text stringlengths 235 3.08k |
|---|
3 −2 −1 −1 −2 −3 −4 1 2 3 4 x 29. 0 33. − 7π 4 37. 180◦ 41. 60◦ 45. t = 5π 6 47. t = 6 30. 34. 4π 3 5π 6 38. −120◦ 42. 300◦ 1 x y 1 y 1 31. 35. 3π 4 π 4 39. 210◦ 43. −30◦ 46. t = −π 48. t = −2 32. − 36. − 3π 2 5π 4 40. 330◦ 44. 90 716 Foundations of Trigonometry 49. t = 12 (between 1 and 2 revolutions) y 1 1 x 50. Abou... |
1 θ 1 x y 1 P (cos(θ), sin(θ)) θ 1 x Example 10.2.1. Find the cosine and sine of the following angles. 1. θ = 270◦ 2. θ = −π 3. θ = 45◦ 4. θ = π 6 5. θ = 60◦ Solution. 1. To find cos (270◦) and sin (270◦), we plot the angle θ = 270◦ in standard position and find the point on the terminal side of θ which lies on the Unit... |
◦ right triangle whose legs have lengths x and y units. From Geometry,2 we get y = x. Since P (x, y) lies on the Unit Circle, we have √ 2 x2 + y2 = 1. Substituting y = x into this equation yields 2x2 = 1, or x = ± 2. √ 2 Since P (x, y) lies in the first quadrant, x > 0, so x = cos (45◦) = 2 and with y = x we have y = si... |
was quite easy to find the cosine and sine of the quadrantal angles, but for In these latter cases, we made good non-quadrantal angles, the task was much more involved. use of the fact that the point P (x, y) = (cos(θ), sin(θ)) lies on the Unit Circle, x2 + y2 = 1. If we substitute x = cos(θ) and y = sin(θ) into x2 + y... |
into The Pythagorean Identity, cos2(θ) + sin2(θ) = 1, we obtain cos2(θ) + 9 5. Since θ is a Quadrant II angle, its terminal side, when plotted in standard position, lies in Quadrant II. Since the x-coordinates are negative in Quadrant II, cos(θ) is too. Hence, cos(θ) = − 4 5. 25 = 1. Solving, we find cos(θ) = ± 4 √ √ 5... |
� = 5π = 5π is called the reference angle for the angle 5π In the above scenario, the angle π 6. In general, for a non-quadrantal angle θ, the reference angle for θ (usually denoted α) is the acute angle made between the terminal side of θ and the x-axis. If θ is a Quadrant I or IV angle, α is the angle between the ter... |
land in Quadrant III. Hence, we obtain θ’s reference angle α by subtracting: α = θ − 180◦ = 225◦ − 180◦ = 45◦. Since θ is a Quadrant III angle, both cos(θ) < 0 and 10.2 The Unit Circle: Cosine and Sine sin(θ) < 0. The Reference Angle Theorem yields: cos (225◦) = − cos (45◦) = − sin (225◦) = − sin (45◦) = − √ 2 2. 723 ... |
3 π 3 1 x Finding cos − 5π 4 and sin − 5π 4 Finding cos 7π 3 and sin 7π 3 724 Foundations of Trigonometry 6 as a reference angle, those with a denominator of 4 have π The reader may have noticed that when expressed in radian measure, the reference angle for a non-quadrantal angle is easy to spot. Reduced fraction mult... |
on the positive. 13, 12 13 y 1 5 13, 12 13 α 1 x Sketching α 2. (a) To find the cosine and sine of θ = π + α, we first plot θ in standard position. We can imagine the sum of the angles π+α as a sequence of two rotations: a rotation of π radians followed by a rotation of α radians.7 We see that α is the reference angle f... |
θ α 1 x 1 x Visualizing 3π − α θ has reference angle α (d) To plot θ = π 2 + α, we first rotate π 2 radians and follow up with α radians. The reference angle here is not α, so The Reference Angle Theorem is not immediately applicable. (It’s important that you see why this is the case. Take a moment to think about this ... |
3, we find θ = π 3 + 2πk for integers k. 3, and since all other Quadrant I solutions must be 3 + 2πk for integers k.9 Proceeding similarly for the Quadrant 3, so our answer in this Quadrant is 2 here is 5π 2. If sin(θ) = − 1 2, then when θ is plotted in standard position, its terminal side intersects the Unit Circle at... |
that is, ‘When in doubt, write it out!’ This is especially important when checking answers to the exercises. For example, another Quadrant IV solution to 2 is θ = − π sin(θ) = − 1 6. Hence, the family of Quadrant IV answers to number 2 above could just have easily been written θ = − π 6 + 2πk for integers k. While on ... |
