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 = _ _ 2 6 2 √ _ 2 3 √ 5π, evaluate sec   = − _ _ 2 6, evaluate tan(45°). — 5π . _ 6 b. Given sin  Solution Because we know the sine and cosine values for these angles, we can use identities to evaluate the other functions. a. b. tan(45°) = sin(45°) _______ cos(45°) = — 5π sec   = _ 6 1 _ 5π   cos ___ — Try It #5 Evaluate csc  7π . _ 6 Example 6 Using Identities to Simplify Trigonometric Expressions Simplify sec t _. tan t Solution We can simplify this by rewriting both functions in terms of sine and cosine. sec t _ tan t = 1 _ cos t _ sin t _ cos t To divide the functions, we multiply by the reciprocal. = 1 cos t _ cos t sin t Divide out the cosines. = 1 _ sin t = csc t Simplify and use the identity. By showing that can be simplified to csc t, we have, in fact, established a new identity. sec t _ tan t sec t _ tan t = csc t Try It #6 Simplify (tan t)(cos t). 48 0 CHAPTER 5 trigonometric Functions Alternate Forms of the Pythagorean Identity We can use these fundamental identities to derive alternative forms of the Pythagorean Identity, cos2 t + sin2 t = 1. One form is obtained by dividing both sides by cos2 t : cos2 t _____ cos2 t + sin2 t ____ cos2 t = 1 ____ cos2 t The other form is obtained by dividing both sides by sin2 t : 1 + tan2 t = sec2 t cos2 t _____ sin2 t + sin2 t ____ sin2 t = 1 ____ sin2 t alternate forms of the pythagorean identity cot2 t + 1 = csc2 t 1 + tan2 t = sec2 t cot2 t + 1 = csc2 t Example 7 Using Identities to Relate Trigonometric Functions If cos(t) = 12 _ 13 and t is in quadrant IV, as shown in Figure 8, find the values of the other five trigonometric functions. y y
Figure 8 12 13 t x Solution We can find the sine using the Pythagorean Identity, cos2 t + sin2 t = 1, and the remaining functions by relating them to sine and cosine. 12 __   13 2 + sin2 t = 1 sin2 t = 1 −  2 12 __  13 sin2 t = 1 − 144 ___ 169 sin2 t = 25 ___ 169 sin t = ± √ ____ 25 ___ 169 sin t = ± — 25 √ _ — 169 √ sin t = ± 5 __ 13 SECTION 5.3 the other trigonometric Functions 481 The sign of the sine depends on the y-values in the quadrant where the angle is located. Since the angle is in quadrant IV, where the y-values are negative, its sine is negative, − 5 _. 13 The remaining functions can be calculated using identities relating them to sine and cosine. tan t = sin t ____ cos t = = − 5 __ 12 sec t = 1 ____ cos t = = 13 __ 12 csc t = 1 ____ sin t = = − 13 __ 5 cot t = 1 ____ tan t = = − 12 __ 5 5 __ − 13 _ 12 _ 13 1 _ 12 _ 13 1 _ − 5 _ 13 1 _ − 5 _ 12 Try It #7 If sec(t) = − 17 _ 8 and 0 < t < π, find the values of the other five functions. As we discussed in the chapter opening, a function that repeats its values in regular intervals is known as a periodic function. The trigonometric functions are periodic. For the four trigonometric functions, sine, cosine, cosecant and secant, a revolution of one circle, or 2π, will result in the same outputs for these functions. And for tangent and cotangent, only a half a revolution will result in the same outputs. Other functions can also be periodic. For example, the lengths of months repeat every four years. If x represents the length time, measured in years, and f (x) represents the number of days in February, then f (x + 4) = f (x). This pattern repeats over and over through time. In other words, every four years, February is guaranteed to have the same number of days as it did 4 years earlier. The positive number 4 is the smallest positive number that
satisfies this condition and is called the period. A period is the shortest interval over which a function completes one full cycle—in this example, the period is 4 and represents the time it takes for us to be certain February has the same number of days. period of a function The period P of a repeating function f is the number representing the interval such that f (x + P) = f (x) for any value of x. The period of the cosine, sine, secant, and cosecant functions is 2π. The period of the tangent and cotangent functions is π. Example 8 Finding the Values of Trigonometric Functions Find the values of the six trigonometric functions of angle t based on Figure 9. y 1 t –1 x 1 –1 Figure 9 48 2 CHAPTER 5 trigonometric Functions Solution — 3 sin t = y = − √ _ 2 cos t = x = − 1 _ 2 tan t = sin t _ cos sec t = 1 _ cos t = csc t = 1 _ sin cot t = 1 _ tan Try It #8 Find the values of the six trigonometric functions of angle t based on Figure 10. y 1 t –1 x 1 (0, –1) Figure 10 Example 9 Finding the Value of Trigonometric Functions If sin(t and cos(t) =, find sec(t), csc(t), tan(t), cot(t). 2 Solution Try It #9 sec t = csc t = 1 _ sin t = tan t = sin t _ cos t = cot t = 1 _ tan cos If sin(t) = — 2 2 √ and cos(t) = √ _ _ 2 2 —, find sec(t), csc(t), tan(t), and cot(t). SECTION 5.3 the other trigonometric Functions 483 evaluating Trigonometric Functions with a Calculator We have learned how to evaluate the six trigonometric functions for the common first-quadrant angles and to use them as reference angles for angles in other quadrants. To evaluate trigonometric functions of other angles, we use a scientific or graphing calculator or computer software. If the calculator has a degree mode and a radian mode, confirm the correct mode is chosen before making a calculation. Evaluating a tangent function with a scientific calculator as opposed to a graphing calculator or computer algebra system is like evaluating a sine or
cosine: Enter the value and press the TAN key. For the reciprocal functions, there may not be any dedicated keys that say CSC, SEC, or COT. In that case, the function must be evaluated as the reciprocal of a sine, cosine, or tangent. If we need to work with degrees and our calculator or software does not have a degree mode, we can enter the degrees multiplied by the conversion factor to convert the degrees to radians. To find the secant of 30°, we could press π _ 180 (for a scientific calculator): 1 _ π _ 30 × 180 COS or (for a graphing calculator): 1 __ 30π cos   _ 180 How To… Given an angle measure in radians, use a scientific calculator to find the cosecant. 1. If the calculator has degree mode and radian mode, set it to radian mode. 2. Enter: 1 / 3. Enter the value of the angle inside parentheses. 4. Press the SIN key. 5. Press the = key. How To… Given an angle measure in radians, use a graphing utility/calculator to find the cosecant. 1. If the graphing utility has degree mode and radian mode, set it to radian mode. 2. Enter: 1 / 3. Press the SIN key. 4. Enter the value of the angle inside parentheses. 5. Press the ENTER key. Example 10 Evaluating the Secant Using Technology Evaluate the cosecant of 5π _. 7 Solution For a scientific calculator, enter information as follows: 1 / ( 5 × π / 7 ) SIN = 5π csc   ≈ 1.279 _ 7 Try It #10 Evaluate the cotangent of − π _. 8 Access these online resources for additional instruction and practice with other trigonometric functions. • Determining Trig Function Values (http://openstaxcollege.org/l/trigfuncval) • More examples of Determining Trig Functions (http://openstaxcollege.org/l/moretrigfun) • Pythagorean Identities (http://openstaxcollege.org/l/pythagiden) • Trig Functions on a Calculator (http://openstaxcollege.org/l/trigcalc) 48 4 CHAPTER 5 trigonometric
Functions 5.3 SeCTIOn exeRCISeS VeRBAl 1. On an interval of [0, 2π), can the sine and cosine values of a radian measure ever be equal? If so, where? 3. For any angle in quadrant II, if you knew the sine of the angle, how could you determine the cosine of the angle? 5. Tangent and cotangent have a period of π. What does this tell us about the output of these functions? 2. What would you estimate the cosine of π degrees to be? Explain your reasoning. 4. Describe the secant function. AlGeBRAIC For the following exercises, find the exact value of each expression. π __ 6. tan 6 π __ 10. tan 4 π __ 14. tan 3 π __ 7. sec 6 π __ 11. sec 4 π __ 15. sec 3 π __ 8. csc 6 π __ 12. csc 4 π __ 16. csc 3 For the following exercises, use reference angles to evaluate the expression. 18. tan 5π ___ 6 22. tan 7π ___ 4 26. tan 8π ___ 3 30. tan 225° 19. sec 7π ___ 6 23. sec 3π ___ 4 27. sec 4π ___ 3 31. sec 300° 20. csc 11π ____ 6 24. csc 5π ___ 4 28. csc 2π ___ 3 32. csc 150° π __ 9. cot 6 π __ 13. cot 4 π __ 17. cot 3 21. cot 13π ____ 6 25. cot 11π ____ 4 29. cot 5π ___ 3 33. cot 240° 34. tan 330° 35. sec 120° 36. csc 210° 37. cot 315° 3 _ 38. If sin t =, and t is in quadrant II, find cos t, sec t, 4 csc t, tan t, cot t. 39. If cos t = − 1 _, and t is in quadrant III, find sin t, sec t, 3 csc t, tan t, cot t. 40. If tan t = π 12 _ _, and 0 ≤ t <, find sin t, cos t, sec t, 5 2 csc t, and cot t. 41.
If sin t = and cot t. — 3 √ and cos t = 1 _ _, find sec t, csc t, tan t, 2 2 42. If sin 40° ≈ 0.643 and cos 40° ≈ 0.766, find sec 40°, csc 40°, tan 40°, and cot 40°. 44. If cos t = 1 _, what is the cos(−t)? 2 43. If sin t =, what is the sin(−t)? — 2 √ _ 2 45. If sec t = 3.1, what is the sec(−t)? 46. If csc t = 0.34, what is the csc(−t)? 47. If tan t = −1.4, what is the tan(−t)? 48. If cot t = 9.23, what is the cot(−t)? SECTION 5.3 section exercises 485 GRAPHICAl For the following exercises, use the angle in the unit circle to find the value of the each of the six trigonometric functions. 49. y 50. y 51. y t x t x t x TeCHnOlOGY For the following exercises, use a graphing calculator to evaluate. 52. csc 56. sec 5π ___ 9 3π ___ 4 60. cot 140° 53. cot 4π ___ 7 π __ 57. csc 4 61. sec 310° 54. sec π __ 10 58. tan 98° 55. tan 5π ___ 8 59. cot 33° exTenSIOnS For the following exercises, use identities to evaluate the expression. 62. If tan(t) ≈ 2.7, and sin(t) ≈ 0.94, find cos(t). 63. If tan(t) ≈ 1.3, and cos(t) ≈ 0.61, find sin(t). 64. If csc(t) ≈ 3.2, and cos(t) ≈ 0.95, find tan(t). 66. Determine whether the function f (x) = 2sin x cos x is even, odd, or neither. 68. Determine whether the function 65. If cot(t) ≈ 0.58, and cos(t) ≈ 0.5, find csc(t). 67. Determine whether the function f (x)
= 3sin2 x cos x + sec x is even, odd, or neither. 69. Determine whether the function f (x) = csc2 x + sec x f (x) = sin x − 2cos2 x is even, odd, or neither. is even, odd, or neither. For the following exercises, use identities to simplify the expression. sec t ____ 70. csc t tan t csc t 71. ReAl-WORlD APPlICATIOnS 1 _ 600 72. The amount of sunlight in a certain city can be d , modeled by the function h = 15 cos  where h represents the hours of sunlight, and d is the day of the year. Use the equation to find how many hours of sunlight there are on February 10, the 42nd day of the year. State the period of the function. 74. The equation P = 20sin(2πt) + 100 models the blood pressure, P, where t represents time in seconds. a. Find the blood pressure after 15 seconds. b. What are the maximum and minimum blood pressures? 76. The height of a piston, h, in inches, can be modeled by the equation y = 2cos x + 5, where x represents the crank angle. Find the height of the piston when the crank angle is 55°. 1 _ 500 73. The amount of sunlight in a certain city can be d , modeled by the function h = 16cos  where h represents the hours of sunlight, and d is the day of the year. Use the equation to find how many hours of sunlight there are on September 24, the 267th day of the year. State the period of the function. 75. The height of a piston, h, in inches, can be modeled by the equation y = 2cos x + 6, where x represents the crank angle. Find the height of the piston when the crank angle is 55°. 48 6 CHAPTER 5 trigonometric Functions leARnInG OBjeCTIVeS In this section, you will: • • • • • Use right triangles to evaluate trigonometric functions. , and 60°  π , 45°  π Find function values for 30 Use cofunctions of complementary angles. Use the definitions of trigonometric functions of any angle. Use right triangle trigonometry to solve applied problems.
5.4 RIGHT TRIAnGle TRIGOnOMeTRY We have previously defined the sine and cosine of an angle in terms of the coordinates of a point on the unit circle intersected by the terminal side of the angle: cos t = x sin t = y In this section, we will see another way to define trigonometric functions using properties of right triangles. Using Right Triangles to evaluate Trigonometric Functions In earlier sections, we used a unit circle to define the trigonometric functions. In this section, we will extend those definitions so that we can apply them to right triangles. The value of the sine or cosine function of t is its value at t radians. First, we need to create our right triangle. Figure 1 shows a point on a unit circle of radius 1. If we drop a vertical line segment from the point (x, y) to the x-axis, we have a right triangle whose vertical side has length y and whose horizontal side has length x. We can use this right triangle to redefine sine, cosine, and the other trigonometric functions as ratios of the sides of a right triangle. (x, y) y 1 t x Figure 1 cos t = x __ = x 1 sin t = y _ = y 1 We know Likewise, we know These ratios still apply to the sides of a right triangle when no unit circle is involved and when the triangle is not in standard position and is not being graphed using (x, y) coordinates. To be able to use these ratios freely, we will give the sides more general names: Instead of x, we will call the side between the given angle and the right angle the adjacent side to angle t. (Adjacent means “next to.”) Instead of y, we will call the side most distant from the given angle the opposite side from angle t. And instead of 1, we will call the side of a right triangle opposite the right angle the hypotenuse. These sides are labeled in Figure 2. SECTION 5.4 right triangle trigonometry 487 Hypotenuse t Adjacent Opposite Figure 2 The sides of a right triangle in relation to angle t. Understanding Right Triangle Relationships Given a right triangle with an acute angle of t, sin(t) = opposite __ hypotenuse cos(t) = adjacent __ hypotenuse tan(t) = opposite _ adjacent A common mnemonic for remembering these relationships is Soh
CahToa, formed from the first letters of “Sine is opposite over hypotenuse, Cosine is adjacent over hypotenuse, Tangent is opposite over adjacent.” How To… Given the side lengths of a right triangle and one of the acute angles, find the sine, cosine, and tangent of that angle. 1. Find the sine as the ratio of the opposite side to the hypotenuse. 2. Find the cosine as the ratio of the adjacent side to the hypotenuse. 3. Find the tangent as the ratio of the opposite side to the adjacent side. Example 1 Evaluating a Trigonometric Function of a Right Triangle Given the triangle shown in Figure 3, find the value of cos α. 8 α 17 15 Figure 3 Solution The side adjacent to the angle is 15, and the hypotenuse of the triangle is 17, so: cos(α) = adjacent _ hypotenuse = 15 __ 17 Try It #1 Given the triangle shown in Figure 4, find the value of sin t. 7 25 24 Figure 4 t 48 8 CHAPTER 5 trigonometric Functions Relating Angles and Their Functions When working with right triangles, the same rules apply regardless of the orientation of the triangle. In fact, we can evaluate the six trigonometric functions of either of the two acute angles in the triangle in Figure 5. The side opposite one acute angle is the side adjacent to the other acute angle, and vice versa. Adjacent to α Opposite β Adjacent to β Opposite α β α Hypotenuse Figure 5 The side adjacent to one angle is opposite the other We will be asked to find all six trigonometric functions for a given angle in a triangle. Our strategy is to find the sine, cosine, and tangent of the angles first. Then, we can find the other trigonometric functions easily because we know that the reciprocal of sine is cosecant, the reciprocal of cosine is secant, and the reciprocal of tangent is cotangent. How To… Given the side lengths of a right triangle, evaluate the six trigonometric functions of one of the acute angles. 1. If needed, draw the right triangle and label the angle provided. 2. Identify the angle, the adjacent side, the side opposite the angle, and the hypotenuse of the right triangle. 3. Find the required function: • sine as the ratio of the opposite
side to the hypotenuse • cosine as the ratio of the adjacent side to the hypotenuse • tangent as the ratio of the opposite side to the adjacent side • secant as the ratio of the hypotenuse to the adjacent side • cosecant as the ratio of the hypotenuse to the opposite side • cotangent as the ratio of the adjacent side to the opposite side Example 2 Evaluating Trigonometric Functions of Angles Not in Standard Position Using the triangle shown in Figure 6, evaluate sin α, cos α, tan α, sec α, csc α, and cot α. 3 α 4 5 Figure 6 Solution sin α = opposite α __ hypotenuse = 4 __ 5 cos α = adjacent to α = 3 __ __ 5 hypotenuse tan α = opposite α = 4 __ __ 3 adjacent to α sec α = hypotenuse = 5 __ __ 3 adjacent to α csc α = hypotenuse __ opposite α = 5 __ 4 cot α = adjacent to α = 3 __ __ 4 opposite α SECTION 5.4 right triangle trigonometry 489 Try It #2 Using the triangle shown in Figure 7, evaluate sin t, cos t, tan t, sec t, csc t, and cot t. 56 t 33 65 Figure 7 Finding Trigonometric Functions of Special Angles Using Side Lengths We have already discussed the trigonometric functions as they relate to the special angles on the unit circle. Now, we can use those relationships to evaluate triangles that contain those special angles. We do this because when we evaluate the special angles in trigonometric functions, they have relatively friendly values, values that contain either no or just one square root in the ratio. Therefore, these are the angles often used in math and science problems. We will use multiples of 30°, 60°, and 45°, however, remember that when dealing with right triangles, we are limited to angles between 0° and 90°. π π π _ _ _, Suppose we have a 30°, 60°, 90° triangle, which can also be described as a triangle. The sides have lengths in the, 2 3 6 π π π 3 s, 2s. The sides of a 45°, 45°, 90° triangle, which can also be described as a _ _ _,, triangle, have lengths in 4 4 2 — — 2 s. These relations are shown in Figure 8. relation s,
√ the relation s, s, √ 2s 6 s s 4 s 2s 4 s We can then use the ratios of the side lengths to evaluate trigonometric functions of special angles. Figure 8 Side lengths of special triangles How To… Given trigonometric functions of a special angle, evaluate using side lengths. 1. Use the side lengths shown in Figure 8 for the special angle you wish to evaluate. 2. Use the ratio of side lengths appropriate to the function you wish to evaluate. Example 3 Evaluating Trigonometric Functions of Special Angles Using Side Lengths π _ Find the exact value of the trigonometric functions of, using side lengths. 3 Solution π __  = sin  3 π __  = cos  3 π __  = tan  3 opp _ hyp adj _ hyp opp _ adj = — 3 s √ ____ 2s = — 3 √ ___ 2 = s __ 2s 1 __ = 2 = — 3 s √ ____ s = √ — 3 49 0 CHAPTER 5 trigonometric Functions π __  = sec  3 π __  = csc  3 π __  = cot  3 = hyp _ adj hyp _ opp = adj _ opp = 2s __ = 2 s 2s √ ____ 3 — 3 √ _ 3 = = Try It #3 π _, using side lengths. Find the exact value of the trigonometric functions of 4 Using Equal Cofunction of Complements If we look more closely at the relationship between the sine and cosine of the special angles relative to the unit circle, π π π _ _ _ we will notice a pattern. In a right triangle with angles of namely, we see that the sine of and cosine of., is also the cosine of, namely, while the sine of 3 2 6 6 — 3 s √ _ 2s, is also the π __ π __ = = cos sin 6 3 π __ π __ = = cos sin 3 6 s __ 2s 1 __ = 2 See Figure 9. 2s 6 s s Figure 9 The sine of π equals the cosine of π _ _ and vice versa 6 3 π _ This result should not be surprising because, as we see from Figure 9,
the side opposite the angle of is also the side 3 π 3 s and 2s. Similarly, cos  _  3 π π π  and cos , so sin  _ _ _  are exactly the same ratio of the same two sides, √ adjacent to 3 6 6 π and sin  _  are also the same ratio using the same two sides, s and 2s. 6 π _ The interrelationship between the sines and cosines of 6 triangle, since in every case, the ratio of the same two sides would constitute the sine of one angle and the cosine of π _ the other. Since the three angles of a triangle add to π, and the right angle is, the remaining two angles must also 2 π π _ _ add up to —in other words, any. That means that a right triangle can be formed with any two angles that add to 2 2 also holds for the two acute angles in any right π _ and 3 two complementary angles. So we may state a cofunction identity: If any two angles are complementary, the sine of one is the cosine of the other, and vice versa. This identity is illustrated in Figure 10. — β α sin α = cos β sin β = cos α Figure 10 Cofunction identity of sine and cosine of complementary angles SECTION 5.4 right triangle trigonometry 491 Using this identity, we can state without calculating, for instance, that the sine of equals the cosine of equals the cosine of. We can also state that if, for a certain angle t, cos t = π _ 12 π _ 12 5π _, 12, then 5 _ 13 and that the sine of π sin  _ − t  = 2 5 _ 13 as well. 5π _ 12 cofunction identities The cofunction identities in radians are listed in Table 1. π __ − t  cos t = sin  2 π __ − t  tan t = cot  2 π __ − t  sec t = csc  2 π __ − t  sin t = cos  2 π __ − t  cot t = tan  2 π __ − t  csc
