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ples of p 2. The range of the tangent function is all real numbers. Because the tangent function has a period of domain, p, for any number t in its tan 1 t Β± p 2 tan t. Example 3 Finding All t-values State all values of t for which tan t is 1. Solution p, The tangent function repeats its pattern of y-values at intervals of so there y tan t are an infinite number t-values for which The graph of shown in Figure 7.1-4 highlights a few points with a y-coordinate of 1. tan t is 1. y y = tan t β 5Ο 4 β2Ο βΟ 2 β Ο 4 0 β2 3Ο 4 Ο 7Ο 4 t 2Ο y = β1 Ο Ο Ο Figure 7.1-4 p 2, p 2 b S On the interval, highlighted in red on the graph above, the graph of y tan t has only one point, p a 4, 1 b, at which the y-coordinate is Technology Tip p Most calculators have a window setting that automatically rescales the horizontal axis in fractional units of when in radian mode. On TI models, select ZTRIG in the ZOOM menu, and on Casio, select F3 (V-Window) then F2 (TRIG) from GRAPH mode. Section 7.1 Graphs of the Sine, Cosine, and Tangent Functions 481 1. t p 4 Therefore, all values of t for which kp, where k is any integer. tan t is 1 can be expressed as β h(t) = 4 cos t f(t) = cos t 4 β2Ο 2Ο β4 Figure 7.1-5 Basic Transformations of Sine, Cosine, and Tangent The graphical transformations (such as shifting and stretching) that were considered in Section 3.4 also apply to trigonometric graphs. Example 4 Vertical Stretch List the transformation needed to change the graph of the graph of Graph both equations on the same screen. 4 cos t. h t f t 1 2 cos t into 1 2 Solution h Because by a factor of 4. Both graphs are identified in Figure 7.1-5. the graph of h is the graph of f after a vertical stretch t t, 2 1 2 1 4 f β Example 5 Reflection and Vertical Stretch Graph g 1 2 t 2 1 sin t on the interval 2p, 2 |
p. 4 3 Solution The graph of g is the graph of t f 1 2 sin t 1 2 compressed vertically by a factor of, as shown in Figure 7.1-6. reflected across the x-axis and y 1 β1 1 g(t) = β sin t 2 t 2Ο Ο f(t) = sin t β2Ο βΟ Figure 7.1-6 β Example 6 Vertical Shift tan t 5 on the interval 3p, 3p. 4 3 Graph h t 2 1 Solution The graph of h is the graph of tan t shifted up 5 units. t f 1 2 482 Chapter 7 Trigonometric Graphs NOTE For a complete discussion of symmetry and odd and even functions, see Excursion 3.4.A. y 8 6 4 2 0 β2 β3Ο β2Ο βΟ t Ο 2Ο 3Ο Figure 7.1-7 β Even and Odd Functions Trigonometric functions can be classified as odd or even as determined by their symmetry. Even Functions A graph is symmetric with respect to the y-axis if the part of the graph on the right side of the y-axis is the mirror image of the part on the left side of the y-axis. Graphing Exploration 1. For each pair of functions f and g below, answer the following questions. cos t sin t tan and and and g g g t cos t 2 1 2 1 t sin t 2 1 2 1 t tan t 2 1 2 1 β’ Is f symmetric with respect to the y-axis? β’ Does the graph of g appear to coincide with the graph of f? 2. If a graph is symmetric with respect to the y-axis, describe the graph after a reflection across the y-axis. A function f whose graph is symmetric with respect to the y-axis is called an even function. Even Function A function f is even if f (x) f(x) for every x in the domain of f. The graph of an even function is symmetric with respect to the y-axis. Section 7.1 Graphs of the Sine, Cosine, and Tangent Functions 483 f For example, t cos 1 1 2 cos t t 2 cos t is an even function because for every t in the domain of f cos t. t 2 1 Odd Functions If a graph is symmetric with respect to the origin, then whenever (x, y) is on is also on the graph. A function |
f whose graph is symthe graph, metric with respect to the origin is called an odd function. x, y 2 1 Odd Function A function f is odd if f(x) f(x) for every x in the domain of f. The graph of an odd function is symmetric with respect to the origin. f For example, t 1 t 1 sin tan 2 2 1 sin t t 2 sin t tan t and g t 2 1 tan t are odd functions because for every t in the domain of f for every t in the domain of g sin t tan t. t t 2 2 1 1 Summary of the Properties of Sine, Cosine, and Tangent Functions Function Symbol Domain Range Period Even/Odd sine cosine tangent sin t all real numbers cos t all real numbers all real numbers from to 1, inclusive all real numbers from to 1, inclusive 1 1 tan t all real numbers except all real numbers odd multiples of p 2 2p 2p p odd even odd Exercises 7.1 In Exercises 1 β 6, graph each function on the given interval. 6. h t 2 1 tan t; 5p 3 S, 3p T 1. 2. h 4. f t t 2 2 1 1 sin t; cos t; tan t; sin t; 3 2p, 6p 4 p, 3p 4 3 3 p, 2p 4 3 5p, 3p 5. g t 2 1 cos t; 7p 6, 7p 2 T S 4 7. For what values of t on the interval sin t 1? 8. For what values of t on the interval cos t 0? 2p, 2p 2p, 2p 3 3 is is 4 4 9. What is the maximum value of g cos t? t 2 1 10. What is the minimum value of f sin t? t 2 1 484 Chapter 7 Trigonometric Graphs 11. For what values of t on the interval tan t 3 have vertical 4 2p, 2p does the graph of asymptotes? h t 2 1 12. What is the y-intercept of the graph of sin t? f t 2 1 13. What is the y-intercept of the graph of g t 2 1 cos t? 14. What is the y-intercept of the graph of tan t? h t 2 1 15. For what values of t on the interval p, p 3 is f t 2 1 4 |
sin t increasing? 16. For what values of t on the interval cos t decreasing? g t 2 1 17. For what values of t on the interval tan t greater than 1? 18. For what values of t on the interval tan t less than 0? 19. For what values of t on the interval tan t increasing? h t 2 1 3p, p is 4 2p, 2p 2p, 2p is is 4 4 p, 2p is 4 3 3 3 3 In Exercises 20β33, find all the exact t-values for which the given statement is true. 20. tan t 0 22. sin t 0 24. tan t 1 26. cos t 0 28. sin t 1 21. sin t 22 2 23. cos t 1 2 25. sin t 23 2 27. cos t 23 2 29. sin t 1 2 30. tan t 23 3 31. cos t 22 2 32. cos t 1 33. tan t 23 In Exercises 34β43, list the transformations that change the graph of f into the graph of g. State the domain and range of g. 34. 35. f t 2 1 f t 2 1 cos t; g cos t 2 t 2 1 cos t; g cos t t 2 1 36. 37. 38. 39. 40. 41. 42. 43 sin t; g 1 tan t; g tan t; g t 1 2 3 sin t t 2 t 1 tan t 5 2 tan t cos t; g t 1 sin t; g t 1 2 3 cos t 2 2 sin t sin t; g 1 cos t; g sin t; g t 1 2 3 sin t 2 t 2 t 1 5 cos t 3 2 sin t 3 In Exercises 44β 48, sketch the graph of each function. 44. 46 cos t 4 tan t 48. f t 2 1 3 sin t 1 2 45. f t 2 1 5 sin t 1 47. f t 2 1 1 4 cos t In Exercises 49β54, match a graph to a function. Only one graph is possible for each function 49. 51. 53. a. 2Ο b. 2Ο 2 tan t sin t 1 3 tan t 1 50. 52. 54.5 cos t 2.5 sin t cos t 1 3 3 3 3 2Ο 2Ο Section 7.1 Graphs of the Sine, Cosine, and Tangent Functions 485 c. 3 2Ο |
2Ο 60. Scientists theorize that the average temperature at a specific location fluctuates from cooler to warmer and then to cooler again over a long period of time. The graph shows a theoretical prediction of the average summer temperature for the last 150,000 years for a location in Alaska. d. e. f. 3 3 Ο Ο 3 3 2Ο 2Ο 3 3 Ο Ο 3 55. Fill the blanks with βevenβ or βoddβ so that the resulting statement is true. Then prove the statement by using an appropriate identity. Excursion 3.4.A may be helpful. a. b. c. d. e. is an ___ function. is an ___ function. is an ___ function. sin t cos t tan t t sin t t tan t is an ___ function. is an ___ function In Exercises 56β59, find tan t, where the terminal side of an angle of t radians lies on the given line. 56. y 1.5x 58. y 0.32x 57. y 1.4x 59. y 11x 85 80 75 70 65 60 55 150000 β100000 β50000 Years Ago a. Find the highest and lowest temperature represented. b. Over what time interval does the temperature repeat the cycle? c. What is the estimated average summer temperature at the present time? 61. A rotating beacon is located at point P, 5 yards from a wall. The distance d, as measured along the wall, where the light shines is given by d 5 tan 2pt where t is time measured in seconds since the beacon began to rotate. When the light is aimed at point A. When the beacon is aimed to the right of A, the distance d is positive, and when it is aimed to the left of A, the value of d is negative. t 0, d A 5 yds P Graph the function and estimate the value of d for the following times. b. a. c. d. e. Determine the position of the beacon when t 0 t 0.7 t 0.5 t 1.4 t 0.25 of d for that value of t. and discuss the corresponding value 486 Chapter 7 Trigonometric Graphs 7.2 Graphs of the Cosecant, Secant, and Cotangent Functions Objectives β’ Graph the cosecant, secant, and cotangent functions β’ Graph transformations of the cosecant, |
secant, and cotangent graphs y sin t, y cos t, and y tan t The graphs of that were developed in Section 7.1 are closely related to the graphs of the reciprocal functions y csc t, y sec t, and y cot t that are studied in this section. Graph of the Cosecant Function The general shape of the graph of the graph of the sine function and the fact that csc t 1 sin t f t 1 2. csc t can be determined by using Graphing Exploration Graph the two functions below on the same screen in a viewing window with 2p t 2p 4 y 4 and sint g f t 2 1 1 sint t 2 1 How are the graphs alike and how are they different? sin t csc t and are reciprocals, csc t Because that is, when t is an integer multiple of t f 1 of csc t t csc t is all real numbers except integer multiples of has vertical asymptotes at integer multiples of p. is not defined when p. sin t 0; Therefore, the domain of and the graph p, 2 f 1 2 Graph of the Cosecant Function β2Ο βΟ y = csc t y = sin t t Ο 2Ο y 4 2 β2 β4 Notice that as the graph of csc t of decreases to a height of 1, f 1 t 2 decreases to a height of 1, and as the graph of increases to a height of 1, the graph y sin t increases to a height csc t f t y sin t the graph of 1 2 Section 7.2 Graphs of the Cosecant, Secant, and Cotangent Functions 487 1. The range of of to 1 or less than or equal to f t 1 2 csc t 1. is all real numbers greater than or equal 2p. The period of the cosecant function is Example 1 Reflection and Vertical Stretch 3 csc t. Graph h t 2 1 Solution y 3 sin t, y sin t First consider the graph of stretched vertically by a factor of 3 and reflected across the horizontal and that of axis. The relationship between the graph of 3 csc t h as shown in Figure 7.2-1. y csc t and y sin t, is similar to that between which is the graph of y 3 sin t t 1 2 y 8 4 0 β4 β8 h(t) = |
β3 csc t y = β3sin t t Ο 2 Ο 2Ο 3Ο 2 β2Ο β 3Ο 2 βΟ β Ο 2 Figure 7.2-1 β Graph of the Secant Function t The graph of 2 that the graph of f 1 sec t f 1 t 2 csc t is related to the cosine graph in the same way is related to the sine graph. Graphing Exploration Graph the two functions below on the same screen in a viewing window with 2p t 2p 4 y 4. and cos t g f t 2 1 1 cos t t 2 1 How are the graphs alike and how are they different? Because cos t and sec t are reciprocals, sec t that is, when t is an odd multiple of is not defined when p 2. Therefore, the domain of cos t 0; f t sec t 2 f 1 1 t of 2 or equal to sec t 1. is all real numbers except odd multiples of p 2, and the range is all real numbers greater than or equal to 1 or less than The period of the secant function is 2p. 488 Chapter 7 Trigonometric Graphs Graph of the Secant Function β2Ο βΟ g(t) = sec t y = cos t Ο t 2Ο y 4 2 β2 β4 Example 2 Vertical Stretch and Vertical Shift 2 sec t 3. Graph g t 2 1 Solution y 2 cos t 3, stretched First graph vertically by a factor of 2 and shifted down 3 units. The graphs of g are related in the same way as the 1 graphs of y 2 cos t 3 y cos t, as shown in Figure 7.2-2. which is the graph of 2 sec t 3 y sec t and and t 2 y cos t β2Ο βΟ y g(t) = 2 sec t β 3 Ο 2Ο t y = 2 cos t β 3 6 4 2 0 β2 β4 β6 β8 β10 Figure 7.2-2 β Graph of the Cotangent Function Because cot t cost sint ing the quotient, the graph of cot t t f 1 2 can be obtained by graph- y cos t sin t. The cotangent function is not defined when whenever t is an integer multiple of f(t) the range of f(t) has vertical asymptotes at integer multiples of p. p. sin t 0, and this occurs |
Therefore the domain of p, and cot t t f 1 2 cot t consists of all real numbers except integer multiples of cot t is the set of real numbers. The graph of Section 7.2 Graphs of the Cosecant, Secant, and Cotangent Functions 489 Graph of the Cotangent Function y 4 2 0 β2 β4 y = tan t f(t) = cot t t Ο 2 Ο 2Ο 3Ο 2 β2Ο β 3Ο 2 βΟ β Ο 2 Notice that as the graph of decreases, and as the graph of increases. The period of the cotangent function is increases, the graph of decreases, the graph of p. y tan t y tan t cot t cot t t t f 1 f 1 2 2 Example 3 Reflection, Vertical Stretch, and Horizontal Shift Graph k t 2 1 3 cot t p 4 b. a Solution The graph of k zontal shift of 2 t 1 p 4 3 cot t p 4 b a is the graph of y cot t after a hori- units to the right, a reflection across the horizontal axis, and a vertical stretch by a factor of 3. The graph of k 3 cot t 2 1 t p 4 b a is shown with the graph of y cot t β2Ο βΟ y 8 6 4 2 0 β2 β4 β6 β8 in Figure 7.2-3 below. k(t) = β3 cot ( y = cot t t β Ο 4 ) Ο t 2Ο Figure 7.2-3 β Even and Odd Functions The fact that the cosecant, secant, and cotangent functions are reciprocals of the sine, cosine, and tangent functions, respectively, can be used to determine whether the functions are even or odd. 490 Chapter 7 Trigonometric Graphs The secant function is an even function, as shown below. sec t 2 1 1 t cos 1 2 1 cos t sec t The cosecant and cotangent functions are odd functions. csc t 1 cot t 2 1 1 t 2 sin 1 t cos 1 t sin 1 2 2 2 1 sin t 1 sin t csc t cos t sin t cos t sin t cot t Summary of the Properties of Secant, Cosecant, and Cotangent Functions Function Symbol Domain Range Period Even |
/Odd secant sec t t f 1 2 all real numbers except odd multiples of p 2 all real numbers less 1 than or equal to greater than or equal to 1 or cosecant csc t t f 1 2 all real numbers except multiples of p all real numbers less 1 than or equal to greater than or equal to 1 or cotangent cot t t f 1 2 all real numbers except multiples of p all real numbers 2p 2p p even odd odd Exercises 7.2 11. 12. t sec t f 1 and shifted 1 unit to the left 2 stretched vertically by a factor of 3 t csc t f 1 and shifted 1 unit up 2 compressed vertically by a factor of 0.5 In Exercises 1β10, describe the transformations that change or h(t) cot t f(t) csc t, g(t) sec t, into the graph of the given function. the graph of 1. s t 1 3. m 1 2 t 3 sec t 2 csc t 2 1 2 4 5. p t 2 1 1 2 sec t 1 2. 4. k t 2 1 t 3 2 5 cot 1 2 cot t 2 1 13. t f 1 2 sec t reflected across the horizontal axis and 2 csc t compressed vertically by a factor of 1 4 7. q t 2 1 sec t 2 1 8 8. s t 2 1 5 cot t 2 1 2 9. v t 2 1 p csc t 10. j t 2 1 1 4 sec t 2 1 14. 15. t cot t f 1 shifted down 2 units 2 reflected across the vertical axis and csc t t f 1 2 down shifted p 2 units to the right and 5 units In Exercises 11β17, state the rule of a function g whose graph is the given transformation of the graph of f. 16. csc t f t 2 1 of 0.75 compressed vertically by a factor Section 7.2 Graphs of the Cosecant, Secant, and Cotangent Functions 491 17. t cot t f 1 across the horizontal axis 2 reflected across the vertical axis and In Exercises 18β21, match graph a, b, c, or d with each function. a. b. c. d. β1Ο β Ο 2 β1Ο β Ο 2 β1Ο β Ο 2 β1Ο β 2 β4 β |
6 6 4 2 β2 β4 β6 6 4 2 β2 β4 β6 6 4 2 β2 β4 β6 0 0 0 0 t 1Ο t 1Ο t 1Ο t 1Ο Ο 2 Ο 2 Ο 2 Ο 2 18. f t 2 1 1 2 cot t 1 19. f t 2 1 1 2 cot t 1 20. t f 1 2 2 cot t 1 21. t f 1 2 2 cot t 1 In Exercises 22 β 25, match graph a or b with each function. a. b. β1Ο β 3Ο 4 β Ο 2 β Ο 4 β1Ο β 3Ο 1 β2 β3 β4 β5 5 4 3 2 1 β1 β2 β3 β4 β5 0 0 t t Ο 4 Ο 2 3Ο 4 1Ο Ο 4 Ο 2 3Ο 4 1Ο 22. 24. f t 2 1 f t 2 1 csc t csc t 2 1 23. 25. f t 2 1 f t 2 1 csc t csc t 2 1 In Exercises 26 β 29, match graph a or b with each function. a1 β2 β3 β4 β5 0 Ο 4 Ο 2 3Ο 4 1Ο 5Ο 4 3Ο 2 t 492 b. Chapter 7 Trigonometric Graphs y 5 4 3 2 1 β1 β2 β3 β4 β5 0 Ο 4 Ο 2 3Ο 4 1Ο 5Ο 4 3Ο 2 β Ο 2 β Ο 4 26. 28. f f t t 2 2 1 1 sec t 2 1 sec t 2 1 27. 29 sec 1 sec t 1 2 2 In Exercises 30 β 33, match graph a or b with each function. a. b. β1Ο β 3Ο 4 β Ο 2 β Ο 4 β1Ο β 3Ο 1 β2 β3 β4 β5 5 4 3 2 1 β1 β2 β3 β4 β5 0 0 t t Ο 4 Ο 2 3Ο 4 1Ο Ο 4 Ο 2 3Ο 4 1Ο 30. 32. f f t t 2 2 1 1 cot t 2 1 cot t 2 1 31. 33. f f t t 2 2 1 1 cot t 2 1 cot t 2 1 t In Exercises 34β38, graph at least |
one cycle of the given function. 34. 36. 38 sec 2t 35. f t 2 1 cot 3t 4 5 csc t a p 2 b 37. f t 2 1 3 4 csc t 2 3 sec t p 2 1 39. Critical Thinking Show graphically that the equation sec t t but none between has infinitely many solutions, p 2 and p 2. 40. Critical Thinking A rotating beacon is positioned 5 yards from a wall at P. In the figure, the distance a is given by a 5 sec 2pt, 0 where t is the number of seconds since the beacon began to rotate. 0 d A 5 yds a. Use the graph of a as a function of t to find a for the following times. t 0 t 0.75 t 1 b. For what values of t is c. How fast is the beacon rotating? a 5? Section 7.3 Periodic Graphs and Amplitude 493 7.3 Periodic Graphs and Amplitude Objectives β’ State the period and amplitude (if any) given the function rule or the graph of a sine, cosine, or tangent function β’ Use the period and amplitude (if any) to sketch the graph of a sine, cosine, or tangent function A surprisingly large number of physical phenomena can be described by functions like the following: 5 sin and g 4 cos 0.5t 1 3t In this section and in the next, the graphs of such functions will be analyzed. All of these functions are periodic and their graphs consist of a series of identical waves. A single wave of the graph is called a cycle. The length of each cycle is the period of the function. 2 0 2 sine cycle 2Ο Figure 7.3-1 Every cycle for the sine function resembles the graph of 0 to 2p, as shown in Figure 7.3-1. t f 1 2 sin t from β’ beginning at a point midway between its maximum and minimum value β’ rising to its maximum value β’ falling to its minimum value β’ returning to the beginning point Every cycle repeats the same pattern. 2 0 2 cosine cycle 2Ο Figure 7.3-2 Similarly, every cycle for the cosine function resembles the graph of g as shown in Figure 7.3-2. from 0 to cos t 2p, t 1 2 494 Chapter 7 Trigonometric Graphs β’ beginning at its maximum value β’ falling to its minimum value β’ returning to the beginning point Again, every cycle |
repeats the same pattern. Period Before proceeding to the discussion about functions that have different periods, it will be helpful to consider functions of the form t f 1 2 sin bt and g 1 t 2 cos bt where b is a constant. The constant b changes the period of the sine or cosine function. Its effect on the graph is to increase or decrease the length of each cycle. Graphing Exploration Graph each function below, one at a time, in a viewing window Answer the questions that follow for each funcwith tion. 0 t 2p. t f 1 2 cos 4t h 1 t 2 sin 5t Determine the number of complete cycles between 0 and 2p. Find the period, or length of one complete cycle. Hint: Use division. The exploration above suggests the following rule. Period of sin bt and cos bt If b 77 0, then the graph of either f(t) sin bt or g(t) cos bt makes b complete cycles between 2P b has a period of. 0 and 2P, and each function y 1 0 β1 g(t) = sin 3t t Ο 3 2Ο 3 Ο 4Ο 3 5Ο 3 2Ο t h The graph of 0 to 2p. on values from 1 g Similarly, the graph of 0 to 2p. 2 sin t 1 2 completes one cycle as t takes on values from completes one cycle as 3t takes sin 3t t When 3t 0, t must be 0. When 3t 2p, t must be 2p 3. Figure 7.3-3 ues from 0 to 2p 3, as shown in Figure 7.3-3. Therefore, the graph of g sin 3t t 2 1 completes one cycle as t takes on val- Section 7.3 Periodic Graphs and Amplitude 495 Example 1 Determining Period Determine the period of each function. a. b. k t 2 1 f t 2 1 cos 3t sin t 2 Solution a. The function Figure 7.3-4. k t 2 1 cos 3t has a period of 2p b 2p 3, as shown in y 1 β1 k(t) = cos 3t Ο 3 2Ο 3 Ο 4Ο 3 5Ο 3 t 2Ο 1 cycle 1 cycle 1 cycle Figure 7.3-4 b. Rewrite f sin t 2 1 has a period of 2p b t as f t 1 2 |
2p 1 2 sin 2 4p, 1 2 a. t b The function f sin t 2 1 1 2 a t b as shown in Figure 7.3-5. β2Ο β1Ο f(t) = sin t 2 1Ο t 2Ο y 1 β1 1 cycle Figure 7.3-5 β 496 Chapter 7 Trigonometric Graphs CAUTION t A calculator may not produce an accurate graph of g cosbt for large values of b. For instance, the graph of sin 50t sinbt or has 50 complete cycles between 0 and your calculator will show. (try it!) 2p, but that is not what Graphing Exploration Graph each function below, one at a time, in a viewing window with p 2 t p 2. Answer the questions that follow for each function. f t 2 1 tan 3t g tan 4t t 2 1 Determine the number of complete cycles between p 2 and p 2. Find the period, that is, the length of one complete cycle. The exploration above suggests the following rule. Period of tan bt If b 77 0, then the graph of f(t) tan bt makes b complete cycles between function has a period of p b. p 2 and p 2, and the y 4 2 0 β2 β4 β Ο 2 β Ο 4 Figure 7.3-6 Ο 4 Ο 2 k(t) = tan 2t Example 2 Determining Period Determine the period of each function. t a. k t 2 1 tan 2t b. f t 2 1 tan t 3 Solution a. The function k cycle between t 2 1 p 4 tan 2t has a period of p b p 2. It completes one and p 4, as shown in Figure 7.3-6. Section 7.3 Periodic Graphs and Amplitude 497 y f(t) = tan 4 2 t 3 t 0 βΟ 3Ο β 2 β Ο 2 Ο 2 Ο 3Ο 2 2 1 3p, of p b p 1 3 Amplitude b. Rewrite as f t tan t 3 as f t 2 1 tan 1 3 a. t b The function has a period as shown in Figure 7.3-7. β β4 Figure 7.3-7 Recall from Section 3.4 that multiplying the rule of a function by a positive constant has the effect of stretching or compressing its graph vertically. Example 3 Vertical and Horizontal St |
