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and whose terminal side is the line-of-sight to an object above the base-line. This is represented schematically below. 6We may choose any values x and y so long as x > 0, y < 0 and x y = −4. For example, we could choose x = 8 and y = −2. The fact that all such points lie on the terminal side of θ is a consequence of ... |
x = 1, or x = h. Substituting this into the second equation gives 3. The result is a linear equation for h, so we proceed to expand the right hand side and gather all the terms involving h to one side. √ 3 3. Clearing fractions, we get 3h = (h + 200) x and tan (30◦) = h h+200 = tan (30◦) = √ h 3h = (h + 200) √ 3h = h ... |
better understanding what set of real numbers we’re dealing with, it pays to write out and graph this set. Running through a few values of k, we find the domain to be {t : t = ± π 2,...}. Graphing this set on the number line we get 2, ± 5π 2, ± 3π 756 Foundations of Trigonometry − 5π 2 − 3π 2 − π 2 0 π 2 3π 2 5π 2 Usin... |
k=−∞ can never actually be ∞ or −∞, but rather, this conveys the idea that k ranges through all of the integers. Now that we have painstakingly determined the domain of F (t) = sec(t), it is time to discuss the range. Once again, we appeal to the definition F (t) = sec(t) = 1 cos(t). The range of f (t) = cos(t) is [−1,... |
functions: csc(t), tan(t) and cot(t). The reader is encouraged to do so. (See the Exercises.) For now, we gather these facts into the theorem below. Theorem 10.11. Domains and Ranges of the Circular Functions • The function f (t) = cos(t) • The function g(t) = sin(t) – has domain (−∞, ∞) – has range [−1, 1] – has doma... |
help solidify the idea that these real numbers are the outputs from the inputs, which we have been calling t. 758 Foundations of Trigonometry We close this section with a few notes about solving equations which involve the circular functions. First, the discussion on page 735 in Section 10.2.1 concerning solving equat... |
= 3 5 with θ in Quadrant II 22. tan(θ) = 12 5 with θ in Quadrant III with θ in Quadrant I 24. sec(θ) = 7 with θ in Quadrant IV 25. csc(θ) = − with θ in Quadrant III 26. cot(θ) = −23 with θ in Quadrant II 27. tan(θ) = −2 with θ in Quadrant IV. 28. sec(θ) = −4 with θ in Quadrant II. 29. cot(θ) = √ 5 with θ in Quadrant I... |
θ) = − √ 3 56. csc(θ) = −2 57. cot(θ) = −1 1 2 54. sec(θ) = −1 In Exercises 58 - 65, solve the equation for t. Give exact values. 58. cot(t) = 1 59. tan(t) = 62. cot(t) = − √ 3 63. tan(t 60. sec(t) = − 64. sec(t) = √ 2 3 61. csc(t) = 0 65. csc(t) = √ 2 3 3 In Exercises 66 - 69, use Theorem 10.10 to find the requested qu... |
the term umbra versa and see what it has to do with the shadow in this problem. 77. The broadcast tower for radio station WSAZ (Home of “Algebra in the Morning with Carl and Jeff”) has two enormous flashing red lights on it: one at the very top and one a few feet below the top. From a point 5000 feet away from the base ... |
for the Sasquatch to reach the firetower from his location at the second sighting? Round your answer to the nearest minute. 79. When I stand 30 feet away from a tree at home, the angle of elevation to the top of the tree is 50◦ and the angle of depression to the base of the tree is 10◦. What is the height of the tree? ... |
Functions and Fundamental Identities 763 94. 96. csc(θ) 1 + cot2(θ) cot(θ) csc2(θ) − 1 = sin(θ) = tan(θ) 95. tan(θ) sec2(θ) − 1 = cot(θ) 97. 4 cos2(θ) + 4 sin2(θ) = 4 98. 9 − cos2(θ) − sin2(θ) = 8 99. tan3(θ) = tan(θ) sec2(θ) − tan(θ) 100. sin5(θ) = 1 − cos2(θ)2 sin(θ) 101. sec10(θ) = 1 + tan2(θ)4 sec2(θ) 102. cos2(θ)... |
. 1 1 − cos(θ) + 1 1 + cos(θ) = 2 csc2(θ) 114. 1 sec(θ) + 1 + 1 sec(θ) − 1 = 2 csc(θ) cot(θ) 115. 1 csc(θ) + 1 + 1 csc(θ) − 1 = 2 sec(θ) tan(θ) 116. 1 csc(θ) − cot(θ) − 1 csc(θ) + cot(θ) = 2 cot(θ) 117. cos(θ) 1 − tan(θ) + sin(θ) 1 − cot(θ) = sin(θ) + cos(θ) 118. 120. 122. 124. 126. 128. 1 sec(θ) + tan(θ) 1 csc(θ) − co... |
