMEVôVOEBO -tan (180° -$ =UBO$ tan45° = 1 ve tany = 5 ise UBO$=UBO$PMVS 5+x tan^ 45° - y h = tan 45° - tan y = x PMVS 1 + tan 45° . tan y 10 + x x3 = PMEVôVOEBOY=PMVS 10 + x 5 7 28. 15 68 7 27. - 29. 16 24

5PQMBN'BSL'PSNÑMMFSJ TEST - 26 1. sin 73° . cos 34° + sin 34° . cos 73° 5. cos55° . sin85° - cos35° . sin5° tan 107° JöMFNJOJOTPO VDVLBÀUS ifadesinJOFöJUJBöBôEBLJMFSEFOIBOHJTJEJS \" TJO # DPT $ TJO \" - 1 # $ 1 % & 3 2 2 2 % DPT & DPU 6. TJOOJOFöJUJBöBôEBLJMFSEFOIBOHJTJEJS sin 3π cos π - sin π cos 3π 77 77 2. \" 2 - 6 # 2 - 6 cos 3π 2 4 C) 3 - 1 14 6- 2 & 6 - 2 4 JGBEFTJOJOFöJUJBöBôEBLJMFSEFOIBOHJTJEJS % 2 \" m # tan 3π C) 1 14 % cot 3π & tan 2π 14 7 7. tan 73° - tan 28° 1 + cot 17° . cot 62° 3. cos 33° . cos 23° + sin 33° . sin 23° ifadesiniOFöJUJBöBôEBLJMFSE FOIBOHJTJEJS sin 100° . sin 70° - cos 100° . cos 70° JGBEFTJOJOFöJUJBöBôEBLJMFSEFOIBOHJTJEJS \" - 3 3 C) -1 % # - & 3 3 \" - # UBO $ UBO % DPU & 8. cos80° . cos50° + sin100° . cos40° 4. DPTOJOFöJUJBöBôEBLJMFSEFOIBOHJTJEJS JöMFNJOJOTPOVDVLBÀUS \" 2 - 6 # 2 - 6 C) 3 - 1 \" - 1 # - 3 C) 0 2 4 2 2 6- 2 6- 2 3 & 1 % & % 2 2 4 2 1. D 2. $ 3. & 4. B 69 5. $ 6. & 7. D 8. D

TEST - 27 5PQMBN'BSL'PSNÑMMFSJ 1. UBOOJOFöJUJBöBôEBLJMFSEFOIBOHJTJEJS 5. \"#$ÑÀHFOJOJOJÀBÀMBSOOÌMÀÑMFSJ\" # $PM- \" - 3 # + 3 C) 3 - 1 NBLÑ[FSF sin ( B + C ) % 3 + & - 3 cos ( A + C ) - cos ( A ) . cos ( C ) JGBEFTJOJOFöJUJBöBôEBLJMFSEFOIBOHJTJEJS \" -TFD $ # -DPTFD $ $ DPTFD $ % TFD( C ) & UBO( C ) 2. tan 52° + tan 8° 6. x - y =PMNBLÑ[FSF tan 52° . tan 8° - 1 ( sinx +TJOZ 2 + ( cosx +DPTZ 2 JGBEFTJOJOFöJUJBöBôEBLJMFSEFOIBOHJTJEJS JGBEFTJOJOFöJUJBöBôEBLJMFSE FOIBOHJTJEJS \" - # - $ % & \" – 3 # - 3 C) -1 3 % 3 & 3 3 7. YWFZEBSBÀPMNBLÑ[FSF cos 5π cos 3π - sin 5π sin 3π UBOYUBOZ= 1 3. 88 88 sin 140° cos 130° + sin 130° cos 140° PMEVôVOBHÌSF TJO Y+Z LBÀUS ifadesinin FöJUJBöBôEBLJMFSEFOIBOHJTJEJS \" 2 C) 3 % 1 & # 22 2 \" - # - 1 C) 1 % & 22 4. cos 67° - 3 . sin 67° 8. 1 - tan2 20° . tan2 10° cos 53° tan2 20° - tan2 10° JGBEFTJOJOFöJUJBöBôEBLJMFSEFOIBOHJTJEJS iöMFNJOJOTPOVDVLBÀUS \" - # - $ % & \" 3 # 3 . tan 10° C) 3 cot 10° % 3 . sin 10° & 3 . cos 10° 1. A 2. A 3. D 4. A 70 5. B 6. D 7. A 8. $

5PQMBN'BSL'PSNÑMMFSJ TEST - 28 1. A 4. A B \"#$%EJLEËSUHFO x _ |&$| = 3|%&| DE |\"#| = 2| #$| B DC m ( A%EB ) = aES C \"#$CJSEJLÐ¿HFO [\"#] m [#$] m( % = x :VLBSEBLJ WFSJMFSF HÌSF DPUa BöBôEBLJMFSEFO DAC ) IBOHJTJOFFöJUUJS | \"#| = | #%| = | |%$ES \" 1 # 1 C) 1 % 1 & 1 10 8 7 54 :VLBSEBLJWFSJMFSFHÌSF UBOYLBÀUS \" 1 # 1 C) 1 % & 3 4 3 2 2 2. A \"#$CJSÐ¿HFO 5. A z m ( A%BC ) = x x m ( A%CB ) = y m ( B%AC ) = z B DC x y UBOZ= 4 || || % B C UBO[=UÐS \"#$FõL FOBSÐ¿H FO #% = 2 %$ m ( DAC ) = x :VLBSEBLJWFSJMFSFHÌSF UBOYBöBôEBLJMFSEFO UJS IBOHJTJEJS :VLBSEBLJWFSJMFSFHÌSF UBOYLBÀUS \" - 7 # - 5 C) 5 % 7 & \" 5 3 # 3 55 3 5 C) 11 11 11 11 3 33 35 % & 5 5 3. D EC \"#$%LBSF Fx [\"$]LËõFHFO 6. D C \"#$%LBSF xE [\"$] a [#&] = { F } CE =3 |%&| = 3 |&$| EB % = x UJS % = x UJS m ( CFB ) m ( DEA ) AB AB :VLBSEBLJWFSJMFSFHÌSF UBOYLBÀUS \" - 3 # - 5 :VLBSEBLJWFSJMFSFHÌSF DPUYLBÀUS 5 3 C) -1 \" - 16 # - 13 C) 16 % 13 & % 3 & 5 13 16 13 16 5 3 1. B 2. D 3. B 71 4. B 5. B 6. $

TEST - 29 5PQMBN'BSL'PSNÑMMFSJ 1. D C 4. y y= x 3 E F D C x x x O B A A BK | | | |\"#$%WF&',#CJSFSLBSF \", = 3 #, WF ôFLJMEF EJL LPPSEJOBU TJTUFNJOEF \"#$% LBSFTJ WF m ( A%EK ) = x UJS y = x EPóSVTVWFSJMNJõUJS 3 :VLBSEBLJWFSJMFSFHÌSF UBOYLBÀUS m % = x PMEVôVOBHÌSF UBOYLBÀUS ( DOC ) \" - # -2 C) - 3 % - & 1 \" 1 # 2 C) 3 % 4 & 5 22 7 7 7 77 D C 5. D C GF F 2. x x B 2E AB E A4 | | | |\"#$%WF#&'(CJSFSLBSF \"# =DN #& =DN \" # &WF% ' &OPLUBMBSLFOEJBSBMBSOEBEPóSV- WFm^ \\FAC h = x PMEVôVOBHÌSF DPUYLBÀUS I I I ITBM \"#$%LBSF #& = 2 \"# WFm(B%DE) = x UJS \" 1 # 1 C) 1 :VLBSEBLJWFSJMFSFHÌSF UBOYLBÀUS 4 2 \" 1 # 1 C) 1 % 4 & % & 5 3 25 3. cosf arcsinf - 3 p + arccosf 5 p p 6. a + i = π PMNBLÑ[FSF 5 13 2 cot a + cot i = 8 3 JGBEFTJOJOFöJUJLBÀUS \" 56 # 39 C) 66 % 1 & 1 PMEVôVOBHÌSF TJOa . siniÀBSQNLBÀUS 65 44 17 9 81 \" 1 # 3 C) 4 % 17 & 2 8 5 20 1. A 2. D 3. A 72 4. A 5. $ 6. B

