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x)) 173. sin(arccos(2x)) 174. sin arccos x 5 175. cos arcsin x 2 176. cos (arctan (3x)) 177. sin(2 arcsin(7x)) 178. sin 2 arcsin √ x 3 3 179. cos(2 arcsin(4x)) 180. sec(arctan(2x)) tan(arctan(2x)) 181. sin (arcsin(x) + arccos(x)) 182. cos (arcsin(x) + arctan(x)) 183. tan (2 arcsin(x)) 184. sin arctan(x) 1 2 185. If sin... |
−0.5637 206. cot(x) = 1 117 207. tan(x) = −0.6109 In Exercises 208 - 210, find the two acute angles in the right triangle whose sides have the given lengths. Express your answers using degree measure rounded to two decimal places. 208. 3, 4 and 5 209. 5, 12 and 13 210. 336, 527 and 625 211. A guy wire 1000 feet long is... |
degree. In Exercises 216 - 221, rewrite the given function as a sinusoid of the form S(x) = A sin(ωx + φ) using Exercises 35 and 36 in Section 10.5 for reference. Approximate the value of φ (which is in radians, of course) to four decimal places. 216. f (x) = 5 sin(3x) + 12 cos(3x) 217. f (x) = 3 cos(2x) + 4 sin(2x) 2... |
that arcsin(x) + arccos(x) = π 2 for −1 ≤ x ≤ 1. 237. Discuss with your classmates why arcsin 1 2 = 30◦. 238. Use the following picture and the series of exercises on the next page to show that arctan(1) + arctan(2) + arctan(3) = π y D(2, 3) A(0, 1) α β γ B(1, 0) x C(2, 0) O(0, 0) 848 Foundations of Trigonometry (a) C... |
ctan √ 3 = √ π 3 28. arccot − = 3 3 26. arccot − √ 3 = 5π 6 2π 3 29. arccot (0) = π 2 31. arccot (1) = 34. arccsc (2) = 37. arcsec √ 2 3 3 π 4 π 6 32. arccot √ 35. arcsec √ 38. arccsc. arcsin 1 2 = π 6 9. arcsin (1) = π 2 12. arccos − √ 2 2 = 3π 4 15. arccos 1 2 = π 3 18. arccos (1) = 0 21. arctan − √ 3 3 = − π 6 24. a... |
2 2 = is undefined. 62. cos arccos 63. cos arccos − 1 2 = − 1 2 65. cos (arccos (−0.998)) = −0.998 67. tan (arctan (−1)) = −1 69. tan arctan 5 12 = 5 12 64. cos arccos 5 13 = 5 13 66. cos (arccos (π)) is undefined. 68. tan arctan √ 3 = √ 3 70. tan (arctan (0.965)) = 0.965 71. tan (arctan (3π)) = 3π 72. cot (arccot (1)) ... |
π 6 92. arccos cos 94. arccos cos 96. arccos cos π 4 3π 2 5π 4 = − tan 98. arctan 100. arctan tan π 4 π 4 = = π 2 3π 4 = − π 4 102. arccot cot π 2 π 3 is undefined = π 3 103. arccot cot 105. arccot cot − π 4 3π 2 = 107. arcsec sec 109. arcsec sec 111. arcsec sec π 4 5π 6 5π 3 = 3π 4 π 2 7π 6 π 3 104. arccot (cot (π)) i... |
ccos csc 136. cos arcsin − 5 = = 4 5 √ 6 6 5 13 = 12 13 √ 3 10 138. cos (arccot (3)) = 140. tan arcsin − √ 2 5 = −2 10 5 = 4 3 141. tan arccos − √ 3 = − 1 2 143. tan (arccot (12)) = 1 12 142. tan arcsec 144. cot arcsin 5 3 12 13 = 5 12 10.6 The Inverse Trigonometric Functions 853 145. cot arccos √ 3 2 √ 3 = 147. cot (a... |
= √ 1 1 − x2 for −1 < x < 1 170. sin (2 arctan (x)) = for all x 2x x2 + 1 √ 171. sin (2 arccos (x)) = 2x 1 − x2 for −1 ≤ x ≤ 1 854 Foundations of Trigonometry 172. cos (2 arctan (x)) = 173. sin(arccos(2x)) = 174. sin arccos 175. cos arcsin x 5 x 2 = = √ 1 − x2 1 + x2 for all x √ 1 − 4x2 for − 1 √ 25 − x2 5 2 ≤ x ≤ 1 2... |
for − π 2 for − π 2 < θ < π 2, then θ + sin(2θ) = arcsin x 2 x + √ 4 − x2 2 < θ < π 2, then 1 2 θ − 1 2 sin(2θ) = 1 2 arctan x 7 − 7x x2 + 49 10The equivalence for x = ±1 can be verified independently of the derivation of the formula, but Calculus is required to fully understand what is happening at those x values. You... |
.0746, 6.2086 194. x = arctan(117) + πk, in [0, 2π), x ≈ 1.56225, 4.70384 195. x = arctan − 1 12 + πk, in [0, 2π), x ≈ 3.0585, 6.2000 196. x = arccos 2 3 + 2πk or x = 2π − arccos 2 3 + 2πk, in [0, 2π), x ≈ 0.8411, 5.4422 197. x = π + arcsin 198. x = arctan − 3 8 199. x = arcsin 200. x = arccos − 7 16 + 2πk or x = 2π − ... |
−0.5637) + 2πk or x = − arccos(−0.5637) + 2πk, in [0, 2π), x ≈ 2.1697, 4.1135 206. x = arctan(117) + πk, in [0, 2π), x ≈ 1.5622, 4.7038 207. x = arctan(−0.6109) + πk, in [0, 2π), x ≈ 2.5932, 5.7348 856 Foundations of Trigonometry 208. 36.87◦ and 53.13◦ 209. 22.62◦ and 67.38◦ 210. 32.52◦ and 57.48◦ 211. 68.9◦ 212. 7.7◦ ... |
