Chapter 7 Analytic Trigonometry
Chapter 7Analytic Trigonometrycosine equals 1.cos 1, 0 0Section 7.11. Domain: x x is any real number ;Range: y 1 y 1 2. x x 1 3. 3, cos 1 1 017. sin 1 1 or x x 1 We are finding the angle , whose sine equals 1 . sin 1, 22 2 sin 1 1 24. True5. 1;3216. ; 12 ,2218. cos 1 1 7. x sin yWe are finding the angle , 0 , whosecosine equals 1 .cos 1, 0 8. 0 x 9. x 10. False. The domain of y sin 1 x is 1 x 1 .cos 1 1 11. True19. tan 1 012. TrueWe are finding the angle , 13. d , whose22tangent equals 0.14. atan 0,15. sin 1 0 0 We are finding the angle , , whose22sine equals 0. sin 0, 22 0 22 1tan 0 020. tan 1 1 We are finding the angle , tangent equals 1 .sin 1 0 0tan 1, 116. cos 1We are finding the angle , 0 , whose 4 tan 1 ( 1) 4 672Copyright 2017 Pearson Education, Inc. 22 , whose22
Section 7.1: The Inverse Sine, Cosine, and Tangent Functions21. sin 1 3 24. sin 1 2 22We are finding the angle , 2.22sin ,2 42 sin 1 24 , whose22We are finding the angle , sine equals22. tan 1sine equals 22We are finding the angle , tangent equals3,3 63 36tan tan 13,2 3 3 sin 1 32 22 3 25. cos 1 2 We are finding the angle , 0 , whose , whose223.23cos ,25 6 35 cos 1 2 63.3 3.2sin 33 , whose22cosine equals 2223. tan 1 30 2 26. sin 1 2 We are finding the angle , , whose22tangent equals 3 . tan 3, 22 3 1tan3 3We are finding the angle , sine equals 2.22,2 4 2 sin 1 4 2 sin 673Copyright 2017 Pearson Education, Inc. , whose22 22
Chapter 7: Analytic Trigonometry 3 41. tan 1 tan 8 27. sin 1 0.1 0.1028. cos 1 0.6 0.93 equation f 1 f x tan 1 tan x x . Since 129. tan 5 1.373 is in the interval , , we can apply8 2 2 the equation directly and get 3 3 .tan 1 tan 8 8 30. tan 1 0.2 0.2031. cos 17 0.51832. sin 11 0.138 3 42. sin 1 sin 733. tan 1 ( 0.4) 0.38 34. tan ( 3) 1.2538. sin 13 0.355 9 43. sin 1 sin 8 follows the form of the cannot use the formula directly since 4 directly and get cos 1 cos5 4 . 5 40. sin 1 sin follows the form of the 10 9 is not8 in the interval , . We need to find an 2 2 angle in the interval , for which 2 2 9 9 sin . The angleis in quadrant IIIsin889 isso sine is negative. The reference angle of8in the interval 0, , we can apply the equation equation f 1 f x sin 1 sin x x , but we4 39. cos 1 cos follows the form of the equation5 4 f 1 f x cos 1 cos x x . Sinceis5 3 is in the interval , , we can apply7 2 2 the equation directly and get 3 3 .sin 1 sin 77 36. cos 1 ( 0.44) 2.032 1.083 35. sin 1 ( 0.12) 0.1237. cos 1 follows the form of the equation f 1 f x sin 1 sin x x . Since 1 follows the form of the equation f 1 f x sin 1 sin x x . Sinceand we want to be in quadrant IV so sine8will still be negative. Thus, we have9 sin . Since is in the intervalsin88 8 is in the interval , , we can apply10 2 2 the equation directly and get sin 1 sin .1010 2 , 2 , we can apply the equation above and 9 1 get sin 1 sin sin sin .8 8 8 674Copyright 2017 Pearson Education, Inc.
