CHAPTER 5 Analytic Trigonometry - KHSPreCalc
C H A P T E R 5Analytic TrigonometrySection 5.1Using Fundamental Identities .397Section 5.2Verifying Trigonometric Identities .404Section 5.3Solving Trigonometric Equations .410Section 5.4Sum and Difference Formulas .425Section 5.5Multiple-Angle and Product-to-Sum Formulas .441Review Exercises .453Problem Solving .464Practice Test .471 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
C H A P T E R 5Analytic TrigonometrySection 5.1 Using Fundamental Identities1. tan u39. sin θ , cos θ 0 θ is in Quadrant IV.42. csc ucos θ 3. cot u4. csc u6. sin u57. sec x , tan x 0 x is in Quadrant II.2sin x 112 5sec x5 2 2 1 5 2 1 4 2521510. cos θ 21sin x215 tan x 2cos x2 51 sin x55 21 2121cot x 122 21 tan x212178. csc x , tan x 0 x is in Quadrant III.6sin x 116 7csc x7 6 6 cos x 1 7 sin x tan x cos xsec x 1 cos xcot x 1 tan x2 1 67 13 7 3613 497 1 9 1674 2, sin θ 0 θ is in Quadrant IV.3 2 sin θ 1 3 csc x 234 3 3 777741144 7sec θ cos θ7774117 cot θ 3tan θ3 7114 csc θ 3sin θ3 4sin θ tan θ cos θ5. 1cos x 3 1 4 2 1 45 9352 5223113 sec θ 2cos θ231122 5cot θ tan θ555 21133 5csc θ sin θ555 3sin θtan θ cos θ 66 13 1313177 13 131313 71 613136 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 397
398Chapter 5Analytic Trigonometry2, cos x 0 x is in Quadrant I.3113 cot x 2tan x2311. tan x sec x 2 1 3 2 1 4 91331 9 41322 3 1 2 11sin x csc x13211cos x sec x13311cot x 2tan x3csc x 33 13 1313327, sin x 0 x is in Quadrant III.4114tan x 7cot x7421665 4 sec x 1 1 497 7 12. cot x 24965 7 csc x 1 1 164 4 1144 65sin x csc x656565 41177 65cos x sec x656565 716. cot x sec x Matches (b).() 1 cos x 2 cos x 1 cos x sec xcos x11 csc xsin x cos xsin xMatches (a).17.sec2 x 1tan 2 xsin 2 x1 sec2 xsin 2 xsin 2 xcos 2 x sin 2 xMatches (e).18.cos 2 (π 2) x sin 2 xsin x sin x tan x sin xcos xcos xcos xMatches (d). 1 tan θ tan θ cot θ tan θ 19. 1sec θcos θ1 1cos θ cos θ π 20. cos x sec x sin x sec x 2 1 sin x cos x tan x21. tan 2 x tan 2 x sin 2 x tan 2 x(1 sin 2 x) tan 2 x cos 2 xsin 2 x cos 2 xcos 2 x sin 2 x 1 13. sec x cos x cos x cos x 1Matches (c).14. cot 2 x csc 2 x (csc 2 x 1) csc2 x 1)Matches (f).22 13 1313 (15. cos x 1 tan 2 x cos x sec2 x()22. sin 2 x sec 2 x sin 2 x sin 2 x sec 2 x 1 sin 2 x tan 2 x23.24.sec 2 x 1(sec x 1)(sec x 1) sec x 1sec x 1 sec x 1cos x 2cos x 2 cos 2 x 4(cos x 2)(cos x 2)1 cos x 2 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 5.125. 1 2 cos 2 x cos 4 x (1 cos 2 x) (sin 2 x)2Using Fundamental Identities39926. sec 4 x tan 4 x (sec 2 x tan 2 x)(sec 2 x tan 2 x) (sec 2 x tan 2 x)(1)2 sec 2 x tan 2 x sin 4 x27. cot 3 x cot 2 x cot x 1 cot 2 x(cot x 1) (cot x 1) (cot x 1)(cot 2 x 1) (cot x 1)csc 2 x28. sec3 x sec 2 x sec x 1 sec2 x(sec x 1) (sec x 1) (sec 2 x 1)(sec x 1) tan 2 x(sec x 1)29. 