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C In Chapter 3, you studied exponential and logarithmic functions, which are two types of transcendental functions. C In Chapter 4, you will: ■ Use trigonometric functions to solve right triangles. ■ Find values of trigonometric functions for any angle. C SATELLITE NAVIGATION Satellite navigation systems operate by receiving signals from satellites in orbit, determining the distance to each of the satellites, and then using trigonometry to establish the location on Earth's surface. These techniques are also used when navigating cars, planes, ships, and spacecraft. ■ Graph trigonometric and inverse trigonometric functions. PREREAD Use the prereading strategy of previewing to make two or three predictions of what Chapter 4 is about. vConnectED.mcgraw-hill.com Animation Vocabulary eGlossary Personal Tutor Graphing Calculator Self-Check Practice Worksheets Get Ready for the Chapter Diagnose Readiness You have tw o options for checking Prerequisite Skills. NewVocabulary English ^ Textbook Option Take the Quick Check below. QuickCheck Find the missing value in each figure. (Prerequisite Skill) 3. 4. Determine whether each of the following could represent the measures of three sides of a triangle. Write yes or no. (Prerequisite Skill) 4 ,8 ,1 2 5. 7. 6. 1 2,15 ,18 ALGEBRA The sides of a triangle have lengths x, x + 17, and 25. If the length of the longest side is 25, what value of x makes the triangle a right triangle? (Prerequisite Skill) Find the equations of any vertical or horizontal asymptotes. 8. f(x) = 10. f(x) = 12. h(x) = 2 x '-4 x2 + 8 x (x - 1 )2 (x — 2)(x + 4) x 2 + x - 20 x+ 5 9. h(x) = 11. g(x) = 13. f(x) = (Lesson 2-5) x 3 - 27 x+ 5 x+ 5 ( x - 3 ) ( x - 5) Espanol trignom etric functions p. 220 funciones trigonom etricas sine p. 220 seno cosine p. 220 coseno tangent p. 220 tangente cosecant p. 220 cosecant secant p. 220 secant cotangent p. 220 funcion reciproca reciprocal function p. 220 cotangente inverse sine p. 223 seno inverso inverse cosine p. 223 coseno inverso inverse tangent p. 223 tangente inversa radian p. 232 radian coterminal angles p. 234 angulos coterminales reference angle p. 244 angulo de referencia unit circle p. 247 circuio de unidad circular function p. 248 funcion circular period p. 250 periodo sinusoid p. 256 sinusoid am plitude p. 257 am plitud frequency p. 260 frecuencia phase shift p. 261 cambio de fase Law of Sines p. 291 ley de senos Law of Cosines p. 295 ley de cosenos ReviewVocabulary reflection p. 48 reflexion the mirror image of the graph of a function with respect to a specific line dilation p. 49 homotecia a nonrigid transformation that has the effect of compressing (shrinking) or expanding (enlarging) the graph of a function vertically or horizontally 2 x 2 + 5x — 12 2x — 3 Online Option Take an online self-check Chapter Readiness Quiz at connectE D .m caraw -hill.com . 219 • You evaluated functions. (Lesson 1-1) Large helium-filled balloons are a tradition of many holiday parades. Long cables attached to the balloon are used by volunteers to lead the balloon along the parade route. Find values of trigonometric functions for acute angles of right triangles. Suppose two of these cables are attached to a balloon at the same point, and the volunteers holding these cables stand so that the ends of the cables lie in the same vertical plane. If you know the measure of the angle that each cable makes with the ground and the distance between the volunteers, you can use right triangle trigonometry to find the height of the balloon above the ground. : Solve right triangles. NewVocabulary trigonometric ratios trigonometric functions sine cosine tangent cosecant secant cotangent reciprocal function inverse trigonometric function inverse sine inverse cosine inverse tangent angle of elevation angle of depression solve a right triangle Values of Trigonometric Ratios The word trigonometry m eans triangle measure. In this chapter, you will study trigonometry as the relationships among the sides and angles of triangles and as a set of functions defined on the real num ber system. In this lesson, you will study right triangle trigonometry. 1 Using the side m easures of a right triangle and a reference angle labeled 9 (the Greek letter theta), we can form the six trigonometric ratios that define six trigonom etric functions. KeyConcept Trigonometric Functions 9 be an acute angle in a right triangle and the abbreviations 9, the length of the side adjacent to 9, and the length of the hypotenuse, Let opp, adj, and hyp refer to the length of the side opposite \h y p opp respectively. Then the six trigonometric functions of e\ 1 9 are defined as follows. adj sine (9) = Sin 9 = ^ cosecant (9) = CSC 9 = hyp opp cosine (9) = COS 9 = 7—!- secant (9) = sec 9 = tangent (9) = tan 9 = cotangent (9) = COt 9 = — hyp adj opp V J The cosecant, secant, and cotangent functions are called reciprocal functions because their ratios are reciprocals of the sine, cosine, and tangent ratios, respectively. Therefore, the following statements are true. csc 9 = „ sin 9 sec 9 = — cos 9 cot 9 = • 1 tan 9 From the definitions of the sine, cosine, tangent, and cotangent functions, you can also derive the following relationships. You will prove these relationships in Exercise 83. , n tan 9 = 220 Lesson 4-1 SUl U j and . n cot 9 = COS s in 9 Memorizing Trigonometric Ratios The mnemonic device SOH-CAH-TOA is most commonly used to remember the ratios for Find the exact values of the six trigonom etric functions of 9. The length of the side opposite 9 is 8, the length of the side adjacent to 9 is 15, and the length of the hypotenuse is 17. sine, cosine, and tangent. — opp = sine = ^ P 8 and hyp = 17 hyp csc i opp hyp COS ( _ adj ’ hyp opp tan 6 adj 1 Find Values of Trigonometric Ratios StudyTip COS I adj 15 °P P 8 tan 9 = —— or — ad 15 opp = 8 a hyp 17 sec 9 = —— or — adj 15 adj = 15 and hyp = 17 hyp 01 17 17 or — 8 and adj = 15 adj cot i opp 15 or — 8 p GuidedPractice 1A. 1B. 13 12 Consider sin 0 in the figure. Using A ABC: sin i S' BC AB Using A AB'C': sin 9 = -^p- Notice that the triangles are sim ilar because they are two right triangles that share a com m on angle, 9. Because the triangles are similar, the ratios BC B'C of the corresponding sides are equal. So, — = Therefore, sin 9 has the same value regardless of the triangle used. The values of the functions are constant for a given angle measure. They do not depend on the size of the right triangle. Use One Trigonometric Value to Find Others If cos 9 — —, find the exact values of the five rem aining trigonom etric functions for the acute angle 9. Begin by drawing a right triangle and labeling one acute angle 9. WatchOut! Common Misconception Because cos f = hypotenuse 5. = j , label the adjacent side 2 and the In Example 2, the adjacent side of the triangle could also have been labeled 4 and the hypotenuse 10. This is because cos 8 = | gives By the Pythagorean Theorem, the length of the leg opposite 9 is V 5 2 - 2 2 or V z f. the ratio of the adjacent side and hypotenuse, not their specific °P P measures. hyp hyp opp V21 5 tan 9 = 5 5V21 -= o r- cot 9 = \Jl1 21 °P P V21 adj 2 adj °P P sec 2 2 V2T -p r r O r -------V5T 21 p GuidedPractice 2. If tan 9 = 1 , find the exact values of the five rem aining trigonom etric functions for the acute angle 9. You will often be asked to find the trigonom etric functions of specific acute angle measures. The table below gives the values of the six trigonom etric functions for three com m on angle measures: 30°, 45°, and 60°. To remember these values, you can use the properties of 30°-60°-90° and 45°-45°-90° triangles. KeyConcept Trigonometric Values of Special Angles 30°-60°-90° Triangle o CO B V3 I I 2 45° 60° V2 2 V | 2 V2 2 1 2 1 tan e y r csc o w rw i ■ N 2 \3 M M 3 45°-45°-90° Triangle 4$° V3 Vs V2 2V 3 3 V2 2 1 V3 3 You w ill verify some of these values in Exercises 57-62. Solving Right Triangles Trigonometric functions can be used to find missing side lengths and angle measures of right triangles. H Q 2 5 J J 3 3 ^ f in d a Missing Side Length TechnologyTip Find the value o f x. Round to the nearest tenth, if necessary. Degree Mode To evaluate a trigonometric function of an angle measured in degrees, first set the calculator to degree mode by selecting DEGREE on the MODE Because you are given an acute angle measure and the length of the hypotenuse of the triangle, use the cosine function to find the length of the side adjacent to the given angle. feature of the graphing calculator. sci Eno cos 6 ■ (' i : J >t E b ? B 5 FiftD iflri m PftR B j l PPL HDRIZ SET CLOCK cos 42° = ^ °SIH U L •3+bl- adj hyp l- 18 cos 42° = x Cosine function 0 = 42°, adj = x, and hyp = 18 Multiply each side by 18. |]-T 13.4 ~ x Use a calculator. Therefore, x is about 13.4. CHECK You can check your answer by substituting x = 13.4 into cos 42° = — . cos 42° = 18 13.4 cos 42° = 18 0.74 = 0.74 ✓ ► GuidedPractice 222 Lesson 4-1 R ight Triangle T rig o n o m e try x = 13.4 Simplify. ^ ;^ B V m p liy '!iT 'l'll'll.-i Finding a Missing Side Length TRIATHLONS A com petitor in a triathlon is running along the course shown. Determine the length in feet that the runner must cover to reach the finish line. A n acute angle m easure and the opposite side length are given, so the sine function can be used to find the hypotenuse. sin 9 = sin 63° = opp Sine function hyp 200 0 = 63°, opp = 200, and hyp = x x x sin 63° = 200 The Ironman Triathlon held in X: Kailua-Kona Bay, Hawaii, consists of three endurance events, including a 2.4-mile swim, a 200 s in 6 3 c Multiply each side by or about 224.47 x. Divide each side by sin 63°. So, the com petitor m ust run about 224.5 feet to finish the triathlon. 112-mile bike ride, and a 26.2-mile marathon. Source: World Triathlon Corporation t GuidedPractice 4. TRIATHLONS Suppose a com petitor in the swim m ing portion of the race is swimming along the course shown. Find the distance the com petitor must swim to reach the shore. W hen a trigonometric value of an acute angle is known, the corresponding inverse trigonometric function can be used to find the measure of the angle. ReadingMath Inverse Trigonometric Ratios KeyConcept Inverse Trigonometric Functions In v e r s e S in e is the measure of angle 9. That is, if sin 9 = x, then sin-1 x = 9. The expression sin-1 x is read the inverse sine ofx. Be careful not to If 9 is an acute angle and the sine of 9 is x then the inverse sine of x In v e r s e C o s in e confuse this notation with the If 9 is an acute angle and the cosine of 9 is x then the inverse cosine of x is the measure of angle 9. That is, if cos 9 = x, then cos-1 x = 9. notation for negative exponents: sin-1 x - J —. Instead, this sin x notation is similar to the notation In v e r s e T a n g e n t If 9 is an acute angle and the tangent of 9 is x, then the inverse tangent of x is the measure of angle 9. That is, if tan 9 = for an inverse function, f~ 1(x). ............. x then tan-1 ........................................................ x = 9. ................................................ ... J ( 2 £ E E S 3 R nc* a Missing Angle Measure Use a trigonometric function to find the measure of 6. Round to the nearest degree, if necessary. Because the measures of the sides opposite and adjacent to 0 are given, use the tangent function. tan 9 = tan 9 = °PP Tangent function adj 26 opp = 26 and adj = 11 11 1= tan 1 ^ or about 67° Definition of inverse tangent 5B. E connectED.mcgraw-hill.com | 223 Some applications of trigonometry use an angle of elevation or depression. An angle of elevation is the angle formed by a horizontal line and an observer's line of sight to an object above. An angle of depression is the angle formed by a horizontal line and an observer's line of sight to an object below. __________________ J angle o f — depression / / / line of sight angle of elevation In the figure, the angles of elevation and depression are congruent because they are alternate interior angles of parallel lines. Real-World Example 6 Use an Angle of Elevation AIRPLANES A ground crew w orker who is 6 feet tall is directing a plane on a runway. If the w orker sights the plane at an angle of elevation of 32°, w hat is the horizontal distance from the worker to the plane? Because the worker is 6 feet tall, the vertical distance from the worker to the plane is 150 — 6, or 144 feet. Because the measures of an angle and opposite side are given in the problem , you can use the tangent function to find x. Real-W orldCareer opp tan 9 = — — adj Airport Ground Crew Tangent function Ground crewpersons operate tan 32° = - ^ - ramp-servicing vehicles, handle 0 = 32°, opp = 144, and adj = x cargo/baggage, and marshal or x tan 32° = 144 tow aircraft. Crewpersons must have a high school diploma, x = a valid driver’s license, and a good driving record. x / 144 Multiply each side by x. —™ Divide each side by tan 32°. 230.4 Use a calculator. tan 32° So, the horizontal distance from the w orker to the plane is approxim ately 230.4 feet. GuidedPractice 6. CAMPING A group of hikers on a cam ping trip clim b to the top of a 1500-foot mountain. W hen the hikers look dow n at an angle of depression of 36°, they can see the campsite in the distance. W hat is the horizontal distance betw een the cam psite and the group to the nearest foot? V 224 | Lesson 4-1 | R ight T riangle T rig o n o m e try Angles of elevation or depression can be used to estim ate the distance betw een two objects, as well as the height of an object w hen given two angles from two different positions of observation. Real-World Example 7 Use Two Angles of Elevation or Depression StudyTip BALLOONING A hot air balloon that is m oving above a neighborhood has an angle of depression of 28° to one house and 52° to a house down the street. If the height of the balloon is 650 feet, estimate the distance between the two houses. Indirect Measurement When calculating the distance between two objects using angles of depression, it is important to Draw a diagram to model this situation. Because the angle of elevation from a house to the balloon is congruent to the angle of depression from the balloon to that house, you can label the angles of elevation as shown. Label the horizontal distance from the balloon to the first house x and the distance betw een the two houses y. remember that the objects must lie in the same horizontal plane. From the sm aller right triangle, you can use the tangent function to find x. opp tan 9 = — — adj Tangent function tan 52° = ^ e = 52°- °PP = 650’ and adi = x x tan 52° = 650 x = Multiply each side by x. 650 Divide each side by tan 52°. tan 52 From the larger triangle, you can use the tangent function to find y . °P P tan 6 = — — adj tan 28° = Tangent function 0 = 28°, opp = 650, and adj = x + y x -+- y (x + y) tan 28° = 650 x + y = TechnologyTip J Using Parentheses When 650 expression using a graphing calculator, be careful to close parentheses. While a calculator + v = tan 52° T y evaluating a trigonometric > u = y Divide each side by tan 28°. tan 28 tan 28° 65P ■ tan 28° y ~ 7 1 4 .6 returns the same value for the Multiply each side by x + y. Substitute ------- 65Q— tan 52° Subtract x= 650 850 . tan 52° tan 52° from each side. Use a calculator. expressions tan(26 and tan(26), it does not for expressions V tan(26 + 50 and tan(26) + 50. Therefore, the houses are about 714.6 feet apart. ______________ p- GuidedPractice 7. BUILDINGS The angle of elevation from a car to the top of an apartm ent building is 48°. If the angle of elevation from another car that is 22 feet directly in front of the first car is 64°, how tall is the building? Trigonometric functions and inverse relations can be used to solve a right triangle, w hich m eans to find the measures of all of the sides and angles of the triangle. ReadingMath Labeling Triangles Throughout this chapter, a capital letter will be used to 1 9 1 Solve a Right Triangle Solve each triangle. Round side lengths to the nearest tenth and angle m easures to the nearest degree. a. represent both a vertex of a triangle and the measure of the angle at that vertex. The same letter in lowercase will be used to represent both the side opposite that angle and Y the length of that side. Find x and z using trigonom etric functions. tan 35° = ^ 10 tan 35° = x 7.0 ^ x cos 35° = — Substitute. Substitute. z z cos 35° = 10 Multiply. Multiply. 10 Use a calculator. 2= • cos 35° Divide. 2 ~ 12.2 Use a calculator. Because the measures of two angles are given, Y can be found by subtracting X from 90°. 35° + Y = 90° Y = 55° Angles Xand Kare complementary. Subtract. Therefore, Y = 55°, x = 7.0, and z 12.2. Because two side lengths are given, you can use the Pythagorean Theorem to find that k = V l8 5 or about 13.6. You can find J by using any of the trigonom etric functions. tan/ = f Substitute. / = ta n -1 ^ - Definition of inverse tangent / ~ 53.97° Use a calculator. Because / is now known, you can find L by subtracting / from 90°. 53.97° + L « 90° L « 36.03° Angles Jand L are complementary. Subtract. Therefore, J ~ 54°, L ~ 36°, and k ~ 13.6. ^ GuidedPractice 8B. 226 | Lesson 4-1 i R ight T riangle T rig o n o m e try Exercises = Step-by-Step Solutions begin on page R29. Find the exact values of the six trigonom etric fu nctions o f 0. (2 7 ) MOUNTAIN CLIMBING A team of climbers m ust determine the w idth o f a ravine in order to set up equipm ent to cross it. If the clim bers w alk 25 feet along the ravine from their crossing point, and sight the crossing point on the far side of the ravine to be at a 35° angle, how wide is the ravine? (Example 4) (Example 1) 28. SNOWBOARDING Brad built a snow boarding ramp w ith a height of 3.5 feet and an 18° incline. (Example 4) a. Draw a diagram to represent the situation. VT65 b. Determ ine the length of the ramp. 7. 29. DETOUR Traffic is detoured from Elwood Ave., left 0.8 mile on M aple St., and then right on Oak St., which intersects Elw ood Ave. at a 32° angle. (Example 4) 10 a. Draw a diagram to represent the situation. b. Determ ine the length of Elw ood Ave. that is detoured. Use the given trigonom etric function value of the acute angle 9 to find the exact values of the five rem aining trigonom etric fu nction values of 6. (Example 2) 9. s in 0 = f 10,) 30. PARACHUTING A paratrooper encounters stronger winds than anticipated while parachuting from 1350 feet, causing him to drift at an 8° angle. H ow far from the drop zone will the paratrooper land? (Example 4) COS 0 = y 11 . tan 9 = 3 13. COS 0 = jr7 15. cot 0 = 5 17. sec 0 = | % Find the m easure o f angle 9. Round to the nearest degree, if necessary. (Example 5) Find the value o f x. Round to the n earest tenth, if necessary. (Example 3) 31. 19. 33. 35. ^connectED^Ticgrav\MTilLcorTj| 227 39. PARASAILING Kayla decided to try parasailing. She was strapped into a parachute towed by a boat. An 800-foot line connected her parachute to the boat, w hich w as at a 32° angle of depression below her. H ow high above the water was Kayla? (Example 6) 3 3 ) LIGHTHOUSE Two ships are spotted from the top of a 156-foot lighthouse. The first ship is at a 27° angle of depression, and the second ship is directly behind the first at a 7° angle of depression. (Example 7) a. Draw a diagram to represent the situation. b. Determ ine the distance betw een the two ships. MOUNT RUSHMORE The faces of the presidents at Mount Rushm ore are 60 feet tall. A visitor sees the top of George W ashington's head at a 48° angle of elevation and his chin at a 44.76° angle of elevation. Find the height of M ount Rushm ore. (Example 7) OBSERVATION WHEEL The London Eye is a 135-meter-tall observation wheel. If a passenger at the top of the wheel sights the London Aquarium at a 58° angle of depression, what is the distance betw een the aquarium and the London Eye? (Example 6) 135 m ( '41, ROLLER COASTER On a roller coaster, 375 feet of track ascend at a 55° angle of elevation to the top before the first and highest drop. (Example 6) Solve each triangle. Round side lengths to the nearest tenth and angle measures to the nearest degree. (Example 8) a. Draw a diagram to represent the situation. b. Determine the height of the roller coaster. 42. SKI LIFT A com pany is installing a new ski lift on a 225-meter-high mountain that will ascend at a 48° angle of elevation. (Example 6) a. Draw a diagram to represent the situation. b. Determine the length of cable the lift requires to extend from the base to the peak of the mountain. 43. BASKETBALL Both Derek and Sam are 5 feet 10 inches tall. Derek looks at a 10-foot basketball goal w ith an angle of elevation of 29°, and Sam looks at the goal w ith an angle of elevation of 43°. If Sam is directly in front of Derek, how far apart are the boys standing? (Example 7) 55. BASEBALL M ichael's seat at a game is 65 feet behind home plate. His line of vision is 10 feet above the field. a. Draw a diagram to represent the situation. 44. PARIS A tourist on the first observation level of the Eiffel Tower sights the M usee D 'O rsay at a 1.4° angle of depression. A tourist on the third observation level located 219 meters directly above the first, sights the Musee D 'O rsay at a 6.8° angle of depression. (Example 7) b. W hat is the angle of depression to home plate? 56. HIKING Jessica is standing 2 miles from the center of the base of Pikes Peak and looking at the sum m it of the m ountain, which is 1.4 m iles from the base. a. Draw a diagram to represent the situation. a. Draw a diagram to represent the situation. b. b. 228 Determine the distance betw een the Eiffel Tower and the Musee D'Orsay. Lesson 4-1 R ight Triangle T rig o n o m e try With w hat angle of elevation is Jessica looking at the sum m it of the mountain? Find the exact value of each expression w ithout using a calculator. 57. sin 60° cot 30° 59. sec 30° 60. cos 45° 61. tan 60° 62) csc 45° 82. a. GRAPHICAL Let P(x, y) be a point in the first quadrant. Graph the line through point P and the origin. Form a right triangle by connecting the points P, (x, 0), and the origin. Label the lengths of the legs of the triangle in terms of x or y. Label the length of the hypotenuse as r and the angle the line m akes with the x-axis 9. Without using a calculator, find the measure of the acute angle 9 in a right triangle that satisfies each equation. 63. tan 9 = 1 65. cot ( ^ . cos 9 = ^y V3 3 67. csc 9 = 2 MULTIPLE REPRESENTATIONS In this problem , you will investigate trigonom etric functions of acute angles and their relationship to points on the coordinate plane. b. ANALYTICAL Express the value of r in terms of x and y. V2 C. ANALYTICAL Express sin 9, cos 9, and tan 9 in terms of x, y, and /or r. sec 9 = 2 d. VERBAL U nder what condition can the coordinates of point P be expressed as (cos 9, sin 9)1 Without using a calculator, determine the value of x. e. ANALYTICAL W hich trigonom etric ratio involving 9 corresponds to the slope of the line? f. ANALYTICAL Find an expression for the slope of the line perpendicular to the line in part a in terms of 9. 71. SCUBA DIVING A scuba diver located 20 feet below the surface of the water spots a shipwreck at a 70° angle of depression. After descending to a point 45 feet above the ocean floor, the diver sees the shipwreck at a 57° angle of depression. Draw a diagram to represent the situation, and determine the depth of the shipwreck. Find the value of cos 6 if 9 is the measure of the smallest angle in each type of right triangle. 72. 3-4-5 73. 5-12-13 74. SOLAR POWER Find the total area of the solar panel shown below. H.O.T. Problem s 83. Use Higher-Order Thinking Skills PROOF Prove that if 9 is an acute angle of a right triangle, then tan 9 = cos 8 and cot 9 = 84. ERROR ANALYSIS Jason and N adina know the value of sin 9 = a and are asked to find csc 9. Jason says that this is not possible, bu t N adina disagrees. Is either of them correct? Explain your reasoning. 85. WRITING IN MATH Explain why the six trigonometric functions are transcendental functions. 86. CHALLENGE W rite an expression in terms of 9 for the area of the scalene triangle shown. Explain. 3.5 ft W ithout using a calculator, insert the appropriate symbol > , < , or = to complete each equation. 75. sin 45° cot 60° 76. tan 60° cot 30‘ 77. cos 30° csc 45° 78. cos 30° sin 60‘ 79. sec 45° csc 60° 80. tan 45° sec 30' 81. ENGINEERING Determine the depth of the shaft at the large end d of the air duct shown below if the taper of the duct is 3.5°. (87) PROOF Prove that if 9 is an acute angle of a right triangle, then (sin 9)2 + (cos 9)2 = 1. REASONING If A and B are the acute angles of a right triangle and m Z A < m Z B , determ ine w hether each statement is true or false. If false, give a counterexam ple. 88. sin A < sin B 89. cos A < cos B 90. tan A < tan B 91. WRITING IN MATH N otice on a graphing calculator that there is no key for finding the secant, cosecant, or cotangent of an angle measure. Explain why you think this m ight be so. & connectEDTrncgraw-hiH^ 229 Spiral Review 92. ECONOMICS The Consumer Price Index (CPI) measures inflation. It is based on the average Year prices of goods and services in the United States, w ith the annual average for the years 1982-1984 set at an index of 100. The table show n gives some annual average CPI values from 1955 to 2005. Find an exponential model relating this data (year, CPI) by linearizing the data. Let x = 0 represent 1955. Then use your model to predict the CPI for 2025. (Lesson 3-5) Solve each equation. Round to the nearest hundredth. (Lesson 3-4) 93. e5x = 24 94.2 ex“ 7 — 6 = 0 CPI 1955 26.8 1965 31.5 1975 53.8 1985 107.6 1995 152.4 2005 195.3 Source: Bureau of Labor Statistics Sketch and analyze the graph of each function. Describe its domain, range, intercepts, asymptotes, end behavior, and where the function is increasing or decreasing. (Lesson 3-1) 95. 96.f{x ) = 2 3j - 4 + 1 f( x ) = - 3 X~ 2 97. fix ) = -4 ~ Solve each equation. (Lesson 2-5) 98 . -----* 2 — 16------ = - A ------------ 1— (x + 4)(2x —1) 101. x+4 99 . ------------- 2x - 1 (x + l)(x —5) = _ J l _ + _ JL _ x+ 1 x —5 100. 2x2 + — 3x2 + 5x + 2 3x + 2 x+ 1 NEWSPAPERS The circulation in thousands of papers of a national newspaper is shown. (Lesson 2-1) Year 2002 2003 2004 2005 2006 2007 2008 Circulation (in thousands) 904.3 814.7 773.9 725.5 716.2 699.1 673.0 a. Let x equal the number of years after 2001. Create a scatter plot of the data. b. Determine a pow er function to model the data. c. Use the function to predict the circulation of the newspaper in 2015. Skills Review for Standardized Tests 104. A person holds one end of a rope 102. SAT/ACT In the figure, what is the value of z? 2y+ 5 Mote: Figure not drawn to scale. A 15 D 30V2 B 15V 2 E 30V3 C 15V3 103. REVIEW Joseph uses a ladder to reach a window 10 feet above the ground. If the ladder is 3 feet away from the wall, how long should the ladder be? F 9.39 ft that runs through a pulley and has a w eight attached to the other end. Assum e that the weight is at the same height as the person's hand. W hat is the distance from the person's hand to the weight? A 7.8 ft B 10.5 ft C 12.9 ft D 14.3 ft 105. REVIEW A kite is being flow n at a 45° angle. The string of the kite is 120 feet long. How high is the kite above the point at w hich the string is held? G 10.44 ft F 60 ft H 11.23 ft G 60V 2 ft J 12.05 ft H 60V 3 ft J 120 ft 230 | Lesson 4-1 I R ight Triangle T rig o n o m e try 12 f t I ** 1 ? Degrees and Radians 4 •Then • You used the measures of acute angles in triangles given in degrees. (Lesson 4-1) j / y NewVocabulary vertex initial side terminal side standard position radian coterminal angles linear speed angular speed sector Now 1 Why? Convert degree measures of angles to radian measures, and vice versa. ) Use angle measures i to solve real-world problems. In Lesson 4-1, you worked only with acute angles, but angles can have any real number measurement. For example, in skateboarding, a 540 is an aerial trick in which a skateboarder and the board rotate through an angle of 540°, or one and a half complete turns, in midair. A ngles and Their M easu res From geometry, you m ay recall an angle being defined as two noncollinear rays that share a com m on endpoint know n as a vertex. An angle can also be thought of as being formed by the action of rotating a ray about its endpoint. From this dynamic perspective, the starting position of the ray forms the initial side of the angle, while the ray's position after rotation forms the angle's term inal side. In the coordinate plane, an angle with its vertex at the origin and its initial side along the positive x-axis is said to be in standard position. 1 A ngle The m easure of an angle describes the am ount and direction of rotation necessary to move from the initial side to the term inal side of the angle. A positive angle is generated by a counterclockwise rotation and a negative angle by a clockwise rotation. P ositive A ngle The most com m on angular unit of measure is the degree (°), which is equivalent to of a full rotation (counterclockwise) about the vertex. From the diagram shown, you can see that 360° corresponds to 1 com plete rotation, 180° to a 1 rotation, 90° to a v rotation, and so on, as 2 4 marked along the circum ference of the circle. StudyTip Base 60 The concept of degree Degree measures can also be expressed using a decimal degree form or a degree-minute-second ^(DMS) form where each degree is subdivided into 60 minutes (') and each m inute is subdivided into 60 seconds ("). measurement dates back to the ancient Babylonians, who made early astronomical calculations J O S H E S ] Convert Between DMS and Decimal Degree Form using their number system, which Write each decimal degree measure in DMS form and each DMS measure in decimal degree form to the nearest thousandth. was based on 60 (sexagesimal) rather than on 10 (decimal) as we do today. a. 56.735° First, convert 0.735° into minutes and seconds. 56.735° = 56° + 0.735° = 5 6 ° + 44.1' 1° = 60' Simplify. Next, convert 0 .1 'in to seconds. 56.735° = 56° + 4 4 '+ 0.1' = 5 6 ° + 4 4 '+ 6 " 1'=60" Simplify. Therefore, 56.735° can be written as 56° 44' 6". TechnologyTip b. 32° 5 '2 8 " 1 1 1 Each m inute is — of a degree and each second is — of a minute, so each second is —— r DMS Form You can use some calculators to convert decimal 60 of a degree. degree values to degrees, minutes, and seconds using the D M S 32- 5' 28" = 32” + 5' ( J j ) + 2 8 " (J £ . ) function under the A n g le menu. 3600 ^ 6u V . ± < n ana I ' . ^ = 32° + 0.083 + 0.008 Simplify. = 32.091° Add. Therefore, 32° 5 ' 28" can be written as about 32.091°. ^ GuidedPractice 1B. 1A. 213.875° 30 56' 7" V___________________ Measuring angles in degrees is appropriate w hen applying trigonometry to solve many real-world problems, such as those in surveying and navigation. For other applications w ith trigonometric functions, using an angle measured in degrees poses a significant problem. A degree has no relationship to any linear measure; inch-degrees or - :*nc^t has no m eaning. M easuring angles in radians provides a solution to this problem. KeyConcept Radian Measure Words The measure 8 in radians of a central angle of a circle is equal to the ratio of the length of the intercepted arc s to the radius r of the circle. Symbols 8 = j , where 8 is measured in radians (rad) Example A central angle has a measure of 1 radian if it intercepts an arc with the same length as the radius of the circle. v. V 7 8 = 1 radian when s = r. _____ , , , ............... Notice that as long as the arc length s and radius r are measured using the same linear units, the ratio j is unitless. For this reason, the word radian or its abbreviation rad is usually om itted when writing the radian measure of an angle. 232 | Lesson 4-2 Degrees and Radians y StudyTip Degree-Radian Equivalences From the equivalence statement shown, you can determine The central angle representing one full rotation counterclockwise about a vertex corresponds to an >arc length equivalent to the circum ference of the circle, 2ixr. From this, you can obtain the following radian measures. -1 1 1 1 1 1 2ttt —rotation = •2 tt —rotation = — •2tt 1 rotation —rotation = — •2 tc r 6 6 4 4 that 1° = 0.017 rad and = 77 rad = 2 tt rad 1 rad k 57.296°. ■rad = f rad Because 2tt radians and 360° both correspond to one com plete revolution, you can write 360° = 2tt radians or 180° = it radians. This last equation leads to the follow ing equivalence statements. ReadingMath i° = 180 Angle Measure If no units of angle measure are specified, radians 1 radian : m \ ° \ it and Using these statements, we obtain the follow ing conversion rules. radian measure is implied. If degrees are intended, the degree symbol (°) must be used. KeyConcept Degree/Radian Conversion Rules 1. To convert a degree measure to radians, multiply by 2. To convert a radian measure to degrees, multiply by •k radians 180° ' 180° irradians' Convert Between Degree and Radian Measure Write each degree measure in radians as a m ultiple of tv and each radian measure in degrees, a. 120° 1 ?n° = 1 ?n° 171radians'\ 180° { radians or = b. 180° ’ Simplify. I tv r a d i a n s \ 180° Multiply by -nr radians I = —^ ra d ia n s or —^ Simplify. ^ radians 6 Multiply by 5 tt 6 = d. i r radians -4 5 ° -45° = -45° c. Multiply by I 6 180° ■k radians (— ^ ) or 150° \7T radians / radians Simplify. 3 ir radians = 2 Multiply by radians ( 18,°° ) or -2 7 0 ° \7T radians/ 180° i t ra dians' Simplify. p GuidedPractice 2A. 210° 2B. -60° 2C. f 2D. conncctED.m cgraw-hill.co m j | 233 ReadingMath Naming Angles In trigonometry, By defining angles in terms of their rotation about a vertex, two angles can have the same initial , and terminal sides but different measures. Such angles are called coterminal angles. In the figures >below, angles a and (3 are coterminal. angles are often labeled using P ositive and N egative C oterm inal A ngles Greek letters, such as a (alpha), /3 (beta), and 6 (theta). P ositive C oterm inal A ngles The two positive coterm inal angles shown differ by one full rotation. A given angle has infinitely many coterm inal angles found by adding or subtracting integer m ultiples of 360° or 27r radians. KeyConcept Coterminal Angles Degrees If a is the degree measure of an angle, then all angles measuring a + 360n°, where n is an integer, are coterminal Radians If a is the radian measure of an angle, then all angles measuring a + 2/77v, where n is an integer, are coterminal with a . with a . and Draw Coterminal Angles Identify all angles that are coterm inal with the given angle. Then find and draw one positive and one negative angle coterm inal with the given angle. a. 45° " I All angles measuring 45° + 360n° are coterm inal w ith a 45° angle. Let n = 1 and —1. All angles measuring —— + 2mr are coterm inal w ith a — ^ angle. Let n = 1 and —1. 45° + 360(1)° = 45° + 360° or 405° ►GuidedPractice 3A. - 3 0 ° 234 Lesson 4 -2 Degrees and Radians 3B. 3t t 4 2 Applications w ith Angle Measure Solving 9 = j for the arc length s yields a convenient formula for finding the length of an arc of a circle. KeyConcept Arc Length If 9 is a central angle in a circle of radius r, then the length of the intercepted arc s is given by 1 s = r9, lv » 1 ' where 9 is measured in radians. ^ N --------/ J J 1 W hen 9 is m easured in degrees, you could also use the equation s = incorporates the degree-radian conversion. Find Arc Length StudyTip Find the length of the intercepted arc in each circle with the given central angle measure and radius. Round to the nearest tenth. Notice in Example 4a that when r = 5 centimeters and 9 = ^ Sir centimeters,4 Sir 4 not ^ centimeter-radians. This > a. -tv r = 5 cm 4 r e Arc length / 7T\ 4 is because a radian is a unitless ratio. I r = 5 and 9 = ^ 4 5( f ) 5ir 4 l The length of the intercepted arc is ^ b. / h 5 cm J Simplify. or about 3.9 centimeters, 60°, r = 2 in. Method 1 Convert 60° to radian measure, and then use s = r9 to find the arc length. 60° = 6 0 ° (™ d0ioa n s ) Multiply by 7T Simplify. 3 II Substitute r = 2 and 0 Arc length s = rO Method 2 ■ k radians 180° ' - 2( i ) r=2and 0 = H Operating with Radians radians, s = w hich already Simplify. Use s = loU 7u 9 180° tt (2)(60°) 180° 2tt 3 to find the arc length. Arc length r = 2 and 0 = 60° Simplify. 2-it The length of the intercepted arc is — or about 2.1 inches. ►GuidedPractice 4A. | , r = 2 m 4B. 135°, r = 0.5 ft 235 & The formula for arc length can be used to analyze circular motion. The rate at which an object moves along a circular path is called its linear speed. The rate at w hich the object rotates about a fixed point is called its angular speed. Linear speed is measured in units like m iles per hour, while angular speed is measured in units like revolutions per minute. KeyConcept Linear and Angular Speed Suppose an object moves at a constant speed along a circular path of radius r. s ' • ' ' ’" ' - x If s is the arc length traveled by the object during time ReadingM ath speed v is given by v= Omega The lowercase Greek If t, then the object’s linear then the / | N' ' J------ - \ >r 6 is the angle of rotation (in radians) through which the object moves during time t, letter omega u> is usually used to denote angular speed. / 1 | 1 angular speed u j of the object is given by j V Find Angular and Linear Speeds BICYCLING A bicycle messenger rides the bicycle shown. a. During one delivery, the tires rotate at a rate of 140 revolutions per m inute. Find the angular speed of the tire in radians per minute. Because each rotation m easures 2tt radians, 140 revolutions correspond to an angle of rotation 9 of 140 x 27V or 2807V radians. lo = = In some U.S. cities, it is — Angular speed 2807V ra d ia n s — ^— :— -— 1 m in u te „ 0 „„„ ,. . = 280 5r radians and t — . . . . 