Section 7.1: The Inverse Sine, Cosine, and Tangent Functions 7 46. cos 1 cos follows the form of the 6 11 44. sin 1 sin follows the form of the 4 equation f 1 f x sin 1 sin x x , but wecannot use the formula directly sinceangle in the interval 0, for which 7 7 cos cos . The angleis in quadrant 6 6III so the we need an angle in the desired interval7 .whose cosine is equal to the cosine of65 7 5 . SinceisThus, we have cos cos 6 66in the interval 0, , we can apply the equationabove and get 5 5 7 cos 1 cos cos 1 cos . 6 6 6 4 47. tan 1 tan 5 11 sin 1 sin .get sin 1 sin 4 4 4 follows the form of the equation f 1 f x tan 1 tan x x , but follows the form of the we cannot use the formula directly since 2 , 2 , we can apply the equation above and 7 is6not in the interval 0, . We need to find an11 is not4 in the interval , . We need to find an 2 2 angle in the interval , for which 2 2 11 11 sin sin . The angleis in quadrant II4411 so sine is positive. The reference angle ofis43 and we need to be in quadrant I so sine4will still be positive. Thus, we have3 is in the interval sin . Sincesin 444 5 45. cos 1 cos 3 equation f 1 f x cos 1 cos x x , butwe cannot use the formula directly since equation f 1 f x cos 1 cos x x , but4 is5 not in the interval , . We need to find an 2 2 angle in the interval , for which 2 2 5 is3not in the interval 0, . We need to find anwe cannot use the formula directly since 4 4 tan is in quadrant tan . The angle5 5 II so tangent is negative. The reference angle of4 isand we want to be in quadrant IV55so tangent will still be negative. Thus, we have 4 tan tan . Since is in the555 angle in the interval 0, for which5 5 is incos cos . The angle 3 3 5 quadrant I so the reference angle of is .33 5 Thus, we have cos cos . Sinceis333 interval , , we can apply the equation 2 2 in the interval 0, , we can apply the equationabove and get 5 1 cos 1 cos cos cos .33 3 675Copyright 2017 Pearson Education, Inc.
Chapter 7: Analytic Trigonometryabove and get 4 tan 1 tan 5 2 is in thewe have tan tan . Since333 interval , , we can apply the equation 2 2 1 tan tan .5 5 10 48. tan 1 tan follows the form of the 9 equation f 1 above and get tan 1 tan 2 tan 1 tan . 3 3 3 f x tan tan x x , but 1we cannot use the formula directly since 4 50. cos 1 cos follows the form of the 3 10 9 is not in the interval , . We need to find 2 2 an angle in the interval , for which 2 2 equation f 1 f x cos 1 cos x x , but10 10 is in tan . The angle tan 9 9quadrant II so tangent is negative. The reference10 isand we want to be inangle of 99quadrant IV so tangent will still be negative. 10 Thus, we have tan tan . Since 9 9 angle in the interval 0, for which4 is3not in the interval 0, . We need to find anwe cannot use the formula directly since 4 4 is in quadrantcos cos . The angle 3 34 is . We want33the angle to be in quadrant II and the cosine to be2 4 negative. Thus, we have cos cos. 3 3III so the reference angle of is in the interval , , we can apply9 2 2 the equation above and get 10 tan 1 tan .tan 1 tan 9 9 9 2 49. tan 1 tan 3 2 is in the interval 0, , we can apply3the equation above and get 2 2 4 cos 1 cos cos 1 cos . 3 3 3 Since follows the form of the 51. cos 1 cos follows the form of the 4 equation f 1 f x tan 1 tan x x . but we 2 cannot use the formula directly since is not3 in the interval , . We need to find an angle 2 2 we cannot use the formula directly since 4not in the interval 0, . We need to find anisangle in the interval 0, for which in the interval , for which2 2 equation f 1 f x cos 1 cos x x , but cos cos . The angle is in 4 4 2 2 is intan tan . The angle 3 3 quadrant III so tangent is positive. The reference2 isand we want to be inangle of 33quadrant I so tangent will still be positive. Thus,quadrant IV so the reference angle of 4is 4 Thus, we have cos cos . Sinceis 4 44676Copyright 2017 Pearson Education, Inc.