consult Sections 2.2 and 6.2 for a review of the properties of absolute value and logarithms before proceeding. 129. ln | sec(θ)| = − ln | cos(θ)| 130. − ln | csc(θ)| = ln | sin(θ)| 131. − ln | sec(θ) − tan(θ)| = ln | sec(θ) + tan(θ)| 132. − ln | csc(θ) + cot(θ)| = ln | csc(θ) − cot(θ)| 133. Verify the domains and ran... |
sin(θ) θ < 1 also holds for − π 2 < θ < 0. 137. Explain why the fact that tan(θ) = 3 = 3 solution to number 6 in Example 10.3.1.) 1 does not mean sin(θ) = 3 and cos(θ) = 1? (See the 766 Foundations of Trigonometry 10.3.3 Answers 1. tan 4. cot π 4 4π 3 = 1 = 7. csc − π 3 = − = 2 √ 3 = 10. sec 13. tan 16. cot 19. csc − ... |
θ) = 7, cot(θ) = − 3 √ 3 12 √ 24. sin(θ) = −4 7 √ 25. sin(θ) = − √ 26. sin(θ) = 91 10, cos(θ) = − 3 530, cos(θ) = − 23 530 10, tan(θ) = √ 91 3, csc(θ) = − 10 91 √ 530, tan(θ) = − 1 530 23, csc(θ) = 3, cot(θ) = 3, sec(θ) = − 10 √ 530 23, cot(θ) = −23 91 91 √ √ 91 √ 27. sin(θ) = − 2 √ 28. sin(θ) = √ 5 5, cos(θ) = 4, cos(... |
�) = 2 5, sec(θ) = √ 2 4 √ 6 5, tan(θ) = − √ 6 12, csc(θ) = 5, sec(θ) = − 5 √ 12, cot(θ) = −2 6 √ 6 33. sin(θ) = − 34. sin(θ) = − √ 110 11, cos(θ) = − √ √ 95 5 10, tan(θ) = − 10, cos(θ) = √ 11 11, tan(θ) = √ √ 10, csc(θ) = − √ 95 19, sec(θ) = 2 √ 110 10, sec(θ) = − √ 19, csc(θ) = − 2 √ 11, cot(θ) = √ 5, cot(θ) = − 19 1... |
� = 50. cot(θ) = 0 when θ = π 6 π 2 51. tan(θ) = −1 when θ = + 2πk or θ = 5π 6 + 2πk for any integer k. + πk for any integer k 3π 4 + πk for any integer k 52. sec(θ) = 0 never happens 53. csc(θ) = − 1 2 never happens 54. sec(θ) = −1 when θ = π + 2πk = (2k + 1)π for any integer k 55. tan(θ) = − √ 3 when θ = 2π 3 + πk fo... |
θ = 43◦, a = 6 cot(47◦) = 6 tan(47◦) ≈ 5.595, c = 6 csc(47◦) = 6 sin(47◦) ≈ 8.204 69. β = 40◦, b = 2.5 tan(50◦) ≈ 2.979, c = 2.5 sec(50◦) = 2.5 cos(50◦) ≈ 3.889 70. The side adjacent to θ has length 4 √ 3 ≈ 6.928 71. The side opposite θ has length 10 sin(15◦) ≈ 2.588 72. The side opposite θ is 2 tan(87◦) ≈ 38.162 73. ... |
the circular functions for angles, they were also useful in simplifying expressions involving the circular functions. In this section, we introduce several collections of identities which have uses in this course and beyond. Our first set of identities is the ‘Even / Odd’ identities.1 Theorem 10.12. Even / Odd Identiti... |
) = cos(−θ0) and sin(−θ) = sin(−θ0). Let P and Q denote the points on the terminal sides of θ0 and −θ0, respectively, which lie on the Unit Circle. By definition, the coordinates of P are (cos(θ0), sin(θ0)) and the coordinates of Q are (cos(−θ0), sin(−θ0)). Since θ0 and −θ0 sweep out congruent central sectors of the Uni... |
and β0, coterminal with α and β, respectively, each of which measure between 0 and 2π radians. Since α and α0 are coterminal, as are β and β0, it follows that α − β is coterminal with α0 − β0. Consider the case below where α0 ≥ β0. y P (cos(α0), sin(α0)) α0 − β0 Q(cos(β0), sin(β0)) y 1 A(cos(α0 − β0), sin(α0 − β0)) α0... |
2(β0) = 1, so (cos(α0) − cos(β0))2 + (sin(α0) − sin(β0))2 = 2 − 2 cos(α0) cos(β0) − 2 sin(α0) sin(β0) Turning our attention to the right hand side of our equation, we find (cos(α0 − β0) − 1)2 + (sin(α0 − β0) − 0)2 = cos2(α0 − β0) − 2 cos(α0 − β0) + 1 + sin2(α0 − β0) = 1 + cos2(α0 − β0) + sin2(α0 − β0) − 2 cos(α0 − β0) O... |
ities cos(α + β) = cos(α − (−β)) = cos(α) cos(−β) + sin(α) sin(−β) = cos(α) cos(β) − sin(α) sin(β) We put these newfound identities to good use in the following example. Example 10.4.1. 1. Find the exact value of cos (15◦). 2. Verify the identity: cos π 2 − θ = sin(θ). Solution. 1. In order to use Theorem 10.13 to find ... |
= sin(θ) = cos(θ) sec csc − θ − θ π 2 π 2 = csc(θ) = sec(θ) tan cot − θ − θ π 2 π 2 = cot(θ) = tan(θ) With the Cofunction Identities in place, we are now in the position to derive the sum and difference formulas for sine. To derive the sum formula for sine, we convert to cosines using a cofunction identity, then expand... |
2 √ 2 6 − 4 = sin − √ − = = π 4 + cos + − 1 2 4π 3 √ 2 2 sin 13 2 = 1, or cos(α) = ± 12 2. In order to find sin(α − β) using Theorem 10.15, we need to find cos(α) and both cos(β) and sin(β). To find cos(α), we use the Pythagorean Identity cos2(α) + sin2(α) = 1. Since 13, we have cos2(α) + 5 sin(α) = 5 13. Since α is a Qu... |