1minute possible for bicycle Therefore, the angular speed of the tire is 280ir or about 879.6 radians per minute. messengers to ride an average of 30 to 35 miles a day while making 30 to 45 deliveries. Source: New York Bicycle Messenger Association b. On part of the trip to the next delivery, the tire turns at a constant rate of 2.5 revolutions per second. Find the linear speed of the tire in miles per hour. A rotation of 2.5 revolutions corresponds to an angle of rotation 9 of 2.5 x 27V or 57V. v = j- Linear speed 15(5tv ) in c h e s = — 3 1 secon d 7517 in c h e s — or — r — 15 inches, 0 = 5 it radians, and — 1 secon d t — 1 second Use dim ensional analysis to convert this speed from inches per second to miles per hour. 7 5 tv i o A e r ~ x 1 seeerrrd" 60 se c e fn fe - x 1 m » w tr 6 0 m itttr f e s - x 1 hour L ie e t12 in e h e s" 1 m i l e _____ 1 3 .4 m ile s 5 2 8 0 fe e t- hour Therefore, the linear speed of the tire is about 13.4 miles per hour. p GuidedPractice MEDIA Consider the DVD shown. 5A. Find the angular speed of the DVD in radians per second if the disc rotates at a rate of 3.5 revolutions per second. 5B. If the DVD player overheats and the disc begins to rotate at a slower rate of 3 revolutions per second, find the disc's linear speed in meters per minute. 236 | Lesson 4 -2 Degrees and Radians Recall from geom etry that a sector of a circle is a region bounded by a central angle and its intercepted arc. For exam ple, the shaded portion in the figure is a sector of circle P . The ratio of the area of a sector to the area of a whole circle is equal to the ratio of the corresponding arc length to the circum ference of the circle. Let A represent the area of the sector. A _ le n g th o f QRS irr2 A area of sector _ arc length area of circle — circum ference o f circle 2m' _ ■nr2 rO The length of QRS is rO. 2~nr Solve for A A = j r 29 KeyConcept Area of a Sector The area A of a sector of a circle with radius rand central angle 9 is 0 / A = \ r2e, where 9 is measured in radians. w ^ ---------^ p r T fT n fffT h Find Areas of Sectors a. Find the area of the sector of the circle. 77T The measure of the sector's central angle 9 is — , and the radius is 3 centimeters. Area of a sector A = - | r 20 M , 63 tt f ) ' 16 r = 3 and 0 ■- 7-k Therefore, the area of the sector is -y -- or about 12.4 square centimeters. b. WIPERS Find the approximate area swept by the w iper blade shown, if the total length of the windshield w iper mechanism is 26 inches. The area swept by the wiper blade is the difference betw een the areas of the sectors with radii 26 inches and 26 — 16 or 10 inches. Convert the central angle measure to radians. 130° = 130° Real-W orldLink ( 7r radians j _ 13tt 18 180° Then use the radius of each sector to find the area swept. Let A 1 = the area of the sector with a 26-inch radius, and let A 2 = the area of the sector w ith a 10-inch radius. A typical wipe angle for a front windshield wiper of a Swept area ■A, A = A, passenger car is about 67°. Windshield wiper blades are = }(2 6 )2 - h Area of a sector generally 12-30 inches long. Source: Car and Driver _ 2197-rr 325 tt 9 9 = 2087V or about 653.5 Simplify. Simplify. Therefore, the swept area is about 653.5 square inches. GuidedPractice Find the area of the sector of a circle with the given central angle 0 and radius r. 6A. 9 = ^ - , r = 1.5 ft 4 6B. 9 = 50°, r = 6 m connectED.mcgraw-hill.com~H 237 * 5 5 /5 (0 Exercises = Step-by-Step Solutions begin on page R29. Write each decimal degree measure in DMS form and each DMS measure in decimal degree form to the nearest thousandth. (Example 1) 1. 11.773° 2. 58.244° 3. 141.549° 4. 273.396° 5. 87° 53' 10" 7. 45° 2 1 '2 5 " Find the length of the intercepted arc with the given central angle measure in a circle with the given radius. Round to the nearest tenth. (Example 4) 28. ■yv r = 3 in. 6. 126° 6 '3 4 " 29. % r = 4 yd 30. 105°, r = 18.2 cm 8. 301° 31. 45°, r = 5 mi '8 " 9. NAVIGATION A sailing enthusiast uses a sextant, an instrument that can measure the angle betw een two objects with a precision to the nearest 10 seconds, to measure the angle betw een his sailboat and a lighthouse. If his reading is 17° 37' 50", what is the measure in decimal degree form to the nearest hundredth? (Example 1) CO N3 27. f , r = 2.5 m 4 2 *£ x 150°, r = 79 mm 33) AMUSEMENT PARK A carousel at an am usem ent park rotates 3024° per ride. (Example 4) a. H ow far would a rider seated 13 feet from the center of the carousel travel during the ride? b. H ow m uch farther would a second rider seated 18 feet from the center of the carousel travel during the ride than the rider in part a? Find the rotation in revolutions per minute given the angular speed and the radius given the linear speed and the rate of rotation. (Example 5) 35. w = 135tt Write each degree measure in radians as a multiple of tv and each radian measure in degrees. (Example 2) 12. -1 6 5 ° 13. 14. 16. 2 tv 15. 3 _7T I 11. 225° Ul o 10. 30° 5-jt 2 17. _7 rr 4 6 Identify all angles that are coterminal with the given angle. Then find and draw one positive and one negative angle coterminal with the given angle. (Example 3) 18. 120° 19. 20. 225° 21. 22. 24. TV 23. 3 7T 25. 12 3tt ’ 4 377 2 26. GAME SHOW Sofia is spinning There are 20 values in equ al-sized spaces around the circum ference of the w heel. The value that Sofia needs to w in is two spaces above the space w here she starts her spin, and the w heel m ust m ake at least one full rotation for the spin to count. D escribe a spin rotation in degrees that w ill give Sofia a w inning result. (Example 3) Space Sofia needs to land on Start of Sofia’s spin 238 Lesson 4-2 Degrees and Radians 36. lo = 104tt rad ra d 37. v = 82.3 — , 131 -=^rs 38. v = 144.2 - 4 - , 10.9 — mm mm mm 39. w = 5 5 3 ^ , 0 . 0 9 ^ h mm 40. MANUFACTURING A com pany m anufactures several circular saws w ith the blade diam eters and motor speeds show n below. (Example 5) Blade Motor Diameter (in.) Speed (rps) 3 2800 5 5500 4 4500 5500 5000 a. Determ ine the angular and linear speeds of the blades in each saw. Round to the nearest tenth. b. How m uch faster is the linear speed of the 6-i-inch saw com pared to the 3-inch saw? 41. CARS On a stretch of interstate, a vehicle's tires range betw een 646 and 840 revolutions per minute. The diam eter of each tire is 26 inches. (Example 5) a. Find the range of values for the angular speeds of the tires in radians per minute. b. Find the range of values for the linear speeds of the tires in miles per hour. ^ g n ° 1. v i 42. TIME A wall clock has a face diam eter of 8 - j inches, th i length of the hour hand is 2.4 inches, the length of the minute hand is 3.2 inches, and the length of the second hand is 3.4 inches. (Example 5) V r o 55. Describe the radian m easure betw een 0 and 27r of an angle 9 that is in standard position w ith a terminal side that lies in: c. Q uadrant I I I d. Quadrant IV a. Q uadrant I b. Q uadrant II 56. If the term inal side of an angle that is in standard position lies on one of the axes, it is called a quadrantal angle. Give the radian m easures of four quadrantal angles. a. Determine the angular speed in radians per hour and the linear speed in inches per hour for each hand. b. If the linear speed of the second hand is 20 inches per m inute, is the clock running fast or slow? How much time would it gain or lose per day? 57. GEOGRAPHY Phoenix, Arizona, and Ogden, Utah, are located on the same line of longitude, w hich means that Ogden is directly north of Phoenix. The latitude of Phoenix is 33° 26' N, and the latitude of Ogden is 41° 12' N. If Earth's radius is approxim ately 3963 miles, about how far apart are the two cities? 110 W 120 W GEOMETRY Find the area of each sector. (Example 6) Find the measure of angle 9 in radians and degrees. 48. 61. 49. GAMES The dart board shown is divided into twenty equal sectors. If the diameter of the board is 18 inches, what area of the board does each sector cover? (Example 6) 62. TRACK The curve of a standard 8-lane track is semicircular as shown. 50. LAWN CARE A sprinkler waters an area that forms one third of a circle. If the stream from the sprinkler extends 6 feet, what area of the grass does the sprinkler water? (Example 6) The area of a sector of a circle and the measure of its central angle are given. Find the radius of the circle. 511 A = 29 ft2, 9 = 68° 53. A = 377 in2, 9 = ^ 52. A = 808 crti2, 9 = 210° 54. A = 75 m 2, 0 = 37V 4 1.22 m a. W hat is the length of the outside edge of Lane 4 in the curve? b. How m uch longer is the inside edge of Lane 7 than the inside edge of Lane 3 in the curve? r-p & c o n n e c tE D .m c g ra w -h ill.c o m | 239 63. DRAMA A pulley with radius r is being used to remove part of the set of a play during intermission. The height of the pulley is 12 feet. a. If the radius of the pulley is 6 inches and it rotates 180°, how high will the object be lifted? b. If the radius of the pulley is 4 inches and it rotates 900°, how high will the object be lifted? 72. SKATEBOARDING A physics class conducted an experiment to test three different wheel sizes on a skateboard with constant angular speed. a. W rite an equation for the linear speed of the skateboard in terms of the radius and angular speed. Explain your reasoning. b. Using the equation you wrote in part a, predict the linear speed in meters per second of a skateboard w ith an angular speed of 3 revolutions per second for w heel diam eters of 5 2 ,5 6 , and 60 millimeters. C. Based on your results in part b, how do you think wheel size affects linear speed? H.O.T. Problem s 64. ENGINEERING A pulley like the one in Exercise 63 is being used to lift a crate in a warehouse. Determ ine w hich of the following scenarios could be used to lift the crate a distance of 15 feet the fastest. Explain how you reached your conclusion. I. The radius of the pulley is 5 inches rotating at 65 revolutions per minute. Use Higher-Order Thinking Skills 73. ERROR ANALYSIS Sarah and M ateo are told that the perim eter of a sector of a circle is 10 times the length of the circle's radius. Sarah thinks that the radian measure of the sector's central angle is 8 radians. M ateo thinks that there is not enough inform ation given to solve the problem. Is either of them correct? Explain your reasoning. 74. CHALLENGE The two circles show n are concentric. If the length of the arc from A to B m easures 87T inches and DB = 2 inches, find the arc length from C to D in terms of 7T. II. The radius of the pulley is 4.5 inches rotating at 70 revolutions per minute. III. The radius of the pulley is 6 inches rotating at 60 revolutions per minute. GEOMETRY Find the area of each shaded region. REASONING D escribe how the linear speed would change for each param eter below. Explain. 75. a decrease in the radius 76. a decrease in the unit of time 67. CARS The speedom eter shown measures the speed of a car in miles per hour. 77. an increase in the angular speed 78. PROOF If y - = prove that 9^ = 62- 79. REASONING W hat effect does doubling the radius of a circle have on each of the follow ing measures? Explain your reasoning. a. If the angle betw een 25 m i/h and 60 mi/h is 81.1°, about how many miles per hour are represented by each degree? b. If the angle of the speedom eter changes by 95°, how much did the speed of the car increase? Find the complement and supplement of each angle, if possible. If not possible, explain your reasoning. 68. 2 *5 240 69. — 6 L esson 4 -2 70. — 8 D e g re e s a n d R a d ia n s 71. 3 a. the perim eter of the sector of the circle w ith a central angle that m easures 9 radians b. the area of a sector of the circle w ith a central angle that measures 9 radians 80. WRITING IN MATH Compare and contrast degree and radian m easures. Then create a diagram sim ilar to the one on page 231. Label the diagram using degree measures on the inside and radian m easures on the outside of the circle. Spiral Review Use the given trigonometric function value of the acute angle 6 to find the exact values of the five remaining trigonometric function values of 0. (Lesson 4-1) 81. 4 V7 82.sec6> = sin 0 = ^ ^ 83.cot 0 = 84. BANKING A n account that H ally's grandm other opened in 1955 earned continuously com pounded interest. The table shows the balances of the account from 1955 to 1959. (Lesson 3-5) a. Use regression to find a function that m odels the am ount in the account. Use the num ber of years after Jan. 1 ,1 9 5 5 , as the independent variable. b. Write the equation from part a in terms of base e. C. W hat was the interest rate on the account if no deposits or withdrawals were m ade during the period in question? Date Balance 1 Jan. 1,1955 $2137.52 2 Jan. 1,1956 $2251.61 3 Jan. 1,1957 $2371.79 4 Jan. 1,1958 $2498.39 5 Jan. 1,1959 $2631.74 Express each logarithm in terms of In 2 and In 5. (Lesson 3-2) 85. I n # 86. 16 87. I n 250 10 In 25 List all possible rational zeros of each function. Then determ ine which, if any, are zeros. (Lesson 2-4) 88. 89.g (x ) = x 3 — 5x2 - 4x + 20 f(x) = x 4 - x 3 - 12x — 144 90.g (x) = 6x4 + 35x3 — x 2 — 7x — 1 Write each set of num bers in set-builder and interval notation if possible. Lesson 1-3) 91 . f( x ) = Ax5 + 2x4 - 3x - 1 92.g(x)= - x 6 + x4 - 5x2 + 4 93.h(x) = — + 2 Describe each set using interval notation. (Lesson 0-1) 94. n > - 7 96. y < 1 or y > 11 95.- 4 < x < 10 Skills Review fo r Standardized Tests 97. SAT/ACT In the figure, C and D are the centers of the two circles w ith radii of 3 and 2, respectively. If the larger shaded region has an area of 9, what is the area of the smaller shaded region? 99. REVIEW If sec 25 , then sin 0 = 7_ A 25 B 24 25 C — or - 2 i 25 25 D 25 7 100. W hich of the follow ing radian m easures is equal Note: Figure not drawn to scale. A 3 C 5 B 4 D 7 E 8 98. REVIEW If cot 0 = 1 , then tan 0 = F -1 GO to 56°? F -2L 15 G — 45 H ^ 45 I -3 J H I J 3 241 Trigonometric Functions on the Unit Circle Then • You found values of trigonometric functions for acute angles using ratios in right triangles. Now * 1 (Lesson 4-1) NewVocabulary quadrantal angle reference angle unit circle circular function periodic function period Why? Find values of trigonometric functions for any angle. 2 A blood pressure of 120 over 80, measured in millimeters of mercury, means that a person’s blood pressure oscillates or cycles between 20 millimeters above and below a pressure of 100 millimeters of mercury for a given time fin seconds. A complete cycle of this oscillation takes about 1 second. Find values of trigonometric functions using the unit circle. If the pressure exerted by the blood at time t = 0.25 second is 120 millimeters of mercury, then at time t — 1.25 seconds the pressure is also 120 millimeters of mercury. Trigonometric Functions of Any Angle In Lesson 4-1, the definitions of the six trigonometric functions were restricted to positive acute angles. In this lesson, these definitions are extended to include any angle. 1 KeyConcept Trigonometric Functions of Any Angle Let 9 be any angle in standard position and point P{x, y) be a point on the terminal side of 9. Let /represent the nonzero distance from Pto the origin. That is, let r= \/x2 + y2 ± 0. Then the trigonometric functions of 9 are as follows. sin 6 = j csc0 = y,y # O cos 9 = j sec 9 = ta n 0 = i xj= 0 cot 9 = ^,yj= 0 xj=0 Evaluate Trigonometric Functions Given a Point Let (8, —6) be a point on the terminal side of an angle 9 in standard position. Find the exact values of the six trigonometric functions of 9. Use the values of x and y to find r. y Pythagorean Theorem r = \]x2 + y 2 .— - x 61 = V s 2 + (—6 )2 x = 8 a n d y = —6 = V lO O or 1 0 Take the positive square root. d * (8, —6 ) V Use x = 8, y = —6, and r sin 0 = — = — ppor csc 9 = 10 r — y = -6 or — — 5 3 = 1 0 to write the six trigonom etric ratios. cos 0 = — = sec 9 = — = ip r x 10 8 or or 5 j 4 tan 9 = — = x 8 or — 3 4 4 cot 9 = — = or — 3 y -6 ►GuidedPractice The given point lies on the terminal side of an angle 0 in standard position. Find the values of the six trigonometric functions of 9. 1A. (4 ,3 ) a 242 L esson 4-3 1B. ( - 2 , - 1 ) In Exam ple 1, you found the trigonom etric values of 9 w ithout know ing the measure of 9. Now we will discuss methods for finding these function values w hen only 9 is known. Consider trigonom etric functions of quadrantal angles. W hen the term inal side of an angle 9 that is in standard position lies on one of the coordinate axes, the angle is called a quadrantal angle. KeyConcept Common Quadrantal Angles StudyTip y Quadrantal Angles There are y y y (0, infinitely many quadrantal angles r)‘ that are coterminal with the quadrantal angles listed at the 0 0 right. The measure of a quadrantal X 0 (r, 0) x angle is a multiple of 90° or y . *(—r, 0) X 0 \ X 0 (0 ,-r ) 9 = 0° or 0 radians 6= 90° or y radians e = 180° or t radians 0 = 270° or -y - radians I You can find the values of the trigonom etric functions of quadrantal angles by choosing a point on the term inal side of the angle and evaluating the function at that point. Any point can be chosen. However, to sim plify calculations, pick a point for w hich r equals 1. H ^ J J S S E iE v a lu a t e Trigonometric Functions of Quadrantal Angles Find the exact value of each trigonom etric function, if defined. If not defined, write u n defin ed. a. sin (-1 8 0 ° ) The terminal side of —180° in standard position lies on the negative x-axis. Choose a point P on the term inal side of the angle. A convenient point is (—1, 0) because r = 1. Sine function sin (-1 8 0 ° ) = f y = 0 a n d r= 1 = y or 0 b. tan The terminal side of in standard position lies on the negative y-axis. Choose a point P(0, —1) on the term inal side of the angle because r = 1. t 3tx V ta n - = I = —jy or undefined Tangent function y = —1 and x —0 C. s e c 4iv The term inal side of 4tt in standard position lies on the positive x-axis. The point (1, 0) is convenient because r = 1. sec 4 tt = ■ = y or 1 Secant function r= 1 a n d x = l p GuidedPractice 2A. cos 270° 2B. c s c f 2C. cot ( - 9 0 ° connectED.mcgraw-hill.com I 243 To find the values of the trigonometric functions of angles that are neither acute nor quadrantal, consider the three cases shown below in w hich a and b are positive real numbers. Compare the values of sine, cosine, and tangent of 9 and 9 '. StudyTip Quadrant II Quadrant III y y i \ Reference Angles Notice that in -.0 < theta a prime) are the same, in other y r some cases, the three trigonometric values of 0 and 0 ' (read Quadrant IV O ' * a i,—^ /jA 0 ^ 0 X X - C V 0 V 0i / r cases, they differ only in sign. - b) (a, - 6 ) ^ sin 9 = j sin 9' = j sin 0 = - j sin 9' = j sin 0 = —j sin 9' = j cos 9 = - j cos 0 ' = | cos 0 = —j cos 9' = j cos 0 = 7 cos 0 ' = j tan 9 = - |a tan 0 ' = |a tan 0 = 4a tan 0 ' = -|a tan 0 = — 4a tan 0 ' = a This angle 9 ', called a reference angle, can be used to find the trigonometric values of any angle I KeyConcept Reference Angle Rules if 9 is an angle in standard position, its reference angle 0 ' is the acute angle formed by the terminal side of 0 and the x-axis. The O'for any angle 0, 0° < 6 < 360° or 0 < 9 < 2ir, is defined as follows. reference angle Quadrant II \y o0 9’ 9'=9 9' = 180° — 9 0 '= i r - 0 9'= 0 - 1 8 0 ° 9' = 9 - tv 0' = 9 360° - 0'= 2tt - 9 To find a reference angle for angles outside the interval 0° < 9 < 360° or 0 < 9 < 2 n , first find a corresponding coterm inal angle in this interval. H 2 E H 3 3 D Find Reference Angles Sketch each angle. Then find its reference angle. a. b. 300° The terminal side of 300° lies in Q uadrant IV. Therefore, its reference angle is 9' = 360° - 300° or 60°. 27T A coterminal angle is 2 tt — terminal side of — lies in Q uadrant III, so J . its reference angle is — tt or . p GuidedPractice 3A. ^ 244 3B. - 2 4 0 ° | Lesson 4 -3 I T r ig o n o m e tric F u n c tio n s o n th e U n it C irc le or 4 ^ . The 3C. 390° B ecause the trig on o m etric valu es of an an gle and its referen ce an gle are equ al or differ only in sign, you can use the fo llow in g steps to fin d the valu e o f a trig on o m etric fu n ction of any angle 9. KeyConcept Evaluating Trigonometric Functions of Any Angle ETHTn Find the reference angle 9'. y PT7!TIW Find the value of the trigonometric function for 9'. ETTBtil Using the quadrant in which the terminal side of 0 lies, determine the sign "" i The signs of the trigonom etric functions in each quadrant can be determ ined using the function definitions given on page 242. it follows that sin 9 is negative when y < 0, w hich occurs in Quadrants III and IV. Using this same logic, you can verify each of the signs for sin 9, cos 9, and tan 9 shown in the diagram. N otice that these values depend only on x and y because r is always positive. StudyTip Memorizing Trigonometric Values To memorize the exact values of sine for 0°, 30°, 45°, 60°, and 90°, consider the following pattern. sin 0° = Quadrant II Quadrant 1 sin 9: + cos 9: tan 9: - sin 9: + cos 9: + tan 9: + Quadrant III Quadrant IV sin ft — cos ft + tan ft - sin 9: cos 9: tan 9: + Because you know the exact trigonom etric values of 30°, 45°, and 60° angles, you can find the exact trigonometric values of all angles for w hich these angles are reference angles. The table lists these values for 9 in both degrees and radians. 9 or 0 30° o r £ 6 45° o r £ 4 1 V2 2 V3 2 V2 2 1 2 1 V3 sin 9 sin 30° = ^ , or± cos 9 Vz sin 45° = y f tan 9 sin 60° = ^ sin 90° = \ ~ 'l y y For exam ple, because sin 9 = 9 ^ 2 V3 2 V3 3 60° or y or 1 A similar pattern exists for the cosine function, except the values are given in reverse order. H 2 J y U J 3 3 Use Reference Angles to Find Trigonometric Values Find the exact value of each expression, a. ( cos 120° l , ° ) Because the terminal side of 9 lies in Q uadrant II, the reference angle 9' is 180° — 120° or 60°. cos 120° = —cos 60° In Quadrant II, cos 0 is negative. ( 0 cos 60° = -1 b. tan 0- 6 Because the terminal side of 9 lies in Q uadrant III, th e r e fe r e n c e a n g le tan 6 = tan ^ 6 y/3 3 'TT 9' i s - — iv i t or or TV —. In Quadrant III, tan 0 is positive. V3 tanf = -3 C= & connectED.mcgraw-hill.com | 245 C. CSC 15tt A coterm inal angle of 9 is — ------ 27T or — , w hich lies in 77T tv Q uadrant IV. So, the reference angle 9' is 27T — — ° r —. Because sine and cosecant are reciprocal functions and sin 9 is negative in Q uadrant IV, it follows that csc 9 is also negative in Q uadrant IV. 15 tv csc - In Quadrant IV, csc 0 is negative. -csc ~ 4 csc 0 = - 1 sin 0 sin — 4 = 7 =- or V2 — \/2 sin i 4 = — ■ 2 2 CHECK You can check your answer by using a graphing calculator. 157T csc —— ' -1 .4 1 4 ✓ 4 -y /2 : -1 .4 1 4 ✓ p GuidedPractice Find the exact value of each expression. 4A. tan 4B. sin 4C. sec(—135°) 6 If the value of one or more of the trigonom etric functions and the quadrant in which the terminal side of 9 lies is known, the remaining function values can be found. E S ! I 3 3 Use One Trigonometric Value to Find Others _5 12 Let tan 9 = rrr, where sin 6 < 0. Find the exact values of the five rem aining trigonometric functions of 9. To find the other function values, you m ust find the coordinates of a point on the terminal side of 9. You know that tan 9 is positive and sin 9 is negative, so 9 must lie in Q uadrant III. This m eans that both x and y are negative. y 5 Because tan 9 = —or — , use the point (—12, —5) to find r. ' = \fx 2 ^ y 2 Pythagorean Theorem = V ( - 1 2 ) 2 + (—5)2 x = — 12 and y = —5 = V l6 9 or 13 Take the positive square root. Use x = —12, y = —5, and r = 13 to write the five rem aining trigonom etric ratios. sin 9 = —or —-^r r 13 csc 9 = — or ■ Rationalizing the Denominator Be sure to rationalize the denominator, if necessary. L esson 4-3 sec i 12 13 cot 9 = ^ or y -X° r ~ r12! p GuidedPractice WatchOut! 246 13 ' 5 cos V = —or r Find the exact values of the five rem aining trigonom etric functions of 6. 5A. sec 9 = V 3 , tan 9 < 0 T r ig o n o m e tric F u n c tio n s o n th e U n it C ircle 5B. sin 9 = cot 9 > 0 Real-World Example 6 Find Coordinates Given a Radius and an Angle ROBOTICS As part of the range of motion category in a high school robotics com petition, a student program m ed a 20-centim eter long robotic arm to pick up an object at point C and rotate through an angle of exactly 225° in order to release it into a container at point D. Find the position of the object at point D, relative to the pivot point O. 225“ W ith the pivot point at the origin and the angle through which the arm rotates in standard position, point C has coordinates (20, 0). The reference angle 9' for 225° is 225° — 180° or 45°. Let the position of point D have coordinates (x, y). The definitions of sine and cosine can then be used to find the values of x and y. The value of r, 20 centim eters, is the length of the robotic arm. Since D is in Q uadrant III, the sine and cosine of 225° are negative. cos 9 = — cos 225° = RoboCup is an international competition in which teams compete in a series of soccer matches, depending on the size and intelligence of their robots. The aim of the project is to advance artificial intelligence and robotics research. Source: RoboCup —cos 45° = _V 2 2 x 20 x 20 x ■ 20 -1 0 V 2 = : Cosine ratio Sine ratio 0 = 225° and r = 20 cos 225° = —cos 45° cos 45° = 2 Solve for x. y_ sin 225° 20 sin 225° = - s i n 45° - s i n 45° = y_ _V 2. 2 20 -10V 2 = 9 = 225° and r = 20 y sin 45' 2 Solve for y. The exact coordinates of D are ( —I 0 V 2 , —I 0 V 2 ). Since 10\/2 is about 14.14, the object is about 14.14 centim eters to the left of the pivot point and about 14.14 centim eters below the pivot point. p GuidedPractice 6. CLOCKWORK A 3-inch-long m inute hand on a clock shows a time of 45 m inutes past the hour. W hat is the new position of the end of the m inute hand relative to the pivot point at 10 m inutes past the next hour? 2 Trigonometric Functions on the Unit Circle a unit circle is a circle of radius 1 centered at the origin. Notice thaton a unit circle, the radian m easure of a central angle 9 = y or s, so the arc length intercepted by 9 corresponds exactly to the angle's radian measure. This provides a way of mapping a real number input value for a trigonometric function to a real num ber output value. StudyTip Wrapping Function The association of a point on the number line with a point on a circle is called the wrapping function, w(t). For example, if w(t) associates a point f on the number line with a point P(x, y) on the unit circle, then w(ir) = ( - 1 , 0 ) and w(2ir) = (1,0). Consider the real num ber line placed vertically tangent to the unit circle at (1, 0) as shown below. If this line were wrapped about the circle in both the positive (counterclockwise) and negative (clockwise) direction, each point t on the line would map to a unique point P(x, y) on the circle. Because r = 1, we can define the trigonom etric ratios of angle t in terms of just x and y. P o s itiv e V a lu e s o f t N e g a tiv e V a lu e s o f t KeyConcept Trigonometric Functions on the Unit Circle Let f be any real number on a number line and let P(x, y) be the point on t when the number line is wrapped onto the unit circle. Then the trigonometric functions of fare as follows. P (x,y) or P (c o s t, sin t) cos t = x ta n f= |,x ^ 0 c s c f= ly = jtO s e c t = j,x j= 0 c o tf= |,y ^ 0 Therefore, the coordinates of P corresponding to the angle f can be written as P(cos t, sin t). ^ - < Ac V x sin f = y y ) X L J Notice that the input value in each of the definitions above can be thought of as an angle measure or as a real number t. W hen defined as functions of the real num ber system using the unit circle, the trigonometric functions are often called circular functions. Using reference angles or quadrantal angles, you should now be able to find the trigonometric function values for all integer m ultiples of 30°, or ^ radians, and 45°, or ^ radians. These special values wrap to 16 special points on the unit circle, as show n below. 16-Point Unit Circle StudyTip 16-Point Unit Circle You have already memorized these values in the first quadrant. The remaining values can be determined using the x-axis, y-axis, and origin symmetry of the unit circle along with the signs of xand yin each quadrant. Using the (x, y ) coordinates in the 16-point unit circle and the definitions in the Key Concept Box at the top of the page, you can find the values of the trigonom etric functions for com m on angle measures. It is helpful to m em orize these exact function values so you can quickly perform calculations involving them. P E E E 0 3 2 Find Trignometric Values Using the Unit Circle Find the exact value of each expression. If undefined, w rite u n d efin ed. a. s in f •j corresponds to the point (x, y) = |y, ^ - J on the unit circle. sin t = y 7T V3 sin y = “ 2 “ 248 Definition of sin t V3 . . Tt y=-y-when f = y - | Lesson 4 -3 j T r ig o n o m e tric F u n c tio n s o n th e U n it C ircle b. co s 135° 135° corresponds to the point (x, y) = Definition ofcost cos t = x cos 135°= C. ° n the unit circle. ~ ~ ^ r when *= 135°' ta n 2 7 0 ° 270° corresponds to the point (x, y) = (0, —1) on the unit circle. y Definition of tan t tan t = j x = 0 and y = -1 , when t= 270°. tan 270° = Therefore, tan 270° is undefined. d. c s c ^ l l i v __________ . \ (V 3 1 corresponds to the point (x, y) = {——, —i j on the unit circle. Definition of csc t csc t = — ' y c s c -^ t— = ~^r 6 i y = —-J-whenf 2 2 = —2 Simplify, p GuidedPractice 7A. cos ~ 4 StudyTip Radians vs. Degrees While we could also discuss one wrapping as corresponding to an angle measure of 360°, this measure is not related to a distance. On the unit circle, one wrapping corresponds to both the angle measuring 2 ir and the distance 2 tt around the circle. 7B.sin 120° 7C.cot 210° 7D.sec ~ 4 As defined by wrapping the num ber line around the unit circle, the dom ain of both the sine and cosine functions is the set of all real numbers (—o o , o o ). Extending infinitely in either direction, the number line can be wrapped m ultiple times around the unit circle, m apping more than one f-value to the same point P(x, y) with each wrapping, positive or negative. Because cos t = x, sin t = y, and one w rapping corresponds to a distance of 2tt, cos (t + 2«7r) = cos t and sin (t + 2 mr) = sin t, for any integer n and real num ber t. 71 connectED.mcgraw-hill.com § 249 StudyTip Periodic Functions The other three circular functions are also periodic. The periods of these functions will be discussed in Lesson 4-5. The values for the sine and cosine function therefore lie in the interval [—1 ,1 ] and repeat for every integer m ultiple of 2-tt on the number line. Functions w ith values that repeat at regular intervals are called periodic functions. V K eyConcept Periodic Functions A function y = f(t) is periodic if there exists a positive real number c such that f(t + c) = f[t) for all values of t in the domain of f. The smallest number c for which f is periodic is called the period of f. The sine and cosine functions are periodic, repeating values after 27T, so these functions have a period of 2 t t . It can be show n that the values of the tangent function repeat after a distance of t t on the number line, so the tangent function has a period of tt and ta n t = t a n ( t 4- H7r), for any integer n and real number f, unless both tan t and tan (t + ntt) are undefined. You can use the periodic nature of the sine, cosine, and tangent functions to evaluate these functions. w m Use the Periodic Nature of Circular Functions m Find the exact value of each expression. _ H it a. cos ——4 cos —r — = cos ( + 2 tt I 4 V 4 = cos ~ I Rewrite -US- as the sum of a number and 2-re. 4 SE and ^ 4 4 4 + 2it map to the same point (x, y) = [ —— , \ 1on 2 2 / the unit circle. = —^ r 2 b. sin cos f = xand x = w h e t ? f = - ^ , 2 4 (-f) sin f — = sin + 2 (—l)T rj = s in 4 ? Rewrite as the sum of a number and an integer multiple of 2x. 4 ? and ™ - 2(—1)ir map to the same point (x, y) = ( - ~ J J \ C. C. / on the unit circle. = — „ , si nf = y a n d y = —^ p w h e n ? = 4 p 19-jt 6 C. tan —— ta n o = ta n (^- + 37t) \o / = tan — 6 Rewrite 6 and 6 6 as the sum of a number and an integer multiple of it. -f 3 it map to points on the unit circle with the same tangent values. = - 7= o r - ^ V3 3 tan t = ~ ; x = ~ and y = ~r when t —^ . x’ 2 ’ 2 6 ~T ^ GuidedPractice 8A. sin 8B.cos 8C.t a n ~ - Recall from Lesson 1-2 that a function/is even if for every x in the dom ain o t f f ( —x) = f( x ) and odd if for every x in the domain o f f , f ( —x) = —f(x ). You can use the unit circle to verify that the cosine function is even and that the sine and tangent functions are odd. That is, cos (—f) = cos t 250 L esson 4 -3 T r ig o n o m e tric F u n c tio n s o n th e U n it C ircle sin (—t) = —sin f tan (—f) = —tan t. Exercises m The given point lies on the term inal side of an angle 9 in standard position. Find the values o f the six trigonom etric functions of 0. (Example 1) 1. (3,4) 21 ( - 6 , 6 ) 3. ( - 4 , - 3 ) 4. (2 ,0 ) 5. (1, - 8 ) 6. (5, - 3 ) 7. ( - 8 ,1 5 ) 8. ( - 1 , - 2 ) 4 1 ) CAROUSEL Zoe is on a carousel at the carnival. The diam eter of the carousel is 8 0 feet. Find the position of her seat from the center of the carousel after a rotation of 2 1 0 ° . (Example 6) J Find the exact value o f each trigonom etric fu nction, if defined. If not d efined, w rite u n d efin ed. (Example 2) 9. s i n y 11. cot (-1 8 0 °) 13. cos (-2 7 0 °) © tan 27T csc 270° sec 180° 15. tan tt SCC( - f Sketch each angle. T h en find its reference ; 17. 135° 19. 18. 210° © 7tr 12 I I tv 3 22. - 7 5 ° 21. -4 0 5 ° s> 13tt 6 5ir 23. 6 Find the exact value o f each expression. (Example 4) 4tv 25. cos — , 7ir 26. tan — 6 27. s i n ^ f 4 28. cot (—45°) 29. csc 390° 30. sec (-1 5 0 ° ) , IItt 31. tan —— 6 32. sin 300° Find the exact values o f the five rem aining trigonom etric functions of 0. (Example 5) 33. tan 9 = 2, where sin 9 > 0 and cos 9 > 0 34. csc 9 = 2, where sin 9 > 0 and cos 9 < 0 35. sin 9 = ——, where cos 9 > 0 5 36. cos 9 = —j j , where sin 9 < 0 37. sec 9 = \[?>, where sin 9 < 0 and cos 9 > 0 38. cot 9 = 1 , where sin 9 < 0 and cos 9 < 0 39. tan 0 = —1, where sin 0 < 0 40. cos 9 - 42. COIN FUNNEL A coin is dropped into a funnel where it spins in sm aller circles until it drops into the bottom of the bank. The diam eter of the first circle the coin makes is 24 centimeters. Before com pleting one full circle, the coin travels 150° and falls over. W hat is the new position of the coin relative to the center of the funnel? (Example 6) 1 , where sin 9 > 0 Find the exact value o f each expression. If un defined, write u n defin ed. (Examples 7 and 8) 43. sec 120° 44. sin 315° 45. c o s ^ 46. tan (-5 2 1 ) 47. csc 390° 48. cot 510° 49. csc 5400° 50. sec 5tt\ 17tv 51. cot (-? ) 52. csc 53. tan 5tt 54. s e c -^p 6 55. sin — 56. cos 57. tan 14tt 58. cos (-¥ ) 7tt (-t ) 59. RIDES Jae and Anya are on a ride at an amusement park. After the first several sw ings, the angle the ride makes with the vertical is m odeled by 9 = 22 cos 7rf, w ith 9 m easured in radians and t measured in seconds. Determ ine the m easure of the angle in radians for t = 0, 0 .5 ,1 ,1 .5 ,2 , and 2.5. (Example 8) 77. Complete each trigonometric expression. cos 60° = s in 60. 61. tan = s in ___ TIDES The depth y in m eters of the tide on a beach varies as a sine function of x, the hour of the day. On a certain day, that function was y = 3 sin 62. sin ~ 3 = cos 63. cos ^ sin (-4 5 ° ) = c o s 64. 65. cos 6 = s in ___ -(x ~ 4) 4- 8, where x = 0 ,1 , 2, ..., 24 corresponds to 12:00 midnight, 1:00 2:00 a . m ., ..., 12:00 m idnight the next night. = s in _ a . m ., a. W hat is the m axim um depth, or high tide, that day? 66. ICECREAM The m onthly sales in thousands of dollars for Fiona's Fine Ice Cream shop can be modeled by TT(t - 4) y = 71.3 + 59.6 s in >where t = 1 represents b. At w hat time(s) does the high tide occur? 78. January, t = 2 represents February, and so on. MULTIPLE REPRESENTATIONS In this problem , you will investigate the period of the sine function. a. a. Estimate the sales for January, March, July, and October. b. Describe why the ice cream shop's sales can be represented by a trigonometric function. o 67. cos(—9) = y -; cos 9 = 1 ; sec 9 = 1 69. sec 9 = 7 iiiii •77 'K 6 4 7T 3 27T sin0 jjj|||j|| sin20 lllll Use the given values to evaluate the trigonometric functions. 68. sin(—9) = TABULAR Copy and com plete a table similar to the one below that includes all 16 angle measures from the unit circle. sin40 jjlllll sin 9 = 1; csc 9 = ? cos 9 = 1 ; cos(—9) = 1 b. VERBAL After w hat values of 9 do sin 9, sin 29, and sin 49, repeat their range values? In other words, what are the periods of these functions? C. VERBAL M ake a conjecture as to how the period of y = sin n9 is affected for different values of n. 70. csc 9 = -jy; sin 9 = 1 ; sin(—9) = 1 71. GRAPHS Suppose the terminal side of an angle 9 in standard position coincides with the graph of y = 2x in Quadrant III. Find the six trigonometric functions of 9. H.O.T. Problem s Use Higher-Order Thinking Skills (79) CHALLENGE For each statem ent, describe n. Find the coordinates of P for each circle with the given radius and angle measure. a. cos (n •y j = 0 b. csc In • y j is undefined. REASONING Determ ine w hether each statement is true or fa l s e . Explain your reasoning. 