3 sin 2 x 5 sin x 2 (3 sin x 1)(sin x 2)cos usin u(sin u ) (cos u)sin ucos u cos u sin u38. cot u sin u tan u cos u 30. 6 cos 2 x 5 cos x 6 (3 cos x 2)( 2 cos x 3)31. cot 2 x csc x 1 (csc 2 x 1) csc x 139. csc2 x csc x 2 (csc x 1)(csc x 2)32. sin x 3 cos x 3 (1 cos x) 3 cos x 32240. cos 2 x 3 cos x 4 (cos 2 x 3 cos x 4) (cos x 1)(cos x 4)33. tan θ csc θ sin θ11 sec θcos θ sin θcos θ34. tan ( x ) cos x tan x cos xsin x cos xcos x sin x1 sin 2 φsin φ 1 sin 2 φ cos 2 φ35. sin φ (csc φ sin φ ) (sin φ )sin β cos βcos βsin 2 βcos 2 β cos βcos βsin 2 β cos 2 βcos β1cos βsec β cos 2 y1 sin 2 y 1 sin y1 sin y(1 sin y )(1 sin y ) 1 sin y 1 sin y41. (sin x cos x) sin 2 x 2 sin x cos x cos 2 x2 (sin 2 x cos 2 x) 2 sin x cos x 1 2 sin x cos x42. ( 2 csc x 2)( 2 csc x 2) 4 csc 2 x 4 4(csc 2 x 1) 4 cot 2 x43. 1 36. cos x(sec x cos x ) cos x cos x cos x 1 cos 2 x sin 2 x37. sin β tan β cos β (sin β )1 sin 2 xcos 2 xsin 2 x222cosxtanxcosx ()csc 2 x 1cot 2 xcos 2 x sin 2 x44.111 cos x 1 cos x 1 cos x 1 cos x(1 cos x)(1 cos x)2 1 cos 2 x2 sin 2 x 2 csc 2 xsec x 1 (sec x 1)11 sec x 1 sec x 1(sec x 1)(sec x 1)sec x 1 sec x 1 sec 2 x 1 2 tan 2 x 1 2 2 tan x 2 cot 2 x 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
40045.Chapter 5Analytic Trigonometrycos x(1 sin x) cos x(1 sin x)cos xcos x 1 sin x 1 sin x(1 sin x)(1 sin x)cos x sin x cos x cos x sin x cos x(1 sin x)(1 sin x) 2 sin x cos x1 sin 2 x 2 sin x cos x cos 2 x 2 sin x cos x 2 tan x 46.sin x(1 cos x) sin x(1 cos x)sin xsin x 1 cos x 1 cos x(1 cos x)(1 cos x) sin x sin x cos x sin x sin x cos x(1 cos x)(1 cos x) 2 sin x1 cos 2 x 2 sin xsin 2 x 2sin x 2 csc x47. tan x sec 2 xtan 2 x sec 2 x tan xtan x 48.50.5( tan x sec x )5tan x sec x tan x sec x tan x sec xtan 2 x sec 2 x 1 cot xtan x cos 2 x (1 sin x)cos x1 sin x 1 sin xcos xcos x(1 sin x) 1 5(sec x tan x )2251. y1 2 cos x 1 2 sin x sin xcos x(1 sin x ) 2 2 sin xcos x(1 sin x) 2(1 sin x)sin x cos xcos x sin x cos xsin x cos xcos x(1 sin x)2cos x sin x cos x sin x cos x cos x sin x cos x sin xsin 2 y1 cos 2 y 1 cos y1 cos y (1 cos y)(1 cos y)1 cos ytan x 1sec x csc xsin x 1cos x 11 cos xsin x 2 sec x49.5( tan x sec x )2 1 cos y 2π2π 2 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 5.152. y1 Using Fundamental Identities 1 1 cos x tan x sin x cos x 5 1 11cos x cos x sin x cos xsinxcosxsin x 2π2π1 cos 2 xsin 2 xsin x tan xsin x cos xsin x cos xcos x53. Let x 3 cos θ .9 x2 401 555. Let x 