cos(β) − sin(α) sin(β) Since tan(α) = sin(α) cos(α) and tan(β) = sin(β) denominator by cos(α) cos(β) we will have what we want cos(β), it looks as though if we divide both numerator and tan(α + β) = sin(α) cos(β) + cos(α) sin(β) cos(α) cos(β) − sin(α) sin(β) · 1 cos(α) cos(β) 1 cos(α) cos(β) sin(α) cos(β) cos(α) cos(β... |
cases for the sum ‘+’ and difference ‘−’ of angles into one formula. The convention here is that if you want the formula for the sum ‘+’ of 776 Foundations of Trigonometry two angles, you use the top sign in the formula; for the difference, ‘−’, use the bottom sign. For example, tan(α − β) = tan(α) − tan(β) 1 + tan(α) t... |
� ≤ π 2, find an expression for sin(2θ) in terms of x. 3. Verify the identity: sin(2θ) = 2 tan(θ) 1 + tan2(θ). 4. Express cos(3θ) as a polynomial in terms of cos(θ). Solution. 1. Using Theorem 10.3 from Section 10.2 with x = −3 and y = 4, we find r = x2 + y2 = 5. Hence, cos(θ) = − 3 5. Applying Theorem 10.17, we get cos(... |
) + x2 = 1, or cos(θ) = ± 2, cos(θ) ≥ 0, and thus √ 1 − x2. cos(θ) = 1 − x2. Our final answer is sin(2θ) = 2 sin(θ) cos(θ) = 2x 1 − x2. Since − π √ √ 3. We start with the right hand side of the identity and note that 1 + tan2(θ) = sec2(θ). From this point, we use the Reciprocal and Quotient Identities to rewrite tan(θ) ... |
(θ) = 2 cos3(θ) − cos(θ) − 2 sin2(θ) cos(θ) Finally, we exchange sin2(θ) for 1 − cos2(θ) courtesy of the Pythagorean Identity, and get cos(3θ) = 2 cos3(θ) − cos(θ) − 2 sin2(θ) cos(θ) = 2 cos3(θ) − cos(θ) − 2 1 − cos2(θ) cos(θ) = 2 cos3(θ) − cos(θ) − 2 cos(θ) + 2 cos3(θ) = 4 cos3(θ) − 3 cos(θ) and we are done. 778 Found... |
8 cos(4θ) cos(4θ) Another application of the Power Reduction Formulas is the Half Angle Formulas. To start, we apply the Power Reduction Formula to cos2 θ 2 cos2 θ 2 We can obtain a formula for cos θ by extracting square roots. In a similar fashion, we may obtain 2 a half angle formula for sine, and by using a quotien... |
one side of the equation and trying to transform it into the other, we will start with the identity we proved in number 3 of Example 10.4.3 and manipulate it into the identity we are asked to prove. The identity we are asked to start with is sin(2θ) = 2 tan(θ) 1+tan2(θ). If we are to use this to derive an identity for... |
research on them as your schedule allows. 10.4 Trigonometric Identities 781 Related to the Product to Sum Formulas are the Sum to Product Formulas, which we will have need of in Section 10.7. These are easily verified using the Product to Sum Formulas, and as such, their proofs are left as exercises. Theorem 10.21. Sum... |
ations of Trigonometry 10.4.1 Exercises In Exercises 1 - 6, use the Even / Odd Identities to verify the identity. Assume all quantities are defined. 1. sin(3π − 2θ) = − sin(2θ − 3π) 2. cos − π 4 − 5t = cos 5t + π 4 3. tan(−t2 + 1) = − tan(t2 − 1) 4. csc(−θ − 5) = − csc(θ + 5) 5. sec(−6t) = sec(6t) 6. cot(9 − 7θ) = − cot... |
) 24. If sin(α) = 3 5, where 0 < α < π 2, and cos(β) = 12 13 where 3π 2 < β < 2π, find (a) sin(α + β) (b) cos(α − β) (c) tan(α − β) 10.4 Trigonometric Identities 783 25. If sec(α) = − 5 3, where π 2 < α < π, and tan(β) = 24 7, where π < β < 3π 2, find (a) csc(α − β) (b) sec(α + β) (c) cot(α + β) In Exercises 26 - 38, ver... |
tan(t + h) − tan(t) h = tan(h) h sec2(t) 1 − tan(t) tan(h) In Exercises 39 - 48, use the Half Angle Formulas to find the exact value. You may have need of the Quotient, Reciprocal or Even / Odd Identities as well. 39. cos(75◦) (compare with Exercise 7) 40. sin(105◦) (compare with Exercise 9) 41. cos(67.5◦) 43. tan(112.... |
(θ) − 1 1 + tan(θ) 62. csc(2θ) = cot(θ) + tan(θ) 2 63. 8 sin4(θ) = cos(4θ) − 4 cos(2θ) + 3 64. 8 cos4(θ) = cos(4θ) + 4 cos(2θ) + 3 65. sin(3θ) = 3 sin(θ) − 4 sin3(θ) 66. sin(4θ) = 4 sin(θ) cos3(θ) − 4 sin3(θ) cos(θ) 67. 32 sin2(θ) cos4(θ) = 2 + cos(2θ) − 2 cos(4θ) − cos(6θ) 68. 32 sin4(θ) cos2(θ) = 2 − cos(2θ) − 2 cos(... |