80. 81. If cos 9 = 0.8, sec 9 — cos (—0)= 0.45. Since tan (—f) = —tan t, the tangent of a negative angle is a negative number. 82. WRITING IN MATH Explain why the attendance at a year-round them e park could be m odeled by a periodic function. W hat issues or events could occur over time to alter this periodic depiction? REASONING Use the unit circle to verify each relationship. 83. sin (—f) = —sin t 84. cos (—f) = cos t 85. tan (—t) = —tan t 76. COMPARISON Suppose the terminal side of an angle 9 } in standard position contains the point (7, —8), and the terminal side of a second angle 92 in standard position contains the point (—7, 8). Compare the sines of 9 ^ and 92. 252 Lesson 4 -3 | T r ig o n o m e tric F u n c tio n s o n th e U n it C ircle 86. WRITING IN MATH Make a conjecture as to the periods of the secant, cosecant, and cotangent functions. Explain your reasoning. Spiral Review Write each decimal degree measure in DM S form and each DMS measure in decimal degree form to the nearest thousandth. (Lesson 4-2) 87.168.35° 88.27.465° 89. 1 4 ° 5 '2 0 " 90. 173° 2 4 '3 5 " 91. EXERCISE A preprogrammed w orkout on a treadmill consists of intervals walking at various rates and angles of incline. A 1% incline m eans 1 unit of vertical rise for every 100 units of horizontal run. (Lesson 4-1) a. At what angle, w ith respect to the horizontal, is the treadm ill bed when set at a 10% incline? Round to the nearest degree. b. If the treadmill bed is 40 inches long, what is the vertical rise w hen set at an 8% incline? Evaluate each logarithm. Lesson 3-3) 92. log8 64 93. log 1 2 5 ' 94. log2 32 95. log4 128 List all possible rational zeros of each function. Then determine which, if any, are zeros. (Lesson 2-4) 96. f( x ) = x3 - Ax2 + x + 2 97. g(x) = x 3 + 6x2 + lOx + 3 98. h(x) = x4 — x 2 + x — 1 99. h(x) = 2 x 3 + 3 x 2 — 8x + 3 100./(x) = 2 x 4 + 3x3 - 6 x 2 - l l x - 3 101. g(x) = 4 x 3 + x 2 + 8x + 2 i I 102. NAVIGATION A global positioning system (GPS) uses satellites to allow a user to determ ine his or her position on Earth. The system depends on satellite signals that are reflected to and from a hand-held transmitter. The tim e that the signal takes to reflect is used to determine the transm itter's position. Radio waves travel through air at a speed of 299,792,458 meters per second. Thus, d(t) = 299,792,458/: relates the time t in seconds to the distance traveled d(t) in meters. (Lesson 1-1) J a. Find the distance a radio wave will travel in 0.05, 0 .2 ,1 .4 , and 5.9 seconds. ^ b. i If a signal from a GPS satellite is received at a transm itter in 0.08 second, how far from the transmitter is the satellite? Skills Review for Standardized Tests 103. SAT/ACT In the figure, AB and AD are tangents to circle C. W hat is the value of m l 105. REVIEW Find the angular speed in radians per second of a point on a bicycle tire if it com pletes 2 revolutions in 3 seconds. F H 104. Suppose 6 is an angle in standard position w ith sin 9 > 0. In which quadrant(s) could the terminal side of 9 lie? A I only C I and III B I and II D I and IV J 106. 2 tt 3 4tt 3 REVIEW W hich angle has a tangent and cosine that are both negative? A 110° B 180° C 210° D 340° 253 Graphing Technology Lab oooo oooo oooo Graphing the Sine Function Parametrically o o o H H H BI Use a graphing calculator and parametric equations to graph the sine function and its inverse. As functions of the real number system, you can graph trigonometric functions on the coordinate plane and apply the same graphical analysis that you did to functions in Chapter 1. As was done in Extend 1-7, parametric equations will be used to graph the sine function. Activity 1 Parametric Graph of y = sin x G raph x = t , y = sin t. CT7?fn Set the mode. In the I MODE j menu, select RADIAN, PAR, and SIMUL. This allows the equations to be graphed simultaneously. N ext, enter the param etric equations. In param etric form, X ,T ,9 ,n | will use f instead of x. SCI P loti Plots PlotJ Eri'3 0i23HS67B9 DEGREE F'DL SEU DDT__ 1~EQLJE TlTIM L S O T \ X itB T V it B s i n ( T ) \ X :t = V ;t = n X jt = VJt= \X ht = af;ii a+bi, H H HDRIZ G-T SET CLOCK W INDOW ET7SBW Set the x- and f-values to range from 0 to 2tt. Set Tm i n = 0 Tstep and x-scale to y^-. Set y to [—1 ,1 ] scl: 0.1. The T n a x = 6 . 2 8 3 1 8 5 3 .. T s t e p = . 2 6 1 7 9 9 3 .. calculator automatically converts to decimal form. X n in = 0 X n a x = 6 . 2 8 3 1 8 5 3 .. X s c l = . 2 6 1 7 9 9 3 8 .. ■ W m n= ‘ 1 Graph the equations. Trace the function to identify points along the graph. Select Trace and use the right arrow to m ove along the curve. Record the corresponding x- and y-values. T:.E2SE?B70 N=.£23E9B7B V=.S ■ [0, 2ir] scl: ^ by [ - 1 ,1 ] scl: 0.1 t: [0, 2it]; fstep The table shows angle measures from 0° to 180°, or 0 to tv , and the corresponding values for sin t on the unit circle. The figures below illustrate the relationship betw een the graph and the unit circle. Degrees StudyTip Decimal Equivalents Below are the decimal equivalents of common trigonometric values. ~ = 0.866 - j - * 0.707 0 30 Radians 0 0.52 y = sin t 0 0.5 ( - 0 . 5 , 0.9) ( - 0 . 7 , 0.7) -0 .9 , 0.5) ( - 1 , 0) Lesson 4 -4 135 150 180 1.571 2.094 2.356 2.618 3.14 1 0.866 0.707 0.5 0 90 0.79 1.05 0.707 0.866 (1, 0) (0.9, -0 .5 ) (0.7, -0 .7 ) (0.5, -0 .9 ) (0, - 1) 120 60 (0.5, 0.9) (0.7, 0.7) (0.9, 0.5) -0 .9 , -0 .5 ) (-0 .7 , -0 . ( - 0 . 5 , -0 .9 ) 254 45 Exercises Graph each function on [0,2 tt]. 1. x = t ,y = c o s t 2. x = f,y = s in 2 t 3. x = f, y = 3 cos t 4. x = t, y = 4 sin t 5. x = t ,y = cos (f + it) 6. x = t, y = 2 sin —- j j By definition, sin f is the y-coordinate of the point P(x, y) on the unit circle to which the real number f on the number line gets wrapped. As shown in the diagram on the previous page, the graph of y = sin t follows the y-coordinate of the point determined by t as it moves counterclockwise around the unit circle. The graph of the sine function is called a sine curve. From Lesson 4-3, you know that the sine function is periodic with a period of 2tt. That is, the sine curve graphed from 0 to 2tt would repeat every distance of 2tt in either direction, positive or negative. Parametric equations can be used to graph the inverse of the sine function. Activity 2 Graph an Inverse Graph x + t, y + sin t and its inverse. Then determ ine a domain for which y — sin t is one-to-one. P T T m Inverses are found by switching x and y. Enter the given equations as X l T and Y lT . To graph the inverse, set X 2T = Y i t and Y 2T = X lT . These are found in the VARS 1menu. Select Y-VARS, parametric, Y u . Repeat for X u . StudyTip Tstep If your graph appears to be pointed, you can change the tstep to a smaller value in order to get a smoother curve. Graph the equations. Adjust the window so that both of the graphs can be seen, as shown. You may need to set the tstep to a smaller value in order to get a sm ooth curve. x = t, y = sin f [ —3 ir, 3 it] s c l : b y [ - 1 0 , 1 0 ] scl: 2 t: [—3tt, 3it]; fstep Because the sine curve is periodic, there are an infinite num ber of domains for w hich the curve will pass the horizontal line test and be one-to-one. One such domain is TT 3TT 2 ' 2 Exercises Graph each function and its inverse. Then determ ine a domain for which each function is one-to-one. 7. x = t ,y = cos 21 9. x = t, y = 2 cos t 11. x = t, y = 2 cos (f — t t ) 8. x = 10. X = 12. X = connectED.mcgraw-hill.com | 255 • (Lesson 1-5) 2 NewVocabulary sinusoid amplitude frequency phase shift vertical shift midline As you ride a Ferris wheel, the height that you are above the ground varies periodically as a function of time. You can model this behavior using a sinusoidal function. Graph transformations of the sine and cosine functions. You analyzed graphs of functions. Use sinusoidal functions to solve problems. Transformations of Sine and Cosine Functions As show n in Explore 4-4, the graph y = sin t follows the y-coordinate of the point determ ined by t as it m oves around the unit circle. Similarly, the graph of y = cos t follow s the x-coordinate of this point. The graphs of these functions are periodic, repeating after a period of 2tt. The properties of the sine and cosine functions are sum marized below. 1 KeyConcept Properties of the Sine and Cosine Functions Cosine Function Sine Function D o m a in : R ange: ( - 00, 00) y -in te r c e p t: [—1 ,1 ] x -in te r c e p ts : rnr, n e Z R a n g e : [—1,1] (— 00, 00) y -in te r c e p t: 0 x -in te rc e p ts : D o m a in : 1 y n, n e Z C o n tin u ity : continuous on ( - 00, 00) C o n tin u ity : continuous on (— 00, 00) S y m m e try : origin (odd function) S y m m e try : y-axis (even function) E x tre m a : E x tre m a : maximum of 1 at X = y + 2 n it, 0 6 Z minimum of - 1 at x = i t + 2 n-rc, minimum of - 1 at neZ 3ir E nd B e h a v io r: O s c illa tio n : lim X — * — OC sin maximum of 1 at x = 2m:, os Z X -+00 between - 1 and 1 E nds B x and lim inehavior: /d o not exist. lim X— * — 0 0 O s c illa tio n : cos xand lim cos xdo not exist. X— »oo between - 1 and 1 The p o rtio n o f each graph on [0, 2tt] rep resents one p eriod or cy cle o f the fu nction . N otice th at the cosine graph is a h o rizo n tal tra n sla tio n o f the sine graph. A n y tran sform atio n o f a sin e fu nction is called a sin u so id . T he g eneral form o f the sin u soid al fu n ction s sine and cosine are y = a sin (bx + c) + d and where a, b, c, and d are constants and neither a nor b is 0. 256 | L esson 4 -4 y = a cos (bx + c) + d Notice that the constant factor a in y = a sin x and y = a cos x expands the graphs of y = sin x and y = cos x vertically if \a\ > 1 and com presses them vertically if \a\ < 1. StudyTip Dilations and x-intercepts Notice that a dilation of a sinusoidal function does not affect where the curve crosses the x-axis, at its x-intercepts. Vertical dilations affect the amplitude of sinusoidal functions. KeyConcept Amplitudes of Sine and Cosine Functions Words M odel The am plitude of a sinusoidal function is half the distance between the maximum and minimum values of the function or half the height of the wave. Symbols V ; .......... r" amplitude \ / Fo ry = asin (bx + c ) + (/and y = a cos (fix + c) + d, amplitude = lal. / jrp p h tu d e ... J To graph a sinusoidal function of the form y = a sin x or y = a cos x, plot the x-intercepts of the parent sine or cosine function and use the am plitude \a\ to plot the new m axim um and minimum points. Then sketch the sine wave through these points. P E 3 J 0 3 J j ^ raPtl Vertical Dilations of Sinusoidal Functions D escribe how the graphs of f i x ) — sin x and gix ) = j sin x are related. T h en find the am plitude of g ix ), and sketch two periods o f both fu nctions on the sam e coordinate axes. The graph of g ix ) is the graph of fi x ) com pressed vertically. The am plitude of g(x ) is |-j |or i . Create a table listing the coordinates of the x-intercepts and extrem a for/(x) = sin x for one period on [0, 2tt]. Then use the amplitude of g ix ) to find corresponding points on its graph. Function x-intercept f(x) = sin x g(x) = (0 ,0 ) ^ sin x (0 ,0 ) M axim um (f’1) I2 ' 4>1 fir 11 x-intercept M inim um ( iv , 0 ) (¥■-’) (iv, 0) I'31 2ir ’ 1 ' 4, x-intercept (2 tv, 0) (2 -tv , 0 ) Sketch the curve through the indicated points for each function. Then repeat the pattern suggested by one period of each graph to com plete a second period on [2 tt, 4 tt]. Extend each curve to the left and right to indicate that the curve continues in both directions. StudyTip Radians Versus Degrees You could rescale the x-axis in terms of degrees and produce sinusoidal graphs that look similar to those produced using radian measure. In calculus, however, you will encounter rules that depend on radian measure. So, in this book, we will graph all trigonometric functions in terms of radians. p GuidedPractice D escribe how the graphs o i f i x ) and gix ) are related. T h en find the am plitude of gix), and sketch two periods o f b oth fu nctions on the same coordinate axes. 1A. fi x ) COS X 1 g ix ) = - c o s x 1B. f i x ) = sin x gix ) = 5 sin x 1C. f i x ) = cos X gix ) = 2 cos x 257 If a < 0, the graph of the sinusoidal function is reflected in the x-axis. Graph Reflections of Sinusoidal Functions Describe how the graphs of f i x ) — cos x and g ix ) = —3 cos x are related. Then find the amplitude of g ix ), and sketch two periods of both functions on the same coordinate axes. WatchOut! The graph of g(x) is the graph of /(x) expanded vertically and then reflected in the x-axis. The am plitude of g(x) is |—3| or 3. Amplitude Notice that Example 2 does not state that the amplitude of <7(x) = - 3 cos x is -3 . Amplitude is a height and is not directional. Create a table listing the coordinates of key points of /(x) = cos x for one period on [0, 2tv]. Use the am plitude of g(x) to find corresponding points on the graph of y = 3 cos x. Then reflect these points in the x-axis to find corresponding points on the graph of gix). f(x ) y = = Extremum /-in te rc e p t Extremum /-in te rc e p t Extremum (0.1) i[?•») (TV, - 1 ) ( f- o ) (2 tv , 1) (0,3) iI f ) (tv, - 3 ) 1(f . o ) (2tv, 3) (0, - 3 ) i( f ° ) (TV, 3) 1I f ' 0) CO I £ CNJ, Function cos X 3 cos x g (x ) = - 3 cos x Sketch the curve through the indicated points for each function. Then repeat the pattern suggested by one period of each graph to com plete a second period on [27V, 4 tv], Extend each curve to the left and right to indicate that the curve continues in both directions. w GuidedPractice Describe how the graphs o f/(.r) a n d g (x ) are related. Then find the amplitude of g i x sketch two periods of both functions on the same coordinate axes. 2A. f{ x ) = cos x g ( x ) = —| -co sx ), and 2B. f( x ) = sin x g (x) = —4 s in x In Lesson 1-5, you learned that if g (x ) = f( b x ) , th eng(x) is the graph of/(x) com pressed horizontally if \b\ > 1 and expanded horizontally if \b\ < 1. H orizontal dilations affect the period of a sinusoidal function— the length of one full cycle. 258 | L esson 4 -4 | G ra p h in g S ine a n d C o s in e F u n c tio n s Determining Period When determining the period of a periodic function from its graph, remember that the period is the smallest distance that contains all values of the function. KeyConcept Periods of Sine and Cosine Functions Words Symbols The period of a sinusoidal function is the distance between any two sets of repeating points on the graph of the function. Model V F o r y = a s i n ( / w + c ) + (fand y = a cos (bx + c) + d, where b=f= 0, period = f y period To graph a sinusoidal function of the form y = sin bx or y = cos bx, find the period of the function period and successively add — - — to the left endpoint of an interval w ith that length. Then use these values as the x-values for the key points on the graph. J E S 3 H 1 3 & GraP*1 Horizontal Dilations of Sinusoidal Functions Describe how the graphs of f i x ) = cos x and g (x ) = cos j are related. Then find the period of g ix ), and sketch at least one period of both functions on the same coordinate axes. y 1 Because cos — = cos —x, the graph of g (x) is the graph of f( x ) expanded horizontally. The period of g(x) is t— or 6tt. I 3| Because the period of g(x) is 6tv, to find corresponding points on the graph of g(x), change the x-coordinates of those key points on/(x) so that they range from 0 to 67V, increasing by increments of or 4 p Function M axim um II O O cn WatchOut! g(x) = cos | (0 ,1 ) (0 ,1 ) x-intercept (w,-1) ( f ’ °) (t ’O M inim um ) (3ir, —1) x-intercept (?■ “ ] (¥■ “ ) M axim um (2ir, 1) (6 tv,1 ) Sketch the curve through the indicated points for each function, continuing the patterns to com plete one full cycle of each. §►GuidedPractice Describe how the graphs of fi x ) an d g (x ) are related. Then find the period o i g ix ), and sketch at least one period of each function on the same coordinate axes. 3A. f( x ) = cos x g (x ) = cos -y 3B. f( x ) = sin x g (x ) = sin 3x 3C. f( x ) = cos x g (x ) = COS-jX Horizontal dilations also affect the frequency of sinusoidal functions. KeyConcept Frequency of Sine and Cosine Functions Words The frequency of a sinusoidal function is the number of cycles the function completes in a one unit interval. The frequency is the reciprocal of the period. Symbols F o r y = a s i n ( / w + c ) + dand y = a cos (bx + c) + oI, Model frequency = — V -r or — . period m mIttiSE - ... ' Because the frequency of a sinusoidal function is the reciprocal of the period, it follows that the period of the function is the reciprocal of its frequency. j Real-World Example 4 Use Frequency to Write a Sinusoidal Function J > Real-WorldLink In physics, frequency is measured in hertz or oscillations per second. For example, the number of sound waves passing a point A in one second would be the wave’s frequency. Source: Science World 2tt MUSIC M usical notes are classified by frequency. In the equal tem pered scale, middle C has a frequency of 262 hertz. Use this inform ation and the inform ation at the left to w rite an equation for a sine function that can be used to model the initial behavior of the sound wave associated with middle C having an amplitude of 0.2. The general form of the equation will be y = a sin bt, where t is the time in seconds. Because the am plitude is 0.2, \a\ = 0.2. This means that a = ± 0 .2 . The period is the reciprocal of the frequency or y y . Use this value to find period = ~ 1 262 2 tv \b\ \b\ = 2tt(262) or 524tt b = ±5247f b. Period formula period = 1 262 Solve for |6j. Solve for b. By arbitrarily choosing the positive values of a and b, one sine function that models the initial behavior is y = 0.2 sin 5247rt. ►GuidedPractice 4. MUSIC In the same scale, the C above middle C has a frequency of 524 hertz. W rite an equation for a sine function that can be used to model the initial behavior of the sound wave associated with this C having an am plitude of 0.1. A phase of a sinusoid is the position of a wave relative to som e reference point. A horizontal translation of a sinusoidal function results in a phase shift. Recall from Lesson 1-5 that the graph of y = f ( x + c) is the graph of y = f( x ) translated or shifted |c| units left if c > 0 and |c| units right if c < 0. 260 Lesson 4 -4 G ra p h in g S ine a n d C o sin e F u n c tio n s KeyConcept Phase Shift of Sine and Cosine Functions Words The phase shift of a sinusoidal function Model is the difference between the horizontal position of the function and that of an otherwise similar sinusoidal function. Symbols For y = a sin (bx + c) + d and y = a cos (bx+ c) + a1, where b =/= 0, phase shift = phase shift 1*1 You will verify the formula for phase shift in Exercise 44. StudyTip Alternative Form The general forms of the sinusoidal functions can also be expressed as y = a sin b(x- h) + k and y = a cos b(x—h) + k. In these forms, each sinusoid has a phase shift of h and a vertical translation of /(in comparison to the graphs of y = a sin bx and y = a cos bx. To graph the phase shift of a sinusoidal function of the form y = a sin (bx + c) 4- d or y = a cos (bx + c) + d, first determ ine the endpoints of an interval that corresponds to one cycle of the graph by adding —j to each endpoint on the interval [0, 27t] of the parent function. B J 2 H uIEI33E! GraPh Horizontal Translations of Sinusoidal Functions State the amplitude, period, frequency, and phase shift of y — sin two periods of the function. — y j . Then graph In this function, a = 1, b = 3, and c = — Period: # f = ^—o or r — ^ |3| 3 Am plitude: \a\ = |1| or 1 Frequency: |3| 2-k ° r 2-k 2tt Phase shift: — \b\ = — rr-o r^ |3| To graph y = sin ^3x— y j , consider the graph of y = sin 3x. The period of this function is -y-. 2tc Create a table listing the coordinates of key points of y = sin 3x on the interval 0, — . To account for a phase shift of ~ , add y to the x-values of each of the key points for the graph of y = sin 3x. x-intercept Function y = sin 3 / y = sin ( 3 * - y ) (0 ,0 ) m Sketch the graph of y = sin com plete two cycles. M axim um /-in te rc e p t Minimum i I( ] M i ( f - 1) It ') I( f ° ) i (T - ~ ' . I /-in te rc e p t 1 (t -° ) — y j through these points, continuing the pattern to ►GuidedPractice State the amplitude, period, frequency, and phase shift of each function. Then graph two periods of the function. 5A. y = c o s (| + ^ ) 5B. y = 3 sin ^2x — y j —-. fHconnectED.mcgraw-hill.com | 261 StudyTip Notation sin ( / + d) j= sin x -t- d The first expression indicates a phase shift, while the second expression indicates a vertical shift. The final w ay to transform the graph of a sinusoidal function is through a vertical translation or Nvertical shift. Recall from Lesson 1-5 that the graph of y = f ( x ) + d is the graph of y = f( x ) translated or shifted \d\ units up if d > 0 and \d\ units dow n if d < 0. The vertical shift is the average of the m axim um and m inimum values of the function. The parent functions y = sin x and y = cos x oscillate about the x-axis. After a vertical shift, a new horizontal axis known as the midline becom es the reference line or equilibrium point about w hich the graph oscillates. For exam ple, the midline of y = sin x + 1 is y = 1, as shown. In general, the midline for the graphs of y = a sin (bx + c) + d and y = a cos (bx + c) + d is y = d. Graph Vertical Translations of Sinusoidal Functions State the amplitude, period, frequency, phase shift, and vertical shift of y = sin (x + 2tt) — 1. Then graph two periods of the function. In this function, a = 1, b = 1, c = 2ir, and d = —1. Am plitude: |<i| = |1| or 1 Period: ^ i> Phase shift: Vertical shift: d or —1 lfel —r r = —2tt 1 or 27r Frequency: Jp- = Jp- or -y1 J 2tt 2tt 2tt Midline: y = d o r y ■ First, graph the midline y = —1. Then graph y = sin x shifted 27T units to the left and 1 unit down. Notice that this transformation is equivalent to a translation 1 unit dow n because the phase shift was one period to the left. f GuidedPractice State the amplitude, period, frequency, phase shift, and vertical shift of each function. Then graph two periods of the function. 6A. y = 2 cos x + l The characteristics of transform ations of the parent functions y : : sin x and y = cos x are sum marized below. TechnologyTip Zoom Trig When graphing a trigonometric function using your graphing calculator, be sure you are in radian mode and use the ZTrig selection under the zoom feature to change your viewing window from the standard window to a more appropriate window of [ - 2 i r , 2ir] scl: -ir/2 by [ - 4 , 4 ] scl: 1. > ConceptSummary Graphs of Sinusoidal Functions The graphs of y = a sin (bx + c) + c/and y = a cos (bx + c) + d, where a # 0 and b have the following characteristics. 0, Amplitude: lal Period: | j L Frequency: — or Phase shift: - Vertical shift: d Midline: y = d 2 6 2 | L esson 4 -4 i G ra p h in g S ine a n d C o sin e F u n c tio n s lol 2i t Period Applications of Sinusoidal Functions M any real-world situations that exhibit periodic mm behavior over time can be m odeled by transform ations of y = sin x or y = cos x. Real-World Example 7 Modeling Data Using a Sinusoidal Function METEOROLOGY Use the inform ation at the left to w rite a sinusoidal function that models the num ber of hours of daylight for New York City as a function of time x, where x = l represents January 15, x = 2 represents February 15, and so on. Then use your model to estimate the num ber of hours of daylight on Septem ber 30 in New York City. R TW n M ake a scatter plot of the data and choose a model. The graph appears wave-like, so you can use a sinusoidal function of the form y = a sin (bx + c) + d or y = a cos {bx + c) + d to model the data. We will choose to use y = a cos (bx + c) + d to model the data. 0 ,1 2 ] scl: 1 by [0, 20] scl: 2 Find the m axim um M and m inim um m values of the data, and use these values to find a, b, c, and d. The m axim um and m inim um hours of daylight are 15.07 and 9.27, respectively. The am plitude a is half of the distance betw een the extrema. Real-WorldLink a = | (M - m) = ± (15.07 - 9.27) or 2.9 The table shows the number of daylight hours on the 15th of each month in New York City. Month The vertical shift d is the average of the m axim um and m inim um data values. Hours of Daylight January 9.58 February 10.67 March 11.9 April 13.3 May 14.43 June 15.07 July 14.8 August 13.8 September 12.48 October 11.15 November 9.9 December 9.27 Source: U.S. Naval Observatory d = | (M + m) = | (15.07 + 9.27) or 12.17 A sinusoid com pletes half of a period in the tim e it takes to go from its maxim um to its minim um value. One period is twice this time. : December 15 or month 12 and Period = 2(xmax - x mw) = 2(12 - 6) or 12 i June 15 or month 6 Because the period equals -^f, you can write \b\ = Therefore, \b\ = —J , or |b| J Period 11 12 6 The m axim um data value occurs when x = 6. Since y = cos x attains its first maximum when x = 0, we m ust apply a phase shift of 6 — 0 or 6 units. Use this value to find c. Phase shift = —— |b| 6 = Phase shift formula Phase shift = 6 and |i>[ = - -~ w 6 C = Solve for c. — 7T ETfnFl Write the function using the values for a, b, c, and d. Use b = ^~. 6 y = 2.9 cos ^ x — 7t| + 12.17 is one model for the hours of daylight Graph the function and scatter plot in the same view ing window, as in Figure 4.4.1. To find the num ber of hours of daylight on Septem ber 30, evaluate the model for x = 9.5. y = 2.9 cos (^ (9 .5 ) — 7rj + 12.17 or about 11.42 hours of daylight p GuidedPractice METEOROLOGY The average m onthly tem peratures for Seattle, W ashington, are shown. Figure 4.4.1 Month Jan Feb Mar Apr May Jun Jul Aug Sept Oct Nov Dec Temp. (°) 41 44 47 50 56 61 65 66 61 54 46 42 7A. W rite a function that models the m onthly tem peratures, using x = 1 to represent January. 7B. According to your m odel, what is Seattle's average monthly temperature in February? connectED.mcgraw-hill.com ] 263 Exercises = Step-by-Step Solutions begin on page R29. Describe how the graphs of fix ) and g(x) are related. Then find the amplitude of gix), and sketch two periods of both functions on the same coordinate axes. (Examples 1 and 2) 21. TIDES The table shown below provides data for the first high and low tides of the day for a certain bay during one day in June. (Example 7) Tide 2. f(x ) = cos x 1. f(x ) = sin x g(x) = ~ gix) = -i sin x first high tide cos x first low tide 4. fix ) = sin x 3. f[x ) = cos x gix) = —8 sin x gix) = 6 cos x Describe how the graphs of fix ) and gix) are related. Then gix) = sin 0.25x 22. METEOROLOGY The average monthly temperatures for Boston, M assachusetts are shown. (Example 7) 9. VOICES The contralto vocal type includes the deepest female singing voice. Some contraltos can sing as low as the E below middle C (E3), w hich has a frequency of 165 hertz. Write an equation for a sine function that models the initial behavior of the sound wave associated with E3 having an amplitude of 0.15. (Example 4) Month Write a sine function that can be used to model the initial behavior of a sound wave with the frequency and amplitude given. (Example 4} 10. / = 440, a = 0.3 11. / = 9 3 2 ,0 = 0.25 12. / = 1245,0 = 0.12 13. / = 623, a = 0.2 (v ^ y = 3 s m ( x - f j 17. y = sin 3 x — 2 T8> y = cos ^x — 19.,, y = sin (x + —1 + 4 VACATIONS The average number of reservations R that a vacation resort has at the beginning of each month is shown. (Example 7) Month R Month R Jan 200 May 121 Feb 173 Jun 175 Mar 113 Jul 198 Apr 87 Aug 168 a. Write an equation of a sinusoidal function that models the average num ber of reservations using x = 1 to represent January. b. According to your model, approximately how many reservations can the resort anticipate in November? L esson 4 -4 Month Temp. (°F) Jan 29 Jul 74 Feb 30 Aug 72 Mar 39 Sept 65 Apr 48 Oct 55 May 58 Nov 45 Jun 68 Dec 34 b. Write an equation of a sinusoidal function that models the m onthly temperatures. C. According to your model, w hat is Boston's average tem perature in August? ( j ^ . y = cos (| + 16. y = 0.25 cos x + 3 Temp. (°F) a. Determ ine the am plitude, period, phase shift, and vertical shift of a sinusoidal function that models the m onthly tem peratures using x = 1 to represent January. State the amplitude, period, frequency, phase shift, and vertical shift of each function. Then graph two periods of the function. (Examples 5 and 6) 264 10:55 a.m. 8. fix ) = sin x gix) = cos 0.2x 20. 2.02 C. According to your m odel, what was the height of the tide at 8:45 p . m . that night? gix) = cos 2x 7. fix ) = cos x 4:25 a.m. b. Write a sinusoidal function that models the data. 6. fix ) = cos x gix) = sin 4x Time 12.95 a. Determ ine the amplitude, period, phase shift, and vertical shift of a sinusoidal function that models the height of the tide. Let x represent the number of hours that the high or low tide occurred after midnight. find the period of g(x), and sketch at least one period of both functions on the same coordinate axes. (Example 3) 5. fix ) = sin x Height (ft) G ra p h in g S ine a n d C o sin e F u n c tio n s GRAPHING CALCULATOR Find the values of x in the interval x < 77that m ake each equation or inequality true. (Hint: Use the intersection function.) -tv < 23. —sin x = cos x 24. sin x — cos x = 1 25. sin x + cos x = 0 26. cos x < sin x 27. sin x cos x > 1 28. sin x cos x < 0 ffl) CAROUSELS A wooden horse on a carousel moves up and dow n as the carousel spins. W hen the ride ends, the horse usually stops in a vertical position different from where it started. The position y of the horse after t seconds can be modeled by y = 1.5 sin (2 1 + c), where the phase shift m ust be continuously adjusted to com pensate for the different starting positions. If during one ride the horse 77T reached a m axim um height after — seconds, find the equation that models the horse's position. 30. AMUSEMENT PARKS The position y in feet of a passenger cart relative to the center of a Ferris wheel over f seconds is shown below. 39. $ MULTIPLE REPRESENTATIONS In this problem , you will investigate the change in the graph of a sinusoidal function of the form y = sin x or y = cos x when m ultiplied by a polynom ial function. a. GRAPHICAL Use a graphing calculator to sketch the graphs of y = 2x, y = —2x, and y = 2 x cos x on the same coordinate plane, on the interval [—20, 20]. b. VERBAL Describe the behavior of the graph of y = 2 x cos x in relation to the graphs of y = 2x and y = - 2x. c. GRAPHICAL Use a graphing calculator to sketch the graphs of y = x 2, y = —x 2, and y = x2 sin x on the same coordinate plane, on the interval [—20, 20]. Side view of Ferris wheel overtime interval [0, 5.5] d. VERBAL Describe the behavior of the graph of y = x 2 sin x in relation to the graphs of y = x 2 and 2 y = - x z. t= 0 f= 2 f = 3.75 t= 5.5 a. Find the time t that it takes for the cart to return to y = 0 during its initial spin. e. ANALYTICAL M ake a conjecture as to the behavior of the graph of a sinusoidal function of the form y = sin x or y = cos x w hen multiplied by polynomial function of the form y = f( x ) . b. Find the period of the Ferris wheel. C. Sketch the graph representing the position of the passenger cart over one period. d. Write a sinusoidal function that models the position of the passenger cart as a function of time t. H.O.T. Problem s Use Higher-Order Thinking Skills 40. CHALLENGE W ithout graphing, find the exact coordinates of the first m axim um point to the right of the y-axis for y = 4 sin ( f * - f ) . Write an equation that corresponds to each graph. REASONING D eterm ine w hether each statement is true or fa ls e . Explain your reasoning. 41. Every sine function of the form y = a sin (bx + c) + d can also be w ritten as a cosine function of the form y = a cos (bx + c) + d. 42. The period of f i x ) = cos 8x is equal to four times the period of g(x) = cos 2x. (43) CHALLENGE How m any zeros does y = cos 1500x have on the interval 0 < x < 2ir? 44. PROOF Prove the phase shift formula. 45. WRITING IN MATH The Pow er Tower ride in Sandusky, Ohio, is show n below. Along the side of each tower is a string of lights that send a continuous pulse of light up and down each tower at a constant rate. Explain why the distance d of this light from the ground over time t cannot be represented by a sinusoidal function. Write a sinusoidal function with the given period and amplitude that passes through the given point. 35. period: 7r; amplitude: 5; point: |-j 36. period: 4tt; amplitude: 2; point: (tt, 2) 37. period: amplitude: 1.5; point: 38. period: 37T; amplitude: 0.5; point: |tt, c o n n e c tiB 'm c g r a w ^ L r o m l 265 Spiral Review The given point lies on the terminal side of an angle 9 in standard position. Find the values of the six trigonometric functions of 9. (Lesson 4-3) (- 4 ,4 ) 46. 47. ( 8 , - 2 ) 48. ( - 5 , - 9 ) 49. (4 ,5 ) Write each degree measure in radians as a multiple of tt and each radian measure in degrees. (Lesson 4-2) 50. 25° -4 2 0 ° 51. 53. ^ 52. 54. SCIENCE Radiocarbon dating is a m ethod of estimating the age of an organic material by calculating the amount of carbon-14 present in the material. The age of a m aterial can be calculated using A = t • where A is the age of the object in years, t is the half-life of carbon-14 or 5700 years, and R is the ratio of the am ount of carbon-14 in the sam ple to the am ount of carbon-14 in living tissue. (Lesson 3-4) a. A sample of organic material contains 0.000076 gram of carbon-14. A living sam ple of the same material contains 0.00038 gram. About how old is the sample? b. A specific sample is at least 20,000 years old. W hat is the m axim um percent of carbon-14 remaining in the sample? State the num ber of possible real zeros and turning points of each function. Then determine all of the real zeros by factoring. (Lesson 2-2) 55. f i x ) = x 3 + 2 x 2 — 8x 56. f( x ) = x 4 - W x2 + 9 f( x ) = x5 + 2 x 4 - 4 x 3 - 8x2 57. 58. f( x ) = x 4 - 1 Determine w h e th e r/h a s an inverse function. If it does, find the inverse function and state any restrictions on its domain. (Lesson 1-7) 59. 60. f( x ) f( x ) = - x - 2 61. f i x ) = (x - 3)2 - 7 62. fi x ) = Skills Review fo r Standardized Tests 63. SAT/ACT If x + y = 90° and x and y are both Identify the equation represented by the graph. 65. nonnegative angles, what is equal to ~^ ~ ? A 0 D 1.5 C 1 E Cannot be determ ined from the inform ation given. BI 64. 10 REVIEW If tan x = — in the figure below , what are sin 24 x and cos x? A B C D y = 21 sin 4x y = 4i sin 2x y = 2 sin 2x y = 4 sin h 66. REVIEW If cos 9 = — and the terminal side of the G sin x = i p and cos x = 26 26 angle is in Q uadrant IV, what is the exact value of sin 6? 15 TT 26 J 26 H sin x = — and cos x = — ] sin x = i^r and cos x = —r J 26 24 266 Lesson 4 -4 | G ra p h in g S ine a n d C osine F u n c tio n s C 17 G “ 15 H JJ 17 15 — Graphing Technology Lab oooo oooo oooo Sums and Differences of Sinusoids CDOO The graphs of the sums and differences of two sinusoids will often have different periods than the • Graph and examine the periods of sums and differences of sinusoids. graphs of the original functions. Activity 1 mm Sum of Sinusoids Determine a com m on interval on which b o th /(x ) = 2 sin 3x and g ix ) = 4 cos j complete a whole num ber of cycles. Then graph h(x) = f i x ) + g ix ), and identify the period of the function. ETHTn Enter/(x) for Y i and g(x) for Y 2 . Then adjust the window until each graph com pletes one or m ore w hole cycles on the same interval. One interval on w hich this occurs is [0, 4tt]. On this interval, g (x) com pletes one w hole cycle and/(x) com pletes six whole cycles. R7!TTO To graph 7z(x) as Y3, under the I VARS I menu, select Y-VARS, function, and Y i to enter Y i. Then press I + I and select Y-VARS, function, and Y 2 to enter Y 2 . W l Graph/(x), g(x) and h(x) on the same screen. To m ake the graph of hix) stand out, scroll to the left of the equals sign next to Y3, and press Ie n t e r |. Then graph the functions using the same w indow as above. pioti Pi*t3 ^ iB 2 s in < 3 X ) \ V 2B 4 c o s ( X / 2 ) V ^ B Y i +Vz \ Y h= \Y s = \Y f i= \Y?= TechnologyTip Hiding Graphs Scroll to the equals sign and select enter to make a graph disappear. CTTSflW By adjusting the x-axis from [0, 4-tt] to [0, 8tt] to observe the full pattern of hix), we can see that the period of the sum of the two sinusoids is 4 - tv . Period - [0, 8 ir] scl: 2 tt by [ - 6 , 6] scl: 1 Exercises Determine a com m on interval on which b o th /(x ) and gix) complete a whole num ber of cycles. Then graph aix) = fix ) + gix) and b(x) = f(x ) — gix), and identify the period of the function. 1. f i x ) = 4 sin 2x gix ) = —2 cos 3x 2. 4. f i x ) = j sin 4x 5- / M = g(x) = 2 sin [x - y ) f i x ) = sin 8x gix ) = cos 6x \ cos f g ix ) = - 2 cos (x - y j 3. f i x ) = 3 sin (x — 7v) gix ) = —2 cos 2x 6. fi x ) = —j sin 2x g(x) = 3 cos 2x 7. MAKE A CONJECTURE Explain how you can use the periods of two sinusoids to find the period of the sum or difference of the two sinusoids. ^j^^^^^E^ricgra^iinxoiJ 267 Mid-Chapter Quiz Lessons 4-1 through 4-4 ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ Find th e e x a c t v a lu e s of th e six trig o n o m e tric 12. TRAVEL A car is traveling a t a speed of 5 5 m iles per hour on tires fu nctio ns of 9. (Lesson 4-1) th a t m e asu re 2 .6 fe e t in diam eter. Find th e a p p ro xim ate angular speed of the tires in radians per m inute. (Lesson 4-2) 1. S k e tc h e a c h a n g le . T h e n fin d its re fe re n c e an g le . (Lesson 4-3) 14. 