B(x, 0) x Not only can we describe the coordinates of Q in terms of cos(θ) and sin(θ) but since the radius of x2 + y2, we can also express cos(θ) and sin(θ) in terms of the coordinates of Q. the circle is r = These results are summarized in the following theorem. Theorem 10.3. If Q(x, y) is the point on the terminal s... |
, y) 41.628◦ 3960 x The terminal side of θ contains Q(4, −2) A point on the Earth at 41.628◦N Using Theorem 10.3, we get x = 3960 cos (41.628◦). Using a calculator in ‘degree’ mode, we find 3960 cos (41.628◦) ≈ 2960. Hence, the radius of the Earth at North Latitude 41.628◦ is approximately 2960 miles. 732 Foundations of... |
π of motion are x = r cos(ωt) = 2960 cos π measured in miles and t is measured in hours. 12 hours. Hence, the equations 12 t, where x and y are 12 t and y = r sin(ωt) = 2960 sin π In addition to circular motion, Theorem 10.3 is also the key to developing what is usually called ‘right triangle’ trigonometry.11 As we sh... |
of the missing angle and the lengths of the missing sides of: 30◦ 7 Solution. The first and easiest task is to find the measure of the missing angle. Since the sum of angles of a triangle is 180◦, we know that the missing angle has measure 180◦ − 30◦ − 90◦ = 60◦. We now proceed to find the lengths of the remaining two si... |
we can spell out this correspondence more precisely. For each real number t, we associate an oriented arc t units in length with initial point (1, 0) and endpoint P (cos(t), sin(t)). cos(t), sin(t)) θ = t 1 x In the same way we studied polynomial, rational, exponential, and logarithmic functions, we will study the tri... |
used is t rather than θ: t = 7π 6 + 2πk for integers, k. We will study the cosine and sine functions in greater detail in Section 10.5. Until then, keep in mind that any properties of cosine and sine developed in the following sections which regard them as functions of angles in radian measure apply equally well if th... |
�) = 28. If cos(θ) = 28 53 √ 2 5 √ 10 10 with θ in Quadrant IV, what is sin(θ)? 5 and π 2 < θ < π, what is cos(θ)? and 2π < θ < 5π 2 3π 2, what is sin(θ)?, what is cos(θ)? 29. If sin(θ) = −0.42 and π < θ < 30. If cos(θ) = −0.98 and π 2 < θ < π, what is sin(θ)? 10.2 The Unit Circle: Cosine and Sine 737 In Exercises 31 -... |
θ, a, and c. c θ b 30◦ 1 45◦ c a θ 3 738 Foundations of Trigonometry 57. Find α, a, and b. 58. Find β, a, and c. b α 8 a 33◦ a 48◦ 6 c β In Exercises 59 - 64, assume that θ is an acute angle in a right triangle and use Theorem 10.4 to find the requested side. 59. If θ = 12◦ and the side adjacent to θ has length 4, how ... |
53 from Section 10.1. Recall: The diameter of the hard disk is 2.5 inches and it spins at 7200 revolutions per minute. 10.2 The Unit Circle: Cosine and Sine 739 72. A passenger on the Big Wheel in Exercise 55 from Section 10.1. Recall: The diameter is 128 feet and completes 2 revolutions in 2 minutes, 7 seconds. 73. C... |
2 2 =, sin π 4 = √ 2 2 = 0, sin, sin, sin 3π 4 7π, sin 4π 3 = − √ 3 2 = 1 2, sin 5π 3 = − √ 3 2 14. cos 23π 6 √ 3 2, sin 23π 6 = − 1 2 = 15. cos − 13π 2 = 0, sin − 13π 2 = −1 16. cos − 43π 6 √ 3 2, sin − 43π 6 = 1 2 = − 17. cos 19. cos − 3π 4 10π 3 = − √ 2 2, sin − 3π 4 √ 2 2 = − 18. cos − π 6 = √ 3 2, sin −, sin 10π ... |
cos(θ) = −0.98 and < θ < π, then sin(θ) = √ 0.0396 ≈ 0.1990. π 2 31. sin(θ) = 1 2 32. cos(θ) = − √ 3 2 when θ = π 6 + 2πk or θ = 5π 6 + 2πk for any integer k. when θ = 5π 6 + 2πk or θ = 7π 6 + 2πk for any integer k. 33. sin(θ) = 0 when θ = πk for any integer k. 34. cos(θ) = 35. sin(θ) = √ 2 2 √ 3 2 when θ = when θ = π... |
47. sin(t) = 1 when t = 48. cos(t) = − √ 2 2 when t = 3π 4 + 2πk or t = 5π 4 + 2πk for any integer k. 49. sin(78.95◦) ≈ 0.981 50. cos(−2.01) ≈ −0.425 51. sin(392.994) ≈ −0.291 52. cos(207◦) ≈ −0.891 53. sin (π◦) ≈ 0.055 54. cos(e) ≈ −0.912 55. θ = 60◦, b = √ 3 3, c = 56. θ = 45◦, a = 3 57. α = 57◦, a = 8 cos(33◦) ≈ 6.... |
minute, x = 1.125 cos(9000π t), y = 1.125 sin(9000π t). Here x and y are measured in inches and t is measured in minutes. 10.2 The Unit Circle: Cosine and Sine 743 70. r = 28 inches, ω = 2π 3 radians second, x = 28 cos 2π 3 t, y = 28 sin 2π 3 t. Here x and y are measured in inches and t is measured in seconds. 71. r =... |