(6θ) 78. sin(3θ) sin(2θ) 79. cos(θ) sin(3θ) In Exercises 80 - 85, write the given sum as a product. You may need to use an Even/Odd or Cofunction Identity. 80. cos(3θ) + cos(5θ) 81. sin(2θ) − sin(7θ) 82. cos(5θ) − cos(6θ) 83. sin(9θ) − sin(−θ) 84. sin(θ) + cos(θ) 85. cos(θ) − sin(θ) 86. Suppose θ is a Quadrant I angle ... |
θ) = 2 cos2(θ) − 1 = 1 − 2 sin2(θ) for all θ. 94. Let θ be a Quadrant III angle with cos(θ) = −. Show that this is not enough information to 1 5 by first assuming 3π < θ < 7π 2 and then assuming π < θ < 3π 2 determine the sign of sin θ 2 and computing sin in both cases. θ 2 786 Foundations of Trigonometry 95. Without us... |
12 = 2 + √ 3 = −(2 + √ 3) √ √ 2 6 − = 18. sin 20. csc π 12 5π 12 8. sec(165◦) = − √ 4 2 + √ 6 √ √ 6 2 − = 10. csc(195◦) = √ √ √ = −( √ 6 + √ 14. sin 11π 12 = 16. cos √ 7π 12 = √ − = 22. (a) cos(α + β) = − √ 2 10 (b) sin(α + β) = √ 2 7 10 √ 2 2 (c) tan(α + β) = −7 (d) cos(α − β) = − (e) sin(α − β) = √ 2 2 23. (a) cos(α... |
√ sin θ 2 = 50. sin(2θ) = sin θ 2 = 51. sin(2θ) = sin θ 2 = 52. sin(2θ) = − 336 625 2 10 2520 2809 √ 5 106 106 120 169 √ 3 13 13 √ 15 8 cos(2θ) = cos θ 2 = − cos(2θ) = − cos θ 2 = 9 527 625 √ 2 7 10 1241 2809 √ 106 106 119 169 √ 2 13 13 cos(2θ) = − cos θ 2 = − cos(2θ) = 7 8 sin θ 2 = √ 15 8 + 2 4 cos θ 2 = √ 15 8 − 2 ... |
2θ) = − cos(2θ) = − sin θ 2 = 50 − 10 10 √ 5 cos θ 2 = − √ 5 50 + 10 10 58. sin(2θ) = − 4 5 cos(2θ) = − 3 5 sin θ 2 = √ 5 50 + 10 10 cos θ 2 = 50 − 10 10 √ tan θ 2 = tan(2θ) = 5 tan θ 2 = tan 10 10 74. 77. cos(2θ) + cos(8θ) 2 cos(4θ) + cos(8θ) 2 80. 2 cos(4θ) cos(θ) 83. 2 cos(4θ) sin(5θ) 90. 1 − x2 2 75. 78. cos(5θ) − ... |
cos(t) for all real numbers t and sin(−t) = − sin(t) for all real numbers t. This means f (t) = cos(t) is an even function, while g(t) = sin(t) is an odd function.1 Another important property of these functions is that for coterminal angles α and β, cos(α) = cos(β) and sin(α) = sin(β). Said differently, cos(t+2πk) = co... |
multiple of 2π is 2π itself, we have the result. Similarly, we can show g(t) = sin(t) is also periodic with 2π as its period.2 Having period 2π essentially means that we can completely understand everything about the functions f (t) = cos(t) and g(t) = sin(t) by studying one interval of length 2π, say [0, 2π].3 One la... |
This allows us to turn our attention to graphing the cosine and sine functions in the Cartesian Plane. To graph y = cos(x), we make a table as we did in Section 1.6 using some of the ‘common values’ of x in the interval [0, 2π]. This generates a portion of the cosine graph, which we call the ‘fundamental cycle’ of y =... |
2 7π 4 2π sin(xx, sin(x)) (0, 0 √ 3π 4, 2 2 0 √ 2 2 −1 √ 2 2 − − 5π 7π (π, 01 3π √ 4, − 2 2 y 1 −1 π 4 π 2 3π 4 π 5π 4 3π 2 7π 4 2π x 0 (2π, 0) The ‘fundamental cycle’ of y = sin(x). As with the graph of y = cos(x), we provide an accurately scaled graph of y = sin(x) below with the fundamental cycle highlighted. y x A... |
onometric function is the expression ‘inside’ the function. Example 10.5.1. Graph one cycle of the following functions. State the period of each. 1. f (x) = 3 cos πx−π 2 + 1 Solution. 1. We set the argument of the cosine, πx−π 2 2. g(x) = 1 2 sin(π − 2x) + 3 2, equal to each of the values: 0, π 2, π, 3π 2, 2π and solve... |
g(x). One cycle was graphed on the interval − π 2, π 2 so the period is π 2 − − π 2 = π. The functions in Example 10.5.1 are examples of sinusoids. Roughly speaking, a sinusoid is the result of taking the basic graph of f (x) = cos(x) or g(x) = sin(x) and performing any of the transformations6 mentioned in Section 1.7... |
oid is exactly the same as the vertical shifts in Section 1.7. In most contexts, the vertical shift of a sinusoid is assumed to be 0, but we state the more general case below. The following theorem, which is reminiscent of Theorem 1.7 in Section 1.7, shows how to find these four fundamental quantities from the formula o... |
