College Trigonometry, 2011a

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832 Foundations of <strong>Trigonometry</strong> Solution. 1. (a) Since 2 ≥ 1, we may invoke Theorem 10.29 to get arcsec(2) = arccos ( ) 1 2 = π 3 . (b) Unfortunately, −2 is not greater to or equal to 1, so we cannot apply Theorem 10.29 to arccsc(−2) and convert this into an arcsine problem. Instead, we appeal to the definition. The real number t = arccsc(−2) lies in ( 0, π ] ( ] 2 ∪ π, 3π 2 and satisfies csc(t) =−2. The t we’re after is t = 7π 6 , so arccsc(−2) = 7π 6 . (c) Since 5π 4 lies between π and 3π 2 , we may apply Theorem 10.29 directly to simplify arcsec ( sec ( )) 5π 4 = 5π 4 . We encourage the reader to work this through using the definition as we have done in the previous examples to see how it goes. (d) To simplify cot (arccsc (−3)) we let t = arccsc (−3) so that cot (arccsc (−3)) = cot(t). We know csc(t) =−3, and since this is negative, t lies in ( π, 3π ] 2 . Using the identity 1+cot 2 (t) = csc 2 (t), we find 1 + cot 2 (t) =(−3) 2 so that cot(t) =± √ 8=±2 √ 2. Since t is in the interval ( π, 3π ] √ 2 , we know cot(t) > 0. Our answer is cot (arccsc (−3)) = 2 2. 2. (a) [ We begin ) [ simplifying ) tan(arcsec(x)) by letting t = arcsec(x). Then, sec(t) =x for t in 0, π 2 ∪ π, 3π 2 , and we seek a formula for tan(t). Since tan(t) is defined for all t values under consideration, we have no additional restrictions on t. To relate sec(t) to tan(t), we use the identity 1+tan 2 (t) =sec 2 (t). This is valid for all values of t under consideration, and when we substitute sec(t) =x, weget1+tan 2 (t) =x 2 . Hence, tan(t) =± √ x 2 − 1. Since t lies in [ 0, π ) [ ) √ 2 ∪ π, 3π 2 , tan(t) ≥ 0, so we choose tan(t) = x 2 − 1. Since we found no additional restrictions on t, the equivalence tan(arcsec(x)) = √ x 2 − 1 holds for all x in the domain of t = arcsec(x), namely (−∞, −1] ∪ [1, ∞). (b) To ( simplify ] ( cos(arccsc(4x)), ] we start by letting t = arccsc(4x). Then csc(t) =4x for t in 0, π 2 ∪ π, 3π 2 , and we now set about finding an expression for cos(arccsc(4x)) = cos(t). Since cos(t) is defined for all t, we do not encounter any additional restrictions on t. From csc(t) =4x, wegetsin(t) = 1 4x , so to find cos(t), we can make use if the identity cos 2 (t)+sin 2 (t) = 1. Substituting sin(t) = 1 4x gives cos2 (t)+ ( 1 2 4x) = 1. Solving, we get √ √ 16x cos(t) =± 2 − 1 16x 16x 2 = ± 2 − 1 4|x| If t lies in ( 0, π ] √ 2 , then cos(t) ≥ 0, and we choose cos(t) = 16x 2 −1 4|x| . Otherwise, t belongs to ( π, 3π ] √ 2 in which case cos(t) ≤ 0, so, we choose cos(t) =− 16x 2 −1 4|x| This leads us to a (momentarily) piecewise defined function for cos(t) ⎧ √ 16x ⎪⎨ 2 − 1 , if 0 ≤ t ≤ π cos(t) = 4|x| 2 √ 16x ⎪⎩ − 2 − 1 , if π
10.6 The Inverse Trigonometric Functions 833 We now see what these restrictions mean in terms of x. Since 4x = csc(t), we get that for 0 ≤ t ≤ π 2 ,4x ≥ 1, or x ≥ 1 4 . In this case, we can simplify |x| = x so cos(t) = √ 16x 2 − 1 4|x| = √ 16x 2 − 1 4x Similarly, for π 0,wecanusethepropertylistedinTheorem10.27 to rewrite arccot(2) as arccot(2) = arctan ( 1 2) . In ‘radian’ mode, we find arccot(2) = arctan ( 1 2) ≈ 0.4636. (b) Since 5 ≥ 1, we can use the property from either Theorem 10.28 or Theorem 10.29 to write arcsec(5) = arccos ( 1 5) ≈ 1.3694.
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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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11.1 Applications of Sinusoids 883
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11.2 The Law of Sines 897 minimizes
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11.2 The Law of Sines 899 a angle-s
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11.2 The Law of Sines 901 determine
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11.2 The Law of Sines 903 Let x den
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11.2 The Law of Sines 905 24. Along
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11.2 The Law of Sines 907 33. The a
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11.2 The Law of Sines 909 25. (a)
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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 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 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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