t = sec  2 Table 1 How To… Given the sine and cosine of an angle, find the sine or cosine of its complement. 1. To find the sine of the complementary angle, find the cosine of the original angle. 2. To find the cosine of the complementary angle, find the sine of the original angle. Example 4 Using Cofunction Identities If sin t = 5 _ 12 π _ − t ., find cos  2 Solution According to the cofunction identities for sine and cosine, π __ − t . sin t = cos  2 π __ − t  = cos  2 5 __. 12 So Try It #4 π π If csc   = 2, find sec  _ _ . 3 6 Using Trigonometric Functions In previous examples, we evaluated the sine and cosine in triangles where we knew all three sides. But the real power of right-triangle trigonometry emerges when we look at triangles in which we know an angle but do not know all the sides. How To… Given a right triangle, the length of one side, and the measure of one acute angle, find the remaining sides. 1. For each side, select the trigonometric function that has the unknown side as either the numerator or the denominator. The known side will in turn be the denominator or the numerator. 2. Write an equation setting the function value of the known angle equal to the ratio of the corresponding sides. 3. Using the value of the trigonometric function and the known side length, solve for the missing side length. 49 2 CHAPTER 5 trigonometric Functions Example 5 Finding Missing Side Lengths Using Trigonometric Ratios Find the unknown sides of the triangle in Figure 11. 30° a c 7 Figure 11 Solution We know the angle and the opposite side, so we can use the tangent to find the adjacent side. We rearrange to solve for a. We can use the sine to find the hypotenuse. Again, we rearrange to solve for c. 7 __ tan(30°) = a a = 7 ______ tan(30°) ≈ 12.1 7 __ sin(30°) = c c = 7 ______ sin(30°) ≈ 14 Try It #5 π _ A right triangle
has one angle of and a hypotenuse of 20. Find the unknown sides and angle of the triangle. 3 Using Right Triangle Trigonometry to Solve Applied Problems Right-triangle trigonometry has many practical applications. For example, the ability to compute the lengths of sides of a triangle makes it possible to find the height of a tall object without climbing to the top or having to extend a tape measure along its height. We do so by measuring a distance from the base of the object to a point on the ground some distance away, where we can look up to the top of the tall object at an angle. The angle of elevation of an object above an observer relative to the observer is the angle between the horizontal and the line from the object to the observer's eye. The right triangle this position creates has sides that represent the unknown height, the measured distance from the base, and the angled line of sight from the ground to the top of the object. Knowing the measured distance to the base of the object and the angle of the line of sight, we can use trigonometric functions to calculate the unknown height. Similarly, we can form a triangle from the top of a tall object by looking downward. The angle of depression of an object below an observer relative to the observer is the angle between the horizontal and the line from the object to the observer's eye. See Figure 12. SECTION 5.4 right triangle trigonometry 493 Angle of depression Angle of elevation Figure 12 How To… Given a tall object, measure its height indirectly. 1. Make a sketch of the problem situation to keep track of known and unknown information. 2. Lay out a measured distance from the base of the object to a point where the top of the object is clearly visible. 3. At the other end of the measured distance, look up to the top of the object. Measure the angle the line of sight makes with the horizontal. 4. Write an equation relating the unknown height, the measured distance, and the tangent of the angle of the line of sight. 5. Solve the equation for the unknown height. Example 6 Measuring a Distance Indirectly To find the height of a tree, a person walks to a point 30 feet from the base of the tree. She measures an angle of 57° between a line of sight to the top of the tree and the ground, as shown in Figure 13. Find the height of the tree. 57° 30 feet Figure 13 Solution We know that the angle of elevation is 57°
and the adjacent side is 30 ft long. The opposite side is the unknown height. The trigonometric function relating the side opposite to an angle and the side adjacent to the angle is the tangent. So we will state our information in terms of the tangent of 57°, letting h be the unknown height. tan θ = opposite _ adjacent tan(57°) = h __ 30 Solve for h. h = 30tan(57°) Multiply. h ≈ 46.2 Use a calculator. The tree is approximately 46 feet tall. 49 4 CHAPTER 5 trigonometric Functions Try It #6 How long a ladder is needed to reach a windowsill 50 feet above the ground if the ladder rests against the building making an angle of with the ground? Round to the nearest foot. 5π _ 12 Access these online resources for additional instruction and practice with right triangle trigonometry. • Finding Trig Functions on Calculator (http://openstaxcollege.org/l/findtrigcal) • Finding Trig Functions Using a Right Triangle (http://openstaxcollege.org/l/trigrttri) • Relate Trig Functions to Sides of a Right Triangle (http://openstaxcollege.org/l/reltrigtri) • Determine Six Trig Functions from a Triangle (http://openstaxcollege.org/l/sixtrigfunc) • Determine length of Right Triangle Side (http://openstaxcollege.org/l/rttriside) SECTION 5.4 section exercises 495 5.4 SeCTIOn exeRCISeS VeRBAl 1. For the given right triangle, label the adjacent side, opposite side, and hypotenuse for the indicated angle. 2. When a right triangle with a hypotenuse of 1 is placed in the unit circle, which sides of the triangle correspond to the x- and y-coordinates? 3. The tangent of an angle compares which sides of the 4. What is the relationship between the two acute right triangle? 5. Explain the cofunction identity. AlGeBRAIC angles in a right triangle? For the following exercises, use cofunctions of complementary angles. 6. cos(34°) = sin(_____°) π __  = sin(______) 7. cos  3 8. csc(21°) = sec(_____°) π __ 
= cot(_____) 9. tan  4 For the following exercises, find the lengths of the missing sides if side a is opposite angle A, side b is opposite angle B, and side c is the hypotenuse. 10. cos B = 4 __, a = 10 5 11. sin B = 1 __, a = 20 2 13. tan A = 100, b = 100 12. tan A =, b = 6 5 __ 12 14. sin √ GRAPHICAl 15. a = 5, ∡ A = 60° 16. c = 12, ∡ A = 45° For the following exercises, use Figure 14 to evaluate each trigonometric function of angle A. A 4 10 Figure 14 17. sin A 19. tan A 21. sec A 18. cos A 20. csc A 22. cot A For the following exercises, use Figure 15 to evaluate each trigonometric function of angle A. 23. sin A 25. tan A 27. sec A 24. cos A 26. csc A 28. cot A 10 A 8 Figure 15 49 6 CHAPTER 5 trigonometric Functions For the following exercises, solve for the unknown sides of the given triangle. 29. B 7 30. 31. A 10 A c a 60° c b 30° 45° TeCHnOlOGY For the following exercises, use a calculator to find the length of each side to four decimal places. 32. A 35. B a 33. B 7 a 62° 10 c 12 b 10° c b 36. 35° 34. B a 10 b 65° A b c 16.5 81° 37. b = 15, ∡ B = 15° 38. c = 200, ∡ B = 5° 39. c = 50, ∡ B = 21° 40. a = 30, ∡ A = 27° 41. b = 3.5, ∡ A = 78° exTenSIOnS 42. Find x. 43. Find x. 82 63° 39° x 85 36° 50° x SECTION 5.4 section exercises 497 44. Find x. 45. Find x. 115 56° 35° x 119 70° 26° x 46. A radio tower is located 400 feet from a building. 47. A radio tower is located 325 feet from a building. From a window in the building, a person determines that the angle of elevation to the top of the tower is 36°,
and that the angle of depression to the bottom of the tower is 23°. How tall is the tower? From a window in the building, a person determines that the angle of elevation to the top of the tower is 43°, and that the angle of depression to the bottom of the tower is 31°. How tall is the tower? 48. A 200-foot tall monument is located in the distance. From a window in a building, a person determines that the angle of elevation to the top of the monument is 15°, and that the angle of depression to the bottom of the tower is 2°. How far is the person from the monument? 49. A 400-foot tall monument is located in the distance. From a window in a building, a person determines that the angle of elevation to the top of the monument is 18°, and that the angle of depression to the bottom of the monument is 3°. How far is the person from the monument? 50. There is an antenna on the top of a building. From a location 300 feet from the base of the building, the angle of elevation to the top of the building is measured to be 40°. From the same location, the angle of elevation to the top of the antenna is measured to be 43°. Find the height of the antenna. 51. There is lightning rod on the top of a building. From a location 500 feet from the base of the buil ding, the angle of elevation to the top of the building is measured to be 36°. From the same location, the angle of elevation to the top of the lightning rod is measured to be 38°. Find the height of the lightning rod. ReAl-WORlD APPlICATIOnS 52. A 33-ft ladder leans against a building so that the angle between the ground and the ladder is 80°. How high does the ladder reach up the side of the building? 53. A 23-ft ladder leans against a building so that the angle between the ground and the ladder is 80°. How high does the ladder reach up the side of the building? 54. The angle of elevation to the top of a building in 55. The angle of elevation to the top of a building in New York is found to be 9 degrees from the ground at a distance of 1 mile from the base of the building. Using this information, find the height of the building. Seattle is found to be 2 degrees from the ground at a distance of 2 miles from the
base of the building. Using this information, find the height of the building. 56. Assuming that a 370-foot tall giant redwood grows vertically, if I walk a certain distance from the tree and measure the angle of elevation to the top of the tree to be 60°, how far from the base of the tree am I? 49 8 CHAPTER 5 trigonometric Functions CHAPTeR 5 ReVIeW Key Terms adjacent side in a right triangle, the side between a given angle and the right angle angle the union of two rays having a common endpoint angle of depression the angle between the horizontal and the line from the object to the observer’s eye, assuming the object is positioned lower than the observer angle of elevation the angle between the horizontal and the line from the object to the observer’s eye, assuming the object is positioned higher than the observer angular speed the angle through which a rotating object travels in a unit of time arc length the length of the curve formed by an arc area of a sector area of a portion of a circle bordered by two radii and the intercepted arc; the fraction multiplied by the θ _ 2π area of the entire circle cosecant the reciprocal of the sine function: on the unit circle, csc t = 1 cosine function the x-value of the point on a unit circle corresponding to a given angle cotangent the reciprocal of the tangent function: on the unit circle, cot t = x coterminal angles description of positive and negative angles in standard position sharing the same terminal side degree a unit of measure describing the size of an angle as one-360th of a full revolution of a circle hypotenuse the side of a right triangle opposite the right angle identities statements that are true for all values of the input on which they are defined initial side the side of an angle from which rotation begins linear speed the distance along a straight path a rotating object travels in a unit of time; determined by the arc length measure of an angle the amount of rotation from the initial side to the terminal side negative angle description of an angle measured clockwise from the positive x-axis opposite side in a right triangle, the side most distant from a given angle period the smallest interval P of a repeating function f such that f (x + P) = f (x) positive angle description of an angle measured counterclockwise from the positive x-axis Pythagorean Identity a corollary of the Pythagorean Theorem stating that the
square of the cosine of a given angle plus the square of the sine of that angle equals 1 quadrantal angle an angle whose terminal side lies on an axis radian the measure of a central angle of a circle that intercepts an arc equal in length to the radius of that circle radian measure the ratio of the arc length formed by an angle divided by the radius of the circle ray one point on a line and all points extending in one direction from that point; one side of an angle reference angle the measure of the acute angle formed by the terminal side of the angle and the horizontal axis secant the reciprocal of the cosine function: on the unit circle, sec t = 1 sine function the y-value of the point on a unit circle corresponding to a given angle _ x, x ≠ 0 standard position the position of an angle having the vertex at the origin and the initial side along the positive x-axis y _ x, x ≠ 0 tangent the quotient of the sine and cosine: on the unit circle, tan t = terminal side the side of an angle at which rotation ends unit circle a circle with a center at (0, 0) and radius 1. vertex the common endpoint of two rays that form an angle CHAPTER 5 review 499 Key equations arc length s = rθ angular speed area of a sector 1 __ A = θr2 2 θ __ t s _ v = t linear speed related to angular speed v = rω linear speed ω = cosine sine cos t = x sin t = y Pythagorean Identity cos2 t + sin2 t = 1 tangent function secant function cosecant function cotangent function cofunction identities Key Concepts 5.1 Angles = sec t = csc t = cot t = tan t = sin t _ cos t 1 _ cos t 1 _ sin t cos t 1 _ _ tan t sin t π __ − t  cos t = sin  2 π __ − t  sin t = cos  2 π __ − t  tan t = cot  2 π __ − t  cot t = tan  2 π __ − t  sec t = csc  2 π __ − t  csc t = sec  2 • An angle is formed from the union of two rays, by
keeping the initial side fixed and rotating the terminal side. The amount of rotation determines the measure of the angle. • An angle is in standard position if its vertex is at the origin and its initial side lies along the positive x-axis. A positive angle is measured counterclockwise from the initial side and a negative angle is measured clockwise. • To draw an angle in standard position, draw the initial side along the positive x-axis and then place the terminal side according to the fraction of a full rotation the angle represents. See Example 1. • In addition to degrees, the measure of an angle can be described in radians. See Example 2. • To convert between degrees and radians, use the proportion θ _ 180 = θR _ π. See Example 3 and Example 4. • Two angles that have the same terminal side are called coterminal angles. • We can find coterminal angles by adding or subtracting 360° or 2π. See Example 5 and Example 6. • Coterminal angles can be found using radians just as they are for degrees. See Example 7. • The length of a circular arc is a fraction of the circumference of the entire circle. See Example 8. 500 CHAPTER 5 trigonometric Functions • The area of sector is a fraction of the area of the entire circle. See Example 9. • An object moving in a circular path has both linear and angular speed. • The angular speed of an object traveling in a circular path is the measure of the angle through which it turns in a unit of time. See Example 10. • The linear speed of an object traveling along a circular path is the distance it travels in a unit of time. See Example 11. 5.2 Unit Circle: Sine and Cosine Functions • Finding the function values for the sine and cosine begins with drawing a unit circle, which is centered at the origin and has a radius of 1 unit. • Using the unit circle, the sine of an angle t equals the y-value of the endpoint on the unit circle of an arc of length t whereas the cosine of an angle t equals the x-value of the endpoint. See Example 1. • The sine and cosine values are most directly determined when the corresponding point on the unit circle falls on an axis. See Example 2. • When the sine or cosine is known, we can use the Pythagorean Identity to find the other. The Pythagorean Identity is also
useful for determining the sines and cosines of special angles. See Example 3. • Calculators and graphing software are helpful for finding sines and cosines if the proper procedure for entering information is known. See Example 4. • The domain of the sine and cosine functions is all real numbers. • The range of both the sine and cosine functions is [−1, 1]. • The sine and cosine of an angle have the same absolute value as the sine and cosine of its reference angle. • The signs of the sine and cosine are determined from the x- and y-values in the quadrant of the original angle. • An angle’s reference angle is the size angle, t, formed by the terminal side of the angle t and the horizontal axis. See Example 5. • Reference angles can be used to find the sine and cosine of the original angle. See Example 6. • Reference angles can also be used to find the coordinates of a point on a circle. See Example 7. 5.3 The Other Trigonometric Functions • The tangent of an angle is the ratio of the y-value to the x-value of the corresponding point on the unit circle. • The secant, cotangent, and cosecant are all reciprocals of other functions. The secant is the reciprocal of the cosine function, the cotangent is the reciprocal of the tangent function, and the cosecant is the reciprocal of the sine function. • The six trigonometric functions can be found from a point on the unit circle. See Example 1. • Trigonometric functions can also be found from an angle. See Example 2. • Trigonometric functions of angles outside the first quadrant can be determined using reference angles. See Example 3. • A function is said to be even if f (−x) = f (x) and odd if f (−x) = −f (x). • Cosine and secant are even; sine, tangent, cosecant, and cotangent are odd. • Even and odd properties can be used to evaluate trigonometric functions. See Example 4. • The Pythagorean Identity makes it possible to find a cosine from a sine or a sine from a cosine. • Identities can be used to evaluate trigonometric functions. See Example 5 and Example 6. • Fundamental identities such as the
Pythagorean Identity can be manipulated algebraically to produce new identities. See Example 7. • The trigonometric functions repeat at regular intervals. • The period P of a repeating function f is the smallest interval such that f (x + P) = f (x) for any value of x. • The values of trigonometric functions of special angles can be found by mathematical analysis. • To evaluate trigonometric functions of other angles, we can use a calculator or computer software. See Example 8. CHAPTER 5 review 501 5.4 Right Triangle Trigonometry • We can define trigonometric functions as ratios of the side lengths of a right triangle. See Example 1. • The same side lengths can be used to evaluate the trigonometric functions of either acute angle in a right triangle. See Example 2. • We can evaluate the trigonometric functions of special angles, knowing the side lengths of the triangles in which they occur. See Example 3. • Any two complementary angles could be the two acute angles of a right triangle. • If two angles are complementary, the cofunction identities state that the sine of one equals the cosine of the other and vice versa. See Example 4. • We can use trigonometric functions of an angle to find unknown side lengths. • Select the trigonometric function representing the ratio of the unknown side to the known side. See Example 5. • Right-triangle trigonometry permits the measurement of inaccessible heights and distances. • The unknown height or distance can be found by creating a right triangle in which the unknown height or distance is one of the sides, and another side and angle are known. See Example 6. 502 CHAPTER 5 trigonometric Functions CHAPTeR 5 ReVIeW exeRCISeS AnGleS For the following exercises, convert the angle measures to degrees. 1. π _ 4 2. − 5π ___ 3 For the following exercises, convert the angle measures to radians. 3. −210° 5. Find the length of an arc in a circle of radius 7 meters subtended by the central angle of 85°. 4. 180° 6. Find the area of the sector of a circle with diameter 3π _ radians. 5 32 feet and an angle of For the following exercises, find the angle between 0° and 360° that is coterminal with the given angle. 7. 420° 8. −80° For the following exercises, find the angle between 0 and
2π in radians that is coterminal with the given angle. 9. − 20π ____ 11 10. 14π ____ 5 For the following exercises, draw the angle provided in standard position on the Cartesian plane. 11. −210° 5π ___ 4 13. 12. 75° 14. − π _ 3 15. Find the linear speed of a point on the equator of the earth if the earth has a radius of 3,960 miles and the earth rotates on its axis every 24 hours. Express answer in miles per hour. 16. A car wheel with a diameter of 18 inches spins at the rate of 10 revolutions per second. What is the car’s speed in miles per hour? UnIT CIRCle: SIne AnD COSIne FUnCTIOnS π _ 17. Find the exact value of sin. 3 π _ 18. Find the exact value of cos. 4 19. Find the exact value of cos π. 20. State the reference angle for 300°. 21. State the reference angle for 3π _. 4 22. Compute cosine of 330°. 23. Compute sine of 5π _. 4 24. State the domain of the sine and 25. State the range of the sine and cosine functions. cosine functions. THe OTHeR TRIGOnOMeTRIC FUnCTIOnS For the following exercises, find the exact value of the given expression. π _ 26. cos 6 π _ 27. tan 4 π _ 28. csc 3 π _ 29. sec 4 For the following exercises, use reference angles to evaluate the given expression. 30. sec 11π _ 3 31. sec 315° 32. If sec(t) = −2.5, what is the sec(−t)? 33. If tan(t) = −0.6, what is the tan(−t)? CHAPTER 5 review 503 1 _ 34. If tan(t) =, find tan(t − π). 3 35. If cos(t) =, find sin(t + 2π). — 2 √ _ 2 36. Which trigonometric functions are even? 37. Which trigonometric functions are odd? RIGHT TRIAnGle TRIGOnOMeTRY For the following exercises, use side lengths to evaluate. π _ 38. cos 4 π _ 39.