shift to the right 1 unit) and the vertical shift is B = 1 (indicating a shift up 1 unit.) All of these match with our graph of y = f (x). Moreover, if we start with the basic shape of the cosine graph, shift it 1 unit to the right, 1 unit up, stretch the amplitude to 3 and shrink the period to 4, we will have reconst... |
the Theorem 10.23, we get one complete cycle of the graph, which means we have completely determined the sinusoid. Example 10.5.2. Below is the graph of one complete cycle of a sinusoid y = f (x). −1, 5 2 y 3 2 1 51 1 2 3 4 5 x −1 −2 2, − 3 2 One cycle of y = f (x). 1. Find a cosine function whose graph matches the gr... |
, 1 2 13 2, 1 2 19 2, 5 2 −1 −2 8, − 3 2 Extending the graph of y = f (x). Note that each of the answers given in Example 10.5.2 is one choice out of many possible answers. taking For example, when fitting a sine function to the data, we could have chosen to start at 1 + 1 6 for an answer of S(x) = −2 sin π A = −2. In t... |
x) − C(x) = A cos(ωx + φ) + B, we get √ √ cos(2x) − 3 sin(2x) = A cos(ωx) cos(φ) − A sin(ωx) sin(φ) + B It should be clear that we can take ω = 2 and B = 0 to get √ cos(2x) − 3 sin(2x) = A cos(2x) cos(φ) − A sin(2x) sin(φ) To determine A and φ, a bit more work is involved. We get started by equating the coefficients of t... |
) cos π 3 cos(2x) 1 − sin(2x) 2 √ 3 sin(2x) = 2 = cos(2x) − − sin(2x) sin π 3 √ 3 2 8This should remind you of equation coefficients of like powers of x in Section 8.6. 10.5 Graphs of the Trigonometric Functions 799 2. Proceeding as before, we equate f (x) = cos(2x) − S(x) = A sin(ωx + φ) + B to get √ 3 sin(2x) with the ... |
5.3 to fit a function into one of the forms in Theorem 10.23, the arguments of the cosine and sine function much match. 3 sin(3x) is not.10 It That is, while f (x) = cos(2x) − is also worth mentioning that, had we chosen A = −2 instead of A = 2 as we worked through Example 10.5.3, our final answers would have looked diffe... |
is periodic with period 2π, it follows that sec(x) is also.11 Below we graph a fundamental cycle of y = sec(x) along with a more complete graph obtained by the usual ‘copying and pasting.’12 +, sec(x) → −∞; as x → 3π 2 2 and x = 3π − + y x 0 π 4 π 2 3π 4 π 5π 4 3π 2 7π 4 2π √ 2 − − √ √ sec(x) 1 cos(x) 1 √ 2 2 0 undefin... |
x) and y = sec(x). y x 0 π 4 π 2 3π 4 π 5π 4 3π 2 7π 4 2π sin(x) √ 1 2 √ √ csc(x) 0 undefined 2 2 1 √ 2 2 0 undefined 2 2 −1 √ 2 2 0 undefined −x, csc(x)) √ 3π 4, 2 5π 2 7π √ 2 4, − 2, −1 3π √ 2 4, − 3 2 1 −1 −2 −3 π 4 π 2 3π 4 π 5π 4 3π 2 7π 4 2π x Once again, our domain and range work in Section 10.3.1 is verified geomet... |
) ∞ k=−∞ (kπ, (k + 1)π) – has domain {x : x = πk, k is an integer} = – has range {y : |y| ≥ 1} = (−∞, −1] ∪ [1, ∞) – is continuous and smooth on its domain – is odd – has period 2π In the next example, we discuss graphing more general secant and cosecant curves. Example 10.5.4. Graph one cycle of the following function... |
the quarter marks and solve for x. One cycle of y = 1 − 2 sec(2x). a π − πx = a x π − πx = 0 1 0 π − πx = π 1 π 2 2 2 π − πx = π 0 π 2 − 1 π − πx = 3π 3π 2 2 2π π − πx = 2π −1 Substituting these x-values into g(x), we generate the graph below and find the period to be 1 − (−1) = 2. The associated sine curve, y = sin(π−... |
and paste’ produces: cos(x) → ∞ producing a vertical asymptote at x = π +, tan(x) → −∞; as x → 3π 2, tan(x) → ∞; and as x → 3π 3π 4 π 5π 4 3π 2 7π 4 2π tan(x) 0 1 (x, tan(x)) (0, 0) 4, 1 π undefined −1 0 1 undefined −1 0 4, −1 3π (π, 0) 4, 1 5π 4, −1 7π (2π, 0) 1 −1 π 4 π 2 3π 4 π 5π 4 3π 2 7π 4 2π x The graph of y = ta... |
). Plotting cot(x) over the interval [0, 2π] results in the graph below. y 1 −1 x 0 π 4 π 2 3π 4 π 5π 4 3π 2 7π 4 2π cot(x) undefined (x, cot(x)) 1 0 −1 undefined 1 0 −1 undefined 1 3π 4, 1 5π 3π 2, 0 4, −1 7π π 4 π 2 3π 4 π 5π 4 3π 2 7π 4 2π x The graph of y = cot(x) over [0, 2π]. From these data, it clearly appears as i... |
(x) – has domain {x : x = πk, k is an integer} = – has range (−∞, ∞) – is continuous and smooth on its domain – is odd – has period π ∞ k=−∞ (kπ, (k + 1)π) 10.5 Graphs of the Trigonometric Functions 807 Example 10.5.5. Graph one cycle of the following functions. Find the period. 1. f (x) = 1 − tan x 2. 2. g(x) = 2 cot ... |