2 sec θ .9 (3 cos θ )2( 2 sec θ )x2 4 2 4 9 9 cos θ 9(1 cos 2 θ )4(sec 2 θ 1) 4 tan 2 θ 9 sin 2 θ 3 sin θ 2 tan θ256. Let 3x 5 tan θ .54. Let x 7 sin θ .9 x 2 25 (3 x )2 2549 x 2 49 (7 sin θ ) 49 49 sin 2 θ (5 tan θ ) 49(1 sin 2 θ ) 25 tan 2 θ 25 49 cos 2 θ 25( tan 2 θ 1) 25 sec 2 θ2 7 cos θ2 25 5 sec θ57. Let x 2 sin θ .4 x2 24 ( 2 sin θ ) 24 4 sin 2 θ 24(1 sin 2 θ ) 24 cos 2 θ 22 cos θ 2cos θ 22sin θ 1 cos 2 θ 22 2 21 22 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
402Chapter 5Analytic Trigonometryx 10 cos θ58. tan t63. ln tan t ln (1 cos 2 t ) ln 2 1 cos5 3 100 x 25 3 100 (10 cos θ )5 3 100(1 cos 2 θ )5 3 100 sin 2 θ25 3 10 sin θsin θ cos θ 59. sin θ 5 3 1032 3 1 2 2 12()(1 cos x , 0 x 2π .y22π66. sec x tan x sin x y1 21 tan 2 θ1and y2 cos xπ21 tan 2 x , 0 x 2π .and()π2 t sin x sin xcos 2 x sin x sin x cos 2 xcos 2 x and1sin x sin xcos x cos x 3π x 2π .21 tan 2 θ for 0 θ 3π θ 2π .21cos t sin t65. μ W cos θ W sin θW sin θμ tan θW cos θ2So, sec θ ln 1 ln cos 2 t 2cos ln (1) 021 cos2 θ for 0 θ π .y1 y2 for 0 x sin t1 cos t sin 2 t ln cos 2 t sec 2 t y1 y2 for 0 x π .Let y1 ln)1 cos θLet y1 sin x and y2 60. sec θ tan tsin 2 t64. ln cos 2 t ln 1 tan 2 t ln cos 2 t 1 tan 2 t 20 ln ln sec t csc t1 sin 2 θ So, sin θ t sin x(1 cos 2 x)cos 2 xsin x sin 2 xcos 2 x sin x tan 2 x467. True.y22π0y1 461. ln sin x ln cot x ln sin x cot xcos x ln sin x sin x ln cos x62. ln cos x ln sin x lncos xsin x ln cot xtan u sin ucos ucot u cos usin usec u 1cos ucsc u 1sin u68. False. A cofunction identity can be used to transform atangent function so that it can be represented by acotangent function.69. As x π 2, tan x and cot x 0. 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 5.170. As x π , sin x 0 and csc x Using Fundamental Identities40372. Let u a tan θ , then1 .sin x71. cos( θ ) cos θa2 u 2 a 2 ( a tan θ ) a 2 a 2 tan 2 θ a 2 1 tan 2 θ a 2 sec 2 θcos( θ ) cos θsin θsin θThe correct identity is cos( θ )cos θ tan θ(2) a sec θ .73. Because sin 2 θ cos 2 θ 1, then cos2 θ 1 sin 2 θ .cos θ 1 sin θtan θ sin θsin θ cos θ 1 sin 2 θcot θ cos θ 1 sin 2 θ sin θsin θsec θ 11 cos θ 1 sin 2 θcsc θ 1sin θa74. To derive sin 2 θ cos 2 θ 1, let sin θ So, sin 2 θ cos 2 θ 2a b2 22a b a2and cos θ b2a b2 .2 a2b2 2 2222a ba b2a b ba2 b2a2 b2 1. To derive 1 tan 2 θ sec 2 θ , let tan θ aand sec θ ba 2 b2.b2a2b2 a 2 a So, 1 tan 2 θ 1 1 2 bb2 b 2 a 2 b2 2 b a 2 b2 b 2 sec 2 θ .To derive 1 cot 2 θ csc 2 θ , let cot θ band csc θ aa 2 b2.a2b2 b So, 1 cot 2 θ 1 1 2a a a 2 b2 2 a a2 b2a2 22a2 b2 csc 2 θ . a Answers will vary. 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