1 y y x The cotangent of θ, denoted cot(θ), is defined by cot(θ) =, provided y = 0., provided x = 0. x y, provided y = 0. While we left the history of the name ‘sine’ as an interesting research project in Section 10.2, the names ‘tangent’ and ‘secant’ can be explained using the diagram below. Consider the acute angle θ... |
two points.2 The line containing the terminal side of θ is a secant line since it intersects the Unit Circle in Quadrants I and III. With the point P lying on the Unit Circle, the length of the hypotenuse of ∆OP A is 1. If we let h denote the length of the hypotenuse of ∆OQB, we have from similar triangles that h x = ... |
value, if it exists. 1. sec (60◦) 2. csc 7π 4 3. cot(3) 4. tan (θ), where θ is any angle coterminal with 3π 2. 5. cos (θ), where csc(θ) = − √ 5 and θ is a Quadrant IV angle. 6. sin (θ), where tan(θ) = 3 and π < θ < 3π 2. 2Compare this with the definition given in Section 2.1. 746 Solution. 1. According to Theorem 10.6,... |
) = 4 √ 5 5. 5, or cos(θ) = ± 2 √ 5 5. Since sin(θ. If tan(θ) = 3, then sin(θ) cos(θ) = 3. Be careful - this does NOT mean we can take sin(θ) = 3 and cos(θ) = 1. Instead, from sin(θ) cos(θ) = 3 we get: sin(θ) = 3 cos(θ). To relate cos(θ) and sin(θ), we once again employ the Pythagorean Identity, cos2(θ) + sin2(θ) = 1. ... |
3 1 √ 3 3 0 Coupling Theorem 10.6 with the Reference Angle Theorem, Theorem 10.2, we get the following. Theorem 10.7. Generalized Reference Angle Theorem. The values of the circular functions of an angle, if they exist, are the same, up to a sign, of the corresponding circular functions of its reference angle. More sp... |
, we get θ = 4π 3 + 2πk for integers k. While these descriptions of the solutions are correct, they can be combined into one list as θ = π 3 + πk for integers k. The latter form of the solution is best understood looking at the geometry of the situation in the diagram below.4 4See Example 10.2.5 number 3 in Section 10.... |
θ), apply Theorem 10.6 once again, and obtain cot2(θ) + 1 = csc2(θ). These three Pythagorean Identities are worth memorizing and they, along with some of their other common forms, are summarized in the following theorem. cos2(θ) = 1 10.3 The Six Circular Functions and Fundamental Identities 749 Theorem 10.8. The Pythag... |
− cos(θ) = 1 + cos(θ) sin(θ) Solution. In verifying identities, we typically start with the more complicated side of the equation and use known identities to transform it into the other side of the equation. 1. To verify 1 csc(θ) = sin(θ), we start with the left side. Using csc(θ) = 1 sin(θ), we get: 1 csc(θ) = 1 1 si... |
(1)(cos(θ)) − 1 sin(θ) cos(θ) (cos(θ)) = 1 cos(θ) − sin(θ), which is exactly what we had set out to show. 5. The right hand side of the equation seems to hold more promise. We get common denomina- tors and add: 3 1 − sin(θ) − 3 1 + sin(θ) = = = = 3(1 + sin(θ)) (1 − sin(θ))(1 + sin(θ)) − 3(1 − sin(θ)) (1 + sin(θ))(1 − ... |
starting with the left hand side and multiplying the numerator and denominator by the quantity 1 + cos(θ): sin(θ) 1 − cos(θ) = = = sin(θ) (1 − cos(θ)) · (1 + cos(θ)) (1 + cos(θ)) = sin(θ)(1 + cos(θ)) (1 − cos(θ))(1 + cos(θ)) sin(θ)(1 + cos(θ)) 1 − cos2(θ) sin(θ)(1 + cos(θ)) sin(θ) sin(θ) = = sin(θ)(1 + cos(θ)) sin2(θ)... |
tan(θ))(sec(θ)+tan(θ)) = sec2(θ)−tan2(θ) = 1 csc(θ) − 1 and csc(θ) + 1: (csc(θ) − 1)(csc(θ) + 1) = csc2(θ) − 1 = cot2(θ) csc(θ) − cot(θ) and csc(θ) + cot(θ): (csc(θ) − cot(θ))(csc(θ) + cot(θ)) = csc2(θ) − cot2(θ) = 1 752 Foundations of Trigonometry Verifying trigonometric identities requires a healthy mix of tenacity a... |
circles of radius r. Using Theorem 10.3 in conjunction with Theorem 10.8, we generalize the remaining circular functions in kind. Theorem 10.9. Suppose Q(x, y) is the point on the terminal side of an angle θ (plotted in standard position) which lies on the circle of radius r, x2 + y2 = r2. Then: r x r y = = x2 + y2 x ... |
, we find cos(θ) = 4√ 17 and tan(θ) = − 1 4. (4)2 + (−1)2 = = − 17, sin(θ) = − 1√ √ √ 17 17 We may also specialize Theorem 10.9 to the case of acute angles θ which reside in a right triangle, as visualized below. b c a θ Theorem 10.10. Suppose θ is an acute angle residing in a right triangle. If the length of the side a... |