3, √ 3] ∪ [ 5, ∞) 2 2 226. (−∞, ∞) 227. (−∞, −3) ∪ (−3, 3) ∪ (3, ∞) 228., ∞ 1 2 230. −∞, − 1 12 ∪ 1 12, ∞ 229., ∞ 1 2 231. (−∞, −6] ∪ [−4, ∞) 232. (−∞, −2] ∪ [2, ∞) 233. [0, ∞) 10.7 Trigonometric Equations and Inequalities 857 10.7 Trigonometric Equations and Inequalities In Sections 10.2, 10.3 and most recently 10.6,... |
π has the form sin(u) = 1 integers k. Since the argument of sine here is 3x, we have 3x = π integers k. To solve for x, we divide both sides2 of these equations by 3, and obtain x = π or x = 5π 6 + 2πk 2? Since this equation 6 + 2πk for 6 + 2πk for 18 + 2π 3 k 3 k for integers k. This is the technique employed in the e... |
calculator, we graph y = cos(2x) and y = − 2 over [0, 2π) and examine where these two graphs intersect. We see that the x-coordinates of the intersection points correspond to the decimal representations of our exact answers. 6 + 2πk = cos 7π 12 and 19π = − 12, 7π 6 2. Since this equation has the form csc(u) = 2, we re... |
� 1 sin( 1 3 x−π) all over the interval [0, 2π). 3Do you see why? 10.7 Trigonometric Equations and Inequalities 859 y = cos(2x) and y = − √ 3 2 y = 1 sin( 1 3 x−π) and y = √ 2 3. Since cot(3x) = 0 has the form cot(u) = 0, we know u = π 2 + πk, so, in this case, 3x = π 2 + πk for integers k. Solving for x yields x = π 3... |
. As a result, these solutions can be combined and we may write our solutions as x = π 3 + πk for integers k. To check the first family of solutions, we note that, depending on the integer. However, it is true that for all integers k, k, sec π sec π 3 = ±2. (Can you show this?) As a result, 3 + πk doesn’t always equal s... |
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 < 2 arctan(−3) < 0... |
the argument of sine here is 2x, we get 2x = arcsin(0.87) + 2πk or 2x = π − arcsin(0.87) + 2πk 2 − 1 which gives x = 1 2 arcsin(0.87) + πk for integers k. To check, 2 arcsin(0.87) + πk or x = π 5Your instructor will let you know if you should abandon the analytic route at this point and use your calculator. But seriou... |
+ πk for integers k. Here, we need to get a better estimate of π 2 arcsin(0.87) < π 4, we first multiply through by −1 and then add π 4, or 4 < π 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 arcsin(0.87) and x = 3π 2 arcsin(0.87).... |
(x) 6. sin(2x) = √ 3 cos(x) 8. cos(x) − √ 3 sin(x) = 2 1. We resist the temptation to divide both sides of 3 sin3(x) = sin2(x) by sin2(x) (What goes wrong if you do?) and instead gather all of the terms to one side of the equation and factor. 3 sin3(x) = sin2(x) 3 sin3(x) − sin2(x) = 0 sin2(x)(3 sin(x) − 1) = 0 Factor ... |
(x). 6Note that we are not counting the point (2π, 0) in our solution set since x = 2π is not in the interval [0, 2π). In the forthcoming solutions, remember that while x = 2π may be a solution to the equation, it isn’t counted among the solutions in [0, 2π). 10.7 Trigonometric Equations and Inequalities 863 This gives... |
1 = 0 (2u − 1)(u − 1) = 0 Let u = cos(x). 2, we get x = π 2 or u = 1. Since u = cos(x), we get cos(x) = 1 This gives u = 1 2 or cos(x) = 1. Solving cos(x) = 1 3 + 2πk or x = 5π 3 + 2πk for integers k. From cos(x) = 1, we get x = 2πk for integers k. The answers which lie in [0, 2π) are x = 0, π 3. Graphing y = cos(2x) ... |
same set of axes over [0, 2π) shows that the graphs intersect at what appears to be (0, 1), as required. y = cos(2x) and y = 3 cos(x) − 2 y = cos(3x) and y = 2 − cos(x) 5. While we could approach cos(3x) = cos(5x) in the same manner as we did the previous two problems, we choose instead to showcase the utility of the ... |
the arguments the same and we proceed as we would solving any nonlinear equation – gather all of the nonzero terms on one side of the equation and factor. sin(2x) = 2 sin(x) cos(x) = √ 2 sin(x) cos(x) − cos(x)(2 sin(x) − 3 cos(x) = 0 3) = 0 √ √ √ 3 cos(x) 3 cos(x) (Since sin(2x) = 2 sin(x) cos(x).) from which we get c... |
much we can do. However, since the arguments of the cosine and sine are the same, we can rewrite the left hand side of this equation as a sinusoid.9 To fit f (x) = cos(x) − 3 sin(x) to the form A sin(ωt + φ) + B, we use what we learned in Example 10.5.3 and find A = 2, B = 0, ω = 1 = 2, 3 sin(x) = 2 as 2 sin x + 5π and ... |