cot 3 π _  = sin(_____°) 41. cos  2 42. csc(18°) = sec(_____°) π _ 40. tan 6 For the following exercises, use the given information to find the lengths of the other two sides of the right triangle. 3 __, a = 6 43. cos B = 5 5 __, b = 6 44. tan A = 9 For the following exercises, use Figure 1 to evaluate each trigonometric function. A 6 B 11 Figure 1 45. sin A 46. tan B For the following exercises, solve for the unknown sides of the given triangle. 47. b 4√2 45º a 48. b 5 a 30º 49. A 15-ft ladder leans against a building so that the angle between the ground and the ladder is 70°. How high does the ladder reach up the side of the building? 50. The angle of elevation to the top of a building in Baltimore is found to be 4 degrees from the ground at a distance of 1 mile from the base of the building. Using this information, find the height of the building. 504 CHAPTER 5 trigonometric Functions CHAPTeR 5 PRACTICe TeST 1. Convert radians to degrees. 5π _ 6 2. Convert −620° to radians. 3. Find the length of a circular arc with a radius 12 centimeters subtended by the central angle of 30°. 5. Find the angle between 0° and 360° that is coterminal with 375°. 7. Draw the angle 315° in standard position on the Cartesian plane. 9. A carnival has a Ferris wheel with a diameter of 80 feet. The time for the Ferris wheel to make one revolution is 75 seconds. What is the linear speed in feet per second of a point on the Ferris wheel? What is the angular speed in radians per second? 4. Find the area of the sector with radius of 8 feet and an angle of radians. 5π _ 4 6. Find the angle between 0 and 2π in radians that is coterminal with − 4π _. 7 8. Draw the angle − π _ in standard position on the 6 Cartesian plane. π _ 10. Find the exact value of sin. 6 11. Compute sine of 240°. 12. State the domain of the sine and cosine functions. 13. State the range
of the sine and cosine functions. π _ 15. Find the exact value of tan. 3 π _ 14. Find the exact value of cot. 4 16. Use reference angles to evaluate csc 7π _. 4 17. Use reference angles to evaluate tan 210°. 18. If csc t = 0.68, what is the csc(−t)? 19. If cos t =, find cos(t − 2π). — 3 √ _ 2 20. Which trigonometric functions are even? π _  = sin(_____) 21. Find the missing angle: cos  6 22. Find the missing sides of the triangle ABC : 3 _, c = 12 sin B = 4 23. Find the missing sides of the triangle. 24. The angle of elevation to the top of a building in Chicago is found to be 9 degrees from the ground at a distance of 2,000 feet from the base of the building. Using this information, find the height of the building. A 9 60º C B Periodic Functions 6 Figure 1 (credit: "Maxxer_", Flickr) CHAPTeR OUTlIne 6.1 Graphs of the Sine and Cosine Functions 6.2 Graphs of the Other Trigonometric Functions 6.3 Inverse Trigonometric Functions Introduction Each day, the sun rises in an easterly direction, approaches some maximum height relative to the celestial equator, and sets in a westerly direction. The celestial equator is an imaginary line that divides the visible universe into two halves in much the same way Earth’s equator is an imaginary line that divides the planet into two halves. The exact path the sun appears to follow depends on the exact location on Earth, but each location observes a predictable pattern over time. The pattern of the sun’s motion throughout the course of a year is a periodic function. Creating a visual representation of a periodic function in the form of a graph can help us analyze the properties of the function. In this chapter, we will investigate graphs of sine, cosine, and other trigonometric functions. 505 506 CHAPTER 6 periodic Functions leARnInG OBjeCTIVeS In this section, you will: • Graph variations of y = sin(x ) and y = cos(x ). • Use phase shifts of sine and cosine curves. 6.1 GRAPHS OF THe SIne
AnD COSIne FUnCTIOnS Figure 1 light can be separated into colors because of its wavelike properties. (credit: "wonderferret"/ Flickr) White light, such as the light from the sun, is not actually white at all. Instead, it is a composition of all the colors of the rainbow in the form of waves. The individual colors can be seen only when white light passes through an optical prism that separates the waves according to their wavelengths to form a rainbow. Light waves can be represented graphically by the sine function. In the chapter on Trigonometric Functions, we examined trigonometric functions such as the sine function. In this section, we will interpret and create graphs of sine and cosine functions. Graphing Sine and Cosine Functions Recall that the sine and cosine functions relate real number values to the x- and y-coordinates of a point on the unit circle. So what do they look like on a graph on a coordinate plane? Let’s start with the sine function. We can create a table of values and use them to sketch a graph. Table 1 lists some of the values for the sine function on a unit circle. x sin(x Table 1 π _ 2 1 2π _ 3 3π 5π _ 6 1 _ 2 π 0 Plotting the points from the table and continuing along the x-axis gives the shape of the sine function. See Figure 2. y 2 1 –1 –2 y = sin (x) π 4 π 2 3π 4 π 5π 4 3π 2 7π 4 x 2π Figure 2 The sine function SECTION 6.1 graphs oF the sine and cosine Functions 507 Notice how the sine values are positive between 0 and π, which correspond to the values of the sine function in quadrants I and II on the unit circle, and the sine values are negative between π and 2π, which correspond to the values of the sine function in quadrants III and IV on the unit circle. See Figure 3. y = sin (x1 Now let’s take a similar look at the cosine function. Again, we can create a table of values and use them to sketch a graph. Table 2 lists some of the values for the cosine function on a unit circle. Figure 3 Plotting values of the sine function x cos(x Table
2 2π _ 3 3π _ 4 5π 1 — 3 √ _ 2 As with the sine function, we can plots points to create a graph of the cosine function as in Figure 4. y 1 0 –1 y = cos (x) 2π 3 3π 4 5π 5π 4 3π 2 7π 4 x 2π Figure 4 The cosine function Because we can evaluate the sine and cosine of any real number, both of these functions are defined for all real numbers. By thinking of the sine and cosine values as coordinates of points on a unit circle, it becomes clear that the range of both functions must be the interval [−1, 1]. In both graphs, the shape of the graph repeats after 2π, which means the functions are periodic with a period of 2π. A periodic function is a function for which a specific horizontal shift, P, results in a function equal to the original function: f (x + P) = f (x) for all values of x in the domain of f. When this occurs, we call the smallest such horizontal shift with P > 0 the period of the function. Figure 5 shows several periods of the sine and cosine functions. y = sin (x) y 1 1 period y = cos (x) 1 period y 1 –3π –2π –π π 2π x 3π –3π –2π –π π 2π 3π x –1 –1 Figure 5 508 CHAPTER 6 periodic Functions Looking again at the sine and cosine functions on a domain centered at the y-axis helps reveal symmetries. As we can see in Figure 6, the sine function is symmetric about the origin. Recall from The Other Trigonometric Functions that we determined from the unit circle that the sine function is an odd function because sin(−x) = −sin x. Now we can clearly see this property from the graph. y 2 1 –1 –2 y = sin (x) π x 2π –2π –π Figure 7 shows that the cosine function is symmetric about the y-axis. Again, we determined that the cosine function is an even function. Now we can see from the graph that cos(−x) = cos x. Figure 6 Odd symmetry of the sine function y 2 1 –1 –2 –2π –π π x 2π y = cos (x) Figure 7
even symmetry of the cosine function characteristics of sine and cosine functions The sine and cosine functions have several distinct characteristics: • They are periodic functions with a period of 2π. • The domain of each function is (−∞, ∞) and the range is [−1, 1]. • The graph of y = sin x is symmetric about the origin, because it is an odd function. • The graph of y = cos x is symmetric about the y- axis, because it is an even function. Investigating Sinusoidal Functions As we can see, sine and cosine functions have a regular period and range. If we watch ocean waves or ripples on a pond, we will see that they resemble the sine or cosine functions. However, they are not necessarily identical. Some are taller or longer than others. A function that has the same general shape as a sine or cosine function is known as a sinusoidal function. The general forms of sinusoidal functions are y = Asin(Bx − C) + D and y = Acos(Bx − C) + D SECTION 6.1 graphs oF the sine and cosine Functions 509 Determining the Period of Sinusoidal Functions Looking at the forms of sinusoidal functions, we can see that they are transformations of the sine and cosine functions. We can use what we know about transformations to determine the period. 2π _. If ∣ B ∣ > 1, then the period is less than 2π and the function In the general formula, B is related to the period by P = ∣ B ∣ undergoes a horizontal compression, whereas if ∣ B ∣ < 1, then the period is greater than 2π and the function undergoes a horizontal stretch. For example, f (x) = sin(x), B = 1, so the period is 2π, which we knew. If f (x) = sin(2x), then B = 2, x 1 _ _ , then B = so the period is π and the graph is compressed. If f (x) = sin , so the period is 4π and the graph is 2 2 stretched. Notice in Figure 8 how the period is indirectly related to ∣ B ∣. y f (x) = sin (x) 1 –1 f (x) = sin( )x 2 x 2π π
2 π 3π 2 f (x) = sin (2x) Figure 8 period of sinusoidal functions If we let C = 0 and D = 0 in the general form equations of the sine and cosine functions, we obtain the forms The period is 2π _. ∣ B ∣ y = Asin(Bx) y = Acos(Bx) Identifying the Period of a Sine or Cosine Function Example 1 π _ Determine the period of the function f (x) = sin  x . 6 Solution Let’s begin by comparing the equation to the general form y = Asin(Bx). π _ In the given equation, B =, so the period will be 6 P = = 2π _ ∣ B ∣ 2π _ π _ 6 6 _ = 2π ⋅ π = 12 Try It #1 x __ Determine the period of the function g(x) =  cos . 3 Determining Amplitude Returning to the general formula for a sinusoidal function, we have analyzed how the variable B relates to the period. Now let’s turn to the variable A so we can analyze how it is related to the amplitude, or greatest distance from rest. A represents the vertical stretch factor, and its absolute value ∣ A ∣ is the amplitude. The local maxima will be a distance ∣ A ∣ above the vertical midline of the graph, which is the line x = D ; because D = 0 in this case, the midline is the x-axis. The local minima will be the same distance below the midline. If ∣ A ∣ > 1, the function is stretched. For example, the amplitude of f (x) = 4sin x is twice the amplitude of f (x) = 2sin x. If ∣ A ∣ < 1, the function is compressed. Figure 9 compares several sine functions with different amplitudes. 510 CHAPTER 6 periodic Functions – 9π 2 – 5π 2 –11π 2 – 7π 2 – 3π 2 f (x) = 4 sin(x) f (x) = 3 sin(x) f (x) = 2 sin(x) f (x) = 1 sin(x) 11π 2 3π 2 7π 2 5π 2 9π 2 y 4 3 2 1 –1 –2
–3 –4 Figure 9 amplitude of sinusoidal functions If we let C = 0 and D = 0 in the general form equations of the sine and cosine functions, we obtain the forms y = Asin(Bx) and y = Acos(Bx) The amplitude is A, and the vertical height from the midline is ∣ A ∣. In addition, notice in the example that 1 ∣ maximum − minimum ∣ __ ∣ A ∣ = amplitude = 2 Example 2 Identifying the Amplitude of a Sine or Cosine Function What is the amplitude of the sinusoidal function f (x) = −4sin(x)? Is the function stretched or compressed vertically? Solution Let’s begin by comparing the function to the simplified form y = Asin(Bx). In the given function, A = −4, so the amplitude is ∣ A ∣ = ∣ −4 ∣ = 4. The function is stretched. Analysis The negative value of A results in a reflection across the x-axis of the sine function, as shown in Figure 10. y – 3π 2 – π 2 f (x) = −4 sin x π 2 3π 2 x 4 3 2 –1 –2 –3 –4 Figure 10 Try It #2 1 __ sin(x)? Is the function stretched or compressed vertically? What is the amplitude of the sinusoidal function f (x) = 2 Analyzing Graphs of Variations of y = sin x and y = cos x Now that we understand how A and B relate to the general form equation for the sine and cosine functions, we will explore the variables C and D. Recall the general form: y = Asin(Bx − C) + D and y = Acos(Bx − C) + D C C __ __   + D   + D and y = Acos  B  x − y = Asin  B  x − B B or SECTION 6.1 graphs oF the sine and cosine Functions 511 C __ B The value for a sinusoidal function is called the phase shift, or the horizontal displacement of the basic sine or cosine function. If C > 0, the graph shifts to the right. If C < 0, the graph shifts to the left. The greater the value of ∣ C �
�, the more the graph is shifted. Figure 11 shows that the graph of f (x) = sin(x − π) shifts to the right by π units, which π π __ __ is more than we see in the graph of f (x) = sin  x − , which shifts to the right by units. 4 4 y 1 f (x) = sin(x) f (x) = sin x − π 4 f (x) = sin(x − π) π 2 π 3π 2 2π 5π 2 x 3π While C relates to the horizontal shift, D indicates the vertical shift from the midline in the general formula for a sinusoidal function. See Figure 12. The function y = cos(x) + D has its midline at y = D. Figure 11 y y = A sin(x) + D Midline π 2π 3π y = D x Figure 12 Any value of D other than zero shifts the graph up or down. Figure 13 compares f (x) = sin x with f (x) = sin x + 2, which is shifted 2 units up on a graph. y 3 2 1 –1 f (x) = sin(x) + 2 f (x) = sin(x) π 2 π 3π 2 2π 5π 2 x 3π Figure 13 variations of sine and cosine functions C __ Given an equation in the form f (x) = Asin(Bx − C) + D or f (x) = Acos(Bx − C) + D, B D is the vertical shift. is the phase shift and Example 3 Identifying the Phase Shift of a Function π __  − 2. Determine the direction and magnitude of the phase shift for f (x) = sin  x + 6 Solution Let’s begin by comparing the equation to the general form y = Asin(Bx − C) + D. 512 CHAPTER 6 periodic Functions π __ In the given equation, notice that B = 1 and C = −. So the phase shift is 6 C __ B = − π __ 6 _ 1 π __ = − 6 π __ or units to the left. 6 Analysis We must pay attention to the sign in the equation for the general form of a sinusoidal function. The equation π __ π __
  − 2.  − 2 can be rewritten as f (x) = sin  x −  − shows a minus sign before C. Therefore f (x) = sin  x + 6 6 If the value of C is negative, the shift is to the left. Try It #3 π Determine the direction and magnitude of the phase shift for f (x) = 3cos  x − _ . 2 Example 4 Identifying the Vertical Shift of a Function Determine the direction and magnitude of the vertical shift for f (x) = cos(x) − 3. Solution Let’s begin by comparing the equation to the general form y = Acos(Bx − C) + D. In the given equation, D = −3 so the shift is 3 units downward. Try It #4 Determine the direction and magnitude of the vertical shift for f (x) = 3sin(x) + 2. How To… Given a sinusoidal function in the form f (x) = Asin(Bx − C) + D, identify the midline, amplitude, period, and phase shift. 1. Determine the amplitude as ∣ A ∣. 2π _ 2. Determine the period as P =. ∣ B ∣ C __. B 3. Determine the phase shift as 4. Determine the midline as y = D. Example 5 Identifying the Variations of a Sinusoidal Function from an Equation Determine the midline, amplitude, period, and phase shift of the function y = 3sin(2x) + 1. Solution Let’s begin by comparing the equation to the general form y = Asin(Bx − C) + D. A = 3, so the amplitude is ∣ A ∣ = 3. Next, B = 2, so the period is P = 2π __ 2 C __ There is no added constant inside the parentheses, so C = 0 and the phase shift is B 2π _ ∣ B ∣ = π. = 0 __ = 0. = 2 Finally, D = 1, so the midline is y = 1. Analysis Figure 14. Inspecting the graph, we can determine that the period is π, the midline is y = 1, and the amplitude is 3. See SECTION 6.1 graphs oF the sine and cosine Functions
513 y 4 3 2 –1 –2 Amplitude: \|A\| = 3 y = 3 sin (2x) + 1 Midline: y = 1 π 2 π 3π 2 x 2π Period = π Figure 14 Try It #5 – π x 1 __ __ __ Determine the midline, amplitude, period, and phase shift of the function y = . cos  3 3 2 Example 6 Identifying the Equation for a Sinusoidal Function from a Graph Determine the formula for the cosine function in Figure 15. y 1 0.5 –2π –π π 2π 3π 4π x Figure 15 Solution To determine the equation, we need to identify each value in the general form of a sinusoidal function. y = Asin(Bx − C) + D y = Acos(Bx − C) + D The graph could represent either a sine or a cosine function that is shifted and/or reflected. When x = 0, the graph has an extreme point, (0, 0). Since the cosine function has an extreme point for x = 0, let us write our equation in terms of a cosine function. Let’s start with the midline. We can see that the graph rises and falls an equal distance above and below y = 0.5. This value, which is the midline, is D in the equation, so D = 0.5. The greatest distance above and below the midline is the amplitude. The maxima are 0.5 units above the midline and the minima are 0.5 units below the midline. So ∣ A ∣ = 0.5. Another way we could have determined the amplitude is 1 __ by recognizing that the difference between the height of local maxima and minima is 1, so ∣ A ∣ = = 0.5. Also, the 2 graph is reflected about the x-axis so that A = −0.5. The graph is not horizontally stretched or compressed, so B = 1; and the graph is not shifted horizontally, so C = 0. Putting this all together, g(x) = −0.5cos(x) + 0.5 Try It #6 Determine the formula for the sine function in Figure 16. y 3 2 1 –2π –π π 2π x Figure 16 514 CHAPTER 6 periodic Functions Example 7 Ident
ifying the Equation for a Sinusoidal Function from a Graph Determine the equation for the sinusoidal function in Figure 17. y 1 –5 –3 –1 1 3 5 7 x –1 –2 –3 –4 –5 Figure 17 Solution With the highest value at 1 and the lowest value at −5, the midline will be halfway between at −2. So D = −2. The distance from the midline to the highest or lowest value gives an amplitude of ∣ A ∣ = 3. The period of the graph is 6, which can be measured from the peak at x = 1 to the next peak at x = 7, or from the distance between the lowest points. Therefore, P = = 6. Using the positive value for B, we find that 2π _ ∣ B ∣ 2π __ = P B = π 2π __ __ = 6 3 π π __ __ x − C  − 2. For the shape and shift, we have more x − C  − 2 or y = 3cos  So far, our equation is either y = 3sin  3 3 than one option. We could write this as any one of the following: • a cosine shifted to the right • a negative cosine shifted to the left • a sine shifted to the left • a negative sine shifted to the right While any of these would be correct, the cosine shifts are easier to work with than the sine shifts in this case because they involve integer values. So our function becomes π π π __ __ __  − 2 or y = −3cos  x + x − y = 3cos  3 3 3 2π __  −2 3 Again, these functions are equivalent, so both yield the same graph. Try It #7 Write a formula for the function graphed in Figure 18. y 8 6 4 2 –9 –7 –5 –3 –1 1 3 5 7 9 11 x Figure 18 SECTION 6.1 graphs oF the sine and cosine Functions 515 Graphing Variations of y = sin x and y = cos x Throughout this section, we have learned about types of variations of sine and cosine functions and used that information to write equations from graphs. Now we can use the same information to create graphs from equations. Instead of focusing on the general form equations we will let C =
0 and D = 0 and work with a simplified form of the equations in the following examples. y = Asin(Bx − C) + D and y = Acos(Bx − C) + D, How To… Given the function y = Asin(Bx), sketch its graph. 1. Identify the amplitude, ∣ A ∣. 2π ___ 2. Identify the period, P =. ∣ B ∣ 3. Start at the origin, with the function increasing to the right if A is positive or decreasing if A is negative. π ____ 2 ∣ B ∣ 4. At x = there is a local maximum for A > 0 or a minimum for A < 0, with y = A. 5. The curve returns to the x-axis at x = π ___. ∣ B ∣ 6. There is a local minimum for A > 0 (maximum for A < 0 ) at x = 7. The curve returns again to the x-axis at x = π ____. 2 ∣ B ∣ 3π ____ 2 ∣ B ∣ with y = −A. Example 8 Graphing a Function and Identifying the Amplitude and Period Sketch a graph of f (x) = −2sin  Solution Let’s begin by comparing the equation to the form y = Asin(Bx). πx __ . 2 Step 1. We can see from the equation that A = −2, so the amplitude is 2. π __ Step 2. The equation shows that B =, so the period is 2 ∣ A ∣ = 2 P = 2π _ π __ 2 2 __ = 2π ⋅ π = 4 Step 3. Because A is negative, the graph descends as we move to the right of the origin. Step 4–7. The x-intercepts are at the beginning of one period, x = 0, the horizontal midpoints are at x = 2 and at the end of one period at x = 4. The quarter points include the minimum at x = 1 and the maximum at x = 3. A local minimum will occur 2 units below the midline, at x = 1, and a local maximum will occur at 2 units above the midline, at x = 3. Figure 19 shows the graph of the function. y πx y = f (x) = −2sin 2
2 1 –2 –1 1 2 3 4 5 6 x –1 –2 Figure 19 516 CHAPTER 6 periodic Functions Try It #8 Sketch a graph of g(x) = −0.8cos(2x). Determine the midline, amplitude, period, and phase shift. How To… Given a sinusoidal function with a phase shift and a vertical shift, sketch its graph. 1. Express the function in the general form y = Asin(Bx − C) + D or y = Acos(Bx − C) + D. 2. Identify the amplitude, ∣ A ∣. 2π ___. ∣ B ∣ C __. B 3. Identify the period, P = 4. Identify the phase shift, C __ 5. Draw the graph of f (x) = Asin(Bx) shifted to the right or left by B and up or down by D. Example 9 Graphing a Transformed Sinusoid π π __ __ x − Sketch a graph of f (x) = 3sin  . 4 4 Solution π π __ __ x − Step 1. The function is already written in general form: f (x) = 3sin  . This graph will have the shape of a sine 4 4 function, starting at the midline and increasing to the right. Step 2. The amplitude is 3. π π  = Step 3. Since ∣ B ∣ =  __ __, we determine the period as follows. 4 4 P = 2π ___ ∣ B ∣ = 4 _ π = 8 = 2π ⋅ 2π _ π __ 4 The period is 8. π __ Step 4. Since C =, the phase shift is 4 The phase shift is 1 unit. Step 5. Figure 20 shows the graph of the function. = = 1. C __ x) = 3 sin 1 –1 –2 –3 –7 –5 –3 Figure 20 A horizontally compressed, vertically stretched, and horizontally shifted sinusoid SECTION 6.1 graphs oF the sine and cosine Functions 517 Try It #9 π π __ __ Draw a graph of g(x) = −2cos  . Determine the midline, amplitude, period, and phase shift. x + 6 3 Example 10 Identifying