retches or Compressions y g(t) = 7 cos 3t Graph each function. k(t) = cos 3t a. g t 2 1 7 cos 3t b. h t 2 1 1 3 sin t 2 7 1 βΟ β 2Ο 3 β Ο 3 Ο 3 2Ο 3 t Ο Solution a. The function is the function by 7. Consequently, the graph of g is the graph of k (see Example 1a) stretched vertically by a factor of 7. multiplied g k t t 2 2 1 1 7 cos 3t cos 3t β7 Figure 7.3-8 y f(t) = sin t 2 t Ο 2Ο 1 h(t) = sin 3 t 2 1 1 3 β2Ο βΟ 0 β 1 3 β1 Figure 7.3-9 As Figure 7.3-8 shows, stretching the graph affects only the height of the waves in the graph, not the period of the function. So the period cos 3t, of g is the same as that of namely k t. 2p b 2p 3 1 2 b. The function h 1 3 t 2 1 sin t 2 is the function f sin t 2 1 t 2 multiplied by 1 3. Consequently, the graph of h is the graph of f (see Example 1b) vertically compressed by a factor of 2p 1 2 as the period of f, namely 2p b The period of h is the same. 1 3 4p. β As the graphs in Example 3 illustrate, vertically stretching or compressing the graph affects only the height, not the period of the function. 7 cos 3t 1 2 t g The graph of in Example 3 reaches a maximum value of 7 units above the horizontal axis and a minimum value of 7 units below a cos bt the horizontal axis. In general, the graph of units above and below the horizontal axis, and reaches a distance of The graph of g(t) 7 cos 3t has an is said to have an amplitude of amplitude of 7. a sin bt g or 498 Chapter 7 Trigonometric Graphs Amplitude and Period If a 0 and b 77 0, f(t) a sin bt then each of the functions or g(t) a cos bt has an amplitude of and a period of a 00 00 2p b. Example 4 Determining Amplitude and Period Determine the amplitude and period of p 2 T the interval p 2,. S |
Solution 2 sin 4t. Then graph f on f t 2 1 The amplitude of 2 sin 4t is t f 1 2 t f 1 2p b 2 sin 4t 2 p 2p 2 4. is a 0 0 2 0 0 2, and the period of So the graph of f consists of cycles that are p 2 long and rise and fall between the heights of 2 and 2. To graph this function, be sure to notice that its graph is the reflection of h across the horizontal axis, as shown in Figure 7.3-10. 2 sin 4t t 1 2 β Although the graph of any function can be vertically stretched or compressed, amplitude only applies to bounded periodic functions. y f(t) = β2 sin 4t 2 h(t) = 2 sin 4t Figure 7.3-10 Exercises 7.3 In Exercises 1β15, state the amplitude (if any) and period of each function. 1. 3. 5. 7. 9 cos t cos 3t 4 cos t 5 tan 2t 0.3 sin t 3 2. 4. 6. 8. 10 sin 2t 2.5 tan t 3 sin t 1.2 cos 0.5t tan 0.4t 11. f t 2 1 1 2 sin 3t 12. f t 2 1 1 2 tan 3t 13. f t 2 1 5 cos 1.7t 15. f t 2 1 1 3 tan pt 4 14. f t 2 1 2 sin 2pt 3 16. a. What is the period of f b. For what values of t (with t 2 1 sin 2pt? 0 t 1 ) is c. For what values of t (with d. For what values of t (with 0 t 1 ) is 0 t 1 ) is 17. a. What is the period of f b. For what values of t (with t 2 1 cos pt? 0 t 2 ) is c. For what values of t (with d. For what values of t (with 0 t 2 ) is 0 t 2 ) is 0? 1? 1? 0? 1? 1 18. a. What is the period of f 2 b. For what values of t (with 1 t tan pt? 1 2 t 1 2 ) is 0? f t 2 1 c. For what values of t (with 1? f t 2 1 d. For what values of t (with 1 ) is 1 2 t 1 2 ) |
is In Exercises 19β38, describe the transformations that change the graph of f into the graph of g. State the amplitude (if any) and the period of g. Section 7.3 Periodic Graphs and Amplitude 499 In Exercises 39 β 44, sketch at least one cycle of the graph of each function. 39. 41. 43 cos t 2 2 tan 3t 3.5 sin 2pt 40. 42. 44. f t 2 1 2 3 sin 2t f t 2 1 f t 2 1 0.8 cos pt tan pt 2 sin 5t t 2 In Exercises 45β50, match a graph to a function. Only one graph is possible for each function. a. 5 2 19. 20. 21. 22. 23. 24. 25. 26. 27. 28 sin t; g 1 tan t; g cos t; g 1 cos t; g 1 tan 3t cos 8t t t 2 2 cos t 2 1 tan t; g tan t 2 1 sin t; g t 1 sin t; g 1 cos t; g 2 2 t sin 2 1 sin 1.6t cos 2.6t t 2 1 sin t; g cos t; g t 2 1 t 2 1 3 sin t 1 2 cos t 29. f t 2 1 tan t; g 1 3 t 2 1 tan t 30. 31. 32. 33. 34. 35 sin t; g t 2 1 4 sin t 2 sin t; g tan t; g t 2 1 t 2 1 tan t; g cos t; g cos t sin 2t 2 tan t 2 2 tan 0.2t 3 cos 6t 2 5 cos 8t 36. f t 2 1 sin t; g t 2 1 2 sin 3pt 5 37. f t 2 1 tan t; g 1 3 t 2 1 tan pt 38. f t 2 1 cos t; g 5 3 t 2 1 cos pt 3 2Ο 2Ο 2Ο 2Ο b. c. d. 2Ο 2Ο 2Ο 2Ο 5 5 5 10 10 5 5 500 e. 2Ο f. Chapter 7 Trigonometric Graphs 2Ο 5 5 10 2Ο 2Ο 58. y 3 0 β3 59. y 5 0 β5 t t 2Ο 3 2Ο 5 In Exercises 60β64, state all local minima and maxima of the function on the given interval. 10 45. 47 sin 2t 3 cos |
t 2 49. f t 2 1 5 tan t 3 46. 48 cos 2t 3 sin t 2 50. f t 2 1 3 tan 2t In Exercises 51β56, write an equation for a sine function with the given information. 60. 61. 62. 63. 64 sin 2t; 0 t p cos 3t; 0 t p cos t 2 ; 2p t p sin t 3 ; 2p t p 3 sin 2pt; 1.5 t 1.5 65. The current generated by an AM radio transmitter 2 1 t A sin 2000 pmt, is given by a function of the form f the location on the broadcast dial and t is measured in seconds. For example, a station at 900 on the AM dial has a function of the form 550 m 1600 where is f t A sin 2000p 900 t A sin 1,800,000pt 1 2 2 1 Sound information is added to this signal by modulating A, that is, by changing the amplitude of the waves being transmitted. AM means amplitude modulation. For a station at 900 on the dial, what is the period of function f? 66. Find the function f, its period, and its frequency for a radio station at 1440 on the dial. (See Exercise 65.) 51. amplitude 2, period 4p 52. amplitude 1 2, period p 2 53. amplitude 1.8, period 3p 2 54. amplitude 1, period 2 55. amplitude 3 2, period 4 56. amplitude 6, period 1 2 In Exercises 57β59, state the rule of a sine function whose graph appears to be identical to the given graph. 57. y 2 0 β2 t Ο 2 Section 7.4 Periodic Graphs and Phase Shifts 501 7.4 Periodic Graphs and Phase Shifts Objectives In Section 7.3, you studied graphs of functions of the form β’ State the period, amplitude, vertical shift, and phase shift given the function rule or graph of a sine or cosine function β’ Use graphs to determine whether an equation could possibly be an identity a sin bt and f t 2 1 a cos bt g t 2 1 and learned how the constants a and b affect the amplitudes and periods of the functions. In this section, you will consider functions of the form f t 2 1 a sin bt c d 2 1 and g t 2 1 a cos bt c d 2 1 where a, b, c, and d |
are constants, and you will determine how these constants affect the graphs of the functions. Vertical Shifts Recall from Section 3.4 that adding a constant to the rule of a function shifts the graph vertically. Example 1 illustrates a vertical shift in combination with a reflection and a change in amplitude. Example 1 Reflection, Vertical Stretch, and Vertical Shift Describe the graph of 2p, 2p. 3 4 2 cos t 3. Then graph k on the interval k t 2 1 Solution k reflected The graph of across the horizontal axis, vertically stretched by a factor of 2, and shifted 3 units upward, as shown in Figure 7.4-1. is the graph of g t t 2 2 1 1 2 cos t 3 cos t y k(t) = β2 cos t + 3 5 4 3 2 1 β2Ο βΟ β1 Figure 7.4-1 t Ο y = cos t 2Ο After the vertical shift, the graph of tered on the horizontal line y 3. k t 2 1 2 cos t 3 is vertically cen- β 502 Chapter 7 Trigonometric Graphs Phase Shifts Recall from Section 3.4 that when the independent variable t in the rule where c is a constant, the graph of a function is replaced by is shifted horizontally. For periodic functions, the number c is the phase shift associated with the graph. t c, t c or Example 2 Phase Shift Describe the graph of each function. t p 2 b a sin cos b. a 2p 3 b a Solution a. The graph of g sin t 2 1 t p a 2 b is the graph of f t 2 1 sin t shifted to the left p 2 units, as shown in Figure 7.4-2. y f(t) = sin t 1 0 β1 Ο 2 β2Ο βΟ g(t) = sin( t + Ο 2 ) Ο 2 Figure 7.4-2 t Ο 2Ο When the graph of g t 2 1 sin t p 2 b a, of g that begins at sin t t f 1 2 is shifted to become the graph of t 0 becomes a cycle the cycle of f that begins at t p 2 t 2p a 3 b is the graph of. cos Thus, g has a phase shift of p 2. cos t shifted to f t 2 1 b. The graph of h t 2 1 the right 2p 3 units, as shown in Figure 7.4-3. y 1 0 |
β1 h(t) = cos( t β 2Ο 3 ) t 2Ο Ο 2Ο 3 Figure 7.4-3 β2Ο βΟ f(t) = cos t Section 7.4 Periodic Graphs and Phase Shifts 503 The cycle of cos h t 1 2 t f 2 1 t 2p 3 b a 2p 3. phase shift of cos t that begins at t 0 becomes a cycle of that begins at t 2p 3. Thus, the function h has a β Combined Transformations Now that you are familiar with the effects of various transformations on the sine and cosine functions, you are ready for some examples that simultaneously include changes in amplitude, period, and phase shift. Example 3 Combined Transformations State the amplitude, period, and phase shift of f t 2 1 3 sin 2t 5. 2 1 Solution Rewrite the function rule. f t 2 1 3 sin 2t 5 1 2 3 sin t 5 a 2 b d 2 c When the rule of f is written in this form, you can see that it is obtained from the rule of k t 2 1 3 sin 2t by replacing t with t 5 2. Therefore, the graph of f can be obtained by horizontally shifting the graph of k to the left units, as shown in Figure 7.4-4. 5 2 f(t) = 3 sin (2t + 5) β6 β4 β 5 2 y k(t) = 3 sin 2t t 2 4 6 4 2 0 β2 β4 Figure 7.4-4 t 0 becomes a cycle of f that begins at The cycle of k that begins at t 5 2 ; so the function f has a phase shift of The amplitude of f is 3 and its period is 2p 2 5 2 p.. β 504 Chapter 7 Trigonometric Graphs The procedure that is used in Examples 1β3 can be used to analyze any function whose rule is of the form a sin bt c d. t f 1 2 1 2 First rewrite the function rule as follows. f t 2 1 a sin bt c 2 1 d a sin t c bb d b a c d Thus, the graph of f is obtained from the graph of k t 2 units horizontally. The cycle of k that 1 a sin bt by shift- ing it d units vertically and c b begins at t 0 becomes the cycle of f that begins at t c b, so f has phase shift c b. The amplitude of |
both f and k is A similar analysis applies to the function g a 0 t 2 0 1 and both have period bt c a cos d. 1 2 2p b. If and b 77 0, a 0 f(t) a sin(bt c) d then each of the functions and g(t) a cos(bt c) d has the following characteristics: amplitude a 00 00 phase shift c b period 2p b vertical shift d Example 4 Combined Transformations Describe the graph of g t 2 1 2 cos 3t 4 1 2 1. Solution Identify the amplitude, period, vertical shift and phase shift. 1 2 cos 3t 4 g t 1 2 1 amplitude a 0 0 3t 4 1 2 cos 1 2 2 period 2p b 3 2 4 2p 3 1 2 phase shift c b 2 cos 4 a 3 b 1, 3t 4 t shown in Figure 7.4-5, is vertically The graph of 2 2 1 y 1. centered on the horizontal line The waves reach a maximum of 2 units above that horizontal line and a minimum of 2 units below that vertical shift 1 4 3 g 1 Combined Transformations 2.5 2Ο 2Ο horizontal line. The graph begins a cosine wave at t 4 3 and completes 3.5 Figure 7.4-5 one cycle in 2 p 3 units. β Section 7.4 Periodic Graphs and Phase Shifts 505 Example 5 Combined Transformations Identify the amplitude, period, vertical shift, and phase shift of 4 sin. Then graph at least one complete cycle of f. Solution The function rule c 1, d 3. and f t 2 1 f(t) = β4 sin y ( t 2 + 1) 3+ is of the form a sin bt c 2 1 d, with a 48 β6 β4 0 β2 2 4 6 8 10 y = 4 sin ( Figure 7.4-6 t 2 + 1) 3+ amplitude a 4 0 0 0 phase shift c b 4 period 2p b 2p 1 2 4p 2 vertical shift 3 0 1 1 2 The waves of the graph are vertically centered on the horizontal line reaching a maximum of 7 and a minimum of wave at y 3 The graph begins a sine units. The graph of 1. 4p 12.6 t 2 and completes one cycle in 1 y 4 sin 3 t 2 a b f is the graph of y 3, as shown in Figure 7.4-6. reflected across the horizontal |
line β β2Ο 4 2 β2 β4 3Ο 4 Ο 4 Figure 7.4-7 Example 6 Identifying Graphs Find a sine function and a cosine function whose graphs look like the graph shown in Figure 7.4-7. Solution 2Ο This graph appears to have an amplitude of 2 and to be centered vertib 1. so cally on the horizontal axis. The period appears to be 2 cos t Therefore, the graph looks like the graph of shifted horizontally. 2 sin t 2p, g t or f 1 t 2 1 2 The graph of 2 Figure 7.4-7 intercepts the x-axis at intercepts the x-axis at t p 4, t f 1 2 sin t t 0. The graph in so it looks like the graph of 2 sin t shifted t f 1 2 p 4 units to the right. Therefore, this graph closely resembles the graph of h 1 At t 0, the graph of g t 2 1 t 2 sin 2 2 cos t t p 4 b a. reaches its maximum of 2. The graph in Figure 7.4-7 reaches its maximum of 2 at t 3p 4, so it looks like the graph of g t 2 1 2 cos t shifted 3p 4 units to the right. Therefore, this graph closely resembles the graph of k 2 cos t 2 1 t 3p 4 b. a β 506 Chapter 7 Trigonometric Graphs NOTE Identities are proved algebraically in Chapter 9. Graphs and Identities Graphing calculators can be used to determine equations that could possibly be identities. A calculator cannot prove that such an equation is an identity, but it can provide evidence that it might be one. On the other hand, a calculator can prove that a particular equation is not an identity. Example 7 Possible Identities Which of the following equations could possibly be an identity? 4 g(t) = sin t a. cos p 2 a t b sin t b. cos p 2 a t b sin t β2Ο 2Ο Ο 2 f(t) = cos ( + t) β4 Figure 7.4-8 Solution a. If cos p 2 Q t R sin t is an identity, then cos f t 2 1 p 2 Q t R and g t 2 1 sin t are equivalent functions and have the same graph. The graphs of f and g, shown in Figure 7.4-8 are obviously different. Therefore, cos p 2 Q t R sin t |
is not an identity. b. In Figure 7.4-9, the graphs of cos f t 2 1 p 2 Q t g and t 1 R 2p, 2p sin t 2 appear to coincide on the interval values for f and g, shown in Figure 7.4-10, also supports the idea Comparing a table of. 3 4 that f t 2 1 cos p 2 Q t R and g t 2 1 sin t are equivalent functions. β2Ο 4 β4 2Ο Figure 7.4-9 Figure 7.4-10 This evidence strongly suggests that the equation cos is an identity, but does not prove it. Therefore, cos could possibly be an identity. Q p 2 Q p t 2 R sin t t R sin t β Section 7.4 Periodic Graphs and Phase Shifts 507 Example 8 Possible Identities Which of the following equations could possibly be an identity? a. cot t cos t sin t b. sin t tan t cos t f(t) = cot t cos t 1.5 g(t) = sin t Solution 2Ο 2Ο a. Rewrite f as sin t. g t 1 2 1 tan t cos t f t 2 1 and compare its graph with the graph of The graphs of f and g, shown in Figure 7.4-11 are obviously different. Therefore, cot t cos t sin t is not an identity. 1.5 Figure 7.4-11 However, it does appear from the graph that cot t cos t csc t could possibly be an identity. b. In Figure 7.4-12a, the graphs of sin t tan t t f 1 2 and g t 2 1 cos t appear to coincide. Comparing a table of values for f and g, shown in Figure 7.4-12b, also supports the idea that sin t tan t t f 1 2 and cos t t g 1 defined; that is, all values of t except those that make tan t 0. are equivalent functions for all values of t for which f is 2 1.5 2Ο 2Ο 1.5 Figure 7.4-12a Figure 7.4-12b Therefore, sin t tan t cos t could possibly be an identity for tan t 0. β CAUTION Do not assume that two graphs that look the same on a calculator screen actually are the same. Depending on the viewing window, two graphs that are actually quite different may appear identical. 508 Chapter |
7 Trigonometric Graphs Exercises 7.4 In Exercises 1β20, state the amplitude, period, phase shift, and vertical shift of the function. 1. h t 2 1 cos t 1 1 2 2. m t 2 1 7 cos t 3 2 1 3. f t 2 1 5. k 7. p t t 2 2 1 1 8. f t 2 1 9. h t 2 1 4. k t 2 1 cos 2pt a 3 b 6. g t 2 1 3 sin 2t p 2 1 5 sin 2t t p 4 2 3p t 1 sin 1 6 cos 1 4.5 sin 2 12t 6 1 2 3t p 6 b a 5 1 4 cos 10. p t 2 1 5 sin t 4 a 3 b 1 11. q t 2 1 7 sin 7t 1 7 B A 12. h t 2 1 16 sin 2t 3 a 4 b 13. d t 2 1 3 sin 2t 5p a 4 b 14. c t 2 1 cos 3t 2 a p 3 b 5 15. 17. g t 2 1 s t 2 1 97 cos 14t 5 1 2 7 cos 2pt 16. 18. f t 2 1 m t 2 1 19. k t 2 1 3 cos p t 3 a 1 b 5 20. h t 2 1 4 sin t 3 a p 4 b 3 2 cos 4t 1 1 t 5 2 2 2 4 cos 1 In Exercises 21β30, state the rule of a sine function with the given amplitude, period, phase shift, and vertical shift, respectively. 21. 3, p 4, p 5, 0 22. 1, 2, 3, 4 23. 2 3, 3p, 2p 3, 2 24. 8, 1 2, 2 3, 4 25. 0.5, 2.5, 1.5, 0.6 26. 1, 5, 0, 3 27. 6, 5p 3, 0, 1 29. 5 2, 1.8, 0.2, 0 28. 2, 8p, 1, 1 30. 1, 1, 1, 1 In Exercises 31β40, a. State the rule of a function of the form t a sin bt c f identical to the given graph. d 1 1 2 2 whose graph appears to be b. State the rule of a function of the form t a cos bt c f identical to the |
given graph. d 1 1 2 2 whose graph appears to be 31. 32. 33. 34. 35. 36. y 12 0 β12 18 0 β18 1 β1 1 β1 1 2 1 2 β 3.5 β3. 3Ο 2 Ο 2 3Ο 4 t t Ο 8 Ο 4 5Ο 16 Ο 3 Ο 37. 38. 39. 40. t 3Ο t Ο 2 4Ο 3 Ο 4 2Ο 2 β4 y 0 β2 β4 β6 β8 β10 10 8 6 4 2 0 βΟ y β2 β4 β6 β8 β10 Section 7.4 Periodic Graphs and Phase Shifts 509 47. f t 2 1 sin 2t 3 1 2 1 48. g t 2 1 5 cos t p a 3 b 2 In Exercises 49β52, graph the function over the interval and determine the location of all local maxima and minima. (0, 2p), 49. f t 2 1 1 2 sin t p 3 b a 50. g t 2 1 2 sin 2t 3 a p 9 b 51. f t 2 1 2 sin 1 52. h t 2 1 1 2 cos p 2 a 3t p 2 t p 8 b 1 53. Describe the graph of f sin2 t cos2 t. t 2 1 In Exercises 54β57, use graphs to determine whether the equation could possibly be an identity or is definitely not an identity. 54. cos t t p a 2 b cos 56. sec t csc t 1 tan t cot t 55. sin t 1 cos t cot t csc t 57. tan t cot p 2 a t b 2p t 2p. In Exercises 58 β 61, graph f in a viewing window Use the trace feature to determine with constants a, b, and c such that the graph of f appears to coincide with the graph of a sin bt c g t. 1 2 1 2 58. 59. 60. 61 sin t 2 cos t 3 sin 4t 2 1 2 2 cos 4t 1 2 sin 3t 5 3 cos 1 2 sin t 5 cos t 2 1 2 1 3t 2 2 In Exercises 41β48, sketch the graph of at least one cycle of the function. 41. k t 2 1 3 sin t 43. p t 2 1 1 2 sin 2t 42. y t 2 1 2 cos |
3t 44. q t 2 1 2 3 cos 3 2 t 45. h t 2 1 3 sin 2t p 2 b a 46. p t 2 1 3 cos 3t p 1 2 In Exercises 62β63, explain why there could not possibly be constants a, b, and c such that the graph of g coincides with the graph of f. a sin bt c t 1 2 62. 63 sin 2 3t 1 1 sin 2t cos 3t 2 3 cos 4t 1 1 2 510 Chapter 7 Trigonometric Graphs 7.4.A Excursion: Other Trigonometric Graphs Objectives β’ Write a sine function whose graph looks like the graph of another given sinusoidal function β’ Find viewing windows for the graphs of other trigonometric functions A graphing calculator enables you to explore with ease a wide variety of trigonometric functions. Graphing Exploration Graph each function on the same screen in a viewing window with 0 t 2p. cos t f g t 2 1 sin t 2 1 t p 2 b a How do the two graphs compare? Do they appear to coincide? The exploration above suggests that the equation cos t sin t p a 2 b is an identity and that the graph of the cosine function can be obtained by horizontally shifting the graph of the sine function. This is true, and it will be proved in Section 9.2. Consequently, every cosine function, such 6, t g as can be expressed as a sine function of the form 2 1 a sin t f The shape of the graph of such a function is called a sinusoid. 1 bt c 3 cos 4t 5 2 d. 2 2 1 1 Other Sinusoidal Graphs The exploration below suggests that other trigonometric functions can be bt c expressed in the form f a sin t. 1 2 1 2 Graphing Exploration 1. Graph with g 2 sin 2p t 2p 3 cos t 7 in a viewing window 2 1. Does the function appear to be periodic? t 2 t 2 1 2 1 2. Using the calculatorβs minimum and maximum finders, deter- mine the approximate amplitude of this function. 3. Using the calculatorβs zero finder, estimate the period of this function (find the length of a complete cycle). 4. What is the smallest positive t-value at which a sine cycle begins? 5. Use the information from Questions 2β4 to write a function of the |
a sin t whose graph looks very much like the form 2 3 cos g t graph of. Graph the new function on the same screen with g. Do the graphs appear to coincide? 1 2 sin 2 t 7 bt Section 7.4.A Excursion: Other Trigonometric Graphs 511 The results of the preceding graphing exploration suggest that the graph of looks like the graph of 4.95 f 2 sin 1 t 2.60. 3 cos t g 2 1 sin These results illustrate the following facts. Sinusoidal Graphs If b, d, k, r, and s are constants, then the graph of the function g(t) d sin(bt r) k cos(bt s) is a sinusoid and there are constants a and c such that d sin(bt r) k cos(bt s) a sin(bt c). Example 1 Sinusoidal Graphs Find a sine function whose graph looks like the graph of g t 2 1 4 sin 3t 2 1 2 2 cos 3t 4. 2 1 5 0.84 3.94 Solution The graph of 7.4.A-1. g t 2 1 4 sin 3t 2 2 1 2 cos 3t 4 1 2 is shown in Figure 2Ο 2Ο 5 Figure 7.4.A-1 By using a graphing calculatorβs minimum and maximum finders with 2 cos 3t 2 t you see that g has an the graph of 2 2 amplitude of approximately 3.94. 4 sin 3t 4 g, 2 1 1 1 The function g has period 2p 3 because this is the period of both sin 3t 4 The function f. 2 t 2 1 a sin bt c 2 1 has period and cos 1 b 3. So 3t 2 1 2p b 2. 2p 3 By using a graphing calculatorβs zero finder, you can see that a sine cycle in the graph of g begins at approximately so the phase shift c b is approximately t 0.84, 0.84. Find c. 0.84 c b c 3 c 2.52 c 2.52 0.84 Substitute 3 for b Therefore, a 3.94, b 3, c 2.52, looks like the graph of f t 2 1 and the graph of 3t 2.52 2 3.94 sin 1 g t 2 1 4 sin 3t 2 2 1 2 cos 3t 4. 2 1 β 512 Chapter 7 |
