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cos π 2 − 4x + 1 11. y = − cos 2x + 3 2 π 3 − 1 2 12. y = 4 sin(−2πx + π) In Exercises 13 - 24, graph one cycle of the given function. State the period of the function. 13. y = tan 16. y = sec x − x − π 3 π 2 14. y = 2 tan 17. y = − csc x − 3 1 4 x + π 3 15. y = 1 3 18. y = − 19. y = csc(2x − π) 20. y = sec(3x − 2π) +... |
34. f (x) = 3 sin x 6 √ − 3 3 cos x 6 35. In Exercises 25 - 34, you should have noticed a relationship between the phases φ for the S(x) and C(x). Show that if f (x) = A sin(ωx + α) + B, then f (x) = A cos(ωx + β) + B where β = α −. π 2 36. Let φ be an angle measured in radians and let P (a, b) be a point on the termi... |
(x). Graph y = x on the same set of axes and describe the behavior of f. 49. f (x) = e−0.1x (cos(2x) + sin(2x)). Graph y = ±e−0.1x on the same set of axes and describe the behavior of f. 50. f (x) = e−0.1x (cos(2x) + 2 sin(x)). Graph y = ±e−0.1x on the same set of axes and describe the behavior of f. 51. Show that a co... |
2 7. y = − 1 3 cos 1 2 x + π 3 Period: 4π Amplitude: 1 3 2π Phase Shift: − 3 Vertical Shift: 0 8. y = cos(3x − 2π) + 4 2π 3 Period: Amplitude: 1 2π Phase Shift: 3 Vertical Shift: 4 −1 y 1 −1 y 1 3 π 2 3π 4 π 5π 4 3π 2 x − 2π 3 π 3 4π 3 7π 3 x 10π 3 − 1 3 y 5 4 3 2π 3 5π 6 π 7π 6 4π 3 x 10.5 Graphs of the Trigonometric... |
= tan x − Period: π π 3 14. y = 2 tan x − 3 1 4 Period: 4π y 1 − π 6 −1 π 12 π 3 7π 12 5π 6 x −2π −π y −1 −3 −5 π 2π x 15. y = tan(−2x − π) + 1 1 3 is equivalent to 1 3 tan(2x + π) + 1 y = − via the Even / Odd identity for tangent. Period − 3π 8 − π 4 x − 3π 4 − 5π 8 10.5 Graphs of the Trigonometric Functions 815 16. ... |
π 2 15π 4 5π x 24. y = 1 3 Period: cot π 2 2x + 3π − 3π 8 − π 4 x − 3π 4 − 5π 8 818 Foundations of Trigonometry 25. f (x) = √ 2 sin(x) + √ 2 cos(x) + 1 = 2 sin x + π 4 + 1 = 2 cos x + 7π 4 + 1 26. f (x) = 3 √ 3 sin(3x) − 3 cos(3x) = 6 sin 3x + 11π 6 = 6 cos 3x + 4π 3 27. f (x) = − sin(x) + cos(x) − 2 = √ 2 sin x + 3π 4... |
nding inverses of the (circular) trigonometric functions. Our immediate problem is that, owing to their periodic nature, none of the six circular functions is one-to-one. To remedy this, we restrict the domains of the circular functions in the same way we restricted the domain of the quadratic function in Example 5.2.3... |
‘arc’ in arccosine. Below are the graphs of f (x) = cos(x) and f −1(x) = arccos(x), where we obtain the latter from the former by reflecting it across the line y = x, in accordance with Theorem 5.3. y 1 −x) = cos(x), 0 ≤ x ≤ π reflect across y = x −−−−−−−−−−−−→ switch x and y coordinates −1 1 x f −1(x) = arccos(x). 1But... |
arcsin(x)) = x provided − – arcsin(sin(x)) = x provided − π – additionally, arcsine is odd 2 2 and sin(t) = x 10.6 The Inverse Trigonometric Functions 821 Everything in Theorem 10.26 is a direct consequence of the facts that f (x) = cos(x) for 0 ≤ x ≤ π and F (x) = arccos(x) are inverses of each other as are g(x) = sin... |
number t in the interval −, we could simply invoke Theorem 10.26 to get arccos cos π 6. However, in order to make sure we understand why this is the case, we choose to work the example through using the definition of arccosine. Working from the inside out, arccos cos π is the real number t with 0 ≤ t ≤ π 6 with sin(t) ... |
(t) = 1 or sin(t) = ± 4 5. Hence, sin arccos − 3 so that cos(ta) We begin this problem in the same manner we began the previous two problems. To help us see the forest for the trees, we let t = arccos(x), so our goal is to find a way to express tan (arccos (x)) = tan(t) in terms of x. Since t = arccos(x), we know cos(t)... |
− π of x. Since cos(2t) is defined everywhere, we get no additional restrictions on t as we did in the previous problem. We have three choices for rewriting cos(2t): cos2(t) − sin2(t), 2 cos2(t) − 1 and 1 − 2 sin2(t). Since we know x = sin(t), it is easiest to use the last form: 2 cos (2 arcsin(x)) = cos(2t) = 1 − 2 si... |