the Properties of a Sinusoidal Function π __ x + π  + 3, determine the amplitude, period, phase shift, and horizontal shift. Then graph the Given y = −2cos  2 function. Solution Begin by comparing the equation to the general form and use the steps outlined in Example 9. y = Acos(Bx − C) + D Step 1. The function is already written in general form. Step 2. Since A = −2, the amplitude is ∣ A ∣ = 2. 2π π _ __ Step 3. ∣ B ∣ =, so the period is P = π 2 __ 2 2π ___ ∣ B ∣ = 2 __ = 4. The period is 4. = 2π ⋅ π C __ Step 4. C = −π, so we calculate the phase shift as B = 2 __ = −2. The phase shift is −2. = −π ⋅ π −π _ π __ 2 Step 5. D = 3, so the midline is y = 3, and the vertical shift is up 3. Since A is negative, the graph of the cosine function has been reflected about the x-axis. Figure 21 shows one cycle of the graph of the function. y = −2 cos 2 π x + π + 3 –6 –4 –2 y Amplitude = 2 Midline: y = 3 Period = 4 2 x 5 4 3 2 1 –1 Using Transformations of Sine and Cosine Functions Figure 21 We can use the transformations of sine and cosine functions in numerous applications. As mentioned at the beginning of the chapter, circular motion can be modeled using either the sine or cosine function. Example 11 Finding the Vertical Component of Circular Motion A point rotates around a circle of radius 3 centered at the origin. Sketch a graph of the y-coordinate of the point as a function of the angle of rotation. Solution Recall that, for a point on a circle of radius r, the y-coordinate of the point is y = r sin(x), so in this case, we get the equation y(x) = 3 sin(x). The constant 3 causes a vertical stretch of the y-values of the function by a factor of 3, which we can see in the graph in Figure 22. y(x) = 3 sin x π 2 3π 2 5π 2 7π
2 x y 3 2 1 –1 –2 –3 Figure 22 518 CHAPTER 6 periodic Functions Analysis Notice that the period of the function is still 2π ; as we travel around the circle, we return to the point (3, 0) for x = 2π, 4π, 6π,... Because the outputs of the graph will now oscillate between –3 and 3, the amplitude of the sine wave is 3. Try It #10 What is the amplitude of the function f (x) = 7cos(x)? Sketch a graph of this function. Example 12 Finding the Vertical Component of Circular Motion A circle with radius 3 ft is mounted with its center 4 ft off the ground. The point closest to the ground is labeled P, as shown in Figure 23. Sketch a graph of the height above the ground of the point P as the circle is rotated; then find a function that gives the height in terms of the angle of rotation. 3 ft P Figure 23 4 ft Solution Sketching the height, we note that it will start 1 ft above the ground, then increase up to 7 ft above the ground, and continue to oscillate 3 ft above and below the center value of 4 ft, as shown in Figure 243 cos x + 4 Midline: y = 4 π 2π 3π 4π x Figure 24 Although we could use a transformation of either the sine or cosine function, we start by looking for characteristics that would make one function easier to use than the other. Let’s use a cosine function because it starts at the highest or lowest value, while a sine function starts at the middle value. A standard cosine starts at the highest value, and this graph starts at the lowest value, so we need to incorporate a vertical reflection. Second, we see that the graph oscillates 3 above and below the center, while a basic cosine has an amplitude of 1, so this graph has been vertically stretched by 3, as in the last example. Finally, to move the center of the circle up to a height of 4, the graph has been vertically shifted up by 4. Putting these transformations together, we find that y = −3cos(x) + 4 SECTION 6.1 graphs oF the sine and cosine Functions 519 Try It #11 A weight is attached to a spring that is then hung from a board, as shown in Figure 25. As the spring oscillates up and down, the position y of the weight
relative to the board ranges from −1 in. (at time x = 0) to −7 in. (at time x = π) below the board. Assume the position of y is given as a sinusoidal function of x. Sketch a graph of the function, and then find a cosine function that gives the position y in terms of x. y Figure 25 Example 13 Determining a Rider’s Height on a Ferris Wheel The London Eye is a huge Ferris wheel with a diameter of 135 meters (443 feet). It completes one rotation every 30 minutes. Riders board from a platform 2 meters above the ground. Express a rider’s height above ground as a function of time in minutes. Solution With a diameter of 135 m, the wheel has a radius of 67.5 m. The height will oscillate with amplitude 67.5 m above and below the center. Passengers board 2 m above ground level, so the center of the wheel must be located 67.5 + 2 = 69.5 m above ground level. The midline of the oscillation will be at 69.5 m. The wheel takes 30 minutes to complete 1 revolution, so the height will oscillate with a period of 30 minutes. Lastly, because the rider boards at the lowest point, the height will start at the smallest value and increase, following the shape of a vertically reflected cosine curve. • Amplitude: 67.5, so A = 67.5 • Midline: 69.5, so D = 69.5 π __ 15 • Period: 30, so B = 2π ___ 30 = • Shape: −cos(t) An equation for the rider’s height would be y = − 67.5cos  t  + 69.5 π __ 15 where t is in minutes and y is measured in meters. Access these online resources for additional instruction and practice with graphs of sine and cosine functions. • Amplitude and Period of Sine and Cosine (http://openstaxcollege.org/l/ampperiod) • Translations of Sine and Cosine (http://openstaxcollege.org/l/translasincos) • Graphing Sine and Cosine Transformations (http://openstaxcollege.org/l/transformsincos) • Graphing the Sine Function (http://openstaxcollege.org/l/graphsinefunc)
520 CHAPTER 6 periodic Functions 6.1 SeCTIOn exeRCISeS VeRBAl 1. Why are the sine and cosine functions called periodic functions? 3. For the equation Acos(Bx + C) + D, what constants affect the range of the function and how do they affect the range? 5. How can the unit circle be used to construct the graph of f(t) = sin t? GRAPHICAl 2. How does the graph of y = sin x compare with the graph of y = cos x? Explain how you could horizontally translate the graph of y = sin x to obtain y = cos x. 4. How does the range of a translated sine function relate to the equation y = Asin(Bx + C) + D? For the following exercises, graph two full periods of each function and state the amplitude, period, and midline. State the maximum and minimum y-values and their corresponding x-values on one period for x > 0. Round answers to two decimal places if necessary. 6. f (x) = 2sin x 9. f (x) = 4sin x 1 __ x  12. f (x) = 2sin  2 7. f (x) = 2 __ cos x 3 10. f (x) = 2cos x 13. f (x) = 4cos(πx) 8. f (x) = −3sin x 11. f (x) = cos(2x) 6 __ x  14. f (x) = 3cos  5 15. y = 3sin(8(x + 4)) + 5 16. y = 2sin(3x − 21) + 4 17. y = 5sin(5x + 20) − 2 For the following exercises, graph one full period of each function, starting at x = 0. For each function, state the amplitude, period, and midline. State the maximum and minimum y-values and their corresponding x-values on one period for x > 0. State the phase shift and vertical translation, if applicable. Round answers to two decimal places if necessary. 18. f(t) = 2sin  t − 5π ___  6 π __  + 1 19. f(t) = −cos  t + 3 π __   − 3 20. f
(t) = 4cos  2  t + 4 1 __ 21. f(t) = −sin  t + 2 5π ___  3 π _ (x − 3)  + 7 22. f (x) = 4sin  2 23. Determine the amplitude, midline, period, and an equation involving the sine function for the graph shown in Figure 26. 24. Determine the amplitude, period, midline, and an equation involving cosine for the graph shown in Figure 27. f(x) –5 –4 –3 –2 5 4 3 2 1 –1–1 –2 –3 –4 –5 21 3 4 5 x – 3π 2 –π – π 2 f(x) 5 4 3 2 1 –1 –2 –3 –4 –5 π 2 π x 3π 2 Figure 26 Figure 27 SECTION 6.1 section exercises 521 25. Determine the amplitude, period, midline, and an equation involving cosine for the graph shown in Figure 28. 26. Determine the amplitude, period, midline, and an equation involving sine for the graph shown in Figure 29. f(x) f(x) –6 –5 –4 –3 –2 5 4 3 2 1 –1–1 –2 –3 –4 –5 21 3 4 5 6 x –20 –16 –12 –8 5 4 3 2 1 –4–1 –2 –3 –4 –5 84 12 16 20 x Figure 28 Figure 29 27. Determine the amplitude, period, midline, and an equation involving cosine for the graph shown in Figure 30. 28. Determine the amplitude, period, midline, and an equation involving sine for the graph shown in Figure 31. f(x2 + π 2 –2 –4 f(x) –5 –4 –3 –2 5 4 3 2 1 –1–1 –2 –3 –4 –5 21 3 4 5 x Figure 30 Figure 31 29. Determine the amplitude, period, midline, and an equation involving cosine for the graph shown in Figure 32. 30. Determine the amplitude, period, midline, and an equation involving sine for the graph shown in Figure 33. f(x) f(x) –5 –4 –3 –2 5 4 3 2 1 –1–1 –
2 –3 –4 –5 21 3 4 5 x –5 –4 –3 –2 5 4 3 2 1 –1–1 –2 –3 –4 –5 21 3 4 5 x Figure 32 Figure 33 522 CHAPTER 6 periodic Functions AlGeBRAIC For the following exercises, let f (x) = sin x. 31. On [0, 2π), solve f (x) = 0. π __ . 33. Evaluate f  2 32. On [0, 2π), solve f (x) = 1 __. 2 34. On [0, 2π), f (x) =. Find all values of x. — 2 √ ____ 2 35. On [0, 2π), the maximum value(s) of the function 36. On [0, 2π), the minimum value(s) of the function occur(s) at what x-value(s)? occur(s) at what x-value(s)? 37. Show that f(−x) = −f (x). This means that f (x) = sin x is an odd function and possesses symmetry with respect to ________________. For the following exercises, let f (x) = cos x. 38. On [0, 2π), solve the equation f (x) = cos x = 0. 40. On [0, 2π), find the x-intercepts of f (x) = cos x. 42. On [0, 2π), solve the equation f (x) = — 3 √ ____. 2 39. On [0, 2π), solve f (x) = 1 __. 2 41. On [0, 2π), find the x-values at which the function has a maximum or minimum value. TeCHnOlOGY 43. Graph h(x) = x + sin x on [0, 2π]. Explain why the 44. Graph h(x) = x + sin x on [−100, 100]. Did the graph appears as it does. graph appear as predicted in the previous exercise? 45. Graph f (x) = x sin x on [0, 2π] and verbalize how the graph varies from the graph of f (x) = sin x. 46. Graph f (x) = x sin x on the window [−10, 10] and explain what the graph shows. 47. Graph
f (x) = on the window [−5π, 5π] and sin x ____ x explain what the graph shows. ReAl-WORlD APPlICATIOnS 48. A Ferris wheel is 25 meters in diameter and boarded from a platform that is 1 meter above the ground. The six o’clock position on the Ferris wheel is level with the loading platform. The wheel completes 1 full revolution in 10 minutes. The function h(t) gives a person’s height in meters above the ground t minutes after the wheel begins to turn. a. Find the amplitude, midline, and period of h(t). b. Find a formula for the height function h(t). c. How high off the ground is a person after 5 minutes? SECTION 6.2 graphs oF the other trigonometric Functions 523 leARnInG OBjeCTIVeS In this section, you will: • Analyze the graph of y = tan x. • Graph variations of y = tan x. • Analyze the graphs of y = sec x and y = csc x. • Graph variations of y = sec x and y = csc x. • Analyze the graph of y = cot x. • Graph variations of y = cot x. 6.2 GRAPHS OF THe OTHeR TRIGOnOMeTRIC FUnCTIOnS We know the tangent function can be used to find distances, such as the height of a building, mountain, or flagpole. But what if we want to measure repeated occurrences of distance? Imagine, for example, a police car parked next to a warehouse. The rotating light from the police car would travel across the wall of the warehouse in regular intervals. If the input is time, the output would be the distance the beam of light travels. The beam of light would repeat the distance at regular intervals. The tangent function can be used to approximate this distance. Asymptotes would be needed to illustrate the repeated cycles when the beam runs parallel to the wall because, seemingly, the beam of light could appear to extend forever. The graph of the tangent function would clearly illustrate the repeated intervals. In this section, we will explore the graphs of the tangent and other trigonometric functions. Analyzing the Graph of y = tan x We will begin with the graph of the tangent function, plotting points as we did for the sine and cosine
functions. Recall that tan x = sin x ____ cos x The period of the tangent function is π because the graph repeats itself on intervals of kπ where k is a constant. If we to π __ graph the tangent function on − π __, we can see the behavior of the graph on one complete cycle. If we look at any 2 2 larger interval, we will see that the characteristics of the graph repeat. We can determine whether tangent is an odd or even function by using the definition of tangent. tan(−x) = Definition of tangent. sin(−x) ______ cos(−x) = −sin x ______ cos x = − sin x ____ cos x Sine is an odd function, cosine is even. The quotient of an odd and an even function is odd. = −tan x Definition of tangent. Therefore, tangent is an odd function. We can further analyze the graphical behavior of the tangent function by looking at values for some of the special angles, as listed in Table 1 tan(x) undefined − √ — 3 −1 − π _ 6 — 3 √ ____ 3 − 0 0 Table 1 π _ 6 — 3 √ ____ undefined These points will help us draw our graph, but we need to determine how the graph behaves where it is undefined. π π π _ _ _ ≈ 1.05 and < x <, we can use a table to look for a trend. Because If we look more closely at values when 3 2 3 π _ ≈ 1.57, we will evaluate x at radian measures 1.05 < x < 1.57 as shown in Table 2. 2 524 CHAPTER 6 periodic Functions x tan x 1.3 3.6 1.5 14.1 Table 2 1.55 48.1 1.56 92.6 π _, the outputs of the function get larger and larger. Because y = tan x is an odd function, we see the As x approaches 2 corresponding table of negative values in Table 3. x tan x −1.3 −3.6 −1.5 −1.55 −1.56 −14.1 −48.1 −92.6 Table 3 We can see that, as x approaches − π _, the outputs get smaller and smaller. Remember that there are some values of x 2 π 3π _ _  = 0.
At these values, the tangent function is undefined,  = 0 and cos  for which cos x = 0. For example, cos  2 2 π 3π _ _ so the graph of y = tan x has discontinuities at x =. At these values, the graph of the tangent has vertical and 2 2 π _ asymptotes. Figure 1 represents the graph of y = tan x. The tangent is positive from 0 to and from π to 2 corresponding to quadrants I and III of the unit circle. 3π _, 2 y y = tan(x) 5 3 1 –1 –3 –5 π x x = π 2 x = 3π 2 –π x = − 3π 2 x = − π 2 Figure 1 Graph of the tangent function Graphing Variations of y = tan x As with the sine and cosine functions, the tangent function can be described by a general equation. y = Atan(Bx) We can identify horizontal and vertical stretches and compressions using values of A and B. The horizontal stretch can typically be determined from the period of the graph. With tangent graphs, it is often necessary to determine a vertical stretch using a point on the graph. Because there are no maximum or minimum values of a tangent function, the term amplitude cannot be interpreted as it is for the sine and cosine functions. Instead, we will use the phrase stretching/compressing factor when referring to the constant A. features of the graph of y = Atan(Bx) • The stretching factor is ∣ A ∣. • The period is P = π ___. ∣ B ∣ • The domain is all real numbers x, where x ≠ • The range is (−∞, ∞). • The asymptotes occur at = Atan(Bx) is an odd function. π _ ∣ B ∣ + π ____ such that k is an integer. k, where k is an integer. SECTION 6.2 graphs oF the other trigonometric Functions 525 Graphing One Period of a Stretched or Compressed Tangent Function We can use what we know about the properties of the tangent function to quickly sketch a graph of any stretched and/ or compressed tangent function of the form f (x) = Atan(Bx). We focus on a single period of the function including
the origin, because the periodic property enables us to extend the graph to the rest of the function’s domain if we π P P wish. Our limited domain is then the interval  − P _ __ __ __  and the graph has vertical asymptotes at ± where On  − π _ __ __ , the graph will come up from the left asymptote at x = −, cross through the origin, and continue to, 2 2 2 π _ increase as it approaches the right asymptote at x =. To make the function approach the asymptotes at the correct 2 rate, we also need to set the vertical scale by actually evaluating the function for at least one point that the graph will pass through. For example, we can use P P __ __  = Atan  B  = Atan  B f  4 4 π __ 4B  = A π __  = 1. because tan  4 How To… Given the function f (x) = Atan(Bx), graph one period. 1. Identify the stretching factor, ∣ A ∣. 2. Identify B and determine the period, P = π _. ∣ B ∣ and x = P 3. Draw vertical asymptotes at. For A > 0, the graph approaches the left asymptote at negative output values and the right asymptote at positive output values (reverse for A < 0)., A , (0, 0), and  − P P __ __, −A , and draw the graph through these points. 5. Plot reference points at  4 4 Example 1 Sketching a Compressed Tangent π __ Sketch a graph of one period of the function y = 0.5tan  x . 2 Solution First, we identify A and B. π __ y = 0.5 tan  x  2 ↑ y = Atan(Bx) π __ Because A = 0.5 and B =, we can find the stretching/compressing factor and period. The period is 2 = 2, so the π _ π _ 2 asymptotes are at x = ±1. At a quarter period from the origin, we have f (
0.5) = 0.5tan  0.5π ____  2 π __ = 0.5tan   4 = 0.5 This means the curve must pass through the points (0.5, 0.5), (0, 0), and (−0.5, −0.5). The only inflection point is at the origin. Figure 2 shows the graph of one period of the function. 526 CHAPTER 6 periodic Functions πx y = 0.5 tan 2 y 4 2 (0.5, 0.5) x (–0.5, –0.5) –2 –4 x = −1 x = 1 Figure 2 Try It #1 π __ Sketch a graph of f (x) = 3tan  x . 6 Graphing One Period of a Shifted Tangent Function Now that we can graph a tangent function that is stretched or compressed, we will add a vertical and/or horizontal (or phase) shift. In this case, we add C and D to the general form of the tangent function. f (x) = Atan(Bx − C) + D The graph of a transformed tangent function is different from the basic tangent function tan x in several ways: features of the graph of y = Atan(Bx − C) + D • The stretching factor is ∣ A ∣. • The period is π ___. ∣ B ∣ C __ • The domain is x ≠ B + π ___ 2 ∣ B ∣ k, where k is an odd integer. • The range is (−∞, ∞). C __ • The vertical asymptotes occur at x = B + π ____ 2 ∣ B ∣ k, where k is an odd integer. • There is no amplitude. • y = Atan(Bx) is an odd function because it is the quotient of odd and even functions (sine and cosine respectively). How To… Given the function y = Atan(Bx − C) + D, sketch the graph of one period. 1. Express the function given in the form y = Atan(Bx − C) + D. 2. Identify the stretching/compressing factor, ∣ A ∣. π ___ 3. Identify B and determine the period, P =. �
� B ∣ C __. B 4. Identify C and determine the phase shift, C __ 5. Draw the graph of y = Atan(Bx) shifted to the right by B + C __ 6. Sketch the vertical asymptotes, which occur at x = B π ____ 2 ∣ B ∣ and up by D. k, where k is an odd integer. 7. Plot any three reference points and draw the graph through these points. SECTION 6.2 graphs oF the other trigonometric Functions 527 Example 2 Graphing One Period of a Shifted Tangent Function Graph one period of the function y = −2tan(πx + π) − 1. Solution Step 1. The function is already written in the form y = Atan(Bx − C) + D. Step 2. A = −2, so the stretching factor is ∣ A ∣ = 2. Step 3. B = π, so the period is P = = π π __ ___ = 1. ∣ B ∣ π −π C ___ __ = −1. Step 4. C = −π, so the phase shift is = π B Step 5-7. The asymptotes are at x = − 3 and x = − 1 __ __ and the three recommended reference points are (−1.25, 1), 2 2 (−1, −1), and (−0.75, −3). The graph is shown in Figure 3. y = −2 tan (πx + π) − 1 (−1.25, 1) –1 (−1, −1) (−0.75, −3) x = −1.5 x = −0.5 Figure 3 y 3 1 –1 –3 –5 x 0.5 Analysis Note that this is a decreasing function because A < 0. Try It #2 How would the graph in Example 2 look different if we made A = 2 instead of −2? How To… Given the graph of a tangent function, identify horizontal and vertical stretches. 1. Find the period P from the spacing between successive vertical asymptotes or x-intercepts. π __ 2. Write f (x) = Atan  x . P 3. Determine a convenient point (x, f (x)) on the given graph and use it to determine A. Example 3 Identifying the Graph of a
Stretched Tangent Find a formula for the function graphed in Figure 4. x = 4 x = 12 x = −12 x = −4 y 6 4 2 –10 –6 –2 2 6 10 x –2 –4 –6 Figure 4 A stretched tangent function 528 CHAPTER 6 periodic Functions Solution The graph has the shape of a tangent function. Step 1. One cycle extends from –4 to 4, so the period is P = 8. Since P = π __ Step 2. The equation must have the form f (x) = Atan  x . 8 Step 3. To find the vertical stretch A, we can use the point (2, 2). π ___ ∣ B ∣ π π __ __, we have B = =. 8 P π __  = 1, A = 2. Because tan  4 π __ This function would have a formula f (x) = 2tan  x . 8 π π __ __ 2 = Atan  · 2  = Atan   4 8 Try It #3 Find a formula for the function in Figure 5. Analyzing the Graphs of y = sec x and y = csc x x = − 3π = 3π π – π 2 –2 –4 –6 Figure 5 The secant was defined by the reciprocal identity sec x =. Notice that the function is undefined when the cosine π __, is 0, leading to vertical asymptotes at 2 3π __, etc. Because the cosine is never more than 1 in absolute value, the secant, 2 1 ____ cos x being the reciprocal, will never be less than 1 in absolute value. We can graph y = sec x by observing the graph of the cosine function because these two functions are reciprocals of one another. See Figure 6. The graph of the cosine is shown as a blue wave so we can see the relationship. Where the graph of the cosine function decreases, the graph of the secant function increases. Where the graph of the cosine function increases, the graph of the secant function decreases. When the cosine function is zero, the secant is undefined. The secant graph has vertical asymptotes at each value of x where the cosine graph crosses the x-axis; we show these in the graph