Section 7.1: The Inverse Sine, Cosine, and Tangent Functionsin the interval 0, , we can apply the equationabove and get cos 1 cos cos 1 cos . 4 4 4 3 54. tan 1 tan follows the form of the 2 equation f3 is . Thus, we have22 3 tan tan . In this case, tan is 2 2 2 f x sin sin x x , but we3 is not4 undefined so tan 1 tan would also be 2 undefined. in the interval , . We need to find an 2 2 angle in the interval , for which 2 2 3 sin sin . The reference angle of 4 1 55. sin sin 1 follows the form of the equation4 1is inf f 1 x sin sin 1 x x . Since4the interval 1,1 , we can apply the equation 3 isand we want to be in quadrant IV44so sine will still be negative. Thus, we have 3 sin sin . Since is in the 4 4 4 2 56. cos cos 1 follows the form of the 3 equation f f 1 x cos cos 1 x x .2is in the interval 1,1 , we can3apply the equation directly and get 2 2 cos cos 1 .33 Since 53. tan 1 tan follows the form of the 2 1 1 directly and get sin sin 1 .4 4 interval , , we can apply the equation 2 2 above and get 3 sin 1 sin sin 1 sin .4 4 4 angle of 1cannot use the formula directly since need to find an angle in the interval , 2 2 3 for which tan tan . The reference 2 3 52. sin 1 sin follows the form of the 4 1 equation f 1 f x tan 1 tan x x . We equation f 1 f x tan 1 tan x x . We need to find an angle in the interval , 2 2 for which tan tan . In this case, 2 57. tan tan 1 4 follows the form of the equation f f 1 x tan tan 1 x x . Since 4 is areal number, we can apply the equation directly tan is undefined so tan 1 tan would 2 2 also be undefined. and get tan tan 1 4 4 .677Copyright 2017 Pearson Education, Inc.
Chapter 7: Analytic Trigonometry 58. tan tan 1 2 follows the form of the equation x 2 15 5 x 2 5 3 x 7The domain of f 1 x is x 3 x 7 , or 1 f f 1 x tan tan 1 x x . Since 2 is areal number, we can apply the equation directly and get tan tan 1 2 2 . 3, 7 in interval notation. Recall that thedomain of a function equals the range of itsinverse and the range of a function equals thedomain of its inverse. Thus, the range of f isalso 3, 7 .59. Since there is no angle such that cos 1.2 ,the quantity cos 1 1.2 is not defined. Thus, cos cos 1 1.2 is not defined.60. Since there is no angle such that sin 2 ,the quantity sin 1 2 is not defined. Thus, 64.y 2 tan x 3x 2 tan y 32 tan y x 3x 3tan y 2x 3y tan 1 f 1 x 2The domain of f x equals the range of f 1 ( x) sin sin 1 2 is not defined. 61. tan tan 1 follows the form of the equation f f 1 x tan tan 1 x x . Since is areal number, we can apply the equation directly and get tan tan 1 .62. Since there is no angle such that sin 1.5 ,the quantity sin 1 1.5 is not defined. Thus, and is y 5sin x 25sin y x 2x 2sin y 565.x 2 f 1 x y sin5The domain of f x equals the range of 12 or , in interval2 2 2 interval notation. Recall that the domain of afunction equals the range of its inverse and therange of a function equals the domain of itsinverse. Thus, the range of f is , .x 5sin y 2 2 x that the argument of the inverse tangent functioncan be any real number. Thus, the domain off 1 x is all real numbers, or , inf x 5sin x 2f 1 ( x ) and is notation. To find the domain of f 1 x we notesin sin 1 1.5 is not defined.63.f x 2 tan x 3 x f x 2 cos 3 x y 2 cos 3 x x 2 cos 3 y or , in2 2 2 cos 3 y interval notation. To find the domain of f 1 x x2 x 3 y cos 1 2 1 x y cos 1 f 1 x 3 2 we note that the argument of the inverse sinex 2and that it must lie in thefunction is5The domain of f x equals the range ofinterval 1,1 . That is,f 1 ( x ) and is 0 x 678Copyright 2017 Pearson Education, Inc. , or 0, in interval3 3