we get denote the height of the tower, then Theorem 10.10 gives tan (60◦) = h h = 30 tan (60◦) = 30 3 ≈ 51.96. Hence, the Clocktower is approximately 52 feet tall. √ h ft. 60◦ 30 ft. Finding the height of the Clocktower 2. Sketching the problem situation below, we find ourselves with two unknowns: the height h of the t... |
associating the real number t with the angle θ = t radians so that the value of the trigonometric function of t coincides with that of θ; lastly, we could simply define them using the Reciprocal and Quotient Identities as combinations of the functions f (t) = cos(t) and g(t) = sin(t). Presently, we adopt the last appro... |
= In order to capture all of the intervals in the domain, k must run through all of the integers, that is, k = 0, ±1, ±2,.... The way we denote taking the union of infinitely many intervals like this is to use what we call in this text extended interval notation. The domain of F (t) = sec(t) can now be written as 2, (2... |
(t) causes a reversal of the inequality so that sec(t) = 1 sec(t) ≤ −1. In this case, as cos(t) → 0−, sec(t) = 1 very small (−) ≈ very big (−), so that as cos(t) → 0−, we get sec(t) → −∞. Since cos(t) ≈ cos(t) ≈ 1 1 10.3 The Six Circular Functions and Fundamental Identities 757 f (t) = cos(t) admits all of the values i... |
1] ∪ [1, ∞) • The function J(t) = tan(t) = sin(t) cos(t) – has domain {t : t = π 2 + πk, for integers k} = ∞ k=−∞ (2k + 1)π 2, (2k + 3)π 2 – has range (−∞, ∞) • The function K(t) = cot(t) = cos(t) sin(t) – has domain {t : t = πk, for integers k} = – has range (−∞, ∞) ∞ k=−∞ (kπ, (k + 1)π) 8Using Theorem 2.4 from Sectio... |
because 1 10.3 The Six Circular Functions and Fundamental Identities 759 10.3.2 Exercises In Exercises 1 - 20, find the exact value or state that it is undefined. 1. tan π 4 5. tan − 11π 6 9. tan (117π) 13. tan 17. tan 31π 2 2π 3 2. sec π 6 6. sec − 10. sec − 3π 2 5π 3 14. sec π 4 18. sec (−7π) 3. csc 7. csc 5π 6 − π 3 ... |
95◦) 36. tan(−2.01) 37. cot(392.994) 38. sec(207◦) 39. csc(5.902) 40. tan(39.672◦) 41. cot(3◦) 42. sec(0.45) 23. csc(θ) = 25 24 10 91 √ 91 760 Foundations of Trigonometry In Exercises 43 - 57, find all of the angles which satisfy the equation. 43. tan(θ) = √ 3 44. sec(θ) = 2 45. csc(θ) = −1 46. cot(θ) = √ 3 3 47. tan(θ)... |
� and the side opposite θ has length 4, how long is the side adjacent to θ? 71. If θ = 15◦ and the hypotenuse has length 10, how long is the side opposite θ? 72. If θ = 87◦ and the side adjacent to θ has length 2, how long is the side opposite θ? 73. If θ = 38.2◦ and the side opposite θ has lengh 14, how long is the hy... |
because they are alternate interior angles. 762 Foundations of Trigonometry (b) From a firetower 200 feet above level ground in the Sasquatch National Forest, a ranger spots a fire off in the distance. The angle of depression to the fire is 2.5◦. How far away from the base of the tower is the fire? (c) The ranger in part 7... |
. csc(θ) cos(θ) = cot(θ) 88. 90. 92. cos(θ) sin2(θ) = csc(θ) cot(θ) 1 − cos(θ) sin(θ) = csc(θ) − cot(θ) sin(θ) 1 − cos2(θ) = csc(θ) 83. tan(θ) cos(θ) = sin(θ) 85. tan(θ) cot(θ) = 1 87. 89. 91. 93. sin(θ) cos2(θ) = sec(θ) tan(θ) 1 + sin(θ) cos(θ) = sec(θ) + tan(θ) cos(θ) 1 − sin2(θ) sec(θ) 1 + tan2(θ) = sec(θ) = cos(θ) ... |
sec(θ) 1 − sec(θ) tan(θ) − 1 tan(θ) + 1 105. sin(θ) + 1 sin(θ) − 1 = 1 + csc(θ) 1 − csc(θ) 107. 1 − tan(θ) 1 + tan(θ) = cos(θ) − sin(θ) cos(θ) + sin(θ) 108. tan(θ) + cot(θ) = sec(θ) csc(θ) 109. csc(θ) − sin(θ) = cot(θ) cos(θ) 110. cos(θ) − sec(θ) = − tan(θ) sin(θ) 111. cos(θ)(tan(θ) + cot(θ)) = csc(θ) 112. sin(θ)(tan(... |
) cot(θ) cos(θ) 1 + sin(θ) = 1 − sin(θ) cos(θ) 1 − sin(θ) 1 + sin(θ) = (sec(θ) − tan(θ))2 119. 121. 123. 125. 1 sec(θ) − tan(θ) 1 csc(θ) + cot(θ) = sec(θ) + tan(θ) = csc(θ) − cot(θ) 1 1 + sin(θ) 1 1 + cos(θ) = sec2(θ) − sec(θ) tan(θ) = csc2(θ) − csc(θ) cot(θ) 127. csc(θ) − cot(θ) = sin(θ) 1 + cos(θ) 764 Foundations of ... |