below with dashed vertical lines, but will not show all the asymptotes explicitly on all later graphs involving the secant and cosecant. Note that, because cosine is an even function, secant is also an even function. That is, sec(−x) = sec x. x = − 3π 2 y x = 3π 2 y = cos (x) –2π y = sec (x) x 2π 8 4 –4 –8 Figure 6 Graph of the secant function, f (x) = sec x = 1 ____ cos x As we did for the tangent function, we will again refer to the constant ∣ A ∣ as the stretching factor, not the amplitude. SECTION 6.2 graphs oF the other trigonometric Functions 529 features of the graph of y = Asec(Bx) • The stretching factor is ∣ A ∣. • The period is π ____ 2 ∣ B ∣ • The domain is x ≠ • The range is (−∞, − ∣ A ∣ ] ∪ [ ∣ A ∣, ∞). • The vertical asymptotes occur at x = 2π ___. ∣ B ∣ • There is no amplitude. k, where k is an odd integer. π ____ 2 ∣ B ∣ k, where k is an odd integer. • y = Asec(Bx) is an even function because cosine is an even function. Similar to the secant, the cosecant is defined by the reciprocal identity csc x = undefined when the sine is 0, leading to a vertical asymptote in the graph at 0, π, etc. Since the sine is never more than 1 in absolute value, the cosecant, being the reciprocal, will never be less than 1 in absolute value.. Notice that the function is 1 ____ sin x We can graph y = csc x by observing the graph of the sine function because these two functions are reciprocals of one another. See Figure 7. The graph of sine is shown as a blue wave so we can see the relationship. Where the graph of the sine function decreases, the graph of the cosecant function increases. Where the graph of the sine function increases, the graph of the cosecant function decreases. The cosecant graph has vertical asymptotes at each value
of x where the sine graph crosses the x-axis; we show these in the graph below with dashed vertical lines. Note that, since sine is an odd function, the cosecant function is also an odd function. That is, csc(−x) = −cscx. The graph of cosecant, which is shown in Figure 7, is similar to the graph of secant. x = −2π y x = −π 10 x = π x = 2π y = csc (x) 6 2 –6 –10 y = sin (x) x Figure 7 The graph of the cosecant function, f (x) = cscx = 1 ____ sinx features of the graph of y = Acsc(Bx) • The stretching factor is ∣ A ∣. • The period is 2π ___. ∣ B ∣ • The domain is x ≠ k, where k is an integer. π ___ ∣ B ∣ • The range is (−∞, − ∣ A ∣ ] ∪ [ ∣ A ∣, ∞). • The asymptotes occur at x = π ___ ∣ B ∣ • y = Acsc(Bx) is an odd function because sine is an odd function. k, where k is an integer. Graphing Variations of y = sec x and y = csc x For shifted, compressed, and/or stretched versions of the secant and cosecant functions, we can follow similar methods to those we used for tangent and cotangent. That is, we locate the vertical asymptotes and also evaluate the functions for a few points (specifically the local extrema). If we want to graph only a single period, we can choose the interval for the 530 CHAPTER 6 periodic Functions period in more than one way. The procedure for secant is very similar, because the cofunction identity means that the secant graph is the same as the cosecant graph shifted half a period to the left. Vertical and phase shifts may be applied to the cosecant function in the same way as for the secant and other functions. The equations become the following. y = Asec(Bx − C) + D y = Acsc(Bx − C) + D features of the graph of y = Asec(Bx − C) +
D • The stretching factor is ∣ A ∣. • The period is 2π ___. ∣ B ∣ C __ • The domain is x ≠ B + π ____ 2 ∣ B ∣ k, where k is an odd integer. • The range is (−∞, − ∣ A ∣ + D] ∪ [ ∣ A ∣ + D, ∞). π C ____ __ • The vertical asymptotes occur at x = 2 ∣ B ∣ B • There is no amplitude. • y = Asec(Bx) is an even function because cosine is an even function. + k, where k is an odd integer. features of the graph of y = Acsc(Bx − C) + D • The stretching factor is ∣ A ∣. • The period is 2π ___. ∣ B ∣ C __ • The domain is x ≠ B + π ____ ∣ B ∣ k, where k is an integer. • The range is (−∞, − ∣ A ∣ + D] ∪ [ ∣ A ∣ + D, ∞). π • The vertical asymptotes occur at x = C ___ __ ∣ B ∣ B + k, where k is an integer. • There is no amplitude. • y = Acsc(Bx) is an odd function because sine is an odd function. How To… Given a function of the form y = Asec(Bx), graph one period. 1. Express the function given in the form y = Asec(Bx). 2. Identify the stretching/compressing factor, ∣ A ∣. 2π ___ 3. Identify B and determine the period, P =. ∣ B ∣ 4. Sketch the graph of y = Acos(Bx). 5. Use the reciprocal relationship between y = cos x and y = sec x to draw the graph of y = Asec(Bx). 6. Sketch the asymptotes. 7. Plot any two reference points and draw the graph through these points. Example 4 Graphing a Variation of the Secant Function Graph one period of f (x) = 2.5sec(0.4x). Solution Step 1. The given function is already written in the general form, y = Asec(Bx). Step 2. A = 2.5
so the stretching factor is 2.5. Step 3. B = 0.4 so P = = 5π. The period is 5π units. Step 4. Sketch the graph of the function g(x) = 2.5cos(0.4x). 2π ___ 0.4 Step 5. Use the reciprocal relationship of the cosine and secant functions to draw the cosecant function. SECTION 6.2 graphs oF the other trigonometric Functions 531 Steps 6–7. Sketch two asymptotes at x = 1.25π and x = 3.75π. We can use two reference points, the local minimum at (0, 2.5) and the local maximum at (2.5π, −2.5). Figure 8 shows the graph. f (x) = 2.5 sec (0.4x) y 6 4 2 –π π 2π 3π 4π x –2 –4 –6 Figure 8 Try It #4 Graph one period of f (x) = −2.5sec(0.4x). Q & A… Do the vertical shift and stretch/compression affect the secant’s range? Yes. The range of f (x) = Asec(Bx − C) + D is (−∞, − ∣ A ∣ + D] ∪ [ ∣ A ∣ + D, ∞). How To… Given a function of the form f (x) = Asec(Bx − C) + D, graph one period. 1. Express the function given in the form y = A sec(Bx − C) + D. 2. Identify the stretching/compressing factor, ∣ A ∣. 3. Identify B and determine the period, 2π ___. ∣ B ∣ 4. Identify C and determine the phase shift, C __ 5. Draw the graph of y = A sec(Bx) but shift it to the right by B π ____ 2 ∣ B ∣ C __ 6. Sketch the vertical asymptotes, which occur at x = B + C __. B and up by D. k, where k is an odd integer. Example 5 Graphing a Variation of the Secant Function π π __ __  + 1. x − Graph one period of y = 4sec  2 3 Solution π �
� __ __ Step 1. Express the function given in the form y = 4sec   + 1. x − 2 3 Step 2. The stretching/compressing factor is ∣ A ∣ = 4. Step 3. The period is Step 4. The phase shift is 2π ___ ∣ B ∣ = = 2π _ π _ 3 3 2π __ ___ · 1 π = 6 π __ 2 _ = π __ 3 π 3 __ __ = · 2 π = 1.5 C __ B 532 CHAPTER 6 periodic Functions C __ Step 5. Draw the graph of y = Asec(Bx), but shift it to the right by B = 1.5 and up by D = 6. Step 6. Sketch the vertical asymptotes, which occur at x = 0, x = 3, and x = 6. There is a local minimum at (1.5, 5) and a local maximum at (4.5, −3). Figure 9 shows the graph. y x = 3 x = 6 10 8 6 4 2 –2 –4 –6 –8 y = 4 sec Figure 9 Try It #5 Graph one period of f (x) = −6sec(4x + 2) − 8. Q & A… The domain of csc x was given to be all x such that x ≠ kπ for any integer k. Would the domain of y = Acsc(Bx − C) + D be x ≠ C + kπ ______? B Yes. The excluded points of the domain follow the vertical asymptotes. Their locations show the horizontal shift and compression or expansion implied by the transformation to the original function’s input. How To… Given a function of the form y = Acsc(Bx), graph one period. 1. Express the function given in the form y = Acsc(Bx). 2. Identify the stretching/compressing factor, ∣ A ∣. 2π ___ 3. Identify B and determine the period, P =. ∣ B ∣ 4. Draw the graph of y = Asin(Bx). 5. Use the reciprocal relationship between y = sin x and y = csc x to draw the graph of y = Acsc(Bx). 6. Sketch the asymptotes. 7. Plot any two reference points and draw the graph through these points. Example 6 Grap
hing a Variation of the Cosecant Function Graph one period of f (x) = −3csc(4x). Solution Step 1. The given function is already written in the general form, y = Acsc(Bx). Step 2. ∣ A ∣ = ∣ −3 ∣ = 3, so the stretching factor is 3. π π __ __ = Step 3. B = 4, so P = units.. The period is 2 2 Step 4. Sketch the graph of the function g(x) = −3sin(4x). 2π __ 4 Step 5. Use the reciprocal relationship of the sine and cosecant functions to draw the cosecant function. π π __ __, and x = Steps 6–7. Sketch three asymptotes at x = 0, x =. We can use two reference points, the local maximum 2 4 π __, −3  and the local minimum at  at  8 3π __, 3 . Figure 10 shows the graph. 8 SECTION 6.2 graphs oF the other trigonometric Functions 533 y 6 4 2 –2 –4 –(x) = –3 csc (4x) π 8 3π 8 x Figure 10 Try It #6 Graph one period of f (x) = 0.5csc(2x). How To… Given a function of the form f (x) = Acsc(Bx − C) + D, graph one period. 1. Express the function given in the form y = Acsc(Bx − C) + D. 2. Identify the stretching/compressing factor, ∣ A ∣. 3. Identify B and determine the period, 2π ___. ∣ B ∣ 4. Identify C and determine the phase shift, C __ 5. Draw the graph of y = Acsc(Bx) but shift it to the right by B π 6. Sketch the vertical asymptotes, which occur at x = C ___ __ ∣ B ∣ B C __. B + and up by D. k, where k is an integer. Graphing a Vertically Stretched, Horizontally Compressed, and Vertically Shifted Cosecant Example 7 π _ x  + 1. What are the domain and range of this function? Sketch
a graph of y = 2csc  2 Solution π __ Step 1. Express the function given in the form y = 2csc  x  + 1. 2 Step 2. Identify the stretching/compressing factor, ∣ A ∣ = 2. 2π _ π __ 2 = 0. Step 4. The phase shift is Step 3. The period is 2 _ π = 4. · 2π ___ ∣ B ∣ 2π _ 1 = = 0 _ π __ 2 Step 5. Draw the graph of y = Acsc(Bx) but shift it up D = 1. Step 6. Sketch the vertical asymptotes, which occur at x = 0, x = 2, x = 4. The graph for this function is shown in Figure 11. y 6 3 –3 –6 –3 y = 2 csc 4 –6 x = −2 x = 2 x = 4 Figure 11 A transformed cosecant function 534 CHAPTER 6 periodic Functions Analysis The vertical asymptotes shown on the graph mark off one period of the function, and the local extrema in this π __ x  + 1, interval are shown by dots. Notice how the graph of the transformed cosecant relates to the graph of f (x) = 2sin  2 shown as the blue wave. Try It #7 π π __ __ x  + 1 on the same axes. x  + 1 shown in Figure 12, sketch the graph of g(x) = 2sec  Given the graph of f (x) = 2cos  2 2 f (x) 6 4 f (x) = 2 cos π x 2 + 1 –5 –3 –1 1 3 5 x –2 –4 –6 Figure 12 Analyzing the Graph of y = cot x The last trigonometric function we need to explore is cotangent. The cotangent is defined by the reciprocal identity cot x = 1 ____ tan x. Notice that the function is undefined when the tangent function is 0, leading to a vertical asymptote in the graph at 0, π, etc. Since the output of the tangent function is all real numbers, the output of the cotangent function is also all real numbers. We can graph y = cot x by observing the graph
of the tangent function because these two functions are reciprocals of one another. See Figure 13. Where the graph of the tangent function decreases, the graph of the cotangent function increases. Where the graph of the tangent function increases, the graph of the cotangent function decreases. The cotangent graph has vertical asymptotes at each value of x where tan x = 0; we show these in the graph below with dashed lines. Since the cotangent is the reciprocal of the tangent, cot x has vertical asymptotes at all values of x where tan x = 0, and cot x = 0 at all values of x where tan x has its vertical asymptotes. f (x) = cot x y 20 10 x – 3π 2 – π 2 3π 2 π 2 –10 –20 x = −π x = −2π Figure 13 The cotangent function x = 2π x = π features of the graph of y = Acot(Bx) • The stretching factor is ∣ A ∣. • The period is P = π ___. ∣ B ∣ π ___ ∣ B ∣ • The range is (−∞, ∞). • The asymptotes occur at x = π ___ ∣ B ∣ • y = Acot(Bx) is an odd function. • The domain is x ≠ k, where k is an integer. k, where k is an integer. SECTION 6.2 graphs oF the other trigonometric Functions 535 Graphing Variations of y = cot x We can transform the graph of the cotangent in much the same way as we did for the tangent. The equation becomes the following. y = Acot(Bx − C) + D features of the graph of y = Acot(Bx − C) + D • The stretching factor is ∣ A ∣. • The period is π ___. ∣ B ∣ C __ • The domain is x ≠ B + π ___ ∣ B ∣ k, where k is an integer. • The range is (−∞, ∞). • The vertical asymptotes occur at x = C __ B • There is no amplitude. • y = Acot(Bx) is an odd function because it is the quotient of even and odd
functions (cosine and sine, respectively) k, where k is an integer. π ___ ∣ B ∣ + How To… Given a modified cotangent function of the form f (x) = Acot(Bx), graph one period. 1. Express the function in the form f (x) = Acot(Bx). 2. Identify the stretching factor, ∣ A ∣. 3. Identify the period, P = π ___. ∣ B ∣ 4. Draw the graph of y = Atan(Bx). 5. Plot any two reference points. 6. Use the reciprocal relationship between tangent and cotangent to draw the graph of y = Acot(Bx). 7. Sketch the asymptotes. Example 8 Graphing Variations of the Cotangent Function Determine the stretching factor, period, and phase shift of y = 3cot(4x), and then sketch a graph. Solution Step 1. Expressing the function in the form f (x) = Acot(Bx) gives f (x) = 3cot(4x). Step 2. The stretching factor is ∣ A ∣ = 3. π __ Step 3. The period is P =. 4 Step 4. Sketch the graph of y = 3tan(4x). Step 5. Plot two reference points. Two such points are  Step 6. Use the reciprocal relationship to draw y = 3cot(4x). π __ 16, 3  and  3π __ 16, −3 . π __ Step 7. Sketch the asymptotes, x = 0, x =. 4 The blue graph in Figure 14 shows y = 3tan(4x) and the red graph shows y = 3cot(4x). y = 3 cot (4x) y = 3 tan (4x) 6 4 2 –2 –4 –6 Figure 14 536 CHAPTER 6 periodic Functions How To… Given a modified cotangent function of the form f (x) = Acot(Bx − C) + D, graph one period. 1. Express the function in the form f (x) = Acot(Bx − C) + D. 2. Identify the stretching factor, ∣ A ∣. 3. Identify the period, P = π ___. ∣ B ∣
C __. B C __ 5. Draw the graph of y = Atan(Bx) shifted to the right by B 4. Identify the phase shift, C __ 6. Sketch the asymptotes x = B π ___ ∣ B ∣ 7. Plot any three reference points and draw the graph through these points. k, where k is an integer. + and up by D. Graphing a Modified Cotangent Example 9 π π __ __  − 2. x − Sketch a graph of one period of the function f (x) = 4cot  2 8 Solution Step 1. The function is already written in the general form f (x) = Acot(Bx − C) + D. Step 2. A = 4, so the stretching factor is 4. π __, so the period is P = Step 3. B = 8 π ___ ∣ B ∣ = π _ π __ 8 = 8. π __ Step 4. C =, so the phase shift is 2 π __ 2 _ π __ 8 π π __ __  − 2. x − Step 5. We draw f (x) = 4tan  2 8 C __ B = = 4. Step 6-7. Three points we can use to guide the graph are (6, 2), (8, −2), and (10, −6). We use the reciprocal relationship of π π __ __  − 2. x − tangent and cotangent to draw f (x) = 4cot  2 8 Step 8. The vertical asymptotes are x = 4 and x = 12. The graph is shown in Figure 15. y 40 30 20 10 –10 –20 –30 –40 f (x) = 4 cot 10 12 14 x x = 4 x = 12 Figure 15 One period of a modified cotangent function Using the Graphs of Trigonometric Functions to Solve Real-World Problems Many real-world scenarios represent periodic functions and may be modeled by trigonometric functions. As an example, let’s return to the scenario from the section opener. Have you ever observed the beam formed by the rotating light on a police car and wondered about the movement of the light beam itself across the wall? The periodic behavior of the distance the light shines as a function of time is obvious, but how
do we determine the distance? We can use the tangent function. SECTION 6.2 graphs oF the other trigonometric Functions 537 Example 10 Using Trigonometric Functions to Solve Real-World Scenarios π __ Suppose the function y = 5tan  t  marks the distance in the movement of a light beam from the top of a police car 4 across a wall where t is the time in seconds and y is the distance in feet from a point on the wall directly across from the police car. a. Find and interpret the stretching factor and period. b. Graph on the interval [0, 5]. c. Evaluate f (1) and discuss the function’s value at that input. Solution π __ a. We know from the general form of y = Atan(Bt) that ∣ A ∣ is the stretching factor and is the period. B π __ y = 5 tan  t  4 ↑ ↑ A B Figure 16 We see that the stretching factor is 5. This means that the beam of light will have moved 5 ft after half the period. The period is π 4 __ __ = = 4. This means that every 4 seconds, the beam of light sweeps the wall. The distance · π 1 π _ π __ 4 from the spot across from the police car grows larger as the police car approaches. b. To graph the function, we draw an asymptote at t = 2 and use the stretching factor and period. See Figure 17 y 6 4 2 –2 –4 –6 π t y = 5 tan 4 1 3 4 5 t t = 2 Figure 17 π __ (1)  = 5(1) = 5; after 1 second, the beam of light has moved 5 ft from the spot across c. Period: f (1) = 5tan  4 from the police car. Access these online resources for additional instruction and practice with graphs of other trigonometric functions. • Graphing the Tangent Function (http://openstaxcollege.org/l/graphtangent) • Graphing Cosecant and Secant Functions (http://openstaxcollege.org/l/graphcscsec) • Graphing the Cotangent Function (http://openstaxcollege.org/l/graphcot) 538 CHAPTER 6 periodic Functions 6.2 SeCTI
On exeRCISeS VeRBAl 1. Explain how the graph of the sine function can be used to graph y = csc x. 2. How can the graph of y = cos x be used to construct the graph of y = sec x? 3. Explain why the period of tan x is equal to π. 4. Why are there no intercepts on the graph of 5. How does the period of y = csc x compare with the period of y = sin x? y = csc x? AlGeBRAIC For the following exercises, match each trigonometric function with one of the graphs in Figure 182π –π x –π 2π – π 2 x π π 2 –2π –π π 2π x x = π II Figure 18 III IV 6. f (x) = tan x 8. f (x) = csc x 7. f (x) = sec x 9. f (x) = cot x For the following exercises, find the period and horizontal shift of each of the functions. π 10. f (x) = 2tan(4x − 32) __ 11. h(x) = 2sec  (x + 1)  4 13. If tan x = −1.5, find tan(−x). π __ 12. m(x) = 6csc  x + π  3 14. If sec x = 2, find sec(−x). 15. If csc x = −5, find csc(−x). 16. If xsin x = 2, find (−x)sin(−x). For the following exercises, rewrite each expression such that the argument x is positive. 17. cot(−x)cos(−x) + sin(−x) 18. cos(−x) + tan(−x)sin(−x) GRAPHICAl For the following exercises, sketch two periods of the graph for each of the following functions. Identify the stretching factor, period, and asymptotes. 19. f (x) = 2tan(4x − 32) π __ 22. j(x) = tan  x  2 π __ 25. f (x) = tan  x +  4 28. f (x) = − 1 __ csc(
x) 4 31. f (x) = 7sec(5x) π __  − 2 34. f (x) = −sec  x − 3 π __ (x + 1)  20. h(x) = 2sec  4 π __ 23. p(x) = tan  x −  2 26. f (x) = π tan(πx − π) − π π __ x + π  21. m(x) = 6csc  3 24. f (x) = 4tan(x) 27. f (x) = 2csc(x) 29. f (x) = 4sec(3x) 30. f (x) = −3cot(2x) csc(πx) 32. f (x) = 9 __ 10 π 35. f (x) = 7 __ __ csc  x −  4 5 π __  − 1 33. f (x) = 2csc  x + 4 π __  − 3  36. f (x) = 5  cot  x + 2 SECTION 6.2 section exercises 539 For the following exercises, find and graph two periods of the periodic function with the given stretching factor, ∣ A ∣, period, and phase shift. π _ 37. A tangent curve, A = 1, period of ; and phase shift 3 π __ (h, k) = , 2  4 π _ 38. A tangent curve, A = −2, period of, and phase 4 shift (h, k) =  − π __, −2  4 For the following exercises, find an equation for the graph of each function. f (x) 8 4 –0.5 0.5 –4 –8 x = −1 x 1 x = 1 39. x = −π f (x) x = π 2 40. 10 6 2 –6 –10 x = − π 2 x = π 41. x = − π 4 f (x) x = π 2 x x 10 6 2 –6 –10 π 2 x = − 42. f (x) 43. f (
x) x = −2π x = π 10 6 2 – 3π 8 π 8 –6 –10 x 5π 8 10 6 2 –6 –10 x = −π x = 2π x x = π 4 x = − 5π 8 x = − π 8 44. f (x) 45. 10 6 2 –2π –π π 2π x −0.01 –10 x = −0.005 TeCHnOlOGY x = 3π 8 f (x) 4 3 2 1 –1 –2 –3 –4 x 0.01 x = 0.005 For the following exercises, use a graphing calculator to graph two periods of the given function. Note: most graphing calculators do not have a cosecant button; therefore, you will need to input csc x as 1 ____. sin x 46. f (x) = \|csc(x)\| 48. f (x) = 2csc(x) 47. f (x) = \|cot(x)\| 49. f (x) = csc(x) _____ sec(x) 50. Graph f (x) = 1 + sec2 (x) − tan2 (x). What is the function shown in the graph? 51. f (x) = sec(0.001x) 52. f (x) = cot(100πx) 53. f (x) = sin2 x + cos2 x 540 CHAPTER 6 periodic Functions ReAl-WORlD APPlICATIOnS π 54. The function f (x) = 20tan  __ 10 x  marks the distance in the movement of a light beam from a police car across a wall for time x, in seconds, and distance f (x), in feet. a. Graph on the interval [0, 5]. b. Find and interpret the stretching factor, period, and asymptote. c. Evaluate f (1) and f (2.5) and discuss the function’s values at those inputs. 55. Standing on the shore of a lake, a fisherman sights a boat far in the distance to his left. Let x, measured in radians, be the angle formed by the line of sight to the ship and a line due north from his position. Assume due north is 0 and x is measured negative to the left and positive to
the right. (See Figure 19.) The boat travels from due west to due east and, ignoring the curvature of the Earth, the distance d(x), in kilometers, from the fisherman to the boat is given by the function d(x) = 1.5sec(x). a. What is a reasonable domain for d(x)? b. Graph d(x) on this domain. c. Find and discuss the meaning of any vertical asymptotes on the graph of d(x). d. Calculate and interpret d  − π __ . Round to the 3 second decimal place. π __ . Round to the e. Calculate and interpret d  6 second decimal place. f. What is the minimum distance between the fisherman and the boat? When does this occur? 56. A laser rangefinder is locked on a comet 57. A video camera is focused on a rocket on a Figure 19 approaching Earth. The distance g(x), in kilometers, of the comet after x days, for x in the interval 0 to 30 days, is given by g(x) = 250,000csc  a. Graph g(x) on the interval [0, 35]. π __ 30 x . b. Evaluate g(5) and interpret the information. c. What is the minimum distance between the comet and Earth? When does this occur? To which constant in the equation does this correspond? d. Find and discuss the meaning of any vertical asymptotes. launching pad 2 miles from the camera. The angle of elevation from the ground to the rocket after x seconds is π ___ x. 120 a. Write a function expressing the altitude h(x), in miles, of the rocket above the ground after x seconds. Ignore the curvature of the Earth. b. Graph h(x) on the interval (0, 60). c. Evaluate and interpret the values h(0) and h(30). d. What happens to the values of h(x) as x approaches 60 seconds? Interpret the meaning of this in terms of the problem. SECTION 6.3 inverse trigonometric Functions 541 leARnInG OBjeCTIVeS In this section, you will: • Understand and use the inverse sine, cosine, and tangent functions. • Find the exact value of expressions involving the inverse sine, cosine, and tangent