1 ≤ 0. Next, we let f (x) = 2 sin(x) − 1 and note that our original inequality is equivalent to solving f (x) ≤ 0. We now look to see where, if ever, f is undefined and where f (x) = 0. Since the domain of f is all real numbers, we can immediately set about finding the zeros of f. Solving f (x) = 0, we have 2 sin(x) − 1... |
� cos(x)(2 sin(x)−1) = 0. From cos(x) = 0, we get x = π 2 and x = 3π 2 which gives x = π 6 + 2πk or x = 5π 6 lie in [0, 2π). Next, we choose 2 lie in [0, 2π). For 2 sin(x) − 1 = 0, we get sin(x) = 1 6 + 2πk for integers k. Of those, only x = π 6 and x = 5π 10See page 214, Example 3.1.5, page 321, page 399, Example 6.3.... |
for integers k and of these, only x = arctan(3) and x = arctan(3) + π lie in [0, 2π). Since 3 > 0, we know 0 < arctan(3) < π 2 which allows us to position these zeros correctly on the sign diagram. To choose test values, we begin with x = 0 and find f (0) = −3. Finding a is a bit more challenging. Keep in mind convenie... |
� 2 (arctan(3) + π) 0 (+) (−) 3π 2 2π y = tan(x) and y = 3 Our next example puts solving equations and inequalities to good use – finding domains of functions. Example 10.7.4. Express the domain of the following functions using extended interval notation.14 1. f (x) = csc 2x + π 3 2. f (x) = sin(x) 2 cos(x) − 1 3. f (x)... |
see that it matches our diagram. 14See page 756 for details about this notation. 10.7 Trigonometric Equations and Inequalities 869 2. Since the domains of sin(x) and cos(x) are all real numbers, the only concern when finding 2 cos(x)−1 is division by zero so we set the denominator equal to zero and 3 +2πk for integers ... |
� (a0, b0) ∪ (b0, a1) ∪ (a1, b1) ∪... If we group these intervals in pairs, (a−2, b−2)∪(b−2, a−1), (a−1, b−1)∪(b−1, a0), (a0, b0)∪(b0, a1) and so forth, we see a pattern emerge of the form (ak, bk) ∪ (bk, ak + 1) for integers k so that our domain can be written as ∞ k=−∞ (ak, bk) ∪ (bk, ak + 1) = ∞ k=−∞ (6k + 1)π 3, (6... |
) term, x = πk for integers k. Next, we recall that for the square root to be defined, we need 1−cot(x) ≥ 0. Unlike the inequalities we solved in Example 10.7.3, we are not restricted here to a given interval. Our strategy is to solve this inequality over (0, π) (the same interval which generates a fundamental cycle of ... |
π arctan(x) − π2 = 0 5. π2 − 4 arccos2(x) < 0 6. 4 arccot(3x) > π Solution. 1. To solve arcsin(2x) = π 3 is in the range of the arcsine function (so a solution exists!) Next, we exploit the inverse property of sine and arcsine from Theorem 10.26 3, we first note that π 10.7 Trigonometric Equations and Inequalities 871 a... |
calculator, we need to graph y = 3 arcsec(2x − 1) + π. To do so, we make from Theorems 10.28 and 10.29.16 We see the graph use of the identity arcsec(u) = arccos 1 u of y = 3 arccos + π and the horizontal line y = 2π intersect at 3 1 2 = 1.5. 2x−1 16Since we are checking for solutions where arcsecant is positive, we k... |
setting f (x) = π2 − 4 arccos2(x) = 0. We get arccos(x) = ± π 2, and since the range of arccosine is = 0 [0, π], we focus our attention on arccos(x) = π as our only zero. Hence, we have two test intervals, [−1, 0) and (0, 1]. Choosing test values x = ±1, we get f (−1) = −3π2 < 0 and f (1) = π2 > 0. Since we are lookin... |
∞. Ideally, we wish only one zero of f, x = 1 to find test values x in these intervals so that arccot(4x) corresponds to one of our oft-used ‘common’ angles. After a bit of computation,18 we choose x = 0 for x < 1 3, we √ 3 = − π choose x = 3 < 0. Since we are looking for where 3. We find f (0) = π > 0 and f. To check g... |
2π) 19. sin (x) = cos (x) 21. sin (2x) = cos (x) 23. cos (2x) = cos (x) 20. sin (2x) = sin (x) 22. cos (2x) = sin (x) 24. cos(2x) = 2 − 5 cos(x) 25. 3 cos(2x) + cos(x) + 2 = 0 26. cos(2x) = 5 sin(x) − 2 27. 3 cos(2x) = sin(x) + 2 29. tan2(x) = 1 − sec(x) 31. sec(x) = 2 csc(x) 33. sin(2x) = tan(x) 35. cos(2x) + csc2(x)... |