Trigonometric Graphs Other Trigonometric Graphs In Example 1, the variable t has the same coefficient b in both the sine and cosine terms of the functionβs rule. When this is not the case, the graph will consist of waves of varying size and shape, as shown in Figure 7.4.A-2. sin 3t cos sin 3 t 5 1 2 4 cos t 2 2 1 h t 2 1 2 sin 2t 3 cos 3t 6 6 6 6 Figure 7.4.A-3 Figure 7.4.A-4 2Ο 2Ο 6 0 6 2Ο 2Ο 6 6 Figure 7.4.A-2 2Ο 2Ο 6 6 2Ο Example 2 Finding a Viewing Window Find a viewing window for one complete cycle of 2Ο Solution 4 sin 100pt 2 cos 40pt. f t 2 1 A graph of f in a viewing window with includes so many cycles that the calculator cannot display an accurate graph, as shown in Figure 7.4.A-3. 2p t 2p Instead, find the period of f by using the following method: Let h t 2 1 4 sin 100pt and g 2 cos 40pt. t 2 1 The period of h is 0.10 2p 100p 1 50 0.02. The period of g is 2p 40p 1 20 0.05. The period of f is the least common multiple of 0.02 and 0.05, which is 0.10. Therefore, the viewing window with one complete cycle of t f 4 sin 100pt 2 cos 40pt. 0 t 0.10 in Figure 7.4.A-4 shows 1 2 β Damped and Compressed Trigonometric Graphs Suppose a weight hanging from a spring is set in motion by an upward push. No spring is perfectly elastic, and friction acts to slow the motion of the weight as time goes on. Consequently, the graph showing the height of the weight above or below its equilibrium point at time t will consist of waves that get smaller and smaller as t gets larger. Many other physical situations can be described by functions whose graphs consist of waves of diminishing or increasing heights. Other situations, such as Section 7.4.A Excursion: Other Trigonometric Graphs 513 sound waves in FM radio transmission, are modeled by functions whose graphs consist of waves of uniform height and varying frequency. 35 Example 3 Analyzing a Damped Graph β35 35 Solution Analyze the graph of f |
t cos t. t 2 1 β35 Figure 7.4.A-5 y y = 0.5t t y = β0.5t not to scale Figure 7.4.A-6 Graph f in a viewing window with shown in Figure 7.4.A-5. 35 t 35 and 35 y 35, as Recall that sidering the cases t 6 0. ) sign when 1 cos t 1. t 0 and Multiply each term of the inequality by t con(Remember to reverse the inequality t 6 0. when when t t cos t t t t cos t t t 0 t 6 0 In graphical terms, this means that the graph of f(t) t cos t lies between the straight lines y t, with the waves growing larger or t and y smaller to fit the space between the lines. The graph touches the lines y t and y 1. This occurs when t k, where k is an integer. t exactly when t cos t t, that is, when cos t p Β± Β± 0 Therefore, the graph of f(t) t cos t consists of waves that diminish in amplitude as t approaches 0 from both negative and positive values, and the waves are bounded by the lines y Β± t. β Graphing Exploration Illustrate the analysis of the graph f(t) t, and y f(t) t cos t t on the same screen. t cos t, y by graphing Example 4 Analyzing a Damped Graph Analyze the graph of g(t) 0.5t sin t. Solution No single viewing window gives a completely readable graph of g. To the left of the y-axis, the graph gets quite large; but to the right, it almost coincides with the t-axis. To get a better mental picture, note that multiply each term of the known inequality for every t. To find the bounds of by 0.5t sin t, 0.5t. 0.5t 7 0 1 sin t 1 sin t 1 1 0.5t 0.5t sin t 0.5t for every t Therefore, the graph of g lies between the graphs of the exponential funcThe graph of g will consist of sine waves tions y 0.5t y 0.5t. and 514 Chapter 7 Trigonometric Graphs rising and falling between the graph of the exponential functions 0.5t y as indicated in Figure 7.4.A-6 (which is not to scale |
). β y 0.5t, and Graphing Exploration Find viewing window ranges that clearly show the graph in Example 4 when t is in the following domains. 2p t 0 0 t 2p 2p t 4p Example 5 Oscillating Behavior Analyze the graph of f(t) sin p t b. a Solution Using a wide viewing window, it is clear that the t-axis is an asymptote of the graph of f(t) sin p t b a, as shown in Figure 7.4.A-7a. Near the ori- gin, however, the graph is not readable, even in a very narrow viewing window like Figure 7.4.A-7b. 1.5 1.5 β76 76 β0.5 0.5 β1.5 Figure 7.4.A-7a β1.5 Figure 7.4.A-7b Consider what happens to the graph between t, and t 0. 1 2 As t goes from 1 2 to 1 4, sin p t b a goes from sin p 1 2 b a to sin a, that is p 1 4 b from sin wave for 2p 1 4 4p. to sin 1 t 2 Therefore, the graph of f makes one complete sine. Similarly, for t 1 6 1 4, the graph of f makes another complete sine wave. The same pattern continues so that the graph of f 1 6 makes a complete wave for, and so on. A sim- t t, for 1 10 1 8 ilar phenomenon occurs as t takes values between and 0. Conse- 1 8 1 2 Section 7.4.A Excursion: Other Trigonometric Graphs 515 quently, the graph of f near 0 oscillates infinitely often between and 1, with the waves becoming more and more compressed as t gets closer to 0, as indicated in Figure 7.4.A-8. Because the function is not defined at t 0, the left and right halves of the graph are not connected. 1 Graph oscillates infinitely often here y 1 β1 1 β1 Figure 7.4.A-8 f(t) = sin (Ο/t) t β 2 Exercises 7.4.A In Exercises 1 β 6, find a sine function whose graph looks like the graph of the given function f. 1. 2. 3. 4. 5. 6 sin t 2 cos t 3 sin t 2 cos t |
2 sin 4t 5 cos 4t 3 sin 1 5 sin 0.3 sin 2t 1 4 cos 2t 3 2 3t 2 1 2t 4 1 2 2 1 2 cos 1 0.4 cos 2 3t 1 2 2t 3 1 2 In Exercises 7β16, find a viewing window that shows a complete graph of the function. 7. g(t) (5 sin 2t)(cos 5t) 8. h(t) esin t 9. f(t) t 2 cos 2t 10. g t 2 1 sin t 3 2 2 cos t 4 2 11. h t 2 1 sin 300t cos 500t Β’ β€ Β’ β€ 12. f 13. g t t 2 2 1 1 3 sin 1 5 sin 14. h t 2 1 4 sin 1 200t 1 2 cos 300t 2 250pt 5 1 600pt 3 2 1 2 cos 2 400pt 7 2 6 cos 1 500pt 3 1 2 2 15. f 16. g t t 2 2 1 1 4 sin 0.2pt 5 cos 0.4pt 6 sin 0.05pt 2 cos 0.04pt In Exercises 1724, describe the graph of the function verbally, including such features as asymptotes, undefined points, amplitude and number of waves between 0 and 2. Find viewing windows that illustrate the main features of the graph. P 17. g(t) sin et 19. f(t) 21. h(t) 2 0 t 0 cos t 1 t sin t 18. h(t) 20. g(t) cos 2t 1 t2 t2 8 e sin 2pt 22. f(t) t sin 1 t 23. h(t) ln 0 cos t 0 24. h(t) ln 0 sin Important Concepts Section 7.1 Section 7.2 Section 7.3 Section 7.4 Graph of the sine function................. 475 Graph of the cosine function................ 477 Domain and range of the sine and cosine functions............................... 477 Graph of the tangent function.............. 480 Domain and range of |
the tangent function..... 480 Even function........................... 482 Odd function............................ 483 Graph of the cosecant function.............. 486 Domain and range of the cosecant function.... 486 Graph of the secant function................ 488 Domain and range of the secant function...... 487 Domain and range of the cotangent function... 488 Graph of the cotangent function............. 489 Period................................. 494 Amplitude.............................. 497 Phase shift.............................. 502 Combined transformations of sine and cosine graphs........................... 503 Graphs and identities..................... 506 Section 7.4.A Sinusoidal graphs........................ 510 Other trigonometric graphs................ 512 Important Facts and Formulas If and, and a cos then each of the functions has: d bt c 1 2 f t 2 1 a sin bt c d 2 1, 0 a amplitude 0 phase shift c b period 2p b, vertical shift d If b 7 0, then the function h tan bt has period t 2 1 p b. 516 Chapter Review 517 Review Exercises Section 7.1 1. Which of the following is not true |
about the graph of f It has no sharp corners. a. b. It crosses the horizontal axis more than once. c. It rises higher and higher as t gets larger. d. It is periodic. e. It has no vertical asymptotes. sin t? t 2 1 In Exercises 2β4, graph each function on the given interval. 2. f t 2 1 sin t 4. h t 2 1 tan t 7p 2, 7p T 2p, 3p 4 S 3 3. g t 2 1 cos t 5p, 7p 2 T S In Exercises 5β7, find all the exact t-values for which the given statement is true. 5. cos t 1 7. tan t 23 6. sin t 1 2 In Exercises 8β10, list the transformations that change the graph of f into the graph of g. State the domain and range of g. 8. f t 2 1 sin t g 1 2 t 2 1 sin t 9. f t 2 1 tan t g t 2 1 tan 2t 10. f t 2 1 cos t g cos t 2 1 1 2 a t b 1 In Exercises 11β13, sketch the graph of each function. 11. 13 cos t 2 sin t 3 12. h t 2 1 tan t 4 14. Which of the following functions has the graph shown below between p and p? a. b. c. d. e sin x, cos x, cos x 1 sin x, sin x e if x 0 if x 6 0 if x 0 if x 6 0, βΟ 2 1 cos x 0 21 sin2 x 0 y 1 β1 β Ο 2 t Ο Ο 2 15. Between (and including) 0 and 2p, the function h a. 3 zeros and is undefined at 2 places b. 2 zeros and is undefined at 3 places c. 2 zeros and is undefined at 2 places d. 3 zeros and is defined everywhere e. no zeros and is undefined at 3 places tan t has?. t 2 1 518 Chapter Review 16. Which of the statements iβiii are true? y 2 Ο 2 β2 βΟ β t Ο Ο 2 Section 7.2 The sine function is an odd function. The cosine function is an odd function. i. ii. iii. The tangent function is an odd function. i and ii only a. b. |
ii only c. d. all of them e. none of them i and iii only 17. Which of the following functions has the graph shown at left? a. b. c. d. e tan t t p tan a 2 b 1 tan t 3 tan t tan t 18. Which of the following is true about sec t? a. b. 0 0 sec 2 1 sec t 1 sin t Its graph has no asymptotes. c. d. It is a periodic function. e. It is never negative. In Exercises 19β21, sketch the graph of each function. 19 g t 2 1 20. f t 2 1 cot t 2 3 sec t 2 21. h t 2 1 csc 1 2 a t b In Exercises 22β23, complete the statement with βoddβ or βeven.β 22. The cosecant function is an function. 23. The secant function is an function. Section 7.3 24. Let f 3 2 t 2 1 sin 5t. a. What is the largest possible value of b. Find the smallest positive number t such that? t f 1 2 0. f t 2 1 25. Sketch the graph of g 26. Sketch the graph of f 27. Sketch the graph of cost. 1 2 sin 2t on the interval 2p t 2p. sin 4t on the interval 0 t 2p. 28. What is the period of the function g sin 4pt? t 2 1 29. If t g 2 1 that g 20 sin 1? t 1 2, for how many values of t with 0 t 2p is it true 200t 2 1 Chapter Review 519 30. What is the period of f tan t 2 1 t 2 b? a 31. Which of the following statements is true? 3 sin 2t 1 is 2. a. The amplitude of b. The period of g t 2 1 c. The period of 2 d. The amplitude of tan 2t cos 2t is 4p. p is 2 3 tan t. is 3 13 cos 14t 15 1 Section 7.4 32. What are the amplitude, period, and phase shift of the function 33. State the rule of a sine function with amplitude 8, period 5, and phase shift 14. 34. State the rule of a sine function with amplitude 3, period p, and phase shift p 3. t 2Ο 5 Ο 4Ο 5 6 |
Ο 5 8Ο 5 2Ο 35. State the rule of a periodic function whose graph from t 0 to t 2p closely resembles the graph at left. y 2 1 β1 β2 In Exercises 36β38, sketch the graph of at least one cycle of each function. 36. f t 2 1 1 2 cos 1 2t p 3 2 37. g t 2 1 sin 1 3 a t p b 38. g t 2 1 4 cos 2t 3 b a 5 In Exercises 39β42, determine graphically whether the given equation could possibly be an identity. 39. cos t sin t p 2 b a 41. sin t sin 3t cos t cos 3t tan t 40. tan t 2 sin t 1 cos t 42. cos 2t 1 1 2 sin2 t Section 7.4.A In Exercises 43 and 44, find a sine function whose graph looks like the graph of the given function. 43. 44 sin 1 5 sin 4t 7 5 cos 4t 8 2 5t 3 1 2 cos 2 5t 2 1 2 1 2 2 In Exercises 45 and 46, find a viewing window that shows a complete graph of the function. 45. f 46. g t t 2 2 1 1 3 sin 1 5 sin 300t 5 2 cos 500t 8 400pt 1 2 1 1 2 cos 2 150pt Approximations with Infinite Series In the Chapter 1 Can Do Calculus, it was shown that the infinite geo- metric series a ar ar 2 ar3... a 1 r when r 0 0 6 1. This section will investigate certain functions that can be represented by an infinite series, a topic considered in depth in calculus. Example 1 Representing a Function as a Series Write f 1 1 x x 1 2 as an infinite series. Solution The expression 1 1 x a ar ar is in the form a 1 r 2 ar3... when r 1 x x2 x3..., when x 0 0 a 1 r 1 1 x where a 1 and r x., Because 0 6 1, 0 6 1. β 5 To confirm that 1 1 x 1 x x2 x3... when 6 1, graph x 0 0 and 1 x y 1 x x2 x3 x4 y 1 Figure 7.C-1 where is drawn with a heavy line. but The graphs of the function and the series are very close when they diverge when When more terms are used |
in graphing the series, the series approximates the function more closely The set of all values of x for which the series converges to when x 6 1 and when x 7 1. y 1 x x2 x3 x4 on the same screen, as shown in 6 1, 6 1. x x 0 0 the function is called the interval of convergence. The function 1 6 x 6 1. 1 1 x x 1, the infinite series does not converge to a single value. interval of convergence when It is not defined when x 1 has and 0 0 Other Types of Series Many interesting functions that can be represented by a series include the product of all the integers from 1 to n. Such a product is written as n!, which is read βn factorial.β n 1 n! 1 2 3 4 n 2 0! is defined to be the number 1. n 2 21 1 p 3 3 5 Figure 7.C-1 Technology Tip The factorial feature is found in the PROB (or PRB) submenu of the MATH or OPTN menu on most calculators. 520 10 Example 2 A Series that Approximates a Function 10 10 10 Figure 7.C-2a 10 10 10 10 Figure 7.C-2b 10 Find a function that is approximated by the following series in the interval p x p. x x3 3! x5 5! x7 7! x9 9! p n1 1 1 2 x2n1 2n 1 1! 2 Solution Begin by graphing the function formed by the first five terms of the series, as shown in Figure 7.C-2a. Next graph the functions formed by first six terms of the series and then the first seven terms, as shown in Figures 7.C-2b and 7.C-2c. The graph of the series is beginning to resemble the graph of the sine function. To test the hypothesis that the series converges to the sine function, graph both the sine function and the function formed by series on the same screen using several terms of the series. In Figure 7.C-2d, the is shown as the series is displayed with the heavier line, and lighter line. In calculus it will be shown that the infinite series converges to the sine function, and the interval of convergence is the entire set of real numbers. y sin x 1 10 10 15 15 10 Figure 7.C-2c Exercises 1 Figure 7.C |
-2d β Find an infinite geometric series that represents the given function, and state the interval of convergence. Find a function that is approximated by the following series. State the interval of convergence. 1. y 2 1 3x 3. y 2 1 x 2. y 3 1 2x 4. y 3 1 2x 5. 1 x2 2! x4 4! x6 6! p x 1 2 6. 1 x 1 2 1 7. 1 x x2 2! 1 x3 3! 2 x4 4 521 C H A P T E R 8 Solving Trigonometric Equations Round and round we go! Trigonometric functions are used to analyze periodic phenomena, because simple harmonic motion models circular motion or any phenomenon that is βback and forth.ββ Some examples of simple harmonic motion include a vibrating prong of a tuning fork, a buoy bobbing up and down in water, seismic and ocean waves, spring-mass systems, a piston in a running engine, a particle of air during the passage of a simple sound wave, or a turning Ferris wheel. See Exercise 1 of Section 8.4. 522 Chapter Outline 8.1 Graphical Solutions to Trigonometric Equations 8.2 Inverse Trigonometric Functions 8.3 Algebraic Solutions of Trigonometric Equations 8.4 Simple Harmonic Motion and Modeling 8.4.A Excursion: Sound Waves Chapter Review can do calculus Limits of Trigonometric Functions Interdependence of Sections 8.1 8.2 > 8.3 > 8.4 There are two kinds of trigonometric equations. Identities, which will be studied more in Chapter 9, are equations that are valid for all val- ues of the variable for which the equation is defined, such as sin2 x cos2 x 1 and cot x 1 tan x. In this chapter, conditional equations will be studied. Conditional equations are valid only for certain values of the variable, such as sin x 0, cos x 1 2, and 3 sin2 x sin x 2. If a trigonometric equation is conditional, solutions are found by using techniques similar to those used to solve algebraic equations. Graphs were used to solve some simple trigonometric equations in Chapter 7. This chapter will extend graphical solution techniques and introduce analytic solution methods. Graphical solution methods are presented in 8.1. Inverse trigonometric functions are discussed in Section 8.2. Methods |
that use inverse functions, basic identities, and algebra to solve trigonometric equations are considered in Section 8.3. Skills from the Sections 8.1 through 8.3 are applied to problem-solving and real-world applications in Section 8.4. NOTE In Chapter 7, the variable t was used for trigonometric functions to avoid confusion with the xβs and yβs that appear in their definitions. Now that you are comfortable with these functions, the letter x, or occasionally y, will be used for the variable. Unless otherwise stated, all trigonometric functions in this chapter are considered as functions of real numbers, rather than functions of angles in degree measure. 523 524 Chapter 8 Solving Trigonometric Equations 8.1 Graphical Solutions to Trigonometric Equations Objectives β’ Solve trigonometric equations graphically β’ State the complete solution of a trigonometric equation Any equation involving trigonometric functions can be solved graphically. To solve trigonometric equations graphically, the same methods of graphical solutions are used here as have been used previously to solve polynomial equations, except that trigonometric equations typically have an infinite number of solutions. These solutions are systematically determined by using the periodicity of the function. Basic Trigonometric Equations An equation that involves a single trigonometric function set equal to a number is called a basic equation. Some examples include the following: sin x 0.39, cos x 0.5, and tan x 3 Examples 1 and 2 show how they can be solved graphically. Example 1 The Intersection Method Solve tan x 2. Solution The equation can be solved by graphing on the same screen and finding intersection points. The x-coordinate of every such point is a number whose tangent is 2; or a solution of Figure 8.1-1 indicates that there are infinitely many intersection points, so the equation has an infinite number of solutions. tan x 2. and Y1 tan x Y2 2 y 4 2 β2Ο β3Ο 2 βΟ β Ο 2 β2 Ο 2 Ο 2Ο 3Ο 2 5Ο 2 3Ο 7Ο 2 4Ο x One period Figure 8.1-1 5 Ο 2 Ο 2 5 Figure 8.1-2 NOTE Solutions in this chapter are often rounded, but the full decimal expansion given by the calculator is used in all computations. The symbol even though these calculator solutions are approximations of the |
actual solutions. is used rather than Section 8.1 Graphical Solutions to Trigonometric Equations 525 1 f x tan x The function p 2 b in this interval. Using the intersecand there is one solution of tion finder on a graphing calculator gives the approximate solution in this interval. completes one cycle on the interval tan.1071 tan x f x repeats its pattern to the left and to the Because the graph of right, the other solutions will differ from this first solution by multiples of the period of the tangent function. The other solutions are 1.1071 Β± p, 1.1071 Β± 2p, and 1.1071 Β± 3p, p, 1 2 and so on. All solutions can be expressed as 1.1071 kp, where k is any integer. Example 2 The x-Intercept Method Solve sin x 0.75. Solution Rewrite the equation sinx 0.75 as sin x 0.75 0. β 3 Recall from Section 2.1 that the solutions of this equation are the x-intercepts of the graph of sin x 0.75. f x 1 2 2Ο 2Ο 1 Figure 8.1-3 2p The graph of f is shown in Figure 8.1--3. The function has a period of and the viewing window can be modified to show one period of sin x 0.75. f Figures 8.1-4a and 8.1-4b show that there are two zeros of f on the interval so the equation has two solutions on that interval. 0, 2p x, 1 2 4 3 The calculatorβs zero finder calculates the zeros: x 3.9897 x 5.4351 3 0 1 3 2Ο 0 2Ο Figure 8.1-4a Figure 8.1-4b 1 526 Chapter 8 Solving Trigonometric Equations Because the graph repeats its pattern every differ from these two by multiples of Therefore, all solutions of 2p, sin x 0.75 2p, the period of are 1 2 the other solutions will sin x 0.75. x f x 3.9897 2kp and x 5.4351 2kp, where k is any integer. β Other Trigonometric Equations The procedures in Examples 1 and 2 can be used to solve any trigonometric equation graphically. Example 3 The x-intercept Method Solve 3 sin2 x cos x 2 0. Solution |
Technology Tip 3 sin2 x Enter culator as on a cal- 2 1 sine and 3 sin2 x cos x 2 cosine have period is at most the period of Both The graph of f, which is shown x f in two viewing windows in Figure 8.1-5, does not repeat its pattern over so you can conclude that f has a period any interval of less than of 2p. 2p, 2p. so 2p, sin x 3 3 1 or 1 sin 2 2 2 2. x 1 22 2Ο 4Ο 0 4 Figure 8.1-5a 2 4 2Ο Figure 8.1-5b The function f makes one complete period on the interval as shown 2p, in Figure 8.1-5b. The equation has four solutions between 0 and namely, the four x-intercepts of the graph in that interval. A graphical zero finder shows these four solutions., 2 3 0, 2p x 1.1216 x 2.4459 x 3.8373 x 5.1616 Because the graph repeats its pattern every tion are given by 2p, all solutions of the equa- x 1.1216 2kp, x 3.8373 2kp, x 2.4459 2kp, x 5.1616 2kp, where k is any integer. β Section 8.1 Graphical Solutions to Trigonometric Equations 527 The solution methods in Examples 1 through 3 depend only on knowing the period of a function and all the solutions of the equation in one period. A similar procedure can be used to solve any trigonometric equation graphically. Solving Trigonometric Equations Graphically 1. Write the equation in the form f(x) 0. 2. Determine the period p of f. 3. Graph f over an interval of length p. 4. Use a calculatorβs zero finder to determine the x-intercepts of the graph in this interval. 5. For each x-intercept u, all of the numbers u kp where k is any integer are solutions of the equation. In Example 1, for example, p was p. In Examples 2 and 3, p was 2p. Example 4 Solving Any Trigonometric Equation 3 Ο 2 Ο 2 Solve tan x 3 sin 2x. Solution First rewrite the equation tan x 3 sin 2x 0 3 Figure 8.1-6 Next, determine the period |