5PQMBN'BSL'PSNÑMMFSJ TEST - 30 1. K = arcsin 1 + arccos 2 5. cosc x - π m 33 4 =2 PMEVôVOBHÌSF DPU,LBÀUS cosc x + π m 3 \" 3 # 2 $ % 1 & 1 4 22 23 PMEVôVOBHÌSF UBOYLBÀUS \" - 1 # - 1 C) 1 % 1 & 1 3 55 3 2 2. D EC \"#$%LBSF F a | |%& = 3| |&$ A B | |#' = 2| CF| 6. x = arctan f 3 pWFY+Z= 315° % 4 m ( EAF ) = a ES PMEVôVOB HÌSF Z BöBôEBLJMFSEFO IBOHJTJOF FöJUUJS \" arctan 4 # arctan 3 5 5 :VLBSEBLJWFSJMFSFHÌSF UBOaLBÀUS C) arctan f - 3 p % arctan f - 4 p 5 5 \" 1 # 7 C) 7 % 6 & 5 2 13 17 17 7 & BSDUBO -7 ) 3. cos^ x + y h = 1 WFcos^ x - y h = 1 7. tancarccos 4 + arcsin 1 m 32 55 PMEVôVOBHÌSF DPUYDPUZLBÀUS JGBEFTJOJOEFôFSJLBÀUS \" - # - $ % & \" # $ % & 4. x + y < π PMNBLÑ[FSF 8. DPU B- 15° ) =LPMNBLÐ[FSF 2 cos ( a + OJOLUÑSÑOEFOFöJUJBöBôEBLJMFS- DPUY+DPUZ=WFTJOYTJOZ= 4 EFOIBOHJTJEJS 5 \") k - 1 # 2 k 1+2 k PMEVôVOBHÌSF tan ( x +Z LBÀUS k+1 C) \" 1 # 1 C) 3 % & 4 % k - 1 2 1 + k2 4 3 4 3 2 + 2k2 & k 2 2 + 2k2 1. B 2. D 3. A 4. & 73 5. B 6. & 7. D 8. D

·/÷7&34÷5&:&)\";*3-*, 5. MODÜL 53÷(0/0.&53÷ www.aydinyayinlari.com.tr ÷,÷,\"5\"¦*'03.·--&3÷ %m/*m ÖRNEK 4 sin ( a + b ) = sina.cosb +DPTBTJOCFõJUMJóJOEF sin x + cos x = 1 CZFSJOFBZB[MSTB 4 sin ( a + a ) = sina.cosa + cosa.sina sin2a =TJOBDPTBGPSNÐMÐFMEFFEJMJS PMEVôVOBHÌSF TJOYJGBEFTJOJOFöJUJOJCVMVOV[ ÖRNEK 1 &öJUMJôJOIFSJLJUBSBGOOLBSFTJOJBMBMN sin 50° JGBEFTJOJOFOTBEFI»MJOJCVMVOV[ ^ sin x + cos x h2 = d 1 2 cos 25° 4 n sin2x + 2sinx . cosx + cos2x = 1 16 1 1 - 15 1 + sin 2x = j sin 2x = - 1 = PMVS 16 16 16 sin50° = 2sin25°.cos25° #VEVSVNEB sin 50° 2 sin 25° . cos 25° = cos 25° cos 25° ÖRNEK 5 =TJOPMVS π < x < π PMNBLÐ[FSF ÖRNEK 2 42 4. cos 80° . cos 10° . cos 340° sin2x - cos2x = 1 cos 50° 1 - sin 2x JGBEFTJOJOFOTBEFI»MJOJCVMVOV[ PMEVôVOBHÌSF sinx +DPTYJGBEFTJOJOEFôFSJOJCV- MVOV[ cos80°= TJO DPT = cos20° ve cos50°= sin40° PMEVôVOEBO sin2x - cos2x = 1 - sin 2x ^ sin x - cos x h^ sin x + cos x h = sin2x + cos2x - 2 sin x.cosx 6 4 4 44sin7204° 4 448 ^ sin x - cos x h^ sin x + cos x h = ^ sin x - cos x h2 2. 2. cos 10° . sin 10° . cos 20° 2. sin 20° . cos 20° ^ sin x - cos x h^ sin x + cos x h = sinx - cosx == sin 40° sin 40° π < x < π JÀJO|sinx - cosx| = sinx - cosx ve sin 40° 42 = = 1PMVS sinx-DPTYâPMBDBôOEBOTJOY+ cosx =PMVS sin 40° ÖRNEK 3 ÖRNEK 6 sin x. cos x. cos 2x = 1 sin 36° + cos 36° ifadesinin FöJUJOJCVMVOV[ 4 sin 12° cos 12° PMEVôVOBHÌSF TJOYJOEFôFSJOJCVMVOV[ &öJUMJôJOIFSJLJUBSBGOJMFÀBSQBMN sin 36° cos 36° sin 36° . cos 12° + cos 36° . sin 12° += 1 sin 12° cos 12° sin 12° . cos 12° 124s4in4x2. c4o4s4x3 · cos 2x = ·2 4 (cos 12° ) (sin 12° ) sin 2x sin 48° 2. sin 24° . cos 24° 1 == = 4 cos 24° sin 2x. cos 2x = jFöJUMJôJOIFSJLJUBSBGOJMFÀBSQBMN sin 24° sin 24° 2 1 22 2.sin2x.cos2x = ·2 & sin 4x = 1PMVS 2 1. 2sin25° 2. 1 3. 1 74 15 5. 1 6. 4cos24° 4. - 16

www.aydinyayinlari.com.tr 53÷(0/0.&53÷ 5. MODÜL ·/÷7&34÷5&:&)\";*3-*, ÖRNEK 7 ÖRNEK 9 cos52° =YPMEVôVOBHÌSF cos4 20° - sin4 20° TJOOJOYUÑSÑOEFOFöJUJOJCVMVOV[ sin 25° . cos 25° A cos52° = sin38° =YUJS JöMFNJOJOTPOVDVOVCVMVOV[. sin76° = 2.sin38° .cos38° 64 4 4 471 4 4 448 1 PMEVôVOEBO a cos220° - sin220° k(cos220° + sin220) x sin76°= 2.x. 1 - x2 PMVS = 38° 2 B 1 – x2 C sin 25° . cos 25° 2 cos 40° 2. cos 40° = 2 PMVS == sin 50° sin 50° 2 %m/*m ÖRNEK 10 cos ( a + b ) = cosa.cosb -TJOBTJOCGPSNÐMÐO- 1 - cos 20° EFCZFSJOFBZB[BSBL 1 + cos 20° cos2a = cos2 a - sin2 aGPSNÐMÐFMEFFEJMJS#V JGBEFTJOJOFOTBEFI»MJOJCVMVOV[ GPSNÐMEFDPT2 a + sin2 a =FõJUMJóJOEFOZBSBS- MBOBSBLDPT2 aZFSJOF- sin2 BWFTJO2 BZFSJ- 1 - a 1 - 2 2 10° k 2 2 OF- cos2 BZB[BSBL cos2a = 2cos2 a - 1WF cos2a = 1 - 2sin2 a GPSNÐMMFSJPMVõUVSVMVS sin 2 sin 10° tan = 2 = 2 = 10° PMVS 1 + 2 cos 10° - 1 2 cos 10° ÖRNEK 8 ÖRNEK 11 sin2 40° - cos2 40° sin10° = x ^ 2 cos2 20° - 1 h. sin 40° PMEVôVOBHÌSF sin2 JGBEFTJOJOYUÑSÑOEFOFöJUJ- JöMFNJOJOTPOVDVOVCVMVOV[. OJCVMVOV[ sin240° - cos240° = -(cos240° - sin240°) = -cos80° sin10° = cos80°= 1 - 2sin240° = x 2cos220° - 1 =DPTPMEVôVOEBO 1 - x = 2sin240° - cos 80° - 2. cos 80° == 1-x 2 cos 40° . sin 40° 2. sin 40° . cos 40° = sin PMVS 2 40° - 2 cos 80° = = - 2 cot 80°PMVS sin 80° 2 8. –2cot80° 75 9. 2 10. tan210° 1-x 11. 7. 2x 1 - x 2

·/÷7&34÷5&:&)\";*3-*, 5. MODÜL 53÷(0/0.&53÷ www.aydinyayinlari.com.tr ÖRNEK 12 ÖRNEK 13 0 < x < π PMNBLÑ[FSF 3cosx = 4sinx 2 PMEVôVOBHÌSF UBOYEFôFSJOJCVMVOV[ 1 + cos x 2 3cosx = 4sinx j sin x 3 = = PMVS cos x 4 sin x 2 PMEVôVOBHÌSF DPTYJOEFôFSJOJCVMVOV[ 2· 3 tan 2x = 2 tan x & tan 2x = 4 1 + 2 cos2 x - 1 2x x 1 - tan 2 1-d 3 2 2. cos 2 cos x n 2 22 = 4 x x= x x 2. sin · cos 2 sin · cos xx 66 2 sin · cos 22 22 4 4 6 16 24 22 tan 2x = = = · = PMVS 21 2 x 9 7 47 7 ·x = ise sin = 1EJS 1- 2 sin 2 2 16 16 2 cosx = 1 - 2sin2 x PMEVôVOEBO cosx = 1 - 2.12 = -PMVS 2 ÖRNEK 14 1 - tan210° = x 2 tan 10° PMEVôVOB HÌSF UBO OJO Y UÑSÑOEFO FöJUJOJ CVMV- OV[ %m/*m 2 tan 10° 1 1 - tan210° = x = tan 20°EJS tan^ a + b h = tan a + tan b 1 - tan a. tan b 2 2 GPSNÐMÐOEFCZFSJOFBZB[BSBL tan 2a = 2 tan a FMEFFEJMJS 2 tan 20° x x 2x 1 - tan2a = tan 40° = == #FO[FSõFLJMEF 2 1 x2 - 1 x2 - 1 cot^ a + b h = cot a. cot b - 1 1 - 20° 1- cot b + cot a tan GPSNÐMÐOEFCZFSJOFBZB[BSBL 2 cot 2a = cot2a - 1 x 2x 2 cot a x FMEFFEJMJS oMVS DPUBUBOB=PMEVóVOEBO ÖRNEK 15 cot 2a = 1 PMVS DPUY-UBOY= 1 tan 2a 2 #V EVSVNEB cot 2a = 1 - tan2 a PMBSBL EB PMEVôVOBHÌSF UBOYEFôFSJOJCVMVOV[ 2 tan a cos x sin x cos2x - sin2x 1 IFTBQMBOBCJMJS = sin x - cos x = = sin x. cos x 2 cos 2x 11 = = 2 cot 2x = j cot2x = sin 2x 24 2 PMEVôVOEBOUBOY=PMVS 12. –1 76 24 2x 15. 4 13. 14. 7 x2 - 1