(x) − √ 2 sin(x) = 1 51. cos(2x) − √ 3 sin(2x) = √ 2 48. sin(x) + √ 3 cos(x) = 1 √ 50. 3 sin(2x) + cos(2x) = 1 √ 52. 3 3 sin(3x) − 3 cos(3x) = 3 √ 3 53. cos(3x) = cos(5x) 54. cos(4x) = cos(2x) 55. sin(5x) = sin(3x) 56. cos(5x) = − cos(2x) 57. sin(6x) + sin(x) = 0 58. tan(x) = cos(x) In Exercises 59 - 68, solve the equa... |
attention to −π ≤ x ≤ π. 81. cos (x) > 84. sin2 (x) < √ 3 2 3 4 82. sin(x) > 1 3 83. sec (x) ≤ 2 85. cot (x) ≥ −1 86. cos(x) ≥ sin(x) In Exercises 87 - 92, solve the inequality. Express the exact answer in interval notation, restricting your attention to −2π ≤ x ≤ 2π. 87. csc (x) > 1 90. tan2 (x) ≥ 1 88. cos(x) ≤ 5 3 ... |
like sin(11x) = 1 2. What do you find? 2, sin(3x) = 1 2 and sin 5x 2, sin 3x = 1 = 1 2 2 = 1 2. Now consider sin x with −1 and repeat the whole exploration. 2 Replace 1 2 10.7 Trigonometric Equations and Inequalities 877 10.7.2 Answers 1. x = πk 5 ; x = 0, π 5, 2π 5, 3π 5, 4π 5, π, 6π 5, 7π 5, 8π 5, 9π 5 2. x = π 9 + 2... |
5π 12 13. No solution 14. x = 5π 8 + πk 2 ; x = π 8, 5π 8, 9π 8, 13π 8 15. x = 16. x = 17. x = 18 + πk or x = + πk or x = 2π 3 5π 6 + πk; x = + πk; x = π 3 π 6,, 2π 3 5π 6,, 4π 3 7π 6,, 5π 3 11π 6 + πk 2 ; x = π 4, 3π 4, 5π 4, 7π 4 + πk or x = 2π 3 + πk; x = π 3, 2π 3, 4π 3, 5π 3 878 19. x = 21. x = π 4 π 6 23. x = 0,... |
7π 4 π 4 3π 2 5π 3 π 2 5π 8, 35. x = 37. x = 39. x = 41,,,,, 5π 6 9π 8, 3π 2 13π 8 4π 7,,, 3π 7 3π 2 13π 48, 43. x = 0, 44. x = 0, π 7 π 2, 2π 7, π, 46. x = π 48, 11π 48 47. x = 0, π 2 49. x = 51. x =, π 12 17π 24 17π 12 41π 24,, 23π 24, 47π 24 42. No solution, 5π 7, 6π 7, π, 8π 7, 9π 7, 10π 7, 11π 7, 12π 7, 13π 7 45.... |
, 3π 5, π,, 7π 5 ≈ 0.6662, π − arcsin 9π 5 −1 + 2 √ 5 ≈ 2.4754 60. x = −1 62. x = − √ 3 2 64. x = 6 66. x = 1 2 √ 68. x = − 3 69. [π, 2π] 71. 73. 0, π 3 ∪ π 4, 3π 4, 2π 3 5π 4 ∪, 7π 4 4π 3 ∪ 5π 3, 2π 0, 75. π 3 ∪ 2π 3, π ∪ π, 4π 3 ∪ 5π 3, 2π 70. 72. 74. 76. π 3, π 2 ∪ 0, 0, 0,, 4π 3 3π 4 11π 6 5π 3, 2π 3π 2 5π 4 ∪ 7π 4... |
7π 4, − 3π 2 −2π, − 91. 5π 3 ∪ 92. − 93. 0, 1 2, − 7π 6 11π 6 ∪ ∪ π 6, 5π 6 ∪,, − π 3π 2 2 2, 1 94. 1 96. (−∞, ∞) 97. [−1, 0) 99. ∞ k=−∞ (2kπ, (2k + 2)π) 100. ∞ k=−∞ (4k − 1)π 2, (4k + 3)π 2 −∞, √ 3 7 95. 98. −14k + 1)π k=−∞ 4 ∞ (6k − 1)π 3 (2k + 1)π 2 ∪ (2k + 1)π 2, (4k + 3)π 4 (6k + 1)π 3 ∪ (4k + 1)π 2, (4k + 3)π 2,... |
section are time-dependent. We reintroduce and summarize all of the important facts and definitions about this form of the sinusoid below. Properties of the Sinusoid S(t) = A sin(ωt + φ) + B The amplitude is |A| The angular frequency is ω and the ordinary frequency is f = ω 2π The period is T = 1 f = 2π ω The phase is ... |
of Sinusoids 883 We know from the equations given on page 732 in Section 10.2.1 that the y-coordinate for counterclockwise motion on a circle of radius r centered at the origin with constant angular velocity (frequency) ω is given by y = r sin(ωt). Here, t = 0 corresponds to the point (r, 0) so that θ, the angle measu... |
127 2 A few remarks about Example 11.1.1 are in order. First, note that the amplitude of 64 in our answer corresponds to the radius of the Giant Wheel. This means that passengers on the Giant Wheel never stray more than 64 feet vertically from the center of the Wheel, which makes sense. 8 = 15.875. This represents the... |
of B is the average of the maximum and minimum values. So here we take B = 3.3+21.8 = 12.55. Next is the amplitude A which is the displacement from the baseline to the maximum (and minimum) values. We find A = 21.8 − 12.55 = 12.55 − 3.3 = 9.25. At this point, we have H(t) = 9.25 sin(ωt + φ) + 12.55. Next, we go after t... |
give us an r2 value like it did for linear regressions in Section 2.5, nor does it give us an R2 value like it did for quadratic, cubic and quartic regressions as in Section 3.1. The reason for this, much like the reason for the absence of R2 for the logistic model in Section 6.5, is beyond the scope of this course. W... |