Section 7.1: The Inverse Sine, Cosine, and Tangent Functionsnotation. To find the domain of f 1 x we note67.y tan x 1 3that the argument of the inverse cosine function xand that it must lie in the interval 1,1 .is2That is,x 1 122 x 2 2 x 2The domain of f 1 x is x 2 x 2 , orx tan y 1 3tan y 1 x 3y 1 tan 1 x 3 y 1 tan 1 x 3 1 tan 1 x 3 f 1 x (note here we used the fact that y tan 1 x is anodd function).The domain of f x equals the range of 2, 2 in interval notation. Recall that thedomain of a function equals the range of itsinverse and the range of a function equals thedomain of its inverse. Thus, the range of f is 2, 2 .66.f 1 ( x) and is 1 2 x 2 1 , ordomain of f 1 x we note that the argument ofy 3sin 2 x x 3sin 2 y the inverse tangent function can be any realnumber. Thus, the domain of f 1 x is all realx32 y sin 1 1 2 , 2 1 in interval notation. To find the f x 3sin 2 x sin 2 y f x tan x 1 3numbers, or , in interval notation. Recallx3that the domain of a function equals the range ofits inverse and the range of a function equals thedomain of its inverse. Thus, the range of f is , .1xy sin 1 f 1 x 23The domain of f x equals the range off 1 ( x) and is 4 x 468. , or , in 4 4 f x cos x 2 1y cos x 2 1x cos y 2 1interval notation. To find the domain of f 1 x cos y 2 x 1we note that the argument of the inverse sinexand that it must lie in the intervalfunction is3y 2 cos 1 x 1 y cos 1 x 1 2 1,1 . That is,x 1 13 3 x 3The domain of f 1 x is x 3 x 3 , orThe domain of f x equals the range off 1 ( x) and is 2 x 2 , or 2, 2 ininterval notation. To find the domain of f 1 x we note that the argument of the inverse cosinefunction is x 1 and that it must lie in the 3,3 in interval notation. Recall that thedomain of a function equals the range of itsinverse and the range of a function equals thedomain of its inverse. Thus, the range of f is 3,3 .interval 1,1 . That is, 1 x 1 10 x 2 1The domain of f x is x 0 x 2 , or 0, 2 in interval notation. Recall that the679Copyright 2017 Pearson Education, Inc.
Chapter 7: Analytic Trigonometrydomain of a function equals the range of itsinverse and the range of a function equals thedomain of its inverse. Thus, the range of f is 0, 2 .69.70.f x 2 cos 3 x 2 y 2 cos 3x 2 x 2 cos 3 y 2 f x 3sin 2 x 1 cos 3 y 2 y 3sin 2 x 1 x 3 y 2 cos 1 2 x 3 y cos 1 2 2 x 3sin 2 y 1 sin 2 y 1 x3x3 x 2 y sin 1 1 3 2 y 1 sin 11 x 2y cos 1 f 1 x 3 2 3The domain of f x equals the range of1 x 1y sin 1 f 1 x 2 3 2f 1 ( x) and is 22 x , or33 3 2 2 3 , 3 3 in interval notation. To find the The domain of f x equals the range of1 1 f 1 ( x) and is x , or2 42 4 1 1 2 4 , 2 4 in interval notation. To find domain of f 1 x we note that the