of y 3 sin 2x f x 1 2 tan x 3 sin 2x. p, Recall from Section which is also the period of 7.3 that y tan x. has a period of 2p 2 p. Therefore, the period of f is p 2 b p 2, a, f on the interval an interval of length p. Figure 8.1-6 shows the graph of Even without the graph, it can be easily verified that there is an x-intercept at 0. f 0 1 2 tan 0 3 sin 2 0 2 1 0 Using the calculatorβs zero finder gives the other x-intercepts of the graph of f on this interval. x 1.1503 and x 1.1503 Because f has a period of are p, all solutions of the equation tan x 3 sin 2x x 1.1503 kp, x 0 kp, and x 1.1503 kp, where k is any integer. β 528 Chapter 8 Solving Trigonometric Equations Trigonometric Equations in Degree Measure Some real-world applications of trigonometric equations require solutions to be expressed as angles in degree measure. The graphical solution procedure is the same, except that you must set the mode of your calculator to βdegree.β Example 5 Trigonometric Equations in Degree Measure Solve 2 sin2u 3 sin u 3 0. Solution 360 The period of the function ure 8.1-7 shows the graph of f on the interval 2 sin2u 3 sin u 3. 0Β°, 360Β° u f 2 1 3 2 is 360Β°, and Fig- A graphical zero finder determines the approximate x-intercepts. u 223.33Β° and u 316.67Β° Figure 8.1-7 Using the fact that the period of f is 360Β°, all solutions of the equation are u 223.33Β° 360Β°k and u 316.67Β° 360Β°k, where k is any integer. β 3 0 5 Exercises 8.1 In Exercises 1β12, solve the equation graphically. 13. Use the graph of the sine function to show the 1. 4 sin 2x 3 cos 2x 2 2. 5 sin 3x 6 cos 3x 1 3. 3 sin32x 2 cos x 4. sin2 2x 3 cos 2x 2 0 5. tan x 5 sin x 1 6. 2 cos2x sin x 1 0 7. cos3 |
x 3 cos x 1 0 8. tan x 3 cos x 9. cos4x 3 cos3x cos x 1 10. sec x tan x 3 11. sin3x 2 sin2 x 3 cos x 2 0 12. csc2x sec x 1 following. a. The solutions of sin x 1 are 5p x p, 2 2 3p 2 x, 9p, 2 7p 2, p and 11p 2,,.... b. The solutions of sin x 1 are 11p 2 x 3p, 2 p 2 x 7p, 2 5p 2,, p and 9p 2,...., 14. Use the graph of the cosine function to show the following. a. The solutions of Β± 4p, x 0, Β± 2p, b. The solutions of x Β± p, Β± 3p, cos x 1 are Β± 6p,.... cos x 1 Β± 5p,.... are Section 8.2 Inverse Trigonometric Functions 529 In Exercises 15β18, approximate all solutions of the given equation in (0, 2p). 15. sin x 0.119 16. cos x 0.958 17. tan x 5 18. tan x 17.65 At the instant you hear a sonic boom from an airplane overhead, your angle of elevation to the plane is given by the equation A sin A 1 m In Exercises 19β28, find all angles with that are solutions of the given equation. U 0 U 66 360 19. tan u 7.95 20. tan u 69.4 21. cos u 0.42 22. cot u 2.4 23. 2 sin2 u 3 sin u 1 0 24. 4 cos2u 4 cos u 3 0 25. tan2u 3 0 26. 2 sin2u 1 27. 4 cos2u 4 cos u 1 0 28. sin2u 3 sin u 10 where m is the Mach number for the speed of the plane (Mach 1 is the speed of sound, Mach 2.5 is 2.5 times the speed of sound, etc.). In Exercises 29β32, find the angle of elevation (in degrees) for the given Mach number. Remember that an angle of elevation must be between 90. and 0 29. m 1.1 31. m 2 30. m 1.6 32. m 2.4 33. Critical Thinking Under what conditions (on the |
constant) does a basic equation involving the sine and cosine function have no solutions? 34. Critical Thinking Under what conditions (on the constant) does a basic equation involving the secant and cosecant function have no solutions? 8.2 Inverse Trigonometric Functions Objectives β’ Define the domain and range of the inverse trigonometric functions β’ Use inverse trigonometric function notation NOTE Other ways of restricting the domains of trigonometric functions are possible. Those presented here for sine, cosine, and tangent are the ones universally agreed upon by mathematicians. Many trigonometric equations can be solved without graphing. Nongraphical solution methods make use of the inverse trigonometric functions that are introduced in this section. Recall from Section 3.6 that a function cannot have an inverse function unless its graph has the following property. No horizontal line intersects the graph more than once. You have seen that the graphs of trigonometric functions do not have this property. However, restricting their domains can modify the trigonometric functions so that they do have inverse functions. Inverse Sine Function, the interval The restricted sine function is p 2 p 2 T ber v in the interval 3 p 2 S, such that sin u v. p 2 T, S 4 sin x, f x 1 2 when its domain is restricted to Its graph in Figure 8.2-1 shows that for each num. 1, 1 there is exactly one number u in the interval 530 Chapter 8 Solving Trigonometric Equations y 1 v (u, v) = (u, sin u) x u Ο 2 β Ο 2 β1 Figure 8.2-1 Because the graph of the restricted sine function passes the Horizontal Line Test, it has an inverse function. This inverse function is called the inverse sine (or arcsine) function and is denoted by g(x) sin1 x g(x) arcsin x. or It is convenient to think of a value of an inverse trigonometric function as an angle; sin 1 23 2 sine is 23 2. Since sin represents an angle in the interval 23 2 p 3 23 2 then 1 sin whose The graph of the inverse sine function, shown in Figure 8.2-2, is readily obtained from a calculator. Because is the inverse of the restricted sine function, its graph is the reflection of the graph of the restricted sine function across the line 2 1 y x. sin Figure |
8.2-2 x 2 sin 1 x is the interval 1, 1 3 4, and its range is the g The domain of p 2 T interval p 2, S 1. Inverse Sine Function For each v with 1 v 1, sin1 v is the unique number u in the interval sine is v; that is, P 2, S P 2 T whose sin1 v u exactly when sin u v. Section 8.2 Inverse Trigonometric Functions 531 Technology Tip The inverse sine function can be evaluated by using the times labeled ASIN) on a calculator. For example, SIN1 key (some- Unless otherwise noted, make sure your calculator is in radian mode. 1 sin 1 0.67 2 0.7342 and sin 1 0.42 0.4334. For many special values, however, you can evaluate the inverse sine function without using a calculator. Technology Tip If you attempt to use a calculator to evaluate the inverse sine function at a number not in its domain, such as sin you will get an error message. 1 2, 2 1 Example 1 Special Values Evaluate: a. sin 1 1 2 Solution b. 1 sin 22 2 b a a. 1 2 sin 1 is the number in the interval p 2 T your study of special values, you know that p 2, S whose sine is 1 2. From sin p 6 1 2. Because p 6 is in the interval b. 1 sin 22 a 2 b p 2 S interval, p 2 S p 4 sin because sin p 4 b a 22 2 and p 4 is in the, p 2 T. β CAUTION 1 x sin The notation 1 1 sinx sin x or 1 2. is not exponential notation. It does not mean For instance, Example 1 shows that sin 1 1 2 p 6 0.5236, but this is not equivalent to sin a 1 2 b 1 1 sin 1 2 1 0.4794 2.0858. Suppose sine function, 1 v u. 1 v 1 p 2 sin and sin u p 2 u and sin sin u v. sin u and 1 v 2 1 2 1 sin 1 Therefore, sin 1 v 1 sin u 2 1 2 v. Then by definition of the inverse This shows that the restricted sine function and the inverse sine function have the usual composition properties of other inverse functions. 532 Chapter 8 Solving Trigonometric Equations Properties of Inverse Sine sin1(sin |
u) u if p 2 u p 2 sin(sin1 v) v if 1 v 1 Example 2 Composition of Inverse Functions Explain why sin 1 sin a p 6 b p 6 is true but sin 1 sin a 5p 6 b 5p 6 is not true. Solution You know that sin p 6 1 2, so by substitution 1 sin sin a p 6 b sin 1 1 2 b a p 6 because p 6 is in the interval p 2 S, p 2 T. Although sin 5p 6 is also 1 2, by substitution 1 sin sin a 5p 6 b sin 1 1 2 b a p 6, not 5p 6, because 5p 6 is not in the interval p 2, S p 2 T. β Inverse Cosine Function The restricted cosine function is 0, p. to the interval 4 3 1, 1 v in the interval such that cos u v. 3 4 cos x, f x when its domain is restricted Its graph in Figure 8.2-3 shows that for each number 0, p, there is exactly one number u in the interval 1 2 3 4 y 1 v β1 u x Ο (u, v) = (u, cos u) Figure 8.2-3 Section 8.2 Inverse Trigonometric Functions 533 Because the graph of the restricted cosine function passes the horizontal line test, it has an inverse function. This inverse function is called the inverse cosine (or arccosine) function and is denoted by g(x) cos1 x or g(x) arccos x. The graph of the inverse cosine function, which is the reflection of the y x, graph of the restricted cosine function (Figure 8.2-3) across the line is shown in Figure 8.2-4. 2 Ο Ο 2 Figure 8.2-4 2 The domain of g x 1 2 cos 1 x is the interval 1, 1 3 4 and its range is 0, p. 4 3 Inverse Cosine Function For each v with 1 v 1, cos1 v cosine is v; that is, is the unique number u in the interval [0, P] whose cos1v u exactly when cos u v. The properties of the inverse cosine function are similar to the properties of the inverse sine function. Properties of Inverse Cosine cos1(cos u) u if 0 u p cos(cos1v) v if 1 v 1 Example |
3 Evaluating Inverse Cosine Expressions Evaluate the following. a. cos 1 1 2 b. cos 1 0 c. cos 1 0.63 2 1 534 Chapter 8 Solving Trigonometric Equations CAUTION 1 cos cos x 1 does not mean 1 1 or cosx. 2 Solution a. cos 1 1 2 p 3 because p 3 whose cosine is 1 2. is the unique number in the interval 0, p 4 3 b. cos 1 0 p 2 COS1 c. The 1 0.63 cos 1 2 because cos p 2 0 and 0 p 2 p. command on a calculator shows that 2.2523. β 1βv2 1 u v Figure 8.2-5 Example 4 Equivalent Algebraic Expressions Write sin cos 1 v 2 1 Solution as an algebraic expression in v. Let cos 1 v u, where 0 u p. Construct a right triangle containing an angle of u radians where cos u adjacent hypotenuse 8.2-5. By the Pythagorean Theorem, the length of the side opposite u is 21 v2, as shown in Figure v. By the definition of sine, sin u opposite hypotenuse 21 v2. 1 v cos sin 1 2 21 v2 1 Therefore, 21 v2 β Inverse Tangent Function The restricted tangent function is p to the interval 2 Q p 2 R,. tan x, f x 1 2 when its domain is restricted Its graph in Figure 8.2-6 shows that for every real number v, there is exactly one number u between tan u v. p 2 and p 2 such that y v (u, v) = (u, tan u) x β Ο 2 u Ο 2 Figure 8.2-6 Section 8.2 Inverse Trigonometric Functions 535 Because the graph of the restricted tangent function passes the horizontal line test, it has an inverse function. This inverse function is called the inverse tangent (or arctangent) function and is denoted g(x) tan1 x g(x) arctan x. or The graph of the inverse tangent function, which is the reflection of the y x, graph of the restricted tangent function (Figure 8.2-6) across the line is shown in Figure 8.2-7. Ο 2 10 10 Ο 2 Figure 8.2-7 The domain of is the interval x g 2 1 p 2 is the |
set of all real numbers and its range 1 x tan p 2 b., a Inverse Tangent Function For each real number v, tan1 v is the unique number u in the interval P 2 a, P 2 b whose tangent is v; that is, tan1v u exactly when tan u v. The properties of the inverse tangent function are similar to the properties of the inverse sine and inverse cosine functions. Properties of Inverse Tangent tan1(tan u) u if P 2 6 u 6 P 2 tan(tan1v) v for every real number v. CAUTION Example 5 Evaluating Inverse Tangent Expressions 1 tan tan x 1 does not mean 1 1 or tan x. 2 Evaluate: 1 1 tan a. b. tan 1 136 536 Chapter 8 Solving Trigonometric Equations Solution a. tan 1 1 p 4 because p 4 such that tan p 4 1. is the unique number in the interval p 2 Q, p 2 R b. The TAN1 key on a calculator shows that tan 1 136 1 2 1.5634. Example 6 Exact Values Find the exact value of cos 1 tan a 25 2 b. Solution Let tan 1 25 2 u. Then tan u 25 2 and p 2 6 u 6 p 2. Because tan u 25 2 is positive, u must be between 0 and p 2. Draw a right triangle containing an angle of u radians whose tangent is 25 2. tan u opposite adjacent 25 2 5 3 u 2 Figure 8.2-8 The hypotenuse has length 1 25 cos tan a 2 b 1 2 cos u adjacent hypotenuse 222 15 2 24 5 3. Therefore, 2 3. β Exercises 8.2 In Exercises 1β14, find the exact value without using a calculator. 11. 1 tan 23 A B 2. cos 10 5. cos 11 3. 6. 1 tan tan 1 1 11 2 13. 1 cos 1 a 2 b 12. 1 cos a 22 2 b 14. sin 1 1 a 2 b 8. cos 1 23 2 10. sin 1 23 2 In Exercises 15β24, use a calculator in radian mode to approximate the functional value. 15. sin 10.35 16. cos 10.76 17. 1 tan 3.256 1 2 18. sin 1 0.795 1 2 1. sin 11 4. sin 7. tan 1 1 2 1 1 23 3 9. sin 1 a |
22 2 b 19. 20. 22. 24. 1 1 1 sin cos sin 7 2 Hint: the answer is not 7. cos 3.5 2 1 sin tan 1 1 2 sin C 1 2D tan 12.4 1 tan cos 1 tan C 1 1 cos C 1 4 2D 8.5 2D 21. 23. 2 25. Given that u sin 1 of cos u and tan u. a 23 2 b, find the exact value Section 8.2 Inverse Trigonometric Functions 537 45. tan sin 1v A B 46. sin A 2 sin 1v B In Exercises 47β50, graph the function. 47. f 48. g 49 cos 1 x 1 A B tan 1 x p sin 1 sin x A B 50. k x 1 2 sin A sin 1 x B 26. Given that u tan 1 4 3 b a sin u and sec u., find the exact value of 51. A model plane 40 feet above the ground is flying away from an observer. In Exercises 27β42, find the exact functional value without using a calculator. 28. 1 cos sin a p 6 b 30. 1 tan cos p 2 1 32. 1 cos tan a 7p 4 b (See Exercise 19.) 27. 1 sin cos 0 1 2 29. 1 cos sin a 4p 3 b 31. 1 sin cos a 7p 6 b 33. 1 sin sin a 2p 3 b 34. 1 cos cos a 5p 4 b x 40 ΞΈ Observer a. Express the angle of elevation of the plane as a function of the distance x from the observer to the plane. u b. What is when the plane is 250 feet from the u observer? 35. 1 cos 36. 1 tan cos tan S S p a 6 b T 4p a 3 b T 37. sin 1 cos S 3 5 b T a (See Example 6.) 38. tan 40. cos 1 sin 1 sin S S 42. sin 1 cos S 3 5 b T 23 5 b T 23 13 b T a a a 39. cos 41. tan 1 tan 3 a 4 b T 1 sin 5 13 b T a S S In Exercises 43β46, write the expression as an algebraic expression in v, as in Example 4. 43. cos sin 1 v A B 44. cot cos 1 v A B 52. Show that the restricted secant function, whose domain consists of all numbers x such that 0 x p |
, has an inverse function. and x p 2 Sketch its graph. 53. Show that the restricted cosecant function, whose domain consists of all numbers x such that p x 0, 2 has an inverse function. x p 2 and Sketch its graph. 54. Show that the restricted cotangent function, whose domain is the interval function. Sketch its graph. 1 0, p 2, has an inverse 55. a. Show that the inverse cosine function actually has the two properties listed in the box on page 533. b. Show that the inverse tangent function actually has the two properties listed in the box on page 535. 538 Chapter 8 Solving Trigonometric Equations 56. Critical Thinking A 15-foot-wide highway sign is placed 10 feet from a road, perpendicular to the road. A spotlight at the edge of the road is aimed at the sign, as shown in the figure below. 57. Critical Thinking A camera on a 5-foot-high tripod is placed in front of a 6-foot-high picture that is mounted 3 feet above the floor, as shown in figure below. 10 A Sign ΞΈ Spotlight ΞΈ 5 ft x 6 ft 3 ft a. Express angle as a function of the distance x u from the camera to the wall. b. The photographer wants to use a particular u 36Β° lens, for which How far. p 5 a radians b a. Express as a function of the distance x from u point A to the spotlight. b. How far from point A should the spotlight be placed so that the angle possible? u is as large as should she place the camera from the wall to be sure the entire picture will show in the photograph? 8.3 Algebraic Solutions of Trigonometric Equations Objective β’ Solve trigonometric equations algebraically Trigonometric equations were solved graphically in Section 8.1. In this section you will learn how to use algebra with inverse trigonometric functions and identities to solve trigonometric equations. Recall from Section 8.1 that equations such as sin x 0.75, cos x 0.6, and tan x 3 are called basic equations. Algebraic solution methods for basic equations are illustrated in Examples 1 through 3. Example 1 Solving Basic Cosine Equations Solve cos x 0.6. Section 8.3 Algebraic Solutions of Trigonometric Equations 539 2 Solution Ο Ο 2 Figure 8.3-1 1.5 Y1 |
cos x The graphs of and just two solutions (intersection points) on the interval one full period of the cosine function. in Figure 8.3-1 show that there are which is p, p, 3 4 Y2 0.6 The definition of the inverse cosine function states that cos 10.6 Using the inverse cosine function, of by using the identity 0, p is the number in the interval 3 10.6 0.9273 x cos p, p. 4 3 cos x, is one solution The second solution can be found with whose cosine is 0.6. x 0.9273. cos x 0.6 cos 4 on the interval x 2 1 0.9273 cos Therefore, the solutions of cos 0.9273 0.6 1 2 cos x 0.6 10.6 0.9273 and x cos on the interval p, p are 10.6 0.9273 3 4 x cos Because the interval tion, all solutions of p, p 3 cos x 0.6 4 are given by is one complete period of the cosine func- x 0.9273 2kp and x 0.9273 2kp, where k is any integer. β Example 2 Solving Basic Sine Equations Solve sin x 0.75. Solution The definition of the inverse sine function states that 1 sin 1 0.75 2 is the number in the interval S, p 2 1 0.75 1 p p on the interval,. S 2 T 2 sin x, p x x sin 2 sin 1 0.8481 2 sin 3.9897 whose sine is 0.75. p 2 T 0.8481 is the solu- A second solution can be x 0.8481. with 0.75 2 4 3.9897 1 2 is also a solution of sin x 0.75, Using the inverse sine function, tion of sinx 0.75 found by using the identity p sin 1 3 0.8481 x p Therefore, 2 and all solutions are given by 1 Ο Ο where k is any integer. x 0.8481 2kp and x 3.9897 2kp, 1.5 Figure 8.3-2 Recall that there are an infinite number of solutions to many trigonometric equations. Figure 8.3-2 indicates that there are two solutions in the interx 0.8481 val can in the solution be found by letting. The solution x |
3.9897 2kp x 2.2935 : x 2.2935 k 1 p, p and. 4 3 β 540 Chapter 8 Solving Trigonometric Equations Example 3 Solving Basic Tangent Equations 5 Solve tan x 3. Solution The definition of the inverse tangent function states that Ο Ο 5 Figure 8.3-3 tan 1 3 is the number in the interval p p a 2 b 2 1 3 1.2490 x tan p a 2 of the tangent function, all solutions are given by Using the inverse tangent function, tan x 3 on the interval p a 2 Because p 2 b p 2 b,,,. whose tangent is 3. is the solution of is one full period where k is any integer. x 1.2490 kp, β The solution method used in Examples 1β3 is summarized in the following table, where k is any integer. Solutions of Basic Trigonometric Equations Equation Possible values of c Solutions sin x c 1 6 c 6 1 x sin 1 c 2kp c 1 c 1 and p sin 1 c 2kp 2 x 1 x p 2 x p 2 2kp 2kp c 7 1 or c 6 1 no solution cos cos 1 c 2kp and 1 c 2kp x cos x 0 2kp 2kp x p 2kp c 7 1 or c 6 1 no solution tan x c all real numbers x tan 1 c kp Section 8.3 Algebraic Solutions of Trigonometric Equations 541 Example 4 Using the Solution Algorithm Solve 8 cos x 1 0. Solution First rewrite the equation as an equivalent basic equation. 8 cos x 1 0 cos x 1 8 Then solve the basic equation using the inverse cosine function. One solution is in Quadrant I. The other solution on the interval is in Quadrant IV. x cos 1.4455 1 1 8 p, p x cos 3 1 1 8 4 1.4455 All solutions are given by x 1.4455 2kp and x 1.4455 2kp, where k is any integer. β Example 5 Solving Basic Equations with Special Values Solve sin u 22 2 Solution exactly, without using a calculator. Because terval p sin 4 p, p 3 4 22 2, u p 4 is one solution of sin u 22 2 on the in-. Another solution is in Quadrant II. u p p 4 3p |
4 Therefore, the exact solution is given by u p 4 2kp and u 3p 4 2kp, where k is any integer. β Sometimes trigonometric equations can be solved by using substitution to make them into basic equations. 542 Chapter 8 Solving Trigonometric Equations Example 6 Using Substitution and Basic Equations Solve sin 2x 22 2 Solution exactly, without using a calculator. First, let u 2x, and solve the basic equation From Exam- ple 5, you know the complete exact solution of is given by sin u 22 2. sin u 22 2 u p 4 2kp and u 3p 4 2kp, each of these solutions leads to a solution of the original for u, and solve for x. where k is any integer. u 2x, Because 2 x equation. Substitute u p 4 2x p 4 x p 8 2kp 2kp kp Therefore, all solutions of x p 8 where k is any integer. and u 3p 4 2x 3p 4 x 3p 8 2kp 2kp kp are given by sin 2x 22 2 kp and x 3p 8 kp, β Algebraic Techniques Many trigonometric equations can be solved algebraically β by using factoring, the quadratic formula, and basic identities to write an equivalent equation that involves only basic equations, as shown in the following examples. Example 7 Factoring Trigonometric Equations Find the solutions of 3 sin2 x sin x 2 0 in the interval p, p. 4 3 Solution Let u sin x. 3 sin2 x sin x 2 0 3u2 u 2 0 Substitution Section 8.3 Algebraic Solutions of Trigonometric Equations 543 This quadratic equation can be solved by factoring. 3u2 u 2 0 u 1 0 3u 2 2 1 2 or u 1 1 u 2 3 Substituting sin x for u results in two basic equations. sin x 2 3 or sin x 1 If sin x 2 3 b a, then 2Ο x sin 1 2 a 3 b 0.7297 2kp x p 2 2kp, then If sin x 1 x p sin 1 2 a 3 b 3.8713 2kp or Therefore, the solutions of x 0.7297 2kp, where k is any integer. 3 sin2 x sin x 2 0 x p 2 2kp, and are x 3.8713 2 |