2 1 t t 13. 1 7 5 ° 13 Find th e e x a c t v a lu e o f e a c h e x p re s s io n . If u n d e fin e d , w rite Find th e v alu e of x. Round to th e n e a re s t te n th if necessary. (Lesson. 4-1) 3. ^ 4. undefined. Lesson 4-3) 15. cos 3 1 5 ° 16. sec-y^ sin 18. tan ^ r - 17. 0 Find th e e x a c t v a lu e s of th e fiv e re m a in in g trig o n o m e tric fu n c tio n s of 9. 5. SHADOWS A pine tree casts a shadow th a t is 7 .9 fe e t long w hen th e Sun is 8 0 ° above the horizon. (Lesson 4-1) a. 19. (Lesson 4-3) cos 9 = — 20. cot 9 = 5 w h e re sin 9 < 0 and tan 9 > 0 w h e re cos 9 > 0 and sin 9 > 0 Find the height of the tree. b. Later th a t s am e day, a person 6 fe e t tall casts a s hadow 6 .7 fe e t long. At w h a t angle is the Sun above the horizon? Find th e m e a s u re o f a n g le 9. R ound to th e n e a re s t d e g re e S ta te th e a m p litu d e , period, frequency, phase shift, and v ertical sh ift of e ach functio n. Th en graph tw o full periods of th e functio n. (Lesson 4-4) 21. if n ecessary. (Lesson 4-1) 23. 6. y = - 3 sin ( * - f ) 22. y = 5 c o s 2 x - 2 MULTIPLE CHOICE W hich of th e functions has th e s am e graph as y = 3 sin (x — it)? (Lesson 4-4) F y = 3 sin (x + tt) G y = 3 cos (x - y j H J y = — 3 sin (x —-tv) y = -3 c o s (x + y j 24. SPRING The m otion of an object attached to a spring oscillating across its original position of rest can be m odeled by x(t) = A cos u t, w h e re A is th e initial dis p la c e m en t of th e o bject from its resting Id e n tify all an g le s th a t a re c o te rm in a l w ith th e given an g le . Th e n find and d ra w one positive and one n e g a tiv e a n g le c o te rm in a l w ith th e given angle. (Lesson 4-2) position, a ; is a constant dep en d en t on the spring and the m ass of the o bject a ttach ed to the spring, and t is tim e m easured in seconds. [Lesson 4-4) a. 10. - 22° D ra w a graph for the m otion of an object attached to a spring and displaced 4 centim e te rs w h e re u = 3. b. H ow long w ill it ta k e fo r the o bject to return to its initial position for the first tim e? 11. MULTIPLE CHOICE Find the approxim ate a re a of the shaded c. The constant uj is equal to region. (Lesson 4-2) w h e re k is the spring constant, and m is the m ass of the object. H ow w ould increasing th e m ass of an o bject a ffe c t the period of its oscillations? Explain your reasoning. 25. BUOY The height above sea level in fe e t of a signal buoy’s tra n s m itte r is m odeled by h = a sin bt + 268 In rough w a te rs , th e height cycles b e tw ee n 1 and A 1 2 .2 in 2 C 85.5 in2 1 0 fe e t, w ith 4 seconds betw een cycles. Find the B 4 2 .8 in 2 D 111.2 in2 values of a and b. C h a p te r 4 M id -C ha pte r Quiz ■ p i {■Graphing Other Trigonometric Functions Then Now Why? You analyzed graphs of trigonometric functions. 1 There are two types of radio transmissions known as amplitude modulation (AM) and frequency modulation (FM). When sound is transmitted by an AM radio station, the amplitude of a sinusoidal wave called the carrier wave is changed to produce sound. The transmission of an FM signal results in a change in the frequency of the carrier wave. You will learn more about the graphs of these waves, known as damped waves, in this lesson. (Lesson 4-4) Graph tangent and reciprocal trigonometric functions. | Graph damped trigonometric functions. AM signal FM signal A A A ir " V V VI 3 NewVocabulary * * damped trigonometric function damping factor damped oscillation damped wave damped harmonic motion Tangent and Reciprocal Functions In Lesson 4-4, you graphed the sine and cosine functions on the coordinate plane. You can use the same techniques to graph the tangent function and the reciprocal trigonom etric functions— cotangent, secant, and cosecant. 1 Since tan x = the tangent function is undefined when cos x = 0. Therefore, the tangent function has a vertical asym ptote whenever cos x = 0. Similarly, the tangent ahd sine functions each have zeros at integer m ultiples of t t because tan x = 0 w hen sin x = 0. 41 ,1\yj/ / 4T \ I T \ |y = sin M 1\ / 1 \ l/ / 1 / / 1 ! \ / h V / 1 ! I /j 1I A\ 1 \ i / y = t a n x I V 1 i " iJ , ---------------1-------------- f -------------- r ...............r t ts CM 1 1 i 1 A a / A / 1 t / / \ 1 i 1 \ L 1 J t rx / l / The properties of the tangent function are sum marized below. KeyConcept Properties of the Tangent Function D o m a in : xe Range: (—oo, oo) x -in te rc e p ts : m t ,n e Z y -in te r c e p t: 0 C o n tin u ity : infinite discontinuity at x = y + me, n e Z A s y m p to te s : x= S y m m e try : origin (odd function) E x tre m a : none E n d B eh a vio r: R, Y y + m r, n e + m r, n e Z lim tan x and lim tan x X — >— o o X — >oo do not exist. The function oscillates between — oo and oo. period: it connectED.mcgraw-hili.com| 269 StudyTip Amplitude The term amplitude does not apply to the tangent or cotangent functions because the heights of these functions are infinite. The general form of the tangent function, w hich is sim ilar to that of the sinusoidal functions, is y = a tan (bx + c) + d, where a produces a vertical stretch or com pression, b affects the period, c produces a phase shift, d produces a vertical shift and neither a or b are 0. KeyConcept Period of the Tangent Function Words The period of a tangent function is the distance between any two consecutive vertical asymptotes. Symbols For y = a tan (bx + c), where b ± 0, period = 1*1 Two consecutive vertical asym ptotes for y = tan x are .r = —y and x = y You can find two consecutive vertical asym ptotes for any tangent function of the form y = a tan (bx + c) + d by solving the equations bx + c — —y and bx + c = You can sketch the graph of a tangent function by plotting the vertical asym ptotes, x-intercepts, and points betw een the asym ptotes and x-intercepts. Graph Horizontal Dilations of the Tangent Function Locate the vertical asymptotes, and sketch the graph of y — tan Zx. The graph of y = tan 2x is the graph of y = tan x com pressed horizontally. The period is -jYj- or Find two consecutive vertical asymptotes. y. bx + C = Tangent asymptote equations —y b = 2 , c —0 2x + 0 = - f x = -JL 4 bx + c = 2x + 0 : Simplify. TT 2 x = * 4 Create a table listing key points, including the x-intercept, that are located betw een the two vertical asym ptotes at x = —y and x = y . Vertical Asymptote /-in te rc e p t * = - f y = tan 2x HM Interm ediate Point (0 .0 ) (?■') (0 .0 ) If'1) Sketch the curve through the indicated key points for the function. Then sketch one cycle to the left on the interval (~ "y v ~ y j ar>d one cycle to the right on the interval |y, - y j . p GuidedPractice Locate the vertical asymptotes, and sketch the graph of each function. 1A. y = tan 4x 270 | L esson 4 -5 G ra p h in g O th e r T r ig o n o m e tric F u n c tio n s Vertical Asymptote II * y = tan x Interm ediate Point * II Function 1B. y = tan y IE H 2 B Graph Reflections and Translations of the Tangent Function Locate the vertical asymptotes, and sketch the graph of each function, a. 2/ = - t a n | The graph of y = —tan y is the graph of y = tan x expanded horizontally and then reflected in the x-axis. The period is j y - or 2tt. Find two consecutive vertical asymptotes. 12 1 b = j .c = f + 0 = - f x = 2 f + ° = f 0 Simplify. y j or —t t x = 2 ^yj or tt Create a table listing key points, including the x-intercept, that are located betw een the two vertical asym ptotes at x = - t t and x = t t . y = tan x — y = -ta n | Interm ediate Point f ( - 7 ’ - 1) Interm ediate Point x-intercept (0 ,0 ) (t (0 ,0 ) X = — TT >) ( f ■- 1) Vertical Asymptote >5 II ro|S Vertical Asymptote Function X = Tt Sketch the curve through the indicated key points for the function. Then repeat the pattern for one cycle to the left and right of the first curve. b. y = tan (x - The graph of y = tan |x — y ^ j is the graph of y = tan x shifted ~ units to the right. The period is jyj- or 7T. Find two consecutive vertical asymptotes. X StudyTip Alternate Method When graphing a function with only a horizontal translation c, you can find the key points by adding cto each of the /-coordinates of the key points of the parent function. 6=1, c= = - y + y y Function y = ta n x y= tan ( x - or Simplify. TT Vertical Asymptote Interm ediate Point x - - - ( - ? • - ') X = TT ( f - 'l * 2 3 tt 7r 2 x = x-intercept (0,0) ( f ,° ) or 2 tt Interm ediate Point Vertical Asymptote ( f - 1) II 37T (t -'I x = 2 -k Sketch the curve through the indicated key points for the function. Then sketch one cycle to the left and right of the first curve. f GuidedPractice 2A. y = tan ^2x + y j 2B. y = - t a n (x - 271 m The cotangent function is the reciprocal of the tangent function, and is defined as cot x = co s* . Like the tangent function, the period of a cotangent function of the form y = a cot (bx + c) + d can be found by calculating y^y. Two consecutive vertical asymptotes can be found by solving the equations bx + c = 0 and bx + c = 7T. The properties of the cotangent function are summarized below. KeyConcept Properties of the Cotangent Function z x e R , x ± m t, n e R an g e : ( - 00, oo) J1 x -in te rc e p ts : ^ + n -K ,n € l 1 ! | y -in te rc e p t: none C o n tin u ity : infinite discontinuity at x = a tt, n e Z S y m m e try : origin (odd function) E nd B eh a vio r: X— 0 2 A1~ \ —4 “ i ^ none \ \ i 11 l 1 —3 r i i i i i ! * ' x \ 3A 2 tt 2 \ i \ i \ i \ I \ I 11 j lim cot x and lim cot x do not exist. ► —OO X — >00 The function oscillates between I i i I i x = n fK ,n e Z \'K\ 1 —7T iy ■1 i 11 2 - " l \y = cot x | \ \ // 1 \ - \V 1 V 1 \ 1 \ V A s y m p to te s : E x tre m a : 3- _ ts D o m a in : period: 7\ 1 and <*>. J You can sketch the graph of a cotangent function using the same techniques that you used to sketch the graph of a tangent function. Sketch the Graph of a Cotangent Function TechnologyTip Locate the vertical asymptotes, and sketch the graph of y = cot j . y= Graphing calculators may produce solid lines where the asymptotes occur. Setting the mode to DOT will eliminate the line. The graph of y = cot j is the graph of y = cot x expanded horizontally. The period is — or Find two consecutive vertical asym ptotes by solving bx + c = 0 and bx + c — i t . M 4 + 0 = 0 ■0 = b= ± c= 0 Simplify. x = 3(0) or 0 x = TT 3 ( tt) or 3 tt Create a table listing key points, including the x-intercept, that are located betw een the two vertical asym ptotes at x = 0 and x = 3 t t . Vertical Asymptote x=0 y = c o t| x=0 Interm ediate Point /-in te rc e p t ( f - ’l (? •» ) / = it (¥ ■ ’ ) (f.o ) £ CO II Function II o o Graphing a Cotangent Function When using a calculator to graph a cotangent function, enter the reciprocal of tangent, Interm ediate Point Following the same guidelines that you used for the tangent function, sketch the curve through the indicated key points that you found. Then sketch one cycle to the left and right of the first curve. GuidedPractice Locate the vertical asymptotes, and sketch the graph of each function. 3A. y = —cot 3x 272 L esson 4 -5 G ra p h in g O th e r T r ig o n o m e tric F u n c tio n s 3B. y = 3 cot |- Vertical Asymptote 3 tt. The reciprocals of the sme and cosine functions are defined as csc x = —— and sec x = >as , , sm x cosx shown below. The cosecant function has asym ptotes when sin x = 0, w hich occurs at integer m ultiples of t t . Likewise, the secant function has asymptotes when cos x = 0, located at odd multiples of y . Notice also that the graph of y = csc x has a relative m inim um at each m axim um point on the sine curve, and a relative m axim um at each m inim um point on the sine curve. The same is true for the graphs of 1/ = sec x and y = cos x. The properties of the cosecant and secant functions are sum m arized below. TechnologyTip Graphing Graphing the cosecant and secant functions on a calculator is similar to graphing the cotangent function. Enter the reciprocals of the sine and cosine functions. KeyConcept Properties of the Cosecant and Secant Functions Secant Function Cosecant Function D o m a in : x e R, D o m a in : x e R , x ± y + me, n e Z R ange: (— 00, — 1] and [1 , 00) R ange: (— 00, —1] and [1,00) x -in te rc e p ts : none x -in te rc e p ts : none y -in te r c e p t: none y -in te r c e p t: 1 C o n tin u ity : infinite discontinuity at x = mt, n e z C o n tin u ity : A s y m p to te s : x = n it, / ) € Z A s y m p to te s : x= y + S y m m e try : origin (odd function) S y m m e try : y-axis (even function) E nd B eh a vio r: ne Z lim csc x and lim csc x do not exist. X — ►— 0 0 X— >0 0 infinite discontinuity at x = y + /?TV, n e l B e h a v io r: n-K, n e Z lim sec x and lim secxdo not exist. X— * — 00 X — >00 The function oscillates between --00 and 00 period: 2 tt period: 2 it Like the sinusoidal functions, the period of a secant function of the form y = a sec (bx + c) + d or cosecant function of the form y = a csc (bx + c) + d can be found by calculating — . Two vertical asym ptotes for the secant function can be found by solving the equations bx + c = —y and bx + c = — and two vertical asym ptotes for the cosecant function can be found by solving bx + c = —t t and bx + c = tt. M connectED.m cgraw-hill.com | 273 To sketch the graph of a cosecant or secant function, locate the asym ptotes of the function and find the corresponding relative m axim um and m inim um s points. B T E T n ffB E l Sketch Graphs of Cosecant and Secant Functions Locate the vertical asymptotes, and sketch the graph of each function, a. i / = c s c ( x + y ) The graph of y = csc^x + y j is the graph of y = csc x shifted y units to the left. The period is or 2 tx. Two vertical asym ptotes occur w hen bx + c 3tv Tt asymptotes are x + — = —tt or x ■ 2 2 = — tv and bx + c = 7V. Therefore, two TV 2' Create a table listing key points, including the relative m axim um and minim um , that are located betw een the two vertical asym ptotes at x = —y ^ and x = y . Function StudyTip Finding Asymptotes and Key Points You can use the periodic nature of trigonometric graphs to help find asymptotes and key points. In Example 4a, notice that the vertical asymptote x = ~ y = cscx Vertical Asymptote Relative M inim um X= — TT (-f'-’l x= 0 I?'1) 3 It T (-Tt, -1 ) — f Vertical Asymptote X = (0,1) * = TT f Sketch the curve through the indicated key points for the function. Then sketch one cycle to the left and right. The graph is show n in Figure 4.5.1 below. is and x = y . Relative M axim um y = esc ( * + f ) equidistant from the calculated asymptotes, x = — y Vertical Asymptote b. y = sec j The graph of y = sec j is the graph of y = sec x expanded horizontally. The period is yyor 8 tv . Two vertical asym ptotes occur when bx + c = —y and bx + c = y k Therefore, 4 two asym ptotes are j + 0 : ■y or x = —2 tt and ^ + 0 = y ^ or x = 6 tt. Create a table listing key points that are located betw een the asymptotes at x = —2 tt and x = 6tt. f Relative Minim um (0,1) *= 7 x = 2tt Relative M axim um (tt, -1 ) (4ir, —1) Vertical Asymptote „ 3it 2 II (0,1) Vertical Asymptote £ CO y= sec j — II I 3 y = sec x Vertical Asymptote ro Function Sketch the curve through the indicated key points for the function. Then sketch one cycle to the left and right. The graph is shown in Figure 4.5.2 below. ^ GuidedPractice 4A. y = csc 2x 274 | Lesson 4 -5 | G ra p h in g O th e r T r ig o n o m e tric F u n c tio n s 4B. y = sec (x + 7r) Damped Trigonometric Functions W hen a sinusoidal function is multiplied by another function /(x), the graph of their product oscillates betw een the graphs of y = f( x ) and y = —f{ x ). W hen this product reduces the am plitude of the wave of the original sinusoid, it is called damped oscillation, and the product of the two functions is know n as a damped trigonom etric function. This change in oscillation can be seen in Figures 4.5.3 and 4.5.4 for the graphs of y = sin x and y = 2x sin x. 2 StudyTip y = sin x Damped Functions Trigonometric functions that are multiplied by constants do not experience damping. The constant affects the amplitude of the function. V —9tt 9/ I Figure 4.5.3 A damped trigonometric function is of the form y = f( x ) sin bx or y = f( x ) cos bx, where f( x ) is the damping factor. Damped oscillation occurs as x approaches ± 0 0 or as x approaches 0 from both directions. ■ I H S H i j S k e t c h Damped Trigonometric Functions >< ii i CO Identify the damping factor f(x ) of each function. Then use a graphing calculator to sketch the graphs of fix ) , —f(x ), and the given function in the same view ing window. Describe the behavior of the graph. a. y = —3x cos x The function y = —3x cos x is the product of the functions y = —3x and y = cos x, so f( x ) — —3x. x l \ w y = -3 x c o s x !/ The am plitude of the function is decreasing as x approaches 0 from both directions. - 3x [—4 ir, 4 ir] scl: tt by [ - 4 0 , 40] scl: 5 b. y — x2 sin x Math HistoryLink The function y = x 2 sin x is the product of the functions y = x 2 and y = sin x. Therefore, the dam ping factor is/(x) = x 2. Cathleen Synge M oraw etz (1923—) A Canadian, Morawetz studied the scattering of sound and magnetic waves and later proved results relating to the nonlinear wave equation. The am plitude of the function is decreasing as x approaches 0 from both directions. -4 ic , 4 ir] scl: it by [ - 1 0 0 ,1 0 0 ] scl: 10 C. y — 2Xcos 3x The function y = 2 T cos 3x is the product of the functions y = 2* and y = cos 3x, so /(x) = 2 X. The am plitude of the function is decreasing as x approaches —0 0 . y GuidedPractice 5A. y = 5x sin x 5B. y = \ cos x 5C. y = 3 Xsin x connectED.m cgraw-hill.com 275 W hen the amplitude of the m otion of an object decreases w ith time due to friction, the m otion is called damped harmonic motion. KeyConcept Damped Harmonic Motion Words An object is in damped harmonic motion when the amplitude is determined by the function a (f) = ke~ct. Symbols F o r y = /fe - c f sin w fan d y = k e ~ a cos ojt, where c > 0, k is the displacement, c is the damping constant, f is time, and w is the period. The greater the dam ping constant c, the faster the am plitude approaches 0. The m agnitude of c depends on the size of the object and the material of w hich it is com posed. S M lk S S M il DamPecl Harmonic Motion MUSIC A guitar string is plucked at a distance of 0.8 centim eter above its rest position and then released, causing a vibration. The dam ping constant for the string is 2.1, and the note produced has a frequency of 175 cycles per second. a. Write a trigonometric function that m odels the m otion of the string. The m axim um displacem ent of the string occurs w hen t = 0, so y = k e~ctcos u>t can be used to model the motion of the string because the graph of y = cos t has a y-in te rc e p t other than 0. The maxim um displacement occurs w hen the string is plucked 0.8 centimeter. The total displacem ent is the m axim um displacem ent M m inus the m inim um displacem ent m, so k = M — m = 0.8 — 0 or 0.8 cm. You can use the value of the frequency to find u). IcjI y = 175 M = 350tt Real-WorldLink Each string on a guitar is stretched to a particular length and tautness. These aspects, along with the weight and type of string, cause it to vibrate with a characteristic frequency or pitch called its fundamental frequency, producing the note we hear. Source: M 2ir : frequency Multiply each side by 2-k . Write a function using the values of k, u;, and c. y = 0.8e_21,cos 350 tt/is one m odel that describes the m otion of the string. b. Determine the amount of time t that it takes the string to be damped so that - 0 .2 8 < y < 0.28. Use a graphing calculator to determ ine the value of t when the graph of y = 0 .8 e "2'1,cos 3507rf is oscillating betw een y = —0.28 and y = 0.28. HowStuffWorks Y1=0.B4''<-2.iK)*C05(3E(HT- From the graph, you can see that it takes approxim ately 0.5 second for the graph of y = 0.8e~2,lfcos 350-Trt to oscillate w ithin the interval —0.28 < y < 0.28. X=.£ [ 0, 1 Y = -.2 7 3 5 5 0 2 scl: 0.5 by [ - 0 .7 5 , 0.75] scl: 0.25 » GuidedPractice 6. MUSIC Suppose another string on the guitar w as plucked 0.5 centim eter above its rest position with a frequency of 98 cycles per second and a dam ping constant of 1.7. A. Write a trigonometric function that models the m otion of the string y as a function of time t. B. Determine the time t that it takes the string to be damped so that —0.15 < y < 0.15. 276 | Lesson 4 -5 G ra p h in g O th e r T r ig o n o m e tric F u n c tio n s Exercises = Step-by-Step Solutions begin on page R29. Locate the vertical asymptotes, and sketch the graph of each function. (Examples 1-4) 1 . y = 2 tan x 2. y = ta n (x + 3. y = cot (x - 4. y = —3 s. y = - \ c o tx 6. y = —ta n 7. y 8. y = c o t| = —2 tan ( 6x — t t ) 9. y = \ esc 2x 11. y = sec (x 4- 7r) ta n 28. DIVING The end of a diving board is 20.3 centimeters above its resting position at the m om ent a diver leaves the board. Two seconds later, the board has moved down and up 12 times. The dam ping constant for the board is 0.901. (Example 6) -| 3x ~ r 20.3 resting ' position I 10. y = csc ^4x + 12. y = —2 csc 3x a. W rite a trigonom etric function that models the motion of the diving board y as a function of time f. 13. y = 4 sec (x — 14. y = sec ( f + f ) 15. y = | c s c ( x - ^ ) 16. y - b. Determ ine the am ount of time f that it takes the diving board to be damped so that —0.5 < y < 0.5. -se c: Locate the vertical asym ptotes, and sketch the graph of each function. Identify the damping factor fix ) of each function. Then use a graphing calculator to sketch the graphs of fix ), —fix ) , and the given function in the same viewing window. Describe the behavior of the graph. (Example 5) 17. y = t * sin x 18. y = 4x cos x 19. y = 2x2 cos x 20. y = — sin x 21. y = i-x sin 2x 22. y = (x — 2 )2 sin x 23. y = e°'5x cos x 24. y = 3 Xsin x 25. y = |x| cos 3x 26. y = In x cos x x3 • 29. y = sec x + 3 30. y = sec t 32. y = csc 31. y = csc 3 - 2 33. y = cot {2x + tt) —3 34. y = cot + 4 ( * +? )+ 3 ( ! + ?) - 1 35) PHOTOGRAPHY Jeff is taking pictures of a haw k that is flying 150 feet above him. The hawk will eventually fly directly over Jeff. Let d be the distance Jeff is from the haw k and 9 be the angle of elevation to the haw k from Jeff's camera. 27. MECHANICS W hen the car shown below hit a bum p in 150 ft the road, the shock absorber was com pressed 8 inches, released, and then began to vibrate in damped harm onic motion with a frequency of 2.5 cycles per second. The damping constant for the shock absorber is 3. (Example 6) Rest Position X (*-? ) a. Write d as a function of 9. b. Graph the function on the interval 0 < 9 < tt . C. Approxim ately how far away is the haw k from Jeff w hen the angle of elevation is 45°? 36. DISTANCE A spider is slowly clim bing up a wall. Brianna is standing 6 feet away from the w all watching the spider. Let d be the distance Brianna is from the spider and 9 be the angle of elevation to the spider from Brianna. a. Write a trigonometric function that models the displacement of the shock absorber y as a function of time f. Let t = 0 be the instant the shock absorber is released. a. W rite d as a function of 9. b. Determine the am ount of tim e t that it takes for the C. amplitude of the vibration to decrease to 4 inches. b. Graph the function on the interval 0 < 9 < y . A pproxim ately how far away is the spider from Brianna w hen the angle of elevation is 32°? to—-...... *............^ fllconnectED.mcgraw-hill.com | 277 GRAPHING CALCULATOR Find the values of 6 on the interval — t v < 0 < 7 T that make each equation true. GRAPHING CALCULATOR Graph each pair of functions on the same screen and m ake a conjecture as to w hether they are equivalent for all real num bers. Then use the properties of the functions to verify each conjecture. 37. cot 9 = 2 sec 9 38. sin 9 = cot i 39. 4 cos 9 = csc 9 40. tan — = sin I 48. f i x ) = sec x cos x; g(x) = 1 41. csc 9 = sec 9 42. tan 9 = sec 49. f{ x ) = sec2 x; gix ) = tan2 x + 1 43. TENSION A helicopter is delivering a large mural that is to be displayed in the center of town. Two ropes are used to attach the mural to the helicopter, as shown. The tension T on each rope is equal to half the downward force times 50. fi x ) = cos x csc x; gix) = cot x 51' = ------z 1 of\' g(x) = sin x sec Ix - —j Write an equation for the given function given the period, phase shift (ps), and vertical shift (vs). sec -• 52. function: sec; period: 3ir; ps: 0; vs: 2 (53) function: tan; period: y ps: vs: —1 54. function: csc; period: y ps: —7r; vs: 0 55. function: cot; period: 3tt; ps: y vs: 4 a. The downward force in newtons equals the m ass of the mural times gravity, which is 9.8 newtons per kilogram. If the mass of the mural is 544 kilograms, find the downward force. b. Write an equation that represents the tension T on each rope. c. Graph the equation from part b on the interval [0,180°]. 56. function: csc; period: y ps: —y vs: —3 H.O.T. Problems Use Higher-Order Thinking Skills 57. PROOF Verify that the y-intercept for the graph of any function of the form y = ke~c> cos u)t is k. REASONING D eterm ine w hether each statement is true or d. Suppose the mural is 9.14 meters long and the ideal angle 9 for tension purposes is a right angle. Determine how much rope is needed to transport the mural and the tension that is being applied to each rope. e. Suppose you have 12.2 meters of rope to use to transport the mural. Find 9 and the tension that is being applied to each rope. fa ls e . Explain your reasoning. 58. If b ^ 0, then y = a + b sec x has extrema of ± {a + b). 59. If x = 9 is an asym ptote of y = csc x, then x = 9 is also an asym ptote of y = cot x. 60. ERROR ANALYSIS Mira and Arturo are studying the graph shown. M ira thinks that it is the graph of y = ~ tan 2x, and Arturo thinks that it is the graph of y = —cot 2x. Is Match each function with its graph. either of them correct? Explain your reasoning. r U 1 lI 1» n 44. y = c s c (| + j ) - 2 45. y = se c(f + f ) - 2 46. y = cot (lx - j j - 2 47. y = ta n (2 x —! ) —2 278 Lesson 4 -5 G ra p h in g O th e r T r ig o n o m e tric F u n c tio n s 61. CHALLENGE Write a cosecant function and a cotangent function that have the same graphs as y = sec x and y = tan x respectively. Check your answers by graphing. 62. WRITING IN MATH A damped trigonom etric function oscillates betw een the positive and negative graphs of the dam ping factor. Explain why a damped trigonometric function oscillates betw een the positive and negative graphs of the dam ping factor and why the am plitude of the function depends on the dam ping factor. Spiral Review State the amplitude, period, frequency, phase shift, and vertical shift of each function. Then graph two periods of the function. (Lesson 4-4) 63. y = 3 sin ( i x — y j + 10 64. y = 2 cos (3x + — 665• y — \ cos (4x — tt) 4-1 Find the exact values of the five rem aining trigonom etric functions of 9. (Lesson 4-3) 66. sin 9 = D cos 9 > 0 67. cos 9 = 0 sin 9 > 0 68. tan 9 = / 7 sin 9 > 0 69. POPULATION The population of a city 10 years ago was 45,600. Since then, the population has increased at a steady rate each year. If the population is currently 64,800, find the annual rate of growth for this city. (Lesson 3-5) 70. MEDICINE The half-life of a radioactive substance is the am ount of time it takes for half of the atoms of the substance to disintegrate. N uclear m edicine technologists use the iodine isotope 1-131, w ith a half-life of 8 days, to check a patient's thyroid function. After ingesting a tablet containing the iodine, the isotopes collect in the patient's thyroid, and a special camera is used to view its function. Suppose a patient ingests a tablet containing 9 microcuries of 1-131. To the nearest hour, how long will it be until there are only 2 .8 microcuries in the patient's thyroid? (Lesson 3-2) Factor each polynomial completely using the given factor and long division. (Lesson 2-3) 71. x 3 + 2x2 — x — 2; x — 1 74. 72. x 3 + x 2 — 16x — 16; x + 4 73. x 3 — x 2 — lOx — 8; x + 1 EXERCISE The Am erican College of Sports M edicine recom m ends that healthy adults exercise at a target level of 60% to 90% of their m axim um heart rates. You can estimate your maximum heart rate by subtracting your age from 220. Write a com pound inequality that models age a and target heart rate r. (Lesson 0-5) Skills Review fo r Standardized Tests 75. 76. SAT/ACT In the figure, A and D are the centers of the two circles, which intersect at points C and E. CE is a diameter of circle D. If AB = CE = 10, w hat is AD? A 5 C 5V 3 B 5V 2 D 10\/2 77. W hich equation is represented by the graph? E 10V 3 B y = cot (0 — C y = ta n ( 9 + j j REVIEW Refer to the figure below. If c = 14, find the value of b. D y = ta n (< 9 -^ ) 78. REVIEW If sin 9 = F V3 F T G 14\/3 H 7 G and tt < 9 < then 1= ? H f 13n 12 7it r 4-k J 7V3 connectlD .m cgfaw -hill.com & | 279 Inverse Trigonometric Functions Then You found and graphed the inverses of relations and functions. Now •1 (Lesson 1-7) S NewVocabulary ® arcsine arnsins fnnntinn function arccosine function arctangent function 2 : Why? Evaluate and graph inverse trigonometric functions. Find compositions of trigonometric functions. Inverse trigonometric functions can be used to model the changing horizontal angle of rotation needed for a television camera to follow the motion of a drag-racing vehicle. | Inverse Trigonometric Functions In Lesson 1-7, you ■ learned that a function has an inverse function if and only if it is one-to-one, meaning that each y-value of the function can be matched with no more than one x-value. Because the sine function fails the horizontal line test, it is not one-to-one. If, however, we restrict the domain of the sine function to the interval 7C TT ■2 , 2 , the restricted function is one-to-one and takes on all possible range values [—1 ,1 ] of the unrestricted function. It is on this restricted dom ain that y = sin x has an inverse function called the inverse sine function y = sin -1 x. The graph of y = sin -1 x is found by reflecting the graph of the restricted sine function in the line y = x. Notice that the dom ain of y = sin 1 x is [—1 ,1 ], and its range is —y y . Because angles and arcs given on the unit circle have equivalent radian m easures, the inverse sine function is sometimes referred to as the arcsine fu nction y = arcsin x. In Lesson 4-1, you used the inverse relationship betw een the sine and inverse sine functions to find acute angle measures. From the graphs above, you can see that in general, y = sin -1 x or y = arcsin x iff sin y = x, when —1 < x < 1 and —y < y < iff means if and only if. This means that sin -1 x or arcsin x can be interpreted as the angle (or arc) between —y and y with a sine o fx . For exam ple, sin -1 0.5 is the angle with a sine of 0.5. 280 | Lesson 4-6 Recall that sin t is the y-coordinate of the point on the unit circle corresponding to the angle or arc length t. Because the range of the inverse sine function is restricted to Inverse S ine Values sin TV TV •2 , 2 , the possible angle m easures of the inverse sine function are located on the right half of the unit circle, as shown. You can use the unit circle to find the exact value of some expressions involving sin x or arcsin x. ■ 2 2 I E E S Q Evaluate Inverse Sine Functions Find the exact value of each expression, if it exists. Technology Tip Evaluate sin-1 You can also use a graphing calculator to find the a. sin -l 1 -1 J .52359 87756 .5235987756 2 Therefore, sin 1 \ 2 6 2 V ___________ h. D 2 6 CHECK If s in " 1 ± = Z then sin £ = i Make sure you select RADIAN on the MODE feature of your graphing calculator. ■2 , 2 T TT w ith a u-coordinate of —. W hen t = —, sin t = —. s i n _1< 0 . 5 ) Jt/6 TV TV Find a point on the unit circle on the interval angle that has a sine of 1 . 6 6 2 ✓ • ( I— — I arcsin Find a point on the unit circle on the interval with a v-coordinate of j Therefore, arcsin CHECK If arcsin C. 2 W hen t = 4 TV TV -2, 2 sin t = -1 2 = —- j. then sin ✓ sin 1 3 Because the dom ain of the inverse sine function is [—1 ,1 ] and 3 > 1, there is no angle w ith a sine of 3. Therefore, the value of sin -1 3 does not exist. ► GuidedPractice *■#) 1A. arcsin 1B. sin 1 (—2tv) Notice in Exam ple la that while sin ^jr- is also not in the interval TV TV ' 2’ 2 2' 6 1C. arcsin (—1) is Therefore, sin -1 \ =f= L 6 co n n e ctE D~m cgraw -hill ~com H 281 StudyTip Principal Values Trigonometric functions with restricted domains are sometimes indicated with capital letters. For example, y = Sin x represents the function W hen restricted to a dom ain of [0, tv ], the cosine function is one-to-one and takes on all of its possible range values on [—1 ,1 ], It is on this restricted dom ain that the cosine function has an inverse function, called the inverse cosine fu n ction y = c o s-1 x or arccosine function y = arccos x. The graph of y = co s-1 x is found by reflecting the graph of the restricted cosine function in the line y = x. In verse C osine Function R estricted C o sin e Function y = sin x, where — y < x < y . 1.5- y The values in these restricted domains are often called principal I y = cos x y values. « 0.5/ ■ \ I \ \ / 0 -0 .5 - \ ! 1 » / / tv I ' 2 tv \ \ I ’ 3 tt ■ A V -1 .5 - Recall that cos t is the x-coordinate of the point on the unit circle corresponding to the angle or arc length t. Because the range of y = co s-1 x is restricted to [0, tv ], the values of an inverse cosine function are located on the upper half of the unit circle. Evaluate Inverse Cosine Functions Find the exact value of each expression, if it exists. a. co s' (-#) Find a point on the unit circle in the interval [0, an x-coordinate of 2 W hen t — 4 tv] cos t = — w ith 2 Therefore, cos 1 j —y p j = CHECK b. If co s” 1 (-#) - 4 then cos ~ V2 4 arccos (—2) Since the domain of the cosine function is [—1 ,1 ] and —2 < —1, there is no angle w ith a cosine of —2. Therefore, the value of arccos (—2) does not exist. c. cos 1 0 Find a point on the unit circle in the interval [0, an x-coordinate of 0. W hen t = y , cos t = 0. tv] w ith (0,1) y ^ t= P \ Therefore, co s-1 0 = y . -il O CHECK If c o s " 1 0 = £ , then cos ^ = 0. ✓ -1, y GuidedPractice 2A. cos 282 | Lesson 4 -6 (-#) In v e rs e T r ig o n o m e tric F u n c tio n s 2B. arccos 2.5 2C. cos - B ) l / ■ * W hen restricted to a dom ain of ( - y , y ) , the tangent function is one-to-one. It is on this restricted StudyTip End Behavior of Inverse Tangent Notice that when the graph of the restricted tangent function is reflected in the line y = x, the vertical asymptotes at >domain that the tangent function has an inverse function called the inverse tangent function y = tan -1 x or arctangent function y = arctan x. The graph of y = ta n -1 x is found by reflecting the graph of the restricted tangent function in the line y = x. N otice that unlike the sine and cosine functions, the dom ain of the inverse tangent function is (—0 0 , 0 0 ). x = ± y become the horizontal asymptotes y = ±~- of the inverse tangent function. Therefore, lim tan_ 1 x = - ^ 2 X-»-oo and lim tan-1 * = ■?• 2 X -*o o You can also use the unit circle to find the value of an inverse tangent expression. On the unit circle, tan t = or —. The values of y = ta n -1 x will be located on the right half of the unit circle, not including —y and y , because the tangent function is undefined at those points. TechnologyTip Evaluate tan ~ 1 You can also use a graphing calculator to find the angle that has a tangent of V 3 . [U ir T O X lJ) m > i Find the exact value of each expression, if it exists. a. tan _1V J Find a point (x, y) on the unit circle in the interval V3 1.047197551 ji/3 Evaluate Inverse Tangent Functions 1.047197551 ( —y , y j such that — = V 3 . W hen t — y , tan t = - ~ ~2 or \/3. Therefore, tan -1 \J?> = y . Make sure you select RADIAN on the MODE feature of your graphing calculator. CHECK If tan -1 V 3 = y , then tan y = \[3. </ b. arctan 0 Find a point (x, y) on the unit circle in the interval ( —y , y ) such that y = 0. W hen t = 0, tan t = y or 0. Therefore, arctan 0 = 0. CHECK If arctan 0 = 0, then tan 0 = 0. ✓" p GuidedPractice 3A. arctan (-#) 3B. ta n ” 1 (—1) W hile inverse functions for secant, cosecant, and cotangent do exist, these functions are rarely used in com putations because the inverse functions for their reciprocals exist. Also, deciding how to restrict the domains of secant, cosecant, and cotangent to obtain arcsecant, arccosecant, and arccotangent is not as apparent. You will explore these functions in Exercise 66. 283 The three most com m on inverse trigonom etric functions are sum m arized below. KeyConcept Inverse Trigonometric Functions Words The angle (or arc) between - y Words and y with a sine of x. Symbols Inverse Tangent of x Inverse Cosine of x Inverse Sine of x y = sin-1 X if and only if sin y = x, for — 1 < x < 1 and - y < y < y . Symbols The angle (or arc) between 0 and tv with a cosine of x. Words y = cos-1 x if and only if cos y = x, for —1 < x < 1 and 0 < y < tv. Symbols The angle (or arc) between - y and y with a tangent of x. y = t a n _1 x if and only if tan y = x , for — oo < x < oo and - y < y < y . Domain: [-1,1 ] Domain: [ - 1 , 1 ] Domain: Range: [ - f , f ] Range: [0 , tv] Range: ( - y f ) 1- 1 < - * ----------- >- -TV y ------------------ .