ot π 2 x + π + 1. We find the period to be 0 − (−2) = 2. As with the secant and cosecant functions, it is possible to extend the notion of period, phase shift and vertical shift to the tangent and cotangent functions as we did for the cosine and sine functions in Theorem 10.23. Since the number of classical applications... |
that the function is a sinusoid by rewriting it in the forms C(x) = A cos(ωx + φ) + B and S(x) = A sin(ωx + φ) + B for ω > 0 and 0 ≤ φ < 2π. 25. f (x) = √ 2 sin(x) + √ 2 cos(x) + 1 26. f (x) = 3 √ 3 sin(3x) − 3 cos(3x) 27. f (x) = − sin(x) + cos(x) − 2 28. f (x) = − 1 2 sin(2x) − √ 3 2 cos(2x) 29. f (x) = 2 √ 3 cos(x)... |
10.5.4 and 10.5.5 using extended interval notation. (We will revisit this in Section 10.7.) In Exercises 38 - 43, verify the identity by graphing the right and left hand sides on a calculator. 38. sin2(x) + cos2(x) = 1 39. sec2(x) − tan2(x) = 1 40. cos(x) = sin 41. tan(x + π) = tan(x) 42. sin(2x) = 2 sin(x) cos(x) 43.... |
�nition of ‘period’. 10.5 Graphs of the Trigonometric Functions 811 10.5.5 Answers 1. y = 3 sin(x) Period: 2π Amplitude: 3 Phase Shift: 0 Vertical Shift: 0 2. y = sin(3x) 2π 3 Period: Amplitude: 1 Phase Shift: 0 Vertical Shift: 0 3. y = −2 cos(x) Period: 2π Amplitude: 2 Phase Shift: 0 Vertical Shift: 0 4. y = cos x − π... |
3 π 8 4x − cos y = Vertical Shift: 1 Phase Shift: 2 3 (You need to use π 2 + 1 to find this.)16 − 9π 4 − 7π 4 − 5π 4 − 3π 4 y π 4 3π 4 5π 4 x 7π 4 − π 4 −1 −2 − 3π 8 π 2 5π 8 x − 3π 8 − π 4 − π 8 11. y = − cos 2x + 3 2 π 3 − 1 2 y 1 Period: π Amplitude: 3 2 Phase Shift: − π 6 Vertical Shift: − 1 2 12. y = 4 sin(−2πx + ... |
x 5π 3 18. y = − 1 3 sec 1 2 x + Start with y = − Period: 4π 1 3 π 3 cos 1 2 x + π 3 y 1 3 − 2π 3 − 1 3 π 3 4π 3 7π 3 x 10π 3 816 Foundations of Trigonometry 19. y = csc(2x − π) Start with y = sin(2x − π) Period: π 20. y = sec(3x − 2π) + 4 Start with y = cos(3x − 2π) + 4 Period: 2π 3 21. y = csc −x − π 4 Start with y ... |
= 2 √ 3 cos(x) − 2 sin(x) = 4 sin x + 30. f (x) = 3 2 cos(2x) − 31. f (x) = − 1 2 cos(5x) − 3 2 √ 3 3 2 √ 2π 3 = 4 cos x + π 6 5π 6 + 6 = 3 cos = cos 5x + 2π 3 sin(2x) + 6 = 3 sin 2x + sin(5x) = sin 5x + 7π 6 2x + π 3 + 6 32. f (x) = −6 √ 3 cos(3x) − 6 sin(3x) − 3 = 12 sin 3x + 4π 3 − 3 = 12 cos 3x + 5π 6 − 3 33. f (x... |
convention of writing (cos(x))2 as cos2(x), (cos(x))3 as cos3(x) and so on. It is far too easy to confuse cos−1(x) with cos(x) = sec(x) so we will not use this notation in our text.1 Instead, we use the notation f −1(x) = arccos(x), read ‘arc-cosine of x’. To understand the ‘arc’ in ‘arccosine’, recall that an inverse... |
) to − π 2, π 2. It should be no surprise that we call g−1(x) = arcsin(x), which is read ‘arc-sine of x’. y 1 − π 2 x π 2 −1 g(x) = sin(x), − 1 1 x reflect across y = x −−−−−−−−−−−−→ switch x and y coordinates − π 2 g−1(x) = arcsin(x). We list some important facts about the arccosine and arcsine functions in the followi... |
cos 11π 6 (h) sin arccos − 3 5 2. Rewrite the following as algebraic expressions of x and state the domain on which the equiv- alence is valid. (a) tan (arccos (x)) (b) cos (2 arcsin(x)) Solution. 1. (a) To find arccos 1 2, we need to find the real number t (or, equivalently, an angle measuring 3 meets these 2. We know ... |
to use Theorem 10.26 directly. Since − the previous problem, we know arccos (g) One way to simplify cos arccos − 3 5 between −1 and 1, we have that cos arccos − 3 5 before, to really understand why this cancellation occurs, we let t = arccos − 3 5 by definition, cos(t) = − 3 in (nearly) the same amount of time. 5 is 5 ... |
t) + sin2(t) = 1 gives x2 + sin2(t) = 1, from which we 1 − x2. Since t corresponds to angles in Quadrants I and II, sin(t) ≥ 0, get sin(t) = ± so we choose sin(t) = 2 from consideration. Hence, either 0 ≤ t < π 1 − x2. Thus, 2 or π √ √ tan(t) = sin(t) cos(t) = √ 1 − x2 x To determine the values of x for which this equi... |