the inverses of tangent and cotangent, which are named arctangent and arccotangent, respectively. First, we restrict f (x) = tan(x) to its fundamental cycle on − π to obtain f −1(x) = arctan(x). Among other things, note that the 2, π 2 vertical asymptotes x = − π 2 and x = π 2 of the graph of f (x) = tan(x) become the... |
2 – arctan(x) = t if and only if − π for x > 0 – arctan(x) = arccot 1 x – tan (arctan(x)) = x for all real numbers x 2 < x < π – arctan(tan(x)) = x provided − π – additionally, arctangent is odd 2 +; as x → ∞, arctan(x and tan(t) = x − Properties of G(x) = arccot(x) – Domain: (−∞, ∞) – Range: (0, π) – as x → −∞, arcco... |
ot(arccot(−5)) = −5. However, working it through provides us with yet another opportunity to understand why this is the case. Letting t = arccot(−5), we have that t belongs to the interval (0, π) and cot(t) = −5. Hence, cot(arccot(−5)) = cot(t) = −5. (a) If we let t = arctan(x), then − π (d) We start simplifying sin ar... |
get tan(2 arctan(x)) = tan(2t) = 2 tan(t) 1 − tan2(t) = 2x 1 − x2 4It’s always a good idea to make sure the identities used in these situations are valid for all values t under consideration. Check our work back in Example 10.6.1. Were the identities we used there valid for all t under consideration? A pedantic point,... |
�, csc(t) > 0, so csc(t) = 4x2 + 1 which gives sin(t) = 1√ √ √ 4x2+1 cos(arccot(2x)) = cos(t) = cot(t) sin(t) = √ 2x 4x2 + 1 Since arccot(2x) is defined for all real numbers x and we encountered no additional restrictions on t, we have cos (arccot(2x)) = 2x√ for all real numbers x. 4x2+1 The last two functions to invert... |
π reflect across y = x −−−−−−−−−−−−→ switch x and y coordinates −1 1 x f −1(x) = arcsec(x) and we restrict g(x) = csc(x) to − π 2, 0 ∪ 0, π 2. y 1 −1 − π 2 x π 2 g(x) = csc(x) on − π 2, 0 ∪ 0, π 2 y π 2 −1 1 x reflect across y = x −−−−−−−−−−−−→ switch x and y coordinates − π 2 g−1(x) = arccsc(x) Note that for both arcse... |
, ∞) – Range: − π – as x → −∞, arccsc(x) → 0−; as x → ∞, arccsc(x) → 0+ – arccsc(x) = t if and only if − π – arccsc(x) = arcsin 1 x – csc (arccsc(x)) = x provided |x| ≥ 1 – arccsc(csc(x)) = x provided − π – additionally, arccosecant is odd 2 ≤ x < 0 or or 0 < t ≤ π provided |x| ≥ 1 2 2 and csc(t) = x a... assuming the ... |
t), so we use the Pythagorean Identity 1 + cot2(t) = csc2(t). Substituting, we have 1 + cot2(t) = (−3)2, or cot(t) = ± 2 ≤ t < 0, cot(t) < 0, so we get cot (arccsc (−3)) = −2 2. Since − π 8 = ±2 √ √ √ 2. ∪ π (a) We begin simplifying tan(arcsec(x)) by letting t = arcsec(x). Then, sec(t) = x for t in 0, π 2, π, and we se... |
for t in, and we now set about finding an expression for cos(arccsc(4x)) = cos(t). − π Since cos(t) is defined for all t, we do not encounter any additional restrictions on t. From csc(t) = 4x, we get sin(t) = 1 4x, so to find cos(t), we can make use if the identity cos2(t) + sin2(t) = 1. Substituting sin(t) = 1 = 1. Sol... |
−1 1 x f −1(x) = arcsec(x) and we restrict g(x) = csc(x) to 0, π 2 ∪ π, 3π 2. y 1 −1 π 2 π 3π 2 x y 3π 2 π π 2 g(x) = csc(x) on 0, π 2 ∪ π, 3π 2 reflect across y = x −−−−−−−−−−−−→ switch x and y coordinates −1 1 x g−1(x) = arccsc(x) Using these definitions, we get the following result. 10.6 The Inverse Trigonometric Fun... |
| ≥ 1 – arccsc(csc(x)) = x provided 0 < x ≤ π 2 or π < t ≤ 3π 2 or π < x ≤ 3π 2 2 and csc(t) = x a... assuming the “Calculus Friendly” ranges are used. bCompare this with the similar result in Theorem 10.28. cCompare this with the similar result in Theorem 10.28. Our next example is a duplicate of Example 10.6.3. The i... |
. Using the identity 2. Since √ 2. We know csc(t) = −3, and since this is negative, t lies in π, 3π 2 1 + cot2(t) = csc2(t), we find 1 + cot2(t) = (−3)2 so that cot(t) = ± t is in the interval π, 3π 2, we know cot(t) > 0. Our answer is cot (arccsc (−3)) = 2 8 = ±2 √ √ ∪ π, 3π 2 (a) We begin simplifying tan(arcsec(x)) by... |
��nd cos(t), we can make use if the identity cos2(t) + sin2(t) = 1. Substituting sin(t) = 1 = 1. Solving, we get 2 cos(t) = ± 4x gives cos2(t) + 1 √ 4x 16x2 − 1 16x2 = ± 16x2 − 1 4|x|, then cos(t) ≥ 0, and we choose cos(t) = in which case cos(t) ≤ 0, so, we choose cos(t) = − If t lies in 0, π 2 to π, 3π 2 (momentarily)... |