functions. • Use a calculator to evaluate inverse trigonometric functions. • Find exact values of composite functions with inverse trigonometric functions. 6.3 InVeRSe TRIGOnOMeTRIC FUnCTIOnS For any right triangle, given one other angle and the length of one side, we can figure out what the other angles and sides are. But what if we are given only two sides of a right triangle? We need a procedure that leads us from a ratio of sides to an angle. This is where the notion of an inverse to a trigonometric function comes into play. In this section, we will explore the inverse trigonometric functions. Understanding and Using the Inverse Sine, Cosine, and Tangent Functions In order to use inverse trigonometric functions, we need to understand that an inverse trigonometric function “undoes” what the original trigonometric function “does,” as is the case with any other function and its inverse. In other words, the domain of the inverse function is the range of the original function, and vice versa, as summarized in Figure 1. Trig Functions Domain: Measure of an angle Range: Ratio Inverse Trig Functions Domain: Ratio Range: Measure of an angle Figure 1 For example, if f (x) = sin x, then we would write f −1(x) = sin−1 x. Be aware that sin−1 x does not mean following examples illustrate the inverse trigonometric functions: 1 ___ sinx. The 1 π 1 π __ __ __ __ . = sin−1   = • Since sin , then 2 6 2 6 • Since cos(π) = −1, then π = cos−1(−1). π π __ __ = tan−1(1).  = 1, then • Since tan  4 4 In previous sections, we evaluated the trigonometric functions at various angles, but at times we need to know what angle would yield a specific sine, cosine, or tangent value. For this, we need inverse functions. Recall that, for a oneto-one function, if f(a) = b, then an inverse function would satisfy f −1(b) = a. Bear in mind that the sine, cosine, and tangent functions are not one-to-one functions. The graph of each
function would fail the horizontal line test. In fact, no periodic function can be one-to-one because each output in its range corresponds to at least one input in every period, and there are an infinite number of periods. As with other functions that are not one-to-one, we will need to restrict the domain of each function to yield a new function that is one-toone. We choose a domain for each function that includes the number 0. Figure 2 shows the graph of the sine function π limited to  − π __ __  and the graph of the cosine function limited to [0, π]., 2 2 y (a) – π 2 – π 4 1 –1 y f(x) = sin x x π 4 π 2 (b) 1 –1 f (x) = cos x π 4 π 2 3π 4 x π Figure 2 (a) Sine function on a restricted domain of  − π __ 2, π __ 2  ; (b) Cosine function on a restricted domain of [0, π] 542 CHAPTER 6 periodic Functions π Figure 3 shows the graph of the tangent function limited to  − π __ __ ., x) = tan x π 4 y 4 3 2 1 –1 –2 –3 –4 Figure 3 Tangent function on a restricted domain of  − π __ 2, π __  2 These conventional choices for the restricted domain are somewhat arbitrary, but they have important, helpful characteristics. Each domain includes the origin and some positive values, and most importantly, each results in a one-to-one function that is invertible. The conventional choice for the restricted domain of the tangent function also has the useful property that it extends from one vertical asymptote to the next instead of being divided into two parts by an asymptote. On these restricted domains, we can define the inverse trigonometric functions. • The inverse sine function y = sin−1 x means x = sin y. The inverse sine function is sometimes called the arcsine function, and notated arcsinx. π y = sin−1 x has domain [−1, 1] and range  − π __ __ , 2 2 • The inverse cosine function y = cos−1 x means
x = cos y. The inverse cosine function is sometimes called the arccosine function, and notated arccos x. y = cos−1 x has domain [−1, 1] and range [0, π] • The inverse tangent function y = tan−1 x means x = tan y. The inverse tangent function is sometimes called the arctangent function, and notated arctan x. π y = tan−1 x has domain (−∞, ∞) and range  − π __ __ , 2 2 The graphs of the inverse functions are shown in Figure 4, Figure 5, and Figure 6. Notice that the output of each of these inverse functions is a number, an angle in radian measure. We see that sin−1 x has domain [−1, 1] and π range  − π __ __ , cos−1 x has domain [−1, 1] and range [0, π], and tan−1 x has domain of all real numbers and range, 2 2  − π π __ __ . To find the domain and range of inverse trigonometric functions, switch the domain and range of the, 2 2 original functions. Each graph of the inverse trigonometric function is a reflection of the graph of the original function about the line y = x. y = sin–1 (x) y = x y = sin (x Figure 4 The sine function and inverse sine (or arcsine) function SECTION 6.3 inverse trigonometric Functions 543 y = cos–1 (x = cos (x) π 2 x π y = tan (x) y = x y = tan–1 (x Figure 5 The cosine function and inverse cosine (or arccosine) function Figure 6 The tangent function and inverse tangent (or arctangent) function relations for inverse sine, cosine, and tangent functions For angles in the interval  − π π , if sin y = x, then sin−1 x = y. __ __, 2 2 For angles in the interval [0, π], if cos y = x, then cos−1 x = y. π For angles in the interval  − π __ __ , if tan y = x, then tan
−1 x = y., 2 2 Example 1 Writing a Relation for an Inverse Function 5π __ 12  ≈ 0.96593, write a relation involving the inverse sine. Given sin  Solution Use the relation for the inverse sine. If sin y = x, then sin−1 x = y. In this problem, x = 0.96593, and y = 5π __. 12 sin−1 (0.96593) ≈ 5π ___ 12 Try It #1 Given cos(0.5) ≈ 0.8776, write a relation involving the inverse cosine. Finding the exact Value of expressions Involving the Inverse Sine, Cosine, and Tangent Functions Now that we can identify inverse functions, we will learn to evaluate them. For most values in their domains, we must evaluate the inverse trigonometric functions by using a calculator, interpolating from a table, or using some other numerical technique. Just as we did with the original trigonometric functions, we can give exact values for the π π π __ __ __ (60°), and their reflections (45°), and (30°), inverse functions when we are using the special angles, specifically 3 4 6 into other quadrants. How To… Given a “special” input value, evaluate an inverse trigonometric function. 1. Find angle x for which the original trigonometric function has an output equal to the given input for the inverse trigonometric function. 2. If x is not in the defined range of the inverse, find another angle y that is in the defined range and has the same sine, cosine, or tangent as x, depending on which corresponds to the given inverse function. Example 2 Evaluating Inverse Trigonometric Functions for Special Input Values Evaluate each of the following. 1 __  a. sin−1  2 2 b. sin−1  − √  ____ 2 — c. cos−1  − — 3 √  ____ 2 d. tan−1(1) 544 Solution CHAPTER 6 periodic Functions 1 1 _ _  is the same as determining the angle that would have a sine value of a. Evaluating sin−1 . In other words, 2 2 1 _ what angle x would satisfy sin(x) =? There
are multiple values that would satisfy this relationship, such 2 π π 5π, but we know we need the angle in the interval  − π 1 _ _ _ _ _  = , so the answer will be sin− and as 6 Remember that the inverse is a function, so for each input, we will get exactly one output. 2 b. To evaluate sin−1  − √ , we know that _ 2 5π _ 4 π interval  − π _ _ . For that, we need the negative angle coterminal with, 2 2 2 _ 7π both have a sine value of − √ _, but neither is in the 4 2 — 2 : sin−1  − √ 7π  = − π _ ____ ___. 2 4 4 and — 3 3 , we are looking for an angle in the interval [0, π] with a cosine value of − √ c. To evaluate cos−1  − √ _ _ 2 2. The — — — — 3 angle that satisfies this is cos−1  − √  = _ 2 5π _. 6 π d. Evaluating tan−1(1), we are looking for an angle in the interval  − π _ _  with a tangent value of 1. The correct, 2 2 π _ angle is tan−1(1) =. 4 Try It #2 Evaluate each of the following. a. sin−1(−1) b. tan−1 (−1) c. cos−1 (−1) 1 d. cos−1  _  2 Using a Calculator to evaluate Inverse Trigonometric Functions To evaluate inverse trigonometric functions that do not involve the special angles discussed previously, we will need to use a calculator or other type of technology. Most scientific calculators and calculator-emulating applications have specific keys or buttons for the inverse sine, cosine, and tangent functions. These may be labeled, for example, SIN-1, ARCSIN, or ASIN. In the previous chapter, we worked with trigonometry on a right triangle to solve for the sides of a triangle given one side and an additional angle. Using the inverse trigonometric functions, we can
solve for the angles of a right triangle given two sides, and we can use a calculator to find the values to several decimal places. In these examples and exercises, the answers will be interpreted as angles and we will use θ as the independent variable. The value displayed on the calculator may be in degrees or radians, so be sure to set the mode appropriate to the application. Example 3 Evaluating the Inverse Sine on a Calculator Evaluate sin−1(0.97) using a calculator. Solution Because the output of the inverse function is an angle, the calculator will give us a degree value if in degree mode and a radian value if in radian mode. Calculators also use the same domain restrictions on the angles as we are using. In radian mode, sin−1(0.97) ≈ 1.3252. In degree mode, sin−1(0.97) ≈ 75.93°. Note that in calculus and beyond we will use radians in almost all cases. Try It #3 Evaluate cos−1 (−0.4) using a calculator. SECTION 6.3 inverse trigonometric Functions 545 How To… Given two sides of a right triangle like the one shown in Figure 7, find an angle. h p θ a Figure 7 1. If one given side is the hypotenuse of length h and the side of length a adjacent to the desired angle is given, use a __ the equation θ = cos−1  . h 2. If one given side is the hypotenuse of length h and the side of length p opposite to the desired angle is given, use p  the equation θ = sin−1  _ h p a . 3. If the two legs (the sides adjacent to the right angle) are given, then use the equation θ = tan−1  _ Example 4 Applying the Inverse Cosine to a Right Triangle Solve the triangle in Figure 8 for the angle θ. 12 θ 9 Figure 8 Solution Because we know the hypotenuse and the side adjacent to the angle, it makes sense for us to use the cosine function. cos θ = 9 __ 12 9 _  θ = cos−1  12 θ ≈ 0.7227 or about 41.4096° Apply definition of the inverse. Evaluate. Try It #4 Solve the
triangle in Figure 9 for the angle θ. 10 6 θ Figure 9 546 CHAPTER 6 periodic Functions Finding exact Values of Composite Functions with Inverse Trigonometric Functions There are times when we need to compose a trigonometric function with an inverse trigonometric function. In these cases, we can usually find exact values for the resulting expressions without resorting to a calculator. Even when the input to the composite function is a variable or an expression, we can often find an expression for the output. To help sort out different cases, let f (x) and g(x) be two different trigonometric functions belonging to the set {sin(x), cos(x), tan(x)} and let f −1(y) and g −1(y) be their inverses. Evaluating Compositions of the Form f (f −1(y )) and f −1(f (x )) For any trigonometric function, f (f −1 (y)) = y for all y in the proper domain for the given function. This follows from the definition of the inverse and from the fact that the range of f was defined to be identical to the domain of f −1. However, we have to be a little more careful with expressions of the form f −1(f (x)). compositions of a trigonometric function and its inverse sin(sin−1 x) = x for −1 ≤ x ≤ 1 cos(cos−1 x) = x for −1 ≤ x ≤ 1 tan(tan−1 x) = x for −∞ < x < ∞ π π __ __ sin−1(sin x) = x only for − ≤ x ≤ 2 2 cos−1(cos x) = x only for 0 ≤ x ≤ π π π __ __ tan−1(tan x) = x only for − < x < 2 2 Q & A… Is it correct that sin−1(sin x) = x? π π __ __ No. This equation is correct if x belongs to the restricted domain  − , but sine is defined for all real input, 2 2 values, and for x outside the restricted interval, the equation is not correct because its inverse always returns a value in π π __ __ . The situation is similar for cosine and tangent and their inverses. For example, sin−1  sin
  −, 2 2 π 3π __ __   =. 4 4 How To… Given an expression of the form f −1(f(θ)) where f(θ) = sin θ, cos θ, or tan θ, evaluate. 1. If θ is in the restricted domain of f, then f −1(f(θ)) = θ. 2. If not, then find an angle ϕ within the restricted domain of f such that f(ϕ) = f(θ). Then f −1(f(θ)) = ϕ. Example 5 Using Inverse Trigonometric Functions Evaluate the following: π __ a. sin−1  sin    3 b. sin−1  sin  2π ___   3 c. cos−1  cos  2π ___   3 d. cos−1  cos  − π __   3 Solution π π π π a. π __ __ __ __ __ , so sin−1  sin  is in  −   =., 3 3 3 2 2 π 2π π π b. 2π ___ __ ___ __ __ , so sin−1  sin   = sin  , but sin  is not in  −, 3 3 2 3 2 2π 2π c. 2π ___ ___ ___ is in [0, π], so cos−1  cos    =. 3 3 3 π is not in [0, π], but cos  − π  because cosine is an even function. π __ d. − π __ __ __  = cos  3 π  − π 3 3 3 __ __   =. 3 3 π 2π __ ___   =. 3 3 is in [0, π], so cos−1  cos SECTION 6.3 inverse trigonometric Functions 547
Try It #5 π __   and tan−1  tan  Evaluate tan−1  tan  8 11π ___ 9  . Evaluating Compositions of the Form f −1(g(x )) Now that we can compose a trigonometric function with its inverse, we can explore how to evaluate a composition of a trigonometric function and the inverse of another trigonometric function. We will begin with compositions of the form f−1(g(x)). For special values of x, we can exactly evaluate the inner function and then the outer, inverse function. However, we can find a more general approach by considering the relation between the two acute angles of a right π __ − θ. Consider the sine and cosine of each angle of the right triangle in triangle where one is θ, making the other 2 Figure 10. π – θ 2 a c θ b Figure 10 Right triangle illustrating the cofunction relationships π b − θ , we have sin−1 (cos θ) = π __ __ __ = sin  Because cos θ = 2 c 2 − θ if 0 ≤ θ ≤ π. If θ is not in this domain, then we need to find another angle that has the same cosine as θ and does belong to the restricted domain; we then subtract π π a π − θ if − π __ __ __ __ __ − θ , so cos−1 (sin θ) = = cos . Similarly, sin θ = this angle from 2 2 c 2 2 π __ ≤ θ ≤. These are just the 2 function-cofunction relationships presented in another way. How To… Given functions of the form sin−1 (cos x) and cos−1 (sin x), evaluate them. π __ 1. If x is in [0, π], then sin−1 (cos x) = − x. 2 2. If x is not in [0, π], then find another angle y in [0, π] such that cos y = cos x. π __ sin−1 (cos x) = − y 2 π π 3. If x is in  − π __ __ __ , then cos−1 (sin
x) = − x., 2 2 2 π , then find another angle y in  − π π 4. If x is not in  − π __ __ __ __  such that sin y = sin x.,, 2 2 2 2 π __ cos−1 (sin x) = − y 2 Example 6 Evaluating the Composition of an Inverse Sine with a Cosine Evaluate sin−1  cos    a. by direct evaluation. 13π ___ 6 b. by the method described previously. Solution a. Here, we can directly evaluate the inside of the composition. cos  13π ___ 6 π _  = cos  + 2π  6 π _  = cos  6 3 √ ____ 2 = — 548 CHAPTER 6 periodic Functions Now, we can evaluate the inverse function as we did earlier. b. We have x = 13π _ 6 π _, y =, and 6 — sin−1  3 π √  = _ ____ 2 3 sin−1  cos  13π ___ Try It #6 Evaluate cos−1  sin  − 11π ___ 4  . Evaluating Compositions of the Form f (g −1(x )) To evaluate compositions of the form f (g −1(x)), where f and g are any two of the functions sine, cosine, or tangent and 1 − x2. When we need to use x is any input in the domain of g −1, we have exact formulas, such as sin(cos−1 x) = √ them, we can derive these formulas by using the trigonometric relations between the angles and sides of a right triangle, together with the use of Pythagorean’s relation between the lengths of the sides. We can use the Pythagorean identity, sin2 x + cos2 x = 1, to solve for one when given the other. We can also use the inverse trigonometric functions to find compositions involving algebraic expressions. — Evaluating the Composition of a Sine with an Inverse Cosine Example 7 4 Find an exact value for sin  cos−1  _  
. 5 4 , which means cos θ = 4 __ __ Solution Beginning with the inside, we can say there is some angle such that θ = cos−1 , 5 5 and we are looking for sin θ. We can use the Pythagorean identity to do this. sin2 θ + cos2 θ = 1 Use our known value for cosine. 4 __  sin2 θ +  5 2 = 1 Solve for sine. sin2 θ = 1 − 16 _ 25 ___ 9 _ sin θ = ± √ 25 = ± 3 __ 5 3 4 __ __  is in quadrant I, sin θ must be positive, so the solution is Since θ = cos−1 . See Figure 11. 5 5 3 5 4 θ 4 3 __ __ Figure 11 Right triangle illustrating that if cos θ =, then sin θ = 5 5 We know that the inverse cosine always gives an angle on the interval [0, π], so we know that the sine of that angle must be positive; therefore sin  cos−1  4 __ 5   = sin θ = 3 __. 5 Try It #7 Evaluate cos  tan−1  5 __ 12  . SECTION 6.3 inverse trigonometric Functions 549 Evaluating the Composition of a Sine with an Inverse Tangent Example 8 7 __  . Find an exact value for sin  tan−1  4 Solution While we could use a similar technique as in Example 6, we will demonstrate a different technique here. 7 __ From the inside, we know there is an angle such that tan θ =. We can envision this as the opposite and adjacent 4 sides on a right triangle, as shown in Figure 12. θ 4 Using the Pythagorean Theorem, we can find the hypotenuse of this triangle. 7 Figure 12 A right triangle with two sides known 42 + 72 = hypotenuse 2 hypotenuse = √ — 65 Now, we can evaluate the sine of the angle as the opposite side divided by the hypotenuse. This gives us our desired composition. sin θ = 7 _ — 65 √ 7 _   = sin θ sin  tan
−1  4 7 _ = — 65 √ = — 65 7 √ ______ 65 Try It #8 7 __  . Evaluate cos  sin−1  9 Finding the Cosine of the Inverse Sine of an Algebraic Expression Example 9 x _ Find a simplified expression for cos  sin−1    for −3 ≤ x ≤ 3. 3 x _ Solution We know there is an angle θ such that sin θ =. 3 sin2 θ + cos2 θ = 1 x _   3 2 + cos2 θ = 1 cos2 θ = 1 − Use the Pythagorean Theorem. Solve for cosine. x2 _ 9 ______ 9 − x2 ______ 9 — 9 − x2 = ± √ _______ 3 π Because we know that the inverse sine must give an angle on the interval  − π __ __ , we can deduce that the cosine of, 2 2 that angle must be positive. cos cos  sin−1  3 — 9 − x2 _______ 3 Try It #9 Find a simplified expression for sin(tan−1 (4x)) for − Access this online resource for additional instruction and practice with inverse trigonometric functions. • evaluate expressions Involving Inverse Trigonometric Functions (http://openstaxcollege.org/l/evalinverstrig) 550 CHAPTER 6 periodic Functions 6.3 SeCTIOn exeRCISeS VeRBAl 1. Why do the functions f (x) = sin−1 x and g(x) = cos−1 x have different ranges? π __ = arcsin(0.5). 3. Explain the meaning of 6 5. Why must the domain of the sine function, sin x, π π __ ___ be restricted to  −  for the inverse sine, 2 2 function to exist? 7. Determine whether the following statement is true or false and explain your answer: arccos(−x) = π − arccos x. AlGeBRAIC For the following exercises, evaluate the expressions. 9. sin−1  − 1 __  2 2 √  ____ 2 — 8. sin−1 �
� 2 11. cos−1  − √  ____ 2 14. tan−1 (−1) — 2. Since the functions y = cos x and y = cos−1 x are π ___ inverse functions, why is cos−1  cos  −   not equal 6 π __ to −? 6 4. Most calculators do not have a key to evaluate sec−1(2). Explain how this can be done using the cosine function or the inverse cosine function. 6. Discuss why this statement is incorrect: arccos (cos x) = x for all x. 1 10. cos−1  __  2 12. tan−1 (1) 15. tan−1  √ — 3  — 3  13. tan−1  − √ 16. tan−1  −1 _  3 √ — For the following exercises, use a calculator to evaluate each expression. Express answers to the nearest hundredth. 17. cos−1 (−0.4) 20. cos−1 (0.8) 18. arcsin(0.23) 21. tan−1 (6) 3 _  19. arccos  5 For the following exercises, find the angle θ in the given right triangle. Round answers to the nearest hundredth. 22. 23. 10 7 12 θ θ 19 For the following exercises, find the exact value, if possible, without a calculator. If it is not possible, explain why. 24. sin−1(cos(π)) 25. tan−1(sin(π)) π _ 27. tan−1  sin    3 30. sin−1  sin  5π _   6 3 _   33. sin  cos−1  5 1 _   36. cos  sin−1  2 28. sin−1  cos  −π _   2 31. tan−1  sin  −5π _ 2   4 _   34. sin  tan−1
 3 π _ 26. cos−1  sin    3 4π _   29. tan−1  sin  3 4 _   32. cos  sin−1  5 35. cos  tan−1  12 __   5 SECTION 6.3 section exercises 551 For the following exercises, find the exact value of the expression in terms of x with the help of a reference triangle. 37. tan(sin−1 (x − 1)) 38. sin(cos−1 (1 − x)) 1 x   39. cos  sin−1  _ 40. cos(tan−1 (3x − 1)) 1 __   41. tan  sin−1  x + 2 exTenSIOnS For the following exercise, evaluate the expression without using a calculator. Give the exact value. 3 2 √ √ 42. sin−1  1 __  − cos−1(1)  + sin−1   − cos−1  _ ___ 2 2 2 3 2 √ 1 √  − sin−1  cos−1   + cos−1  __ ___ _  − sin−1(0) 2 2 2 — — — — For the following exercises, find the function if sin t = x ____. x + 1 43. cos t 44. sec t 45. cot t 46. cos  sin−1  x ____   x + 1 GRAPHICAl 47. tan−1  x _  2x + 1 √ — 48. Graph y = sin−1 x and state the domain and range of 49. Graph y = arccos x and state the domain and range the function. of the function. 50. Graph one cycle of y = tan−1 x and state the domain and range of the function. 52. For what value of x does cos x = cos−1 x? Use a graphing calculator to approximate the answer. ReAl-WORlD APPlICATIOnS 51.