► i 1 1 1 y ( - o o , oo) y=cos~1x y = sin—1 x y = tan-1 x L I - -J 1x 0 1 -1 ^ —TV- TT 2 You can sketch the graph of one o f the inverse trigonom etric fu nctions show n above by rew riting the function in the form sin y = x, cos y = x, or tan y = x, assigning values to y and m aking a table of values, and then plotting the poin ts and connecting the points w ith a sm ooth curve. K 5 3 j j 3 3 I a D Sketch Graphs of Inverse Trigonometric Functions Sketch the graph of y = arccos 2x. By definition, y = arccos 2x and cos y = 2x are equivalent on 0 < y < 7V, so their graphs are the same. Rewrite cos y = 2x as x = j cos y and assign values to y on the interval [0, table of values. WatchOut! Remember that tv = 3.14 radians or 180°. o TT 4 IV 6 7T 2 1 V2 V3 2 o 4 4 57T 6 3 TV 4 7V V3 V2 1 4 4 2 Then plot the points (x, y) and connect them with a sm ooth curve. Notice that this curve has endpoints at j , 7vj and |-i, oj, indicating that the entire graph of y — arccos 2x is shown. p GuidedPractice Sketch the graph of each function. 4A. y = arcsin 3x 284 | L esson 4 -6 j In v e rs e T r ig o n o m e tric F u n c tio n s 4B. y = tan 1 2x tv ] to make a Real-World Example 5 Use an Inverse Trigonometric Function MOVIES In a movie theater, a person's view ing angle for watching a movie changes depending on where he or she sits in the theater. a. Write a function modeling the view ing angle 9 for a person in the theater whose eye-level w hen sitting is 4 feet above ground. Draw a diagram to find the m easure of the viewing angle. Let 9 1 represent the angle form ed from eye-level to the bottom of the screen, and let 02 represent the angle formed from eye-level to the top of the screen. In the late 19th century, Thomas Edison began work on a device to record moving images, called the kinetoscope, which would later become the film projector. The earliest copyrighted motion picture is a film of one of Edison’s employees sneezing. 32 ft So, the view ing angle is 9 = 02 — 9V You can use the tangent function to find 9 Xand 92. Because the eye-level of the person when seated is 4 feet above the floor, the distance opposite 9l is 8 — 4 feet or 4 feet long. Source: The Library of Congress tan 0, = — a opp = 4 and adj = d = tan -1 4 Inverse tangent function The distance opposite 92 is (32 + 8) — 4 feet or 36 feet. tan 9? = ^ a = tan opp = 36 and adj = d -l 36 Inverse tangent function So, the viewing angle can be m odeled by 9 = tan 1 ~ b. — tan 1 j . D eterm ine the distance that corresponds to the m axim um view ing angle. The distance at w hich the m axim um view ing angle occurs is the m axim um point on the graph. You can use a graphing calculator to find this point. From the graph, you can see that the m axim um view ing angle occurs approxim ately 12 feet from the screen. -.Y = S 3 .i3 0 1 0 £ „ [0 ,1 0 0 ] scl: 10 by [0, 60] scl: 5 p GuidedPractice 5. TELEVISION Tucas has purchased a new flat-screen television. So that his fam ily will be able to see, he has decided to hang the television on the w all as shown. ■■■■■■■■■■■■■■■■■■■■■■■■■■■■■■■■■■HI □ 7.5 ft I A. Write a function m odeling the distance d of the m axim um view ing angle 9 for Lucas if his eye level w hen sitting is 3 feet above ground. B. Determ ine the distance that corresponds to the m axim um view ing angle. V ................ connectED.m cgraw-hill.com I 285 2 Compositions of Trigonometric Functions In Lesson 1-7, you learned that if x is in the domain of f( x ) and f ~ \ x ), then / [/ “ '(*)] = x and f ~ l lf(x )] = x. Because the domains of the trigonom etric functions are restricted to obtain the inverse trigonometric functions, the properties do not apply for all values of x. For example, while sin x is defined for all x, the dom ain of sin -1 x is [—1 ,1 ]. Therefore, sin (sin-1 x) = x is only true w hen —1 < x < 1. A different restriction applies for the com position sin 1 (sin x). Because the dom ain of sin x is restricted to the interval TT TT ■2 , 2 , sin (sin x) = x is only true w hen - ~ < x < ^ - . These dom ain restrictions are sum m arized below. KeyConcept Domain of Compositions of Trigonometric Functions f[f-Hx)) = x f- '[ f(x ) ] = x If - 1 < x < 1, then sin (sin-1 x) = x. If - y < x < y , then sin-1 (sin x) = x. If - 1 < x < 1, then cos (cos-1 x) = x. If 0 < x < tt, then cos-1 (cos x) = x. If - o o < x < oo, then tan (tan-1 x) = x. If - y < x < y , then tan-1 (tan x) = x. V -j Use Inverse Trigonometric Properties Find the exact value of each expression, if it exists. H) a. sin The inverse property applies because —i lies on the interval [—1 ,1 ]. --H ) Therefore, sin b. i ~4' arctan |tan y j Because tan x is not defined w hen x = y , arctan |tan y j does not exist. c. arcsin 7 77 WatchOut! Compositions and Inverses When computing M [ f ( x ) ] with trigonometric functions, the domain appears to be (— oo, oo). However, because the ranges of the inverse functions are restricted, coterminal angles must sometimes be found. (sinZr ) Notice that the angle — does not lie on the interval w ith 4 —2 tt or — 4 whi ch is on the interval sin ^sin -y -j = arcsin j^sin ( ~ y j j TT > Since - ' 4 Therefore,, arcsin |sin 77T . However, ——is coterminal 4 TT TT •2 , 2 • 7-ir sm ^ = sin 4 TT TT '2 ' 2 (-7) ■< —^ arcsin (sin x) = x. JL '4 ' G u id ed P ra ctice 6A. tan ^tan-1 y j 286 | L esson 4 -6 i In v e rs e T r ig o n o m e tric F u n c tio n s 6B. cos 6C. arcsin in (sin f ) You can also evaluate the com position of two different inverse trigonom etric functions. s .^ ii Evaluate Compositions of Trigonometric Functions Find the exact value of cos To sim plify the expression, let u = ta n -1 -|J, so tan u = Because the tangent function is negative in Quadrants II and IV, and the dom ain of the inverse tangent function is restricted to Quadrants I and IV, u m ust lie in Q uadrant IV. Using the Pythagorean Theorem , you can find that the length of the hypotenuse is 5. Now, solve for cos u. cos u adj hyp Cosine function 4 5 adj = 4 and hyp = 5 So, cos tan 4 5' - H ) ►GuidedPractice Find the exact value of each expression. 7A, co s-1 (sin y j 7B. sin |arctan Som etimes the com position of two trigonom etric functions reduces to an algebraic expression that does not involve any trigonom etric expressions. Evaluate Compositions of Trigonometric Functions StudyTip Decomposing Algebraic Functions The technique used to convert a trigonometric expression into an algebraic expression can be reversed. Decomposing an algebraic function as the composition of two trigonometric functions is a technique used frequently in calculus. Write tan (arcsin a) as an algebraic expression of a that does not involve trigonometric functions. > Let u = arcsin a, so sin u = a. Because the domain of the inverse sine function is restricted to Quadrants I and IV, u m ust lie in Quadrant I or IV. The solution is sim ilar for each quadrant, so we will solve for Q uadrant I. J y / y a From the Pythagorean Theorem , you can find that the length of the side adjacent to u is V l — a 2. Now, solve for tan u. opp ta n u = — 77" adj = n V l-fl2 ----- A X Tangent function opp = a and adj = V i — a2 or a — 1-« So, tan (arcsin a) = - —1 ■ a . 1 —u p GuidedPractice Write each expression as an algebraic expression of x that does not involve trigonometric functions. 8A. sin (arccos x) - 8B.cot [sin-1 x] V________________________________________________________ c o n n e c tE D jn c ^ i 287 Exercises = Step-by-Step Solutions begin on page R29. Find the exact value of each expression, if it exists. DRAG RACE A television camera is film ing a drag race. The camera rotates as the vehicles m ove past it. The cam era is 30 m eters aw ay from the track. Consider 9 and x as show n in the figure. (Example 5) (Examples 1 -3 ) 1. sin-1 0 . V3 2. arcsm - r - . V2 3. arcsin - y 4. sin -1 j 5. 6. arccos 0 - 1 V2 7. cos 1 — 8. arccos (—1) 'A 9. arccos V3 x 10. c o s-1 j 11. arctan 1 12. arctan (—V 3 ) V3 13. tan -1J — 14. tan-1 0 a. Write 9 as a function of x. b. Find 9 when x = 6 meters and x = 14 meters. 15. ARCHITECTURE The support for a roof is shaped like two right triangles, as show n below. Find 9. (Example 3) 28. SPORTS Steve and Ravi w ant to project a pro soccer game on the side of their apartment building. They have placed a projector on a table that stands 5 feet above the ground and have hung a 12-foot-tall screen that is 10 feet above the ground. (Example 5) 16. RESCUE A cruise ship sailed due west 24 miles before turning south. W hen the cruise ship becam e disabled and the crew radioed for help, the rescue boat found that the fastest route covered a distance of 48 miles. Find the angle 9 at w hich the rescue boat should travel to aid the cruise ship. (Example 3) a. Write a function expressing 9 in terms of distance d. b. Use a graphing calculator to determ ine the distance for the m axim um projecting angle. Find the exact value of each expression, if it exists. (Examples 6 and l)r 29. sin |sin_1 30. sin 1 |sin y j 31. cos ^cos-1 32. 33. tan |tan-1 34. tan _ 1 | ta n -jj 35. cos (tan-1 1) 36. sin -1 |cos y j CO S 1 (cos 7v) Sketch the graph of each function. (Example 4) 17. y = arcsin x 18. y = sin -1 2x 19. y = sin -1 (x + 3) 20. y = arcsin x — 3 21. y = arccos x 22. y = cos-1 3x 23. y = arctan x 24. y = tan -1 3x 25. y = tan -1 (x + 1) 26. y = arctan x — 1 288 Lesson 4 -6 ] In v e rs e T r ig o n o m e tric F u n c tio n s sin (tan-1 1 — sin -1 1) 37. sin ^2 cos 1 38. 39. cos (tan- 1 1 — sin -1 1) 40. cos |cos-1 0 + sin -1 Write each trigonometric expression as an algebraic expression of x. (Example 8) Write each algebraic expression as a trigonometric function of an inverse trigonom etric function of x. (4 l) tan (arccos x) 42. csc (cos 1 x) 63. 43. sin (cos 1 x) 44. cos (arcsin x) 45. csc (sin 1 x) 46. sec (arcsin x) 47. cot (arccos x) 48. cot (arcsin x) 64. Vl -X 2 65. «§l MULTIPLE REPRESENTATIONS In this problem , you will explore the graphs of com positions of trigonometric functions. a. ANALYTICAL Consider/(x) = sin x and / -1 (x) = arcsin x. Describe the dom ain and range o f / 0 / - 1 and/-1 o f . Describe how the graphs of g(x) a n d /(x ) are related. b. GRAPHICAL Create a table of several values for each com posite function on the interval [—2, 2], Then use the table to sketch the graphs o f / 0 / - 1 and/-1 o f . Use a graphing calculator to check your graphs. 49. /(x) = sin -1 x and g (x ) = sin -1 (x — 1) — 2 50. /(x) = arctan x and g(x ) = arctan 0.5x — 3 51. f( x ) = cos-1 x a n d g (x ) = 3 (cos-1 x — 2) c. ANALYTICAL Consider g(x) = cos x and g 1 (x) = arccos x. Describe the dom ain and range of g o g -1 and g ~1 o g and m ake a conjecture as to what the graphs of g o g _1 and g -1 o g will look like. Explain your reasoning. 52. /(x) = arcsin x and g (x) = ^ arcsin (x + 2) 53. /(x) = arccos x and g (x) = 5 + arccos 2x 54. /(x) = tan-1 x and g (x ) = tan-1 3x — 4 d. GRAPHICAL Sketch the graphs of g o g -1 and g -1 o g. Use a graphing calculator to check your graphs. 55. SAND W hen piling sand, the angle form ed betw een the pile and the ground remains fairly consistent and is called the angle o f repose. Suppose Jade creates a pile of sand at the beach that is 3 feet in diam eter and 1.1 feet high. e. VERBAL M ake a conjecture as to what the graphs of the two possible com positions of the tangent and arctangent functions will look like. Explain your reasoning. Then check your conjecture using a graphing calculator. H.O.T. Problems Use Higher-Order Thinking Skills 66. ERROR ANALYSIS Alisa and Trey are discussing inverse a. W hat is the angle of repose? trigonom etric functions. Because tan x = b. If the angle of repose remains constant, how m any feet in diameter would a pile need to be to reach a height of 4 feet? conjectures that ta n -1 x = sm cos of them correct? Explain. Give the domain and range of each composite function. Then use your graphing calculator to sketch its graph. 56. y = cos (tan-1 x) 57. y = sin (cos-1 x) 58. y = arctan (sin x) 59. y = sin -1 (cos x) 60. y = cos (arcsin x) 61. y = tan (arccos x) 62. 67. 68. , Alisa x . Trey disagrees. Is either x CHALLENGE Use the graphs of y = sin - x and y — cos 1 . to find the value of sin -1 x + co s-1 x on the interval [—1 ,1 ]. Explain your reasoning. REASONING Determ ine w hether the follow ing statem ent is true o r false: If cos ■— = - y - , then co s-1 -^y- = -y-. Explain your reasoning. INVERSES The arcsecant function is graphed by restricting the domain of the secant function to the intervals [o, y j and ( y , i r j, and the arccosecant function is graphed by REASONING D eterm ine w hether each function is odd, even, or neither. Justify your answer. y = sin -1 x restricting the domain of the cosecant function to the 69. intervals [—y , 70. y — cos 1 x 0 ) and ( 0 , y j . a. State the domain and range of each function. 71. y = ta n -1 x b. Sketch the graph of each function. C. Explain why a restriction on the dom ain of the secant and cosecant functions is necessary in order to graph the inverse functions. 72. WRITING IN MATH Explain how the restrictions on the sine, cosine, and tangent functions dictate the domain and range of their inverse functions. ^ c oconnectED.m n n e c tE cgraw-hill.com 1 289 Spiral Review Locate the vertical asymptotes, and sketch the graph of each function. Lesson 4-5) 75. y = 3 csc - 73. y = 3 tan 9 74. y = cot 5 9 76. WAVES A leaf floats on the water bobbing up and down. The distance betw een its highest and lowest points is 4 centimeters. It m oves from its highest point dow n to its low est point and back to its highest point every 10 seconds. Write a cosine function that models the movement of the leaf in relationship to the equilibrium point. (Lesson 4-4) Find the value of x. Round to the nearest tenth, if necessary. (Lesson 4-1) 79. 77. 17.8 For each pair of functions, find [ f o g](x), [g o f](x ), and [ / o gl(4). (Lesson 1-6) 80. 81. f( x ) = 6 - 5 x f( x ) = x 2 + 3x - 6 g(x) = 4x + l 82. f( x ) = \]x + 3 g {x ) = x 2 + 1 gW = j 83. EDUCATION Todd has answered 11 of his last 20 daily quiz questions correctly. His baseball coach told him that he must raise his average to at least 70% if he w ants to play in the season opener. Todd vows to study diligently and answer all of the daily quiz questions correctly in the future. How m any consecutive daily quiz questions m ust he answer correctly to raise his average to 70%? (Lesson 0-8) Skills Review for Standardized Tests 84. SAT/ACT To the nearest degree, what is the angle of depression 9 betw een the shallow end and the deep end of the swimming pool? 86. REVIEW The hypotenuse of a right triangle is 67 inches. If one of the angles has a m easure of 47°, what is the length of the shortest leg of the triangle? A 45.7 in. C 62.5 in. B 49.0 in. D 71.8 in. 87. REVIEW Two trucks, A and B, start from the intersection C of two straight roads at the same time. Truck A is traveling twice as fast as truck B and after 4 hours, the two trucks are 350 m iles apart. Find the approxim ate speed of truck B in miles per hour. A 25° C 41° B 37° D 53° E 73° AK :1 \ 85. Which of the following represents the exact value of sin |tan~1-^-)'? \ : \ 90“ iff1 F -- 2V5 H J 290 \ 1 Vf 350mi \ „ 5 2V5 5 | Lesson 4 -6 | In v e rs e T r ig o n o m e tric F u n c tio n s F 39 H 51 G 44 J 78 LrfWi The Law of Sines and the Law of Cosines :Then • You solved right triangles using trigonometric functions. : Why? • (Lesson 4-1) ® NewVocabulary oblique triangles Law of Sines ambiguous case Law of Cosines Heron’s Formula 1 Solve oblique • triangles by using the Law of Sines or the Law of Cosines. I Find areas of oblique ■triangles. Triangulation is the process of finding the coordinates of a point and the distance to that point by calculating the length of one side of a triangle, given the measurements of the angles and sides of the triangle formed by that point and two other known reference points. Weather spotters can use triangulation to determine the location of a tornado. Solve Oblique Triangles In Lesson 4-1, you used trigonom etric functions to solve right triangles. In this lesson, you will solve oblique triangles— triangles that are not right triangles. 1 You can apply the Law of Sines to solve an oblique triangle if you know the measures of two angles and a nonincluded side (AAS), two angles and the included side (ASA), or two sides and a nonincluded angle (SSA). KeyConcept Law of Sines If A ABC has side lengths a, b, and c representing the lengths of the sides opposite the angles with measures A fl,an d C ,th en ^ = ^ ® = ^ . a b c You will derive the Law of Sines in Exercise 69. H S S E E D 3 E I Apply the Law of Sines (AAS) Solve A A B C . Round side lengths to the nearest tenth and angle m easures to the nearest degree. Because two angles are given, C = 180° — (103° 4- 35°) or 42°. Use the Law of Sines to find a and c. sin B sin A sin 103° _ sin 35° 20 ~ a ! sin 103° = 20 sin 35° 20 sin 35° sin 103° a = 11.8 sin B Law of Sines Substitution Multiply. sin 103° 20 sin C sin 42° c c sin 103° = 20 sin 42° 20 sin 42° sin 103° Divide. Use a calculator. c ~ 13.7 Therefore, a = 11.8, c = 13.7, and ZC = 42°. |fe GuidedPractice Solve each triangle. Round side lengths to the nearest tenth and angle measures to the nearest degree. 1A. 1B. I ............. ........ .............. ........,.,,.1.1,...-........ fl|connectE D . m cgraw -hill.com | ] 291 J j g E S E E i a S S ® Apply the Law of Sines (ASA) SATELLITES An Earth-orbiting satellite is passing betw een the Oak Ridge Laboratory in Tennessee and the Langley Research Center in Virginia, which are 446 miles apart. If the angles of elevation to the satellite from the Oak Ridge and Langley facilities are 58° and 72°, respectively, how far is the satellite from each station? To gain an improved understanding of the atmosphere, land surface changes, and ecosystem processes, NASA uses a series of satellites as part of its Earth Observing System (EOS) to study the air, land, and water on Earth. Oak Ridge -• 446 m i ► Langley Because two angles are given, C = 180° — (58° + 72°) or 50°. Use the Law of Sines to find the distance to the satellite from each station. Source: NASA sin C _ shiB Law of Sines sin 50° sin 72° —7 7 7 — = ;—446 b . . Substitution b sin 50° = 446 sin 72° u Multiply. 446 sin 72° sin 50° b = 553.72 n. .;J„ sin C _ sin A . si n 50° sin 58° —7T7— —---------446 a a sin 50° = 446 sin 58° 446 sin 58° a _= ■ “ sin50' Use a calculator. « ~ 493.74 So, the satellite is about 554 miles from O ak Ridge and about 494 m iles from Langley. p GuidedPractice 2. SHIPPING Two ships are 250 feet apart and traveling to the same port as shown. Find the distance from the port to each ship. _ 2 5 0 ft_ _ Ship A N '~ 39° StudyTip 4 1 V ' S h ip ! From geometry, you know that the measures of two sides and a nonincluded angle (SSA) do not necessarily define a unique triangle. Consider the angle and side measures given in the figures below. Alternative Representations The Law of Sines can also be written in reciprocal form as a _ b _ c sin /I sin 6 sin C ‘ In general, given the measures of two sides and a nonincluded angle, one of the follow ing will be true: (1) no triangle exists, (2) exactly one triangle exists, or (3) two triangles exist. In other words, when solving an oblique triangle for this am biguous case, there m ay be no solution, one solution, or two solutions. 292 Lesson 4 -7 | T h e L a w o f Sines a n d th e L a w o f C osines KeyConcept The Ambiguous Case (SSA) Consider a triangle in which a, b, and Aare given. For the acute case, sin A = | , so h = b sin A. A is Acute. (A < 90°) a < b and a < h no solution a < b and a = h one solution a < b and a > h two solutions one solution To solve an ambiguous case oblique triangle, first determ ine the num ber of possible solutions. If the triangle has one or two solutions, use the Law of Sines to find them. E S S E E S * The Ambiguous Case— One or No Solution StudyTip Make a Reasonable Sketch When solving triangles, a reasonably accurate sketch can help you determine whether your answer is feasible. In your sketch, check to see that the longest side is opposite the largest angle and that the shortest side is opposite the smallest angle. Find all solutions for the given triangle, if possible. If no solution exists, write no solution. Round side lengths to the nearest tenth and angle m easures to the nearest degree. a. a = 15, c = 12, A = 94° Notice that A is obtuse and a > c because 15 > 12. Therefore, one solution exists. A pply the Law of Sines to find C. sin C _ sin 94° 12 15 sin C ; Law of Sines 12 sin 94° 15 Multiply each side by 12. c = s in -i (12sm 941) or about 53° Definition of sin-1 Because two angles are now known, B ~ 180° — (94° + 53°) or about 33°. Apply the Law of Sines to find b. Choose the ratios w ith the few est calculated values to ensure greater accuracy. sin 94° 15 sin 33° 15 sin 33° or about 8.2 sin 94° Law of Sines Solve for b. Therefore, the rem aining m easures of A A BC are B ~ 33°, C ~ 53°, and b ~ 8.2. b. a = 9, b = 11, A — 61° ( Notice that A is acute and a < b because 9 < 11. Find h. sin 61° = h = 11 sin 61° or about 9.6 Definition of sine h = b sin A Because a < h, no triangle can be formed with sides a = 9, b = 11, and A = 61°. Therefore, this problem has no solution. ^ GuidedPractice 3A. a = 12, b = 8, B = 61° 3B. a = 13, c = 26, A = 30° 1 293 E 2 J J 3 H 3 3 The Ambiguous Case-Two Solutions Find two triangles for which A = 43°, a = 25, and b = 28. Round side lengths to the nearest tenth and angle measures to the nearest degree. C A is acute, and h = 28 sin 43° or about 19.1. N otice that a < b because 25 < 28, and a > h because 25 > 19.1. Therefore, two different triangles are possible w ith the given angle and side measures. Angle B will be acute, while angle B' will be obtuse. A Make a reasonable sketch of each triangle and apply the Law of Sines to find each solution. Start with the case in which B is acute. Solution 1 ZB is acute. Find B. B s in s in 4 3 ° 28 Law of Sines 25 sin B = 2 8 s in 4 3 ° Solve for sin S. 25 Use a calculator. sin B = 0.7638 TechnologyTip B = s in -1 0.7638 or about 50° Using sin-1 Notice that when calculating sin-1 of a ratio, your calculator will never return two possible angle measures because sin-1 is a function. Also, your calculator will never return an obtuse angle measure for sin- ' because the inverse sine function has a range o f - y to y or - 9 0 ° to 90°. Definition of s i n - 1 Find C. C = 180° - (43° + 50°) ~ 87° Apply the Law of Sines to find c. s in 8 7 ° s in 4 3 ° c Law of Sines 25 „ 2 5 s in 87° or about 36.6 s in 4 3 ° Solution 2 Solve for c. Z B' is obtuse. Note that mZCB'B = mZCBB'. To find B', you need to find an obtuse angle w ith a sine that is also 0.7638. To do this, subtract the measure given by your calculator to the nearest degree, 50°, from 180°. Therefore, B' is approxim ately 180° — 50° or 130°. Find C. C ~ 180° - (43° + 130°) or 7° Apply the Law of Sines to find c. s in 7 ° c s in 4 3 ° Law of Sines 25 „ 2 5 s in 7 ° s in 4 3 ° or about 4.5 Solve for c. Therefore, the missing measures for acute A A BC are B ~ 50°, C = 87°, and c = 36.6, while the missing measures for obtuse A AB'C are B' ~ 130°, C ~ 7°, and c ~ 4.5. f GuidedPractice Find two triangles with the given angle measure and side lengths. Round side lengths to the nearest tenth and angle measures to the nearest degree. 4A. A = 38°, a = 8, b = 10 294 L esson 4 -7 | T h e L a w o f Sines a n d th e L a w o f C osines 4B. A = 65°, a = 55, b = 57 StudyTip Law of Cosines Notice that the You can use the Law of Cosines to solve an oblique triangle for the rem aining two cases: when >you are given the m easures of three sides (SSS) or the m easures of two sides and their included angle (SAS). angle referenced in each equation of the Law of Cosines corresponds to the side length on the other side of the equation. KeyConcept Law of Cosines In A A B C, if sides with lengths a, 6, and c are opposite angles with measures A, B, and C, respectively, then the following are true. a 2 = b2 + c 2 - 2bc cos A b 2 = a2 + c2 - 2accos B c 2 = a 2 + b 2 - 2aficos C B a 2 = b 2 + c 2 - 2bc cos A y b 2 = a 2 + c 2 - 2accos B A c 2 = a 2 + b 2 ~ 2a6cos C \a -------------bH---------- J You will derive the first formula for the Law of Cosines in Exercise 70. Real-World Example 5 Apply the Law of Cosines (SSS) HOCKEY W hen a hockey player attempts a shot, he is 20 feet from the left post of the goal and 24 feet from the right post, as shown. If a regulation hockey goal is 6 feet wide, what is the player's shot angle to the nearest degree? 20 ft / shot angle StudyTip 24 ft Since three side lenghts are given, you can use the Law of Cosines to find the player's shot angle, A. Check for Reasonableness Because a triangle can have at most one obtuse angle, it is wise to find the measure of the largest angle in a triangle first, which will be the angle opposite the longest side. If the largest angle is obtuse, then you know that the other two angles must be acute. If the largest angle is acute, the remaining two angles must still be acute. a 2 — b 2 + c 2 — 2 b e cos A Law of Cosines 62 = 242 + 202 - 2(24) (20) cos A a = 6, b = 24, and c = 20 36 = 576 + 400 - 960 cos A Simplify. 36 = 976 - 960 cos A Add. Subtract 976 from each side. —940 = —960 cos A 940 960 cos Divide each side by —960. ■cos -1 / 9 4 0 \ = A \9 6 0 / Use the cos-1 function. Use a calculator. 11.7° as A So, the player's shot angle is about 12°. y GuidedPractice 5. HIKING A group of friends who are on a cam ping trip decide to go on a hike. According to the map show n, what is the angle that is form ed by the two trails that lead to the camp? Trail 1 I Checkpoint Camp 2 mi Trail 2 Checkpoint Apply the Law of Cosines (SAS) Solve A A BC . Round side lengths to the nearest tenth and angle measures to the nearest degree. EfljflTl Use the Law of Cosines to find the m issing side measure. c2 = a 2 + b 2 — l a b cos C Law of Cosines c2 = 5 2 + 8 2 - 2(5)(8) cos 65° a = 5 , b = 8, and C = 65° c2 « 55.19 Use a calculator. Take the positive square root of each side. c ~ 7.4 ETTSffW Use the Law of Sines to find a m issing angle measure. sin A 5 sin 65° 7.4 sin ^ _ sin C a ~ c sin A = 5 sin 65° 7.4 Multiply each side by 5. Definition of sin-1 A = 38° ETTflin Find the measure of the remaining angle. B ~ 180° - (65° + 38°) or 77° Therefore, c = 7.4, A « 38°, and B ~ 77°. y GuidedPractice 6. Solve A HJK if H = 34 ° ,j = 7, and k = 10. l Find Areas of Oblique Triangles W hen the measures of all three sides of a triangle are i known, the Law of Cosines can be used to prove Heron's Form ula for the area of the triangle. KeyConcept Heron’s Formula StudyTip If the measures of the sides of AABCare a, fi, and c, then the area of the triangle is Semiperimeter The measure s used in Heron’s Formula is called the semiperimeter oi the triangle. B where s = ^ ( a + b + c). A V , V / Area = y /s ( s - a )( s - b)(s — c), b C J ............. ..... ... .............,........... You will prove this formula in Lesson 5-1, 51 in. Find the area of A XYZ. The value of s is 2-(45 + 51 + 38) or 67. Area = yjs(s ~ *)(s — 2/)(s — z) p Lesson 4 -7 = V 67(67 - 45)(67 - 51)(67 - 38) s = 67, x = 45, y = 51, and z —38 = V 683,936 Simplify. « 827 in2 Use a calculator. GuidedPractice 7A. 296 Heron’s Formula x = 24 cm , y = 53 cm, z = 39 cm T h e L a w o f Sines a n d th e L a w o f C osines 7B. x = 61 ft, y = 70 ft, 2 = 88 ft In the am biguous case of the Law of Sines, you com pared the length of a to the value h = b sin A. In the triangle shown, h represents the length of the altitude to side c in A ABC. You can use this expression for the height of the triangle to develop a formula for the area of the triangle. Area = —c/z Formula for area of a triangle : — c(b sin A) Replace h with b sin A. Simplify, = —be sin A By a sim ilar argument, you can develop the form ulas Area = —ab sin C and Area = —ac sin \ Notice that in each of these form ulas, the inform ation needed to find the area of the triangle is the measures of two sides and the included angle. KeyConcept Area of a Triangle Given SAS StudyTip Area of an Obtuse Triangle This formula works for any type of triangle, including obtuse triangles. You will prove this in Lesson 5-3. W o rd s The area of a triangle is one half the product of the lengths of two sides and the sine of their included angle. S y m b o ls Area = — 6c sin A C Area = j a c sin B A Area = ^ab sin C c B J V Because the area of a triangle is constant, the form ulas above can be w ritten as one formula. I l l Area = —be sin A = —ab sin C = —ac sin B If the included angle m easures 90°, notice that each formula sim plifies to the formula for the area of a right triangle, i(b a se )(h eig h t), because sin 90° = 1. B S I i n F i n c l the Area of a Triangle Given SAS Find the area of A G H J to the nearest tenth. H In A GHJ, g = 7 , h = 10, and J = 108°. Area = ^ g h sin J = j ( 7 ) ( l 0 ) sin 108° = 3 3 .3 Area of a triangle using SAS Substitution Simplify. So, the area is about 33.3 square centimeters. (►GuidedPractice Find the area of each triangle to the nearest tenth. 8A. 25 yd 8B. 297 Exercises = Step-by-Step Solutions begin on page R29. Solve each triangle. Round to the nearest tenth, if necessary. (Examples 1 and 2) 18. SKIING A ski lift rises at a 28° angle during the first 20 feet up a m ountain to achieve a height of 25 feet, w hich is the height m aintained during the rem ainder of the ride up the m ountain. Determ ine the length of cable needed for this initial rise. (Example 3) 1. ~~ ^ 25 ft 3. X o j 28' Ar J O ft Find two triangles with the given angle measure and side lengths. Round side lengths to the nearest tenth and angle measures to the nearest degree. (Example 4) 5. T 7. GOLF A golfer m isses a 12-foot p u tt b y p u tting 3° off course. The hole now lies at a 129° angle b etw een the b all and its spot before the putt. W hat d istance does the golfer need to p u tt in order to m ake the shot? (Examples 1 and 2) 8 . ARCHITECTURE An architect's client wants to build a home based on the architect Jon Lautner's Sheats-Goldstein House. The length of the patio will be 60 feet. The left side of the roof will be at a 49° angle of elevation, and the right side will be at an 18° angle of elevation. Determine the lengths of the left and right sides of the roof and the angle at which they will meet. (Examples 1 and 2) 19. A = 39°, a = 12, b = 17 20. A = 26° , a = 5 ,b = 9 21. A — 61°, a = 14, b = 15 22. A = 47°, a = 25, b = 34 23. A = 54°, a = 31, b = 36 24. A = 18°, a = 8, b = 13 25. BROADCASTING A radio tow er located 38 m iles along Industrial Parkw ay transm its radio broadcasts over a 30-m ile radius. Industrial Parkw ay intersects the interstate at a 41° angle. How far along the interstate can vehicles pick up the broadcasting signal? (Example 4) Industrial Parkway \ Tower \ 41° Y __ Interstate : 9 j TRAVEL For the initial 90 miles of a flight, the pilot heads 8° off course in order to avoid a storm. The pilot then changes direction to head toward the destination for the remainder of the flight, making a 157° angle to the first flight course. (Examples 1 and 2) a. Determine the total distance of the flight. b. Determine the distance of a direct flight to the destination. 10. a = 9, b = 7, A = 108° 12. a = 1 8 ,b = 12,A = 27° 14. a = 14, b = 6, ,4 = 145° 15. a= 19,b = 38, A 16. a = 5 ,b = 6 ,A = 63° 17. a= 10,b = V 2 0 0 , A = 45° 298 Lesson 4 -7 a= 14,b = 15, A < Solve each triangle. Round side lengths to the nearest tenth and angle measures to the nearest degree. (Examples 5 and 6) 27. A A BC, if A = 42°, b = 12, and c = 19 Find all solutions for the given triangle, if possible. If no solution exists, write no so lu tio n . Round side lengths to the nearest tenth and angle measures to the nearest degree. (Example 3) 11. 26. BOATING The light from a lighthouse can be seen from an 18-mile radius. A boat is anchored so that it can just see the light from the lighthouse. A second boat is located 25 miles from the lighthouse and is headed straight toward it, making a 44° angle with the lighthouse and the first boat. Find the distance betw een the two boats when the second boat enters the radius of the lighthouse light. (Example 4) = 117° . 13. a = 35, b = 24, A = 92° = 30° T h e L a w o f Sines a n d th e L a w o f C osines 28. A X Y Z , if x = 5, y = 18, and z = 14 29. A PQR, if P = 73°, q = 7, and r = 15 30. A JKL, if / = 125°, k = 24, and I = 33 31. A RST, if r = 35, s = 22, and t = 25 32. A FGH, if/ = 39, g = 50, and h = 64 33. A BCD, if B = 16°, c = 27, and d = 3 34. A LM N, i f £ = 12, m = 4, and n = 9 35. AIRPLANES D u rin g h er shift, a p ilo t flies from C o lu m bu s 51. DESIGN A free-stan d in g a rt p ro ject requ ires a triang u lar to A tlan ta, a d istan ce o f 448 m iles, and th e n on to su p p o rt p iece fo r stability. Tw o sid es o f the trian g le m u st Phoenix, a d istan ce o f 1583 m iles. From P h o en ix, she m easu re 18 and 15 feet in len g th and a n o n in clu d ed angle returns h o m e to C o lu m b u s, a d istan ce o f 1667 m iles. m u st m easu re 42°. If su p p o rt p u rp o ses requ ire the D eterm ine the ang les o f the trian g le created b y h er flig h t path. (Examples 5 and 6) triang le to h av e an area o f at le ast 75 squ are feet, w h at is the m easu re o f the third sid e? (Example 8) 36. CATCH L ola rolls a b all on the grou nd at an an g le o f 23° to the righ t o f h e r d o g B u ttons. If the b a ll ro lls a total d istance o f 48 feet, and sh e is stan d in g 30 feet aw ay, ho w far w ill B u tto n s h av e to ru n to retriev e the Use Heron's Form ula to find the area of each figure. Round answers to the nearest tenth. 52. (5 3 s 17 mm b all? (Examples 5 and 6) 34 ml 79.7 m Use Heron's Formula to find the area of each triangle. Round to the nearest tenth. (Example 7) QI U 2 4 .3 m 4 3 .2 m / 37. x = 9 cm , y = 11 cm , 2 = 16 cm 1.1 m 21 mm ' 5 8 .9 m mm 27 mm 7 H 20 mm \4 3 m \ / 3 9 . 6 m 38. x = 29 in., y = 25 in., 2 = 2 7 in. 39. x = 58 ft, y = 40 ft, 2 = 63 ft 40. x = 37 m m , y = 10 m m , 2 = 34 m m 41 . x = 8 yd, y= 15 yd , 2 = 8 yd 54. 8 cm 42. x = 133 m i, y = 82 m i, 2 = 77 m i 43. LANDSCAPING T h e S teele fam ily w an ts to exp an d their backyard b y p u rch asin g a v acan t lo t ad jacen t to their property. To g et a ro u gh m easu rem en t o f the area o f the lot, Mr. Steele cou n ted the step s n eed ed to w alk arou nd the b o rd er and d iag o n al o f the lot. (Example 7) 56. ZIP LINES A to u rist attractio n cu rren tly h as its base co n n ected to a tree p latfo rm 150 m eters aw ay by a zip lin e. T h e o w n ers n o w w an t to co n n ect the b ase to a seco n d p latfo rm lo cated acro ss a can y o n and then co n n e ct the p latfo rm s to e ach other. T h e bearings from the b a se to e ach p latfo rm and from platform 1 to p latfo rm 2 are giv en. F in d the d istan ces from the base to p latfo rm 2 and from p latfo rm 1 to p latfo rm 2 . a. E stim ate the area o f the lo t in steps. b. If Mr. S teele m easu red his step to b e 1.8 feet, d eterm in e the area o f the lo t in squ are feet. 44. DANCE D u rin g a p erfo rm an ce, a d an cer rem ain ed w ith in a triang u lar area o f the stage. (Example 7) 57. LIGHTHOUSES T h e b e a rin g fro m the S o u th B ay lig hthou se ♦ a. Find the area o f stage u sed in the p erform an ce. b. If the stage is 250 squ are feet, d eterm in e the p ercen tag e o f the stage u sed in the p erform an ce. F in d the area o f each tria n g le to the n e a re st te n th . (Example 8) 45. A ABC, if A = 98°, b = 13 m m , and c = 8 m m 46. A/XL, if L = 67°, j = 11 yd , and k = 24 yd to the S teep R o ck lig h th o u se 25 m iles aw ay is N 28° E. A sm all b o a t in d istress sp o tted o ff the coast b y each lig h th o u se h as a b e a rin g o f N 50° W from S o u th Bay and S 80° W from S teep R o ck. H o w far is each to w er from the b o at? ■ Steep m Rock^ - — r ' ^ ,'j 47. A RST, if R = 35°, s = 42 ft, and t = 26 ft 48. A X Y Z , if Y = 124°, x = 16 m, and 2 = 18 m 49. A FGH, if F = 41°, g = 22 in., and h = 36 in. 50. A PQR, if Q = 153°, p = 27 cm , an d r = 21 cm 50-1 7 25 sn n / X i/S o u t h ^ Bay mi | ■ 299 Find the area of each figure. Round answers to the nearest tenth. H.O.T. Problem s Use Higher-Order Thinking Skills 65. ERROR ANALYSIS M o n iq u e and R o g elio are solv in g an acu te trian g le in w h ich Z A = 34°, a = 16, and b = 21. M o n iq u e th in k s th at the trian g le has one solu tio n , w h ile R o g elio th in k s th at the trian g le h as n o solu tion. Is eith er o f th em correct? E xp lain you r reasoning. 66. WRITING IN MATH E xp lain the d ifferent circu m stan ces in w h ich yo u w o u ld u se the L aw o f C o sin es, the L aw o f Sin es, the P y th ag o rean T heo rem , and the trig on o m etric ratios to so lv e a triang le. 67. REASONING W h y d oes an o b tu se m easu rem en t ap p ear on the g rap h in g calcu lato r fo r in v erse cosin e w h ile n eg ativ e m easu res ap p ear fo r in v erse sine? 62. BRIDGE DESIGN In the figure below , Z F D E = 45°, Z.CED = 55°, ZFDE = ZFGE, B is the m id p o in t o f A C, and DE = EG. If AD = 4 feet, DE = 12 feet, and CE = 14 feet, find BF. 68. PROOF S h o w for a g iv en rh om b u s w ith a sid e len g th o f s and an in clu d ed an g le o f 6 th at the area can b e foun d w ith the fo rm u la A = s 2 sin 6. 69. PROOF D eriv e the L aw o f Sines. 