. In the sections to come, we will have need to approximate the values of the inverse circular functions. On most calculators, only the arcsine, arccosine and arctangent functions are available and they are usually labeled as sin−1, cos−1 and tan−1, respectively. If we are asked to approximate these values, it is a sim... |
Consider α, the reference angle for θ, as pictured below. By definition, α is an acute angle so 0 < α < π 2, and the Reference Angle Theorem, Theorem 10.2, tells us that cot(α) = 2. This means α = arccot(2) radians. Since the argument of arccotangent is now a positive 2, we can use Theorem 10.27 to get α = arccot(2) = ... |
letting 2. Then t is a real number with csc(t) = − 3 t = arccsc − 3 2. Since csc(t) < 0, we have 2 that π < θ ≤ 3π 2, so t corresponds to a Quadrant III angle, θ. As above, we let α be 2, which means α = arccsc 3 the reference angle for θ. Then 0 < α < π 2 radians. Since the argument of arccosecant is now positive, we... |
argument of the arctangent, in this case, 4x. Since 4x is defined for all real numbers, we have established that the domain of f is all real numbers. To determine the range of f, we can, once again, appeal to Theorem 1.7. Choosing our ‘key’ point to be (0, 0) and tracking the horizontal asymptotes y = − π 2 and y = π 2... |
a cue from number 1c, we attempt to rewrite +π in terms of the arctangent function. Using Theorem 10.27, we have g(x) = arccot x 2 when x = arctan 2 that arccot x 2 > 0, or, in this case, when x > 0. Hence, for x > 0, x 2 + π. When x we have g(x) = arctan 2 2 < 0, we can use the same argument in number x. 1c that gave... |
down the middle, we find that the roof line forms the hypotenuse of a right triangle with legs of length 6 feet and 12 feet. Using Theorem 10.10, we 6Without Calculus, of course... 7The authors would like to thank Dan Stitz for this problem and associated graphics. 838 Foundations of Trigonometry find the angle of incli... |
the following example. √ √ Example 10.6.7. Solve the following equations. 1. Find all angles θ for which sin(θ) = 1 3. 2. Find all real numbers t for which tan(t) = −2 3. Solve sec(x) = − 5 3 for x. Solution. 1. If sin(θ) = 1 3, then the terminal side of θ, when plotted in standard position, intersects the Unit Circle... |
the ‘common angles’, we need to use the arctangent function to express our answers. The real number t = arctan(−2) satisfies tan(t) = −2 and − π 2 < t < 0. If we let β = arctan(−2) radians, we see that all of the Quadrant IV solutions 840 Foundations of Trigonometry to tan(θ) = −2 are coterminal with β. Moreover, the s... |
cos(θ) = − 3 + 2πk for integers k. Switching back to the variable x, we θ = −β + 2πk = − arccos − 3 5 + 2πk record our final answer to sec(x) = − 5 for integers k. + 2πk or x = − arccos − 3 5 2 < t < π with cos(t) = − 3 3 as x = arccos − 3 is defined so that π 5 5 5 y 1 y 1 β = arccos − 3 5 radians β = arccos − 3 5 radi... |
. arctan (1) 27. arccot (−1) 28. arccot − √ 3 3 31. arccot (1) 32. arccot √ 3 35. arcsec √ 2 36. arccsc √ 2 39. arcsec (1) 40. arccsc (1) In Exercises 41 - 48, assume that the range of arcsecant is 0, π 2 arccosecant is 0, π 2 when finding the exact value. ∪ π, 3π 2 ∪ π, 3π 2 and that the range of 41. arcsec (−2) 45. ar... |
72. cot (arccot (1)) 73. cot arccot − √ 3 75. cot (arccot (−0.001)) 76. cot arccot 78. sec (arcsec (−1)) 79. sec arcsec 17π 4 1 2 82. csc arccsc √ 2 81. sec (arcsec (117π)) 84. csc arccsc √ 2 2 74. cot arccot − 7 24 77. sec (arcsec (2)) 80. sec (arcsec (0.75)) 83. csc arccsc − √ 2 3 3 85. csc (arccsc (1.0001)) 86. csc... |
ccsc csc 116. arccsc csc − π 2 5π 4 11π 6 sec π 4 119. arcsec 122. arcsec sec 125. arccsc csc 128. arccsc csc − π 2 5π 4 11π 6 108. arcsec sec 111. arcsec sec 114. arccsc csc 120. arcsec sec 123. arcsec sec 126. arccsc csc 4π 3 5π 3 2π 3 11π 12 4π 3 5π 3 2π 3 11π 12 ∪ π, 3π 2 and that the range of 109. arcsec sec 5π 6 ... |