to a spring as depicted below. The weight of the object will stretch the spring. The system is said to be in ‘equilibrium’ when the weight of the object is perfectly balanced with the restorative force of the spring. How far the spring stretches to reach equilibrium depends on the spring’s ‘spring constant’. Usually d... |
, we use feet (ft.) to measure displacement. If we second2 and 1 Newton = 1 kg meter second2. are in the SI system, we measure displacement in meters (m). Time is always measured in seconds (s). 11.1 Applications of Sinusoids 887 object and initial velocity v0 of the object. As with x(t), x0 = 0 means the object is rel... |
11.1, we first need to determine the spring constant k and the mass of the object m. To find k, we use Hooke’s Law F = kd. We know the object weighs 64 lbs. and stretches the spring 8 ft.. Using F = 64 and d = 8, we get 64 = k · 8, or k = 8 lbs. s2. We get m = 2 slugs. We can now proceed to apply Theorem 11.1. ft.. To fi... |
. To check this answer, we graph one cycle of x(t). Since our applied domain in this situation is t ≥ 0, and the period of x(t) is T = 2π 2 = π, we graph x(t) over the interval [0, π]. Remembering that x(t) > 0 means the object is below the equilibrium position and x(t) < 0 means the object is above the equilibrium pos... |
+ π − arcsin 3 on a graphing utility 5 and confirm the coordinates of the first relative minimum to be approximately (1.107, −5). + π 2 arcsin 3 + π 5 x π 4 π 2 3π 4 π t 3 2 1 −1 −2 −3 x(t) = 3 sin 2t + π 2 y = 5 sin 2x + π − arcsin 3 5 It is possible, though beyond the scope of this course, to model the effects of frict... |
so that x(t) = 10e−t/5 sin t + π cos(t) +. Graphing this on the 3 sin(t) = 2 sin t + π 3 3 calculator as y = 10e−x/5 sin x + π reveals some interesting behavior. The sinusoidal nature 3 continues indefinitely, but it is being attenuated. In the sinusoid A sin(ωx + φ), the coefficient, we can think A of the sine function ... |
t + 3) 2 cos(2t) + (t + 3) √ √ √ 890 Applications of Trigonometry as x → ∞, this is hardly surprising. The phenomenon illustrated here is ‘forced’ motion. That is, we imagine that the entire apparatus on which the spring is attached is oscillating as well. In this case, we are witnessing a ‘resonance’ effect – the frequ... |
and y = ±10 sin(x) over [0, 2π]. This is an example of the ‘beat’ phenomena, and the curious reader is invited to explore this concept as well.17 y = 5 sin(6x) − 5 sin(8x) over [0, π] y = 5 sin(6x) − 5 sin(8x) and y = ±10 sin(x) over [0, 2π] 16The reader is invited to investigate the destructive implications of resona... |
.1, we introduced the yo-yo trick ‘Around the World’ in which a yo-yo is thrown so it sweeps out a vertical circle. As in that exercise, suppose the yo-yo string is 28 inches and it completes one revolution in 3 seconds. If the closest the yo-yo ever gets to the ground is 2 inches, find a sinsuoid which models the heigh... |
st of the month for each month during the years 1971 – 2000.19 For example, t = 3 represents the average of the temperatures recorded for Lake Erie on every March 1 for the years 1971 through 2000. Month Number, t Temperature (◦ F), 10 11 12 36 33 34 38 47 57 67 74 73 67 56 46 (a) Using the techniques discussed in Exam... |
can find for them and share your results with your class. 20The computed average is 41◦F for April 15th and 71◦F for September 15th. 21See this website: http://www.usno.navy.mil/USNO/astronomical-applications/data-services/frac-moon-ill. 22You may want to plot the data before you find the phase shift. 23The listed fract... |
42.43◦F and the average temperature on September 15th to be approximately 70.05◦F. This model appears to be more accurate. 11.1 Applications of Sinusoids 895 9. (a) Based on the shape of the data, we either choose A < 0 or we find the second value of t which closely approximates the ‘baseline’ value, F = 0.505. We choo... |
inverse trigonometric functions. To decrease the chances of propagating error, however, we stick to using the data given to us in the problem. In this case, the lengths 4 and 7 were given, so we want to relate these to α. According to Theorem 10.4, cos(α) = 4 7. Since α is an acute angle, α = arccos 4 radians. Convert... |