For what value of x does sin x = sin−1 x? Use a graphing calculator to approximate the answer. 53. Suppose a 13-foot ladder is leaning against a 54. Suppose you drive 0.6 miles on a road so that the building, reaching to the bottom of a second-floor window 12 feet above the ground. What angle, in radians, does the ladder make with the building? 55. An isosceles triangle has two congruent sides of length 9 inches. The remaining side has a length of 8 inches. Find the angle that a side of 9 inches makes with the 8-inch side. 57. A truss for the roof of a house is constructed from two identical right triangles. Each has a base of 12 feet and height of 4 feet. Find the measure of the acute angle adjacent to the 4-foot side. 3 _ 59. The line y = − x passes through the origin in the 7 x,y-plane. What is the measure of the angle that the line makes with the negative x-axis? 61. A 20-foot ladder leans up against the side of a building so that the foot of the ladder is 10 feet from the base of the building. If specifications call for the ladder’s angle of elevation to be between 35 and 45 degrees, does the placement of this ladder satisfy safety specifications? vertical distance changes from 0 to 150 feet. What is the angle of elevation of the road? 56. Without using a calculator, approximate the value of arctan(10,000). Explain why your answer is reasonable. 3 __ 58. The line y = x passes through the origin in the x,y5 plane. What is the measure of the angle that the line makes with the positive x-axis? 60. What percentage grade should a road have if the angle of elevation of the road is 4 degrees? (The percentage grade is defined as the change in the altitude of the road over a 100-foot horizontal distance. For example a 5% grade means that the road rises 5 feet for every 100 feet of horizontal distance.) 62. Suppose a 15-foot ladder leans against the side of a house so that the angle of elevation of the ladder is 42 degrees. How far is the foot of the ladder from the side of the house? 552 CHAPTER 6 periodic Functions CHAPTeR 6 ReVIeW Key Terms amplitude the vertical height of a function; the constant A appearing in the definition of a sinusoidal function
arccosine another name for the inverse cosine; arccos x = cos−1 x arcsine another name for the inverse sine; arcsin x = sin−1 x arctangent another name for the inverse tangent; arctan x = tan−1 x inverse cosine function the function cos−1 x, which is the inverse of the cosine function and the angle that has a cosine equal to a given number inverse sine function the function sin−1 x, which is the inverse of the sine function and the angle that has a sine equal to a given number inverse tangent function the function tan−1 x, which is the inverse of the tangent function and the angle that has a tangent equal to a given number midline the horizontal line y = D, where D appears in the general form of a sinusoidal function periodic function a function f (x) that satisfies f (x + P) = f (x) for a specific constant P and any value of x phase shift the horizontal displacement of the basic sine or cosine function; the constant sinusoidal function any function that can be expressed in the form f (x) = Asin(Bx − C) + D or f (x) = Acos(Bx − C) + D C __ B Key equations Sinusoidal functions f (x) = Asin(Bx − C) + D f (x) = Acos(Bx − C) + D Shifted, compressed, and/or stretched tangent function y = A tan(Bx − C) + D Shifted, compressed, and/or stretched secant function y = A sec(Bx − C) + D Shifted, compressed, and/or stretched cosecant function y = A csc(Bx − C) + D Shifted, compressed, and/or stretched cotangent function y = A cot(Bx − C) + D Key Concepts 6.1 Graphs of the Sine and Cosine Functions • Periodic functions repeat after a given value. The smallest such value is the period. The basic sine and cosine functions have a period of 2π. • The function sin x is odd, so its graph is symmetric about the origin. The function cos x is even, so its graph is symmetric about the y-axis. • The graph of a sinusoidal function has the same
general shape as a sine or cosine function. • In the general formula for a sinusoidal function, the period is P = See Example 1. 2π _ ∣B∣ • In the general formula for a sinusoidal function, ∣A∣ represents amplitude. If ∣A∣ > 1, the function is stretched, whereas if ∣A∣ < 1, the function is compressed. See Example 2. • The value in the general formula for a sinusoidal function indicates the phase shift. See Example 3. C __ B • The value D in the general formula for a sinusoidal function indicates the vertical shift from the midline. See Example 4. • Combinations of variations of sinusoidal functions can be detected from an equation. See Example 5. • The equation for a sinusoidal function can be determined from a graph. See Example 6 and Example 7. • A function can be graphed by identifying its amplitude and period. See Example 8 and Example 9. • A function can also be graphed by identifying its amplitude, period, phase shift, and horizontal shift. See Example 10. • Sinusoidal functions can be used to solve real-world problems. See Example 11, Example 12, and Example 13. CHAPTER 6 review 553 6.2 Graphs of the Other Trigonometric Functions • The tangent function has period π. • f (x) = Atan(Bx − C) + D is a tangent with vertical and/or horizontal stretch/compression and shift. See Example 1, Example 2, and Example 3. • The secant and cosecant are both periodic functions with a period of 2π. f (x) = Asec(Bx − C) + D gives a shifted, compressed, and/or stretched secant function graph. See Example 4 and Example 5. • f (x) = Acsc(Bx − C) + D gives a shifted, compressed, and/or stretched cosecant function graph. See Example 6 and Example 7. • The cotangent function has period π and vertical asymptotes at 0, ±π, ±2π,... • The range of cotangent is (−∞, ∞), and the function is decreasing at each point in its range. • The cotangent is zero at ± π __ 2 • f (x) = Acot(
Bx − C) + D is a cotangent with vertical and/or horizontal stretch/compression and shift. See Example, ± 3π __ 2,... 8 and Example 9. • Real-world scenarios can be solved using graphs of trigonometric functions. See Example 10. 6.3 Inverse Trigonometric Functions • An inverse function is one that “undoes” another function. The domain of an inverse function is the range of the original function and the range of an inverse function is the domain of the original function. • Because the trigonometric functions are not one-to-one on their natural domains, inverse trigonometric functions are defined for restricted domains. • For any trigonometric function f (x), if x = f−1(y), then f (x) = y. However, f (x) = y only implies x = f−1(y) if x is in the restricted domain of f. See Example 1. • Special angles are the outputs of inverse trigonometric functions for special input values; for example, 1 π π __ __ __ . See Example 2. = sin−1  = tan−1(1) and 2 6 4 • A calculator will return an angle within the restricted domain of the original trigonometric function. See Example 3. • Inverse functions allow us to find an angle when given two sides of a right triangle. See Example 4. • In function composition, if the inside function is an inverse trigonometric function, then there are exact expressions; for example, sin(cos−1 (x)) = √ 1 − x2. See Example 5. — π _ • If the inside function is a trigonometric function, then the only possible combinations are sin−1 (cos x) = − x if and cos−1 (sin x) = ≤ x ≤ − x if −. See Example 6 and Example 7. 2 2 2 • When evaluating the composition of a trigonometric function with an inverse trigonometric function, draw a reference triangle to assist in determining the ratio of sides that represents the output of the trigonometric function. See Example 8. • When evaluating the composition of a trigonometric function with an inverse trigonometric function, you may use trig identities to assist in determining the ratio of sides. See Example 9. 554 CHAPTER 6 periodic Functions CHAPTeR 6 ReVIeW exe
RCISeS GRAPHS OF THe SIne AnD COSIne FUnCTIOnS For the following exercises, graph the functions for two periods and determine the amplitude or stretching factor, period, midline equation, and asymptotes. 1. f (x) = −3cos x + 3 2π _ 4. f (x) = −2sin . f (x) = 6sin  3x − 6 1 _ 2. f (x) = sin x 4 π _ 5. f (x) = 3sin  x −  − 4 4 8. f (x) = −100sin(50x − 20) GRAPHS OF THe OTHeR TRIGOnOMeTRIC FUnCTIOnS π _  3. f (x) = 3cos  x + 6 4π _  + 1  6. f (x) = 2  cos  x − 3 For the following exercises, graph the functions for two periods and determine the amplitude or stretching factor, period, midline equation, and asymptotes. 9. f (x) = tan x − 4 12. f (x) = 0.2cos(0.1x) + 0.3 π _ 10. f (x) = 2tan  x −  6 11. f (x) = −3tan(4x) − 2 For the following exercises, graph two full periods. Identify the period, the phase shift, the amplitude, and asymptotes. 1 _ 13. f (x) = sec x 3 1 _ x  16. f (x) = 8sec  4 14. f (x) = 3cot x 15. f (x) = 4csc(5x) 2 1 _ __ x  csc  17. f (x) = 3 2 18. f (x) = −csc(2x + π) For the following exercises, use this scenario: The population of a city has risen and fallen over a 20-year interval. Its population may be modeled by the following function: y = 12,000 + 8,000sin(0.628x), where the domain is the years since 1980 and the range is the population of the
city. 19. What is the largest and smallest population the city 20. Graph the function on the domain of [0, 40]. may have? 21. What are the amplitude, period, and phase shift for 22. Over this domain, when does the population reach the function? 18,000? 13,000? 23. What is the predicted population in 2007? 2010? For the following exercises, suppose a weight is attached to a spring and bobs up and down, exhibiting symmetry. 24. Suppose the graph of the displacement function is shown in Figure 1, where the values on the x-axis represent the time in seconds and the y-axis represents the displacement in inches. Give the equation that models the vertical displacement of the weight on the spring. y 5 4 3 2 1 –1 –2 –3 –4 –5 2 4 68 1 0 x Figure 1 CHAPTER 6 review 555 25. At time = 0, what is the displacement of the weight? 26. At what time does the displacement from the equilibrium point equal zero? 27. What is the time required for the weight to return to its initial height of 5 inches? In other words, what is the period for the displacement function? InVeRSe TRIGOnOMeTRIC FUnCTIOnS For the following exercises, find the exact value without the aid of a calculator. 28. sin−1(1) 1 _  31. cos−1  — 2 √ 34. cos−1  tan  3π _   4 37. tan  cos−1  5 _ 13   29. cos−1  — 3 √  _ 2 32. sin− 35. sin  sec−1    5 38. sin  cos−1  x _   x + 1 30. tan−1(−1) π _ 33. sin−1  cos    6 3 _ 36. cot  sin−1    5 39. Graph f (x) = cos x and f (x) = sec x on the interval [0, 2π) and explain any observations. 40. Graph f (x) = sin x and f (x) =
csc x and explain any observations. x _ 41. Graph the function f (x) = − 1 x3 _ 3! x5 _ 5! x7 _ 7! + − on the interval [−1, 1] and compare the graph to the graph of f (x) = sin x on the same interval. Describe any observations. 556 CHAPTER 6 periodic Functions CHAPTeR 6 PRACTICe TeST For the following exercises, sketch the graph of each function for two full periods. Determine the amplitude, the period, and the equation for the midline. 1. f (x) = 0.5sin x 4. f (x) = sin(3x) 1 _ 7. f (x) = 3cos  x − 3 5π _  6 2. f (x) = 5cos x 3. f (x) = 5sin x π _  + 1 5. f (x) = −cos  x + 3 8. f (x) = tan(4x) π _   + 4 6. f (x) = 5sin  3  x − 6 9. f (x) = −2tan  x − 7π _  + 2 6 π _ 12. f (x) = πsec  x  2 10. f (x) = πcos(3x + π) 11. f (x) = 5csc(3x) π _  − 3 13. f (x) = 2csc  x + 4 For the following exercises, determine the amplitude, period, and midline of the graph, and then find a formula for the function. 14. Give in terms of a sine function. y 15. Give in terms of a sine function. y 16. Give in terms of a tangent function. y 6 4 2 6 4 2 12 8 4 –3 –2 –1 1 2 3 x –3 –2 –1 1 2 3 x – 3π 2 –π – π 2 π 2 π 3π 2 x –2 –4 –6 –2 –4 –6 –4 –8 –12 For the following exercises, find the amplitude, period, phase shift, and midline. π _ x + π
 − 3 17. y = sin  6 18. y = 8sin  7π _ 6 x + 7π _  + 6 2 19. The outside temperature over the course of a day can be modeled as a sinusoidal function. Suppose you know the temperature is 68°F at midnight and the high and low temperatures during the day are 80°F and 56°F, respectively. Assuming t is the number of hours since midnight, find a function for the temperature, D, in terms of t. 20. Water is pumped into a storage bin and empties according to a periodic rate. The depth of the water is 3 feet at its lowest at 2:00 a.m. and 71 feet at its highest, which occurs every 5 hours. Write a cosine function that models the depth of the water as a function of time, and then graph the function for one period. For the following exercises, find the period and horizontal shift of each function. 21. g(x) = 3tan(6x + 42) 22. n(x) = 4csc  5π _ 3 x − 20π _  3 23. Write the equation for the graph in Figure 1 in terms of the secant function and give the period and phase shift. y 6 4 2 –3 –2 –1 1 2 3 x –2 –4 –6 Figure 1 CHAPTER 6 practice test 557 24. If tan x = 3, find tan(−x). 25. If sec x = 4, find sec(−x). For the following exercises, graph the functions on the specified window and answer the questions. 26. Graph m(x) = sin(2x) + cos(3x) on the viewing window [−10, 10] by [−3, 3]. Approximate the graph’s period. 28. Graph f (x) = observations. sin x _____ x on [−0.5, 0.5] and explain any 27. Graph n(x) = 0.02sin(50πx) on the following domains in x: [0, 1] and [0, 3]. Suppose this function models sound waves. Why would these views look so different? 3 __ For the following exercises, let f (x) = cos(6x). 5 29. What is the largest possible value for f (x)? 31. Where is the function increasing on
the interval [0, 2π]? 30. What is the smallest possible value for f (x)? For the following exercises, find and graph one period of the periodic function with the given amplitude, period, and phase shift. π _ 32. Sine curve with amplitude 3, period, and phase 3 π _, 2  shift (h, k) =  4 π _ 33. Cosine curve with amplitude 2, period, and phase 6 shift (h, k) =  − π _ 4, 3  For the following exercises, graph the function. Describe the graph and, wherever applicable, any periodic behavior, amplitude, asymptotes, or undefined points. 34. f (x) = 5cos(3x) + 4sin(2x) 35. f (x) = esint — For the following exercises, find the exact value. 3  36. sin−1  √ _ 2 39. cos−1(sin(π)) 37. tan−1  √ — 3  40. cos−1  tan  7π _   4 42. cos−1(−0.4) 43. cos(tan−1(x2)) For the following exercises, suppose sin t = x _. x + 1 44. tan t 45. csc t 46. Given Figure 2, find the measure of angle θ to three decimal places. Answer in radians. For the following exercises, determine whether the equation is true or false. 47. arcsin  sin  5π _   = 6 5π _ 6 48. arccos  cos  5π _   = 6 5π _ 6 — 3  38. cos−1  − √ _ 2 41. cos(sin−1(1 − 2x)) 12 θ 19 49. The grade of a road is 7%. This means that for every horizontal distance of 100 feet on the road, the vertical rise is 7 feet. Find the angle the road makes with the horizontal in radians. Trigonometric Identities and Equations 7 Figure 1 A sine wave models disturbance. (credit: modification of work by Mikael Altemark, Flickr). CHAPTeR OUTlI
ne 7.1 Solving Trigonometric equations with Identities 7.2 Sum and Difference Identities 7.3 Double-Angle, Half-Angle, and Reduction Formulas 7.4 Sum-to-Product and Product-to-Sum Formulas 7.5 Solving Trigonometric equations 7.6 Modeling with Trigonometric equations Introduction Math is everywhere, even in places we might not immediately recognize. For example, mathematical relationships describe the transmission of images, light, and sound. The sinusoidal graph in Figure 1 models music playing on a phone, radio, or computer. Such graphs are described using trigonometric equations and functions. In this chapter, we discuss how to manipulate trigonometric equations algebraically by applying various formulas and trigonometric identities. We will also investigate some of the ways that trigonometric equations are used to model real-life phenomena. 559 560 CHAPTER 7 trigonometric identities and eQuations leARnInG OBjeCTIVeS In this section, you will: • • Verify the fundamental trigonometric identities. Simplify trigonometric expressions using algebra and the identities. 7.1 SOlVInG TRIGOnOMeTRIC eQUATIOnS WITH IDenTITIeS Figure 1 International passports and travel documents In espionage movies, we see international spies with multiple passports, each claiming a different identity. However, we know that each of those passports represents the same person. The trigonometric identities act in a similar manner to multiple passports—there are many ways to represent the same trigonometric expression. Just as a spy will choose an Italian passport when traveling to Italy, we choose the identity that applies to the given scenario when solving a trigonometric equation. In this section, we will begin an examination of the fundamental trigonometric identities, including how we can verify them and how we can use them to simplify trigonometric expressions. Verifying the Fundamental Trigonometric Identities Identities enable us to simplify complicated expressions. They are the basic tools of trigonometry used in solving trigonometric equations, just as factoring, finding common denominators, and using special formulas are the basic tools of solving algebraic equations. In fact, we use algebraic techniques constantly to simplify trigonometric expressions. Basic properties and formulas of algebra, such as the difference of squares formula and the perfect squares formula, will simplify the work involved with trigonometric expressions and equations. We already know that
all of the trigonometric functions are related because they all are defined in terms of the unit circle. Consequently, any trigonometric identity can be written in many ways. To verify the trigonometric identities, we usually start with the more complicated side of the equation and essentially rewrite the expression until it has been transformed into the same expression as the other side of the equation. Sometimes we have to factor expressions, expand expressions, find common denominators, or use other algebraic strategies to obtain the desired result. In this first section, we will work with the fundamental identities: the Pythagorean identities, the even-odd identities, the reciprocal identities, and the quotient identities. We will begin with the Pythagorean identities (see Table 1), which are equations involving trigonometric functions based on the properties of a right triangle. We have already seen and used the first of these identifies, but now we will also use additional identities. sin2 θ + cos2 θ = 1 Pythagorean Identities 1 + cot2 θ = csc2 θ Table 1 1 + tan2 θ = sec2 θ SECTION 7.1 solving trigonometric eQuations with identities 561 The second and third identities can be obtained by manipulating the first. The identity 1 + cot2 θ = csc2 θ is found by rewriting the left side of the equation in terms of sine and cosine. Prove: 1 + cot2 θ = csc2 θ Rewrite the left side. Write both terms with the common denominator. cos2 θ  1 + cot2 θ =  1 + _____ sin2 θ cos2 θ sin2 θ   +  _____ _____ sin2 θ sin2 θ sin2 θ + cos2 θ ___________ sin2 θ =  = = 1 ____ sin2 θ = csc2 θ Similarly, 1 + tan2 θ = sec2 θ can be obtained by rewriting the left side of this identity in terms of sine and cosine. This gives Rewrite left side. Write both terms with the common denominator. 2 sin θ 1 + tan2 θ = 1 +   ____ cos θ 2 2 cos θ sin θ   ____ ____ cos θ cos