in 5 13 + π 4 156. cos (arcsec(3) + arctan(2)) 157. tan arctan(3) + arccos − 3 5 158. sin 2 arcsin − 4 5 159. sin 2arccsc 13 5 161. cos 2 arcsin 3 5 163. cos 2arccot − √ 5 160. sin (2 arctan (2)) 162. cos 2arcsec 25 7 164. sin arctan(2) 2 10.6 The Inverse Trigonometric Functions 845 In Exercises 165 - 184, rewrite the ... |
ercises 188 - 207, solve the equation using the techniques discussed in Example 10.6.7 then approximate the solutions which lie in the interval [0, 2π) to four decimal places. 188. sin(x) = 7 11 189. cos(x) = − 2 9 190. sin(x) = −0.569 191. cos(x) = 0.117 192. sin(x) = 0.008 193. cos(x) = 194. tan(x) = 117 195. cot(x) ... |
a right triangle under the shelf to support it. What angle, to the nearest degree, will the rod make with the wall? 214. A parasailor is being pulled by a boat on Lake Ippizuti. The cable is 300 feet long and the parasailor is 100 feet above the surface of the water. What is the angle of elevation from the boat to the... |
f (x) = arctan(4x) 227. f (x) = arccot 2x x2 − 9 228. f (x) = arctan(ln(2x − 1)) 229. f (x) = arccot( √ 2x − 1) 230. f (x) = arcsec(12x) 231. f (x) = arccsc(x + 5) 232. f (x) = arcsec x3 8 233. f (x) = arccsc e2x 234. Show that arcsec(x) = arccos of f (x) = arcsec(x). 1 x for |x| ≥ 1 as long as we use 0, ∪ π 2, π π 2 ... |
arctan(1) + arctan(2) + arctan(3) = π. 10.6 The Inverse Trigonometric Functions 849 10.6.6 Answers 1. arcsin (−1) = − π 2 2. arcsin − √ 3 2 = − π 3 3. arcsin − √ 2 2 = − π 4 4. arcsin 7. arcsin −. arcsin (0) = 0 8. arcsin √ 3 2 10. arccos (−1) = π 11. arccos − √ 3 2 = π 3 = 5π 6 13. arccos 16. arccos = 2π 14. arccos (... |
(−1) = π 45. arccsc (−2) = 47. arccsc − √ 3 2 3 = 4π 3 48. arccsc (−1) = 7π 6 3π 2 50. arcsec − 2 = √ 51. arcsec 2 − √ 3 3 = 5π 6 54. arccsc − √ 2 = − π 4 3π 4 π 6 π 2 52. arcsec (−1) = π 53. arccsc (−2) = − 55. arccsc − √ 3 2 = − π 3 56. arccsc (−1) = − 43. arcsec − √ 3 2 3 = 7π 6 46. arccsc − √ 2 = 5π 4 49. arcsec (... |
arccsc 2 − √ 3 3 √ 3 2 3 = − 85. csc (arccsc (1.0001)) = 1.0001 87. arcsin sin 89. arcsin sin 91. arcsin sin 93. arccos cos π 6 3π 4 4π 3 2π 3 95. arccos − cos π 6 = π 4 = − π 3 = 2π 3 = π 6 = π 6 97. arctan tan π 3 = π 3 99. arctan (tan (π)) = 0 101. arctan tan 2π 3 = − π 3 80. sec (arcsec (0.75)) is undefined. 82. cs... |
csc 129. arcsec sec − π 2 11π 12 = 3π 2 = 13π 12 π 4 = = 5π 5π 6 5π 3 5π 4 − π 2 11π 12 = 11π 12 131. sin arccos − = 1 2 133. sin (arctan (−2)) = − 135. sin (arccsc (−3)) = − 137. cos arctan √ 7 = 139. cos (arcsec (5)) = 1 5 114. arccsc csc 116. arccsc csc 118. arccsc csc 120. arcsec sec 2π 3 11π 6 9π 8 4π 3 = π 3 = 7... |
arcsec(3) + arctan(2)) = √ 10 √ 5 − 4 15 √ 17 2 26 3 5 = 1 3 158. sin 2 arcsin − 4 5 = − 24 25 159. sin 2arccsc 161. cos 2 arcsin 13 5 3 5 = 120 169 = 7 25 160. sin (2 arctan (2)) = 4 5 162. cos 2arcsec 25 7 = − 163. cos 2arccot − √ 5 = 2 3 165. sin (arccos (x)) = √ 1 − x2 for −1 ≤ x ≤ 1 164. sin arctan(2) 2 = 5 − 10 5... |
ctan(2x)) tan(arctan(2x)) = 2x 1 + 4x2 for all x 181. sin (arcsin(x) + arccos(x)) = 1 for −1 ≤ x ≤ 1 182. cos (arcsin(x) + arctan(x)) = √ 1 − x2 − x2 √ 1 + x2 for −1 ≤ x ≤ 1 183. 10 tan (2 arcsin(x)) = √ 2x 1 − x2 1 − 2x2 for x in −1 184. sin arctan(x) = 1 2 √ x2 + 1 − 1 √ x2 + 1 2 √ x2 + 1 − 1 √ x2 + 1 2... |
2πk, in [0, 2π), x ≈ 1.7949, 4.4883 190. x = π + arcsin(0.569) + 2πk or x = 2π − arcsin(0.569) + 2πk, in [0, 2π), x ≈ 3.7469, 5.6779 191. x = arccos(0.117) + 2πk or x = 2π − arccos(0.117) + 2πk, in [0, 2π), x ≈ 1.4535, 4.8297 192. x = arcsin(0.008) + 2πk or x = π − arcsin(0.008) + 2πk, in [0, 2π), x ≈ 0.0080, 3.1336 1... |
, 2π), x ≈ 3.3316, 6.0932 201. x = arctan(0.03) + πk, in [0, 2π), x ≈ 0.0300, 3.1716 202. x = arcsin(0.3502) + 2πk or x = π − arcsin(0.3502) + 2πk, in [0, 2π), x ≈ 0.3578, 2.784 203. x = π + arcsin(0.721) + 2πk or x = 2π − arcsin(0.721) + 2πk, in [0, 2π), x ≈ 3.9468, 5.4780 204. x = arccos(0.9824) + 2πk or x = 2π − arc... |
) 218. f (x) = cos(x) − 3 sin(x) = √ 10 sin x + arccos − 10 219. f (x) = 7 sin(10x) − 24 cos(10x) = 25 sin 10x + arcsin − √ 3 10 √ ≈ 10 sin(x + 2.8198) 24 25 ≈ 25 sin(10x − 1.2870) 220. f (x) = − cos(x) − 2 √ 2 sin(x) = 3 sin x + π + arcsin 221. f (x) = 2 sin(x) − cos(x) = √ 5 sin x + arcsin − 1 3 ≈ 3 sin(x + 3.4814) √... |