40475.Chapter 5Analytic Trigonometrysec θ (1 tan θ )sec θ csc θsin θ 1 1 cos θ cos θ 11 cos θsin θcos θ sin θcos 2 θ sin θ cos θsin θ cos θ sin θ cos θ sin θ cos θ cos 2 θ sin θ cos θ sin θcos θSection 5.2 Verifying Trigonometric Identities1. identity16.2. conditional equation3. tan u 1 π 17. sin t csc t sin t sec t sin t 2 cos t sin t tan tcos t4. cot u5. sin u6. cot 2 u π 18. sec 2 y cot 2 y sec2 y tan 2 y 12 7. csc u8. sec u9. tan t cot t 10.cos (π 2) x sin x tan xcos xsin (π 2) x 19.sin t cos t 1cos t sin tcot x tan x1 tan x cot x tan x cot x1 sec xcos xcos x11. (1 sin α )(1 sin α ) 1 sin α cos α2(12. cos 2 β sin 2 β cos 2 β 1 cos 2 β220.)13. cos 2 β sin 2 β 1 sin 2 β sin 2 β 1 2 sin 2 β14. sin 2 α sin 4 α sin 2 α (1 sin 2 α ) (1 cos 2 α )(cos 2 α ) cos α cos α24 π 15. tan θ tan θ cot θ tan θ 2 1 tan θ tan θ 111csc x sin x sin x csc xsin x csc xcsc x sin x1 csc x sin x ) 2 cos 2 β 1(11cot x tan x tan xcot xtan x cot x(1 sin θ ) cos2 θ1 sin θcos θ cos θ1 sin θcos θ (1 sin θ )221. 1 2 sin θ sin 2 θ cos 2 θcos θ (1 sin θ ) 2 2 sin θcos θ (1 sin θ ) 2(1 sin θ )cos θ (1 sin θ )2cos θ 2 sec θ 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 5.222.cos θ cot θ (1 sin θ )cos θ cot θ 1 1 sin θ1 sin θ cos θ cos θ 1 sin θsin θ sin θ 1 sin θsin θcos 2 θ sin θ sin 2 θ sin θ (1 sin θ ) 1 sin θsin θ (1 sin θ ) 1sin θ 125. sec y cos y cos(27.(sin θ cos θ ) tan θ sin θ tan θtan 2 θ sec θ1 cos θ28.cot 3 tcot t cot 2 t csc tcsc t 2 cos x sin 2 x 2 csc tcos t sin t(csc2 t 1)sin t cos t (csc 2 t 1)1cos x sin x sin x 2 csc x cot x24. cos x cot t (csc 2 t 1)cos t(csc2 t 1)sin t 1sin t11cos x 1 cos x 1 cos x 1 cos x 1(cos x 1)(cos x 1)2 cos xcos 2 x 129.11 tan 2 β tan β tan βtan β cos x(1 tan x) cos xcos x 1 tan x1 tan x cos y 1y ) 40526. cot 2 y sec2 y 1 cot 2 y tan 2 y 1 csc θ23.Verifying Trigonometric Identitiessec 2 βtan β cos x tan x1 tan x cos x(sin x cos x)1 (sin x cos x) cos xcos x sin x cos x cos x sin x sin x cos xsin x cos x30.sec θ (sec θ 1)sec θ 1sec θ 1sec θ sec θ1 cos θ1 (1 sec θ ) sec θsec θ 131.cot 2 tcos 2 t sin 2 tcos2 t1 sin 2 t csc t1 sin tsin tsin t sin x 32. cos x sin x tan x cos x sin x cos x 22cos x sin xcos x1 cos x sec x33. sec x cos x 1 cos xcos x 1 cos 2 xcos x sin 2 xcos x sin x sin xcos x sin x tan x 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