diagram from the beginning of the section, partially reproduced below, to answer the following. sin(θ(1, 0) x (a) Show that triangle OP B has area 1 2 sin(θ). (b) Show that the circular sector OP B with central angle θ has area (c) Show that triangle OQB has area 1 2 tan(θ). (d) Comparing areas, show that sin(θ) < θ <... |
= −1 17. tan 20. cot 21. sin(θ) = 3 5, cos(θ) = − 4 5, tan(θ) = − 3 4, csc(θ) = 5 3, sec(θ) = − 5 4, cot(θ) = − 4 3 22. sin(θ) = − 12 13, cos(θ) = − 5 13, tan(θ) = 12 5, csc(θ) = − 13 12, sec(θ) = − 13 5, cot(θ) = 5 12 23. sin(θ) = 24 25, cos(θ) = 7 25, tan(θ) = 24 24, sec(θ) = 25 7, cot(θ) = 7 24 7, csc(θ) = 25 √ 3, ... |
5 √ 5 5, tan(θ) = −2, csc(θ) = − √ 4, tan(θ) = − √ 30 6, tan(θ) = √ 15, csc(θ) = 4 √ 5 5, csc(θ) = − 530, sec(θ) = − √ 5, cot(θ) = − 1 2 √ 5 2, sec(θ) = √ 15 15, sec(θ) = −4, cot(θ) = − √ √ 30 5, cot(θ) = 6, sec(θ) = − √ 15 15 √ 5 3, tan(θ) = 2 √ 5, tan(θ) = 1 5 2, csc(θ) = 3 √ 2, csc(θ) = √ 2 4, sec(θ) = 3, cot(θ) = ... |
081 43. tan(θ) = √ 3 when θ = π 3 36. tan(−2.01) ≈ 2.129 38. sec(207◦) ≈ −1.122 40. tan(39.672◦) ≈ 0.829 42. sec(0.45) ≈ 1.111 + πk for any integer k 44. sec(θ) = 2 when θ = π 3 + 2πk or θ = 5π 3 + 2πk for any integer k 45. csc(θ) = −1 when θ = 46. cot(θ) = √ 3 3 when θ = 3π 2 π 3 + 2πk for any integer k. + πk for any ... |
= − √ 2 3 3 when t = 5π 6 + 2πk or t = 7π 6 61. csc(t) = 0 never happens + 2πk for any integer k 62. cot(t) = − √ 3 when t = 5π 6 5π 6 + πk for any integer k + πk for any integer k when t = √ 3 3 63. tan(t) = − 64. sec(t) = 65. csc(t when t = π 6 + 2πk or t = 11π 6 + 2πk for any integer k when t = + 2πk or t = 2π 3 + ... |
581 feet from the base of the tower. (c) The Sasquatch ran 200 cot(6◦) − 200 cot(6.5◦) ≈ 147 feet in those 10 seconds. This translates to ≈ 10 miles per hour. At the scene of the second sighting, the Sasquatch was ≈ 1755 feet from the tower, which means, if it keeps up this pace, it will reach the tower in about 2 minu... |
(We can construct the angle θ0 by rotating counter-clockwise from the positive x-axis to the terminal side of θ as pictured below.) Since θ and θ0 are coterminal, cos(θ) = cos(θ0) and sin(θ) = sin(θ0). y 1 θ0 y 1 θ0 P (cos(θ0), sin(θ0)) 1 x θ Q(cos(−θ0), sin(−θ0)) 1 x −θ0 We now consider the angles −θ and −θ0. Since θ... |
same as those for θ and −θ, respectively, we get cos(−θ) = cos(θ) and sin(−θ) = − sin(θ), as required. The Even / Odd Identities are readily demonstrated using any of the ‘common angles’ noted in Section 10.2. Their true utility, however, lies not in computation, but in simplifying expressions involving the circular f... |
cos2(α0) − 2 cos(α0) cos(β0) + cos2(β0) + sin2(α0) − 2 sin(α0) sin(β0) + sin2(β0) = cos2(α0) + sin2(α0) + cos2(β0) + sin2(β0) −2 cos(α0) cos(β0) − 2 sin(α0) sin(β0) 2In the picture we’ve drawn, the triangles P OQ and AOB are congruent, which is even better. However, α0 − β0 could be 0 or it could be π, neither of whic... |
ifies to: cos(α0 − β0) = cos(α0) cos(β0) + sin(α0) sin(β0). Since α and α0, β and β0 and α − β and α0 − β0 are all coterminal pairs of angles, we have cos(α − β) = cos(α) cos(β) + sin(α) sin(β). For the case where α0 ≤ β0, we can apply the above argument to the angle β0 − α0 to obtain the identity cos(β0 − α0) = cos(β0)... |
(θ)) cos (θ) + sin sin (θ) = sin(θ) The identity verified in Example 10.4.1, namely, cos π 2 − θ = sin(θ), is the first of the celebrated ‘cofunction’ identities. These identities were first hinted at in Exercise 74 in Section 10.2. From sin(θ) = cos π 2 − θ, we get: π 2 which says, in words, that the ‘co’sine of an angle... |
(β) sin(α − β) = sin(α) cos(β) − cos(α) sin(β) 774 Example 10.4.2. 1. Find the exact value of sin 19π 12 Foundations of Trigonometry 2. If α is a Quadrant II angle with sin(α) = 5 13, and β is a Quadrant III angle with tan(β) = 2, find sin(α − β). 3. Derive a formula for tan(α + β) in terms of tan(α) and tan(β). Solutio... |
could use The Pythagorean Identity cos2(β) + sin2(β) = 1, but we opt instead to use a quotient identity. From tan(β) = sin(β) cos(β), we have sin(β) = tan(β) cos(β) √ 5 5. We now have all the pieces needed to find sin(α − β): so we get sin(β) = (2) 5 so cos(β) = 1 sec(β sin(α − β) = sin(α) cos(β) − cos(α) sin(β) √ 5 5 ... |