> 1, there are no real solutions. To solve sec(u) = c or csc(u) = c for c ≤ −1 or c ≥ 1, convert to cosine or sine, respectively, and solve as above. If −1 < c < 1, there are no real solutions. To solve tan(u) = c for any real number c, first solve for u in the interval − π 2, π 2 and add integer multiples of the perio... |
2πk for integers k. Since 6 + 2πk for integers 12 + πk for integers k. To check these answers 6 + 2πk or 2x = 7π 12 + πk or x = 7π 6 + 2πk or u = 7π 2 are u = 5π 3 cos 2 5π 12 + πk = cos 5π = cos 5π 6 √ = − 3 2 6 + 2πk (the period of cosine is 2π) 1See the comments at the beginning of Section 10.5 for a review of this... |
to get 1 3 x = 5π 4 + 2πk x = 3 5π 4 + 2πk = 15π 4 + 6πk Solving the other equation, 1 check the first family of answers, we substitute, combine line terms, and simplify. 4 + 2πk produces x = 21π 3 x − π = 3π 4 + 6πk for integers k. To csc 1 3 15π 4 + 6πk − π = csc 5π = csc π = csc π √ 4 2 = 4 + 2πk − π 4 + 2πk (the pe... |
quotient identity cot(3x) = cos(3x) sin(3x) and y = 0 (the x-axis), we see that the x-coordinates of the intersection points approximately match our solutions. 6, 3π 6, 7π 2, 5π 6, π 4. The complication in solving an equation like sec2(x) = 4 comes not from the argument of secant, which is just x, but rather, the fact... |
appears identical, but what happens when you try to find the intersection points? 860 Foundations of Trigonometry a reciprocal identity to rewrite the secant as cosine. The x-coordinates of the intersection points of y = (cos(x))2 and y = 4 verify our answers. 1 y = cos(3x) sin(3x) and y = 0 y = 1 cos2(x) and y = 4 5. ... |
produces x = 2 arctan(−3) + 4π. Once again, we get from −π < 2 arctan(−3) < 0 that 3π < 2 arctan(−3) + 4π < 4π. Since this is outside the interval [0, 2π), we discard x = 2 arctan(−3) + 4π and all solutions of the form x = 2 arctan(−3) + 2πk for k > 2. Graphically, we see y = tan x and y = −3 intersect only 2 once on ... |
period of sine is 2π) (sin(π − t) = sin(t)) (See Theorem 10.26) 2 arcsin(0.87) < π 2 so that multiplying through by 1 4. Starting with the family of solutions x = 1 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 calcu... |
3 y = sin(2x) and y = 0.87 862 Foundations of Trigonometry Each of the problems in Example 10.7.1 featured one trigonometric function. If an equation involves two different trigonometric functions or if the equation contains the same trigonometric function but with different arguments, we will need to use identities and ... |
(sin(x))3 and and x = π − arcsin 1 to be x = arcsin 1 3 3 y = (sin(x))2 and find the x-coordinates of the intersection points of these two curves. Some extra zooming is required near x = 0 and x = π to verify that these two curves do in fact intersect four times.6 3, we use the arcsine function to get x = arcsin 1 3 2. ... |
y = (cos(x))2 and y = tan(x) + 3 to find the x-values of the points where they intersect. 4 and x = 7π 1 y = 3(sin(x))3 and y = (sin(x))2 y = 1 (cos(x))2 and y = tan(x) + 3 3. In the equation cos(2x) = 3 cos(x) − 2, we have the same circular function, namely cosine, on both sides but the arguments differ. Using the iden... |
= 2 − cos(x) 4 cos3(x) − 3 cos(x) = 2 − cos(x) 2 cos3(x) − 2 cos(x) − 2 = 0 4u3 − 2u − 2 = 0 Let u = cos(x). 864 Foundations of Trigonometry To solve 4u3 − 2u − 2 = 0, we need the techniques in Chapter 3 to factor 4u3 − 2u − 2 into (u − 1) 4u2 + 4u + 2. We get either u − 1 = 0 or 4u2 + 2u + 2 = 0, and since the discri... |
0 is x = πk for integers k. The second set of solutions is contained in the first set of solutions,8 so our final solution to cos(5x) = cos(3x) is x = π 4 k for integers k. There are eight of these answers which lie in [0, 2π): x = 0, π 4, 3π 4, π, 5π 4. Our plot of the graphs of y = cos(3x) and y = cos(5x) below (after... |
√ 3 cos(x) and, y = cos(3x) and y = cos(5x) y = sin(2x) and y = √ 3 cos(x) 7. Unlike the previous problem, there seems to be no quick way to get the circular functions or their arguments to match in the equation sin(x) cos x = 1. If we stare at 2 it long enough, however, we realize that the left hand side is the expan... |