kp, p, p, Figure 8.3-4 indicates that there are three solutions in the interval is outside which is marked with vertical lines. The solution the interval, but the corresponding solution within the interval can be x 3.8713 2kp. found by letting, the solutions are x 3.8713 k 1 p, p Within in 4 3 4 3 x 3.8713 2p 2.4119, x 0.7297, and x p 2. β Example 8 Factoring Trigonometric Equations Solve tan x cos2x tan x. Solution Write an equivalent equation as an expression equal to zero, and factor. tan x cos2x tan x 0 0 cos2 x 1 0 cos2 x 1 2cos2x 21 cos x Β± 1 x 0 kp tan x or cos2x 1 tan x 0 1 2 x 0 2kp or x p 2kp β2Ο 3 β3 Figure 8.3-4 CAUTION cos2 x 1 sin2 x 544 Chapter 8 Solving Trigonometric Equations 2 0 2 More simply stated, the solution of tan x cos2 x tan x is x 0 kp kp, 2Ο where k is any integer. The graphs of shown in Figure 8.3-5. Y1 tan x cos2 x and Y2 tan x are β Figure 8.3-5 Many trigonometric equations can be solved if trigonometric identities are used to rewrite the original equation, as shown in Examples 9 and 10. Example 9 Identities and Factoring Solve 10 cos2 x 3 sin x 9 0. Solution Use the Pythagorean identity to rewrite the equation in terms of the sine function. 10 cos2x 3 sin x 9 0 3 sin x 9 0 1 sin2x 10 10 10 sin2x 3 sin x 9 0 10 sin2x 3 sin x 1 0 1 2 Factor the left side and solve. 2 sin x 1 0 sin x 1 2 2 sin x 1 1 2 1 5 sin x 1 0 2 5 sin x 1 0 or sin x 1 5 x sin 1 1 2 b a x sin 1 1 a 5 b or x 0.2014 2kp or 2kp x p 6 x p p 6 5p 6 2kp 2kp x p 0.2014 1 2 2kp 3.3430 2kp 15 5 Ο 2 Figure 8.3-6 Therefore, all solutions of x p 6 10 |
cos2 x 3 sin x 9 0 2 kp, x 5p 6 2kp, are x 0.2014 2kp, and x 3.3430 2kp, where k is any integer. The graph of in Figure 8.3-6 confirms the solution. Y1 10 cos2x 3 sin x 9 shown β 3Ο 2 Example 10 Identities and Quadratic Formula Solve sec2 x 5 tan x 2. Section 8.3 Algebraic Solutions of Trigonometric Equations 545 Solution Use a Pythagorean identity to rewrite the equation in terms of the tangent function. sec2x 5 tan x 2 1 tan2x sec2x 5 tan x 2 0 5 tan x 2 0 tan2x 5 tan x 3 0 2 1 Use the quadratic formula to solve for tan x. tan x 5 Β± 252 4 2 1 1 2 1 3 2 21 1 5 Β± 213 2 0.6972 or tan x 4.3028 tan x 5 213 2 0.6972 1 1 0.6088 kp 2 x tan Therefore, the solution set of x 0.6089 pk and sec2x 5 tan x 2 5 213 2 4.3028 1 1 1.3424 kp 2 x tan is x 1.3424 pk, sec2 x 5 tan x and 20 Ο 2 3Ο 2 10 Figure 8.3-7 where k is any integer. The graphs of in Figure 8.3-7 confirm the solution. Y1 Y2 2 β Exercises 8.3 In Exercises 1β8, find the exact solutions. 1. sin x 23 2 3. tan x 23 5. 2 cos x 23 2. 2 cos x 22 4. tan x 1 6. sin x 0 7. 2 sin x 1 0 8. csc x 22 In the following exercises, find exact solutions if possible and approximate solutions otherwise. When a calculator is used, round to four decimal places. Use the following information in Exercises 9β12. When a light beam passes from one medium to another (for instance, from air to water), it changes both its speed and direction. According to Snellβs Law of Refraction, sin u1 sin u2 v1 v2, v1 where speed in the second medium, and is the speed of light in the first medium, its u1 the angle of incidence, the angle of refraction, as shown in the |
figure. v2 u2 The number v1 v2 is called the index of refraction. Angle of incidence Incident ray, speed v 1 ΞΈ 1 Refracted ray, speed v 2 ΞΈ 2 Angle of refraction 9. The index of refraction of light passing from air to, find water is 1.33. If the angle of incidence is the angle of refraction. 38Β° 546 Chapter 8 Solving Trigonometric Equations 10. The index of refraction of light passing from air to 40. tan x sec x 3 tan x 0 ordinary glass is 1.52. If the angle of incidence is 17Β°, find the angle of refraction. 11. The index of refraction of light passing from air to dense glass is 1.66. If the angle of incidence is, 24Β° find the angle of refraction. 12. The index of refraction of light passing from air to, find quartz is 1.46. If the angle of incidence is the angle of refraction. 50Β° In Exercises 13β32, find all the solutions of each equation. 13. sin x 0.465 14. sin x 0.682 15. cos x 0.564 16. cos x 0.371 17. tan x 0.237 18. tan x 12.45 19. Hint: cot x 1 cot x 2.3 S tan x. T 20. cot x 3.5 21. sec x 2.65 22. csc x 5.27 23. sin 2x 23 2 24. cos 2 x 22 2 26. 2 sin x 3 1 25. 2 cos x 2 22 27. tan 3 x 23 28. 5 sin 2x 2 29. 5 cos 3x 3 30. 2 tan 4x 16 32. 5 sin x 4 4 31. 4 tan x 2 8 41. 4 sin x tan x 3 tan x 20 sin x 15 0 Hint: One factor is tan x 5. 42. 25 sin x cos x 5 sin x 20 cos x 4 43. sin2x 2 sin x 2 0 44. cos2x 5 cos x 1 45. tan2x 1 3 tan x 46. 4 cos2x 2 cos x 1 47. 2 tan2x 1 3 tan x 48. 6 sin2x 4 sin x 1 49. sec2x 2 tan2x 0 50. 9 12 sin x 4 cos2x 51. sec2 x tan x 3 52. cos2x sin |
2x sin x 0 53. 2 tan2x tan x 5 sec2x 54. The number of hours of daylight in Detroit on t 0 day t of a non-leap year (with 1) is given by the following function. being January 3 sin d t 2 1 2p 365 1 S t 80 2 T 12 a. On what days of the year are there exactly 11 hours of daylight? b. What day has the maximum amount of daylight? 55. A weight hanging from a spring is set into motion moving up and down. Its distance d (in centimeters) above or below the equilibrium point at time t seconds is given by d 5 1 sin 6 t 4 cos 6 t. 2 At what times during the first 2 seconds is the weight at the equilibrium position d 0? 1 2 In Exercises 33β53, use factoring, the quadratic formula, or identities to solve the equation. Find all solutions in the interval [0, 2p). 33. 3 sin2x 8 sin x 3 0 34. 5 cos2x 6 cos x 8 35. 2 tan2x 5 tan x 3 0 36. 3 sin2x 2 sin x 5 37. cot x cos x cos x 38. tan x cos x cos x 39. cos x csc x 2 cos x In Exercises 56β59, use the following information. When a projectile (such as a ball or a bullet) leaves its starting point at angle of elevation with velocity v, the horizontal distance d it travels is given by the equation u d v2 32 sin 2u, where d is measured in feet and v in feet per second. Note that the horizontal distance traveled may be the same for two different angles of elevation, so that some of these exercises may have more than one correct answer. Section 8.4 Simple Harmonic Motion and Modeling 547 Vmax is the maximum voltage, f is the where frequency (in cycles per second), t is the time in seconds, and is the phase angle. a. If the phase angle is 0, solve the voltage f equation for t. f 0, Vmax b. If 20, V 8.5, and f 120, find the smallest positive value of t. 61. Critical Thinking Find all solutions of sin2x 3 cos2x 0 in the interval 0, 2p. 2 3 62. Critical Thinking What is wrong with this βsolutionββ? sin x tan x sin x tan |
x 1 or 5p 4 x p 4 Hint: Solve the original equation by moving all terms to one side and factoring. Compare your answers with the ones above. 63. Critical Thinking Let n be a fixed positive integer. Describe all solutions of the equation sin nx 1 2. ΞΈ d a d b 56. If muzzle velocity of a rifle is 300 feet per second, at what angle of elevation (in radians) should it be aimed in order for the bullet to hit a target 2500 feet away? 57. Is it possible for the rifle in Exercise 56 to hit a target that is 3000 feet away? At what angle of elevation would it have to be aimed? 58. A fly ball leaves the bat at a velocity of 98 miles per hour and is caught by an outfielder 288 feet away. At what angle of elevation (in degrees) did the ball leave the bat? 59. An outfielder throws the ball at a speed of 75 miles per hour to the catcher who is 200 feet away. At what angle of elevation was the ball thrown? 60. In an alternating current circuit, the voltage is given by the formula V Vmax sin 1 2pft f, 2 8.4 Simple Harmonic Motion and Modeling Objective In Section 7.4, graphs of functions of the form β’ Write a sinusoidal function whose graph resembles a given graph β’ Write a sinusoidal function to represent a given simple harmonic motion, and use the function to solve problems β’ Find a sinusoidal model for a set of data, and use the model to make predictions f t 2 1 a sin bt c 2 1 d and g t 2 1 a cos bt c 2 1 d, were studied; and the constants a, b, c, and d were examined to see how they affect the graphs of the functions. In this section, trigonometric functions of this form are used to model real-world phenomena. Recall that if a 0 and b 7 0, bt c a sin 1 2 f t 2 1 then each of the functions d and g t 2 1 a cos bt c d 2 1 has the following characteristics. a amplitude 0 phase shift c b 0 period 2p b vertical shift d 548 Chapter 8 Solving Trigonometric Equations Recall the shapes of sine and cosine waves. 2 0 2 2Ο 0 2Ο 2 Figure 8.4-1 2 Figure 8.4-2 The wave shape of the graphs of these functions is called a sin |
usoid and the functions are called sinusoidal functions. a sin a 1 2 f t bt, is the number Recall that the amplitude of the function 0 1 units above the horand that its graph consists of waves that rise to 0 units below the horizontal axis. In other izontal axis and fall to words, the amplitude is half the distance from the maximum value to the minimum value of the function. This number remains the same when the graph is shifted vertically or horizontally. Thus the amplitude of the sinusoidal function is the number d or g a cos a sin bt c bt fmax fmin2, where fmax and fmin denote the maximum and minimum values of f. The vertical shift d of a sinusoidal function can be determined by averaging the maximum and minimum values as shown below. vertical shift d fmax fmin 2 Example 1 Constructing Sinusoidal Functions Write a sine function and a cosine function whose graph resembles the sinusoidal graph below. y 4 2 0 β2 Figure 8.4-3 β2Ο β1Ο t 2Ο 1Ο 3Ο 4 NOTE The function that 2 1 t t represents the graph, 3 cos 2t 1, g is not unique because infinitely many functions can name the graph, such as 1. 3 cos h 2 1 1 They can be written using different phase shifts. The first function above is the representative answer because it uses the phase shift that is closest to zero. t p 2 2 2 1 Section 8.4 Simple Harmonic Motion and Modeling 549 Solution The graph shows that the function has a maximum of 4 and a minimum of Therefore, the amplitude is 2. a 1 2 1 fmax fmin2 1 2 1 4 2 1 22 3. The graph shows one complete cosine cycle between 0 and is Therefore, p. p, so the period p 2p b b 2 The vertical shift d is one unit. d fmax fmin 2 4 2 1 2 2 1 The variable c depends on whether the function is described in terms of g is used, there is no phase shift because a sine or cosine. If cosine waveβs maximum occurs at So the phase shift is 0, and a function for the sinusoidal graph in Figure 8.4-3 has a value of 0 for c. cos t t 0. t 2 1 3 cos 2t 1 g t 2 1 The graph indicates that a sine wave begins at for a function using f sin t is |
t 2 1 3p 4. t 3p 4, so the phase shift 3p 4 3p 4 c b c 2 c 3p 2 phase shift is c b b 2 Solve for c. Therefore, a function for the sinusoidal graph in Figure 8.4-3 is 3 sin f t 2 1 2t 3p 2 b a 1. β Simple Harmonic Motion Motion that can be described by a function of the form f t 2 1 a sin bt c 1 2 d or g t 2 1 a cos bt c d 2 1 is called simple harmonic motion. Many kinds of physical motion are simple harmonic motions. 550 Chapter 8 Solving Trigonometric Equations Example 2 Rotating Wheel P(2, 0) A wheel with a radius of 2 centimeters is rotating counterclockwise at 3 radians per second. A free-hanging rod 10 centimeters long is connected to the edge of the wheel at point P and remains vertical as the wheel rotates (Figure 8.4-4). a. Assuming that the center of the wheel is at the origin and that P is find a function that describes the y-coordinate at (2, 0) at time of the tip E of the rod at time t. t 0, b. What is the first time that the tip E of the rod will be at a height of 9 centimeters? Solution a. The wheel is rotating at 3 radians per second. After t seconds, the point P has moved through an angle of from the origin. 3t radians and is 2 units E Figure 8.4-4 To find the time t that it takes to complete one revolution (i.e., 2p), solve 3t 2p. t 2p 3 β2 β4 β6 β8 β10 β12 y 0 t Ο 6 Ο 3 Ο 2 2Ο 3 5Ο 6 Ο 1 4 After of a revolution, or 1 4 maximum of 2 centimeters, so the y-coordinate of E is 1 2 After of a revolution, or 2p 3 p 3 2p 3 p 6 1 2,, the height of P reaches its 2 10 8. the height of P is at 0, so the y-coordinate of E is 10. Continuing this process, it can be found that the y-coordinate of E is 12 and at 2p 3 the y-coordinate of E is p at 2 10. Figure 8.4-5 Plotting these key points |
shows the main features of the graph. The amplitude is a 1 2 1 fmax fmin2 1 2 1 8 12 1 22 2. Use the period to find b. 2p 2p 3 b b 3 so there is no phase shift and c 0. The sine wave begins at The vertical shift, d, is d fmax t 0, 10 units. fmin 2 8 1 2 12 2 10 Thus, the function giving the y-coordinate of E at time t is 2 sin 3t 10. f t 2 1 Section 8.4 Simple Harmonic Motion and Modeling 551 b. To find the first time that the tip E of the rod will be at a height of 2 sin 3t 10 9 centimeters, solve for t. 9 2 sin 3t 10 9 2 sin 3t 1 sin 3t 1 2 3t sin 1 1 2 b a 3t p 6 t p 18 2kp 1 2kp 3 k is any integer 2 0.1745 2kp 3 The first time that the tip E of the rod will be at a height of 9 cm is when k 0, that is, t p 18 0.1745 seconds. β Example 3 Bouncing Spring Suppose that a weight hanging from a spring is set in motion by an upward push (Figure 8.4-6). It takes 5 seconds for it to complete one cycle of moving from its equilibrium position to 8 centimeters above, then dropping to 8 centimeters below, and finally returning to its equilibrium position. (This is an idealized situation in which the spring has perfect elasticity, and friction, air resistance, etc., are negligible.) Equilibrium position Figure 8.4-6 a. Find a sinusoidal function to represent the motion of the moving weight. b. Sketch a graph of the function you wrote in part a. c. Use the function from part a to predict the height of the weight after 3 seconds. d. In the first 5 seconds, when will the height of the weight be 6 centimeters below the equilibrium position? 552 Chapter 8 Solving Trigonometric Equations Solution a. Let h t 1 2 denote the distance of the weight above or below t h equilibrium position at time t. Then from 0 to 5, increases from 0 to 8, decreases to again to 0. In the next 5 seconds it repeats the same pattern, and so on. Therefore, the graph of h is periodic and has some kind of wave shape. Two possibilities are shown in Figures 8 |
.4-7a and 8.4-7b. is 0 when t is 0. As t increases and increases h t 1 2 1 2 1 2 8, 1 2 its h(t) 8 h(t) 8 β8 t or β8 t Figure 8.4-7a Figure 8.4-7b Careful physical experiment suggests that the curve in Figure 8.4-7a, which resembles the sine graphs you have studied, is a reasonably accurate model of this process. Facts from physics and calculus show that the rule of the function h is the form for some constants a, b, and c. a sin bt c h t 2 2 1 1 The function h has an amplitude of 8, a period of 5, and a phase shift 0, so the constants a, b, and c must satisfy 8, 2p b 5, and c b 0, a 0 0 y 8 4 0 β4 β8 or equivalently, a 8, b 2p 5, and c 0. 1 2 3 4 5 t Therefore, the motion of the moving weight can be described by this function: 8 sin h t 2 1 2p 5 t Figure 8.4-8 b. The graph of h 8 sin t 2 1 2p 5 t is shown in Figure 8.4-8. c. The value of h 3 1 2 gives the height of the weight after 3 seconds. 8 sin h 3 1 2 2p 5 a 3 b 8 sin 6p 5 4.7 The height of the weight after 3 seconds is approximately 4.7 centimeters below the equilibrium point. d. To find the times in the first 5 seconds when the weight is 6 centimeters below the equilibrium, you must solve the equation 10 0 10 10 0 Figure 8.4-9 5 5 10 Figure 8.4-10 Section 8.4 Simple Harmonic Motion and Modeling 553 8 sin 2p 2p 5 Y2 6 Y1 8 sin a t b and are shown in Figure 8.4-9 and 8.4-10. The points of The graphs of 0 t 5 intersection show that the weight will be 6 centimeters below the equilibrium at two times between 0 and 5 seconds. t 3.1749 and t 4.3251 in a window with β Modeling Trigonometric Data Periodic data that appears to resemble a sinusoidal curve when plotted can often be modeled by a sine function, as shown in Example 4. Example 4 |
Temperature Data The following table shows the average monthly temperature in Cleveland, Ohio, based on data from 1971 to 2000. Since average temperatures are not likely to vary much from year to year, the data essentially repeats the same pattern in subsequent years. So, a periodic model is appropriate. Month Temperature Β°F 1 2 Month Temperature Β°F 2 1 Jan. Feb. Mar. Apr. May June 25.7 28.4 37.5 47.6 58.5 67.5 July Aug. Sep. Oct. Nov. Dec. 71.9 70.2 63.3 52.2 41.8 31.1 [Source: National Climatic Data Center] a. Make a scatter plot of the data. b. Find a sinusoidal function that models the temperature data. c. Use the sine regression feature on a calculator to find another sinusoidal model for the data. d. How do the models in parts b and c differ from one another? e. Use one of the models to predict time(s) of year in which the average temperature is 45Β°F. 100 0 0 13 Solution a. Let t 1 represent January. Enter 1 through 12 in List 1 and the Figure 8.4-11a temperatures in List 2. The scatter plot is shown in Figure 8.4-11a. 554 Chapter 8 Solving Trigonometric Equations b. To find a sinusoidal function to represent the temperature data, examine the properties of the scatter plot. The minimum value is 25.7 and the maximum value is 71.9, so the amplitude is a 1 2 1 fmax fmin2 1 2 1 71.9 25.7 23.1. 2 One complete cycle is 12 months, so the period, 2p b, is 12. 2p b 12 b p 6 12 Figure 8.4-11b The vertical shift is d fmax fmin 2 71.9 25.7 2 48.8. A sine wave begins close to the data point (4, 47.6), as shown in Figure 8.4-11b, so the phase shift c b is approximately 4. Find c. 4 4 c b c p 6 c 2p 3 Therefore, a sinusoidal function to represent the temperature data is approximately 23.1 sin f t 2 1 p 6 t 2p 3 b a 48.8. The graph of this function is shown with the scatter plot of the data in Figure 8.4-12. c. Using |
the 12 given data points, the regression feature on a calculator 13 Figure 8.4-12 produces the following model. 23.1202 sin t f 1 2 0.5018t 2.0490 48.6927 2 1 The period of this function is approximately 2p 0.5018 12.52, which is not a very good approximation for a 12-month cycle. Because the data repeats the same pattern from year to year, a more accurate model can be obtained by using the same data repeated for 100 0 0 100 0 0 Section 8.4 Simple Harmonic Motion and Modeling 555 the second year. The data for a two-year period is plotted in Figure 8.4-13a. The sine regression feature on a calculator produces this model from the 24 data points: 22.7000 sin 0.5219t 2.1842 49.5731 t f 2 1 The period of this function is 2 1 2p 0.5219 12.04, which is slightly off from the expected 12-month period. However, its graph in Figure 8.4-13b appears to fit the data well. d. Using the decimal approximation of p, the rule of the function found in part b becomes f t 2 1 23.1 sin 1 0.5236 t 2.0944 48.8. 2 This model differs only slightly from the second model in part c. f t 2 1 22.7000 sin 1 0.5219 t 2.1842 49.5731 2 Visually, however, the model shown in Figure 8.4-12 does not seem to fit the data points quite as well as the model shown in Figure 8.4-13b. Nevertheless, considering that the model found in part b can be obtained without technology, it is remarkedly close. e. The model from part c will be used to predict the times of year in which the average temperature is 45Β°F. There are two points of intersection of the graphs of 0.5219 t 2.1842 22.7000 sin y 45 and t f 1 2 1 49.5731, 2 as shown in Figure 8.4-14. Their approximate coordinates are (3.8, 45) and (10.6, 45). Therefore, according to this model, the temperature would be around in late March and late October. 45Β°F β 80 0 0 80 0 0 100 0 0 25 25 13 Figure 8.4-13a Figure 8 |
.4-13b Figure 8.4-14 Exercises 8.4 1. The original Ferris wheel, built by George Ferris for the Columbian Exposition of 1893, was much larger and slower than its modern counterparts: It had a diameter of 250 feet and contained 36 cars, each of which held 40 people; it made one revolution every 10 minutes. Suppose that the Ferris wheel revolves counterclockwise in the x-y plane with its center at the origin. Car D in the figure had coordinates (125, 0) at time the rule of a function that gives the y-coordinate of car D at time t. t 0. Find y 125 β125 125 x D 556 Chapter 8 Solving Trigonometric Equations 2. Do Exercise 1 if the wheel turns at 2 radians per minute and car D is at 1 0, 125 at time 2 t 0. 3. A circular wheel with a radius of 1 foot rotates counterclockwise. A 4-foot rod has one end attached to the edge of this wheel and the other end to the base of a piston (see figure). It transfers the rotary motion of the wheel into a back-andforth linear motion of the piston. If the wheel is rotating at 10 revolutions per second, point W is and point P is always on at x-axis, find the rule of a function that gives the x-coordinate of P at time t. at time t 0, 1, 0 1 2 from the pendulum to the (dashed) center line at time t seconds (with distances to the right of the line measured by positive numbers and distances to the left by negative ones). Assume that the pendulum is on the center line at time moving to the right. Assume the motion of the pendulum is simple harmonic motion. Find the rule of the function t 0 and 1 β1 A B 4. Do Exercise 3 if the wheel has a radius of 2 feet, rotates at 50 revolutions per second, and is at 2, 0 t 0. when 1 2 In Exercises 5β8, suppose a weight is hanging from a spring (under the same conditions as in Example 3). The weight is pushed to start it moving. At time t, let h(t) be the distance of the weight above or below its equilibrium point. Assume the maximum distance the weight moves in either direction from the equilibrium point is 6 cm and that it moves through a complete cycle every 4 seconds. Express |