= Naturally, this formula is limited to those cases where all of the tangents are defined. The formula developed in Exercise 10.4.2 for tan(α +β) can be used to find a formula for tan(α −β) by rewriting the difference as a sum, tan(α+(−β)), and the reader is encouraged to fill in the details. Below we summarize all of the... |
� squares of cosine and sine via the Pythagorean Identity cos2(θ) + sin2(θ) = 1 and we leave the details to the reader. It is interesting to note that to determine the value of cos(2θ), only one piece of information is required: either cos(θ) or sin(θ). To determine sin(2θ), however, it appears that we must know both s... |
(θ) = x!’ then you have indeed learned something, and we take comfort in that. However, context is everything. Here, ‘x’ is just a variable - it does not necessarily represent the x-coordinate of the point on The Unit Circle which lies on the terminal side of θ, assuming θ is drawn in standard position. Here, x represe... |
cos(2θ) = 2 cos2(θ) − 1, expresses cos(2θ) as a polynomial in terms of cos(θ). We are now asked to find such an identity for cos(3θ). Using the sum formula for cosine, we begin with cos(3θ) = cos(2θ + θ) = cos(2θ) cos(θ) − sin(2θ) sin(θ) Our ultimate goal is to express the right hand side in terms of cos(θ) only. We su... |
2(θ) results in the aptly-named ‘Power Reduction’ formulas below. Theorem 10.18. Power Reduction Formulas: For all angles θ, cos2(θ) = sin2(θ) = 1 + cos(2θ) 2 1 − cos(2θ) 2 Example 10.4.4. Rewrite sin2(θ) cos2(θ) as a sum and difference of cosines to the first power. Solution. We begin with a straightforward application ... |
.4.3 to derive the identity tan θ 2 = sin(θ) 1 + cos(θ) Solution. 1. To use the half angle formula, we note that 15◦ = 30◦ 2 and since 15◦ is a Quadrant I angle, its cosine is positive. Thus we have cos (15◦) = + 1 + cos (30◦) = = Back in Example 10.4.1, we found cos (15◦) by using the difference formula for cosine. In ... |
θ) 1 + cos(θ) 2 = tan cos2 θ 2 1 + cos 2 θ 2 2 (1 + cos(θ)) Our next batch of identities, the Product to Sum Formulas,3 are easily verified by expanding each of the right hand sides in accordance with Theorem 10.16 and as you should expect by now we leave the details as exercises. They are of particular use in Calculus,... |
cosine, cos(−4θ) = cos(4θ). 2. Identifying α = θ and β = 3θ yields sin(θ) − sin(3θ) = 2 sin θ − 3θ 2 cos θ + 3θ 2 = 2 sin (−θ) cos (2θ) = −2 sin (θ) cos (2θ), where the last equality is courtesy of the odd identity for sine, sin(−θ) = − sin(θ). The reader is reminded that all of the identities presented in this sectio... |
(105◦) 12. tan(375◦) 15. tan 13π 12 18. sin π 12 21. sec − π 12 22. If α is a Quadrant IV angle with cos(α) = √ 5 5, and sin(β) = √ 10 10, where π 2 < β < π, find (a) cos(α + β) (b) sin(α + β) (c) tan(α + β) (d) cos(α − β) (e) sin(α − β) (f) tan(α − β) 23. If csc(α) = 3, where 0 < α < π 2, and β is a Quadrant II angle w... |
) cos(β) 32. cos(α + β) − cos(α − β) = −2 sin(α) sin(β) 33. sin(α + β) sin(α − β) = 1 + cot(α) tan(β) 1 − cot(α) tan(β) 34. 36. 37. cos(α + β) cos(α − β) = 1 − tan(α) tan(β) 1 + tan(α) tan(β) 35. tan(α + β) tan(α − β) = sin(α) cos(α) + sin(β) cos(β) sin(α) cos(α) − sin(β) cos(β) sin(t + h) − sin(t) h = cos(t) sin(h) h ... |
. tan(θ) = 53. cos(θ) = 55. cos(θ) = 57. sec(θ) = 12 5 3 5 12 13 √ where π < θ < 3π 2 where 0 < θ < π 2 52. csc(θ) = 4 where π 2 < θ < π 54. sin(θ) = − 4 5 where π < θ < 3π 2 where 5 where 3π 2 3π 2 < θ < 2π 56. sin(θ) = 5 13 where < θ < 2π 58. tan(θ) = −2 where In Exercises 59 - 73, verify the identity. Assume all qua... |
(θ) + 1 (HINT: Use the result to 69.) 71. sec(2θ) = cos(θ) cos(θ) + sin(θ) + sin(θ) cos(θ) − sin(θ) 72. 73. 1 cos(θ) − sin(θ) 1 cos(θ) − sin(θ) + − 1 cos(θ) + sin(θ) 1 cos(θ) + sin(θ) = = 2 cos(θ) cos(2θ) 2 sin(θ) cos(2θ) 10.4 Trigonometric Identities 785 In Exercises 74 - 79, write the given product as a sum. You may ... |