x). Since arcsin(x) is defined only for −1 ≤ x ≤ 1, the equivalence cos (2 arcsin(x)) = 1−2x2 is valid only on [−1, 1]. A few remarks about Example 10.6.1 are in order. Most of the common errors encountered in dealing with the inverse circular functions come from the need to restrict the domains of the original function... |
its fundamental cycle on (0, π) to obtain g−1(x) = arccot(x). Once again, the vertical asymptotes x = 0 and x = π of the graph of g(x) = cot(x) become the horizontal asymptotes y = 0 and y = π of the graph of g−1(x) = arccot(x). We show these graphs on the next page and list some of the basic properties of the arctang... |
for x > 0 – arccot(cot(x)) = x provided 0 < x < π 10.6 The Inverse Trigonometric Functions 825 Example 10.6.2. 1. Find the exact values of the following. √ (a) arctan( 3) (c) cot(arccot(−5)) √ 3) (b) arccot(− (d) sin arctan − 3 4 2. Rewrite the following as algebraic expressions of x and state the domain on which the ... |
sc2(t), since this relates the reciprocals of tan(t) and sin(t) and is valid for all t under consideration.4 From tan(t) = − 3 4, we get cot(t) = − 4 3. Since = − 3 − π 5. = csc2(t) so that csc(t) = ± 5 5. Hence, sin arctan − 3 2 and tan(t) = x. We look for a way to express tan(2 arctan(x)) = tan(2t) in terms of x. Bef... |
= ±1. Hence, the equivalence tan(2 arctan(x)) = 2x (b) To get started, we let t = arccot(2x) so that cot(t) = 2x where 0 < t < π. In terms of t, cos(arccot(2x)) = cos(t), and our goal is to express the latter in terms of x. Since cos(t) is always defined, there are no additional restrictions on t, so we can begin using... |
graph which covers its entire range of (−∞, −1] ∪ [1, ∞) and restricts the domain of the function so that it is one-to-one. The same is true for cosecant. Thus in order to define the arcsecant and arccosecant functions, we must settle for a piecewise approach wherein we choose one piece to cover the top of the range, n... |
so we include it here for completeness. Using these definitions, we get the following properties of the arcsecant and arccosecant functions. 828 Foundations of Trigonometry Theorem 10.28. Properties of the Arcsecant and Arccosecant Functionsa Properties of F (x) = arcsec(x) ∪ π – Domain: {x : |x| ≥ 1} = (−∞, −1] ∪ [1, ... |
algebraic expressions of x and state the domain on which the equiv- alence is valid. (a) tan(arcsec(x)) (b) cos(arccsc(4x)) 10.6 The Inverse Trigonometric Functions 829 Solution. 1. 2. (a) Using Theorem 10.28, we have arcsec(2) = arccos 1 2 (b) Once again, Theorem 10.28 comes to our aid giving arccsc(−2) = arcsin − 1 ... |
and when we substitute sec(t) = x, we get 1 + tan2(t) = x2. Hence, tan(t) = ± x2 − 1. 2, π then then tan(t) ≥ 0; if, on the the other hand, t belongs to π If t belongs to 0, π 2 tan(t) ≤ 0. As a result, we get a piecewise defined function for tan(t) tan(t) = √ √ − x2 − 1, x2 − 1, if 0 ≤ t < π 2 if π 2 < t ≤ π Now we ne... |
−x2 4|x| 2 √ 830 Foundations of Trigonometry which this equivalence is valid, we look back at our original substution, t = arccsc(4x). Since the domain of arccsc(x) requires its argument x to satisfy |x| ≥ 1, the domain of arccsc(4x) requires |4x| ≥ 1. Using Theorem 2.4, we rewrite this inequality and solve to get x ≤ ... |
�) – Range: 0, π 2 – as x → −∞, arcsec(x) → 3π 2 – arcsec(x) = t if and only if 0 ≤ t < π – arcsec(x) = arccos 1 for x ≥ 1 onlyb x – sec (arcsec(x)) = x provided |x| ≥ 1 – arcsec(sec(x)) = x provided 0 ≤ x < π − 2 or π ≤ x < 3π 2 ; as x → ∞, arcsec(x) → π 2 2 or π ≤ t < 3π − 2 and sec(t) = x Properties of G(x) = arccsc... |
)) 832 Solution. Foundations of Trigonometry 1. 2. (a) Since 2 ≥ 1, we may invoke Theorem 10.29 to get arcsec(2) = arccos 1 2 (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. and satisfies ... |
t) = sec2(t). This is valid for all values of t under consideration, and when we substitute sec(t) = x, we get 1 + tan2(t) = x2. Hence, tan(t) = ± x2 − 1. Since t lies in 0, π, tan(t) ≥ 0, so we choose tan(t) = x2 − 1. Since we found √ 2 x2 − 1 holds for all x no additional restrictions on t, the equivalence tan(arcsec... |