www.aydinyayinlari.com.tr 53÷(0/0.&53÷ 5. MODÜL ·/÷7&34÷5&:&)\";*3-*, ÖRNEK 16 ÖRNEK 18 tanc x + π m = 2 \"õBóEBLJõFLJMEFFOÐTULBUZBONBLUBPMBONZÐL- 4 TFLMJóJOEFLJCJSCJOBZB JUGBJZFFSJZFSMFaEFSFDFMJLB¿ ZBQBDBLõFLJMEFZBOHONFSEJWFOJLPZVZPS PMEVôVOBHÌSF DPUYEFôFSJOJCVMVOV[ 2 tand x + π n π 4 2.2 tand 2x + n = = 2 π 2 1 - 2 d x + n 1-2 tan 4 ÷UGBJZFFSJ tand 2x + π n = - cot 2x = 4 4 j cot 2x = PMVS 2 -3 3 ÖRNEK 17 2a 4BIJM õFSJEJOJO \" WF # OPLUBMBSOEBO TSBTZMB TBIJMMF a ÷UGBJZF FSJOJO I[ TBOJZFEF ZBSN NFUSF tan a = 1 WFaEFSFDFB¿ZBQBSBLJLJUFLOFIBSFLFUFCBõMZPSWF 2 $OPLUBTOEBLBSõMBõZPSMBS PMNBLÑ[FSF JUGBJZFFSJOJOZBOHOZFSJOFLBÀTOTPn- C SBVMBöBDBôOCVMVOV[ A 2 tan a tan2a = 1 - tan2 a 200 1 160 2. 2a 24 B x = 120m C == 12 3 1-d n 2 Aa 2a B 160 4 PMEVôVOEBO tan2a = = ise x = 120 NPMVS TBIJMõFSJEJ x 3 120 - 160 - 200 Ì[FMÑÀHFOJOdeONFSEJWFOJOCPZVN EJSTBOJZFEF NZPMBMBOJUGBJZFFSJNZPMVTO EFBMS | |\"$ =LNWe tan a = 1 PMEVôVOBHÌSF CBöMBO- ÖRNEK 19 2 HÀUBJLJUFLOFBSBTV[BLMLLBÀLNEJS y :BOEBLJ õFLJMEF 0 NFS- C \"%$ÑÀHFOJ1JTBHPS LF[MJ¿FZSFLCJSJN¿FNCFS UFPSFNJOEFO a2 + 4a2= 25 B WF \"0# Ð¿HFOJ WFSJMNJõ- j 5a2 = 25 5 a= 5 UJS\"OPLUBTOEBCVMVOBO 2a j a = 5 PMVS a D xB A IBSFLFUMJ # OPLUBTOB A 2a _ EPóSV IBSFLFU FEFSLFO 1 O x õFLJMEFLJ HJCJ a EFSFDFMJL 2 tan a 2· 24 | |B¿ ZBQUóOB HËSF AB tan 2a = = = ve nin aUÑSÑOEFOFöJUJOJCVMVOV[ 2 1-d 1 2 3 1 - tan a n 2 54 O õFLJMEFLJHJCJ0\"#JLJ[LFOBS aa %$#ÑÀHFOJOEFO tan 2a = x = 1 22 a 3 ÑÀHFOJOEFO sin = x oM- AD 1 EVôVOEBO 2 B 35 | AB | = 2x = 2.sin a x = PMVS PMVS 2 4 #VEVSVNEB| AB | = 2 5 + 3 5 = 11 5 44 4 11 5 a 16. 17. 77 18. 400 19. 2 sin 2 3 4

TEST - 31 ÷LJ,BU\"À'PSNÑMMFSJ 1. cos345° . sin165° 5. sin x - cos x = 2 JGBEFTJOJOFöJUJBöBôEBLJMFSE FOIBOHJTJEJS 3 PMEVôVOB HÌSF TJOY JGBEFTJOJO FöJUJ BöBô \" 1 # 1 C) 1 % & 8 4 2 EBLJMFSEFOIBOHJTJEJS \" 2 # 4 C) 5 % 2 & 8 9 9 9 39 2. UBO DPU 6. 2cos2 105° JGBEFTJOJOFöJUJBöBôEBLJMFSEFOIBOHJT JEJS JGBEFTJOJOFöJUJBöBôEBLJMFSEFOIBOHJTJEJS \" 1 - 3 # 1 + 3 2+ 3 2 2 C) \" - # - 2 2 C) -2 2 % - 2 & -1 2- 3 3- 2 % & 2 2 3. sin 72° - cos 72° 7. 1 + cos 4x sin 24° cos 24° 1 - cos 4x JGBEFTJOJOFöJUJBöBôEBLJMFSE FOIBOHJTJEJS JGBEFTJOJOFöJUJBöBôEBLJMFSEFOIBOH JTJEJS \" 1 # 1 \" UBOY # DPUY $ UBO2 x 4 2 $ % & % UBO2 Y & DPU2 2x 4. DPU -UBO 8. 1 + sin 2x JGBEFTJOJOFöJUJBöBôEBLJMFSE FOIBOHJTJEJS sin x + cos x \" - # -2 C) - % & JGBEFTJOJOFöJUJBöBôEBLJMFSEFOhanHJTJEJS 1. B 2. B 3. D 4. B \" # -1 C) sinx -DPTY % TJOY+ cosx & DPTY- sinx 78 5. $ 6. D 7. & 8. D

÷LJ,BU\"À'PSNÑMMFSJ TEST - 32 1. 8. sin 3x. cos 6x. cos 12x 5. cos12° =NPMEVôVOBHÌSF sin 24x sin 36° cos 12° + sin 12° cos 36° sin 12° cos 12° JGBEFTJOJOFöJUJBöBôEBLJMFSEFOIBOHJTJEJS JGBEFTJOJO N UÑSÑOEFO FöJUJ BöBôEBLJMFSEFO \" DPTY # DPUY $ UBOY IBOH JTJEJS % TFDY & DPTFDY \" N2 - # -N2 C) 4 -N2 % N2 - & N 2. tan x = 1 6. UBO= x 2 PMEVôVOB HÌSF DPU OJO Y UÑSÑOEFO FöJUJ PMEVôVOB HÌSF UBOY JGBEFTJOJO FöJUJ BöBôEB BöBô EBLJMFSEFOIBOHJTJEJS LJMFSE FOIBOHJTJEJS # x2 - 1 x \" 2 # 3 $ % 3 & 4 \" x C) 2x 3 4 23 x2 - 1 x2 - 1 % x2 - 1 & 1 - x2 2x 2x 7. TFD= x 3. tan 2x = 7 PMEVôVOBHÌSF DPTOJOYUÑSÑOEFOFöJUJBöB ôEBLJMFSEFOIBOHJTJEJS 24 \" x2 - 2 # 2 - x2 C) x2 - 2 PMEVôVOBHÌSF UBOYJGBEFTJOJOFöJUJBöBôEBLJ x x x2 MFSEFOIBOHJTJPMBCJMJS \" 1 # 1 C) 1 % 1 & 1 % 2 - x2 & 1 - x2 7 6 5 43 x2 x2 8. sin55° = x 3 cos 25° - sin 25° 3 ifadesi- 4. sin x - cosec x = 27 PMEVôVOB HÌSF sin 110° cos x - sec x OJO Y UÑSÑOEFO FöJUJ BöBôEBLJMFSEen hanHJTJ- PMEVôVOBHÌSF tanx ifadesiOJOFöJUJBöBôEBLJ- MFSEFOIBOHJTJEJS EJS \" - 1 # - 1 C) - % 1 & 1 \" Y # x 3 C) 1 93 39 % x 2 x3 2 & x 1. D 2. & 3. A 4. D 79 5. D 6. D 7. $ 8. $