of nonlinear equations in section 8.7 – when it comes to solving equations involving the trigonometric functions, it helps to just try something. 9We are essentially ‘undoing’ the sum / difference formula for cosine or sine, depending on which form we use, so this problem is actually closely related to the previous one... |
ically by seeing where the graph of y = 2 sin(x) crosses or is below the graph of y = 1. 6 + 2πk and x = 5π 6 and x = 5π ∪ 5π 2 (−) 0 (+) π 6 0 (−) 5π 6 2π 0 y = 2 sin(x) and y = 1 2. We first rewrite sin(2x) > cos(x) as sin(2x) − cos(x) > 0 and let f (x) = sin(2x) − cos(x). Our original inequality is thus equivalent to... |
2 2. We see f (x) > 0 on π 2 √ 2 2 = 2 = −2− 4 we get f 7π = −1 − 2 = 2− ∪ 5π 6, 3π 4 we get ; √ √ (−) 0 (+) π 6 0 (−) π 2 0 (+) 5π 6 0 (−) 3π 2 2π 0 y = sin(2x) and y = cos(x) 2 and 3π 3. Proceeding as in the last two problems, we rewrite tan(x) ≥ 3 as tan(x) − 3 ≥ 0 and let f (x) = tan(x) − 3. We note that on [0, 2π)... |
ctan(117) + π yields f (arctan(117) + π) = tan(arctan(117) + π) − 3 = tan(arctan(117)) − 3 = 114. We = −4. Since we want f (x) ≥ 0, we 4 and find f 7π choose our last test value to be x = 7π. Using the graphs of y = tan(x) ∪ arctan(3) + π, 3π see that our answer is arctan(3), π 2 2 and y = 3, we see when the graph of th... |
pattern. Graphing the situation on a numberline, we have = 0, we get x = − π 6 + π 6, − 7π 6, − 4π 6 + π 6, 8π 6, 5π 6, 2π. − 7π 6 − 4π 6 − π 6 2π 6 5π 6 8π 6 Proceeding as we did on page 756 in Section 10.3.1, we let xk denote the kth number excluded from the domain and we have xk = − π for integers k. The intervals ... |
in our domain work to describe the intervals. To that end, we let ak = π for integers k. The goal now is to write the domain in terms of the a’s an b’s. We find a0 = π 3, a1 = 7π 3 and b−2 = − 7π 3. Hence, in terms of the a’s and b’s, our domain is 3 + 2πk = (6k+5)π 3 + 2πk = (6k+1)π 3, a−2 = − 11π 3, a−1 = − 5π 3, b−1... |
and right on the x-axis by adding integer multiples of 2π. One such interval that arises from our domain work is π. The portion. Adding integer multiples of 2π, we get the family of of the domain here is π 3, 7π 3 + 2πk for integers k. We leave it to the reader intervals π 3 + 2πk, 7π to show that getting common denom... |
π = 1. = 1 − (−) 0 0 (+) π 4 π We find g(x) ≥ 0 on π the intervals π express our final answer as 4 + πk, π + πk = 4, π. Adding multiples of the period we get our solution to consist of. Using extended interval notation, we, (k + 1)π (4k+1)π 4 ∞ k=−∞ (4k + 1)π 4, (k + 1)π We close this section with an example which demon... |
os(x)) = cos 3π 4 x = − √ 2 2 Since cos(arccos(u)) = u The calculator confirms y = 4 arccos(x) − 3π crosses y = 0 (the x-axis) at − √ 2 2 ≈ −0.7071. y = arcsin(2x) and y = π 3 y = 4 arccos(x) − 3π 3. From 3 arcsec(2x − 1) + π = 2π, we get arcsec(2x − 1) = π 3. As we saw in Section 10.6, there are two possible ranges for... |
we only get solutions from the first equation. Using Theorem 10.27, we get arctan(x) = − π 4 tan(arctan(x)) = tan − π 4 x = −1 Since tan(arctan(u)) = u. The calculator verifies our result. y = 3 arcsec(2x − 1) + π and y = 2π y = 4 arctan2(x) − 3π arctan(x) − π2 5. Since the inverse trigonometric functions are continuous... |
begin, we rewrite 4 arccot(3x) > π as 4 arccot(3x) − π > 0. We let f (x) = 4 arccot(3x) − π, and note the domain of f is all real numbers, (−∞, ∞). To find the zeros of f, we set f (x) = 4 arccot(3x) − π = 0 and solve. We get arccot(3x) = π 4 is in the range of arccotangent, we may apply Theorem 10.27 and solve 4, and ... |
18, find all of the exact solutions of the equation and then list those solutions which are in the interval [0, 2π). 1. sin (5x) = 0 2. cos (3x) = 1 2 4. tan (6x) = 1 5. csc (4x) = −1 √ 3 3 = 0 7. cot (2x) = − 10. cos x + 5π 6 13. csc(x) = 0 16. sec2 (x) = 4 3 8. cos (9x) = 9 11. sin 2x − π 3 = − 1 2 14. tan (2x − π) =... |