70. PROOF C o n sid er the figu re below . C F 63. BUILDINGS B arbara w an ts to kn ow the d istan ce b etw een the top s o f tw o bu ild in g s R and S. O n the top o f h er B a. U se the fig u re and h in ts b elo w to d eriv e the first b u ild in g , she m easu res the d istance b etw een the p oints T and U and find s the giv en an g le m easu res. Find the fo rm u la a 2 = b 2 + c 2 — 2 be cos A in the L aw o f C osines. d istan ce b etw een the tw o build ings. • U se the P y th ag o rean T h eo rem for A D B C . • In A A D B, c 2 — x 2 + h 2. • cos A = jr b. E xp lain h o w y o u w o u ld go ab o u t d eriv in g the o th er tw o fo rm u las in the L aw o f C osin es. 71) CHALLENGE A satellite is o rb itin g 850 m iles ab ov e M ars and is n o w p o sitio n e d d irectly ab ov e on e o f the poles. The rad iu s o f M ars is 2110 m iles. If the satellite w as p o sitio n ed at p o in t X 14 m in u tes ago, ap p ro x im ately how m an y h o u rs d o es it take for the satellite to com p lete a full orbit, assu m in g th at it trav els at a con stan t rate arou nd a circu lar orbit? 64. DRIVING A fter a h ig h school football gam e, D ella left the park in g lo t trav elin g at 35 m iles p er h ou r in the d irection N 55° E. If D ev on left 20 m inu tes after D ella at 45 m iles p er h o u r in the d irection S 10° W, how far ap art are D ev on and D ella an h ou r and a h a lf after D ella left? 72. WRITING IN MATH D escrib e w h y solv in g a trian g le in w h ich h < a < b u sin g the L aw o f S in es resu lts in tw o solu tions. Is this also tru e w h e n u sin g the L aw o f C osin es? E xp lain yo u r reasoning. 300 Lesson 4-7 The Law o f Sines and the Law o f Cosines Spiral Review Find the exact value of each expression, if it exists. (Lesson 4-6) 73. c o s ^ - i 74. s in ” 1 ^ -jC ■ _i V3 76. sin 1 — 75. arctan 1 Identify the damping factor/(.v) of each function. Then use a graphing calculator to sketch the graphs of f(x ), —fix ), and the given function in the same view ing window. D escribe the behavior of the graph. (Lesson 4-5) 77. 78. y = -|x cos x y = —2x sin x 79. y — (x — l ) 2 sin x 80. y = —4 x 2 cos x 81. CARTOGRAPHY T h e d istan ce arou nd E arth along a g iv en latitu d e can b e fou n d u sing C = 27Tr cos L, w here r is the rad iu s o f E arth and L is the latitu d e. T h e rad iu s o f E arth is app roxim ately 3960 m iles. M ak e a table o f v alu es for the latitu d e and co rresp on d in g d istance arou nd E arth th at in clu d es L = 0°, 30°, 45°, 60°, and 90°. U se the table to d escrib e the d istances alo ng the latitu d es as yo u go fro m 0° at the e q u ato r to 90° at a pole. (Lesson 4-3) 82. RADIOACTIVITY A scien tist starts w ith a 1-gram sam p le o f lead-211. T h e am o u n t o f the sam p le rem ain in g after v ariou s tim es is sh o w n in the tab le below . (Lesson 3-5) Time (min) Pb-211 present(g) 10 20 30 40 0.83 0 .6 8 0.56 0.46 a. Find an exp o n en tial reg ression eq u atio n for the am o u n t y o f lead as a fu n ctio n o f tim e x. b. W rite the reg ressio n eq u atio n in term s o f b ase e. C. U se the eq u atio n from p art b to e stim ate w h en there w ill b e 0.01 g ram o f lead-211 present. Write a polynomial function of least degree with real coefficients in standard form that has the given zeros. (Lesson 2-4) 83. - 1 ,1 ,5 84. - 2 , - 0 . 5 , 4 85. - 3 , - 2 i , 2 i 86. - 5 z , - i , i, 5i Skills Review for Standardized Tests f;. IT / / 89. FREE RESPONSE T h e p en d u lu m 87. SAT/ACT W h ich o f the fo llo w in g is the at the rig h t m o v es acco rd in g to p erim eter o f the trian g le show n? 9 = — cos 12f, w h e re 9 is the A 49.0 cm a n g u la r d is p la c e m e n t in B 66.0 cm rad ian s and t is the tim e in second s. C 71.2 cm D 91.4 cm 22 cm E 93.2 cm f 6 — 4 a. S et the m od e to rad ian s and g rap h the fu n ctio n for 0 < t< 2 . 88. In A DEF, w h at is the v alu e o f 9 to the n earest d egree? b. W h a t are the p erio d , am p litu d e, and freq u en cy of the fu n ctio n ? W h a t do th ey m ean in the con te x t o f this situ ation? C. W h a t is the m ax im u m an g u lar d isp lacem en t of the p e n d u lu m in d eg rees? VTT5 F 26° G 74° d. W h a t d o es the m id lin e o f the g rap h represent? H 80° e. A t w h at tim es is the p e n d u lu m d isp laced 5 d egrees? J 141° tor--1 .. -....... ■j connectED.m cgraw-hill.com | 301 Study Guide and Review Chapter Summary KeyVocabulary Right Triangle Trigonometry sin0 = ^P hyp 'Lesson 4-1) adj hyp csc 6 ■ opp sec 9 = Degrees and Radians (Lesson 4 2) • • • tane = ^ cos 9 = — L hyp adj adj cot 9 = opp adj 7t radians 180° ' 180° To convert from radians to degrees, multiply by ■k radians' To convert from degrees to radians, multiply by angular speed (p. 236) phase shift (p. 261) circular function (p. 248) quadrantal angle (p. 243) cosecant p. 220) radian (p. 232) cosine (p. 220) reciprocal function (p. 220) cotangent (p. 220) reference angle (p. 244) coterminal angles p. 234) secant ip. 220) sector (p. 237) sine (p. 220) w = j , where 9 is the angle of rotation damped wave (p. 275) sinusoid (p. 256) (in radians) moved during time t damping factor (p. 275) standard position (p. 231) frequency (p. 260) tangent (p. 220) initial side (p. 231) terminal side (p. 231) inverse trigonometric function (p. 223) trigonometric functions (p. 220) (Lesson 4-3) For an angle 9 in radians containing (x, y), cos 9 = j , sin 9 = j , For an angle t containing (x, y) on the unit circle, cos 9 = x, Law of Cosines (p. 295) Law of Sines (p. 291) linear speed (p. 236) (Lesson 4-4) trigonometric ratios (p. 220) unit circle (p. 247) vertical shift (p. 262) midline (p. 262) A sinusoidal function is of the form y = a sin (bx + c) + d or y = a cos (b x- 1- c) + d, where amplitude = \a\, period = Ih\ frequency = V - , phase shift = — and vertical shift = d. 27T ID] Graphing Other Trigonometric Functions (Lesson 4-5) A damped trigonometric function is of the form y = f(x) sin bx or y = f(x) cos bx, where f(x) is the damping factor. Inverse Trigonometric Functions (Lesson 4-6) • y = s i n -1 x iff sin y = x, for —1 < x < 1 and - y < y < y . • y = cos-1 x iff cos y = x, for - 1 < x < 1 and 0 < y < tt. • y = tan~1 x iff tan y = x, for -o o < x< oo The Law of Sines and the Law of Cosines and (Lesson 4-7) Let A A B C be any triangle. • periodic function (p. 250) Angular speed: Graphing Sine and Cosine Functions • angle of elevation p 224) damped trigonometric function (p. 275) sin 9 = y, and tan 0 = | - • period (p. 250) v = ~t , where s is the arc length traveled during time t and tan Q = \ , where r = \J x 2 + y 2. • oblique triangles (p. 291) Linear speed: Trigonometric Functions on the Unit Circle • amplitude (p. 257) angle of depression (p. 224) The Law of Sines: The Law of Cosines: sin C c 2be cos A a2 + c 2 ■ la c cos B c 2 = a2 + b2 - la b cos C C h a p te r 4 State whether each sentence is true or false. If false, replace the underlined term to make a true sentence. 1. The sine of an acute angle in a right triangle is the ratio of the lengths of its opposite leg to the hypotenuse. 2. The secant ratio is the reciprocal of the sine ratio. 3. An angle of elevation is the angle formed by a horizontal lineand an observer’s line of sight to an object below the line. 4. The radian measure of an angle is equal to the ratio of the length of its intercepted arc to the radius. 5. The rate at which an object moves along a circular path is called its linear speed. 6. 0°, tt , and - y are examples of reference angles. 7. The period of the graph of y = 4 sin 3x is 4. sin A _ sin B a b a2 = b2 + c2 b2 302 VbcabularyGheck S tu d y G u id e a n d R e v ie w 8. For f(x) = cos bx, as b increases, the freouencv decreases. 9. The 10. range of the arcsine function is [0, -tt], The Law of Sines can be used to determine or angle measures of some triangles. unknown sidelengths Lesson-by-Lesson Review c ^ ^ "f (j/V Right Triangle Trigonometry (pp. 220 - 230 ) Find the exact values of the six trigonometric functions of 0. 12 . 11. ' Example 1 Find the value of x. Round to the nearest tenth, if necessary. 41 Tangent function tane = ^ ad] tan 38° = Find the value of x. Round to the nearest tenth, if necessary. 10 0 = 38°, opp = 10, and adj = x 13. Multiply each side by x. X tan 38° ^ Find the measure of angle 9. Round to the nearest degree, if necessary. ( Use a calculator. X s: 12.8 K Divide each side by tap 38° / 15. Degrees and Radians (pp. 231 - 241 ) Write each degree measure in radians as a multiple of radian measure in degrees. 17. 135° 18. 450° 19' -74t 20. tt and each Exam ple 2 Identify all angles coterminal w ith . Then find and draw one positive and one negative coterminal angle. 1 3 tt All angles measuring 10 5 tt 12 2n-rr are coterminal with a ^ angle. Let n = 1 and - 1 . Identify all angles coterminal w ith the given angle. Then find and draw one positive and one negative angle coterminal w ith the given angle. 21. 342° 22. 6 ^ + 2tt(1) = ^ - 2 tt(—1) = 2-7l_ IT T t Find the area of each sector. 23. z' ' — 24. / _____ \ [3 -1 V i4 \ 1 in. / / \ / \ 10 m A \ 47tt 30 *r j / / 303 ' . ■ . " Study Guide and Review ' . . . : ' ' . • . . ... :: " . . 'Vs'*- . .. Continued Trigonometric Functions on the Unit Circle (pp. 242-253) Sketch each angle. Then find its reference angle. 25. 240° 26. 75° 28. 27. Example 3 Let cos 9 = 11 TT where sin 9 < 0. Find the exact values of the five remaining trigonom etric functions of 9. 18 Find the exact values of the five remaining trigonometric functions of 0 . Since cos 9 is positive and sin 9 is negative, 9 lies in Quadrant IV. This means that the x-coordinate of a point on the terminal side of 9 is positive and the y-coordinate is negative. 29. Since cos 9 = j = cos 0 = • where sin 9 > 0 and tan 9 > 0 5 ( 30. tan 0 = ~ | , where sin 9 > 0 and cos 9 < 0 31. sin 9 = where cos 9 > 0 and cot 9 < 0 32> cot 9 = |, where sin 9 < 0 and tan 9 > 0 i= use x = 5 and r = 13 to find y. Pythagorean Theorem V r2- x2 = V169 - 25 or 12 r — 13 and sin 6» = 7 o r t a n 0 = | o r ^ | '1 3 * CSC 6 - L or™ y Find the exact value of each expression. If undefined, write undefined. 12 o x — 5 sec0 = ^ o r^ |A o cot0 = | o r A 11 TT 33. sin 180° 34. cot- 35. sec 450° 36. cos - Hr) Graphing Sine and Cosine Functions (pp. 256-266) Describe how the graphs of f(x ) and g(x) are related. Then find the amplitude and period of g(x), and sketch at least one period of both functions on the same coordinate axes. 37. 38. f(x) = cos x f(x) = sin x g(x) =.cos 2x g(x) = 5 sin x 39. 40. f(x) = sin x 9 (x) = Tj-sin x f ( x ) = cos x Example 4 State the amplitude, period, frequency, phase shift, and vertical shift of y = 4 sin — y j — 4. Then graph two periods of the function. In this function, a = 4, b = 1, c = and d = - 4 . Amplitude: \a\ = |4| or 4 Period: t t = g[x) = -c o s \b\ \b\ |1 | 1 F re q u e n c y :- = - o r State the amplitude, period, frequency, phase shift, and vertical shift of each function. Then graph two periods of the function. 41. y = 2 cos (x - tv) 42. y = - s in 2x + 1 43. y = lc o s ( x + ^ ) 44. y = 3 s in ( x + ^ ) 304 Chapter 4 Study G uide and Review Phase shift: — t t | 1| o r 2 tv Vertical shift: rfo r - 4 2 \b\ |1 | ° r 2 First, graph the midline y = - 4 . Then graph y = 4 sin xshifted • units to the right and 4 units down. Other Trigonometric Functions (pp. 269-279) Locate the vertical asymptotes, and sketch the graph of each function. Exam ple 5 45. y = 3 tan x 46. y = ^ta n ( * - y ) y = 2 sec ( x + j ) . 47. y = c o t ( x + y j 48. y = - c o t (x - -it) Because the graph of y = 2 sec I x + y j is the graph of 49. y = 2 sec |- |j 50. y = - c s c ( 2 x ) y = 2 sec xshifted to the left y units, the vertical asymptotes for one 51 . y = sec (x - -tt) 52. y = | c s c ( x + | ) Locate the vertical asymptotes, and sketch the graph of period are located at - ¥ . T . a n d ^ . 4 the interval inverse irigonometric Functions 4 4 Graph two cycles on 3 tt 1 3 tt 4 4 (pp. 280-290) Find the extict value of each expression, if it exists. 53. s in "1 ( - 1 ) 54. c o s "1 ~ - 55. ta n " 1 56. arcsin 0 57. arctan ( - 1 ) 58. arccos - y - 59. sin-1 sin ( - y ) V2 Find the exact val ue of arctan - V 3 . Find a point on the unit circle in the interval ( - y . y ) w ith a tangent of -\/3 .W h e n t == - f . t a n f = - V 3 . Therefore, arctan -V 3 = - f . 60. c o s -1 [cos ( - 3 * ) ] V , -J The Law of Sines and the Law of Cosines (pp. 291-301) Find all solutions for the given triangle, if possible. If no solution exists, write no solution. Round side lengths to the nearest tenth and angle measurements to the nearest degree. 61. a = 11, b = 6, AY 22° 62. a = 9 , b = 1 0 , A = 42° 63. a = 20, 6 = 1 0 , 4 = 78° 64. a = 2 , 6 = 9 ,4 = 88° Solve each triangle. Round side lengths to the nearest tenth and angle measures to the nearest degree. 65. a = 1 3 , 6 = 1 2 , c = I 66. Exam ple 7 Solve the triangle if a = 3, b = 4, and A = 71°. In the figure, h = 4 sin 71 ° or about 3.I Because a < h, there is no triangle that can be formed with sides a = 3, b = 4, and A = 71 °. Therefore, this problem has no solution. a = 4, 6 = 5 , C = 9 6 ° connectED.m cgraw-hill.com 305 Applications and Problem Solving 67. CONSTRUCTION A construction company is installing a three-foot- high wheelchair ramp onto a landing outside of an office. The angle of the ramp must be 4°. (Lesson 4-1) a. What is the length of the ramp? b. What is the slope of the ramp? 68. NATURE For a photography project, Maria is photographing deer from a tree stand. From her sight 30 feet above the ground, she spots two deer in a straight line, as shown below. How much farther away is the second deer than the first? (Lesson 4-1) 72. AIR CONDITIONING An air-conditioning unit turns on and off to maintain the desired temperature. On one summer day, the air conditioner turns on at 8:30 a .m . when the temperature is 80° Fahrenheit and turns off at 8:55 a . m . when the temperature is 74°. (Lesson 4-4) a. Find the amplitude and period if you were going to use a trigonometric function to model this change in temperature, assuming that the temperature cycle will continue. b. Is it appropriate to model this situation with a trigonometric function? Explain your reasoning. 73. TIDES In Lewis Bay, the low tide is recorded as 2 feet at 4:30 a . m ., and the high tide is recorded as 5.5 feet at 10:45 a . m . (Lesson 4-4) a. Find the period for the trigonometric model. b. At what time will the next high tide occur? j Ll 69. FIGURE SKATING An Olympic ice skater performs a routine in which she jumps in the air for 2.4 seconds while spinning 3 full revolutions. (Lesson 4-2) a. Find the angular speed of the figure skater. b. Express the angular speed of the figure skater in degrees per minute. 74. MUSIC When plucked, a bass string is displaced 1.5 inches, and its damping factor is 1.9. It produces a note with a frequency of 90 cycles per second. Determine the amount of time it takes the string’s motion to be dampened so that —0.1 < y < 0.1. (Lesson 4-5) 75. PAINTING A painter is using a 15-foot ladder to paint the side of a house. If the angle the ladder makes with the ground is less than 65°, it will slide out from under him. What is the greatest distance that the bottom of the ladder can be from the side of the house and still be safe for the painter? Lesson 4-6) 70. TIMEPIECES The length of the minute hand of a pocket watch is 1.5 inches. What is the area swept by the minute hand in 40 minutes? (Lesson 4-2) 76. NAVIGATION A boat is 20 nautical miles from a port at a bearing 30° north of east. The captain sees a second boat and reports to the port that his boat is 15 nautical miles from the second boat, which is located due east of the port. Can port personnel be sure of the second boat’s position? Justify your answer. (Lesson 4-7) 71. WORLD’S FAIR The first Ferris wheel had a diameter of 250 feet and took 10 minutes to complete one full revolution. (Lesson 4-3) a. How many degrees would the Ferris wheel rotate in 100 seconds? a. Find C. b. How far has a person traveled if he or she has been on the Ferris wheel for 7 minutes? c. How long would it take for a person to travel 200 feet? 306 C h a p te r 4 77. GEOMETRY Consider quadrilateral ABCD. (Lesson 4-7) Study G uide and Review b. Find the area of ABCD. Find the value of x. Round to the nearest tenth, if necessary. 16. TIDES The table gives the approximate times that the high and low tides occurred in San Azalea Bay over a 2-day period. 1- 2X High 1 Low 1 High 2 2:35 a . m . 8:51 a . m . 3:04 p . m . 9:19 p . m . 9:48 a . m . 3:55 p . m . 1 0 :2 0 Tide Day 1 Day 2 Find the measure of angle 9. Round to the nearest degree, if necessary. 3:30 a . m . Low 2 p .m . a. The tides can be modeled with a trigonometric function. 4. Approximately what is the period of this function? b. The difference in height between the high and low tides is 7 feet. What is the amplitude of this function? c. Write a function that models the tides where f is measured in hours. Assume the function has no phase shift or vertical shift. 5. MULTIPLE CHOICE What is the linear speed of a point rotating at an angular speed of 36 radians per second at a distance of 12 inches from the center of the rotation? A 420 in./s C 439 in./s B 432 in./s D 444 in./s Write each degree measure in radians as a multiple of ir and each radian measure in degrees. 6. 200 ° 8. Find the area of the sector of the circle shown. Locate the vertical asymptotes, and sketch the graph of each function. 17. y = ta fr(x + | ) 18. y = l s e c 4 x Find all solutions for the given triangle, if possible. If no solution exists, w rite no solution. Round side lengths to the nearest tenth and angle measurements to the nearest degree. 19. a = 8, * = 1 6 , A = 22° 21. a = 3, b = 5, c = 7 20. a = 9 , b = 7, A = 84° 22. a = 8, 6 = 1 0 , C = 4 6 ° Find the exact value of each expression, if it exists. 23. Sketch each angle. Then find its reference angle. 9. 165° 10. 21 TV 13 cos” 1 24. s in - 1 ( - 1 ) 25. NAVIGATION A boat leaves a dock and travels 45° north of west averaging 30 knots for 2 hours. The boat then travels directly west averaging 40 knots for 3 hours. Find the exact value of each expression. 11. s e c ^ 6 13. 12. cos (-2 4 0 ° ) MULTIPLE CHOICE An angle 9 satisfies the following inequalities: csc 9 < 0, cot 9 > 0, and sec 9 < 0. In which quadrant does 9 lie? F I H III G II J IV State the amplitude, period, frequency, phase shift, and vertical shift of each function. Then graph tw o periods of the function. 14. y = 4 cos-| - 5 15. y = - s i n ( x + y ) a. How many nautical miles is the boat from the dock after 5 hours? b. How many degrees south of east is the dock from the boat’s present position? ^connectEDjncgraw^^^^ 307 If air is being pumped into a balloon at a given rate, can we find the rate at which the volume of the balloon is expanding? How does the rate a company spends money on advertising affect the rate of its sales? Related rates problems occur when the rate of change for one variable can be found by relating that to rates of change for other variables. # Model and solve related rates problems. Suppose two cars leave a point at the same time. One car is traveling 40 miles per hour due north, while the second car is traveling 30 miles per hour due east. How far apart are the two cars after 1 hour? 2 hours? 3 hours? We can use the formula d = r t and the Pythagorean Theorem to solve for these values. In this situation, we know the rates of change for each car. What if we want to know the rate at which the distance between the two cars is changing? Activity 1 Rate of Change Two cars leave a house at the same time. One car travels due north at 35 miles per hour, while the second car travels due east at 55 miles per hour. Approxim ate the rate at which the distance betw een the two cars is changing. ETTTT1 M ak e a sk etch o f the situ atio n . k W '.H W rite eq u ation s fo r the d istan ce traveled b y e ach car after t hou rs. Find the d istan ce trav eled b y each car after 1, 2, 3, and 4 h ou rs. E T7m U se the P y th ag orean T h eo rem to find the N 105» n 70 3 5*. ‘X X f= 3 t=2 \ d istan ce b etw een the tw o cars at each p o in t in tim e. EVM I1 Find the av erage rate o f ch an g e o f the d istan ce b etw een the tw o cars for 1 < t < 2 , 2 < f < 3, and 3 < t < 4. 55 11C pAnalyze the Results 1. M ak e a scatter p lo t d isp lay in g the total d istan ce b etw een the tw o cars. L et tim e t b e the in d ep en d en t v ariab le and total d istan ce d b e the d ep en d en t v ariab le. D raw a lin e throu gh the points. 2. W h at typ e o f fu n ctio n does the grap h seem to m o d el? H o w is y o u r co n jectu re su p p o rted by the v alu es foun d in Step 5? 3. W h at w o u ld h ap p en to the av erag e rate o f ch an g e o f the d istan ce b e tw e e n the tw o cars if one o f the cars slow ed d ow n? sp ed up ? E xp lain y o u r reasoning. ^ V ............................................................................................................................................................................................ The rate that the distance between the two cars is changing is related to the rates of the two cars. In calculus, problems involving related rates can be solved using im plicit differentiation. However, before we can use advanced techniques of differentiation, we need to understand how the rates involved relate to one another. Therefore, the first step to solving any related rates problem should always be to model the situation with a sketch or graph and to write equations using the relevant values and variables. 308 C h a p te r 4 Activity 2 Model Related Rates A rock tossed into a still body of water creates a circular ripple that grows at a rate of 5 centim eters per second. Find the area of the circle after 3 seconds if the radius of the circle is 5 centim eters at t = 1. M ake a sk etch o f the situ ation. . t= 3 W rite an eq u atio n fo r the rad iu s r o f the circle after f second s. / r= 5 t . t= 2 / ' t= 1 i / / Find the rad iu s for f = 3, and then fin d the area. '\ '\ H 5 cm ^ Analyze the Results \ \ ) ~~ / / ' ’i 1 1 / 4. Find an eq u atio n for th e area A o f the circle in term s o f f. 5. Find the area o f the circle fo r t = 1, 2 , 3 , 4 , and 5 second s. 6 . M ak e a g rap h o f the v alu es. W h at typ e o f fu n ctio n d o es the g rap h seem to m od el? You can use the difference quotient to calculate the rate of change for the area of the circle at a certain point in time. Activity 3 Approximate Related Rate Approxim ate the rate of change for the area of the circle in Activity 2. StudyTip Difference Quotient Recall that the difference quotient for calculating the slope of the line tangent to the graph of f(x) at the point (x, f(x)) is m S u b stitu te the exp ressio n for the area o f the m■ circle in to the d ifferen ce qu otient. it[5 ( f + h)]2 —TT(5f)2 Ti A p p ro x im ate the rate o f ch an g e o f the circle at 2 secon d s. L et h = 0.1, 0.01, and 0.001. f(x+tl)~f(x) R ep eat Step s 1 and 2 for t = 3 seco n d s and t = 4 second s. y Analyze the Results 7. W h at do the rates o f ch an g e ap p ear to ap p roach for e ach v a lu e o f t ? 8 . W h at h ap p en s to the rate o f ch an g e o f the area o f the circle as the rad iu s in creases? Explain. 9. H o w d o es this ap p roach d iffer from the ap p roach y o u u sed in A ctiv ity 1 to fin d the rate of ch an g e fo r the d istan ce b e tw e e n the tw o cars? E xp lain w h y th is w as necessary. Model and Apply 10. A 13-fo ot lad d er is lean in g ag ain st a w all so th at the b a se o f th e lad d er is exactly 5 feet from the b a se o f the w all. If the b o tto m o f the lad d er starts to slid e aw ay fro m the w all at a rate of 2 feet p er secon d , h o w fast is the top o f the lad d er slid in g d o w n the w all? a. S k etch a m o d el o f th e situ atio n . L et rf b e the d istan ce from the top o f the lad d er to the g ro u n d and m b e the rate at w h ich the top o f the lad d er is slid in g d o w n the w all. b. W rite an e xp ressio n fo r the d istan ce from the b a se o f the lad d er to the w all after t second s. c. F in d an eq u atio n for the d istan ce d from the top o f the lad d er to th e grou nd in term s o f t b y su b stitu tin g the e xp ressio n fou n d in p art b in to the P y th ag o rean T heorem . d. U se the P y th ag o rean T h eo rem to fin d the d istan ce d from the top o f the lad d er to the grou nd fo r t = 0 ,1 , 2, 3, 3.5, and 3.75. e. M ak e a g rap h o f the v alu es. W h a t typ e o f fu n ctio n d o es the g rap h seem to m od el? f. U se the d ifferen ce q u o tien t to ap p rox im ate the rate o f ch an g e m fo r the d istan ce from the top o f the lad d er to the grou nd at t = 2. L et h = 0 .1 ,0 .0 1 , and 0.001. A s h ap p roaches 0, w hat d o the v alu es for m ap p ear to ap p roach ? hp= § [connectED^mcgraw-h^ 309 Get Ready for the Chapter Diagnose Readiness You have tw o options fo r checking Prerequisite Skills. NewVocabulary English ^ Textbook Option Take the Quick Check below. QuickCheck trig o n o m e tric ide n tity p. 312 identidad trigonom etrica reciprocal ide n tity p. 312 identidad reciproca qu o tie n t ide n tity p. 312 cociente de identidad Pythagorean ide n tity p. 313 P itagoras identidad o d d -e ve n -id e n tity p. 314 im p a r-in clu so -d e identidad cofu nction p. 314 co fun cion v e rify an ide n tity p. 320 v e rific a r una identidad sum ide n tity p. 337 sum a de identidad redu ctio n ide n tity p. 340 identidad de reduccion d ou ble -a ng le ide n tity p. 346 d ob le-a n gu lo de la identidad pow e r-re d u cin g ide n tity p. 347 p od er-re du cir la identidad h a lf-a n g le ide n tity p. 348 m edio angulo identidad Solve each equation by factoring. [Lesson 0-3! 1. x2 + 5 x - 2 4 = 0 2. x2 — 11 x + 28 = 0 3. 2x2 — 9x — 5 = 0 4. 15x2 + 26x + 8 = 0 5. 2x3 — 2x2 — 12x = 0 6. 12x3 + 78x2 — 42x = 0 7. ROCKETS A rocket is projected vertically into the air. Its distance in feet after t seconds is represented by s(t) = - 1 6 f2 + 192f. Find the amount of time that the rocket is in the air. Lesson 0-3} Espanol Find the missing side lengths and angle measures of each triangle. Lessons 4-1 and 4-7) 8. ReviewVocabulary e xtraneous solu tio n p. 91 solucion extraha a solution that does not satisfy the original equation q uadrantal angle p. 243 angulo cua dran ta an angle 0 in standard position that has a terminal side that lies on one of the coordinate axes u n it circle p. 247 c irc u lo u nitario a circle of radius 1 centered at the origin p eriodic fu n c tio n p. 250 fu n cio n perio dica a function with range values that repeat at regular intervals trig o n o m e tric fu n c tio n s p. 220 fu n cio n e s trig o n o m e trica s Let e be any angle in standard position and point P(x, y) be a point on the terminal side of 0. Let /represent the nonzero distance from Pto the Find the exact value of each expression. (Les: 12. cot 420° 14. sec 10tv 16. csc 2n 2 3 13. cos ~~4 15. tan 480° origin or |r| = V * 2 + y 2 # as follows. Then the trigonometric functions of 9 are sin 0 = cos 6 = j tan 0 = ^ x ^ 0 csc 0 = -jj, y =£ 0 sec 0 = -j, x ^ 0 co t0 = ^ O 17. sin 510° Online Option Take an online se lf-ch e ck Chapter I Readiness Quiz at c o n n e c tE D .m c a ra w -h ill.c o m . W amI 311 • You found trigonom etric values using the unit circle. (Lesson 4-3) NewVocabulary identity trigonometric identity cofunction 1 Identify and use basic trigonom etric identities to find trigonom etric values. Use basic trigonom etric ■identities to sim plify and rewrite trigonometric expressions. • Many physics and engineering applications, such as determining the path of an aircraft, involve trigonometric functions. These functions are made more flexible if you can change the trigonometric expressions involved from one form to an equivalent but more convenient form. You can do this by using trigonometric identities. Basic Trigonometric Identities 1 A n eq u atio n is an identity if the left sid e is equ al to the rig h t sid e for all v alu es o f the v ariab le fo r w h ic h b o th sid es are d efined . C o n sid er the equ atio n s below . x —9 = x + 3 x —3 This is an identity since both sides of the equation are defined and equal for all x such that x 3. This is not an identity. Both sides of this equation are defined and equal for certain values, such as sin x = 1 — cos x when x —0, but not for other values for which both sides are defined, such as when x = Trigonometric identities are id en tities th at in v o lv e trig o n o m etric fu n ction s. You alread y k n o w a few b asic trig o n o m etric identities. T h e recip ro cal an d q u o tie n t id e n tities b e lo w fo llow d irectly fro m the d efin itio n s o f the six trig o n o m etric fu n ctio n s in tro d u ced in L esso n 4-1. KeyConcept Reciprocal and Quotient Identities Reciprocal Identities sin 9 = — CSC 0 CSC V 6 = -4 -r sin 6 Quotient Identities cos e = — sec 6 tan 0 = 4 cot 9 tan 0 = ^ 4 sec 6 = — cos 9 cot 9 = — tan 9 cot 9 = ^ sin 9 COS 9 You can u se these basic trig o n o m etric id en tities to find trig o n o m etric v alu es. A s w ith an y fraction , the d en o m in ator can n o t equ al zero. Use Reciprocal and Quotient Identities a. If csc 6 = 4 b. find sin 8. 1 csc 9 1 2 V5 If cot x = — — and sin x = — , find cos x. 5 V5 Reciprocal Identity csc 0 = c o tx = t 5V5 COS X s in x cos x V5 Quotient Identity cot X= — 2 5 V5" • V5 si n x = - y - 3 Divide. 2 V5 5V 5 3 Vs = COS X — = cos x f Lesson 5-1 Simplify. GuidedPractice 1A. If sec x = —, find cos x. 312 Multiply each side by - y - . 1B. If csc /3 = -y- and sec fi = find tan /3. R ecall from L esso n 4-3 th at trig on o m etric fu n ctio n s can b e d efined on a u n it circle as show n. N o tice th at for an y an g le 0, sin e and cosin e are the d irected len g th s o f the legs o f a rig h t trian g le w ith h y p o ten u se 1. W e can ap p ly the P y th ag o rean T h eo rem to th is rig h t trian g le to estab lish an o th er b asic trig o n o m etric identity. (sin 0 )2 + (cos 0 )2 = l 2 Pythagorean Theorem s in 2 0 + c o s2 0 = 1 Simplify, W h ile the sig n s o f th ese d irected len g th s m a y ch an g e d ep en d in g on the q u ad ran t in w h ich the trian g le lies, n o tice th at b e cau se th ese len g th s are sq u ared , the eq u atio n abo v e h o ld s true for any v alu e o f 0. T h is eq u atio n is one o f th ree Pythagorean identities. ReadingMath m a KeyConcept Pythagorean Identities Powers of Trigonometric Functions sin2 9 is read as sine squared theta and interpreted as the square of the quantity sin 9. sin2 9 + cos2 9 = 1 cot2 0 + tan2 9 + 1 = sec2 9 1= t i csc2 6 V. J You will prove the remaining two Pythagorean Identities in Exercises 69 and 70. N otice the shorth an d n o tatio n u sed to rep resen t p o w ers o f trig o n om etric fu n ctions: s in 2 0 = (sin 0 )2, co s2 0 = (cos 0)2, ta n 2 0 = (tan 0) 2, and so on. Use Pythagorean Identities If tan 0 = —8 and sin 0 > 0, find sin 0 and cos 0. U se the P y th ag o rean Id en tity th at in v o lv es tan 0. tan2 0 + 1 = sec2 0 Pythagorean Identity tan 9 = —8 (—8)2 + 1 = sec2 0 65 = sec2 0 Simplify. ± V 6 5 = sec 0 Take the square root of each side. ±V&5 = —i — Reciprocal Identity COS 0 ±- V65 Solve for cos 9. : CO S I 65 S in ce tan 9 = cos 9 is n e g ativ e and sin 0 is p o sitiv e, cos 0 m u st b e rie You can th en use this q u o tien t id en tity ag ain to find sin 0. tan 0 = -8 9 s in Quotient Identity cos 9 9 V65 s in V65 tan 9 = —8 and cos 9 = 65 65 8 \ /6 5 65 StudyTip Checking Answers It is beneficial to confirm your answers using a different identity than the ones you used to solve the problem, as in Example 2, so that you do not make the same mistake twice. / CHECK = sin I Multiply each side by ■ s in 2 0 + c o s 2 i = 1 V 65 )2 ± 1 65 I ff 65 = 1 V65 65 ‘ Pythagorean Identity sin 0 = 8V65 65 and cos 0 - V65 65 Simpiify. p GuidedPractice Find the value of each expression using the given information. 2A. csc 0 and tan 0; co t 0 = —3, cos 0 < 0 2B. cot x and sec x; sin x = —, cos x > 0 6 i ......... ................... nl.connectED .m cgraw -hill.com | 313 A n o th er set o f b asic trig o n om etric id en tities in v olv e cofu n ctio n s. A trig o n o m etric fu n ctio n / is a cofunction o f an o th er trig o n o m etric fu n ctio n g if f ( a ) = g(J3) w h e n a and (3 are com p lem en tary angles. In the rig h t trian g le sh ow n , an g les a and (3 are co m p le m e n tary angles. U sin g the righ t trian g le ratio s, yo u can show th at the fo llo w in g statem en ts are true. sin a = co s /3 = cos (90° — a ) = j tan a = cot (3 = co t (90° — a ) = — sec a = csc /3 = csc (90° — a ) = — From these statem en ts, w e can w rite the fo llo w in g co fu n ctio n id en tities, w h ich are v alid for all real nu m bers, n o t ju st acu te angle m easures. StudyTip Writing Cofunction Identities Each of the cofunction identities can also be written in terms of degrees. For example, sin 9 = cos (90° - 9). KeyConcept Cofunction Identities sin 9 = cos - flj sec 6 = csc ( j - 0 j cot 9 = tan | y - 0 j csc 9 = sec ( y - e j tan 9 = cot - ej cos 9 = sin ( y - 0 j J I You w ill prove these identities for any angle in Lesson 5-3, You h av e also seen th at each o f the b asic trig o n o m etric fu n ctio n s— sin e, cosine, tan g en t, cosecan t, secan t, and cotan g en t— is eith er odd or even. U sin g the u n it circle, y o u can sh o w th at the fo llo w in g statem ents are true. sin (—a ) = —y cos (—a ) = x sin a = i/ cos a = x R ecall from L esso n 1-2 th at a fu n ctio n / is e v en if for e v ery x in the d o m ain o f /,/(—* ) = f ( x ) and od d if fo r ev ery x in the d o m ain of/, f ( —x) = —f(x ). T h ese relatio n sh ip s lead to the fo llo w in g o d d -ev en id entities. KeyConcept Odd-Even Identities sin (-9 ) = - s in 9 cos (-9 ) = cos 9 tan ( - 0 ) = - t a n 0 csc (—9) = —csc 9 Sec (— 0) = sec 0 cot ( - 0 ) = - c o t 0 J You can use co fu n ctio n and o d d -ev en id en tities to fin d trig o n o m etric valu es. Use Cofunction and Odd-Even Identities If tan 0 = 1.28, find cot — y j. co t ^0 - y j = cot [ ~ ( y ~ ^ ) ] Factor. Odd-Even Identity f = —tan 0 Cofunction Identity = - 1 .2 8 tan 0 = 1 .2 8 GuidedPractice 3. If sin x = —0.37, fin d cos 314 Lesson 5-1 | Trigo no m etric Identities — y j. 2 Simplify and Rewrite Trigonometric Expressions To sim p lify a trigo n o m etric exp ressio n , sta rt b y rew ritin g it in term s o f o n e trig o n o m etric fu n ctio n or in term s o f sine and cosin e only. Simplify by Rewriting Using Only Sine and Cosine Simplify csc 0 sec 9 — cot 9. Solve Algebraically csc 9 sec 9 — co t 9 = 1 s in cos 1 8 cos 9 s in cos 1 8 s in 8c o s 8 s in s in 0 c o s 9 9c o s ( Rewrite fractions using a common denominator. Subtract. s in TechnologyTip Reciprocal and Quotient Identities. Multiply. 1 —cos2 0 8c o s 9 sin2 9 s in 9 c o s 9 s in 8 o r tan cos 8 Rewrite in term s of sine and cosine using 9 9 cos2 1 s in 8 8 Pythagorean Identity Divide the numerator and denominator by sin 9. 9 Support Graphically T h e graphs o f y = csc 9 sec 9 — cot 9 and y = tan 9 appear to be identical. Graphing Reciprocal Functions When using a calculator to graph a reciprocal function, such as y = csc x, you can enter the reciprocal of the function. P lo ti M o t£ Plots \Y iB l/s in < X > \V i= \V j = \V h= [ —2tv, 2tv] scl: f \V e= Guided Practice sVb= nV?= 4. by [ - 2 , 2] scl: 0.5 S im p lify sec x — tan x sin x. S om e trig o n o m etric exp ressio n s can b e sim p lified b y ap p ly in g id en tities an d factorin g. W W H simolifv by Factoring Simplify sin 2 x cos x — sin ( y — *)• Solve Algebraically Cofunction Identity s in 2 x cos x — s in ( Y ~ x ) = s*n2 x cos * ~ cos x WatchOut! Graphing While the graphical approach shown in Examples 4 and 5 can lend support to the equality of two expressions, it cannot be used to prove that two expressions are equal. It is impossible to show that the graphs are identical over their entire domain using only the portion of the graph shown on your calculator. Support Graphically = —co s x ( —s in 2 x + 1) Factor —cos / from each term. = —cos x (1 — sin2 x) Commutative Property = —cos x (cos2 jc) or —cos3 x Pythagorean Identity T h e grap h s b elo w ap p ear to b e id en tical. y = sin2 x c o s x - s i n | y - x j zz: A . . A ■ / A , s ----------------h | y = - c o s 3x [—2 ir, 2 tt] scl: f by [ - 2 , 2] scl: 0.5 [—2 ir, 2 tt] scl: y by [ - 2 , 2] scl: 0.5 ^ Guided Practice 5. Sim p lify —c sc ^ y — x) — ta n 2 x sec ; lconnectED .m cgraw -hill.com I 315 You can simplify some trigonom etric expressions by com bining fractions. S M 3 W Simplify s in 1 x cos — s in Simplify by Combining Fractions I sin x cos x 1 —sin x x + s in 1 x _ 1 x sin a cos x + x cos x s in x cos (1 (c o s x) (1 — s in x ) ( c o s x ) s in cos x + s in x ) ( l — s in x ) (c o s 1 —sin2 x x —s in x c o s x x cos x x cos x —s i n x cos2 x x — s in x c o s x cos x — s in x cos Multiply. cos s in cos Common denominator x)(\ —s i n x) cos x cos2 x x cos x Pythagorean Identity — s in Subtract. x (c o s 2 x )( s in x — 1 ) Factor the numerator and denominator. ( — c o s x )(s in x — 1 ) Divide out common factors. = —cos x f GuidedPractice Simplify each expression. cos X 6A. 1 + s in x + 1 + s in x 6B. cos x CSC X 1 + sec x ■4 - CSC X 1 - sec x In calcu lu s, you w ill som etim es n eed to rew rite a trig o n o m etric exp ressio n so it does n o t in v o lv e a fraction . W h e n the d en o m in ato r is o f the fo rm 1 ± u or u ± 1, y o u can som etim es do so b y m u ltip lyin g the n u m erato r and d en o m in ato r b y the con ju g ate o f the d en o m in ato r and ap p lyin g a P yth ag o rean identity. Review Vocabulary BT# conjugate a binomial factor which when multiplied by the original binomial factor has a product that is the difference of two squares (Lesson 0-3) R ew rite 1 + cos x 1 Rewrite to Eliminate as an expression that does not involve a fraction. 