TEST - 33 ÷LJ,BU\"À'PSNÑMMFSJ 1. ( sin2105° - cos2105°) . ( sin4105° - cos4105° ) 5. aEBSBÀPMNBLÑ[FSF ifaEFTJOJOFöJUJBöBôEBLJMFSEFOIBOHJTJEJS x2 + ( sina + cosa ) x + sin 2a = 0 2 \" - 1 # - 3 $ % 1 & 3 44 44 EFOLMFNJOJOLÌLMFSJBWFCPMEVôVOBHÌSF a2 +C2LBÀUS \" # $ - % & -2 2. 2 - 4 sin2 2x + 2 sin 2x 6. aEBSB¿PMNBLÐ[FSF SFFMTBZMBSEBUBONMG B C cos 4x. cos 2x + sin 2x. cos 2x GPOLTJZPOV JGBEFTJOJOFöJUJBöBôEBLJMFSEFOIBOHJTJEJS G B C = a2+ b2 - 3ab \" UBOY # DPUY $ TFDY % DPTFDY & TFDY PMBSBLWFSJMJZPS G TJOa DPTa) = 1 2 PMEVôVOBHÌSF TJOaLBÀUS \" - # - 1 $ % 1 & 3 3 3. sin π . cos π . cos π . cos π 7. 1 Y CJSQPMJOPN aEBSB¿PMNBLÐ[FSF 16 16 8 4 P ( x ) = x2 - 3x + sin2 aQPMJOPNVUBONMBOZPS JGBEFTJOJOFöJUJBöBôEBLJMFSEFOIBOHJTJEJS \" 1 # 1 C) 1 % 1 & 16 8 42 1 Y QPMJOPNVOVO Y+ cosa JMFCÌMÑNÑOEFO LBMBOPMEVôVOBHÌSF DPTaLBÀUS \" - 1 # - 1 $ % 1 & 1 12 9 9 12 4. cos3 x + sin3 x 8. 3 tanf 2. arcsin p sin 2x - 1 5 2 JGBEFTJOJOFöJUJBöBôEBLJMFSEFOIBOHJTJEJS \" -cosx -TJOY # DPTY- sinx JGBEFTJOJOEFôFSJLBÀUS C) sinx -DPTY % TJOY \" 7 # 5 C) 24 % 12 & 24 12 75 & TJOY+ cosx 1. & 2. & 3. B 4. A 80 5. A 6. D 7. B 8. $

÷LJ,BU\"À'PSNÑMMFSJ TEST - 34 1. 10x =ÖPMNBLÑ[FSF 4. 1 + sin x + cos x sin 45° . 1 - sin 4x 1 + sin x - cos x sin x ifadesinin en sade I»MJ BöBôEBLJMFSEFO IBOHJ- JGBEFTJBöBôEBLJMFSEFOIBOHJTJOFFöJUUJS TJEJS \" 1 cos x # 1 sin x \" cot x # tan x C) sin x 22 22 2 2 2 C) 1 sec x % 1 cosec x % cos x & sec x 22 22 2 2 & 1 tan x 22 2. 5. A A a y 3 2a 2 C B x | | \"#$Ð¿HFOJOEF #$ =DN BC | | | |ôFLJMEF\"#$JLJ[LFOBSÐ¿HFO \"# = \"$ | |\"$ =DN m ( % ) = a WFm ( % ) = 2a ES BAC ABC %% 1 EJS :VLBSEBLJWFSJMFSFHÌSF DPTaLBÀUS m ( BAC ) = y m ( ABC ) = x WF tan y = 2 \" 4 $ 3 % 3 & 1 :VLBSEBLJWFSJMFSFHÌSF UBOYLBÀUS 3 # 4 23 \" 3 # 4 C) - 3 % - 4 & -1 4 3 43 C 6. A 3. 4 35 a D 2a A 11 B BD C | | | |\"# % | |\"#$Ð¿HFOJOEF [\"%]B¿PSUBZ \"# =DNWF =DN $% =DN m ( CAD ) = a WF m ( C%BD ) = 2aES | |\"$ =DNWFm ( % ) = a ES DAC | | #VOBHÌSF DB LBÀDNEJS :VLBSEBLJWFSJMFSFHÌSF UBOaLBÀUS \" # $ % & \" 1 # 1 C) 3 % 4 & 2 23 1. $ 2. D 3. B 81 4. A 5. $ 6. A

·/÷7&34÷5&:&)\";*3-*, 5. MODÜL 53÷(0/0.&53÷ www.aydinyayinlari.com.tr 53÷(0/0.&53÷,%&/,-&.-&3 DPTYB%FOLMFNJOJO¦Ì[ÑN,ÑNFTJ TJOYB%FOLMFNJOJO¦Ì[ÑN,ÑNFTJ %m/*m %m/*m -1 # a #PMNBLÐ[FSF DPTY=BEFOLMFNJ- -1 # a #PMNBLÐ[FSF TJOY=BEFOLMFNJ- nin [ Õ BSBMóOEBCJSLËLÐiJTFEFOLMFNJ- nin [ Õ BSBMóOEBCJSLËLÐiJTFEFOLMFNJ- OJO¿Ë[ÐNLÐNFTJ OJO¿Ë[ÐNLÐNFTJ |Ç = {x x = i +LÕWFZBY= -i +LÕ L` Z} |Ç = {x x = i +LÕWFZBY=Õ- i +LÕ L` Z} PMVS PMVS sin sin 1 P' 1 P P –1 1 a i r–i cos O –i a 1 –1 cos ii O P' –1 –1 ÖRNEK 1 ÖRNEK 3 2 sin x = 1 cos x = 2 2 EFOLMFNJOJOÀÌ[ÑNLÑNFTJOJCVMVOV[ dFOLMFNJOJOÀÌ[ÑNLÑNFTJOJCVMVOV[ sin x = 1 EFOLMFNJOJO [ Ö OEB CJS LÌLÑ π ise 26 cos x = 2 EFOLMFNJOJO[ Ö OEBCJSLÌLÑ π PM- EFOLMFNJOÀÌ[ÑNLÑNFTJ 24 Ç = ( x: x = π + 2πk veya x = π - π + 2πk , k ! Z 2 EVôVOEBO EFOLMFNJOÀÌ[ÑNLÑNFTJ 66 Ç = (x: π veya x = - π + 2πk , k ! Z 2 PMVS x = + 2πk 44 ÖRNEK 2 ÖRNEK 4 3 2 cos x = - sin x = - 2 2 EFOLMFNJOJOÀÌ[ÑNLÑNFTJOJCVMVOV[ EFOLMFNJOJOÀÌ[ÑNLÑNFTJOJCVMVOV[ cos x = - 3 EFOLMFNJOJO[ Ö OEBCJSLÌLÑ 5π PM- sin x = - 2 EFOLMFNJOJO[ Ö OEBCJSLÌLÑ 5π PMEV- 26 24 ôVOEBOEFOLMFNJOÀÌ[ÑNLÑNFTJ EVôVOEB EFOLMFNJOÀÌ[ÑNLÑNFTJ 5π x = π - 5π + 2πk , k ! Z 2 + 2πk 5π x = - 5π + 2πk, k ! Z 2 Ç = ( x: x= veya + 2πk 44 Ç = ( x: x = veya 66 PMVS PMVS ππ 5π - 5π + 2πk 82 π 5π 5π -π 1. + 2πk veya - + 2πk 2. + 2πk veya 3. + 2πk veya + 2πk 4. + 2πk veya + 2πk 66 6 64 4 44