406Chapter 5Analytic Trigonometry34. cot x tan x 35.36.cos xsin x sin xcos xcos 2 x sin 2 xsin x cos x1 sin 2 x sin 2 xsin x cos x1 2 sin 2 xsin x cos x1 1 2 sin 2 x cos x sin x 1 12 sin 2 x cos x sin xsin x sec x(csc x 2 sin x )cot xcos x sin xcos2 x1 sin 2 x1sin 2 x csc x sin xsec x1 cos xsin xsin xsin xsin xcsc( x )1 sin ( x ) sec( x )1 cos( x)cos( x ) sin ( x)cos x sin x cot x()37. sin1 2 x cos x sin 5 2 x cos x sin1 2 x cos x 1 sin 2 x sin1 2 x cos x cos 2 x cos3 x sin x()38. sec6 x(sec x tan x) sec 4 x(sec x tan x) sec 4 x(sec x tan x) sec 2 x 1 sec 4 x(sec x tan x) tan 2 x sec5 x tan 3 x39. (1 sin y ) 1 sin ( y ) (1 sin y )(1 sin y )1 sin θ 1 sin θ41. 1 sin 2 y cos 2 y 11 tan x tan ycot x cot ycot xcot y 40.111 tan x tan ycot x cot y 1 cot x cot ycot y cot x cot x cot y 142. 1 sin θ 1 sin θ 1 sin θ 1 sin θ(1 sin θ )2(1 sin θ )21 sin 2 θcos θ21 sin θcos θ(cos x cos y )(cos x cos y ) (sin x sin y )(sin x sin y )cos x cos ysin x sin y sin x sin ycos x cos y(sin x sin y )(cos x cos y )cos 2 x cos 2 y sin 2 x sin 2 y (sin x sin y )(cos x cos y ) (cos2 x sin 2 x) (cos2 y sin 2 y) 0(sin x43. cot ( x ) cot xThe correct substitution is cot ( x ) cot x.1 cot ( x ) cot x cot x 0tan x sin y )(cos x cos y )44. The first line claims that sec( θ ) sec θ andsin ( θ ) sin θ . The correct substitutions aresec( θ ) sec θ and sin ( θ ) sin θ . 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 5.245. (a)Verifying Trigonometric Identities407(b)3 2π2πIdentity 1Identity(c) (1 cot 2 x)(cos 2 x ) csc 2 x cos 2 x 346. (a)1 cos 2 x cot 2 xsin 2 x(b) 2π2π 1IdentityIdentitysin x cos xcos x cot x csc2 x csc x sin x 1 cot xsin xsin x(c) csc x(csc x sin x) csc2 x 1 1 cot x cot x csc2 x547. (a)(b)y2y1 2π2π 1Not an identityNot an identity(c) 2 cos 2 x 3 cos 4 x 1 cos 2 x 2 3 cos 2 x sin 2 x 2 3 cos 2 x sin 2 x 3 2 cos 2 x(48. (a)5 π)()()()(b)y1 y2πNot an identity 5Not an identity(c) tan 4 x tan 2 x 3 sin 4 xsin 2 x 3cos 4 xcos 2 x sin 4 x sin 2 x 3 x cos 2 x 1 sin 4 x sin 2 x cos 2 x 3cos 2 x cos 2 x 1cos 2 sin 2 x cos 21 sin 2 cos 2 x cos 2 1cos 2x 22 (sin x cos x) 3x x 1 3x sec 2 x tan 2 x 3 sec 2 x( 4 tan 2 x 3) 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
408Chapter 549. (a)Analytic Trigonometry50. (a)33y1 2π 22π2y2 3 5IdentityNot an identity(b)(b)Identity(c)Not an identity(1 cos x)(1 cos x)1 cos x sin xsin x(1 cos x) 1 cos 2 xsin x(1 cos x) sin 2 xsin x(1 cos x) sin x1 cos x(c)cot αcsc α 1is the reciprocal of.csc α 1cot αThey will only be equivalent at isolated points intheir respective domains. So, not an identity.51. tan 3 x sec 2 x tan 3 x tan 3 x(sec 2 x 1) tan 3 x tan 2 x tan 5 x sin 2 xsin 4 x 1 52. ( tan 2 x tan 4 x) sec2 x 2cos 4 x cos 2 x cos x 1cos 4 2sin 4 x sin x cos 2 x x 1cos 4 sin 2 x cos 2 x sin 4 x cos 2 xx 1cos 4 sin 2 x(cos 2 x sin 2 x) cos 2 xx 1 sin 2 x 1 sec 4 x tan 2 x cos 4 x cos 2 x 53. (sin 2 x sin 4 x ) cos x sin 2 x(1 sin 2 x) cos x54. sin 