h(t) in terms of the sine or cosine function under the given conditions. 5. There is an initial push upward from the equilibrium point. 6. There is an initial pull downward from the equilibrium point. Hint: What does the graph of y a sin bt look like when a 6 0? 7. The weight is pulled 6 cm above the equilibrium ) is point, and the initial movement (at downward. Hint: Think of the cosine graph. t 0 8. The weight is pulled 6 cm below its equilibrium point, and the initial movement is upward. 9. A pendulum swings uniformly back and forth, taking 2 seconds to move from the position directly above point A to the position directly above point B, as shown in the figure. The distance from A to B is 20 centimeters. Let be the horizontal distance d t 1 2 10. The following figure shows a diagram of a merrygo-round that is turning counterclockwise at a constant rate, making 2 revolutions in 1 minute. On the merry-go-round are horses A, B, C, and D 4 meters from the center and horses E, F, and G t a 8 meters from the center. There is a function 1 that gives the distance the horse A is from the y-axis (this is the x-coordinate of Aβs position) as a function of time t measured in minutes. Similarly, gives the x-coordinate for B as a function of b time, and so on. Assume the diagram shows the situation at time t 0. Which of the following function rules does a have? 4 cos t, 4 cos pt, 4 cos 2t, 4 cos 2pt, t 2 1 1 2 4 cos, 4 cos t t b b b. Describe the function a a p 2, 4 cos 4pt t b 2 using the cosine function and g t 2 1 Section 8.4 Simple Harmonic Motion and Modeling 557 c. Suppose the x-coordinate of a horse S is given 4 cos by the function and the s t 4pt 5p 6 b a 1 2 x-coordinate of another horse R is given by 2. t r 8 cos Where are these horses 4pt p a 1 3 b located in relation to the rest of the horses? Copy the diagram and mark the positions of R and S at t 0. 11. The following table shows the number, in millions, of unemployed people in the labor force |
for 1991β2002. Year Unemployed Year Unemployed 1991 1992 1993 1994 1995 1996 8.628 9.613 8.940 7.996 7.404 7.236 1997 1998 1999 2000 2001 2002 6.739 6.210 5.880 5.655 6.742 8.234 a. Sketch a scatter plot of the data, with x 0 corresponding to 1990. b. Does the data appear to be periodic? If so, find an appropriate model. c. Do you think this model is likely to be accurate much beyond the year 2002? Why? In Exercises 12 and 13, do the following: a. Use 12 data points (with x 1 corresponding to January) to find a periodic model of the data. b. What is the period of the function found in part a? Is this reasonable? c. Plot 24 data points (two years) and graph the function from part a on the same screen. Is the function a good model in the second year? d. Use the 24 data points in part c to find another periodic model for the data. e. What is the period of the function in part d? Does its graph fit the data well? 12. The table shows the average monthly temperature in Chicago, IL, based on data from 1971 to 2000. Month Temperature Β°F 2 1 Jan. Feb. Mar. Apr. May June July Aug. Sep. Oct. Nov. Dec. 22.0 27.0 37.3 47.8 58.7 68.2 73.3 71.7 63.8 52.1 39.3 27.4 [Source: National Climatic Data Center] 13. The table shows the average monthly precipitation, in inches, in San Francisco, CA, based on data from 1971 to 2000. Month Precipitation Jan. Feb. Mar. Apr. May June July Aug. Sep. Oct. Nov. Dec. 4.45 4.01 3.26 1.17 0.38 0.11 0.03 0.07 0.20 1.04 2.49 2.89 [Source: National Climatic Data Center] 558 Chapter 8 Solving Trigonometric Equations 14. Critical Thinking A grandfather clock has a pendulum length of k meters. Its swing is given (as in Exercise 9) by the function f t 0.25 sin vt, where v 2 1 a. Find k such that the period of the pendulum is B 1 2 9.8 k. b. The temperature |
in the summer months causes the pendulum to increase its length by 0.01%. How much time will the clock lose in June, July, and August? Hint: These three months have a total of 92 days (7,948,800 seconds). If k is increased by 0.01%, what is f 2? 1 2 2 seconds. 8.4.A Excursion: Sound Waves Objectives β’ Find the frequency of a sound wave using a tuning fork and a data collection device β’ Model sound wave data with sinusoidal functions and graphs Sound travels through the air like small ripples traveling across a body of water. Throwing a rock into a calm body of water causes the water to begin to move up and down around the entry point. The movement of the water causes ripples to move outward. If a flower is floating in this body of water, the ripples will cause the flower to move up and down on the water. Sound is produced by a vibrating object that disturbs the surrounding air molecules and causes them to vibrate. These vibrations cause a periodic change in air pressure that travels through the air much like the ripples in a body of water. When the air pressure waves reach the eardrum, they cause it to vibrate at the same frequency as the source. These air pressure waves are more commonly called sound waves. The voices and sounds heard each day are generally a combination of many different sound waves. The sound from a tuning fork is a single tone that can be described mathematically using a sinusoidal function, or d. g a cos a sin bt c bt Recall from Sections 7.4 and 8.4 that if functions a 0 and b 7 0, then each of the f t 2 1 a sin bt c 2 1 d and g t 2 1 a cos bt c d 2 1 has the following characteristics: amplitude 0 phase shift c b a period 2p b 0 vertical shift d The period of a sound wave determines the soundβs frequency. The frequency f of a sound wave is the reciprocal of its period. f b 2p Section 8.4.A Excursion: Sound Waves 559 Technology Tip Data obtained may differ from one collection to another, and specific values produced by a calculator may vary. Frequency gives the number of cycles (periods) that the sound wave completes in one second. Frequencies are measured in units called Hertz (Hz), where one Hertz is one cycle per second. Most tuning forks have their |
frequency and corresponding musical note listed on them. A tuning fork, a microphone connected to a data collection device (such as a CBL2), and a calculator with a tuning program can be used to simulate the sounds heard through the eardrum. NOTE A tuning program can be obtained by downloading or entering the DataMate application available from Texas Instruments. data collection device calculator tuning fork (-) microphone Figure 8.4.A-1 Example 1 Frequency Confirm with a sinusoid graph that the frequency of a sound wave formed by striking a C tuning fork is 262 Hz. Solution Connect the calculator to the data collection device, connect the microphone to the data collection device, and run the program needed to calculate the pressure for the sound waves. Follow the directions on the screen of your calculator to obtain a graph like the one shown in Figure 8.4.A-2. 0.05 0 β0.05 0.05 Figure 8.4.A-2a 0.05 0 β0.05 0.05 0 Figure 8.4.A-2b 0.05 0.05 β0.05 Figure 8.4.A-2c The period of the data can be found by dividing the differences between the x-value of the first maximum and the x-value for the last maximum by the number of complete cycles between the maximums. For the data 560 Chapter 8 Solving Trigonometric Equations shown in Figure 8.4.A-2a, Figures 8.4.A-2b and 8.4.A-2c show that the first maximum occurs when, the last maximum occurs when x 0.0208608, and there are five complete cycles between the first and last maximum. Therefore, the period of the data is given by the following. x 0.0020768 0.0208608 0.0020768 5 0.0037568 cycle length The frequency is the reciprocal of the period. 1 0.0037568 266.184 Therefore, the graph indicates a frequency of about 266.184, which is very close to the actual frequency of 262 Hz. β 0.06 0.0001 Example 2 Sinusoidal Model Find a sinusoidal model to fit the sound waves produced by a tuning fork with the note G. 0.024 Solution β0.06 Figure 8.4.A-3 Use a data collection device, a G tuning fork, and a tuning program to obtain a graph similar to the one shown in Figure 8 |
.4.A-3. Find the amplitude a by finding half the difference between the maximum value and minimum value. The data graphed in Figure 8.4-3 has a maximum value of 0.043942 and a minimum value of 0.045618. a 0.043942 1 2 0.045618 2 0.04478 amplitude To find the period, find the x-value of the first maximum of the graph, find the x-value of the last maximum of the graph, and divide the difference between the two x-values by 8 (the number of cycles between the first and last maximum). The first maximum shown on the graph occurs x 0.0221088. when, and the last maximum occurs when x 0.0020768 p 0.0221088 0.0020768 8 0.002504 period Use the period to find b. 2p b 0.002504 The graph has a maximum at tion using cosine is 0.0020768., so the phase shift for a func- b 2p 0.002504 x 0.0020768 Section 8.4.A Excursion: Sound Waves 561 Use the phase shift and b to find c. c b phase shift c 0.0020768 2p 0.002504 b a 1.658786p Find the vertical shift d by finding the average of the maximum and minimum values. In this example, d 0. Therefore, the sinusoidal model to fit the sound waves produced by a tuning fork with the note G is y 0.04478 cos 2p 0.002504 a x 1.658786p b. Notice that the frequency, the reciprocal of the period, is 399.361, which is very close to the actual frequency of 392 Hz for the note G. β When two sounds of slightly different frequency are produced simultaneously, a beat is heard. A beat is a single sound that gets louder and softer at periodic intervals. By using more than one tuning fork, a beat can be displayed. Example 3 Chord Frequency Place tuning forks with notes of C and G close to a microphone. a. Find a graphical representation for the sound waves produced by playing the two notes simultaneously. b. Find the period of the function. c. How does the frequency of this sound compare with the actual frequencies of C and G? Solution a. Using a data collection device, the graph in Figure 8.4.A-4 is obtained |
. b. The graph appears to be periodic. By finding the length of one complete cycle, the period of this graph appears to be approximately 0.0077632. c. The frequency of this sound is the reciprocal of the period of the Figure 8.4.A-4 function. 1 0.0077632 128.8128607 Hz The frequency of this sound is very close to the difference between the actual frequency of C, 262 Hz, and the actual frequency of G, 392 Hz. 392 262 130 β 562 Chapter 8 Solving Trigonometric Equations NOTE The intensity of the strike on the tuning fork determines the amplitude. Graphing Exploration Find a sinusoidal function to model the graph produced by the C tuning fork. Add the sinusoidal models for C and G notes. (Use the model from Example 2.) Graph this sum. What is the frequency of the graph? How does the frequency of this sound compare with the sum of the actual frequencies of C and G? Most sounds are more complex than those produced by tuning forks. A tuning fork produces a graph of a single note. Most musical instruments produce a sound that is a combination of several different sounds. A C-G chord was produced with the tuning forks in Example 3. The exploration above indicates that this sound can be modeled by the sum of the models for the C note and the G note. Exercises 8.4.A Bottles of water can be tuned using a microphone and data collection device. Place some water in the bottle and blow air over the top of the bottle to produce a sound. Use the frequency of the graph formed from the sound to approximate the frequency. If a higher note is needed, place more water in the bottle and calculate the frequency again. Display the graph of the following notes using the chart below. Note Frequency in Hz G# or Ab A A# or Bb B 415 440 466 494 524 Note Frequency in Hz C (next octave) C C# or Db D D# or Eb E F F# or Gb G 262 277 294 311 330 349 370 392 1. D 2. B 3. A 4. C# 5. Using only one tuning fork at a time and the sum of three functions, sketch a graph of the C-major chord C E G. 1 2 6. Using three tuning forks for the notes C, E, and G at one time, find a graph of the C-major chord Important Concepts Section 8.1 Section |
8.2 Section 8.3 Section 8.4 Basic trigonometric equations............... 524 Intersection method...................... 524 x-intercept method....................... 525 Inverse sine function...................... 530 Inverse cosine function.................... 533 Inverse tangent function................... 535 Solution algorithm for basic trigonometric equations............................... 540 Writing sinusoidal functions................ 548 Simple harmonic motion................... 549 Modeling trigonometric data............... 553 Section 8.4.A Frequency of a sound..................... 558 Sound waves............................ 558 Important Facts and Formulas sin cos 1 v u exactly when sin u v a 1 v u exactly when cos, 1 v 1, 1 v 1 b tan 1 v u exactly when tan real number b a 0 Let teristics. and b 7 0. The following functions have the given charac- f t 2 1 a sin bt c 2 1 d and g t a cos bt c 0 2 1 a amplitude fmax 1 2 1 1 fmin2 0, phase shift c b fmin2 fmax period 2p b vertical shift d 1 2 1 d 2 563 564 Chapter Review Review Exercises In Exercises 1β6, solve each equation graphically. Section 8.1 1. 5 tan x 2 sin 2x 2. sin3 x cos2 x tan x |
2 3. sin x sec2 x 3 4. cos2 x csc2 x tan x p 2 R Q 5 0 5. cos 2x sin x 6. 3 sin 2 x cos x 7. A weight hanging from a spring is set into motion, moving up and down. (See Figure 8.4-6 in Example 3 of Section 8.4.) Its distance in centimeters above or below the equilibrium point at time t seconds is given by d 5 sin 3t 3 cos 3t. weight at the equilibrium position At what times during the first two seconds is the d 0? 1 2 In Exercises 8β17, find the exact value without using a calculator. Section 8.2 8. 1 cos 22 2 b a? 9. 1 sin 23 2 b a? 10. tan 123? 12. 1 cos sin a 5p 3 b? 14. 1 sin sin 0.75 1? 2 16. 1 sin sin a 8p 3 b? 11. 1 sin cos a 11p 6 b? 13. 1 tan cos a 7p 2 b? 15. 1 cos cos 2 1 2? 17. 1 cos cos a 13p 4 b? 18. Sketch the graph of f 19. Sketch the graph of g x 2 1 x 2 1 tan 1 x p. sin 1 x 2. 2 1 20. Find the exact value of sin 1 cos a 1 4 b. 21. Find the exact value of sin 1 tan a 1 2 cos 1 4 5 b. Section 8.3 22. Find all angles with u 0Β° u 360Β° such that sin u 0.7133. 23. Find all angles with u 0Β° u 360Β° such that tan u 3.7321. In Exercises 24β38, solve the equation by any means. Find exact solutions when possible and approximate solutions otherwise. 24. 2 sin x 1 26. tan x 1 28. sin x 0.7 30. tan x 13 25. cos x 23 2 27. sin 3x 23 2 29. cos x 0.8 31. cot x 0.4 32. 2 sin2 x 5 sin x 3 33. 4 cos2 x 2 0 Chapter Review 565 34. 2 sin2 x 3 sin x 2 35. sec2 x 3 tan2 x 13 36. sec2 x 4 tan x 2 37. 2 sin2 x sin x 2 0 38. cos2 x 3 cos x 2 0 39. A cannon has a muzzle velocity of 600 |
feet per second. At what angle of elevation should it be fired in order to hit a target 3500 feet away? Hint: Use the projectile equation given for Exercises 56β59 of Section 8.3. Section 8.4 40. The following table gives the average population, in thousands, of a southern town for each month throughout the year. The population is greater in the winter and smaller in the summer, and it repeats this pattern from year to year. Month Population Month Population Jan. Feb. Mar. Apr. May June 10.5 9.3 7.8 6.0 4.9 4.5 July Aug. Sep. Oct. Nov. Dec. 4.7 5.8 7.6 9.4 10.6 10.9 a. Make a scatter plot of the data. b. Find a sinusoidal function to represent the population data. c. Use the sine regression feature on a calculator to find a periodic model for the data. d. Use the model from part c to predict time(s) of year in which the average population is 6200. 41. The paddle wheel of a steamboat is 22 feet in diameter and is turning at 3 revolutions per minute. The axle of the wheel is 8 feet above the surface of the water. Assume that the center of the wheel is at the origin and that a point P on the edge of the paddle wheel is at (0, 19) at time a. What maximum height above the water does point P reach? b. How far below the water does point P reach? c. d. Write a cosine function for the height of point P at time t. e. Write a sine function for the height of point P at time t. f. Use the function from part d or e to find the time(s) at which point P In how many seconds does the wheel complete one revolution? t 0 seconds. will be at a height of 10 feet. Section 8.4.A 42. Confirm with a sinusoid graph that the frequency of a sound wave formed by striking an F tuning fork is 349 Hz. 43. Find a sinusoidal model to fit the sound waves produced by striking an E tuning fork(x) k lim f(x) = k x n x n x Figure 8.C-1 Limits of Trigonometric Functions A main focus of calculus is the behavior of the output of a function as the input approaches a specific value. The value that the |
function approaches, if it exists, is called a limit. In this section an informal description of a limit is illustrated with some interesting trigonometric functions, but the discussion is not intended to be complete. See Chapter 14 for a detailed discussion of limits. x If the output of a function approaches a single real number k as the input approaches the real number n, then the function is said to have a limit of k as the input approaches n. This is written as lim xSn f 1 x 2 k, and is read βthe limit of f x 1 2 as x approaches n is kβ. See Figure 8.C-1. If the outputs of the function do not approach a single real number as the inputs approach n, the limit does not exist. Calculus is needed to find limits analytically, but a calculatorβs table feature and a graph can approximate a limit, if it exists. A table or a graph will also indicate when a limit does not exist. The following two limits are very important in calculus and are used in future can do calculus features. Example 1 Limit of sin x x Find lim xS0 sin x x, if it exists, by using a table and a graph. Solution Enter Y1 sin x x into the function editor of a calculator and produce the Figure 8.C-2a table shown in Figure 8.C-2a and the graph shown in Figure 8.C-2b. 2 The table confirms that sin 0 0 0 0 is undefined, but it also suggests that the values of itive and negative. sin x x are approaching 1 for x-values near 0, both pos- β2Ο 2Ο proaches 0 from the positive side and from the negative side. Therefore, The graph confirms that the values of sin x x are approaching 1 as x ap- β0.5 Figure 8.C-2b 566 sin x x lim xS0 1 β There is another trigonometric limit that is often used in calculus. Example 2 Limit of cos x 1 x Find lim xS0 cos x 1 x Solution, if it exists, by using a table and a graph. Figure 8.C-3a Figure 8.C-3a confirms that cos x 1 x 0 0 is undefined, but that the val- β2Ο 2 β2 ues of cos x 1 x are approaching 0 when x is near 0. The graph shown in Figure 8.C-3b also illustrates this. Therefore |
, 2Ο cos x 1 x lim xS0 0. β There are two ways in which a limit may not exist, as shown in the next examples. The first example illustrates a function that does not have a Figure 8.C-3b limit as x approaches p 2 because function values on either side of p 2 do not approach a single real number. The second example illustrates a function that does not have a limit as x approaches 0 because the function values oscillate wildly. NOTE p 2 1.57. when x 7 p 2 Example 3 Determining the Behavior of a Function Near an x-Value Discuss the behavior of f 1 cos x x 2 1 as x approaches p 2. Find lim xSp 2 1 cos x, if it exists, by using a table and a graph. Solution As shown in Figure 8.C-4a, the values of f 1 cos x x 1 2 when x 6 p 2 are large positive numbers, and the values are large negative numbers. Figure 8.C-4b shows that the graph of f 1 cos x x 1 2 has a p 2 vertical asymptote at. Because the values of f do not approach a x 1 2 single real number as x approaches p 2, lim xSp 2 1 cos x does not exist. 567 10 β2Ο 2Ο Figure 8.C-4a β10 Figure 8.C-4b β Example 4 Oscillating Function Values Discuss the behavior of f cos 1 xb a x 2 1 near x 0 and find lim xS0 cos 1 xb, a if it exists. Solution A table of values of f cos 1 xb a x 1 2 near x 0 is shown in Figure 8.C-5a. 1.1 β2Ο 2Ο Figure 8.C-5a β1.1 Figure 8.C-5b The table suggests that the function values near may be near 0.86232, but using the trace feature on the graph shown in Figure 8.C-5b indicates that the function value is near 0.365 when x is near 0. x 0 Using a window where lates wildly around x 0. 0.01 x 0.01, you can see that the graph oscil- See Figure 8.C-5c. 1 β0.01 0.01 β1 Figure 8.C-5c Because the values of x f 1 2 cos 1 xb a oscillate as x approaches 0, lim |
xS0 cos 1 xb a does not exist. β 568 Exercises Discuss the behavior of the function around the given x-value by using a table and a graph. Find the limit of each function, if it exists. 1. lim xS0 1 cos x 3. lim xSp 2 tan x x 5. lim xS0 sin 3x 3x 7. lim xS0 x tan2 x 9. lim xS0 3x sin 3x 11. lim xS0 sin 3x sin 4x 2. lim xSp 2 x tan x 4. lim xS0 tan x x 6. lim xS0 sin 2x 2x 8. lim xS0 x sin2 x 10. lim xS0 x sin x 12. lim xS0 sin 8x sin 7x 13. lim xS0 sin a 1 xb 15. lim xSp 2 x cos x 17. lim xS2 3x 19. lim xS1 x 1 x 14. lim xS0 2x sin x x 16. lim xS0 tan 1 xb a 18. 2x lim xS2 20. lim xS1 x 1 x 21. lim xS3 x 2 x 6 x 3 22. lim xS1 x 2 2x 3 x 1 23. lim xS1 3 1 x x 1 24. lim xS1 x 3 x2 x 1 x 1 25. Make a conjecture about the and c are real numbers. lim xS0 sin bx sin cx, where b 569 C H A P T E R 9 Trigonometric Identities and Proof Time. To find the exact period of the oscillations of a simple pendulum, a trigonometric expression must be written in an alternate form, which is obtained by using trigonometric identities. See Exercise 75 in Section 9.3. 570 Chapter Outline 9.1 Identities and Proofs 9.2 Addition and Subtraction Identities 9.2.A Excursion: Lines and Angles 9.3 Other Identities 9.4 Using Trigonometric Identities Chapter Review can do calculus Rates of Change in Trigonometry Interdependence of Sections 9.3 9.1 9.4 9.2 > > > T he basic trigonometric identities, which were discussed in Chapter 6 and used in Chapter 8, are not the only identities that are useful in rewriting trigonometric expressions and in solving trigonometric equ |