www.aydinyayinlari.com.tr 53÷(0/0.&53÷ 5. MODÜL ·/÷7&34÷5&:&)\";*3-*, UBOYB%FOLMFNJOJO¦Ì[ÑN,ÑNFTJ DPUYB%FOLMFNJOJO¦Ì[ÑN,ÑNFTJ %m/*m %m/*m a `3PMNBLÐ[FSF UBOY=BEFOLMFNJOJO a `3PMNBLÐ[FSF DPUY=BEFOLMFNJOJO Õ BSBMóOEBCJSLËLÐiJTFEFOLMFNJO¿Ë[ÐNLÐ- [ Õ BSBMóOEBCJSLËLÐi c i ≠ π mJTFEFOL- NFTJ 2 |Ç = {x x = i +LÕ L` Z }PMVS MFNJO¿Ë[ÐNLÐNFTJ |Ç = {x x = i +LÕ L` Z }PMVS sin 1 sin PK 1 aK r+i a cot P i cos –1 r+i –1 O 1 i O 1 cos P' P' –1 –1 tan ÖRNEK 5 ÖRNEK 7 3 cot x = 3 tan x = EFOLMFNJOJOÀÌ[ÑNLÑNFTJOJCVMVOV[ 3 EFOLMFNJOJOÀÌ[ÑNLÑNFTJOJCVMVOV[ cot x = 3 EFOLMFNJOJO[ Ö OEBCJSLÌLÑ π PMEV- tan x = 3 EFOLMFNJOJO[ Ö OEBCJSLÌLÑ π PMEV- 6 ôVOEBOEFOLMFNJOÀÌ[ÑNLÑNFTJ 36 Ç = ( x: x = π + πk , k ! Z 2PMVS ôVOEBOEFOLMFNJOÀÌ[ÑNLÑNFTJ Ç = ( x: x = π + πk , k ! Z 2PMVS 6 6 ÖRNEK 6 ÖRNEK 8 tan x = - 3 3 EFOLMFNJOJOÀÌ[ÑNLÑNFTJOJCVMVOV[ cot x = - tan x = - 3 EFOLMFNJOJO[ Ö OEBCJSLÌLÑ 2π 3 PM- EFOLMFNJOJOÀÌ[ÑNLÑNFTJOJCVMVOV[ 3 cot x = - 3 EFOLMFNJOJO[ Ö OEBCJSLÌLÑ 2π PM- EVôVOEBOEFOLMFNJOÀÌ[ÑNLÑNFTJ 33 Ç = ( x: x = 2π + πk , k ! Z 2PMVS EVôVOEBOEFOLMFNJOÀÌ[ÑNLÑNFTJ 3 Ç = ( x: x = 2π + πk , k ! Z 2PMVS 3 π 2π π 2π 5. + πk 6. + πk 83 7. + πk 8. + πk 63 63

·/÷7&34÷5&:&)\";*3-*, 5. MODÜL 53÷(0/0.&53÷ www.aydinyayinlari.com.tr ÖRNEK 9 ÖRNEK 12 sinc 4x + π m = sinc x - π m 0 #U#PMNBLÐ[FSF CJS¿FLJSHFQPQVMBTZPOVOEBU 63 IBGUBEBLJ¿FLJSHFTBZT 1 U = 7500 + 3000 sin f πt pPMBSBLNPEFMMFONJõUJS EFOLMFNJOJOÀÌ[ÑNLÑNFTJOJCVMVOV[ 8 ππ ππ 4x + = x - + 2πk ya da 4x + = π - x + + 2πk #VOBHÌSF QPQVMBTZPOVOVOFOGB[MBOÑGVTBVMBöU- 63 63 ôIBGUBZWFIBOHJUEFôFSMFSJJÀJOOÑGVTVOPM- EVôVOVCVMVOV[ ππ ππ 3x = - - + 2πk 5x = π + - + 2πk 63 36 π 7π a) 1PQVMBTZPOVONBLTJNVNOÑGVTBVMBöNBTJÀJO 3x = - + 2πk 5x = + 2πk sind πt n = 1PMNBMES#VEVSVNEB 8 2 6 πt π π 2πk PMVS 7π 2πk PMVS = + 2πk PMBDBôOEBU= 4 +LPMVSL=JÀJO x =- + x= + 63 30 5 82 cevap t =IBGUBPMBSBLCVMVOVS ÖRNEK 10 C + 3000 sind πt n = 9000 8 cosc 2x + π m = cosc x + π m 34 3000 sind πt n = 1500 8 EFOLMFNJOJOÀÌ[ÑNLÑNFTJOJCVMVOV[ sind πt n = 1500 = 1 = sind π nPMEVôVOEBO 8 3000 2 6 πt π πt 5π = + 2πk ya da = + 2πk PMVS ππ ππ 86 86 2x + = x + + 2πk ya da 2x + = - x - + 2πk 34 34 πt π 4 = + 2πk & t = + 16k k ! Z 86 3 ππ -π π x = - + 2πk 3x = - + 2πk 4 k =JÀJO t = ve 43 34 π 3 - 7π x = - + 2πk PMVS 3x = + 2πk 12 12 πt 5π 20 j = + 2πk & t = + 16k k ! Z 86 3 - 7π 2πk PMVS x= + 36 3 4 20 t = ve t = EFôFSMFSJJÀJOOÑGVTPMVS 33 ÖRNEK 11 ÖRNEK 13 cosc π - π m = sinc 2x - π m .BSNBSB %FOJ[JhOEF PDBL HFDF ZBSTOEB CBõMBZBO 34 ËM¿ÐNMFSEF EFOJ[EFLJ EBMHBMBSO ZÐLTFLMJLMFSJ IFS TBBU EFOLMFNJOJOÀÌ[ÑNLÑNFTJOJCVMVOV[ CBõ ËM¿ÐMNÐõUÐS < U < PMNBL Ð[FSF EBMHB ZÐL- TFLMJóJh^ t h = 3 cosf πt p + 1 NFUSFPMBSBLNPEFMMFO- sinx = cos d π - x nPMEVôVJÀJO 62 2 NJõUJS cosd x - π n = cosd π - 2x + π nPMVS %BMHBZÑLTFLMJôJOJOFOGB[MBPMEVôVTBBUJCFMJSMFZJOJ[ 3 24 π 3π π - 3π %BMHB CPZVOVO FO ZÑLTFL PMNBT JÀJO cosd πt n = 1 x - = - 2x + 2πk ya da x - = + 2x + 2πk 34 34 πt 6 3π π π 3π PMNBTHFSFLJS cosd n = cos 0°PMEVôVOEBO 3x = + + 2πk - x = - + 2πk 6 43 34 πt = 0 + 2πk & t = 12k k ! Z PMVS 3x = 13π + 2πk 5π - x = - + 2πk 6 12 12 0 < t <JÀJOL= 1 ve t =PMVS#VEVSVNEBHFDF 13π 2πk PMVS 5π ZBSTOEBOTBBUTPOSBÌôMFOEBFOZÑLTFLEBM- x= + x = - 2πk PMVS 36 3 HBÌMÀÑMÑS 36 9. - π /6 + 2πk/3, 7π /30 + 2πk/5 10. - π /12 + 2πk, - 7π /36 + 2πk/3 84 4 20 13. 12.00 11. 13π /36 + 2πk/3, 5π /36 - 2πk 12. IBGUB ve hafta 33

www.aydinyayinlari.com.tr 53÷(0/0.&53÷ 5. MODÜL ·/÷7&34÷5&:&)\";*3-*, %m/*m B C` R - {0}PMNBLÐ[FSF asinx +CDPTY= cCJ¿JNJOEFLJEFOLMFNMFSF TJOYWFDPTYFHÌSFMJOFFS EPôSVTBM EFOLMFNEFOJS asinx + bcosx =D FõJUMJóJOIFSJLJUBSBGEBBJMFCËMÐOÐSTF sin x + b · cos x = c f b = tan a ve tan a = sin a p a aa cos a sin x + sin a · cos x = c cos a a cos a. sin x + sin a. cos x = c j sin^ x + a h = c · cos a cos a a a EFOLMFNJOJO¿Ë[ÐMFCJMNFTJJ¿JO- 1 # c · cos a # 1PMNBMES a - 1 # c · cos a # 1& 0 # c2 · cos2 a # 1 a a2 & c2 # a2 f 1 = 1 + tan2 a p cos2 a cos2 a j c2 # a2 . (1 +UBO2a) & c2 # a2·f 1 + b2 p a2 j c2 # a2 + b2FMEFFEJMJS #VFõJUTJ[MJóJOTBóMBONBTEVSVNVOEBEFOLMFNJO¿Ë[ÐNLÐNFTJCVMVOBCJMJS\"LTJEVSVNEBEFOLMFNJO¿Ë[ÐN LÐNFTJqPMVS ÖRNEK 14 ÖRNEK 15 2 sin x - 3 cos x = A EFOLMFNJO¿Ë[ÐNLÐNFTJOJO 3 cos x + sin x = 1 i) #PöLÑNFPMNBTJÀJO\"OOBMBCJMFDFôJFOLÑ- EFOLMFNJOJO[ ]BSBMôOEBLJLÌLMFSJOJCVMVOV[ ÀÑLQP[JUJGUBNTBZLBÀPMNBMES 3 = tan 60°PMNBLÑ[FSF ii) %FOLMFNJO ÀÌ[ÑN LÑNFTJOJO CPö LÑNF PMNB- tan60°.cosx + sinx = 1 NBTJÀJO\"OOEFôFSBSBMôOFPMNBMES sin 60° i) c2> a2+C2 j¦,= q · cos x + sin x = 1FöJUMJôJDPTJMFÀBSQBMN A2 > ^ 2 h2 + ^ - 3 h2 j A2 >PMEVôVOEBO ANJO= 3 cos 60° sin60°.cosx + sinx.cos60° = cos60° ii) c2ãB2+C2 sin(60° + x) = cos60° = sin30° A2ã^ 2 h2 + ^ - 3 h2 j A2ã 60°+ x = 30° + 360°k veya 60° + x = 150° + 360°k - 5 # A # 5 PMVS x = -30 + 360k veya x = 90° + 360k k =JÀJOY= 330° veya k =JÀJOPMVS 14. J JJ 7 - 5, 5 A 85 15. \\ ^