4 x cos 4 x sin 2 x sin 2 x cos 4 x sin 2 x cos 2 x cos x (1 cos 2 x)(1 cos 2 x) cos 4 x sin 2 x cos3 x 1 2 cos 2 x cos 4 x cos 4 x 1 2 cos 2 x 2 cos 4 x55. sin 2 25 sin 2 65 sin 2 25 cos 2 (90 65 ) sin 2 25 cos 2 25 156. tan 2 63 cot 2 16 sec 2 74 csc 2 27 tan 2 63 cot 2 16 csc2 (90 74 ) sec2 (90 27 ) tan 2 63 cot 2 16 csc 2 16 sec2 63 ( tan 2 63 sec 2 63 ) (cot 2 16 csc 2 16 ) 1 ( 1) 2 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 5.2x.157. Let θ sin 1 x sin θ x 1Verifying Trigonometric Equations60. Let θ cos 1x 1x 1 cos θ .222xθ4094 (x 1) 2θx 11 x2From the diagram,From the diagram,tan (sin 1 x) tan θ x1 x258. Let θ sin 1 x sin θ x 14 ( x 1)x 1 tan cos 1 tan θ 2 .x.161. cos x csc x cot x cos x 2x 1.1 cos xsin x sin x 1 cos x 1 sin 2 x x cos x(1 csc 2 x) cos x(csc 2 x 1)θ cos x cot 2 x1 x2From the diagram,cos(sin 1x) cos θ 59. Let θ sin 11 x2 121 x .62. (a)(b)x 1x 1 sin θ .44h sin (90 θ )sin θ h cos θ h cot θsin θθ15 30 45 60 75 90 s18.668.6652.891.340(c) Maximum: 15 4Minimum: 90 x 1(d) Noon63. False. tan x 2 tan ( x x) andθtan 2 x ( tan x)( tan x), tan x 2 tan 2 x.16 (x 1) 2From the diagram,x 1 tan sin 1 tan θ 4 64. True. Cosine is an even function,x 116 ( x 1)2. ππ cos θ cos θ 2 2 π cos θ 2 sin θ .65. False. For the equation to be an identity, it must be truefor all values of θ in the domain. 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
410Chapter 566. If sin θ Analytic Trigonometryac, sec θ , andcb68. tan θ sec2 θ 1True identity: tan θ sec 2 θ 1a 2 b 2 c 2 a 2 c 2 b 2 , then2sec2 θ 1 is not true for π 2 θ πtan θ c 1sec θ 1b 22sec θ c b 2or 3π 2 θ 2π . So, the equation is not true forθ 3π 4.1 cos θ sin θ69.c2 12 b 2cb2(1 cos θ )2 (sin θ )21 2 cos θ cos 2 θ sin 2 θ1 2 cos θ cos 2 θ 1 cos 2 θc2 b2b2 c2b22 cos 2 θ 2 cos θ 02 cos θ (cos θ 1) 0 c2 b2 b2 2b2cThe equation is not an identity because it is only truewhen cos θ 0 or cos θ 1. So, one angle for which c2 b2c2the equation is not true is .2 a2c2 a c π1 tan θ sec θ70.(1 2tan θ ) (sec θ )21 2 tan θ tan 2 θ sec2 θ1 2 tan θ tan 2 θ 1 tan 2 θ sin 2 θ .2 tan θ 0tan θ 067. Because sin θ 1 cos θ , then222sin θ 1 cos 2 θ ; sin θ This equation is not an identity because it is only truewhen tan θ 0. So, one angle for which the equation1 cos 2 θ if θlies in Quadrant III or IV.7π.One such angle is θ 4is not true isπ6.Section 5.3 Solving Trigonometric Equations6. sec x 2 01. isolate2. general(a) x 3. quadraticsec(a) x tanπ1 2cos(π 3) 2 1 2 2 2 0125π35π1 2 2sec3cos(5π 3)(b) x 333 3 0π3π4. extraneous5. tan
Analytic Trigonometry Section 5.1 Using Fundamental Identities 1. tan u 2. csc u 3. cot u 4. csc u 5. 1 6. sin u 7. 5 sec , tan 0 2 x xx is in Quadrant II. 2 11 2 cos sec 55 2 2421 sin 1 1