1 1 + cos X 1 — 1 + cos X _ Fractions cosX 1 — cos X 1 — COS X Multiply numerator and denominator by the conjugate of 1 + cos x, which is 1 — cos x. Multiply. 1 — COS2 X _ 1 — COS X Pythagorean Identity s in 2 x 1 COS X 1 s in = x Write as the difference of two fractions. COS X 1 s in x s in x C SC2 X — c o t XCSC X Factor. Reciprocal and Quotient Identities ► GuidedPractice Rewrite as an expression that does not involve a fraction. 7A. 316 Lesson 5-1 c o s '- x 1 — s in x Trigo n o m etric Identities 70 7B. 4 sec x + tan x m Exercises = Step-by-Step Solutions begin on page R29. Find the value of each expression using the given information. (Example 1) 1. If cot I 5 Simplify each expression. (Example 6) 32. cos X sec x + l gg 1 — cos x tan x 34. 1 ■+ ■ 1 sec x + l sec x — 1 35. cos x cot x , sin x sec x + tan x sec x —tan x -, find tan 0. 2. If cos x = —, find sec x. 3. If tan a = 5 find cot a . 4. If sin /3 = —§> find csc /3. 6 5. If cos x = -r and sin x = 6 6. If sec ip = 2 and tan 6 30 fin d cot x. 7. If csc Q; = — and co t a = fin d sec s in csc = V 3 , find sin <p. x x+ . COS X sec x — 1 sin x 1 + cos x _ j_ l s in csc x x —1 ( 3 7 | SUNGLASSES M an y su n g lasses are m ad e w ith polarized len ses, w h ich red u ce the in te n sity o f light. T he intensity o f lig h t em erg in g fro m a sy ste m o f tw o p o larizin g lenses a. 8. If sec 0 = 8 and tan 9 = 3 V 7 , find csc 9. I can b e calcu lated b y I = /0 ------ —— , w here I0 is the csc2 Axis 1 9. sec 0 and cos 0; tan 0 == —5, cos 0 > 0 10. cot 0 and sec 0; sin 0 ==j , tan 0 < 0 Si* Unpolarized 11. tan 0 and sin 0; sec 0 == 4, sin 0 > 0 li®ht from a sy ste m o f tw o p o larized lenses. b. If a p air o f su n g lasses co n tain s a system o f tw o p o larizin g len ses w ith ax e s at 30° to one another, w h a t p ro p o rtio n o f th e in ten sity o f lig h t enterin g the su n g lasses em erg es? 15. cot 0 and sin 0; sec 0 == - j , sin 0 > 0 16. tan 0 and csc 0; cos 0 == - j , sin 0 < 0 Find the value of each expression using the given information. (Example 3) —y j . 19. If tan 9 = - 1 .5 2 , find co t ( f l - f ) . 20. 21 . If sin 6= 0.18, find cos (0 - f ). If cot x = 1.35, find tan |x —y j . 22. csc x sec x — tan x 23. csc x — co s x co t x 24. sec x co t x — sin x 25. 1 — sin2 x CSC 28. x —1 sec x csc x —tan x sec x csc x 30. cot x — csc2 x co t x 27. 29. sin x cot X 39. CSC X — 40. cot X sec x — tanx 42. 3 tan x 1 - cos x 44. sin x 1 — sec x 46. Simplify each expression. (Examples 4 and 5) 26. Rewrite as an expression that does not involve a fraction. (Example 7) 38. If csc 9 =—1.24, find sec ^0 — If cos x = 0.61, find sin Lens 2 a. S im p lify th e fo rm u la fo r th e in te n sity o f lig ht em ergin g 14. sin 0 and cos 0; cot 0 == 8, csc 0 < 0 18. Lens1 sin 0 < 0 13. cos 0 and tan 0; csc 0 == -|, tan 0 >•0 17. 9 in ten sity o f lig h t en terin g the sy ste m o f lenses and 0 is the an g le o f the axis o f th e seco n d lens in relatio n to that o f the first len s. (Example 6) Find the value of each expression using the given information. (Example 2) 12. sin 0 and cot 0; cos 0 == -v 41. 43. 45. 5 47. sec x + l CSC X 1 — s in cot 1 + s in 2 cot x x x s in x+ x csc c o t " x cos csc x x x —1 s in x ta n x cos x + l Determine w hether each parent trigonom etric function shown is odd or even. Explain your reasoning. tan x + sin x sec x csc x tan x H i J csc x cos x + cot x I sec x cot x n cot2 x + l 31 . co t x — cos3 X n CSC X connectED.mcgraw-hill.com 1 & 317 y n 50. SOCCER W h en a soccer b all is kick ed from the grou nd , its 61. zEU y- 2v02 cos2 0 LIGHT WAVES W h e n lig h t sh in es thro u gh tw o n arro w slits, a series o f lig h t an d d ark frin g es appear. T h e an g le 9, in h eigh t y and h o rizo n tal d isp lacem en t x are related b y rad ian s, lo catin g the m th frin g e can b e calcu lated b y + T s'n , w here i>0 is the in itial v elo city of cos 6 sin 9 = the b all, 9 is the angle at w h ich it w as k ick ed , and g is the acceleration d u e to gravity. R ew rite this eq u atio n so th at a w h ere d is the d istan ce b etw een the tw o slits, and A is the w av e le n g th o f light. tan 9 is the o n ly trigo n o m etric fu n ctio n th at ap p ears in the equation. Write each expression in terms of a single trigonometric function. 51 . tan x — csc x sec x 52. cos x + tan x sin x 53. csc x tan 2 x — s e c 2 x csc x a. R ew rite the fo rm u la in term s o f csc 9. 54. sec x csc x — cos x csc x 55. b. D eterm in e the an g le lo catin g the 100th fring e w h en lig h t h a v in g a w av e le n g th o f 550 n an om eters is shined th rou g h d o u b le slits sp aced 0.5 m illim eters apart. * MULTIPLE REPRESENTATIONS In this p ro blem , y o u w ill in v estig ate the v erificatio n o f trig o n om etric id entities. C o n sid er the fu n ctions show n. i. i/j = tan x + 1 H.O.T. Problems y 2 = sec x cos x — sin x sec x ii. y3 = tan x sec x — sin x y4 = sin x ta n 2 x 62. g rap h in g the functions. —2tt —IT PROOF P rove th at the area o f the trian g le is A = a. TABULAR C o p y and com p lete the table below , w ith ou t Use Higher-Order Thinking Skills s(s — a)(s — b)(s — c) w h ere s = j ( a + b + c). (Hint: T h e area o f an o bliqu e trian g le is A = j be sin A.) 0 7T 27T B 63. ERROR ANALYSIS Je n e lle and C h lo e are sim p lify in g b. GRAPHICAL G rap h e ach fu n ctio n on a grap h in g — \ — sin x —. Je n e lle th in k s th at the exp ression sin2 x — cos x calculator. c. VERBAL M ake a con jectu re ab ou t the relatio n sh ip betw een y 1 and y 2. R ep eat fo r y 3 and y 4. sim p lifies to c sc 2 x — ta n 2 x. Is eith er o f th em correct? E xp lain y o u r reasoning. d. ANALYTICAL A re the con jectu res th at you m ad e in p art c valid fo r the entire d o m ain o f e ach fun ctio n ? E xplain y ou r reasoning. Rewrite each expression as a single logarithm and simplify the answer. 56. In |sin x\ — In |cos x\ 2 sim p lifies to — cos , and C h lo e th in k s th at it 1 —2 cos x CHALLENGE Write each of the basic trigonometric functions in terms of the following functions. 64. sin x 65. cos x 66. tan x REASONING Determ ine w hether each statem ent is true or 57. In |sec x\ — In |cos x\ fa ls e . Explain your reasoning. 58. In (co t2 x + 1) + In |sec x\ 67. c sc 2 x tan x = csc x sec x is tru e fo r all real nu m bers. 59. In (sec2 x — ta n 2 x ) — In (1 — c o s2 x ) 68. T h e o d d -ev en id en tities can b e u sed to p ro v e th at the g rap h s o f y = co s x and y = sec x are sy m m etric w ith resp ect to th e y-axis. 60. ELECTRICITY A cu rren t in a w ire in a m ag n etic field cau ses a force to act o n the w ire. T h e stren gth o f the m ag n etic field can be d eterm in ed u sin g the fo rm u la B = — —— , w here F is the force on the w ire, I is the cu rren t in the w ire, £ is the len gth o f the w ire, and 9 is the angle the w ire m akes w ith the m ag n etic field. S o m e p h y sics b o o k s give the form u la as F = I£B sin 9. S h ow th at the tw o fo rm u las are equivalent. 318 Lesson 5-1 Trigo no m etric Identities PROOF Prove each Pythagorean identity. (69| ta n 2 9 + 1 = s e c 2 9 71. 70. c o t2 9 + 1 = c sc 2 6 PREWRITE U se a ch art or a tab le to help y o u org an ize the m ajo r trig o n o m etric id en tities fo u n d in L esso n 5-1. Spiral Review Solve each triangle. Round to the nearest tenth, if necessary. (Lesson 4-7) 73. 75. 76. Find the exact value of each expression, if it exists. (Lesson 4-6) H) 78. cot (s in -1 79. tan (arctan 3) 80. cos 81 . cos — — cos 82. cos 1 (s in " 1 f ) 83. sin (c o s -1 84. ANTHROPOLOGY Allometry is the stu d y o f the relation sh ip b e tw e e n the size Growth of the Average American M ale (0 -3 years of age) o f an org an ism and the size o f any o f its p arts. A research er d ecid ed to test for an allom etry b e tw e e n the size o f the h u m an h ead com p ared to the Head Circum ference (in.) h u m an b o d y as a p erso n ages. T h e data in the table rep resen t the av erage A m erican m ale. (Lesson 3-5) a. Find a qu ad ratic m o d el relatin g th ese d ata b y lin earizin g the d ata and find ing the lin ear reg ressio n equ ation. b. U se the m od el fo r the lin earized d ata to find a m o d el fo r the o rigin al data. C. U se y o u r m o d el to p red ict the h e ig h t o f an A m e rica n m ale w h o se h ead circu m ference is 24 inches. Let U = { 0 ,1 , 2, 3, 4, 5 }, A = {6 , 9 }, B = {6 , 9 ,1 0 } , C = { 0 ,1 , 6, 9 ,1 1 }, D = {2 , 5 ,1 1 }. Determine whether each statem ent is true or fa l s e . Explain your reasoning. (Lesson 0-1) 85. A C i Height (in.) 14.1 19.5 18.0 26.4 18.3 29.7 18.7 32.3 19.1 34.4 19.4 36.2 19.6 37.7 Source: National Center for Health Statistics 86. D C U Skills Review for Standardized Tests 87. SAT/ACT If x > 0, then *2 _ 1 (X + x+ l l ) 2 x+ 2 1 89. W h ich o f the fo llo w in g is e q u iv ale n t to 1 - sin2 ( (* + 2)2 x + 3 1 — cos2 t •tan 91 A (x + l )2 A tan 0 C B (x — l )2 B D cos i co t 9 sin t C 3 x -l 90. REVIEW R efer to the figure. If cos D = 0.8, w h a t is the D 3x len g th o f D F? E 3(x — l )2 88. REVIEW If sin x = m and 0 < x < 90°, then tan x : H 1 - m2 J F 5 G 4 «V1 - m 2 H 3.2 1 —m2 J I F 4 m D 1 — m2 connectED.m cgraw-hill.com 3 & 3 1 9 y i • You simplified trigonometric expressions. (Lesson 5-1) Verify trigonometric identities. • Determine whether equations are identities. Two firew orks travel at the same speed v. The fireworks technician wants to explode one firew ork higher than another by adjusting the angle 0 of the path each rocket makes with the ground. To calculate the maximum height h of each rocket, the form ula h = v tan„ e could 2 g sec2 9 be used, but would h - v sin e give the same 2g result? NewVocabulary verify an identity Verify Trigonometric Identities 1 In L esso n 5-1, you u sed trig o n o m etric id en tities to rew rite exp ressio n s in eq u iv alen t and so m etim es m ore u sefu l form s. O n ce v e rifie d , th ese new id en tities can also b e u sed to solv e p ro b lem s o r to rew rite o th er trig o n o m etric expression s. To verify an identity m ean s to prove th at b o th sid es o f the e q u atio n are equ al fo r all v alu es o f the v ariable for w h ich b o th sid es are d efined . T h is is d on e b y tran sfo rm in g the exp ressio n o n one sid e o f the id en tity in to the exp ressio n on the oth er sid e th ro u g h a seq u en ce o f in term ed iate exp ression s that are each eq u iv alen t to the first. A s w ith o th er ty p es o f p ro o fs, e ach step is ju stifie d b y a reason, u su ally an o th er v erified trig o n o m etric id e n tity o r an alg eb raic op eratio n . You w ill find th at it is often easier to start the v erification o f a trigo nom etric identity b y b eg in n in g on the side w ith the m ore com plicated exp ression and w ork in g tow ard the less com plicated expression. Verify a Trigonometric Identity Verify that csc * — - = cos2 x. csc x The left-h an d sid e o f this id en tity is m ore com p licated , so start w ith th at exp ressio n first. = “ 4 ^ Pythagorean Identity = c o t2 x s in 2 x Reciprocal Identity s if te r Quotient Identity Pythagorean Identity N o tice th at the v erificatio n end s w ith the e xp ressio n on the o th er sid e o f the identity. ► GuidedPractice V erify each identity. 1A. s e c 2 9 c o t2 9 — 1 = c o t2 9 1B. tan a = sec a csc a tan a — 1 Th ere is u su ally m o re th an one w ay to v e rify a n identity. F o r e xam p le, the id e n tity in E xam p le 1 can also b e v erified as follow s. csc x —1 Write as the difference of two fractions. sin2 x Simplify and apply a Reciprocal Identity, Pythagorean Identity 320 I Lesson 5-2 W h en there are m u ltip le fractio n s w ith d ifferen t d en o m in ato rs in an exp ressio n , y ou can find a com m on d en o m in ato r to red u ce the e xp ressio n to o n e fraction . J j J E E E E Verify a Trigonometric Identity by Combining Fractions Verify that 2 csc x = — csc x + cot x -\--------------------------- . csc x — cot x The rig h t-h an d sid e o f th is id en tity is m o re co m p licated , so start th ere, rew ritin g e ach fraction u sin g the co m m o n d en o m in ato r (csc x + cot x)(csc x — co t x). 1_______ ,_______ 1 csc x + cot x Start with the right-hand side of the identity. csc x — cot x csc x — cot x , (csc x + cot x)(csc x — cot x) (csc X csc x + cot x + cot x)(csc X — cot x) Common denominator __________ 2 csc x__________ (csc x + cot x)(csc X — cot x) Add. 2 csc x csc x — c o t- x Multiply. Pythagorean Identity = 2 csc x ✓ ►GuidedPractice 2. V erify th at - - ^ S.Q + J 1 + sin a 1 cos a 0 — = 2 sec a . To elim in ate a fractio n in w h ich the d en o m in ato r is o f the form 1 ± u or u + 1, rem em ber to try m u ltip ly in g the n u m erato r and d en o m in ato r b y the co n ju g ate o f the d en om inator. T h en yo u can p o ten tially ap p ly a P y th ag o rean Id e n tity ■ S E H U p e r i f y a Trigonometric Identity by Multiplying StudyTip Alternate Method You do not always have to start with the more complicated side of the equation. If you start with the right-hand side in Example 3, you can still prove the identity. csc a + cot a Verify that - - s‘n Q— = csc a + cot a . 1 —cos a B ecau se the left-h an d sid e o f th is id e n tity in v o lv es a fractio n , it is slig h tly m ore com p licated th an the rig h t side. So, start w ith the left side. sin a _ 1 — cos a 1 + cos a _ 1 + cos a sin 1 — cos2 a a sin a (1 + cos a) 1 - cos a sin a sin a 1 - cos a 1 + cos a 1 + cos a sin a (1 + cos a) _ 1 + cos a — sin a sin a _ sin a 1 — cos a 1 - cos sin2 a a , _ 1 + cos a sin a sin a sin a = csc a + co t a V Multiply numerator and denominator by the conjugate of 1 — cos a , which is 1 + cos a . Multiply. Pythagorean Identity Divide out the common factor of sin a . Write as the sum of two fractions. Reciprocal and Quotient identities ^ GuidedPractice 3. tan x V erify that = csc x - cot x. sec x + 1 U n til an id e n tity h as b e e n v erified , yo u can n o t assu m e th at b o th sid es o f the eq u atio n are equal, th e re fo re , y o u can n o t u se the p ro p erties o f eq u ality to p erfo rm alg eb raic o p eratio n s on each side o f an identity, su ch as ad d in g the sam e q u an tity to e ach sid e o f the equ ation. c o n n e c tE tU iic g r a ^ iilU o m 1 321 W hen the more com plicated expression in an identity involves powers, try factoring. B f T f T H T T i Verify a Trigonometric Identity by Factoring Verify that cot 0 sec 6 csc2 6 —cot3 6 sec 6 = csc 0. co t 9 sec 9 c sc 2 9 — c o t3 9 sec 9 Start with the left-hand side of the identity. = co t 9 sec 9 (csc 2 9 — c o t2 9) Factor. = co t 9 sec 9 Pythagorean identity cos 9 sin 9 1 cos i Reciprocal and Quotient Identities 1 sin 8 Multiply. Reciprocal Identity = csc 9 y/ p GuidedPractice 4. V erify th at s in 2 x ta n 2 x c sc 2 x + c o s 2 x ta n 2 x c sc2 x = s e c 2 x. It is som etim es h elp fu l to w o rk e ach sid e o f an id e n tity sep arately to o b tain a co m m o n in term ed iate expression . StudyTip Additional Steps When verifying an identity, the number of steps that are needed to justify the verification may be obvious. However, if it is unclear, it is usually safer to include too many steps, rather than too few. Verify an Identity by Working Each Side Separately > Verify that — + sec x B o th sid es lo ok com p licated , b u t the left-h an d sid e is slig h tly m ore co m p licated sin ce its d en o m in ato r in v o lv es tw o term s. So, start w ith the exp ressio n on the left. / ta n 2 * - sec2 * - 1 1 + sec x 1 + sec x Pythagorean Identity (sec x — l)(s e c x + l ) 1 + sec x Factor. Divide out common factor of sec x + 1. = sec x — 1 From h ere, it is u n clear h o w to tran sfo rm sec x — 1 into 1 COs°^'V>so start w ith the righ t-h an d sid e and w o rk to tran sfo rm it into th e in term ed iate fo rm sec x — 1. 1 ~coscos i — x * = — cos x - !!§ cos!■x Write as the difference of two fractions, Use the Quotient Identity and simplify. = sec x — 1 To com p lete the proof, w o rk b ack w ard to co n n e ct th e tw o p arts o f the proof. ta n 2 x _ sec2 x — 1 1 + sec x 1 + sec x Pythagorean Identity (sec x — l)(s e c x + l ) 1 + sec x Factor. = sec x — 1 Divide out common factor of sec x + 1. = cosx “ f § f f Usethe Quotient ^ < 1% and write 1 a s |3 § £ 1 — cos X , COS X Combine fractions. p GuidedPractice 5. 322 V erify th at sec4 x — se c 2 x = ta n 4 x + ta n 2 x. | Lesson 5-2 | V e rifyin g Trigo n o m etric Identities ConceptSum m ary Strategies for Verifying Trigonometric Identities • Start with the more complicated side of the identity and work to transform it into the simpler side, keeping the other side of the identity in mind as your goal. • Use reciprocal, quotient, Pythagorean, and other basic trigonometric identities. • Use algebraic operations such as combining fractions, rewriting fractions as sums or differences, multiplying expressions, or factoring expressions. • Convert a denominator of the form 1 ± u or u + 1 to a single term using its conjugate and a Pythagorean Identity. • Work each side separately to reach a common intermediate expression. • If no other strategy presents itself, try converting the entire expression to one involving only sines and cosines. v __________ ___ __________________________ _____ __________________________ ____________________________________ ______________ Identifying Identities and Nonidentities You can u se a g rap h in g calcu lator to in v estigate w h eth er an eq u atio n m ay b e an id en tity b y g rap h in g the fu n ctions related to each sid e o f the equation. 2 WatchOut! Using a Graph You can use a graphing calculator to help confirm a nonidentity, but you cannot use a graphing calculator to prove that an equation is an identity. You must provide algebraic verification of an identity. K > I B E Determine Whether an Equation is an Identity Use a graphing calculator to test w hether each equation is an identity. If it appears to be an identity, verify it. If not, find a value for which both sides are defined but not equal. a. cos 0 + 1 _ c o sff sec 0 + 1 ta n 2 0 T h e grap h s o f th e related fu n ctio n s d o n o t coin cid e fo r all v alu es o f x fo r w h ich the b oth fu n ctio n s are d efined . W h e n x = Yx = 1.7 b u t Yz ~ 0.3. T h e eq u atio n is n o t an identity. [ —2-jt, 2 it ] scl: |j COS 0 + 1 _ ta n 2 0 tt by [ - 1 , 3 ] scl: 1 [ — 2 t t , 2 tt] scl: it by [ - 1 , 3 ] scl: 1 COS 0 sec 0 — 1 T h e eq u atio n appears to b e an id e n tity b e cau se the g rap h s of the related fu n ctio n s co in cid e. V erify th is algebraically. cos 0 _ cos 0 sec 0 — 1 sec 0 + 1 sec 0 — 1 sec 0 + 1 cos 0 sec 0 + cos 0 sec2 0 —1 cos 0\(— \cos 0 ]\ 1 + cos 0 sec 0 —1 i tan 0 conjugate of sec 0 —1. Multiply. + cos 0 sec 2 0 —1 cos 0 + 1 Multiply numerator and denominator by the v Reciprocal Identity Simplify. [—2-rr, 2-it] scl: -k by [ - 1 , 3] scl: 1 Commutative Property and Pythagorean Identity Guided Practice CB ,, cot 8 tan2 8 + cot ( 6A. csc 0 = - sec 9 6B. cos X + 1 sec x — 1 $ connectED.mcgraw-hill.com 1 BP ! 323 Exercises = Step-by-Step Solutions begin on page R29. Verify each identity. (Examples 1 -3 ) Verify each identity. (Examples 4 and 5) 1. (sec2 9 — 1) c o s2 9 = sin 2 9 \2o] (csc 9 + cot 0)(1 — cos 9) = sin 9 2 . ' s e c 2 9( 1 — c o s 2 9) = tan 2 9 . 2 < sin 2 9 ta n 2 9 = ta n 2 9 - s in 2 9 3. sin r^2 1 — tan2 9 _ cos2 9 —1 1 —cot2 1 cos 9 ^ 1 + csc 6 ■sin 9 c o s2 9 = s in 3 1 (ffl, 4. csc 9 — cos 9 co t 9 = sin 9 5. c o t2 9 csc 2 9 — c o t2 9 = co t4 { 6. tan 9 csc 2 9 — tan 9 = co t 9 7." sec e sin 9 (a 9. lo . 11. 12. v j sin 9 _ CQt g cos i sin 9 1 — cos , 1 — cos 9 H :— ; — sin 9 1 + — cot l 1 1 — tan 1 -+ 1 — tan2 9 1 oc 7 1 + tan2 9 1 la . < = 1 — tan2 9 1 2 cos2 I 2 CSC 2 26. ta n 2 9 cos cos2 9 = 1 —cos2 _ 2 f . , sec 9 — cos 9 = tan 9 sin 9 - = sin 9 + cos 9 -= 1 1 — cot2 o ■H-------- \— - = 2 s e c 2 9 sin 9 csc 9 + 1 2 _ 1 — cos 9 + cos 9 24. (csc 9 - cot 9 )2 = j r, cos 9 + tan 9 — sec i 1 + sin 9 , = co s 9 + co t 9 28. 1 — ta n 4 9 = 2 se c 2 9 — sec 4 9 29. (csc 9 - co t 9 )2 = } •••cos I j/' 30. 1 + tan 9 sin 9 + cos 9 w 2 + csc 9 sec 6 = (sin 9 + cos 9 )2 csc 9 sec csc 9 — 1 13. (csc 9 — cot #)(csc 9 + co t 9) = 1 1 + cos 9 1 : sec 9 14. co s4 9 — sin 4 9 = c o s 2 9 — s in 2 9 15. ljg\ 1 . r+ ■ 1 — sin 9 1 1 + sin i cos 9 , cos 9 1 + sin 9 1 — sin l - 2 se c 2 i 32. OPTICS If tw o p rism s o f the sam e p o w er are p laced n ex t to each other, th eir total p o w er can be d eterm in ed u sing z = 2 p cos 9, w h ere 2 is the com b in ed p o w er o f the =2 sec 9 p rism s, p is the p o w er o f the in d iv id u al p rism s, and 9 is th e an g le b e tw e e n the tw o p rism s. V erify th at 2 p co s 9 = 2 p (l — s in 2 9) sec 9. (Example 4) 17. csc4 9 — co t4 9 = 2 c o t2 9 + 1 18. csc2 9 + 2 csc 9 —3 _ csc 9 + 3 CSC2 9 -1 csc 9 + 1 33. PHOTOGRAPHY T h e a m o u n t o f lig h t p assin g th ro u g h a level, the m axim u m h eig h t th at it reach es is giv en by p o larizatio n filter can b e m o d eled u sin g I = I m c o s2 9, w h ere I is the am o u n t o f lig h t p assin g th rou gh the filter, I m is the am o u n t o f lig h t shined on the filter, and 9 is the an g le o f rotation b etw een the lig h t sou rce and the h = 0 ^ filter. V erify th at (l9 ) FIREWORKS If a rocket is lau nched from ground ®, w h ere 9 is the an g le b etw een the ground and the in itial p ath o f the rocket, v is the ro ck e t's in itial speed , and g is the acceleration d u e to gravity, 9.8 m eters p er second squared. (Example 3) cos2 9 = l m tan x + l _ 1 + cot x tan x — 1 1 —cot x 35. sec x + tan x 1 sec x —tan x 36. sec x — 2 sec x tan x + tan x 37. b. 324 v2 tan2 9 2g sec2 9 Su p p ose a second ro cket is fired at an angle o f 80° from the ground w ith an in itial sp eed o f 110 m eters p er second . Find the m ax im u m h eig h t o f the rocket. | Lesson 5-2 | V e rifyin g T rigo n o m etric Identities (Example4) GRAPHING CALCULATOR Test w hether each equation is an identity by graphing. If itappears to be an identity, verify it. If not, find a value for which both sides are defined but not equal. (Example 6) 2^ v2 sin2 9 a. Verify that 2g ^ ------. cor 9+ 1 38. on cot2 j — 1 1 + cot2 X = 1 — 2 s in 2 x tan x —sec x tan x - tan x + sec x 2 -2 39. cos x — sin x = cot x —tan x — tan x + cot x 1 — cos x 1 + cos x Verify each identity. 40. GRAPHING CALCULATOR Graph each side of each equation. If the equation appears to be an identity, verify it algebraically. /sin x tan x I = |sin x\ sec x sec x —1 sec z + 1 41. sec x 55. cos X I sec x — 1 I I tan x I tan x sec x CSC X 1 56. sec x — cos.2 x. csc x = tan x sec x 42. In |csc x + co t x\ + In |csc x — co t x\ = 0 57. (tan x + sec x )( l — sin x) = cos x 43. In |cot x\ + In |tan x cos x\ = In |cos x| 58. Verify each identity. 44. s e c 2 9 + ta n 2 9 = sec4 45. —2 co s2 9 = sin 4 tan * 6 sec x cos x 1 cot2 x tan2 x — sin2 x tan2 x = -1 59. i § l MULTIPLE REPRESENTATIONS In this p ro blem , you w ill in v estig ate m eth od s u sed to so lv e trigo n o m etric equ atio n s. C o n sid er 1 = 2 sin x. c o s4 9 — 1 46. sec2 9 sin 2 9 = s e c 4 — (tan 44 9, + sec 2 9) a. NUMERICAL Iso late the trig o n o m etric fu n ctio n in the eq u atio n so th at sin x is the o n ly exp ression o n one 47. 3 sec2 9 ta n 2 9 + 1 = s e c 6 0 — ta n 6 9 sid e o f the equ atio n . 48. s ec4 x = 1 + 2 ta n 2 x + ta n 4 x b. GRAPHICAL G rap h the left and rig h t sid es o f the eq u atio n y o u fo u n d in p a rt a on the sam e grap h over 49. sec2 x csc2 x = sec2 x + csc 2 x [0, 27t). L ocate an y p o in ts o f in tersection and express the v alu es in term s o f rad ian s. 50. ENVIRONMENT A b io lo g ist stu d y in g p o llu tio n situ ates a net across a riv er and positio n s in stru m en ts at tw o d ifferent statio n s on the riv er b an k to collect sam ples. In the d iag ram sh ow n , d is the d istan ce b e tw e e n the c. GEOMETRIC U se the u n it circle to v erify the answ ers stations and w is w id th o f the river. d. GRAPHICAL G rap h the le ft and rig h t sid es o f the e q u atio n yo u fou n d in p a rt a o n the sam e grap h over —27T < x < 277. L o cate an y p o in ts o f in tersection and y o u fo u n d in p a rt b . exp ress the v alu es in term s o f rad ians. e. VERBAL M ake a co n jectu re as to the solu tio n s of 1 = 2 sin x. E xp lain y o u r reasoning. a. H.O.T. Problems D eterm in e an eq u atio n in term s o f tan g en t a th at can b e used to find the d istan ce b e tw e e n the stations. b. Verify th at d ■ 60. w cos (90° — a) Use Higher-Order Thinking Skills REASONING C an su b stitu tio n b e u sed to d eterm in e w h eth er an eq u atio n is an id en tity ? E xp lain you r reasoning. C. C om p lete the table sh o w n for d = 40 feet. w 20 40 60 80 100 120 a d. If a > 60° or a < 20°, the in stru m en ts w ill n o t fu n ctio n properly. U se the table from p a rt c to d eterm in e w h eth er sites in w h ich the w id th o f the riv er is 5, 35, 61) CHALLENGE V erify th at the area A o f a trian g le is g iv en b y a 2 sin /3 sin 7 2 sin (f3 + 7) w h ere a, b, and c rep resen t the sid es o f the trian g le and a , (3, and 7 are the resp ectiv e o p p o site angles. 62. WRITING IN MATH U se the p ro p erties o f lo g arith m s to or 140 feet cou ld b e u sed fo r the exp erim en t. e xp lain w h y the su m o f th e n a tu ra l lo garith m s o f the six b asic trig o n o m etric fu n ctio n s for any angle 9 is 0. HYPERBOLIC FUNCTIONS The hyperbolic trigonom etric functions are defined in the following ways. sin h x = i ( e * - e - * ) csch x = —7—-— , x ± 0 s in h x cosh x = ~ (ex + e x) sech x = tanh x - s in h co sh x x coth c o s h ,Y 1 X — ; ta n h x , X ± 0 63. OPEN ENDED C reate id e n tities for sec x and csc x in term s o f tw o o r m o re o f the oth er b asic trig o n o m etric functions. 64. REASONING If tw o an g les a and (3 are com plem entary, is c o s 2 a + c o s 2 f3 = 1? E xp lain y o u r reasoning. Ju stify your answ ers. Verify each identity using the functions shown above. 51. co sh 2 x — sin h 2 x = 1 52. s in h (—x) = —sin h x 53. sech 2 x = 1 — ta n h 2 x 54. c o s h (—x) = cosh x 65. WRITING IN MATH E xp lain h o w you w o u ld v erify a trig o n o m etric id e n tity in w h ich b o th sid es o f the equ ation are eq u ally com plex. 325 Spiral Review Simplify each expression. (Lesson 5-1) 69. 6 8 . sin 9 co t i 67. tan 9 cot I 66. cos 9 csc i cos 9 csc 9 tan 9 70. sin 9 csc 9 cot 9 71. 1 - cos2 9 sin2 9 72. BALLOONING A s a h o t-air b allo o n crosses ov er a straig h t p o rtio n o f in terstate highw ay, its p ilo t eyes tw o co n secu tiv e m ilep o sts on the sam e sid e o f the ballo o n. W h en v ie w in g the m ilep osts, the an g les of d ep ression are 64° and 7°. H ow h ig h is the b a llo o n to the n earest foot? (Lesson 4-7) Locate the vertical asymptotes, and sketch the graph of each function. (Lesson 4-5) 1 y = — tan x 73. 1 75. y = — sec 3x 74. y = csc 2x Write each degree measure in radians as a multiple of it and each radian measure in degrees. (Lesson 4-2) 76. 660° 79. — 4 77. 570° 78. 158° 80. iZlL 6 81. 9 Solve each inequality. (Lesson 2-6) 82. x 2 - 3x - 18 > 0 83. x 2 + 3x - 28 < 85. x 2 + 2x > 24 86. - x 2 - 88. 84. x 2 - 4x < 5 0 87. - x 2 - 6x + 7 < 0 x + 12 > 0 FOOD T he m an ager o f a b ak ery is ran d o m ly ch eck in g slices o f cake prep ared b y em p lo y ees to ensure th at the correct am ou n t o f flav o r is in each slice. E ach 12-o u n ce slice should con tain h a lf ch o colate and h alf v an illa flavo red cream . T h e a m o u n t o f ch o co late b y w h ich each slice varies can b e rep resented b y g(x) = j| x — 12|. D escrib e the tran sfo rm atio n s in the function. T h en grap h the fun ction. (Lesson 1-5) Skills Review for Standardized Tests 89. SAT/ACT a, b, a, b, b, a, b, b, b, a, b, b, b ,b , a , ... 91. REVIEW W h ich o f the fo llo w in g is n o t eq u iv alen t to cos 9 w h e n 0 < 9 < y ? cos 9 If the sequ ence con tin u es in this m an ner, h o w m an y bs are there b etw een the 44th and 4 7 th ap p earan ces of the letter a ? A 91 C 138 B 135 D 182 B 1 - sin2 cos 9 E 230 90. W h ich exp ression can be used to form an id en tity sec 9 + csc I -w h e n tan 8 =/= —1? w it h : 92. F 2 sin 9 1 sin 9 F sin ( H cos2 6 H tan i J csc ( 326 | Lesson 5-2 | V e rifyin g Trigo n o m etric Identities D tan 9 csc i REVIEW W h ich o f the fo llo w in g is eq u iv alen t to sin 9 + cot 9 cos 9 ? 1 + tan f G cos i C co t 9 sin i J sin 9 + cos ( sin2 9 ■§ Solve trigonom etric • I equations using algebraic techniques. 2 Solve trigonom etric equations using basic identities. A baseball leaves a bat at a launch angle 0 and returns to its initial batted height after a distance of d meters. To find the velocity v0 of the ball as it leaves the bat, you can solve the 2Kn2 sin 0 cos 0 trigonom etric equation d = — — — ---------- . ■4 Use Algebraic Techniques to Solve In L esso n 5-2, y o u v erified trig o n om etric equations 1 called id en tities th at are tru e fo r all v alu es o f the v ariab le for w h ich b o th sid es are d efined . In th is lesson w e w ill co n sid er conditional trig o n o m etric e q u atio n s, w h ich m a y be tru e fo r certain v alu es o f the v ariab le b u t false fo r others. C o n sid er the grap h s o f b o th sid es o f the con d itio n al trig o n o m etric eq u atio n cos x = j . X = -52L - 27r 3 1 ^ V V - ---------------------* i \ I / 1 \ v \y ^ x = y= ! r 2 1 II — (Lesson 5-2) • 1 You verified trigonometric identities. i 1 tt / y - T - 2* T h e grap h sh o w s th at cos y Sin ce y= cos x h as 3 i 2 ir , y ^ ! rS\ 2 T r ^ / ' X x= f + 2. ^ h as tw o solu tio n s o n the in terv al [0, a p erio d o f i I JT i 2 tt y = cos x x= x = - 5 f + 2iT I o * • 2 tt), x= -j and x= cos x = \ h as in fin itely m an y solu tio n s o n the in terv al (—oo, oo). A d d itio n al solu tio n s are fo u n d b y ad d in g in teg er m u ltip les o f the p erio d , so w e exp ress all solu tio n s b y w ritin g x= y + 2 « tt and x= + 2 mr, w h ere n is an integer. To solv e a trig o n om etric eq u atio n th at in v o lv es o n ly o n e trig o n o m etric exp ressio n , b e g in b y iso latin g th is exp ression . Solve by Isolating Trigonometric Expressions Solve 2 tan x — V 3 = tan x. Original equation 2 tan x — V 3 = tan x tan x — \ / 3 = 0 Subtract tan xfrom each side to isolate the trigonometric expression, Add V3 to each side. tan x = \/3 The p erio d o f tan g e n t is so you o n ly n eed to find solu tio n s on the in terv al [0, tt, so lu tio n on th is in terv al is ad d in g in teg er m u ltip les x= tt. T h erefore, the g en eral form o f the so lu tio n s is x= p GuidedPractice 1. Solv e 4 sin tt). The only -j. T h e solu tio n s on the in terv al (—oo, oo) are th en foun d b y x = 2 sin x + \[2. —■+ H7t, w h e re n is an integer. p i M i T l f f t Solve by Taking the Square Root of Each Side StudyTip Find Solutions Using the Unit Circle Since sine corresponds to the y-coordinate on the unit circle, you can find the solutions of sin Solve 4 sin2 x + 1 = 4. • 2 3 sm x = — using the unit circle, as shown. 1 V3 \ 2’ 2 'K\ 4it 3 O \ 3 5tv J 2 fl and sin x = — wh e n x = ^ and x = S in ce sin e has a p erio d o f 27T, the solu tio n s o n the in terv al (— oo, oo) h av e the gen eral form x + 2 = y m it , x = ■ 2htt, x = + 2 n7r, an d x = - ^ + 2« ir, w here n is an integer. _V 3\ 1 V3 w h e n x = y an d x = X / r'H 1 2’ Take the square root of each side. O n the in terv al [0, 2ir), sin x = \ 3y f^l \ \ V3 \ 2’ 2 ) 3 Divide each side by 4. sin x = +■ \y(l /2 it / Subtract 1 from each side. 4 sin 2 x = 3 x = ± ^ - on the interval [0 ,2it] I Original equation 4 sin 2 x + 1 = 4 Any angle coterminal with these angles will also be a solution of the equation. ►GuidedPractice 2. S olv e 3 co t2 x + 4 = 7. W h en trig o n om etric fu n ctio n s can n o t be co m b in ed on one sid e o f an equ ation , try facto rin g and ap p lyin g the Z ero P ro d u ct Property. If the eq u atio n h as q u ad ratic form , facto r if p ossible. If n o t p o ssib le, ap p ly the Q u ad ratic Fo rm u la. Solve by Factoring Find all solutions of each equation on the interval [0, 2ir). a. Watch Out! cos x sin x = 3 cos x Original equation cos x sin 9 = 3 co s x Dividing by Trigonometric Factors Do not divide out the cos x in Example 3a. If you were to do this, notice that you might conclude that the equation had no solutions, when in fact, it has two on the interval [0, 2-k). isolate the trigonometric expression. cos x sin x — 3 cos x = 0 Factor, cos x(sin x — 3) = 0 cos x = 0 or sin x — 3 = 0 Zero Product Property sin x = 3 Solve for x on [0 , 2it]. T h e eq u atio n sin x = 3 h as n o so lu tio n sin ce the m ax im u m v alu e the sin e fu n ctio n can attain is 1. T herefore, o n the in terv al [0, 2 7 t ) , the solu tio n s o f the o rigin al eq u atio n TT j 3TV are — and — b. cos4 x + cos2 x — 2 = 0 Original equation c o s4 x + c o s2 x — 2 = 0 Write in quadratic form. (co s2 x) 2 + c o s2 x — 2 = 0 Factor. (cos2 x + 2 )(co s2 x — 1) = 0 c o s2 x + 2 = 0 or Zero Product Property cos x — 1 = 0 c o s 2 x = —2 cos x = + V —2 Solve for cos2 x. c o s2 x = 1 cos x = ± V l o r + 1 Take the square root of each side. The eq u atio n cos x = ± V —2 h as n o real solu tio n s. O n the in terv al [0, 2ir), the eq u atio n cos x = ± 1 has solu tio n s 0 and 7r. i» GuidedPractice 3A. 2 sin x cos x = \ fl cos x V______________________________ 328 | Lesson 5-3 So lving Trigo n o m etric Equations 3B. 4 cos2 x + 2 cos x — 2\/2 cos x = \ fl So m e trig o n o m etric eq u atio n s in v olv e fu n ctio n s o f m u ltip le an g les, su ch as cos 2x = these e q u atio n s, first solv e for the m u ltip le angle. StudyTip R eal-W orld E x a m p le 4 Exact Versus Approximate Solutions When solving trigonometric equations that are not in a real-world context, write your answers using exact values rather than decimal approximations. For example, the general solutions of the equation tan x = 2 should be expressed as x = t a n _1 2 + m vor x = arctan 2 + mr. Tri To solve ic Functions of Multiple Angles BASEBALL A baseball leaves a bat with an initial speed of 30 meters per second and clears a fence 90.5 meters away. The height of the fence is the same height as the initial height of the v02s in 26 batted ball. If the distance the ball traveled is given by d = ------- 9 .0 -, where 9.8 is in meters per second squared, find the interval of possible launch angles of the ball. vg2s in 29 <0i s in 29 9 0 0 s in 29 30 9 0 .5 : 90.5 = 886.9 = 900 sin 29 8 8 6 .9 900 = sin 29 -,n - l 8 8 6 .9 _ Sln ~ 9 0 0 ~ -29 Original formula d = 90.5 and i/0 = 30 Simplify. Multiply each side by 9.8. Divide each side by 900. Definition of inverse sine R ecall from L esso n 4-6 th at the ran ge o f the in v e rse sin e fu n ctio n is restricted to acute an g les o f I in the in terv al [—90°, 90°]. S in ce w e are fin d in g the in v e rse sin e o f 2 9 in stead o f 9, w e n eed to co n sid er an g les in the in terv al [ - 2 ( 9 0 ° ) , 2(90°)] o r [ - 1 8 0 ° , 180°]. U se y o u r calcu lato r to find the acu te an g le and the referen ce an g le relatio n sh ip sin (180° — 9) = sin 9 to find the o btu se angle. „ i n - l 8 8 6 .9 . sin W Definition of inverse sine " 20 80.2° or 99.8° = 29 s in - 1 40.1° or 49.9° = 9 Divide by 2. a 80.2° and sin (180° - 80.2°) = sin 99.8° T h e in terv al is [40.1°, 49.9°]. T h e b a ll w ill clear the fen ce if the an g le is b e tw e e n 40.1° and 49.9°. CHECK S u b stitu te the an g le m easu res in to the o rigin al eq u atio n to co n firm the solu tion. v02s i n 20 9 .8 StudyTip Optimal Angle Ignoring wind resistance and other factors, a baseball will travel the greatest distance when it is hit at a 45° angle. This is because sin 2(45) = 1, which maximizes the distance formula in the example. 3 0 2 s in (2 • 4 0 .1 ° ) 90.5 : 90.5 ~ 90.497 ✓ 9 .8 Original formula 0 = 40.1° or 0 = 49.9° Use a calculator. vn2s in 29 d = ■ 9 .8 n n r ? 3 0 2 s in ( 2 - 4 9 . 9 ° ) 9 0 5 = ---------------- 9£ --------- 90.5 ~ 90.497 ✓ p GuidedPractice > 4. BASEBALL Find the in terv al o f p o ssib le lau n ch an g les requ ired to clear the fence if: A. th e in itial sp eed w as in creased to 35 m eters p er second . B. th e in itial sp eed rem ain ed th e sam e, b u t the d istan ce to the fen ce w as 80 m eters. connectED.m cgraw-hill.com j 329 2 Use Trigonometric Identities to Solve You can u se trig o n o m etric id en tities alo ng w ith algebraic m eth o d s to solv e trig o n o m etric equ atio n s. Example 5 Solve by Rewriting Using a Single Trigonometric Function Find all solutions of 2 cos2 x — sin x — 1 = 0 on the interval [0, 2 ji). 2 (1 — sin2 x ) — 1 ( 2 s in bo and L x x — s in x—1 s in x—1= s in x —1 s in x + — l) ( s in x + s in x — 1 = 0 1 = Pythagorean Identity 0 = 1 ) = Multiply. Simplify. 0 Factor. 0 s in x + or 1 = Zero Product Property 0 sin x = —1 3 tt — x = £ o r^ 6 6 on Original equation 0 = 0 s in x = j Alternate Method An alternate way to check Example 5 is to graph y = 2 cos^ x - sin x - 1, it has zeros ■£, — s in 2 —2 2 — — 2 s in 2 x — 2 StudyTip x cos2 2 2 Solve for sin x. Solve for x on [ 0 ,2 ir]. CHECK T h e grap h s o f Y i = 2 c o s 2 x — sin x and Y 2 = 1 in tersect the interval [0 ,2 it) as shown. at 6 an d 4 ? on the in terv al [0, 2 ttJ as sh o w n v'' 2 6 p Guided Practice [0, 2-rc] scl: f by [ - 2 , 4] scl: 1 [0, 2 ir] scl: f by [ - 2 , 4 ] scl: 1 Find all solutions of each equation on the interval [0, 2 it). 