·/÷7&34÷5&:&)\";*3-*, 5. MODÜL 53÷(0/0.&53÷ www.aydinyayinlari.com.tr ÖRNEK 16 ÖRNEK 18 2 3 sin2x + 5 sin 2x = 3 3 x ` Ö PMNBLÑ[FSF EFOLMFNJOEFUBOYEFôFSMFSJOJCVMVOV[ 3 sin x + 3 cos x = 0 EFOLMFNJOEFDPTYJOEFôFSJOJCVMVOV[ 2 3 2 + 10. sin x. cos x = 3 3 . 2 x + 3 2 sin x sin 3 cos x 0= 3 2 - 10. sin x. cos x + 3 2 sin x - 3 sin x 3 cos x 0 = ^ 3 sin x - cos x h.^ sin x - 3 3 cos x h 3 sin x = - 3 cos x j cos x = =- 3 3 3 sin x - cos x = 0 sin x - 3 3 . cos x = 0 2π 2π 1 j tanx = - 3 j tan x = tan j cos = - 3 sin x = cos x sin x = 3 3 cos x 3 32 1 tan x = 3 3 PMVS PMVS tan x = 3 ÖRNEK 17 ÖRNEK 19 Õ BSBMóOEBTJOY- 4cosx = 5 sinc 2x + π m + cosc 2x + π m = 0 EFOLMFNJOEFDPTYJOEFôFSJOJCVMVOV[ 33 45 4 EFOLMFNJOJO [ Ö] BSBMôOEB LBÀ UBOF LÌLÑ PMEV- sin x - cos x = tan y = PMNBLÑ[FSF 33 3 ôVOVCVMVOV[ 5 - sind 2x + π n = cosd 2x + π n sinx - tany.cosx = 33 3 tand 2x + π n = - 1 3 sin y 5 sin x - cos y · cos x = 3 π 3π 2x + = + kπ 5 cos y. sin x - sin y. cos x = · cos y 34 3 x = 5π + kπ PMVQ[ Ö]BSBMôOEB ππ 24 2 sin(x - y) = 1 j x - y = & x = + y k = JÀJOGBSLMEFôFSWBSES 22 cos x = cosd π + y n = - sin y = - 4 PMVS 25 %m/*m B C` R - {0}PMNBLÐ[FSF asinx + bcosx = CJ¿JNJOEFLJ EFOLMFNMFSF ÖRNEK 20 CJSJODJ EFSFDFEFO IPNPKFO USJHPOPNFUSJL EFOLMFNEFOJS 5cos2 x + 3cosx.sinx - 1 = 0 asinx + bcosx =EFOLMFNJOEFFõJUMJóJOIFSJLJ EFOLMFNJOEFUBOYEFôFSMFSJOJCVMVOV[ ZBODPTYáPMNBLÐ[FSF DPTYFCËMÐOÐQ sin2x + cos2x =PMEVôVOEBO a· sin x + b· cos x = 0 5cos2x + 3sinx.cosx - sin2x - cos2x = 0 cos x cos x 4cos2x + 3sinx.cosx - sin2x = 0 (4cosx - sinx) (cosx + sinx) = 0 BUBOY+ b = 0 4cosx = sinx veya cosx = -sinx 4 = tanx veya tanx = -PMVS tan x = - b a EFOLMFNJOFEËOÐõUÐSÐMFSFL¿Ë[ÐNZBQMS 1 4 86 1 19. 4 20. \\m ^ 16. ( , 3 3 2 17. - 18. - 2 3 5

5SJHPOPNFUSJL%FOLMFNMFS TEST - 35 1. DPU Y - 2 3 = 0 5. UBO - x) =DPU Y- 60° ) EFOLMFNJOJOFOLÑÀÑLQP[JUJGJLJLÌLÑOÑOUPQMB- EFOLMFNJOJOFOLÑÀÑLQP[JUJGLÌLÑLBÀEFSFDF- NLBÀEFSFDFEJS EJS \" # $ % & \" # $ % & 2. cos ( 3x - 40° ) = cos ( 2x - 50° ) 6. 2cos6x = 3 EFOLMFNJOJOFOLÑÀÑLQP[JUJGLÌLÑLBÀEFSFDF- EFOLMFNJOJO [ Ö] BSBMôOEB LBÀ GBSLM LÌLÑ EJS WBSES \" # $ % & \" # $ % & 3. cosc 2x - π m = sinc x + π m 7. sec 3x = 3 12 12 cosec 3x EFOLMFNJOJOFOLÑÀÑLQP[JUJGLÌLÑLBÀEFSFDF- EJS EFOLMFNJnin [ Ö]BSBMôOEBLJFOCÑZÑLLÌLÑ \" # $ % & LBÀEFSFDFEJS \" # $ % & 4. sin 5x = 1 8. UBOYUBOY= 1 2 EFOLMFNJOJO[ ]BSBMôOEBLBÀGBSLMLÌ- EFOLMFNJOJO [ Ö] BSBMôOEBLBÀGBSLMLÌLÑ LÑWBSES WBSES \" # $ % & \" # $ % & 1. D 2. B 3. B 4. & 87 5. & 6. $ 7. & 8. B

TEST - 36 5SJHPOPNFUSJL%FOLMFNMFS 1. sin2x - 3 sinx = 0 5. \"öBôEBLJMFSEFO IBOHJTJ DPTY + 3cosx = 1 EFOLMFNJOJO[ Ö]BSBMôOEBLBÀGBSLMLÌLÑ EFOLMFNJOJOLÌLMFSJOEFOCJSJEJS WBSES \" # $ % & \" # $ % & 2. UBOY+DPU π = 0 6. \"öBôEBLJMFSEFOIBOHJTJ 12 cos2 x + sinx = sin2 x + cosx EFOLMFNJOJOLÌLMFSJOEFOCJSJEJS EFOLMFNJnin [ Ö]BSBMô OEBLBÀGBSLMLÌLÑ \" # $ % & WBSES \" # $ % & 3. \"öBôEBLJMFSEFOIBOHJTJ 7. sin3x + cos3x = 2 UBO2 x - ^ 3 + 1 hUBOY+ 3 = 0 1 - sin x cos x EFOLMFNJOJOLÌLMFSJOEFOCJSJ EFôJMEJS EFOLMFNJOJO [ Ö]BSBMôOEBLBÀGBSLMLÌLÑ \" # $ % & WBSES \" # $ % & 8. #JMHJ cosx -DPTZ= -2sin ^x+yh ^x+yh · sin 4. DPU2 x -DPUY+ 5 = 0 22 EFOLMFNJOJO[ Ö]BSBMô OEBLBÀGBSLMLÌLÑ PMBSBLCVMVOVS WBSES cos6x - sin5x = cos4x \" # $ % & EFOLMFNJOJO[ Ö]BSBMô OEBLBÀGBSLMLÌLÑ WBSES \" # $ % & 1. D 2. & 3. $ 4. $ 88 5. D 6. D 7. A 8. &

5SJHPOPNFUSJL%FOLMFNMFS TEST - 37 1. sin10x . cos6x = sin6x . cos10x 5. 4 cos2 x - ^ 2 2 + 2 hcos x + 2 = 0 EFOLMFNJOJOFOLÑÀÑLQP[JUJGLÌLÑLBÀEFSFDF- EFOLMFNJOJOFOLÑÀÑLQP[JUJGJLJLÌLÑOÑOUPQMB- EJS NLBÀEFSFD FE JS \" # $ % & \" # $ % & 2. cos x + 3 1 cos x + cos y sin x = 33 6. = 1 EFOLMFNJOJ TBôMBZBO FO LÑÀÑL QP[JUJG Y BÀT sin x + sin y LBÀEFSFDFEJS FöJUMJôJOEFYZUPQMBNOOFOLÑÀÑLQP[JUJGEF- ôFSJLBÀEFSFDFEJS \" # $ % & \" # $ % & 3. cos4x + 2sin2 2x - sinx = 0 7. x ` [ ]PMNBLÑ[FSF EFOLMFNJOJO LÌLMFSJOEFO CJSJ BöBôEBLJMFSEFO 2sin2 x - 3sinx + 1 = 0 IBOHJTJEJS EFOLMFN JOJO LÌLMFS UPQMBN BöBôEBLJMFSEFO \" # $ % & IBOHJTJEJS \" # $ % & 8. #JMHJ sinx +TJOZ= 2sin ^x+yh · cos ^x-yh 22 PMBSBLCVMVOVS 4. 8sinx + 6cosx - 10 = 0 0 # x <PMNBLÑ[FSF PMEVôVOBHÌSF TJOYBöBôEBLJMFSEFOIBOHJTJ- sin6x + sin5x + sin4x = 0 EJS PMEVôVOBHÌSF YJOBMBCJMFDFôJLBÀGBSLMEFôFS \" 2 # 3 C) 3 % 4 & 5 WBSES 3 4 5 5 12 \" # $ % & 1. D 2. & 3. & 4. D 89 5. A 6. & 7. D 8. &