| 16x2 − 1 4|x|,, if 0 ≤ t ≤ π 2 if π < t ≤ 3π 2 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) = 16x2 − 1 4|x| = 16x2 − 1 4x Similarly, for π < t ... |
3 arctan (4x). (c) f (x) = arccot x 2 + π Solution. 1. (a) Since 2 > 0, we can use the property listed in Theorem 10.27 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 ... |
) < 0, we know 2 more specifically that − π 2 < t < 0, so t corresponds to an angle β in Quadrant IV. To find the value of arccot(−2), we once again visualize the angle θ = arccot(−2) radians and note that it is a Quadrant II angle with tan(θ) = − 1 2. This means it is exactly π units away from β, and we get θ = π + β = ... |
the domain of F (x) = arccos(x) is −1 ≤ x ≤ 1, we can find the domain of 2 − arccos x f (x) = π 5, between −1 and 1. Solving −1 ≤ x 5 ≤ 1 gives −5 ≤ x ≤ 5, so the domain is [−5, 5]. To determine the range of f, we take a cue from Section 1.7. Three ‘key’ points on the graph of and (1, 0). Following the procedure outlin... |
ctan (4x) π 2 x 5 (c) To find the domain of g(x) = arccot x + π, we proceed as above. Since the domain of 2 G(x) = arccot(x) is (−∞, ∞), and x 2 is defined for all x, we get that the domain of g is (−∞, ∞) as well. As for the range, we note that the range of G(x) = arccot(x), like that of F (x) = arctan(x), is limited by... |
involving arctangent will produce this result.6 Hence, in order to graph y = g(x) on our calculators, we need to write it as a piecewise defined function: + π = arctan 2 x g(x) = arccot x 2 + π = We show the input and the result below. arctan 2 x arctan 2 x + 2π, π, + π, if x < 0 if x = 0 if x > 0 y = g(x) in t... |
the fact that the answers corresponded to a set of ‘common angles’ listed on page 724. If, on the other hand, we had been asked to find all angles with sin(θ) = 1 3 or solve tan(t) = −2 for real numbers t, we would have been hard-pressed to do so. With the introduction of the inverse trigonometric functions, however, w... |
for all of the solutions to this equation in Quadrant II = arcsin 1 3 radians α 1 x Since 1 3 isn’t the sine of any of the ‘common angles’ discussed earlier, we use the arcsine functions to express our answers. The real number t = arcsin 1 is defined so it satisfies 3 radians. Since the solutions in Quadrant I 0 < t < π... |
, sec(x) = − 5 3, poses two immediate problems. First, we are not told whether or not x represents an angle or a real number. We assume the latter, but note that we will use angles and the Unit Circle to solve the equation regardless. Second, as we have mentioned, there is no universally accepted range of the arcsecant... |
41 10.6.5 Exercises In Exercises 1 - 40, find the exact value. 1. arcsin (−1) 2. arcsin − √ 3 2 5. arcsin (0) 6. arcsin 1 2 3. arcsin − √ 2 2 7. arcsin √ 2 2 4. arcsin − 1 2 8. arcsin √ 3 2 9. arcsin (1) 10. arccos (−1) 11. arccos − √ 3 2 12. arccos − √ 2 2 13. arccos − 1 2 17. arccos √ 3 2 21. arctan − √ 3 3 14. arccos... |
0, π 2 arccosecant is − π when finding the exact value. 2, 0 ∪ 0, π 2 ∪ π 2, π and that the range of 49. arcsec (−2) 53. arccsc (−2) 50. arcsec − √ 54. arccsc − √ 2 2 51. arcsec − 55. arccsc − √ 2 3 3 √ 2 3 3 52. arcsec (−1) 56. arccsc (−1) In Exercises 57 - 86, find the exact value or state that it is undefined. 57. sin... |
arccos 96. arccos cos cos 11π 6 2π 3 5π 4 91. arcsin sin 94. arccos 97. arctan 4π 3 3π 2 cos 92. arccos 95. arccos cos π 4 cos − π 6 tan tan π 3 π 2 98. arctan tan 101. arctan tan − π 4 2π 3 99. arctan (tan (π)) 100. arctan 102. arccot 105. arccot cot cot π 3 π 2 103. arccot cot 106. arccot cot − π 4 2π 3 104. arccot ... |
ercises 131 - 154, find the exact value or state that it is undefined. 131. sin arccos − 1 2 132. sin arccos 3 5 134. sin arccot √ 5 137. cos arctan √ 7 135. sin (arccsc (−3)) 140. tan arcsin − √ 2 5 5 141. tan arccos − 1 2 143. tan (arccot (12)) 144. cot arcsin 12 13 138. cos (arccot (3)) 139. cos (arcsec (5)) 133. sin ... |
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