) cos(θ) = √ 1 x2 + 1 (c) sin(2θ) = 2x x2 + 1 (b) sin(θ) = √ x x2 + 1 (d) cos(2θ) = 1 − x2 x2 + 1 89. Discuss with your classmates how each of the formulas, if any, in Exercise 88 change if we change assume θ is a Quadrant II, III, or IV angle. for − for − 90. If sin(θ) = 91. If tan(θ) = 92. If sec(θ) = x 2 x 7 x 4, fin... |
do the same for sin(5θ)? What about for sin(4θ)? If not, what goes wrong? 98. Verify the Even / Odd Identities for tangent, secant, cosecant and cotangent. 99. Verify the Cofunction Identities for tangent, secant, cosecant and cotangent. 100. Verify the Difference Identities for sine and tangent. 101. Verify the Produc... |
30 2 −4 + 7 30 √ √ 28 + 4 − 7 2 2 = − 24. (a) sin(α + β) = 16 65 (b) cos(α − β) = 33 65 (c) tan(α − β) = 63 + 100 √ 2 41 56 33 788 Foundations of Trigonometry 25. (a) csc(α − β) = − 5 4 (b) sec(α + β) = 125 117 (c) cot(α + β) = 117 44 39. cos(75◦) = 41. cos(67.5◦) = √ 40. sin(105◦) = 42. sin(157.5◦) = √ 1 − √ 2 44. co... |
√ 5 5 24 7 tan(2θ) = − tan θ 2 = 1 2 10.4 Trigonometric Identities 789 54. sin(2θ) = sin θ 2 = 24 25 √ 2 5 sin(2θ) = − √ sin θ 2 = sin(2θ) = − sin θ 2 = 5 5 120 169 26 26 120 169 √ 26 26 4 5 55. 56. 57. cos(2θ) = − cos θ 2 = − cos(2θ) = cos θ 2 = − 26 26 cos(2θ) = cos θ 2 = 7 25 √ 5 5 119 169 5 √ 119 169 √ 26 26 3 5 t... |
� 85. − θ − 2 sin π 4 √ 92. ln |x + x2 + 16| − ln(4) 790 Foundations of Trigonometry 10.5 Graphs of the Trigonometric Functions In this section, we return to our discussion of the circular (trigonometric) functions as functions of real numbers and pick up where we left off in Sections 10.2.1 and 10.3.1. As usual, we beg... |
an exercise for you.) Returning to the circular functions, we see that by Definition 10.3, f (t) = cos(t) is periodic, since cos(t + 2πk) = cos(t) for any integer k. To determine the period of f, we need to find the smallest real number p so that f (t + p) = f (t) for all real numbers t or, said differently, the smallest... |
we go to graph these functions. 4In some advanced texts, the interval of choice is [−π, π). 10.5 Graphs of the Trigonometric Functions 791 corners or cusps. As we shall see, the graphs of both f (t) = cos(t) and g(t) = sin(t) meander nicely and don’t cause any trouble. We summarize these facts in the following theorem... |
aesthetics. Below is an accurate-to-scale graph of y = cos(x) showing several cycles with the ‘fundamental cycle’ plotted thicker than the others. The 5The use of x and y in this context is not to be confused with the x- and y-coordinates of points on the Unit Circle which define cosine and sine. Using the term ‘trigon... |
orem 1.7 in Section 1.7 to graph more complicated curves. To do so, we need to keep track of 2 units. A visual inspection confirms this. 10.5 Graphs of the Trigonometric Functions 793 the movement of some key points on the original graphs. We choose to track the values x = 0, π 2, π, 3π 2 and 2π. These ‘quarter marks’ c... |
cycle is graphed on [1, 5] so the period is the length of that interval which is 4. One cycle of y = f (x). 2. Proceeding as above, we set the argument of the sine, π − 2x, equal to each of our quarter marks and solve for x. 794 Foundations of Trigonometry a π − 2x = a x π π − 2x = 0 0 2 π π π − 2x = π 2 4 2 π − 2x = ... |
x), so of course, the reverse is true: f (x) = cos(x) can be written as a transformed version of g(x) = sin(x). The authors have seen some instances where sinusoids are always converted to cosine functions while in other disciplines, the sinusoids are always written in terms of sine functions. We will discuss the appli... |
. We can always ensure ω > 0 using the Even/Odd Identities.7 We now test out Theorem 10.23 using the functions f and g featured in Example 10.5.1. First, we write f (x) in the form prescribed in Theorem 10.23, ω induced by φ. f (x) = 3 cos πx − π 2 + 1 = 3 cos π 2 x + − π 2 + 1, 7Try using the formulas in Theorem 10.23... |
is then 2π Instead of the graph starting at x = π 2, ω = 2, φ = −π and B = 3 2 = π We find A = − 1 2 = π, the amplitude is − 1 2 (indicating a shift right π 2 units) and the vertical shift is up 2 3 2. Note that, in this case, all of the data match our graph of y = g(x) with the exception of the phase shift. 2, it ends... |