TEST - 38 5SJHPOPNFUSJL%FOLMFNMFS 1. cos ( x + 45° ) = sinx . cos30° + cosx . sin30° 5. sin 8x = 2 FöJUMJô JOJTBôMBZBOFOLÑÀÑLQP[JUJGYBÀTLBÀ 3 EFS FDFEJS \" # $ % & EFOLMFNJOJO[ Ö]BSBMôOEBLBÀGBSLMLÌLÑ WBSES \" # $ % & 2. sin3 x - 1 sin x = 0 6. sin2x - cos2x = 0 16 EFOLMFNJOJOFOLÑÀÑLQP[JUJGLÌLÑLBÀEFSFDF- EJS EFOLMFNJOJO [ Ö] BSBMô OEB LBÀ LÌLÑ WBS- \" # $ % & ES \" # $ % & 3. sin x + 3 cos x = 2 EFOLMFNJOJO [ Ö BSBMôOEBLJ LÌLMFSJ UPQ 7. sinx - 2cosx = 0 MBNLBÀUS EFOLMFNJOJO [ Ö] BSBMôOEB LBÀ GBSLM LÌLÑ \" 5π # 23π 5π WBSES 12 12 C) 3 % Ö & 7π \" # $ % & 3 4. cos2x - cosx + 1 = 0 EFOLMFNJOJO[ Ö BSBMô OE BLJLÌLMFSJUPQMBN 8. 2sin2 x + sin2x - cos2x = 0 LBÀUS EFOLMFNJJÀJOUBOYEFôFSMFSJOJOÀBSQNLBÀUS \" π # π C) 5π % Õ & 2π \" - # - 1 C) 1 % & 3 2 6 3 33 1. A 2. D 3. & 4. $ 90 5. & 6. B 7. $ 8. B

Trigonometri KARMA TEST - 1 1. sin x 5. sin8° = x 1 + cos x PMEVôVOBHÌSF DPT2 BöBôEBLJMFSEFOIBO- JGBEFTJOJOFöJUJBöBôEBLJMFSEFOIBOHJTJEJS HJTJEJS \" cos x # sin x C) tan x \" 1- x # x + 1 C) x + 1 2 2 2 2 2 2 % sec x & cosec x % Y & x 2 2 2 2. x ! d 0 , π nPMNBLÑ[FSF 6. π < x < 3π PMNBLÑ[FSF 2 2 1 + cos x . 1– cos x BöBôEBLJMFSEFOIBOHJTJOFFöJUUJS \" TJOY # DPTY $ -cosx sin x – cos x = - 5 % -TJOY & 2 25 PMEVôVOBHÌSF DPUYLBÀUS \" 3 # 1 C) 1 % 2 & 1 7. \"#$%ZBNVóVOEB[%$] // [\"#] m ( % ) = a 4 2 3 75 BCD I I I I I I%$ =DN \"# =DN #$ =DN I I\"% =DNEJS #VOBHÌSF cosa kBÀUS 3. \"#$Ð¿HFOJOEFC=DN B= 4 3 cmWF \" 2 # 3 C) 3 % - 2 & - 3 3 4 5 34 \" \"#$ =DN2EJS #VOBHÌSF \"#$ÑÀHFOJOEF \"$#BÀTOOÌMÀÑ- 8. 0 # x <ÖPMNBLÑ[FSF TÑLBÀEFSFDFPMBCJMJS 1 - 4cos2 2x = 0 \" # $ % & EFOLMFNJOJO ÀÌ[ÑN LÑNFTJ BöBôEBLJMFSEFO 4. tan x = 2 IBOHJTJEJS 2 \") * π , π , 5π 4 # * π , π , 2π 4 PMEVôVOBHÌSF UBO 45 +Y LBÀUS 63 6 63 3 \" - 1 # - 1 C) 3 % 4 & 5 C) ( π , π , π 2 % * π , π , 2π , 5π 4 2 77 5 4 6 43 63 3 6 & * π , 5π , 5π 4 66 3 1. $ 2. A 3. $ 4. B 91 5. A 6. D 7. & 8. D

KARMA TEST - 2 5SJHPOPNFUSJ 1. tan x = 1 PMEVôVOBHÌSF 4. G Y Z = x4 - y4WFH Y = 2x2PMEVôVOBHÌSF 2 G DPTa TJOa ) +H TJOa ) cosc π + x m. cosc π - x m UPQMBNOOEFôFSJLBÀUS 66 \" TJOa # DPTa C) 2cos2a ÀBSQNOOEFôFSJLBÀUS \" # 1 C) 11 % & 11 % & 2 20 10 5. ( 1 + cos15° + sin15° ) . ( sin15° - 1 + cos15° ) 2. i = π PMEVôVOBHÌSF JGBEFTJOJOFöJUJBöBôEBLJMFSEFOIBOHJTJEJS 7 \" # 1 3 DPT Õ+ 4i ) -DPT Õ+ 3i ) 2 C) 4 JGBEFTJOJOFöJUJLBÀUS % 2 3 & 3 3 2 \" m # - 1 $ 3 2 % & 2 6. E F ôFLJMEFLJLÐQUF K 3. C D m ( D%AK ) = a C F % 2 m ( AKC ) = b DE G 1 H A6 B AB | | | | | | | |\"#$CJSÐ¿HFO CF = '& = &# \"% =DN :VLBSEBLJ WFSJMFSF HÌSF TJOa + cosb UPQMBN LBÀUS | | | |$% =DN m % =DN \"# ( FDE ) = a \" 2 + 6 2+ 3 6 3 # C) :VLBSEBLJWFSJMFSFHÌSF tanaLBÀUS 3 3 \" 1 # 1 C) 1 % & 3 2 4 3 2 % & 3 3 1. $ 2. $ 3. $ 92 4. D 5. B 6. D

5SJHPOPNFUSJ KARMA TEST - 3 1. 4 cos2 a - 2 sin2 a - 2 5. #JSCJSJOFFõJUWFEõUBOUFóFU¿FNCFSJOPMVõUVS- 8 cos2 a - sin2 a - 5 EVóVEBJSFTFMCJS[JODJS õFLJMEFHËSÐMEÐóÐHJCJZBS- ¿BQCSPMBOCJS¿FNCFSFEõUBOUFóFUUJS JGBEFTJOJOFöJUJBöBôEBLJMFSEFOIBOHJTJEJS \" 4 # 3 C) 2 % 1 & 1 5 4 3 23 2. sin^330° – xh + sin^330° + xh + sin x JGBEFTJOJOFöJUJBöBôEBLJMFSEFOIBOHJTJEJS #VOBHÌSF LÑÀÑLÀFNCFSMFSEFOCJSJOJOZBSÀB- QBöBôEBLJMFSEFOIBOHJTJOFFöJUUJS \" DPTY # TJOY $ DPTY \" 2 5 % m & TJOYmDPTY 5+1 # sin 10° 1- 2 sin 10° C) cos 20° % cos 10° 1- cos 10° 1- cos 10° 3. A & 2 sin 10° 1- sin 10° 45° 6 4 B DC I I I I I I\"#$Ð¿HFO #% = 2 %$ \"# = 6 br 6. y I I\"$ =CS m ( B%AD ) = 45° m ( % ) = aES E DAC A :VLBSEBLJWFSJMFSFHÌSF sinaLBÀUS B \" 3 #) 3 2 33 x 8 8 C) O CD % 3 8 4 & 3 5 ôFLJMEFLJ EJL LPPSEJOBU TJTUFNJOEF WFSJMFO 0 NFS- 8 LF[MJCSZBS¿BQM¿FZSFL¿FNCFSEF[\"$] m [0%] % = a ES m ( AOD ) 4. \"#$ÑÀHFOJOJOLFOBSMBSB C DPMNBLÑ[FSF I I#VOB HÌSF AB BöBôEBLJMFSEFO IBOHJsine a2 + b2 + c2 = 14a + 16b + 10c - 138 FöJUUJS PMEVôVOBHÌSF m (WA)LBÀEFSFDFEJS \" - cosa # TJOa - cosa + 1 ) \" # $ % & C) 3 ( sina + cosam % TJOa + 1) & TJOa + cosa + 1 ) 1. $ 2. & 3. B 93 4. D 5. & 6. $