argument ofxand that it must2lie in the interval 1,1 . That is,the inverse cosine function isthe domain of f 1 x we note that the argumentof the inverse sine function isx2x 12 2 x 2The domain of f 1 x is x 2 x 2 , or 1 xand that it must3lie in the interval 1,1 . That is,x 1 13 3 x 3The domain of f 1 x is x 3 x 3 , or 2, 2 in interval notation. Recall that thedomain of a function equals the range of itsinverse and the range of a function equals thedomain of its inverse. Thus, the range of f is 2, 2 . 3,3 in interval notation. Recall that thedomain of a function equals the range of itsinverse and the range of a function equals thedomain of its inverse. Thus, the range of f is 3,3 .71. 4sin 1 x sin 1 x 4 22 2 The solution set is . 2 x sin680Copyright 2017 Pearson Education, Inc.4
Section 7.1: The Inverse Sine, Cosine, and Tangent Functions77. 4 cos 1 x 2 2 cos 1 x72. 2 cos 1 x cos 1 x 2 cos 1 x 2 02x cos2 cos 1 x 2 02The solution set is {0} .cos 1 x x cos 1The solution set is { 1} .73. 3cos 1 2 x 2 78. 5sin 1 x 2 2sin 1 x 3 2 cos 2 x 3 13sin 1 x 2 x cos2 3sin 1 x 1 The solution set is . 4 79. Note that 29 45 29.75 .74. 6sin 1 3x a. 6 3 x sin 6 13x 21x 6 1 The solution set is . 6 b.c.75. 3 tan x a. 3x tan 3 3The solution set isb. 3 .76. 4 tan 1 x tan 1 x cos 1 tan 23.5 180 tan 29.75 180 D 24 1 13.92 hours or 13 hours, 55 minutes cos 1 tan 0 180 tan 29.75 180 D 24 1 12 hours cos 1 tan 22.8 180 tan 29.75 180 D 24 1 13.85 hours or 13 hours, 51 minutes80. Note that 40 45 40.75 . 1tan 1 x 33 x sin 2 3 3 The solution set is . 2 121x 42x sin 1 3x c. 4 x tan 1 4 The solution set is { 1} . cos 1 tan 23.5 180 tan 40.75 180 D 24 1 14.93 hours or 14 hours, 56 minutes cos 1 tan 0 180 tan 40.75 180 D 24 1 12 hours cos 1 tan 22.8 180 tan 40.75 180 D 24 1 14.83 hours or 14 hours, 50 minutes681Copyright 2017 Pearson Education, Inc.
Chapter 7: Analytic Trigonometry81. Note that 21 18 21.3 .b. cos tan 23.5 180 tan 21.3 180 D 24 1 13.30 hours or 13 hours, 18 minutes 1a.b.c. cos 1 tan 0 180 tan 21.3 180 D 24 1 12 hoursc. cos 1 tan 22.8 180 tan 21.3 180 D 24 1 13.26 hours or 13 hours, 15 minutesb.c.83. a.b.c. cos 1 tan 23.5 180 tan 61.167 180 D 24 1 18.96 hours or 18 hours, 57 minutes85. Let point C represent the point on the Earth’saxis at the same latitude as Cadillac Mountain,and arrange the figure so that segment CQ liesalong the x-axis (see figure). cos 1 tan 0 180 tan 61.167 180 D 24 1 12 hoursyP cos 1 tan 22.8 180 tan 61.167 180 D 24 1 18.64 hours or 18 hours, 38 minutesC cos 1 tan 23.5 180 tan 0 180 D 24 1 12 hours cosD 24 1 12 hours 1180 180 2710 mix2710
Analytic Trigonometry Section 7.1 1. Domain: xx is any real number ; Range: yy 11 2. xx 1 or xx 1 3. 3, 4. True 5. 1; 3 2 6. 1 2 ; 1 7. x sin y 8. 0 x 9. x 10. False. The domain of yx sin 1 is 11x. 11. True 12. T
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
COMPLEX ANALYTIC GEOMETRY AND ANALYTIC-GEOMETRIC CATEGORIES YA’ACOV PETERZIL AND SERGEI STARCHENKO Abstract. The notion of a analytic-geometric category was introduced by v.d. Dries and Miller in [4]. It is a category of subsets of real analytic manifolds which extends the c
