College Trigonometry, 2011a

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926 Applications of <strong>Trigonometry</strong> with its terminal side along the negative y-axis, so our answer is ( 3, 3π ) 2 . To check, we note x = r cos(θ) =3cos ( ) ( 3π 2 = (3)(0) = 0 and y = r sin(θ) =3sin 3π ) 2 =3(−1) = −3. 4. The point S(−3, 4) lies in Quadrant II. With x = −3 andy =4,wegetr 2 =(−3) 2 +(4) 2 =25 so r = ±5. As usual, we choose r =5≥ 0 and proceed to determine θ. Wehavetan(θ) = y x = 4 −3 = − 4 3 , and since this isn’t the tangent of one the common angles, we resort to using the arctangent function. Since θ lies in Quadrant II and must satisfy 0 ≤ θ
11.4 Polar Coordinates 927 Solution. 1. One strategy to convert an equation from rectangular to polar coordinates is to replace every occurrence of x with r cos(θ) and every occurrence of y with r sin(θ) and use identities to simplify. This is the technique we employ below. (a) We start by substituting x = r cos(θ)andy =sin(θ)into(x−3) 2 +y 2 = 9 and simplifying. With no real direction in which to proceed, we follow our mathematical instincts and see where they take us. 5 (r cos(θ) − 3) 2 +(r sin(θ)) 2 = 9 r 2 cos 2 (θ) − 6r cos(θ)+9+r 2 sin 2 (θ) = 9 r 2 ( cos 2 (θ)+sin 2 (θ) ) − 6r cos(θ) = 0 Subtract 9 from both sides. r 2 − 6r cos(θ) = 0 Since cos 2 (θ)+sin 2 (θ) =1 r(r − 6 cos(θ)) = 0 Factor. We get r =0orr = 6 cos(θ). From Section 7.2 we know the equation (x − 3) 2 + y 2 =9 describes a circle, and since r = 0 describes just a point (namely the pole/origin), we choose r = 6 cos(θ) for our final answer. 6 (b) Substituting x = r cos(θ) andy = r sin(θ) intoy = −x gives r sin(θ) = −r cos(θ). Rearranging, we get r cos(θ)+r sin(θ) =0orr(cos(θ)+sin(θ)) = 0. This gives r =0or cos(θ)+sin(θ) = 0. Solving the latter equation for θ, wegetθ = − π 4 + πk for integers k. As we did in the previous example, we take a step back and think geometrically. We know y = −x describes a line through the origin. As before, r = 0 describes the origin, but nothing else. Consider the equation θ = − π 4 . In this equation, the variable r is free, 7 meaning it can assume any and all values including r = 0. If we imagine plotting points (r, − π 4 ) for all conceivable values of r (positive, negative and zero), we are essentially drawing the line containing the terminal side of θ = − π 4 which is none other than y = −x. Hence, we can take as our final answer θ = − π 4 here.8 (c) We substitute x = r cos(θ) andy = r sin(θ) intoy = x 2 and get r sin(θ) =(r cos(θ)) 2 ,or r 2 cos 2 (θ) − r sin(θ) = 0. Factoring, we get r(r cos 2 (θ) − sin(θ)) = 0 so that either r =0 or r cos 2 (θ) =sin(θ). We can solve the latter equation for r by dividing both sides of the equation by cos 2 (θ), but as a general rule, we never divide through by a quantity that may be 0. In this particular case, we are safe since if cos 2 (θ) = 0, then cos(θ) =0, and for the equation r cos 2 (θ) =sin(θ) to hold, then sin(θ) would also have to be 0. Since there are no angles with both cos(θ) = 0 and sin(θ) = 0, we are not losing any 5 Experience is the mother of all instinct, and necessity is the mother of invention. Study this example and see what techniques are employed, then try your best to get your answers in the homework to match Jeff’s. 6 Note that when we substitute θ = π into r =6cos(θ), we recover the point r = 0, so we aren’t losing anything 2 by disregarding r =0. 7 See Section 8.1. 8 We could take it to be any of θ = − π + πk for integers k. 4
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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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10.6 The Inverse Trigonometric Func
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10.7 Trigonometric Equations and In
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11.6 Hooked on Conics Again 981 1.
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11.6 Hooked on Conics Again 987 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 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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