. tan3 (x) = 3 tan (x) 38. cos3 (x) = − cos (x) 40. csc3(x) + csc2(x) = 4 csc(x) + 4 41. 2 tan(x) = 1 − tan2(x) 42. tan (x) = sec (x) 10.7 Trigonometric Equations and Inequalities 875 In Exercises 43 - 58, solve the equation, giving the exact solutions which lie in [0, 2π) 43. sin(6x) cos(x) = − cos(6x) sin(x) 44. sin(... |
− π2 = 0 67. 8 arccot2(x) + 3π2 = 10π arccot(x) 68. 6 arctan(x)2 = π arctan(x) + π2 In Exercises 69 - 80, solve the inequality. Express the exact answer in interval notation, restricting your attention to 0 ≤ x ≤ 2π. 69. sin (x) ≤ 0 72. cos2 (x) > 75. cot2 (x) ≥ 1 2 1 3 78. cos(3x) ≤ 1 70. tan (x) ≥ √ 3 73. cos (2x) ≤... |
page 756 in Section 10.3.1 for details.) 99. f (x) = 1 cos(x) − 1 100. f (x) = cos(x) sin(x) + 1 101. f (x) = tan2(x) − 1 102. f (x) = 2 − sec(x) 103. f (x) = csc(2x) 104. f (x) = sin(x) 2 + cos(x) 105. f (x) = 3 csc(x) + 4 sec(x) 106. f (x) = ln (| cos(x)|) 107. f (x) = arcsin(tan(x)) 108. With the help of your class... |
π 8, 15π 8 + 2πk 3 or x = 7π 12 + 2πk 3 ; x = π 12, 7π 12, 3π 4, 5π 4, 17π 12, 23π 12 3. x = 4. x = 5. x = 6. x = 2π 3 π 24 3π 8 π 12 7. x = π 3 + πk 2 ; x = π 3, 5π 6, 4π 3, 11π 6 8. No solution 9. x = 3π 4 10. x = − π 3 11. x = 3π 4 + 6πk or x = 9π 4 + 6πk; x = 3π 4 + πk; x = 2π 3, 5π 3 + πk or x = 13π 12 + πk; x = π... |
(2), π + arctan(2) Foundations of Trigonometry 20. x = 0, 22. x = 24. x = 26, 5π 3 3π 2, π 3 5π 6 5π 3 5π 6,,, 28. x = 3π 4, 7π 4, arctan 1 2, π + arctan 1 2 30. x = 32. x = 34. x = π 6 π 6 π 6 36. x = 0, 38. x = 40. x = π 2 π 6,,,,,, 5π 6 7π 6 π 4 π 3 3π 2 5π 6,,, π 2 5π 6 3π 4 2π 3,, 11π 6 5π 6 4π 3,, 7π 6 5π 3,, π,,... |
, 3π 2, 29π 18 10.7 Trigonometric Equations and Inequalities 879 53. x = 0, 55. x = 0 3π 8 3π 4 5π 8,, π,, 5π 4 7π 8, 3π 2 9π 8 7π 4 11π 8,, 13π 8, 15π 8 54. x = 0, π 3, 2π 3, π, 4π 3, 5π 3,, π, 56. x = π 7, π 3, 3π 7, 5π 7, π, 9π 7, 11π 7, 5π 3, 13π 7 57. x = 0, 2π 7, 58. x = arcsin 59. x = − 1 2 61. x = 2 3 63. x = 2... |
4), 2π) 82. arcsin 1 3, π − arcsin 1 3 84. − 2π 3, − π 3 ∪ π 3, 2π 3 880 Foundations of Trigonometry −π, − 85. π 4 ∪ 0, 3π 4 −2π, − 87. 3π 2 ∪ − 3π 2, −π ∪ 0, π 2 ∪ π 2, π 86. − 3π 4, π 4 88. [−2π, 2π] 89. (−2π, arccot(5) − 2π] ∪ (−π, arccot(5) − π] ∪ (0, arccot(5)] ∪ (π, π + arccot(5)] − 3π 2, − 5π 4 − ∪ 3π 4, − −π, −... |
)π 2, (2k + 1)π 2 101. 102. 103. 105. 107. k=−∞ ∞ k=−∞ ∞ k=−∞ ∞ k=−∞ Chapter 11 Applications of Trigonometry 11.1 Applications of Sinusoids In the same way exponential functions can be used to model a wide variety of phenomena in nature,1 the cosine and sine functions can be used to model their fair share of natural be... |
oid has over the un-shifted sine function. The figure below is repeated from Section 10.5. amplitude baseline period In Section 10.1.1, we introduced the concept of circular motion and in Section 10.2.1, we developed formulas for circular motion. Our first foray into sinusoidal motion puts these notions to good use. Exam... |
shift upward,4 we add 72 to our formula for y to obtain the new formula h = y + 72 = 64 sin 4π 127 t + 72. Next, we need to adjust things so that t = 0 corresponds to the point P instead of the point Q. This is where the phase comes into play. Geometrically, we need to shift the angle θ in the figure back π 2 radians. ... |
4 5 6 7 8 9 10 11 12 5.8 9.3 12.4 15.9 19.4 21.8 19.4 15.6 12.4 9.1 5.6 3.3 1. Find a sinusoid which models these data and use a graphing utility to graph your answer along with the data. 2. Compare your answer to part 1 to one obtained using the regression feature of a calculator. Solution. 1. To get a feel for the d... |
= 2π 6. The last quantity to find is the phase φ. Unlike the previous example, it is easier in this case to find the phase shift − φ ω. Since we picked A > 0, the phase shift corresponds to the first value of t with H(t) = 12.55 (the baseline value).7 Here, we choose t = 3, since its corresponding H value of 12.4 is clos... |
isn’t subjected to relativistic speeds... 886 Applications of Trigonometry An object which weighs 6 pounds on the surface of the Earth would weigh 1 pound on the surface of the Moon, but its mass is the same in both places. In the English system of units, ‘pounds’ (lbs.) is a measure of force (weight), and the corresp... |