a- tions. This chapter presents many widely used trigonometric identities and specific methods for solving particular forms of trigonometric equa- tions. 9.1 Identities and Proofs Objectives β’ Identify possible identities by using graphs β’ Apply strategies to prove identities Recall that an identity is an equation that is true for all values of the variable for which every term of the equation is defined. Several trigonometric identities have been discussed in previous sections. This section will introduce other identities and discuss techniques used to verify that an equation is an identity. Trigonometric identities can be used for simplifying expressions, rewriting the rules of trigonometric functions, and performing numerical calculations. There are no hard and fast rules for dealing with identities, but some suggestions follow. The phrases βprove the identityβ and βverify the identityβ mean βprove that the given equation is an identity.β 571 572 Chapter 9 Trigonometric Identities and Proof 4 Graphical Testing 2Ο 2Ο 4 Figure 9.1-1 When presented with a trigonometric equation that might be an identity, it is a good idea to determine graphically whether or not this is possible. x For instance, the equation can be tested to determine sin t cos p 2 a b if it is possibly an identity by graphing Y1 cos p 2 a x b and Y2 sin x on the same screen, as shown in Figure 9.1-1 where the graph of darker than the graph of concluded that the equation is not an identity. is. Because the graphs are different, it can be Y2 Y1 Any equation can be tested by simultaneously graphing the two functions whose rules are given by the left and right sides of the equation. If the graphs are different, the equation is not an identity. If the graphs appear to be the same, then it is possible that the equation is an identity. However, The fact that the graphs of both sides of an equation appear identical does not prove that the equation is an identity, as the following exploration demonstrates. Graphing Exploration In the viewing window with both sides of the equation p x p and 2 y 2, graph cos x 1 x2 2 x4 24 x6 720 x8 40,320 Do the graphs appear identical? Now change the viewing window so that Is the equation an identity? 2p x 2p. Example 1 Graphical Identity Testing Is either of the following equations an identity? a. b. |
2 sin2x cos x 2 cos2x sin x 1 sin x sin2x cos x cos x tan x 4 Solution Test each equation graphically to see if it might be an identity by graphing each side of the equation. 2 sin2x cos x 2 cos2x sin x a. Graph and Y2 Y1 on the same is shown darker screen, as shown in Figure 9.1-2a. The graph of Y2. than 2 sin2x Because the graphs are not the same, the equation cos x 2 cos2x sin x is not an identity. Y1 2Ο 2Ο 4 Figure 9.1-2a Section 9.1 Identities and Proofs 573 b. The graph shown in Figure 9.1-2b suggests that 1 sin x sin2x cos x cos x tan x may be an identity, but the proof that it actually is an identity must be done algebraically. 4 Ο Ο Figure 9.1-2b 4 β Example 2 Finding an Identity Find an equation involving 2 sin x cos x that could possibly be an identity. Solution y 2 sin x cos x is shown in Figure 9.1-3a. Does it look familThe graph of iar? At first it looks like the graph of but there is an important difference. The function graphed in Figure 9.1-3a has a period of As was shown in Section 7.3, the graph of looks like the sine graph but has a period of y sin 2x y sin x, p. Y1 The graphs and 9.1-3b. Because the graphs appear identical, may be an identity. and Y2 sin 2x are shown in Figures 9.1-3a 2 sin x cos x sin 2x p. 2 sin x cos x β Proving Identities CAUTION Be sure to use parentheses correctly when entering each function to be graphed. 3 y = 2 sin x cos x 2Ο 2Ο 3 Figure 9.1-3a 3 y = sin 2x 2Ο 2Ο A useful feature of trigonometric functions is that they can be written in many ways. One form may be easier to use in one situation, and a different form of the same function may be more useful in another. 3 Figure 9.1-3b The elementary identities that were given in Section 6.5 are summarized for your reference on the following page. Memorizing these identities will benefit you greatly in the future |
. NOTE The definitions of the basic trigonometric ratios may help you remember the quotient and reciprocal identities. The shapes of the graphs of sine, cosine, and tangent may help you remember the periodicity and negative angle identities. Also, if you can remember the first of the Pythagorean identities, which is based on the Pythagorean Theorem, the other two can easily be derived from it. 574 Chapter 9 Trigonometric Identities and Proof Basic Trigonometry Identities Quotient Identities tan x sin x cos x cot x cos x sin x Reciprocal Identities sin x 1 csc x cos x 1 sec x csc x 1 sin x cot x 1 tan x sec x 1 cos x tan x 1 cot x Periodicity Identities sin(x 2P) sin x csc(x 2P) csc x tan(x P) tan x cos(x 2P) cos x sec(x 2P) sec x cot(x P) cot x sin2x cos2x 1 Pythagorean Identities tan2x 1 sec2x 1 cot2x csc2x sin( x) sin x Negative Angle Identities cos(x) cos x tan(x) tan x Just looking at the graphs of the two expressions that make up the equation is not enough to guarantee that it is an identity. Although there are no exact rules for simplifying trigonometric expressions or proving identities, there are some common strategies that are often helpful. Strategies for Proving Trigonometric Identities 1. Use algebra and previously proven identities to transform one side of the equation into the other. 2. If possible, write the entire equation in terms of one trigonometric function. 3. Express everything in terms of sine and cosine. 4. Deal separately with each side of the equation A B. First use identities and algebra to transform A into some expression C, then use (possibly different) identities and algebra to transform B into the same expression C. Conclude that 5. Prove that with B 0 and D 0. You can then A B. AD BC, C A B D. conclude that Section 9.1 Identities and Proofs 575 There are often a variety of ways to proceed, and it will take some practice before you can easily decide which strategies are likely to be the most efficient in a particular case. Keep these two purposes of working with trigonometric identities in mind: β’ to |
learn the relationships among the trigonometric functions β’ to simplify an expression by using an equivalent form CAUTION Proving identities is not the same as solving equations. Properties that apply to equations, such as adding the same value to both sides, are not valid when verifying identities because the beginning statement (to be verified) may not be true. In the following example, the Pythagorean identity is used to replace 1 sin2x Consider using one of the Pythagorean identities whenever a squared trigonometric function appears. cos2x. with replace sin2x cos2x tan2x with 1 cos2x 1 sin2x sec2x 1 replace csc2x sec2x cot2x with 1 cot2x tan2x 1 csc2x 1 Example 3 Transform One Side into the Other Side Verify that 1 sin x sin2x cos x cos x tan x. Solution The graph of each side of the equation is shown in Figure 9.1-2b of Example 1, where it was noted that the equation might be an identity. Begin with the left side of the equation. 1 sin x sin2x cos x sin x 1 sin2x 1 2 cos x cos2x sin x cos x cos2x sin x cos x cos x cos x sin x cos x cos x tan x regrouping terms Pythagorean identity a b c a c b c a2 a a quotient identity β Strategies for proving identities can also be used to simplify complex expressions. 576 Chapter 9 Trigonometric Identities and Proof Example 4 Write Everything in Terms of Sine and Cosine Simplify 1 csc x cot x 1 cos x. 2 2 1 Solution csc x cot x 1 2 1 2 cos x sin x b 2 1 cos x Λ 1 1 cos x 2 2 2 Λ1 sin x 1 cos x 1 a sin x 1 cos x sin x 1 cos x 1 1 1 cos2x sin x sin2x sin x sin x 1 cos x 1 2 reciprocal and quotient identities ab c a b a b 2 2 Λ1 1 a2 b2 Pythagorean identity x2 x x β The strategies presented above and those to be considered are βplans of attack.β By themselves they are not much help unless you also have some techniques for carrying out these plans. In the previous examples, the techniques of basic algebra and the use of known identities were used to change trigonometric expressions into |
equivalent expressions. There is another technique that is often useful when dealing with fractions. Rewrite a fraction in equivalent form by multiplying its numerator and denominator by the same quantity. Example 5 Transform One Side into the Other Side Prove that sin x 1 cos x 1 cos x sin x. Solution Beginning with the left side, multiply the numerator and denominator by 1 cos x. sin x 1 cos x sin x 1 cos x sin x 1 cos x 1 1 cos x 1 cos x 1 cos x 2 1 cos x 1 sin x 1 2 1 1 cos x 2 1 cos2x 2 1 cos x sin x 1 sin2x 1 cos x sin x 2 Pythagorean identity β NOTE If a denominator 1 cos x, is of the form 1 cos x multiplying by 1 cos2x sin2x. gives Similarly, if a denominator is of the form multiplying by gives Compare this with earlier techniques used to rationalize denominators and simplify numbers with complex denominators. 1 sin x, 1 sin x 1 sin2x cos2x. Section 9.1 Identities and Proofs 577 Alternate Solution The numerators of the given equation, look similar to the Pythagorean identityβexcept the squares are missing. So begin with the left side and introduce some squares by multiplying it by sin x sin x 1. and sin x 1 cos x, sin x 1 cos x sin x sin x sin x 1 cos x sin xΛ1 sin2x 1 cos x 2 1 cos2x 1 cos x sin xΛ1 1 cos x 1 sin xΛ1 1 cos x sin x 2 1 cos x 2 Λ1 1 cos x 2 Pythagorean identity 2 a2 b2 a b a b 2 21 1 Example 6 Dealing with Each Side Separately Prove that csc x cot x sin x 1 cos x. Solution Begin with the left side. csc x cot x 1 sin x 1 cos x sin x cos x sin x β [1] Example 5 shows that the right side of the identity to be proved can also be transformed into this same expression. sin x 1 cos x 1 cos x sin x Combining the equalities [1] and [2] proves the identity. csc x cot x 1 cos x sin x sin x 1 cos x [2] β Proving identities involving fractions can sometimes be quite complicated. It often helps to approach a fractional identity indirectly, as in the following example. 5 |
78 Chapter 9 Trigonometric Identities and Proof Example 7 Proving Identities that Involve Fractions Prove the first identity below, then use the first identity to prove the second identity. a. sec xΛ1 sec x cos x tan2x 2 b. sec x tan x tan x sec x cos x Solution a. Begin by transforming the left side. sec xΛ1 sec x cos x 2 cos x Λcos x sec2x sec x cos x sec2x 1 sec2x 1 tan2x tan2x. reciprocal identity Pythagorean identity Therefore, sec xΛ1 b. By part a, sec x cos x 2 sec x cos x sec xΛ1 Divide both sides of this equation by sec xΛ1 tan xΛ1 sec xΛ1 tan xΛ1 sec x cos x 2 sec x cos x 2 sec x cos x 2 sec x cos x 2 sec x tan x. tan2x 2 sec x cos x tan xΛ1 tan2x sec x cos x tan x tan x sec x cos x tan xΛ1 2 tan xΛ1 tan x sec x cos x 2 2 β Look carefully at how identity b was proved in Example 7. First prove AD BC B tan x, identity a, which is of the form C tan x, D sec x cos x Then divide both sides by BD, that. A sec x, (with and sec x cos x 2 is, by tan xΛ1 to conclude that, 2 A B C D. This property pro- vides a useful strategy for dealing with identities involving fractions. Example 8 If AD BC, with B 0 and D 0, then A B C D. Prove that cot x 1 cot x 1 1 tan x 1 tan x. Solution Use the same strategy used in Example 7. First prove A cot x 1, B cot x 1, and D 1 tan x. AD BC, C 1 tan x, AD BC cot x 1 1 2 Λ1 1 tan x cot x 1 1 tan x 2 2 Λ1 2 1 with [3] Section 9.1 Identities and Proofs 579 Multiply out the left side of [3]. cot x 1 1 2 Λ1 1 tan x 2 cot x cot x tan x 1 tan x cot x 1 tan x cot x 1 1 tan x cot |
x tan x. tan x 1 tan x Similarly, multiply the right side of [3]. cot x 1 1 2 Λ1 1 tan x 2 cot x cot x tan x 1 tan x cot x 1 1 tan x cot x tan x. Because the left and right sides are equal to the same expression, [3] has been proven to be an identity. Therefore, conclude that cot x 1 cot x 1 1 tan x 1 tan x is also an identity. CAUTION β C D and cross multiply to eliminate the fractions. If you did Strategy 5 does not say that you begin with a fractional equation A B that, you would be assuming that the statement was true, which is what has to be proved. What the strategy says is that to prove an identity involving fractions, you need only prove a different identity that does not involve fractions. In other words, if you prove that then you can conclude whenever D 0, B 0 and that Note that you do not assume that AD BC; you use AD BC C A B D. some other strategy to prove this statement. It takes a good deal of practice, as well as much trial and error, to become proficient at proving identities. The more practice you have, the easier it will become. Because there are many correct methods, your proofs may be quite different from those of your classmates, instructor, or text answers. If you do not see what to do immediately, try something and see where it leads: multiply out, factor, or multiply numerator and denominator by the same nonzero quantity. Even if this does not lead anywhere, it may give you some ideas on other strategies to try. When you do obtain a proof, check to see if it can be done more efficiently. Do not include the βside tripsβ in your final proofβthey may have given you some ideas, but they are not part of the proof. 580 Chapter 9 Trigonometric Identities and Proof Exercises 9.1 In Exercises 1β4, test the equation graphically to determine whether it might be an identity. You need not prove those equations that appear to be identities. 1. sec x cos x sec x sin2x 2. tan x cot x sin x cos x 3. 4. 1 cos 2x 2 sin2x tan x cot x csc x sec x In Exercises 5β8, insert one of aβf on the right of the equal sign so |
that the resulting equation appears to be an identity when you test it graphically. You need not prove the identity. a. cos x b. sec x c. sin2x d. sec2 x e. sin x cos x f. 1 sin x cos x 5. csc x tan x ____ 6. 7. sin x tan x ____ sin4x cos4 Λx sin x cos x ____ 8. tan2 x 2 1 sin x 1 sin x ____ 2 In Exercises 9β18, prove the identity. 9. tan x cos x sin x 10. cot x sin x cos x 11. cos x sec x 1 12. sin x csc x 1 13. tan x csc x sec x 14. sec x cot x csc x 15. 17. 18. tan x sec x sin x 16. cot x csc x cos x 1 cos x csc x 1 1 cos x 2 1 csc x 1 sin2x 2 cot2x 2 2 1 1 1 In Exercises 19β48, state whether or not the equation is an identity. If it is an identity, prove it. 19. sin x 21 cos2x 20. cot x csc x sec x 21. sin cos x 2 x 2 1 1 tan x 22. tan x 2sec2x 1 23. cot x 2 1 cot x 24. 25. sec x sec x 1 2 1 sec2x tan2x 26. sec4 Λx tan4 Λx 1 2 tan2x 27. sec2x csc2x tan2x cot2x 28. sec2x csc2x sec2x csc2x sin2x 1 cos2x cot x 1 2 sec x 1 2 2 cos2x tan x 1 2 2 1 1 cos x 2 2 1 2 sin2x tan2x sin2x tan2x 1 29. 30. 31. 33. 34. 35. 36. 32. cot2x 1 csc2x tan2x tan2x 1 2 1 csc x sin x 2 cos2x 1 1 2 1 cos2x 1 tan x sec x csc x cos sin x 1 x 1 2 2 cot x 37. cos4x sin4x cos2x sin2x 38. cot2x cos2x cos2x cot2x 39. 40. 41 |
. 42. 43. 44. sin x cos x 2 sin2x cos2x 1 1 1 tan x 2 2 2 sec2x sec x csc x sin x cos x 2 tan x 1 cos x sin x sin x 1 cos x 2 csc x sec x csc x 1 tan x csc x cot x 1 1 tan x csc x sec x Section 9.2 Addition and Subtraction Identities 581 54. sin x 1 cot x cos x 1 tan x cos x sin x 55. cos x 1 sin x sec x tan x 45. 1 csc x sin x sec x tan x 46. 47. 48. 1 csc x csc x cos2x 1 sin x sin x cos x tan x tan x sin x cos x cot x csc x 1 csc x 1 cot x In Exercises 49β52, half of an identity is given. Graph and this half in a viewing window with write a conjecture as to what the right side of the identity is. Then prove your conjecture. 2P x 2P 49. 50. 51. 52. 1 sin2x has a graph that looks like this? 1 cos x? Hint: What familiar function 1 cos x cos2x sin x cot x? sin x cos x sec x csc x cot x 2? 1 cos3x 21 1 tan4x sec4x 2? 2 1 In Exercises 53β66, prove the identity. 53. 1 sin x sec x cos3x 1 sin x 56. 57. 58. 59. 60. 61. 62. 63. 64. 65. 66. 1 sec x tan x sin x csc x cos x cot x cot x cos x cot x cos x cos x cot x cos3x sin3x cos x sin x 1 sin x cos x 2 2 cot x sec x tan x cos x log101 log101 2 csc x cot x log101 log101 log101 log101 log101 log101 tan x tan y tan x tan y 2 sec x tan x 2 2 csc x cot x 2 sec x tan x cot x cot y 1 2 2 tan x tan y cot x cot y tan x tan y cos x sin y cos y sin x cos y sin x cos x sin y tan x tan y cot x cot y tan x tan y 1 1 cot x cot y 9 |
.2 Addition and Subtraction Identities Objectives β’ Use the addition and subtraction identities for sine, cosine, and tangent functions β’ Use the cofunction identities Many times, the input, or argument, of the sine or cosine function is the sum or difference of two angles, and you may need to simplify the expression. Be careful not to make this common student error. x p 6 b is not sin x sin p 6 sin a Graphing Exploration Verify graphically that the expressions above do NOT form an iden- tity by graphing Y1 sin x p a 6 b and Y2 sin x sin p 6. 582 Chapter 9 Trigonometric Identities and Proof The exploration shows that sin x y sin x sin y because it is false when y p 6. 1 So, is there an identity involving sin 2 x y 1? 2 Graphing Exploration Graph Y1 sin x p a 6 b and Y2 23 2 Λ sin x 1 2 Λ cos x on the same screen. Do the graphs appear identical? The exploration suggests that sin x p 6 b a 23 2 sin x 1 2 cos x may be an identity. Furthermore, note that the coefficients on the right side can 23 2 be expressed in terms of In other words, 1 2 p 6 the following equation appears to be an identity. x p 6 b sin x cos cos x sin : cosΛ and sin Λ p 6 p 6 p 6 p 6 sin a. Graphing Exploration 1 Y1 sin x 5 Graph on the same screen. Do the graphs appear identical? What identity does this suggest? Repeat the process with some other number in place of 5. Are the results the same? sin x cos 5 cos x sin 5 and Y2 2 The equations examined in the discussion and exploration above are examples of the first identity listed below. Each identity can be confirmed by assigning a constant value to y and then graphing each side of the equation, as in the Graphing Exploration above. Addition and Subtraction Identities for Sine and Cosine sin(x y) sin x cos y cos x sin y sin(x y) sin x cos y cos x sin y cos(x y) cos x cos y sin x sin y cos(x y) cos x cos y sin x sin y The addition and subtraction identities are important trigonometric identities. You should become familiar with the examples and special cases that follow. NOTE In |
order to use addition or subtraction identities to find exact values, first write the argument as a sum or difference of two terms for which exact values are known, such as, and p. Section 9.2 Addition and Subtraction Identities 583 Example 1 Addition Identities Use the addition identities to find the exact values of sin 5p 12 and cos 5p 12. Solution Because x p 6 5p 12 2p 12 y p 4. and 3p 12 p 6 p 4, apply the addition identities with sin 5p 12 sin p a 6 22 2 p 4 b 23 2 p 6 22 2 1 2 p 4 22 sin cos cos p 6 sin p 4 A 1 23 4 B cos sin p 4 p 4 22 sin p 6 23 1 4 A B cos 5p 12 cos a 23 2 p 4 b p 6 22 2 1 2 cos p 6 22 2 Example 2 Subtraction Identity Find sin p y. 2 1 Solution Apply the subtraction identity for the sine function with 1 sin p cos y cos p sin y 0 cos y p y sin x p. sin y sin y 1 2 1 2 β β Example 3 Addition Identity Prove that cos x cos y 1 2 3 cos 1 x y 2 cos x y 1. 2 4 Solution Begin with the more complicated right side and use the addition and subtraction identities for cosine to transform it into the left side. 1 2 3 cos x y cos x y 1 2 4 2 1 1 cos x cos y sin x sin y 2 2 3 1 cos x cos y sin x sin y 2 4 cos x cos y cos x cos y 1 2 1 cos x cos y 2 2 cos x cos y 2 β 1 1 2 1 584 Chapter 9 Trigonometric Identities and Proof NOTE Recall that the difference quotient of a function f is. Simplifying the Difference Quotient of a Trigonometric Function The difference quotient is very important in calculus, and the addition identities are needed to simplify difference quotients of trigonometric functions. Example 4 The Difference Quotient of f(x) sin x Show that for the function Solution sin x and any number h 0, x 2 1 sin x cos h 1 h a b cos x sin h a h b. Use the addition identity for sin sin y h. with sin sin x cos h cos x sin h sin x h cos x sin h sin x 1 cos h 1 2 h cos h 1 h a sin x cos x b sin h a |
h b β Addition and Subtraction Identities for the Tangent Function The addition and subtraction identities for sine and cosine can be used to obtain the addition and subtraction identities for the tangent function. Addition and Subtraction Identities for Tangent tan(x y) tan(x y) tan x tan y 1 tan x tan y tan x tan y 1 tan x tan y A proof of these identities is outlined in Exercise 36. Example 5 Addition and Subtraction Identities for Tangent Find the exact values of such that 0 6 x 6 p 2, and x y sin 2 1 p 6 y 6 3p 2, x y tan 1 sin x 3 4, 2 and if x and y are numbers cos y 1 3. Deter- NOTE See Figure 6.4-7 for the signs of the functions in each quadrant. y ( 7, 3) x x 4 3 x 42 β 32 = 7 sin x = 3 4 Figure 9.2-1a y 1 y 3 32 β 12 = 8 = 2 2 (β1, β2 2) cos y = β 1 3 Figure 9.2-1b Section 9.2 Addition and Subtraction Identities 585 mine in which of the following intervals x y lies: p 2 b, 0, a 3p 2 b, or 3p 2 a, 2p. b Solution Use the Pythagorean identity and the fact that cos x and tan x are positive in the first quadrant to obtain the following. See Figure 9.2-1a. 27 4 cos x 21 sin2x 1 9 16 3 4 b 7 16 1 B B B a 2 tan x sin x cos x 3 4 27 4 3 27 327 7 Because y lies between p and 3p 2, its sine is negative. See Figure 9.2-1b. sin y 21 cos2y 28 3 222 3 tan y sin y cos y 222 3 1 3 222 3 3 1 222 The addition identities for sine and tangent show exact values. x y sin 1 x y tan 1 2 2 sin x cos y cos x sin y 222 1 27 3 3 3 4 4 tan x tan y 1 tan x tan y 3 12 2214 12 3 2214 12 222 327 7 327 222 327 1422 7 7 6214 7 B 1 7 bA a 327 1422 7 6214 2727 3222 65 must be in the interval Both the sine and tangent of 3p |
