College Trigonometry, 2011a

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860 Foundations of <strong>Trigonometry</strong> a reciprocal identity to rewrite the secant as cosine. The x-coordinates of the intersection 1 points of y = and y = 4 verify our answers. (cos(x)) 2 y = cos(3x) sin(3x) and y =0 y = 1 cos 2 (x) and y =4 5. The equation tan ( ) x 2 = −3 has the form tan(u) =−3, whose solution is u =arctan(−3)+πk. Hence, x 2 =arctan(−3) + πk, sox = 2 arctan(−3) + 2πk for integers k. To check, we note tan ( ) 2 arctan(−3)+2πk 2 = tan (arctan(−3) + πk) = tan (arctan(−3)) (the period of tangent is π) = −3 (See Theorem 10.27) To determine which of our answers lie in the interval [0, 2π), we first need to get an idea of the value of 2 arctan(−3). While we could easily find an approximation using a calculator, 5 we proceed analytically. Since −3 < 0, it follows that − π 2 < arctan(−3) < 0. Multiplying through by 2 gives −π
10.7 Trigonometric Equations and Inequalities 861 sin ( 2 [ 1 2 arcsin(0.87) + πk]) = sin (arcsin(0.87)+2πk) = sin (arcsin(0.87)) (the period of sine is 2π) = 0.87 (See Theorem 10.26) For the family x = π 2 − 1 2 arcsin(0.87) + πk ,weget sin ( 2 [ π 2 − 1 2 arcsin(0.87) + πk]) = sin(π − arcsin(0.87)+2πk) = sin(π − arcsin(0.87)) (the period of sine is 2π) = sin (arcsin(0.87)) (sin(π − t) =sin(t)) = 0.87 (See Theorem 10.26) To determine which of these solutions lie in [0, 2π), we first need to get an idea of the value of x = 1 2 arcsin(0.87). Once again, we could use the calculator, but we adopt an analytic route here. By definition, 0 < arcsin(0.87) < π 2 so that multiplying through by 1 2 gives us 0 < 1 2 arcsin(0.87) < π 4 . Starting with the family of solutions x = 1 2 arcsin(0.87) + πk, weuse the same kind of arguments as in our solution to number 5 above and find only the solutions corresponding to k = 0 and k = 1 lie in [0, 2π): x = 1 2 arcsin(0.87) and x = 1 2 arcsin(0.87) + π. Next, we move to the family x = π 2 − 1 2 arcsin(0.87) + πk for integers k. Here, we need to get a better estimate of π 2 − 1 2 arcsin(0.87). From the inequality 0 < 1 2 arcsin(0.87) < π 4 , we first multiply through by −1 and then add π 2 to get π 2 > π 2 − 1 2 arcsin(0.87) > π 4 ,or π 4 < π 2 − 1 2 arcsin(0.87) < π 2 . Proceeding with the usual arguments, we find the only solutions which lie in [0, 2π) correspond to k = 0 and k = 1, namely x = π 2 − 1 2 arcsin(0.87) and x = 3π 2 − 1 2 arcsin(0.87). All told, we have found four solutions to sin(2x) =0.87 in [0, 2π): x = 1 2 arcsin(0.87), x = 1 2 arcsin(0.87) + π, x = π 2 − 1 2 arcsin(0.87) and x = 3π 2 − 1 2 arcsin(0.87). By graphing y =sin(2x) andy =0.87, we confirm our results. y =tan ( x 2 ) and y = −3 y =sin(2x) andy =0.87
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College Trigonometry Version ⌊π
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Table of Contents vii 10 Foundation
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Preface Thank you for your interest
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xi text and we have gone to great l
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Chapter 10 Foundations of Trigonome
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10.1 Angles and their Measure 695 k
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10.1 Angles and their Measure 697 2
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10.1 Angles and their Measure 701 T
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10.1 Angles and their Measure 703 y
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10.1 Angles and their Measure 705 y
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10.1 Angles and their Measure 707 h
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10.1 Angles and their Measure 711 I
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10.2 The Unit Circle: Cosine and Si
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30 ◦ 1 10.2 The Unit Circle: Cosi
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10.3 The Six Circular Functions and
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10.4 Trigonometric Identities 771 f
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10.4 Trigonometric Identities 773 2
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10.4 Trigonometric Identities 779 T
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10.4 Trigonometric Identities 781 R
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10.4 Trigonometric Identities 785 I
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10.5 Graphs of the Trigonometric Fu
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Page 197 and 198: Chapter 11 Applications of Trigonom
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11.3 The Law of Cosines 911 of B ar
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11.3 The Law of Cosines 913 the pre
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11.3 The Law of Cosines 915 A 2 = =
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11.3 The Law of Cosines 917 22. A n
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11.4 Polar Coordinates 919 11.4 Pol
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11.4 Polar Coordinates 921 The poin
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11.4 Polar Coordinates 923 4. We mo
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11.4 Polar Coordinates 925 Solution
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11.4 Polar Coordinates 927 Solution
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11.4 Polar Coordinates 929 algebrai
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11.4 Polar Coordinates 931 45. ( )
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11.4 Polar Coordinates 933 5. ( 12,
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11.4 Polar Coordinates 935 13. (
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11.4 Polar Coordinates 937 73. r =7
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11.5 Graphs of Polar Equations 939
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11.6 Hooked on Conics Again 973 11.
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11.6 Hooked on Conics Again 975 Exa
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11.6 Hooked on Conics Again 977 The
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11.6 Hooked on Conics Again 979 We
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11.6 Hooked on Conics Again 983 The
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11.6 Hooked on Conics Again 985 In
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11.6 Hooked on Conics Again 989 2 9
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11.7 Polar Form of Complex Numbers
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11.8 Vectors 1013 Q ′ (1, 7) up 4
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11.8 Vectors 1015 ⃗v + ⃗w = 〈
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11.8 Vectors 1019 The remaining pro
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11.8 Vectors 1021 ‖k⃗v‖ = ‖
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11.8 Vectors 1023 at the origin, th
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11.8 Vectors 1025 Geometrically, th
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11.8 Vectors 1027 11.8.1 Exercises
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11.8 Vectors 1029 44. ⃗v = 〈−
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11.8 Vectors 1031 11.8.2 Answers 1.
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11.8 Vectors 1033 47. ‖⃗v‖ =2
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11.9 The Dot Product and Projection
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11.10 Parametric Equations 1049 obj
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11.10 Parametric Equations 1051 2.
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11.10 Parametric Equations 1053 Now
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11.10 Parametric Equations 1057 Our
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11.10 Parametric Equations 1061 Sup
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Index n th root of a complex number
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Index 1071 central angle, 701 chang
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Index 1073 reflective property, 523
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Index 1075 matrix, multiplicative,
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Index 1077 midpoint definition of,
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Index 1079 instantaneous, 161, 472
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Index 1081 inconsistent, 553 indepe
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College Trigonometry, 2011a

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