�uences on the system except gravity and the spring force, then Physics tells us that gravity and the spring force will battle each other forever and the object will oscillate indefinitely. In this case, we describe the motion as ‘free’ (meaning there is no external force causing the motion) and ‘undamped’ (meaning we i... |
0. It is a great exercise in ‘dimensional analysis’ to verify that the formulas given in Theorem 11.1 work out so that ω has units 1 s and A has units ft. or m, depending on which system we choose. Example 11.1.3. Suppose an object weighing 64 pounds stretches a spring 8 feet. 1. If the object is attached to the spring... |
object would fall towards the ground, which is the ‘natural’ or ‘positive’ direction. Since the spring force acts in direct opposition to gravity, any movement upwards is considered ‘negative’. 888 Applications of Trigonometry = 0. Going through the usual analysis we find t = − π satisfy this condition, so we pick14 th... |
= 3 5. This means that φ is a Quadrant II angle which we can describe in terms of either arcsine or arccosine. Since x(t) is expressed in terms of sine, we choose to express φ = π − arcsin 3. Hence, 5. Since the amplitude of x(t) is 5, the object will travel x(t) = 5 sin 2t + π − arcsin 3 5 at most 5 feet above the eq... |
A(t) sin(ωt + φ). Graph x(t) 2. Write x(t) = (t + 3) √ 2 cos(2t) + (t + 3) √ x(t) using a graphing utility. 2 sin(2t) in the form x(t) = A(t) sin(ωt + φ). Graph 3. Find the period of x(t) = 5 sin(6t) − 5 sin (8t). Graph x(t) using a graphing utility. Solution. √ √ √ 1. We start rewriting x(t) = 5e−t/5 cos(t) + 5e−t/5 ... |
smaller amplitude. y = 10e−x/5 sin x + π 3 y = 10e−x/5 sin x + π 3, y = ±10e−x/5 √ 2. Proceeding as in the first example, we factor out (t + 3) 2 from each term in the function 2(cos(2t) + sin(2t)). We find x(t) = (t + 3). Graphing this on the (cos(2t) + sin(2t)) = calculator as y = 2(x + 3) sin 2x + π, we find the sinus... |
the period of f (t) (which is g makes 4. four times the period of g(t)), or π. We graph y = 5 sin(6x) − 5 sin(8x) over [0, π] on the calculator to check this. This equation of motion also results from ‘forced’ motion, but here the frequency of the external oscillation is different than that of the object on the spring.... |
and makes one revolution (counterclockwise) every 30 minutes. It is constructed so that the lowest part of the Eye reaches ground level, enabling passengers to simply walk on to, and off of, the ride. Find a sinsuoid which models the height h of the passenger above the ground in meters t minutes after they board the Ey... |
, θ, can be modeled as a sinusoid.18 θ The amplitude of the sinusoid is the same as the initial angular displacement, θ0, of the pendulum and the period of the motion is given by T = 2π l g where l is the length of the pendulum and g is the acceleration due to gravity. (a) Find a sinusoid which gives the angular displa... |
in the table below.21 Day of June, t Fraction Illuminated, F 3 6 9 12 15 18 21 24 27 30 0.81 0.98 0.98 0.83 0.57 0.27 0.04 0.03 0.26 0.58 (a) Using the techniques discussed in Example 11.1.2, fit a sinusoid to these data.22 (b) Using a graphing utility, graph your model along with the data set to judge the reason- able... |
is heading upwards. (c) x(t) = √ 2 2 sin 4t + 7π 4. The object passes through the equilibrium point heading down- wards for the third time when t = 17π 16 ≈ 3.34 seconds. 7. (a) θ(t) = θ0 sin g l t + π 2 (b) θ(t) = π 12 sin 4πt + π 2 8. (a) T (t) = 20.5 sin π (b) Our function and the data set are graphed below. The si... |
main goal of this section and the next is to develop theorems which allow us to ‘solve’ triangles – that is, find the length of each side of a triangle and the measure of each of its angles. In Sections 10.2, 10.3 and 10.6, we’ve had some experience solving right triangles. The following example reviews what we know. E... |
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