2 intervals in which both sine and tangent are negative., 2p b a x y are negative numbers. Therefore x y because it is the only one of the four β Cofunction Identities Special cases of the addition and subtraction identities are the cofunction identities. 586 Chapter 9 Trigonometric Identities and Proof Cofunction Identities sin x cos P 2 a x b tan x cot P 2 a x b sec x csc P 2 a x b cos x sin cot x tan csc x sec P 2 a x b The first cofunction identity is proved by using the identity for cos with p 2 in place of x and x in place of y. x y 2 1 cos p 2 a x b cos p 2 cos x sin p 2 sin x 0 cos x 1 sin x sin x Because the first cofunction identity is valid for every number x, it is also valid with the number p 2 x in place of x. sin p 2 a x b cos p 2 c p 2 a x b d cos x Thus, the second cofunction identity is proved. The others now follow from these previous two. For instance, Also, tan p 2 a x b sin a cos a p 2 p 2 x b x b cos x sin x cot x csc p 2 a x b 1 p 2 sin a x b 1 cos x sec x Example 6 Cofunction Identities cos x p 2 b a cos x tan x. Verify that Solution Beginning with the left side, the term cos x p 2 b a looks almost, but not quite, like the term cos. Therefore, in the cofunction identity. But note that Section 9.2 Addition and Subtraction Identities 587 cos x p a 2 b cos x cos c x p a cos x 2 b d cos x p a 2 cos x b sin x cos x tan x negative angle identity with x p 2 in place of x cofunction identity quotient identity β Exercises 9.2 In Exercises 1β12, find the exact value. 1. sin 4. sin 7. tan p 12 5p 12 7p 12 2. cos 5. cot p 12 5p 12 8. cos 11p 12 3. tan 6. cos p 12 7p 12 9. cot 11p 12 10. sin 75Β° Hint: 75Β° 45Β° 30Β°. 11. sin 105Β° 12. cos 165Β° 25. If sin x 1 3 and |
0 6 x 6 p 2, then sin p 4 a x b? 26. If cos x 1 4 and p 2 6 x 6 p, then cos 27. If cos x 1 5 and p 6 x 6 3p 2, then sin? 28. If sin x 3 4 and 3p 2 6 x 6 2p, then cos p 4 a x b? In Exercises 13β18, rewrite the given expression in terms of cos x. sin x and 13. sin p 2 a x b 15. cos x 3p 2 b a 17. sec x p 1 2 14. cos x p 2 b a 16. csc x p a 2 b 18. cot 1 x p 2 In Exercises 19β24, simplify the given expression. 19. sin 3 cos 5 cos 3 sin 5 20. sin 37Β° sin 53Β° cos 37Β° cos 53Β° x y x y 2 sin y 2 sin y 21. cos 1 sin 22. 23. 1 cos 1 24. sin 1 x y x y x y x y cos y sin 1 2 2 cos y cos 1 2 cos 1 x y 2 sin 1 x y 2 2 In Exercises 29β34, assume that sin y 20.75 and that x and y lie between 0 and sin x 0.8 and P 2. Evaluate the given expressions. x y x y 2 2 30. sin 32. sin 34. tan 1 29. cos 31. cos 33. tan 1 1 1 35. If f x 1 2 x y 2 cos x prove that the difference quotient is and h is a fixed nonzero number cos x cos h 1 h b a sin x sin h a h b. 36. Prove the addition and subtraction identities for the tangent function. Hint: tan 1 x y 2 sin 1 cos 1 x y 2 x y 2 Use the addition identities on the numerator and denominator; then divide both numerator and denominator by, and simplify. cos x cos y x y 1 x y 1 cos 2 sin 2 1 x y 1 x y cos2 Λx cos2 Λy sin2 Λx sin2 Λy 2 sin2x cos2y cos2x sin2y 2 56. sin In Exercises 57β66, determine graphically whether the equation could not possibly be an identity (by choosing a numerical value for y and graphing both sides), or write a proof that it is. |
588 Chapter 9 Trigonometric Identities and Proof 37. If x is in the first quadrant and y is in the second quadrant, sin x 24 25 x y value of and tan sin 2 x y quadrant in which and, 1 1 lies. sin y 4 5 x y and the, 2 find the exact 50. cos 51. tan x p x p 2 2 1 1 cos x tan x 38. If x and y are in the second quadrant, and, cos y 3 4 x y and lies., 2 x y cos 1 which find the exact value of tan x y 2 1 and the quadrant in, sin x 1 3 x y sin 1 52. sin x cos y 1 2 3 sin 53. sin x sin y 1 2 3 cos 54. cos x sin y 1 2 3 sin sin 1 x y 2 4 cos x y 1 2 4 sin 1 x y 2 4, 2 39. If x is in the first quadrant and y is in the second 55. cos quadrant, sin x 4 and, 5 x y 2 x y exact value of 1 quadrant in which cos cos y 12, 13 x y tan and lies. find the and the 2 40. If x is in the fourth quadrant and y is in the first quadrant, cos x 1 3 x y and value of sin 2 x y quadrant in which, 1 and tan 1 lies. cos y 2 3 x y and the, find the exact 41. Express sin u v w 1 cosines of u, v, and w. Hint: First apply the addition identity with y w. and 2 x u v in terms of sines and 1 2 42. Express x y z cos cosines of x, y, and z. 1 in terms of sines and 2 60. cos 43. If x y p 2, show that sin2x sin2y 1. 44. Prove that cot x y 1 cot x cot y 1 cot x cot y. 2 In Exercises 45β56, prove the identity. 61. 62. 63. 45. sin 1 46. cos 47. cos 1 1 48. tan 49. sin sin x cos x cos x tan x 2 2 2 sin x 2 57. 58. x y cos 2 sin x cos y 1 x y cos 2 sin x cos y 1 59. sin 1 x y 2 x y 2 cot x tan y cot x tan y sin x |
sin y cos x cos y 1 1 1 1 1 sin sin sin sin cos cos 1 1 x y x y 2 2 tan x tan y tan x tan cot y cot x cot y cot x cot x tan y cot x tan y 64. cos cos 65. tan 66. cot cot y tan x cot y tan x 2 2 tan x tan y 2 cot x cot y 2 Section 9.2.A Excursion: Lines and Angles 589 9.2.A Excursion: Lines and Angles Objectives β’ Find the angle of inclination of a line with a given slope β’ Find the angle between two lines Several interesting concepts dealing with lines are defined in terms of trigonometry. They lead to useful facts whose proofs are based on the addition and subtraction identities for sine, cosine, and tangent. If L is a nonhorizontal straight line, the angle of inclination of L is the formed by the part of L above the x-axis and the x-axis, positive angle as shown in Figure 9.2.A-1 Figure 9.2.A-1 The angle of inclination of a horizontal line is defined to be Thus, the radian measure of the angle of inclination of any line satisfies 0 u 6 p. Furthermore, u 0. Angle of Inclination Theorem If L is a nonvertical line with angle of inclination tan U slope of L. U, then Proof If L is horizontal, then L has slope 0 and angle of inclination tan u tanΛ 0 0, tan u slope L 0. so of u 0. Hence, If L is not horizontal, then it intersects the x-axis at some point shown for two possible cases in Figure 9.2.A-2. 1 x1, 0, as 2 (x2, y2) (x2, y2) L y2 y2 L (x1, 0) ΞΈ x ΞΈΞΈΟ β x x2 β x1 Figure 9.2.A-2a x1 β x2 Figure 9.2.A-2b (x1, 0) 590 Chapter 9 Trigonometric Identities and Proof The right triangle in Figure 9.2.A-2a shows that slope of L y2 x2 0 x1 y2 x2 x1 opposite adjacent tan u. The right triangle in Figure 9.2.A-2b shows that slope of |
L 0 y2 x2 x1 y2 x1 x2 opposite adjacent tan p u. 2 1 [1] Use the fact that the tangent function has period and the negative angle identity for tangent to obtain tan tan p u u p tan u. 2 L tan Λu shows that slope of tan u 2 1 1 2 Combining this fact with also. 1 1 3 4 in this case β Example 1 Angle of Inclination Find the angle of inclination of a line of slope 5 3. Solution By the Angle of Inclination Theorem, culator shows that u 1.0304 radians, or. The TAN1 tan u 5 3 u 59.04Β°. key on a cal- β Example 2 Angle of Inclination Find the angle of inclination of a line L with slope 2. Solution Because line L has slope tan u 2 that lies between 1.1071. solution p, another solution is needed. tanΛ t tan Recall that 2, p 2 its angle of inclination is a solution of and A calculator gives the approximate p. Because an angle of inclination must be between 0 and 1 t p, for every t. So u 1.1071 p 2.0344 2 is the solution of angle of inclination is approximately 2.03 radians, or about in the interval from 0 to tan u 2 p. Therefore, the 116.57Β°. β ΞΈΟ β ΞΈ ΞΈ ΞΈΟ β Figure 9.2.A-3 Angles Between Two Lines If two lines intersect, then they determine four angles with vertices at the point of intersection, as shown in Figure 9.2.A-3. If one of these angles Section 9.2.A Excursion: Lines and Angles 591 u measures p u vertical angle theorem from plane geometry. radians, then each of the two angles adjacent to it measures radians by the radians. (Why?) The fourth angle also measures u The angles formed by intersecting lines can be determined from the angles a of inclination of the lines. Suppose L and M have angles of inclination Basic facts about parallel lines, as and respectively, such that b a is one angle between L and illustrated in Figure 9.2.A-4, show that M, and is the other. b a. b a p bΞ² β Ξ±) L Ξ± M Ξ² x Figure 9.2.A-4 The angle between two lines can also be found from their slopes by |
using the following fact. Angle Between Two Lines If two nonvertical, nonperpendicular lines have slopes m and k, then one angle between them satisfies U tanΛ U m k 1 mk ` `. Proof Suppose L has slope k and angle of inclination and that M has slope m and angle of inclination a b. m k 1 mk By the definition of absolute value m k m k 1 mk 1 mk ` or ` m k 1 mk ` ` whichever is positive. It will be shown that one angle between L and M has tangent, and that the other has tangent. Thus, one m k 1 mk m k 1 mk of them necessarily has tangent. If b a, then b a is one m k 1 mk ` ` angle between L and M. By the subtraction identity for tangent, tan 1 b a 2 tan b tan a 1 tan b tan a p m k 1 mk b a The other angle between L and M is 2 negative angle identity, and the addition identity 1 and by periodicity, the 592 Chapter 9 Trigonometric Identities and Proof p tan tan 1 3 b a tan 1 tan b tan a 1 tan b tan a m k 1 mk. This completes the proof when similar. b a. The proof in the case a b is β Example 3 The Angle Between Two Lines If the slopes of lines L and M are 8 and angle between them. 3, respectively, then find one Solution Substitute 8 for m and 3 ` tan u for k to find the tangent values. 3 2 3 2 1 11 23 8 1 1 8 11 23 ` ` u 0.4461 radians, or ` Solving the equation yields 25.56Β°. β Exercises 9.2.A In Exercises 1β6, find the angle of inclination of the straight line through the given points. 1. 1 1, 2 2, (3, 5) 3. (1, 4), (6, 0) 5. 1 3, 7 2, (3, 5) 2. (0, 4), 4. (4, 2), 6. (0, 0), 5, 1 2 3, 2 4, 5 1 1 1 2 2 In Exercises 7β12, find one of the angles between the straight lines L and M. 7. L has slope 3 2 and M has slope 1. 8. L has slope 1 and M has slope 3. 9. L has slope 10. L |
has slope 1 2 and M has slope 0. and M has slope 3. 11. (3, 2) and (5, 6) are on L; (0, 3) and (4, 0) are on M. 12. 1, 2 1 on M. 2 and 1 3, 3 2 are on L; 3, 3 2 1 and (6, 1) are Section 9.3 Other Identities 593 9.3 Other Identities Objectives β’ Use the following identities: double-angle power-reducing half-angle product-to-sum sum-to-product Double-Angle Identities A variety of identities that are special cases of the addition and subtraction identities of Section 9.2 are presented in this section. These identities include double-angle identities, power-reducing identities, half-angle identities, product-to-sum identities, and sum-to-product identities. Double-Angle Identities Special cases of the addition identities occur when two angles have the same measure. These identities are called the double-angle identities. sin Λ2x 2 sin x cos x cosΛ 2x cos2x sin2x tan 2x 2 tan x 1 tan2x Proof Substitute x for y in the addition identities. sin 2x sin 1 cos 2x cos 1 x x 2 x x 2 tan 2x tan x x 2 1 sinΛ x cos Λx cos Λx sin Λx 2 sinΛ x Λcos x cos Λx Λcos Λx sin Λx ΛsinΛ x cos2x sin2 tan Λx tan Λx 1 tan Λx tan Λx 2 tan x 1 tan2 Λx Λx Example 1 Use Double-Angle Identities If p 6 x 6 3p 2 and cos Λx 8 17, find sin 2x and cos 2x, and show that 5p 2 6 2x 6 3p. Solution In order to use the double-angle identities, first determine sin x, which can be found by using the Pythagorean identity. sin2 Λx 1 cos2 Λx 1 8 2 1 64 289 225 289 Thus, sin x Β± B rant, and sin x is negative there.. Since 225 289 a 17 b p 6 x 6 3p 2, x must be |
in the third quad- sin Λx B 225 289 15 17 y x 8 x 15 17 (β8, β15) Figure 9.3-1 594 Chapter 9 Trigonometric Identities and Proof Now substitute these values in the double-angle identities to find sin 2x and cos 2x. sin Λ2x 2Λ sinΛ x cos Λx 2 15 a 17 ba 8 17 b 240 289 0.8304 cos 2x cos2x sin2x 8 a 17 b 2 2 15 a 17 b 64 289 225 289 161 289 0.5571 You know that x lies in the third quadrant. Multiply the inequality p 6 x 6 3p 2 first or second quadrant. The calculations above show that at 2x, sine is positive and cosine is negative. This can occur only if 2x lies in the sec- That is, 2x is in either the 2p 6 2x 6 3p. by 2 to find that ond quadrant, so 5p 2 6 2x 6 3p. β Example 2 Use Double-Angle Identities Express the rule of the function and constants. f x 1 2 sin 3x in terms of powers of sin x Solution First use the addition identity for f x 1 2 sin 3x sin x 2x 1 2 2 sin x cos x 2 1 identity for sin 2x 1 2 sin with y 2x. x y sin x cos 2x cos x sin 2x sin xΛ1 cos x cos2x sin2x 2 identity for cos 2x sin x cos2x sin3x 2 sin x cos2x 3 sin x cos2x sin3x 3 sin x sin3x 2 1 1 sin2x Pythagorean identity 3 sin x 3 sin3x sin3x 3 sin x 4 sin3x β Forms of cos 2x The double-angle identity for can be rewritten in several useful ways. For instance, we can use the Pythagorean identity in the form of cos2x 1 sin2x cos 2x cos 2x cos2x sin2x to obtain the following. 1 sin2x 1 2 sin2x 1 2 sin2x sin2x 1 cos2x to Similarly, use the Pythagorean identity in the form obtain the following. cos 2x cos2x sin2x cos2x 1 cos2x 2 1 2 cos2x 1 Forms of cos 2x cos 2x |
1 2 sin2x cos 2x 2 cos2x 1 Section 9.3 Other Identities 595 Example 3 Use Forms of cos 2x Prove that 1 cos 2x sin 2x tan x. Solution Use the first identity in the preceding box and the double-angle identity for sine. 1 cos 2x sin 2x 1 1 2 sin2x 2 sin x cos x 1 2 2 sin2x 2 sin x cos x sin x cos x tan x β Power-Reducing Identities If the first equation in the preceding box is solved for ond one for new forms are called the power-reducing identities. and the secalternate forms for these identities are obtained. The cos2 x, sin2x Power-Reducing Identities sin2x 1 cos 2x 2 cos2x 1 cos 2x 2 Example 4 Use Power-Reducing Identities Express the function of cosine functions. f x 1 2 Solution sin4 x in terms of constants and first powers Begin by applying the power-reducing identity. f x 1 2 sin4x sin2x sin2x 1 cos 2 x 2 1 2 cos 2 x cos2 2 x 4 1 cos 2x 2 596 Chapter 9 Trigonometric Identities and Proof NOTE To write cos2 2x in terms of first powers of cosine functions, use 2x in place of x in the powerreducing identity for cosine. 2x cos2 2x 1 cos 2 1 2 1 cos 4x 2 2 Half-Angle Identities Next apply the power-reducing identity for cosine to (See Note.) 1 2 cos 2 x cos22 x 4 2 cos22x. 1 2 cos 2 x 1 cos 4x 4 cos 2x 1 8 Λ cos 2x 1 8 1 2 1 2 cos 4x 1 1 4 3 8 1 cos 4 x 2 β Half-Angle Identities The power-reducing identities with the half-angle identities. x 2 in place of x can be used to obtain 1 cos2 x 2 b a 2 Β± 1 cos x 2 B sin2 sin x 2 b a x 2 b a 1 cos x 2 This proves the first of the half-angle identities. sin x 2 1 cos x 2 B cos x 2 Β±Β± 1 cos x 2 B tan x 2 Β±Β± 1 cos x 1 cos x B The sign in front of the radical depends upon the quadrant in which x 2 lies. The half-angle identity for cos |
ine is derived from a power-reducing identity, as was the half-angle identity for sine. The half-angle identity for tangent then follows immediately since tan x 2 b a sin x 2 b a cos x 2 b a. Example 5 Half-Angle Identities Find the exact value of a. cos 5p 8 b. sin p 12 Section 9.3 Other Identities 597 use the half-angle identity with x 5p 4 and Solution a. Because 5p 8 1 2 a, 5p 4 b 22 2 5p 4 5p 8 the fact that cos. The sign chart given in Section 6.4 shows that cos is negative because 5p 8 is in the second quadrant. So, use the negative sign in front of the radical. cos 5p 8 cos 5p 4 2 R 1 cos 5p 4 b a R R 2 12 2 b 1 a 2 2 12 A B 2 2 2 12 4 B 22 12 2 p 12 1 2 a p 6 b and p 12 b. Because positive, is in the first quadrant, where sine is sin p 12 sin p 6 2 R S 1 cos p 6 b a 2 1 23 2 2 2 23 2 2 2 23 4 S C 32 23 2 β The problem of determining signs in the half-angle formulas can be eliminated for the tangent by using the following formulas. 598 Chapter 9 Trigonometric Identities and Proof Half-Angle Identities for Tangent tan tan x 2 x 2 1 cos x sin x sin x 1 cos x Proof The proof of the first of these identities follows from the identity tan x 1 cos 2 x this identity. which was proved in Example 3. Replace x by sin 2x in x 2, tan x 2 1 cos 2 x 2 b a sin 2 x 2 b a 1 cos x sin x The second identity in the box is proved in Exercise 71. Example 6 Use Half-Angle Identity for Tangent If tan x 3 2 and p 6 x 6 3p 2, find tanΛ x 2. Solution The terminal side of an angle of x radians in standard position lies in the third quadrant, as shown in Figure 9.3-2. The tangent of the angle in standard position whose terminal side passes through the point is 3 2 Because there is only one angle in the third quadrant with tan- 2, 3 3 2. 2 1 gent 3 2, the point 2, 3 1 2 radians. must lie |
on the terminal side of the angle of x The distance from triangle. 1 2, 3 2 to the origin is the hypotenuse of the Therefore 213 sin x 3 213 cos x 2 213 By the first of the half-angle identities for tangent, tan Λ x 2 1 cos x sin x 1 2 213 b a 3 213 213 2 213 3 213 213 2 3 β y x 2 x 3 13 (β2, β3) Figure 9.3-2 Section 9.3 Other Identities 599 Product-to-Sum Identities Use the addition and subtraction identities to rewrite x y. 2 sin x cos y cos x sin y sin x cos y cos x sin y 2 sin x cos y sin x y sin 2 1 1 sin 1 x y sin x y 1 2 2 Dividing both sides of the equation by 2 produces the first of the following identities. Product-to-Sum Identities sin x cos y 1 sin x sin y 1 cos x cos y 1 2 cos x sin y 1 2 2 Λ [sin(x y) sin(x y)] 2 Λ [cos(x y) cos(x y)] [cos(x y) cos(x y)] [sin(x y) sin(x y)] The proofs of the second and fourth product-to-sum identities are similar to the proof of the first. The third product-to-sum identity was proved in Example 3 of Section 9.2. Sum-to-Product Identities Use the first product-to-sum identity with x y 2 in place of x and Λ x y 2 Λ in place of y to obtain the first sum-to-product identity sin Λ cos sin sin Λ sin x sin y 2 Multiplying both sides of the equation by 2 produces the identity. Sum-to-Product Identities sin x sin y 2 sin sin x sin y 2 cos cos x cos y 2 cos Λ cos Λ sin Λ cos cos x cos y 2 sin Λ sin 600 Chapter 9 Trigonometric Identities and Proof The other sum-to-product identities are proved the same way as the first. Example 7 Use Sum-to-Product Identities Prove the identity below. sin t sin 3t cos t cos 3t tan 2t Solution Use the first sum-to-product identity with sin t sin 3t 2 sin t 3t 2 a b Λ |
cos a Similarly, x t t 3t 2 b and y 3t. 2 sin 2t cos t 2 1 cos t cos 3t 2 cos Λa t 3t 2 b Λ cos a t 3t 2 b 2 cos 2t cos t. 2 1 Therefore, sin t sin 3t cos t cos 3t 2 sin 2t cos 2 cos 2t cos t 2 t 2 1 1 sin 2t cos 2t tan 2t. β Exercises 9.3 In Exercises 1β12, use the half-angle identities to evaluate the given expression exactly. In Exercises 23β30, find under the given conditions. sin 2x, cos 2x, and tan 2 x 1. cos 5. tan p 8 p 12 2. tan p 8 6. sin 5p 8 9. sin 7p 8 10. cos 7p 8 11. tan 3. sin 3p 8 7. cos p 12 7p 8 4. cos 8. tan 12. cot 3p 8 5p 8 p 8 In Exercises 13β18, write each expression as a sum or difference. 13. sin 4x cos 6x 14. sin 5x sin 7x 15. cos 2x cos 4x 16. sin 3x cos 5x 17. sin 17x sin 3x 1 2 18. cos 13x cos 5x 2 1 In Exercises 19β22, write each expression as a product. 19. sin 3x sin 5x 20. cos 2x cos 6x 21. sin 9x sin 5x 22. cos 5x cos 7x 23. sin x 5 13, for 0 6 x 6 p 2 24. sin x 4 5, for p 6 x 6 3p 2 25. cos x 3 5, for p 6 x 6 3p 2 26. cos x 1 3, for p 2 6 x 6 p 27. tan x 3 4, for p 6 x 6 3p 2 28. tan x 3 2, for p 2 6 x 6 p 29. csc x 4, for 0 6 x 6 p 2 30. sec x 5, for p 6 x 6 3p 2 5910ac09_570-613 9/21/05 2:30 PM Page 601 In Exercises 31β36, find sin the given conditions. x 2, cos x 2, and tan x 2 under 31. cos x 0.4, for 0 6 x 6 p 2 |
32. sin x 0.6, for p 2 6 x 6 p 33. sin x 3 5, for 3p 2 6 x 6 2p 34. cos x 0.8, for 3p 2 6 x 6 2p 35. tan x 1 2, for p 6 x 6 3p 2 36. cot x 1, for p 6 x 6 p 2 In Exercises 37β42, assume sin x 0.6 and evaluate the given expression. Section 9.3 Other Identities 601 53. cos4x sin4x cos 2x 54. sec 2x 1 1 2 sin2x 55. cos 4x 2 cos 2x 1 56. sin2x cos2x 2 sin x 57. 1 cos 2x sin 2x cot x 58. sin 2x 2 cot x csc2x 59. sin 3x sin x 1 2 Λ1 60. 61. 4 cos x sin x sin 4x 1 cos 2x 2 tan x sec2x 2 Λ1 3 4 sin2x 2 1 2 sin2x 2 and 0 66 x 66 P 2 62. cos 3x cos x 1 2 Λ1 3 4 cos2x 2 37. sin 2x 38. cos 4x 39. cos 2x 40. sin 4x 41. sin x 2 42. cos x 2 63. csc2 Λa x 2 b 2 1 cos x 64. sec2 x 2 b Λa 2 1 cos x 43. Express cos 3x in terms of cos x. In Exercises 65β70, prove the identity. 44. a. Express the function f cos3x in terms of x 1 2 constants and first powers of the cosine function, as in Example 4. b. Do the same for f x 2 1 cos4x. In Exercises 45β50, simplify the given expression. 45. sin 2x 2 sin x 47. 2 cos 2y sin 2y 46. 1 2 sin2 Λa x 2 b 48. 49. 50. cos2 Λa x 2 b sin2 Λa x 2 b sin x cos x 2 sin 2x 2 1 2 sin x cos3x 2 sin3x cos x In Exercises 51β64, determine graphically whether the equation could not possibly be an identity, or write a proof showing that it is. 51. sin 16x 2 sin 8x cos |
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