sin(α) a = sin(β) b = sin(γ) c a sin(α) = b sin(β) = c sin(γ) The proof of the Law of Sines can be broken into three cases. For our first case, consider the triangle ABC below, all of whose angles are acute, with angle-side opposite pairs (α, a), (β, b) and (γ, c). If we drop an altitude from vertex B, we divide the tr... |
Sines to solve a triangle, we need at least one angle-side opposite pair. The next example showcases some of the power, and the pitfalls, of the Law of Sines. Example 11.2.2. Solve the following triangles. Give exact answers and decimal approximations (rounded to hundredths) and sketch the triangle. 1. α = 120◦, a = 7... |
found using the difference identity for sine or a half-angle formula, but that becomes unnecessarily messy for the discussion at hand. Thus “exact” here means 7 sin(15◦) sin(120◦). 11.2 The Law of Sines 899 angle-side pair available is (γ, c). The Law of Sines gives rearrangement, we get a = 5.25 sin(85◦) which yields ... |
and 150◦ in order to fit inside the triangle with α. The only angle that satisfies this requirement and has sin(γ) = 1 is γ = 90◦. In other words, we have a right triangle. We find the measure of β to be β = 180◦ − 30◦ − 90◦ = 60◦ and then determine b using the Law of Sines. We find b = 2 sin(60◦) 3 ≈ 3.46 units. In this ... |
two triangles on our hands. In the case γ = arcsin 2 3 find6 β ≈ 180◦ − 30◦ − 41.81◦ = 108.19◦. Using the Law of Sines with the angle-side opposite radians pair (α, a) and β, we find b ≈ 3 sin(108.19◦) ≈ 138.19◦, we repeat the exact same steps and find β ≈ 11.81◦ and b ≈ 1.23 units.7 Both triangles are drawn below. sin(3... |
180◦ = π radians. Hence, β = π − π 7An exact answer for β in this case is β = arcsin 2 6 − arcsin 2 − π 3 6 − arcsin 2 = 5π 6 radians ≈ 11.81◦. 3 3 radians ≈ 108.19◦. 11.2 The Law of Sines 901 determined using the equation γ = 180◦ − α − β. Knowing the measures of all three angles of a triangle completely determines i... |
h < a < c, then two distinct triangles exist which satisfy the given criteria. If a ≥ c, then γ is acute and exactly one triangle exists which satisfies the given criteria Theorem 11.3 is proved on a case-by-case basis. If a < h, then a < c sin(α). If a triangle were c = sin(α) to exist, the Law of Sines would have sin... |
◦ − γ0. We need to argue that each of these angles ‘fit’ into a triangle with α. Since (α, a) and (γ0, c) are angle-side opposite pairs, the assumption c > a in this case gives us γ0 > α. Since γ0 is acute, we must have that α is acute as well. This means one triangle can contain both α and γ0, giving us one of the tria... |
that the angles γ and 45◦ are supplemental, so that γ = 180◦ − 45◦ = 135◦. We can now 5 find β = 180◦ − 30◦ − γ = 180◦ − 30◦ − 135◦ = 15◦. By the Law of Sines, we have sin(15◦) which gives d = 5 sin(30◦) sin(15◦) ≈ 9.66 miles. Next, to find the point on the coast closest to the island, which we’ve labeled as C, we need ... |
ac sin(β) = 1 2 ab sin(γ) Example 11.2.4. Find the area of the triangle in Example 11.2.2 number 1. Solution. From our work in Example 11.2.2 number 1, we have all three angles and all three sides to work with. However, to minimize propagated error, we choose A = 1 2 ac sin(β) from Theorem 11.4 because it uses the mos... |
, c = 3, a = 3.05 15. β = 102◦, b = 16.75, c = 13 16. β = 102◦, b = 16.75, c = 18 17. β = 102◦, γ = 35◦, b = 16.75 18. β = 29.13◦, γ = 83.95◦, b = 314.15 19. γ = 120◦, β = 61◦, c = 4 20. α = 50◦, a = 25, b = 12.5 21. Find the area of the triangles given in Exercises 1, 12 and 20 above. (Another Classic Application: Gra... |
��nd the height of the tree. (Hint: First show that the tree makes a 94◦ angle with the road.) (Another Classic Application: Bearings) In the next several exercises we introduce and work with the navigation tool known as bearings. Simply put, a bearing is the direction you are heading according to a compass. The classi... |