Part One: Heir of Ash Chapter 1 Chapter 2 Chapter 3 Chapter 4 Chapter 5 Chapter 6 Chapter 7 Chapter 8 Chapter 9 Chapter 10 Chapter 11 Chapter 12 Chapter 13 Chapter 14 Chapter 15 Chapter 16 Chapter 17 Chapter 18 Chapter 19 Chapter 20 Chapter 21 Chapter 22 Chapter 23 Chapter 24 Chapter 25 Chapter 26 Chapter 27 Chapter 28 Chapter 29 Chapter 30 .
(Answers for Chapter 5: Analytic Trigonometry) A.5.1 CHAPTER 5: Analytic Trigonometry SECTION 5.1: FUNDAMENTAL TRIGONOMETRIC IDENTITIES 1) Left Side Right Side Type of Identity (ID) csc(x) 1 sin(x) Reciprocal ID tan(x) 1
TO KILL A MOCKINGBIRD. Contents Dedication Epigraph Part One Chapter 1 Chapter 2 Chapter 3 Chapter 4 Chapter 5 Chapter 6 Chapter 7 Chapter 8 Chapter 9 Chapter 10 Chapter 11 Part Two Chapter 12 Chapter 13 Chapter 14 Chapter 15 Chapter 16 Chapter 17 Chapter 18. Chapter 19 Chapter 20 Chapter 21 Chapter 22 Chapter 23 Chapter 24 Chapter 25 Chapter 26
DEDICATION PART ONE Chapter 1 Chapter 2 Chapter 3 Chapter 4 Chapter 5 Chapter 6 Chapter 7 Chapter 8 Chapter 9 Chapter 10 Chapter 11 PART TWO Chapter 12 Chapter 13 Chapter 14 Chapter 15 Chapter 16 Chapter 17 Chapter 18 Chapter 19 Chapter 20 Chapter 21 Chapter 22 Chapter 23 .
CHAPTER 1 1 Angles and Applications 1.1 Introduction Trigonometry is the branch of mathematics concerned with the measurement of the parts, sides, and angles of a triangle. Plane trigonometry, which is the topic of this book, is restricted to triangles lying in a plane. Trigonometry is based on certain ratio
Trigonometry Standard 12.0 Students use trigonometry to determine unknown sides or angles in right triangles. Trigonometry Standard 5.0 Students know the definitions of the tangent and cotangent functions and can graph them. Trigonometry Standard 6.0 Students know the definitions of the secant and cosecant functions and can graph them.
Analytic Continuation of Functions. 2 We define analytic continuation as the process of continuing a function off of the real axis and into the complex plane such that the resulting function is analytic. More generally, analytic continuation extends the representation of a function
FSA ELA Reading Practice Test Questions Now answer Numbers 1 through 6. Base your answers on the passages “from The Metamorphoses” and “from Romeo and Juliet.” 1. Fill in a circle before two phrases Ovid uses in Passage 1 to show that Pyramus and Thisbe experience a shared love. “A A thing which they could not forbid, B they were both inflamed, with minds equally captivated. C There .