KARMA TEST - 4 Trigonometri 1. 0 # x #ÖPMNBLÑ[FSF 4. arccosf 1 p = arc cotf 12 - 4x p x2 + 1 9 cos x + 2 sin π 6 =1 FöJUMJôJOJTBôMBZBOYEFôFSJLBÀUS cos 5π - sin 3π \" 3 # $ % 7 & 32 2 2 PMEVôVOBHÌSF YLBÀUS \" π # π C) 5π % 2π & 5π 6 3 12 3 6 2. C 5. %JLLPPSEJOBUTJTUFNJOEFWFSJMFOBõBóEBLJHSBGJL y Z= a +CTJO DY GPOLTJZPOVOBBJUUJS x Dy AB 2 \"#%$EËSUHFOJOEF [\"$] m [$%] [$%] m [%#] O Õx –2 % (BCD) I I[$#] m[\"#] m = a \"$ =YCS #VOBHÌSF B+C+DUPQMBNLBÀUS I I$% =ZCSEJS \" # $ % & #VOBHÌSF ZOJOYWFa DJOTJOEFOFöJUJBöB- ôEBLJMFSEFOIBOHJTJEJS \" YTJOa # YDPTa $ YUBOa & x sin 2a % YTJOa 2 6. E 3. A O A 20 B 4C D x O F B x ôFLJMEFLJ0NFSLF[MJ¿FNCFSEF [&'] m [\"%] C % ADF I I'UFóFUOPLUBT m =DN % ( ) = x \"0 BAC 0NFSLF[MJ¿FNCFSJOZBS¿BQDN m ( ) = A I I#$ = 4 cNEJS 3 #VOBHÌSF UBOYLBÀUS sin 2A = EJS 2 I I#VOBHÌSF #$ =YLBÀDNEJS \" 4 # 5 $ % 7 & 3 3 3 \" # $ % & 1. B 2. & 3. $ 94 4. A 5. A 6. A

Trigonometri KARMA TEST - 5 1. 1 - 1 = 2 2 4. a + i = π PMNBLÑ[FSF sin a cos a 17 denkMFNJOJO Ö BSBMôOEBLBÀLÌLÑWBSES cos^ 23a - 8i h + cos^ 25i - 6a h \" # $ % & UPQMBNOOTPOVDVLBÀUS \" sin π # cos 2π C) 1 17 17 % & cot π 17 J 11π . cos 7π N–1 K cos O K 12 12 O 2. arcsinf a + 1 p = arccosc a m 5. K O cos π . sin 3π a+2 a+2 K O K O L 8 8P EFOLMFNJOJ TBôMBZBO B EFôFSMFSJOJO UPQMBN LBÀUS JöMFNJOJOTPOVDVLBÀUS \" # $ % & \") 2 + 2 # 2 C) 2+2 & 6 3 % 2 + 3 3 2 3. A 6. A 12 8 BC D B 12 D C \"#%CJSÐ¿HFO m ( % ) = a WFUBOa = 2 CAD I I I I I I\"% = $% = 2 #$ EJS I I I I\"#$CJSÐ¿HFO [\"#] m [\"$] \"# = #% =DN II % % m ( BAC ) = i PMNBL Ñ[FSF ZVLBSEBLJ WFSJMFSe \"$ =DNWFm ( CAD ) = aES :VLBSEBLJWFSJMFSFHÌSF UBOaLBÀUS HÌSF taniLBÀUS \" # 2 C) 1 % 2 & 1 \" 3 # 2 C) 1 % & 4 3 2 53 4 3 2 3 1. B 2. $ 3. D 95 4. D 5. A 6. B

KARMA TEST - 6 Trigonometri 1. 0 < x <PMNBLÑ[FSF 4. 0 < x < π PMEVôVOBHÌSF cos2 ( 2x ) = -1 + sin2x 4 FöJUMJôJOJTBôMBZBOYBÀTLBÀEFSFDFEJS 1 + sin 2x + 1 - sin 2x \" # $ % & JGBEFTJBöBôEBLJMFSEFOIBOHJTJOFFöJUUJS \" TJOY+DPTY # TJOY- cosx ) $ DPTY % DPTY & TJOY 2. #JMHJ sina + sinb = 2.sin ^a+bh . cos ^a-bh 5. 0 # x #ÖPMNBLÑ[FSF 22 2cos2 x + cosx - 1 = 0 cos10° =YPMEVôVOBHÌSF EFOLMFNJOJO ÀÌ[ÑN LÑNFTJ BöBôEBLJMFSEFO IBOHJTJEJS sin10° + sin50° UPQMBNOO Y UÑSÑOEFO EFôFSJ BöBôEBLJMFSEFO \" ( π , π 2 # * π , 2π 4 3 33 IBOHJTJEJS \" Y2 - # - 2x2 C) x2 - 1 C) * π , π , 11π 4 % * π , π , 5π 4 2 66 33 % 1 - x2 & Y & * π , π , 5π 4 2 36 3 6. A 8 B 2 3. A \"#$CJSÐ¿HFO L K x | \"#| = 4 br 4 6 | \"$| = 6 br DC 4 a m % = a \"#$%EJLEËSUHFO [\"$] a [%,] = { L } C ( ACB ) 2a B % | | | | | |\"# =CS br m % m (ABC ) = 2a #, =CS KC = 4 ( KLC ) = x :VLBSEBLJ WFSJMFSF HÌSF DPTa ifadesinin de- :VLBSEBLJWFSJMFSFHÌSF cos 2x - 1 + sin 2x ifa- 1 - tan x ôFSJLBÀUS EFTJOJOEFôFSJLBÀUS \" 1 # 3 C) 1 % 3 & 7 \" 2 # 1 C) 2 % 3 & 4 4 8 8 16 16 55 5 55 1. $ 2. A 3. $ 96 4. D 5. D 6. &

Trigonometri KARMA TEST - 7 1. 16sinx = a 4. sin2c π + 2a m = 1- sin2c 3a + π m 8cos x 16 8 FöJUMJôJOJTBôMBZBOLBÀBEPôBMTBZTWBSES EFOLMFNJOJO Ö BSBMôOEBLJFOLÑÀÑLLÌLÑ \" # $ % & BöBôEBLJMFSEFOIBOHJTJEJS \" 23π # 21π C) 7π % π & π 16 80 16 16 32 2. sin2 x - 5sinxcosx + 2 = 0 PMEVôVOBHÌSF UBOY JO EFôFSJ BöBôEBLJMFSEFO 5. #JS\"#$Ð¿HFOJOEFB C DLFOBSV[VOMVLMBSWF IBOHJTJPMBCJMJS \" # $J¿B¿MBSOOËM¿ÐMFSJPMNBLÐ[FSF \" 1 # 2 C) 3 % & 5 a + b =DNWFsin A + sin B = 2 EJS 3 3 4 2 3 #VOB HÌSF \"#$ ÑÀHFOJOJO ÀFWSFM ÀFNCFSJOJO ÀBQLBÀDNEJS \" # $ % & 3. O C D 3 K 6. x = Arc sinf - 2 p Z= Arc sinf 1 p A4 x2 34 B ôFLJMEFLJ 0 NFSLF[MJ ¿FNCFSJO [%$] ¿BQ \"#$% [= Arc cosf 1 p u = Arc cosf - 1 p 42 | | | |QBSBMFMLFOBSOLFOBS \"% = 3CS \"# = 4 br | |#, =CSWFm ( % ) = x UJS TBZMBSOO CÑZÑLUFO LÑÀÑôF EPôSV TSBMBOö ABC BöBôEBLJMFSEFOIBOHJTJEJS #VOBHÌSF UBOYLBÀUS \" V>[>Z>Y # V>Z>[> x \" - 15 # - 11 C) - 3 C) x >Z>[>V % Y>[>Z> x % 3 & 15 & [>Z> x > u 1. & 2. B 3. D 97 4. D 5. & 6. A

KARMA TEST - 8 Trigonometri 1. sin^ 2i + 60° h = 1 4. 1 + sin 2x = cos 2x 3 sin i + 3 cos i 2 3 FöJUMJôJOJ TBôMBZBO Ö BSBMôOEBLJ FO LÑ- FöJUMJôJOJ TBôMBZBO FO LÑÀÑL QP[JUJG i BÀT LBÀ ÀÑLYBÀTLBÀSBEZBOES EFSFDFEJS \" π # π C) π % 3π & Õ \" # $ % & 4 3 24 2. x ! ^ –2π, 2π h, sin 3x = 1 5. cot x - 3 = 2 6 3 cot x - 1 3 PMEVôVOB HÌSF EFOLMFNJO ÀÌ[ÑN LÑNFTJ LBÀ EFOLMFNJOJO BSBMôOEBLJ LÌLMFSJOJO FMFNBOMES topMBNLBÀEFSFDFEJS \" # $ % & \" # $ % & H G C 3. 6. D C D E x F 6 45 A8 B AB ôFLJMEFLJEJLEËSUHFOMFSQSJ[NBTOEB ôFLJMEFLJ\"#$Ð¿HFOJOEF%J¿UFóFU¿FNCFSJONFS- | | | | | |&\" = 4 5 DN \"# =DN $# =DN % 4 % ACB ADB % LF[J m ( ) = C sin C = m ( ) = x UJS HCA m ( ) = a ES 5 #VOBHÌSF DPTaLBÀUS #VOBHÌSF UBOYLBÀUS \" 8 # 8 C) 15 % 7 & 24 \" m # - 1 C) - 1 % 1 & 15 17 17 25 25 2 32 1. B 2. B 3. A 98 4. D 5. & 6. A

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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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