The word ‘plumb’ here means that the tree is perpendicular to the horizontal. 906 Applications of Trigonometry (e) N31.25◦W (f) S72◦4112W15 (g) N45◦E (h) S45◦W 26. The Colonel spots a campfire at a of bearing N42◦E from his current position. Sarge, who is positioned 3000 feet due east of the Colonel, reckons the bearing... |
. A hiker determines the bearing to a lodge from her current position is S40◦W. She proceeds to hike 2 miles at a bearing of S20◦E at which point she determines the bearing to the lodge is S75◦W. How far is she from the lodge at this point? Round your answer to the nearest hundredth of a mile. 31. A watchtower spots a ... |
angle between them are given. (Said another way, explain why the Law of Sines cannot be used in the SSS and SAS cases.) 37. Given α = 30◦ and b = 10, choose four different values for a so that (a) the information yields no triangle (b) the information yields exactly one right triangle (c) the information yields two dis... |
Information does not produce a triangle 2. 4. 6. 8. 10. 12. 14. 16. 18. 20. α = 73.2◦ β = 54.1◦ γ = 52.7◦ c ≈ 97.22 b ≈ 99.00 a = 117 β = 62◦ α = 95◦ a = 33.33 b ≈ 29.54 c ≈ 13.07 γ = 23◦ α = 117◦ β ≈ 56.3◦ γ ≈ 6.7◦ c ≈ 5.89 b = 42 a = 45 b = 23.5 α = 42◦ β ≈ 67.66◦ γ ≈ 70.34◦ c ≈ 23.93 a = 17 α = 42◦ β ≈ 112.34◦ γ ≈ ... |
909 25. (a) θ = 180◦ (b) θ = 353◦ (c) θ = 84.5◦ (d) θ = 270◦ (e) θ = 121.25◦ (f) θ = 197◦1848 (g) θ = 45◦ (h) θ = 225◦ 26. The Colonel is about 3193 feet from the campfire. Sarge is about 2525 feet to the campfire. 27. The distance from the Muffin Ridge Observatory to Sasquach Point is about 7.12 miles. The distance from ... |
+ c2 − 2ac cos(β) c2 = a2 + b2 − 2ab cos(γ) or, solving for the cosine in each equation, we have cos(α) = b2 + c2 − a2 2bc cos(β) = a2 + c2 − b2 2ac cos(γ) = a2 + b2 − c2 2ab To prove the theorem, we consider a generic triangle with the vertex of angle α at the origin with side b positioned along the positive x-axis. ... |
2 − 2bc cos(α) Since cos2(α) + sin2(α) = 1 The remaining formulas given in Theorem 11.5 can be shown by simply reorienting the triangle to place a different vertex at the origin. We leave these details to the reader. What’s important about a and α in the above proof is that (α, a) is an angle-side opposite pair and b an... |
whereas the cosine of an obtuse angle is negative. Since the sine of both acute and obtuse angles are positive, the sine of an angle alone is not 53 − 28 cos (50◦) ≈ 5.92 units. 2This shouldn’t come as too much of a shock. All of the theorems in Trigonometry can ultimately be traced back to the definition of the circul... |
�) 53−28 cos(50◦) 2ab radians ≈ 15.01◦. We β = 50◦ a = 7 c = 2 α ≈ 114.99◦ γ ≈ 15.01◦ b ≈ 5.92 As we mentioned earlier, once we’ve determined b it is possible to use the Law of Sines to find the remaining angles. Here, however, we must proceed with caution as we are in the ambiguous (ASS) case. It is advisable to first fi... |
α = arccos 29 35 radians ≈ 34.05◦. β ≈ 101.54◦ c = 5 a = 4 α ≈ 34.05◦ γ ≈ 44.42◦ b = 7 We note that, depending on how many decimal places are carried through successive calculations, and depending on which approach is used to solve the problem, the approximate answers you obtain may differ slightly from those the autho... |
6 using Theorem 11.4. Using the convention that the angle γ is opposite the side c, we have A = 1 2 ab sin(γ) from Theorem 11.4. In order to simplify computations, we start by manipulating the expression for A2. A2 = 2 ab sin(γ) 1 2 a2b2 sin2(γ) = = 1 4 a2b2 4 1 − cos2(γ) since sin2(γ) = 1 − cos2(γ). The Law of Cosines... |
a+b+c 2. To (s − a − 2a 2 = b + c − a 2 Similarly, we find (s − b) = a+c−b 2 and (s − c) = a+b−c 2. Hence, we get A2 = (b + c − a) 2 · (a + c − b) 2 · (a + b − c) 2 · (a + b + c) 2 = (s − a)(s − b)(s − c)s so that A = s(s − a)(s − b)(s − c) as required. We close with an example of Heron’s Formula. Example 11.3.3. Find ... |
and angle(s), if possible, using any appropriate technique. 11. a = 18, α = 63◦, b = 20 12. a = 37, b = 45, c = 26 13. a = 16, α = 63◦, b = 20 14. a = 22, α = 63◦, b = 20 15. α = 42◦, b = 117, c = 88 16. β = 7◦, γ = 170◦, c = 98.6 17. Find the area of the triangles given in Exercises 6, 8 and 10 above. 18. The hour ha... |