B = 1 2. Our final answer is 2 + 1 C(x) = 2 cos π 2. 3. To find the amplitude, note that the range of the sinusoid is − 3 2 + − 3 5 2 (1. Most of the work to fit the data to a function of the form S(x) = A sin(ωx + φ) + B is done. The period, amplitude and vertical shift are the same as before with ω = π 3, A = 2 and B =... |
cos(φ) + A cos(ωx) sin(φ) + B. Making these observations allows us to recognize (and graph) functions as sinusoids which, at first glance, don’t appear to fit the forms of either C(x) or S(x). Example 10.5.3. Consider the function f (x) = cos(2x) − √ 3 sin(2x). Find a formula for f (x): 1. in the form C(x) = A cos(ωx + ... |
know cos2(φ) + sin2(φ) = 1, so multiplying this by A2 gives A2 cos2(φ)+A2 sin2(φ) = A2. Since A cos(φ) = 1 and A sin(φ) = 3)2 = 4 or A = ±2. Choosing A = 2, we have 2 cos(φ) = 1 and 2 sin(φ) = 3 or, after some √ 3 rearrangement, cos(φ) = 1 2. One such angle φ which satisfies this criteria is. We can easily φ = π check ... |
2(φ) = A2 as before, we get A = ±2, and again we choose A = 2. √ 3 This means 2 sin(φ) = 1, or sin(φ) = 1 2.. One such angle which meets these criteria is φ = 5π 6 Checking our work analytically, we have 3, which means cos(φ) = − 6. Hence, we have f (x) = 2 sin 2x + 5π 2, and 2 cos(φ) = − √ √ f (x) = 2 sin 2x + 5π 6 + ... |
y = sec(x). Since sec(x) = 1 cos(x), we can use our table of values for the graph of y = cos(x) and take reciprocals. We know from Section 10.3.1 that the domain of F (x) = sec(x) excludes all odd multiples of π 2, and sure enough, we run into trouble at 2 and x = 3π x = π 2 since cos(x) = 0 at these values. Using the... |
β) are defined, sec(α) = sec(β) if and only if cos(α) = cos(β). Hence, sec(x) inherits its period from cos(x). 12In Section 10.3.1, we argued the range of F (x) = sec(x) is (−∞, −1] ∪ [1, ∞). We can now see this graphically. 10.5 Graphs of the Trigonometric Functions 801 As one would expect, to graph y = csc(x) we begin... |
the secant and cosecant functions. Note that 13Just like the rational functions in Chapter 4 are continuous and smooth on their domains because polynomials are continuous and smooth everywhere, the secant and cosecant functions are continuous and smooth on their domains since the cosine and sine functions are continuo... |
5 Graphs of the Trigonometric Functions 803 Next, we substitute these x values into f (x). If f (x) exists, we have a point on the graph; otherwise, we have found a vertical asymptote. In addition to these points and asymptotes, we have graphed the associated cosine curve – in this case y = 1 − 2 cos(2x) – dotted in th... |
oid forms mentioned in Theorem 10.23. Since these quantities match those of the corresponding cosine and sine curves, we do not spell this out explicitly. Finally, since the ranges of secant and cosecant are unbounded, there is no amplitude associated with these curves. 10.5.3 Graphs of the Tangent and Cotangent Functi... |
(x))(0) = tan(x), which tells us the period of tan(x) is at most π. To show that it is exactly π, suppose p is a positive real number so that tan(x + p) = tan(x) for all real numbers x. For x = 0, we have tan(p) = tan(0 + p) = tan(0) = 0, which means p is a multiple of π. The smallest positive multiple of π is π itself... |
icking the proof that the period of tan(x) is an option; for another approach, consider transforming tan(x) to cot(x) using identities. 806 Foundations of Trigonometry y x The graph of y = cot(x). The properties of the tangent and cotangent functions are summarized below. As with Theorem 10.24, each of the results belo... |
� 2 π 2 π x We see that the period is π − (−π) = 2π. One cycle of y = 1 − tan x 2. 2. The ‘quarter marks’ for the fundamental cycle of the cotangent curve are 0, π To graph g(x) = 2 cot π and solving for x. 2 x + π + 1, we begin by setting π 4 and π. 2 x + π equal to each quarter mark 4, π 2, 3π 808 Foundations of Trig... |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.