θ cos2 θ + sin2 θ ___________ cos2 θ +  =  = = 1 _____ cos2 θ = sec2 θ The next set of fundamental identities is the set of even-odd identities. The even-odd identities relate the value of a trigonometric function at a given angle to the value of the function at the opposite angle and determine whether the identity is odd or even. (See Table 2). tan(−θ) = −tan θ cot(−θ) = −cot θ Even-Odd Identities sin(−θ) = −sin θ csc(−θ) = −csc θ Table 2 cos(−θ) = cos θ sec(−θ) = sec θ Recall that an odd function is one in which f (−x) = −f (x) for all x in the domain of f. The sine function is an odd function because sin(−θ) = −sin θ. The graph of an odd function is symmetric about the origin. For example, consider  is opposite the output of sin  − π π π π __ __ __ __ and − . Thus,. The output of sin  corresponding inputs of 2 2 2 2 π __  = 1 sin  2 π sin  − π __ __  = −sin   2 2 = −1 and This is shown in Figure 2. y 2 π, 12 f(x) = sin x x π 2π –2π – π, 2 –π –1 –2 Figure 2 Graph of y = sin θ Recall that an even function is one in which f (−x) = f (x) for all x in the domain of f 562 CHAPTER 7 trigonometric identities and eQuations The graph of an even function is symmetric about the y-axis. The cosine function is an even function because π π π __ __ __ cos(−θ) = cos θ. For example, consider corresponding inputs and −  is the same as the. The output of cos  4 4 4 output of cos  − π __ . Thus, 4 π cos  −
π __ __  = cos   4 4 ≈ 0.707 See Figure 3. y – π, 4 0.707 2 π, 4 0.707 –2π –π –2 x π 2π f(x) = cos x For all θ in the domain of the sine and cosine functions, respectively, we can state the following: Figure 3 Graph of y = cos θ • Since sin(−θ) = −sin θ, sine is an odd function. • Since, cos(−θ) = cos θ, cosine is an even function. The other even-odd identities follow from the even and odd nature of the sine and cosine functions. For example, consider the tangent identity, tan(−θ) = −tan θ. We can interpret the tangent of a negative angle as = −tan θ. Tangent is therefore an odd function, which means that tan(−θ) = −tan(θ) tan(−θ) = sin(−θ) ______ = cos(−θ) −sin θ ______ cos θ for all θ in the domain of the tangent function. The cotangent identity, cot(−θ) = −cot θ, also follows from the sine and cosine identities. We can interpret the = −cot θ. Cotangent is therefore an odd function, cotangent of a negative angle as cot(−θ) = which means that cot(−θ) = −cot(θ) for all θ in the domain of the cotangent function. cos(−θ) _______ = sin(−θ) cos θ _____ −sin θ The cosecant function is the reciprocal of the sine function, which means that the cosecant of a negative angle will be interpreted as csc(−θ) = = −csc θ. The cosecant function is therefore odd. 1 ______ = sin(−θ) 1 _____ −sin θ Finally, the secant function is the reciprocal of the cosine function, and the secant of a negative angle is interpreted as sec(−θ) = = sec θ. The secant function is therefore even. 1 ______ = cos(−�
�) 1 ____ cos θ To sum up, only two of the trigonometric functions, cosine and secant, are even. The other four functions are odd, verifying the even-odd identities. The next set of fundamental identities is the set of reciprocal identities, which, as their name implies, relate trigonometric functions that are reciprocals of each other. See Table 3. Reciprocal Identities 1 ____ csc θ 1 ____ sec θ 1 ____ cot θ csc θ = sec θ = cot θ = 1 ____ sin θ 1 ____ cos θ 1 ____ tan θ sin θ = cos θ = tan θ = The final set of identities is the set of quotient identities, which define relationships among certain trigonometric functions and can be very helpful in verifying other identities. See Table 4. Table 3 Quotient Identities tan θ = sin θ ____ cos θ cot θ = cos θ ____ sin θ Table 4 SECTION 7.1 solving trigonometric eQuations with identities 563 The reciprocal and quotient identities are derived from the definitions of the basic trigonometric functions. summarizing trigonometric identities The Pythagorean identities are based on the properties of a right triangle. cos2 θ + sin2 θ = 1 1 + cot2 θ = csc2 θ 1 + tan2 θ = sec2 θ The even-odd identities relate the value of a trigonometric function at a given angle to the value of the function at the opposite angle. tan(−θ) = −tan θ cot(−θ) = −cot θ sin(−θ) = −sin θ csc(−θ) = −csc θ cos(−θ) = cos θ sec(−θ) = sec θ The reciprocal identities define reciprocals of the trigonometric functions. sin θ = cos θ = tan θ = csc θ = sec θ = cot θ = 1 ____ csc θ 1 ____ sec θ 1 ____ cot θ 1 ____ sin θ 1 ____ cos θ 1 ____ tan θ The quotient identities define the relationship among the trigonometric functions. tan θ = cot θ = sin θ ____
cos θ cos θ ____ sin θ Example 1 Graphing the Equations of an Identity Graph both sides of the identity cot θ =. In other words, on the graphing calculator, graph y = cot θ and y = 1 ____ tan θ 1 ____. tan θ Solution See Figure 4. y = cot θ = 1 tan θ y 10 6 2 – 5π 2 – 3π 2 π 2– π 2 3π 2 5π 2 θ –6 –10 Figure 4 564 CHAPTER 7 trigonometric identities and eQuations Analysis We see only one graph because both expressions generate the same image. One is on top of the other. This is a good way to prove any identity. If both expressions give the same graph, then they must be identities. How To… Given a trigonometric identity, verify that it is true. 1. Work on one side of the equation. It is usually better to start with the more complex side, as it is easier to simplify than to build. 2. Look for opportunities to factor expressions, square a binomial, or add fractions. 3. Noting which functions are in the final expression, look for opportunities to use the identities and make the proper substitutions. 4. If these steps do not yield the desired result, try converting all terms to sines and cosines. Example 2 Verifying a Trigonometric Identity Verify tan θ cos θ = sin θ. Solution We will start on the left side, as it is the more complicated side: tan θcos θ =  sin θ ____ cos θ sin θ ____ cos θ  cos θ  cos θ =  = sin θ Analysis This identity was fairly simple to verify, as it only required writing tan θ in terms of sin θ and cos θ. Try It #1 Verify the identity csc θ cos θ tan θ = 1. Example 3 Verifying the Equivalency Using the Even-Odd Identities Verify the following equivalency using the even-odd identities: (1 + sin x)[1 + sin(−x)] = cos2 x Solution Working on the left side of the equation, we have (1 + sin x)[1 + sin(−x)] = (1 + sin x)(1 − sin x) Example 4 Verifying a Tr
igonometric Identity Involving sec2 θ = 1 − sin2 x = cos2 x Since sin(−x) = −sin x Difference of squares cos2 x = 1 − sin2 x Verify the identity sec2 θ − 1 ________ sec2 θ = sin2 θ Solution As the left side is more complicated, let’s begin there. sec2 θ − 1 ________ = sec2 θ (tan2 θ + 1) − 1 _____________ sec2 θ sec2 θ = tan2 θ + 1 = tan2 θ _____ sec2 θ 1 ____  = tan2 θ  sec2 θ = tan2 θ(cos2 θ) =  =  = sin2 θ  (cos2 θ)  (cos2 θ) sin2 θ _____ cos2 θ sin2 θ _____ cos2 θ cos2 θ = tan2 θ = 1 ____ sec2 θ sin2 θ _____ cos2 θ SECTION 7.1 solving trigonometric eQuations with identities 565 There is more than one way to verify an identity. Here is another possibility. Again, we can start with the left side. sec2 θ _____ sec2 θ = 1 − cos2 θ = sin2 θ sec2 θ − 1 ________ = sec2 θ − 1 ____ sec2 θ In the first method, we used the identity sec2 θ = tan2 θ + 1 and continued to simplify. In the second method, Analysis we split the fraction, putting both terms in the numerator over the common denominator. This problem illustrates that there are multiple ways we can verify an identity. Employing some creativity can sometimes simplify a procedure. As long as the substitutions are correct, the answer will be the same. Try It #2 Show that cot θ ____ csc θ = cos θ. Example 5 Creating and Verifying an Identity Create an identity for the expression 2tan θ sec θ by rewriting strictly in terms of sine. Solution There are a number of ways to begin, but here we will use the quotient and reciprocal identities to rewrite the expression: 2 tan θ sec θ = 2  sin θ   ____ cos �
� 1 ____  cos θ Thus, = 2 sin θ _____ cos2 θ = 2 sin θ ________ 1 − sin2 θ 2tan θ sec θ = 2 sin θ ________ 1 − sin2 θ Substitute 1 − sin2 θ for cos2 θ Example 6 Verifying an Identity Using Algebra and Even/Odd Identities Verify the identity: sin2(−θ) − cos2(−θ) _________________ sin(−θ) − cos(−θ) = cos θ − sin θ Solution Let’s start with the left side and simplify: sin2(−θ) − cos2(−θ) _________________ = sin(−θ) − cos(−θ) [sin(−θ)]2 − [cos(−θ)]2 __________________ sin(−θ) − cos(−θ) sin(−x) = −sin x and cos(−x) = cos x = (−sin θ)2 − (cos θ)2 ________________ −sin θ − cos θ = (sin θ)2 − (cos θ)2 −sin θ − cos θ ______________ Difference of squares = (sin θ − cos θ)(sin θ + cos θ) ______________________ −(sin θ + cos θ) = (sin θ − cos θ)(sin θ + cos θ) ______________________ −(sin θ + cos θ) = cos θ − sin θ Try It #3 Verify the identity sin2 θ − 1 _____________ tan θ sin θ − tan θ = sin θ + 1 _______. tan θ 566 CHAPTER 7 trigonometric identities and eQuations Example 7 Verifying an Identity Involving Cosines and Cotangents Verify the identity: (1 − cos2 x)(1 + cot2 x) = 1. Solution We will work on the left side of the equation (1 − cos2 x)(1 + cot2 x) = (1 − cos2 x)  1 + sin2 x _____ sin2 x = (1 − cos2 x)  cos2 x  _____ sin2 x cos2 x  _____ sin2 x
sin2 x + cos2 x  ___________ sin2 x + = (1 − cos2 x)  1 ____  = (sin2 x)  sin2 x = 1 Find the common denominator. Using Algebra to Simplify Trigonometric expressions We have seen that algebra is very important in verifying trigonometric identities, but it is just as critical in simplifying trigonometric expressions before solving. Being familiar with the basic properties and formulas of algebra, such as the difference of squares formula, the perfect square formula, or substitution, will simplify the work involved with trigonometric expressions and equations. For example, the equation (sin x + 1)(sin x − 1) = 0 resembles the equation (x + 1)(x − 1) = 0, which uses the factored form of the difference of squares. Using algebra makes finding a solution straightforward and familiar. We can set each factor equal to zero and solve. This is one example of recognizing algebraic patterns in trigonometric expressions or equations. Another example is the difference of squares formula, a2 − b2 = (a − b)(a + b), which is widely used in many areas other than mathematics, such as engineering, architecture, and physics. We can also create our own identities by continually expanding an expression and making the appropriate substitutions. Using algebraic properties and formulas makes many trigonometric equations easier to understand and solve. Example 8 Writing the Trigonometric Expression as an Algebraic Expression Write the following trigonometric expression as an algebraic expression: 2cos2 θ + cos θ − 1. Solution Notice that the pattern displayed has the same form as a standard quadratic expression, ax2 + bx + c. Letting cos θ = x, we can rewrite the expression as follows: 2x2 + x − 1 This expression can be factored as (2x − 1)(x + 1). If it were set equal to zero and we wanted to solve the equation, we would use the zero factor property and solve each factor for x. At this point, we would replace x with cos θ and solve for θ. Example 9 Rewriting a Trigonometric Expression Using the Difference of Squares Rewrite the trigonometric expression: 4 cos2 θ − 1. Solution Notice that both the coefficient and the trigonometric expression in the first term are squared, and the square of the number 1 is 1
. This is the difference of squares. Thus, 4 cos2 θ − 1 = (2 cos θ)2 − 1 = (2 cos θ − 1)(2 cos θ + 1) Analysis If this expression were written in the form of an equation set equal to zero, we could solve each factor using the zero factor property. We could also use substitution like we did in the previous problem and let cos θ = x, rewrite the expression as 4x2 − 1, and factor (2x − 1)(2x + 1). Then replace x with cos θ and solve for the angle. Try It #4 Rewrite the trigonometric expression: 25 − 9 sin2 θ. SECTION 7.1 solving trigonometric eQuations with identities 567 Example 10 Simplify by Rewriting and Using Substitution Simplify the expression by rewriting and using identities: Solution We can start with the Pythagorean Identity. Now we can simplify by substituting 1 + cot2 θ for csc2 θ. We have 1 + cot2 θ = csc2 θ csc2 θ − cot2 θ csc2 θ − cot2 θ = 1 + cot2 θ − cot2 θ = 1 Try It #5 Use algebraic techniques to verify the identity: cos θ _______ = 1 + sin θ 1 − sin θ _______ cos θ. (Hint: Multiply the numerator and denominator on the left side by 1 − sin θ.) Access these online resources for additional instruction and practice with the fundamental trigonometric identities. • Fundamental Trigonometric Identities (http://openstaxcollege.org/l/funtrigiden) • Verifying Trigonometric Identities (http://openstaxcollege.org/l/verifytrigiden) 568 CHAPTER 7 trigonometric identities and eQuations 7.1 SeCTIOn exeRCISeS VeRBAl 1. We know g(x) = cos x is an even function, and f (x) = sin x and h(x) = tan x are odd functions. What about G(x) = cos2 x, F (x) = sin2 x, and H(x) = tan2 x? Are they even, odd, or neither? Why? 2. Examine the graph of f (x
) = sec x on the interval [−π, π]. How can we tell whether the function is even or odd by only observing the graph of f (x) = sec x? 3. After examining the reciprocal identity for sec t, explain why the function is undefined at certain points. 4. All of the Pythagorean identities are related. Describe how to manipulate the equations to get from sin2 t + cos2 t = 1 to the other forms. AlGeBRAIC For the following exercises, use the fundamental identities to fully simplify the expression. 5. sin x cos x sec x 7. tan x sin x + sec x cos2 x 9. cot t + tan t __________ sec(−t) 11. −tan(−x)cot(−x) 6. sin(−x)cos(−x)csc(−x) 8. csc x + cos x cot(−x) 10. 3 sin3 t csc t + cos2 t + 2 cos(−t)cos t 12. −sin(−x)cos x sec x csc x tan x ________________________ cot x + sin2 θ + 1 ____ sec2 θ 13. 1 + tan2 θ ________ csc2 θ 15. 1 − cos2 x ________ tan2 x + 2 sin2 x 14.  tan x ____ csc2 x + tan x ____   sec2 x 1 + tan x ________  − 1 + cot x 1 _____ cos2 x For the following exercises, simplify the first trigonometric expression by writing the simplified form in terms of the second expression. 16. tan x + cot x __________ csc x ; cos x 17. sec x + csc x __________ ; sin x 1 + tan x 18. cos x _______ 1 + sin x + tan x; cos x 19. 1 ________ sin x cos x − cot x; cot x 20. 1 _______ − 1 − cos x cos x _______ 1 + cos x ; csc x 21. (sec x + csc x)(sin x + cos x) − 2 − cot x; tan x 22. 1 __________ csc x − sin x ; sec x and tan x 23. 1 − sin x _______ − 1 + sin x 1 + sin x _______ 1
− sin x ; sec x and tan x 24. tan x; sec x 26. sec x; sin x 28. cot x; csc x 25. sec x; cot x 27. cot x; sin x For the following exercises, verify the identity. 29. cos x − cos3 x = cos x sin2 x 30. cos x(tan x − sec(−x)) = sin x − 1 31. 1 + sin2 x ________ = cos2 x 1 _____ cos2 x + sin2 x _____ cos2 x = 1 + 2 tan2 x 33. cos2 x − tan2 x = 2 − sin2 x − sec2 x 32. (sin x + cos x)2 = 1 + 2 sin x cos x SECTION 7.1 section exercises 569 exTenSIOnS For the following exercises, prove or disprove the identity. 34. 1 _______ − 1 + cos x 1 ___________ 1 − cos( − x) = −2 cot x csc x 35. csc2 x(1 + sin2 x) = cot2 x 36.  sec2(−x) − tan2 x   ______________ tan x 2 + 2 tan x _________ 2 + 2 cot x  − 2 sin2 x = cos 2x 37. tan x ____ sec x sin(−x) = cos2 x 38. sec(−x) __________ tan x + cot x = −sin(−x) 39. 1 + sin x _______ = cos x cos x _________ 1 + sin(−x) For the following exercises, determine whether the identity is true or false. If false, find an appropriate equivalent expression. 40. cos2 θ − sin2 θ ___________ 1 − tan2 θ = sin2 θ 42. sec θ + tan θ __________ cot θ + cos θ = sec2 θ 41. 3 sin2 θ + 4 cos2 θ = 3 + cos2 θ 570 CHAPTER 7 trigonometric identities and eQuations leARnInG OBjeCTIVeS In this section, you will: • • • • • Use sum and difference formulas for cosine. Use sum and difference formulas for sine. Use sum and difference formulas for tangent. Use sum and difference formulas for co
functions. Use sum and difference formulas to verify identities. 7. 2 SUM AnD DIFFeRenCe IDenTITIeS Figure 1 Mount McKinley, in Denali National Park, Alaska, rises 20,237 feet (6,168 m) above sea level. It is the highest peak in North America. (credit: Daniel A. Leifheit, Flickr) How can the height of a mountain be measured? What about the distance from Earth to the sun? Like many seemingly impossible problems, we rely on mathematical formulas to find the answers. The trigonometric identities, commonly used in mathematical proofs, have had real-world applications for centuries, including their use in calculating long distances. The trigonometric identities we will examine in this section can be traced to a Persian astronomer who lived around 950 AD, but the ancient Greeks discovered these same formulas much earlier and stated them in terms of chords. These are special equations or postulates, true for all values input to the equations, and with innumerable applications. In this section, we will learn techniques that will enable us to solve problems such as the ones presented above. The formulas that follow will simplify many trigonometric expressions and equations. Keep in mind that, throughout this section, the term formula is used synonymously with the word identity. Using the Sum and Difference Formulas for Cosine Finding the exact value of the sine, cosine, or tangent of an angle is often easier if we can rewrite the given angle in terms of two angles that have known trigonometric values. We can use the special angles, which we can review in the unit circle shown in Figure 2.,, 1 2 135° 3π 4 120° 2π 3 90° (0, 1) π 2 60° π 3, 1 2 150° 5π 6 (–1, 0) 180° π, 1 2, 210° 7π 6 225° 5π 4, 1 2, 45° π 4 π 6 30°, 1 2 0°, (1, 0) 330°, 1 2 2π 11π 6 7π 4 315°,,1 2 3π 2 4π 3 5π 3 240° 270° (0, –1) Figure 2 The Unit Circle 300°,1 2 SECTION 7.2 sum and diFFerence identities 571 We will begin with the sum and difference formulas for cosine, so that we can find the cosine of a given angle if

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