from the lodge at this point. Round your answer to the nearest hundredth of a mile. What bearing should she follow to return to the lodge? Round your angle to the nearest degree. 23. The HMS Sasquatch leaves port on a bearing of N23◦E and travels for 5 miles. It then changes course and follows a heading of S41◦E for 2... |
is formed and if the equation has two distinct positive real zeros then two distinct triangles are formed. Apply the Law of Cosines to Exercises 11, 13 and 14 above in order to demonstrate this result. 27. Discuss with your classmates why Heron’s Formula yields an area in square units even though four lengths are bein... |
◦ β ≈ 89.40◦ γ ≈ 35.30◦ a = 37 c = 26 b = 45 α = 63◦ β ≈ 54.1◦ γ ≈ 62.9◦ c ≈ 21.98 a = 22 b = 20 α ≈ 3◦ γ = 170◦ β = 7◦ a ≈ 29.72 b ≈ 69.2 c = 98.6 √ 2. 4. 6. 8. 10. 12. 14. 16. √ √ 17. The area of the triangle given in Exercise 6 is The area of the triangle given in Exercise 8 is The area of the triangle given in Exer... |
corner diagonally across the rectangle from the origin.1 For this reason, the Cartesian coordinates of a point are often called ‘rectangular’ coordinates. In this section, we introduce a new system for assigning coordinates to points in the plane – polar coordinates. We start with an origin point, called the pole, and... |
.5, π 4 If we interpret the angle first, we rotate π Here we are locating a point 3.5 units away from the pole on the terminal side of 5π 4 radians, then move back through the pole 3.5 units. 4, not π 4. θ = π 4 Pole θ = π 4 Pole Pole Q −3.5, π 4 As you may have guessed, θ < 0 means the rotation away from the polar axis... |
pole, r = ±2. Next, we choose angles θ for each of the r values. The given representation for P is (2, 240◦) so the angle θ we choose for the r = 2 case must be coterminal with 240◦. (Can you see why?) One such angle is θ = −120◦ so one answer for this case is (2, −120◦). For the case r = −2, we visualize our rotation... |
from the pole, any representation (r, θ) for P satisfies r = ±117. For the r = 117 case, we can take θ to be any angle coterminal with − 5π 2. In this case, we choose as one answer. For the r = −117 case, we visualize moving left 117 θ = 3π units from the pole and then rotating through an angle θ to reach P. We find tha... |
if one of the following is true: r = r and θ = θ + 2πk for some integer k r = −r and θ = θ + (2k + 1)π for some integer k All polar coordinates of the form (0, θ) represent the pole regardless of the value of θ. The key to understanding this result, and indeed the whole polar coordinate system, is to keep in mind that... |
2k + 1)π for some integer k. To plot P, we first move a directed distance r from the pole; to plot P, our first step is to move the same distance from the pole as P, but in the opposite direction. At this intermediate stage, we have two points equidistant from the pole rotated exactly π radians apart. Since θ = θ + (2k +... |
. The remaining case is r = 0, in which case (r, θ) = (0, θ) is the pole. Since the pole is identified with the origin (0, 0) in rectangular coordinates, the theorem in this case amounts to checking ‘0 = 0.’ The following example puts Theorem 11.7 to good use. Example 11.4.2. Convert each point in rectangular coordinate... |
Q lies in Quadrant III, 4, which satisfies the requirement that 0 ≤ θ < 2π. Our final answer is. To check, we find x = r cos(θ) = (3 = −3 so r = ± tan(θ) = −3 we choose θ = 5π (r, θ) = 3 −3 = 1, which means θ has a reference angle of π 2. Since we are asked for r ≥ 0, we choose r = 3 = (3 2) √ √ √ √ − √ 2) cos 5π 4 2, 5π... |
, 4) lies in Quadrant II. With x = −3 and y = 4, we get r2 = (−3)2 +(4)2 = 25 so r = ±5. As usual, we choose r = 5 ≥ 0 and proceed to determine θ. We have tan(θ) = y x = 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 satis... |
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