5A. 1 — cos x = 2 s in 2 x 5B. co t2 x c sc 2 x + 2 c sc 2 x — c o t2 x = 2 S o m etim es y o u can o b tain an eq u atio n in o n e trig o n o m etric fu n ctio n b y sq u arin g each sid e, b u t this tech n iq u e m ay p rod u ce extran eo u s solu tions. M ' i S S S Q Solve bV Squaring Find all solutions of csc CSC X — c o t X = CSC X = (csc x) 2 - CS C 2 X : 1 + c o t2 x = = Original equation 1 + (1 Add cot x to each side. cot X + cot 1 + 2 1 + cot cot *= ? “ CHECK — cot x = 1 on the interval [0, 2n]. 1 0 = 2 0 x cot 2 Square each side. x )2 x + cot x+ x c o t2 x Multiply. c o t2 x Pythagorean Identity Subtract 1 + cot2 xfrom each side. Divide each side by 2. x Solve for x on [ 0 ,2-rr]. ¥ x —co t x = 1 It . TT ? h CSC —---------- C O t — = 1 2 2 csc 1 — 0 = 1*''' Original equation CSC 3 ir Substitute. Simplify. 3 3 0 | Lesson 5-3 | So lving Trigo n o m etric Equations 1 - 1 - 0 # IX Guided Practice 6A. sec x + 1 = ta n x . 3 ir ? csc —------ co t — = 1 2 2 Therefore, the o n ly v alid so lu tio n is ^ on the in te rv al [0, 2 tt]. y x — co t x = 1 6B. co s x = sin x — 1 Exercises = Step-by-Step Solutions begin on page R29. Solve each equation for all values of x. (Examples 1 and 2) Find all solutions of each equation on the interval [0, 2 tt]. (Examples 5 and 6) 1 . 5 sin x + 2 = sin x 2. 5 = s e c 2 x + 3 3. 2 = 4 c o s 2 x + 1 4. 4 tan x — 7 = 3 ta n x — 6 5. 9 + c o t2 x = 12 6. 2 — 10 sec x = 4 — 9 sec x 22. sec 7. 3 csc x = 2 csc x + \[2 8. 11 = 3 c sc 2 x + 7 23. ta n 2 10. 9 + sin 2 x = 10 24. csc 12. 7 cos x = 5 cos x + \/3 25. 2 9. 6 ta n 2 x — 2 = 4 11. 7 co t x — V 3 = 4 co t x 21. 1 = c o t2 x= ta n x= 1 x+ 2 — x+ cot cos2 x —4 x csc x+ 1 — sec x x = = x 1 x+ s in x 1 Find all solutions of each equation on [0, 2iv]. (Example 3) 26. 13. sin 4 x + 2 s in 2 x — 3 = 0 27. 3 14. —2 sin x = —sin x cos x 28. c o t2 x csc2 x —c o t 2 x = 29. sec2 x —1 30. sec2 x ta n 2 15. 4 cot x = co t x s in 2 x 16. csc 2 x — csc x + 9 = 11 cos s in x = = s in 3 —3 — x cos + ta n x+3 4 x 9 — V 3 ta n x = sec2 V 3 x —2 t a n 2 x = 3 17. c os3 x + c o s 2 x — cos x = 1 18. 2 sin 2 x = sin x + 1 19) TENNIS A ten n is b all leav es a racq u et and head s tow ard a 31. OPTOMETRY O p to m etrists so m etim es jo in tw o o bliqu e or tilted p rism s to co rrect v isio n . T h e resu ltan t refractiv e po w er P R o f jo in in g tw o o b liq u e p rism s can be calcu lated net 40 feet away. T h e h eig h t o f the n e t is the sam e h eig h t as the in itial h eig h t o f the ten n is ball. (Example 4) b y first reso lv in g e ach p rism in to its h o rizo n tal and v ertical com p o n en ts, P H an d P v . P o s itio n o f B a s e O p h th a lm ic P ris m R e s o lv in g P ris m P o w e r P v = P R sin 9 a. If the b all is h it at 50 feet p er seco n d , n eg lectin g air P H= P R cos 9 Base resistance, use d = -^ v 02 sin 29 to find the in terv al of po ssible an g les o f the b a ll n eed ed to clear th e net. b. Find 9 if the in itial v elo city rem ain ed the sam e b u t the U sin g the e q u atio n s ab o v e, d eterm in e fo r w h at v alu es of 9 P v a n d P H are equ iv alen t. d istance to the n e t w as 50 feet. 20. SKIING In the O ly m p ic aerial skiin g com p etitio n , skiers speed d o w n a slop e th at lau n ch es th em into the air, as show n. T h e m ax im u m h eig h t a sk ier can reach is g iv e n by V r? s i n 2 ft_eak = — 9 2 -------' w “ ere o S *s acceleratio n d u e to g rav ity or 9.8 m eters p er secon d squared . (Example 4) Find all solutions of each equation on the interval [0, 2 tv]. 32. 33. 34. 35. ta n 2 x sec x + co s x = 2 cos x 1 + s in x 1 + sin x + ■ = -4 sin x + cos x , 1 - sin x ta n x C O t x CO S X + H : = CO S X 1 = sec x — 1 ta n x a. If a skier obtain s a h eig h t o f 5 m eters abo v e th e end o f the ram p, w h at w as the s k ie r's in itial speed ? b. U se you r answ er from p a rt a to d eterm in e ho w long it to o k the skier to reach the m ax im u m heigh t •f 1 1 peak — vn Us i n 9 g ' GRAPHING CALCULATOR Solve each equation on the interval [0, 2ir] by graphing. Round to the nearest hundredth. 36. 3 cos 2x = ex + 1 37. sin 38. x 2 = 2 cos x + x 39. x lo g x + 5x cos x = —2 tt x + cos 7rx = 3x ^connectECUricgm^^^^^J 331 8 40. METEOROLOGY T h e av erage d aily tem p eratu re in d egrees F ah ren h eit for a city can be m o d eled b y t= 8.05 cos (~x — Trj 56. REFRACTION W h e n lig h t trav els fro m one tran sp aren t m ed iu m to an o th er it b e n d s o r refracts, as show n. + 66.95, w here x is a fu n ctio n of tim e, x = 1 represents Jan u ary 15, x = 2 represents Febru ary 15, and so on. a. U se a grap h in g calcu lator to estim ate the tem p eratu re o n Jan u ary 31. b. A p p roxim ate the n u m b er o f m on ths th at the av erage d aily tem p eratu re is greater th an 70° th rou g h ou t the entire m onth. c. E stim ate the h ig h e st tem p eratu re o f the y ear and the m on th in w h ich it occurs. lig h t is e xitin g , 9 1 is the an g le o f in cid en ce, and 92 is the an g le o f refractio n . Find the x-intercepts of each graph on the interval [0, 2 ttI. 41. R efractio n is d escrib ed b y n 2 sin 9 1 = n 1 sin 02, w h ere n 1 is the in d ex o f refractio n o f the m ed iu m the lig h t is e n terin g , n2 is the in d ex o f refractio n o f the m ed iu m the a. F in d 92 fo r e ach m aterial sh ow n if the an g le of y = s ir r x c o s x - c o s x M aterial Index of Refraction glass 1.52 ice 1.31 plastic 1.50 water 1.33 in cid en ce is 40° and the in d ex o f refractio n for air is 1 .0 0 . b. If the an g le o f in cid en ce is d ou bled to 80°, w ill the resu ltin g an g les o f refractio n b e tw ice as large as th o se fou n d in p a rt a? H.O.T. Problems Use Higher-Order Thinking Skills 57. ERROR ANALYSIS V ijay and A licia are solv in g ta n 2 x — tan x + V 3 = V 3 tan x. V ijay th in k s th at the so lu tio n s are x = ~ + H7T, x = Find all solutions of each equation on the interval [0, 4 tv]. 45. 4 tan x = 2 sec2 x 46. 2 sin 2 x + 1 = —3 sin x 47. csc x co t2 x = csc x 48. sec x + 5 = 2 sec x + 3 47T 4 4 + W7T, x = ^ + nix, 3 and x = — + « 7r. A licia th in k s th at the so lu tio n s are x = y + 777T an d x = y + nix. Is eith er o f th em correct? E xp lain y o u r reasoning. 49. GEOMETRY C o n sid er the circle below . CHALLENGE Solve each equation on the interval [0, 2 tt]. 58. 16 sin 5 x + 2 sin x = 12 sin 3 x ( 5 9 ) 4 c o s 2 x — 4 s in 2 x c o s 2 x + 3 s in 2 x = 3 a. T he len gth s o f arc AB is giv en b y s = r(29) w here 60. REASONING A re the solu tio n s o f csc x = \ fl and c o t2 x + 1 = 2 e q u iv alen t? If so, v erify y o u r an sw er a lg e b ra ica lly If no t, e xp lain y o u r reasoning. 0 < 9 < 77. W h en s = 18 and AB = 14, the rad iu s is s in -. U se 8’ a grap h in g calcu lator to fin d the m easu re o f 29 in rad ians, OPEN ENDED Write a trigonom etric equation that has each of the following solutions. b. The area o f the shad ed reg io n is g iv en by A = r~(8 - s in 8) . U se a grap h in g calcu lato r to fin d the rad ian m easu re o f 9 if the rad iu s is 5 in ch es and the area is 36 square inches. R o u n d to the n earest hund redth. Solve each inequality on the interval [0, 2 tt). 50. 1 > 2 sin x 52. cos 51. 0 < 2 V3 (!-* )* ■ * V| 54. cos x < — 2 332 | Lesson 5-3 COS X—V2 53. sin ^x — y j < tan x co t x 55. \fl sin x — 1 < 0 So lving Trigo n o m etric Equations 63. WRITING IN MATH E xp lain the d ifference in the tech n iq u es th at are u sed w h e n solv in g eq u atio n s and v erify in g identities. Spiral Review Verify each identity. (Lesson 5-2) 1 + s in s in 8_ 9 9— 1 c o t2 9 csc gc 1 + ta n 1 + cot 8_ 8 s in 9 9 eg ' cos 1 sec 2 , 9 1 csc2 _ 8 Find the value of each expression using the given information. (Lesson 5-1) 67. tan 6} sin 6 = j , tan 8 > 0 68. csc 8, cos 6 = — 69. csc 8 < 0 sec 8; tan 8 = —1, sin 8 < 0 70. POPULATION The p o p u latio n o f a certain sp ecies o f d eer can be m o d eled b y p = 30,000 + 20,000 cos |-^f j, w here p is the p o p u latio n and t is the tim e in years. (Lesson 4-4) a. W h at is the am p litu d e of the fun ctio n ? W h at d o es it rep resent? b. W h at is the p erio d o f the fu n ctio n ? W h a t d oes it rep resent? c. G rap h the fun ction. Given f ( x ) = l x 2 — 5 x + 3 and g(x) = 6 x + 4, find each. (Lesson 1-6) 71. ( f - g ) ( x ) 72. (f.g ) ( x ) 73. (£ )(x ) 74. BUSINESS A sm all b u sin ess o w n er m u st h ire season al Number of Additional Employees Needed Month w orkers as the n eed arises. T h e fo llo w in g list sh o w s the nu m b er o f em p loy ees hired m o n th ly fo r a 5-m o n th p eriod. August September October November December If the m ean o f these d ata is 9, w h at is the p o p u latio n standard d ev iation for these d ata? R ou n d to the n earest U tenth. (Lesson 0-8) Skills Review fo r Standardized Tests 75. SAT/ACT For all p o sitiv e valu es o f m and n, if = 2 , th en x = — 77. W h ich o f the fo llo w in g is not a so lu tio n of 0 = sin 8 + cos 6 ta n 2 91 3 tt A B C D E 4 3 llI 4 3 + 2« 2m 2m - 3 2n C 2 tt D 5 tv 2 2m 3 + 2n 3 2m — 2 n 78. REVIEW T h e g rap h o f y = 2 cos 8 is sh ow n . W h ich is a solu tio n for 2 cos 0 = 1 ? 76. If cos x = —0.45, w h at is sin F - 0 .5 5 G - 0 .4 5 H 0.45 J 0.55 F G 8tt H 3 IO tv 1 3 tv I JJZIL J 3 J j 3 connectED.mcgraw-hill.com I 333 ........... Graphing Technology Lab no o o o ^Solving Trigonometric Inequalities Objective • Use a graphing calculator to solve trigonometric inequalities. oooo oooo V C D O O y You can use a graphing calculator to solve trigonometric inequalities. Graph each inequality. Then locate the end points of each intersection in the graph to find the intervals within which the inequality is true. Activity 1 Graph a Trigonometric Inequality Graph and solve sin 2x > cos x. ETEfffl R ep lace each sid e o f th is in eq u ality w ith y to fo rm the n ew inequ alities. sin 2x > Yi and Y 2 > cos x R E ffW G rap h e ach inequality. M ak e e ach in eq u ality sy m b o l b y scro llin g to the le ft o f the equal sig n and selectin g Ie n t e r j u n til the sh ad ed trian g les are flashing . T h e trian g le abo v e rep resen ts greater than, and the trian g le b e lo w rep resen ts less than (Figu re 5.3.1). In the MODE 1m en u , select RADIAN. G rap h the eq u atio n s in the ap p rop riate w ind ow . U se the d o m ain and ran ge o f each trig on o m etric fu n ctio n as a g u id e (Figu re 5.3.2). Ploti Plot2 Plot3 tY iB s in (2 X) ''Y z B c o s C X ) \Yj = \Yh = \Ys = \Ys = \Y? = Figure 5.3.1 Figure 5.3.2 ETTSm T h e d ark ly sh ad ed area in d icates the in te rsectio n o f the grap h s and the so lu tio n o f the sy ste m o f in eq u alities. U se CALC: in te rse c t to lo cate th ese in tersection s. M o v e the cu rso r o v er the in tersection and select e n t e r 13 tim es. [— 2 tt, 2 t v ] scl: j by [ - 1 , 1 ] scl: 0.1 ETHTI3 T h e first in tersection is w h en y = 0.866 o r in tersection is at x = solu tio n in terv als is O S in ce cos -j- = -^r-, the T h e n ex t in tersectio n is at x = TT TT 6' 2 . A n o th er in terv al is 5 tt 3 tt 6 ' 2 Z T h erefore, one o f the . T h ere are an infin ite n u m b er o f in terv als, so th e solu tio n s for all v alu es o f x are ^ + 6 and 6 + 2«tt, ^ 1 2m tt , ^ 2 + + 2mr Exercises Graph and solve each inequality. 334 Lesson 5-3 1. sin 3x < 2 co s x 2. 3 cos x > 0.5 sin 2x 3. sec x < 2 co s x 4. csc 2x > sin 8,r 5. 2 tan 2x < 3 sin 2x 6 . ta n x > cos x 2m r Mid-Chapter Quiz Lessons 5-1 through 5-3 Find the value of each expression using the given information. Find all solutions of each equation on the interval [0 ,2 ir]. (Lesson 5-1) (Lesson 5-3) 1. sin 0 and cos 0, cot 0 = 4, cos 9 > 0 16. 4 sec 9 + 2 V 3 = sec 9 2. sec 9 and sin 9, tan 9 = —| , csc 9 > 0 3. tan 9 and csc 9 , cos 9 = 17. 2 tan 0 + 4 = tan 0 + 5 sin 9 > 0 18. 4 cos2 0 + 2 = 3 19. cos 9 - 1 = sin 9 Simplify each expression. (Lesson 5-1) sin (- x ) tan (-x ) 5. sin (90° - x) cot2 (90° - x) + 1 7. cot2 X + 1 20. sin x 1 + sec x MULTIPLE CHOICE Which of the following is the solution set for cos 9 tan 9 - sin2 0 = 0 ? (Lesson 5-3) F yn , where n is an integer G y + nix, where n is an integer 8. ANGLE OF DEPRESSION From his apartment window, Tim can see the top of the bank building across the street at an angle of elevation of 9, as shown below. (Lesson 5-1) H 7r + 2/77T, where n is an integer J nix, where n is an integer (employee Solve each equation for all values of 9. (Lesson 5-3) 21. 3 sin2 0 + 6 = 2 sin2 9 + 7 a. If a bank employee looks down at Tim’s apartment from the top of the bank, what identity could be used to conclude that sin 9 = cos 9’? b. If Tim looks down at a lower window of the bank with an angle of depression of 35°, how far below his apartment is the bank window? 22. sin 9 + cos 9 = 0 23. sec 9 + tan 9 = 0 24. 2 tan2 0 = 1 25. PROJECTILE MOTION The distance tfth a t a kick ball travels (in feet) is given by d = 9. MULTIPLE CHOICE Which of the following is not equal to csc 91 vn2 sin 29 ^ — , where i/0 is the object s initial speed, 0 is the angle at which the object is launched, and 32 isin feet per second squared. If the ball is kicked with aninitial speed of82 feet per second, and the ball travels 185 feet, what is the launch angle of the ball? (Lesson 5-3) (Lesson 5-1) A sec (90° - 9) B Vcot2 9 + 1 1 C sin 9 26. 1 sin (90° - 9) FERRIS WHEEL The height h of a rider in feet on a Ferris wheel after f seconds is shown below. (Lesson 5-3) Verify each identity. (Lesson 5-2) 10. cos 9 1 + sin 8 cos ‘ = - 2 tan i 1 - sin 6 /7(f) = 7 5 + 70 sin ( £ t - f ) 11. csc2 9 — sin2 9 — cos2 9 — cot2 9 = 0 12. sin 9 + ^ 4 tan 8 = csc i 13. cos 8 = sec 9 - tan 9 1 + sin 8 14. csc| sin 8 15. 1 + sin 8 sin 8 cote = cot2 £ + csc £ + 1 cos 9 sin 8 csc ( 1 - sin 8 1 - sin 8 a. If the Ferris wheel begins at f = 0, what is the initial height of a rider? b. When will the rider first reach the maximum height of 145 feet? c o n n e c t^ m c g ra w -h ilL c o m | 335 You found values of trigonometric functions by using the unit circle. * 1 (Lesson 4-3) NewVocabulary reduction identity When the picture on a television screen is blurry or a radio station w ill not tune in properly, the problem is too often due to interference. Interference results when waves pass through the same space at the same tim e. You can use trigonom etric identities to determine the type of interference that is taking place. Use sum and difference identities to evaluate trigonometric functions. Use sum and difference identities to solve trigonometric equations. Evaluate Trigonometric Functions In L esso n 5-1, y o u used b asic id en tities inv o lv ing o n ly one v ariable. In this lesso n , w e w ill co n sid er id en tities in v o lv in g tw o variab les. O n e o f th ese is the cosine o f a difference identity. 1 (3 b e an g les on the in terv al [0, 27t], a > (3 as sh ow n in Figu re 5.4.1. B ecau se e ach p o in t is lo cated o n the u n it circle, x 2 + i/,2 = xi + Vi = 1' and x 3 2 + J/3 2 = 1- N o tice also th at the m easu re o f arc C D = a — (a — (3) or /3 and the m easu re o f arc AB = (3. L et p o in ts A, B, C, and D b e located o n th e u n it circle, a and and 1, _ 0 (x2, y.) D(x,,y.) Figure 5.4.1 Since arcs AB and CD hav e the sam e m easure, chords AC and BD show n in Figure 5.4.2 are congruent. Chords AC and BD are congruent. AC = BD Distance Formula V (x2 - l )2 + (y 2 - 0)2 = (x2 - l )2 + (y 2 - 0 )2 = (x3 - X j) 2 + ( 1/3 - y j ) 2 Square each side. y 22 = Square each binomial. 2Z - 2x2 + 1+ \ j ( x 3 - X j ) 2 + ( 1/3 - y x 3z - 2 x 3^ + t )2 X jz + y 32 - 2y3y 1 + y xl (*22 + 1/22) ~ 2x2 + 1 = (x l 2 + 3/l2) + <x 32 + J/32) - 2*3*1 ~ 23/33/l 1 — 2x2 + 1 = 1 + 1 Substitution — 2 X3X2 — 2 y3yj Add. 2 - 2 x 2 = 2 - 2 x 3x j - 2 y 3 y j - 2 x 2 = - 2 x 3x j - Group similar squared terms. Subtract 2 from each side. 2y3y x Divide each side by —2. x 2 = x 3x j + y 3 y j In Figu re 5.4.1, n o tice th at b y the u n it circle d efin itio n s fo r cosin e and sin e, x , = co s /3, x 2 = cos ( a — /3), x 3 = cos a , y x = sin (3, and y 3 = sin a . S u b stitu tin g , x 2 = x 3x x + y 3y x beco m es (a co s B y rew ritin g a — (3 as cosin e of a sum . cos [a a — 13) = co s a cos 336 Lesson 5-4 sin a sin (3. Cosine Difference Identity + (—f3) and ap p ly in g the abo v e identity, w e o b tain the id e n tity for the + (—/3)] = cos a cos = co s a cos llll (3 + {—(3) + sin a sin (3 — sin a sin (3 (—f3) Cosine Difference Identity Even-Odd Identity T h ese tw o cosin e id en tities can be u sed to e stab lish e ach o f the oth er su m and d ifferen ce id entities listed below . KeyConcept Sum and Difference Identities Sum Difference Identities Identities cos ( a + /3) = cos a cos /3 - sin a sin /3 sin ( a + 0) = sin a cos (3 + cos a sin /3 , tan a + tan 0 tan a + /3) = - — r------- r~ ^ 1 - t a n a tan 0 cos [ a - 13) = cos a cos /3 + sin a sin f3 sin ( a - / 3 ) = sin a cos /3 - cos a sin f3 . , tan a - tan 0 tan (a - /3 = - — ;------- 7— ^ 1 + tan a tan 0 r-J \ You will prove the sine and tangent sum and difference identities in Exercises 5 7 -6 0 . B y w ritin g an g le m easu res as the su m s or d ifferen ces o f sp ecial an g le m easu res, you can use these su m and d ifferen ce id en tities to find e xact v alu es o f trig o n o m etric fu n ctio n s o f ang les th at are less com m on. J ^ ^ ^ ^ j E v a l u a t e a Trigonometric Expression Find the exact value of each trigonom etric expression, a. sin 15° W rite 15° as the su m o r d ifferen ce o f an g le m easu res w ith sin es th at you know. StudyTip 4 4 5 ° -30° = 15° sin 15° = sin (45° - 30°) Check Your Answer You can check your answers by using a graphing calculator. In Example 1a, sin 1 5 ° « 0.259 and = sin 45° co s 30° — sin 30° cos 45° V2 2 ~ 0.259. Make sure ’ V | _i 2 2 ' V2 2 = V6 _ V| that the calculator is in the correct mode. 4 sin 30° = 1 sin 45° = cos 45° = cos 30° = -y- Multiply. 4 V6 - V2 b. Sine Difference Identity Combine the fractions. ta n £ . Z3L ta n l f = t a n ( f + f ) ta n y : 12 + ta n y i - t a n f tan ^ 4 V3+ 1 tan i = V3 and tan ~ = 1 3 1 - V3(l) V3 + 1 V3+ 1 1 + V3 1 - V3 1 + V3 V3 + 1 + 3 + V3 1 -V 3 + V 3 -3 -2 = - 2 - V3 4 Simplify. 1 - V3 4 + 2V3 Tangent Sum identity Rationalize the denominator. Multiply. Simplify. Simplify. ► GuidedPractice 1A. cos 15° IB . s i n f 337 Sum and difference identities are often used to solve real-world problems. -> M D m B Use a Sum or Difference Identity ELECTRICITY An alternating current i in amperes in a certain circuit can be found after t seconds using i = 3 (sin 165 )t, where 165 is a degree measure. a. Rewrite the formula in terms of the sum of two angle measures. Original equation i = 3 (sin 1 65)f = 3 sin b. (120 + 4 5 )f 120 + 45 = 165 Use a sine sum identity to find the exact current after Rewritten equation i = 3 sin (120 + 4 5 )t = 3 sin (120 + 45) f= 1 = 3[(sin 120)(cos 45) + (cos 120)(sin 45)] Sine Sum Identity 1 -atm Real-WorldCareer Lineworker Lineworkers are responsible for the building and upkeep of electric power transmission and distribution facilities. The term also applies to tradeworkers who install and maintain telephone, cable TV, and fiber optic lines. 1 second. Substitute. Multiply. 3V 6 - 3V 2 Simplify. T h e e xact cu rren t after 1 secon d is am p eres. p GuidedPractice 2. ELECTRICITY A n altern atin g cu rren t i in am p eres in an o th e r circu it can b e fo u n d after t seco n d s u sin g i = 2 (sin 2 85)f, w h ere 285 is a d eg ree m easu re. A. R ew rite the fo rm u la in term s o f the d ifferen ce o f tw o an g le m easures. B. U se a sin e d ifference id en tity to find the e xact cu rren t after 1 second . If a trig on o m etric e xp ressio n h as the form o f a su m o r d ifferen ce identity, y o u can use the id en tity to fin d an e xact v alu e or to sim p lify an e xp ressio n b y rew ritin g the exp ressio n as a fu n ctio n o f a sin gle angle. B P IE f f f f T T i Rewrite as a Single Trigonometric Expression a. Find the exact value of ta n 3 2 ° + ta n 13° 1 — ta n 32° ta n 1 3 £ b. tan 32° + tan 13° 1 - t a n 32° tan 13° ‘ = tan (32° + 13°) Tangent Sum Identity = tan 45° o r 1 Simplify. Simplify sin x sin 3x — cos x cos 3x. sin x sin 3x — cos x c o s 3x = —(cos x cos 3x — sin x sin 3x) p GuidedPractice 8 3B. S im p lify ,tan 6x ~ tan 7* . r 1 + tan 6x tan 7x 338 | Lesson 5-4 | Sum and D ifference Identities cos + sin — ■sin 24 8 Properties Tangent Sum Identity = —cos (x + 3x) o r —cos 4x 3A. Find the e xact v alu e o f cos ^ Distributive and Commutative 24 Sum and difference identities can be used to rewrite trigonometric expressions as algebraic expressions. Write as an Algebraic Expression Write sin (arctan y/3 + arcsin x) as an algebraic expression of x that does not involve trigonom etric functions. A p p ly in g the S in e o f a Su m Identity, w e find that sin (arctan V 3 + arcsin x) = sin (arctan V 3 ) cos (arcsin x) + cos (arctan V 3 ) sin (arcsin x). If w e let a = arctan V 3 and /3 = arcsin x, th e n tan a = \f?> an d sin (3 = x. S k etch one rig h t trian g le w ith an acu te an g le a and an o th er w ith an acu te an g le [3. L a b el the sid es su ch that tan a = \/3 and the sin (3 = x. T h en u se the P y th ag o rean T h eo rem to exp ress the len gth o f each third side. U sin g th ese trian g les, w e fin d th at sin (arctan V 3 ) = sin a o r cos (arctan \/3) = cos a or j , c o s (arcsin x) = co s (3 o r V l — x 2, and sin (arcsin x) = sin /3 o r x. N ow ap p ly su b stitu tio n and sim plify. sin (arctan \/3 + arcsin x) = sin (arctan V 3 ) cos (arcsin x) + cos (arctan V 3 ) sin (arcsin x) V3 | 2 ( V 1 - x 2 ) + jX V3 - 3x2 p Guided Practice Write each trigonom etric expression as an algebraic expression. 4A. cos (arcsin 2x + arccos x) 4B. sin ^arctan x — arccos - jj S u m and d ifference id en tities can b e u sed to v e rify o th er identities. ReadingMath Cofunction Identities The “co” in cofunction stands for “complement.” Therefore, sine and cosine, tangent and cotangent, and secant and cosecant are all complementary functions or cofunctions. B Q 2 G 2 3 3 S Ver'fy Cofunction Identities — x j = cos x. Verify sin ■ ITX — x\\ = sm (y = • sin —TTc o s l( c o s = cos X x) — 0 ✓ 'Jt s in ; x — cos — (s in x) Sine Difference identity sin 2 = 1 and cos 2 = Multiply. Guided Practice Verify each cofunction identity using a difference identity. 5A. co s — x j = sin x 5B. csc ^ = sec 0 339 S u m and d ifference id en tities can b e u sed to rew rite trig o n o m etric e xp ressio n s in w h ich o n e o f the ang les is a m u ltip le o f 90° o r y rad ian s. T h e re su ltin g id e n tity is called a reduction identity b ecau se it reduces the co m p lexity o f the exp ression . P E S u H S Q Verify Reduction Identities Verify each reduction identity, a. sin ^0 + = —cos 9 = sin 0 cos ^ sin ^0 + b. Sine Sum Formula + cos ® sin = sin 0 (0 ) + cos 0 (—1) cos ~ = —cos 0 ✓ Simplify. 2 = 0 and sin 2 = -1 tan (x — 180°) = tan x ta n tan (x — 180)° : 1 ta n 1 x- + ta n ta n 1 8 0 ° x ta n x— Tangent Sum Formula 180° 0 tan 180° = 0 + ta n x (0 ) Simplify. = tan x ✓ ► GuidedPractice Verify each cofunction identity. 6B. sin 6A. cos(360°\— 0) = cos 0 I Solve Trigonometric Equations ( f + «) = COS X You can solv e trig o n o m etric eq u atio n s u sin g the sum i and d ifference id en tities and the sam e tech n iq u es th at y o u u sed in L esson 5-3. j E a J H E H Solve a Trigonometric Equation + x j + cos Find the solutions of cos — xj = j on the interval [0, 2ir]. C0S(f+x)+C os(f-*)=I \ cos y cos x — sin y sin x + cos -j cos x + sin -j sin x = ~ ■|(cos x) - ~ ( s i n x) + i ( c o s x) + ^ - ( s i n x) = -| 1 2 cos x = TechnologyTip Viewing Window When checking your answer on a graphing calculator, remember that one period for y = sin x or y —cos x is 2tt and the amplitude is 1. This will help you define the viewing window. O n the in terv al [0, 2 tt], co s x = — w h e n x = — and x — > CHECK T h e g rap h o f y = cos h as zeros at -j and + cos [~J ~ x) ~ 2 o n the in terv al [0 , 2 tv] . ✓ ► GuidedPractice 7. 340 Find the solu tio n s o f cos (x + tt) — sin (x — 7r) = 0 o n the in terv al [0, 2 tt], | Lesson 5 -4 | Sum and D ifference Identities 5 tv Original equation Cosine Sum Identities Substitute. Simplify. Exercises Find the exact value of each trigonometric expression. Find the exact value of each expression. (Example 3) (Example 1) 11. 1. cos 75° 3. sin 5. II tt 12 tan 23tt 12 2 . sin ( - 2 1 0 °) 4. cos ta n 4 3 ° — ta n 13° 1 + ta n 4 3° ta n 13° " ^ c o s 4 ? cos v + sin ~ 1 7 tt 12 12 13. 4 sin ^ 12 4 sin 15° cos 75° + cos 15° sin 75° 6. ta n y j ( l £ ) sin J cos 7. VOLTAGE A n aly sis o f the v o ltag e in a h aird ry er inv o lv es term s o f the form sin (nzvt — 90°), w h ere n is a p o sitiv e integer, w is the freq u en cy o f the v o ltag e, and t is tim e. U se an id en tity to sim p lify th is exp ressio n . (Example 2) - cos y sin 15. cos 40° cos 20° — sin 40° sin 20° 167) ta n 4 8 ° + ta n 12° 1 - ta n 4 8° ta n 12° Simplify each expression. (Example 3) 8. BROADCASTING W h en the su m o f the am p litu d es o f tw o 17. w av es is greater th a n th at o f th e co m p o n e n t w av es, the resu lt is constructive interference. W h e n the co m p o n en t 8. 28 — t a n 8 29 t a n ( ta n 1 + ta n cos y cos x + sin y sin x w av es com bin e to h av e a sm aller am p litu d e, destructive 19. sin 3y cos y + cos 3y sin y interference occurs. f o ) cos 2x sin x — sin 2x co s x 21 ■ cos x co s 2x + sin x sin 2x <2 ^ ^ 58 + ta n ta n 5 0 ta n 9 9—1 ta n (23) SCIENCE A n electric circu it co n tain s a capacitor, an ind uctor, and a resistor. T h e v o ltag e d rop across the in d u cto r is g iv en b y VL = IivL cos (zvt + y j , w here I is C on sid er tw o sig nals m o d eled b y y = 10 sin (It + 30°) the p e a k cu rren t, w is the frequ ency, L is the ind u ctance, and y = 10 sin (2 1 + 210°). (Example 2) and t is tim e. U se the cosin e su m id en tity to exp ress VL as a fu n ctio n o f sin wt. (Example 3) a. Find the su m o f the tw o fu n ctions. b. W h at typ e o f in terferen ce results w h e n the sig nals m o d eled b y the tw o eq u atio n s are com b in ed ? 9. WEATHER T he m o n th ly h ig h tem p eratu res fo r M in n eap o lis can be m od eled b y / (x ) = 31.65 sin — 2.09 j + 52.35, w here x rep resen ts th e m o n th s in w h ich Ja n u a ry = 1, Febru ary = 2, and so on. T h e m o n th ly low tem p eratu res for M in n eap o lis can b e m od eled b y g(x) = 31.65 sin Write each trigonom etric expression as an algebraic expression. (Example 4) 24. sin (arcsin x + arccos x) 25. cos (sin 1 x + cos 1 2x) 26. cos |sin_1 x — ta n - 1 -^y-J 27. sin |sin_1 -^y- — ta n " 1 xj — 2.09 j + 32.95. (Example 2) a. W rite a n ew fu n ctio n h(x ) b y ad d in g the tw o fu n ction s and d iv id in g th e re su lt b y 2 . b. W h at d oes the fu n ctio n y o u w ro te in p a rt a rep resen t? 28. co s (arctan V 3 — arcco s x) 29. tan (cos 1 x + tan 1 x) 30. tan |sin-1 — c o s -1 xj 31. tan |sin-1 x + y j 10. TECHNOLOGY A blin d m ob ility aid u ses the sam e id ea as a b a t's son ar to en ab le p eo p le w h o are v isu ally im p aired to detect o bjects arou nd them . T h e sou n d w av e em itted by the d ev ice for a certain p atie n t can b e m o d eled b y b = 30 (sin 195°)f, w here t is tim e in seco n d s and b is air pressu re in pascals. (Example 2) a. R ew rite the form ula in term s o f the d ifferen ce o f tw o Verify each cofunction identity using one or more difference identities. (Example 5) 32. ta n (f - * ) = cot X 33. sec (f - * ) = CSC 34. cot | angle m easures. b. W h at is the p ressu re after 1 seco nd ? (? - * ) = ta n X x § connectED.m cgraw-hill.com | 341 Verify each reduction identity. (Example 6) 35. cos 56. — 9) = —cos 9 ( tt |§&. cos (2tt + 9) = cos 9 37. sin ( tt ANGLE OF INCLINATION T h e angle o f inclination 9 o f a lin e is the an g le fo rm ed b e tw e e n the p o sitiv e x -ax is and the lin e, w h ere 0° < 9 < 180°. a. P rove th at the slop e m y o f lin e £ sh o w n at the rig h t is g iv en b y m = tan 9. — 9) = sin 9 38. sin (90° + 9) = cos 9 /\ 0 39. cos (270° — 9) = —sin 9 b. C o n sid er lin es (Example 7) 41. COS (TT (ffij cos 43. + x) + X and i 2 b elo w w ith slo p es and m 2, respectively. D eriv e a form ula for the an g le 7 form ed b y the tw o lines. + x\ = 0 + x j - sin e / / Find the solution to each expression on the interval [0, 2 tt], ^ST cos t/ (TT + x) = 1 COS + x j + sin + xj = 0 s in (f+ x ) + s in ( f - x ) = i 4?, sin 45. + x j + sin tan ( tt + x) + tan + xj = -2 ( tt + x) = 2 c. U se the fo rm u la yo u fo u n d in p art b to find the angle fo rm ed b y y = ^ - x and y = x. Verify each identity. (x —y) cos xcos y c o s (a + /3 ) s in 46. tan x — tan y 47. cot a — tan (3 (ta n 48. (ta n uu+ ta n ta n v) v) s in a s in s in (u- v) (u + v) cos H.O.T. Problems f3 PROOF Verify each identity. __ , , ta n (3)= 1 —t a n a tan ( a — f3) 57. GRAPHING CALCULATOR Graph each function and make a conjecture based on the graph. Verify your conjecture algebraically. 58. 59. sin (a + 50. y 60. sin ( a — (3)= --[sin (x + 2 tt) + sin (x — 2 t t ) ] 51. y = c o s2 (x + f ) + c o s2 (x - f ' j a + ta n B tan (a + 49. 2 sin a cos b = sin (a + b) + sin (a — b) = Use Higher-Order Thinking Skills f3)= ta n ta n (3 a —t a n f3 a ta n 1 + ta n /3 sin a cos /3 + cos a sin /3 sin a cos [3 — cos a sin (3 61. REASONING U se the su m id en tity fo r sin e to d eriv e an id e n tity fo r sin (x + y + z) in term s o f sin es and cosines. PROOF Consider A X Y Z . Prove each identity. (Hint: x + y + z = tt) CHALLENGE If sin x = and cos y = find each of the following if x is in Quadrant IV and y is in Quadrant I. 62. cos (x + y) (63) sin (x — y) 64. tan (x + y) y Y 65. REASONING C o n sid er sin 3x cos 2x = cos 3x sin 2x. a. Find the solu tio n s o f the eq u atio n ov er [0, 2tt] 52. cos(x + y) — —cos z algebraically. 53. sin z = sin x cos y + cos x sin y b. S u p p ort y o u r an sw er graphically. 54. tan x + tan y + tan z = tan x tan y tan z 55. CALCULUS The difference quotient is given by f{x + h) -f(x) h a. L et/ (x) = sin x. W rite and exp an d an exp ressio n for PROOF Prove each difference quotient identity. 66. the d ifference quotient. b. Set you r answ er from p art a equ al to y. U se a grap h in g calcu lato r to grap h the fu n ctio n for the fo llow in g v alu es o f h : 2 , 1, 0 .1, and 0 .0 1 . c. W h at fu n ctio n d o es the grap h in p a rt b resem b le as h app roach es zero? 342 Lesson 5-4 Sum an d D ifference Identities __ 67. 68. {x + h) —s i n x = h c o s (x + h) — c o s x s in ; . I c o s h —1 \ s in h + c o s x —— \ h I h c o s h— 1 s in h sin x -— = cos x sm x WRITING IN MATH C an a tan g en t su m o r d ifferen ce id en tity b e u sed to solv e an y tan g en t red u ctio n fo rm u la? E xp lain y o u r reaso n in g. Spiral Review 69. PHYSICS A cco rd in g to S n e ll's law , th e an g le at w h ich lig h t en ters w ater a is related to the angle at w h ich lig h t trav els in w ater (3 b y sin a = 1.33 sin (3. A t w h at an g le d o es a b eam of lig ht en ter the w ater if the beam trav els at an an g le o f 23° th ro u g h the w ater? (Lesson 5-3) Verify each identity. (Lesson 5-2) 70. cos 9 ■= sec i 1 - sin2 71. - = co t I Find the exact value of each expression, if it exists. (Lesson 4-6) 72. 73. ta n -1 V 3 s in -1 (—1) 74. tan (arcsin |-j 75. MONEY S u p p ose yo u d ep o sit a p rin cip al am o u n t o f P d ollars in a b a n k accou n t th at pays com poun d interest. If the an n u al in terest rate is r (exp ressed as a d ecim al) and the b an k m akes in terest p ay m en ts n tim es e v ery year, the a m o u n t o f m o n ey A y o u w o u ld h av e after t years is giv en b y A(t) = P ( l + ■£)” . (Lesson 3-1) a. If the p rin cip al, in terest rate, and n u m b er o f in terest p ay m en ts are k n o w n , w h a t typ e o f fu n ctio n is A(t) = P ( l + £ )" * ? E xp lain y o u r reasoning. b. W rite an eq u atio n giv in g the am o u n t o f m o n ey yo u w ou ld h av e after t y ears if you d ep o sit $1000 into an accou n t p ay in g 4% an n u al in terest com p o u n d ed q u arterly (four tim es p er year). C. Find the acco u n t b alan ce after 20 years. List all possible rational zeros of each function. Then determ ine which, if any, are zeros. (Lesson 2-4) 76. p(x) = x4 + x3 — l l x — 5x + 30 77. d(x) = 2x4 — x 3 — 6x2 + 5x — 1 78. f(x ) = x3 — 2x2 — 5x — 6 Skills Review for Standardized Tests 79. SAT/ACT T h ere are 16 green m arbles, 2 red m arbles, 81 . Find the e xact v alu e o f sin 6. and 6 y ellow m arb les in a jar. H o w m an y y ellow m arb les n eed to b e ad d ed to the ja r in ord er to d ou ble the p ro b ab ility o f selectin g a y ello w m arble? A 4 C 8 B 6 D 12 E 16 A B 80. REVIEW R efer to the figure below . W h ich eq u ation cou ld b e u sed to find m ZG ? C H D 82. V2 + V6 4 V 2 -V 6 4 2 + V3 4 2 - V3 REVIEW W h ich o f the fo llo w in g is eq u iv alen t to cos 0 (cot2 6 + 1) ? csc 9 F sin G = ■ G cos G = — 4 H co t G = -7 4 I tan G = ■ F, tan 9 G co t 0 H sec 8 J csc 8 E c o n n ec tiD ~m cg ra w -h ilL co n ^ 343 Graphing Technology Lab oooo oooo oooo CDOO Reduction Identities ftJLH ■ M r Another reduction identity involves the sum or difference of the measures of an angle and a quadrantal • Use Tl-Nspire technology and quadrantal angles to reduce identities. angle. This can be illustrated by comparing graphs of functions on the unit circle with Tl-Nspire technology. Activity 1 Use the Unit Circle Use the unit circle to explore a reduction identity graphically. R T f l l l A dd a Graphs page. S elect Zoom-Trig from the W indow m en u , and select Show Grid from the View m enu. From the File m en u u n d er Tools, ch o o se Docum ent Settings, set the Display Digits to Float 2 , and con firm th at the an g le m easu re is in rad ians. PT7TW S elect Points & Lines an d th en Point fro m the m en u . P lace a p o in t at (1 ,0 ). N ext, select Shapes, and th en Circle from the m en u. To co n stru ct a circle cen tered at the origin th ro u gh (1 ,0 ) , click o n the screen and d efin e th e cen ter p o in t at the origin. M o v e the cu rso r aw ay from the center, and the circle w ill appear. S top w h en yo u get to a rad iu s o f 1 and (1, 0 ) lies on the circle. E T T lT lfl P lace a p o in t to the rig h t o f the circle on the y d s x-axis, and lab el it A. C h o o se Actions and then Coordinates and Equations fro m the m en u , and th e n d o u b le-click th e p o in t to d isp lay its coord inates. From the C onstruction m en u , f ‘ '•"■ a 2*71*23*9 r a d AO B A -n 0.858407 -6 * 7B . (4 ,0 ) . . . t v. \A > 6.2 f A B ch o o se Measurement transfer. S e le ct the x-co o rd in ate o f A , the circle, and the p o in t at (1 ,0 ). L abel the p o in t created on the circle as B, ESXSSSSH I (-0.653644,-0.7 56802) . 4 a and d isp lay its coord inates. W ith O as the o rig in , calcu late and lab el the m easu re o f ZAO B. S e le ct Text from the Actions m en u to w rite the exp ressio n a — 7V. T h en sele ct Calculate from the Actions m en u to calcu late the d ifferen ce o f the x -co o rd in ate o f A and tv. StudyTip M ov e A alo ng the x-axis, and o bserve the Angle Measure The Tl-Nspire only measures angles between 0 and tt. effect o n the m easu re o f /.A O B. m m F ro m the C onstruction m en u , ch o o se Location of A mZAOB (radians) (4,0) 2.7124 (3,0) 3.0708 (2,0) 2.5708 (5,0) 2.2123 (-2 ,0 ) 0.5708 1.16 RAD A U T O REAL Measurement transfer. S elect the x-ax is and the v alu e o f a — tv. L abel th e p o in t as C and d isp lay its coo rd in ates. U sin g Measurement transfer again , select the x-co o rd in ate o f C, , ‘4: n . , 0.858407 the circle, and the p o in t at (1, 0). L abel the p o in t as D and d isp lay its coord inates. ( "p. 653644,-p. 456802 j ^ tO B 2'.71'239rarf * [ 0.653644,0,756802) , D ( ~y { 0.85^407.0,) (4>o) 6,2^ , V___________________________________________________________________________________ Analyze the Results 1A. In Step 5, h o w are the lo catio n o f A and th e m easu re o f /.A O B related ? IB . C o n sid er the lo catio n s o f p o in ts B and D. W h at red u ctio n id en tity or id en tities d o es this relation ship su g g est are true? IC . MAKE A CONJECTURE If yo u ch an g e the exp ressio n a — tt to a + 7V, w h at red u ctio n id en tities do you think w o u ld result? 344 | Lesson 5-4 Activity 2 Use Graphs Use graphs to identify equal trigonom etric functions. O p e n a n ew Graphs p ag e. S e le ct Zoom-Trig fro m the W in d ow m en u. EflSffEl G rap h / (x) = cos x ,f(x ) = cos (x — tt) , and f(x ) = sin x. U sin g the A ttributes featu re from the Actions m en u , m ak e the lin e w e ig h t o f f(x ) = cos (x — tt) m e d iu m and the lin e style o f/ (x) = sin x dotted. 4' * y=sin(x) -------------- «(r)-cos(x f6U’=cosU_nJ I B S E U se tran slation s, reflectio n s, o r d ilatio n s to tran sfo rm / (x ) = sin x so th at the graph coin cid es w ith / (x) = cos x. S elect the g rap h and d rag it o v er/ (x ) = cos x. A s you m ov e the g rap h , its fu n ctio n w ill ch an g e o n the screen. E S B U se tran slatio n s, reflectio n s, o r d ilatio n s to tran sfo rm / (x ) = cos (x — tt) s o th at the grap h co in cid es w ith the o th er tw o grap h s. A g ain , as you m o v e th e g rap h , its fu n ctio n w ill ch an g e o n th e screen. I ________________________________________________________________________ Analyze the Results 2A. W rite the id e n tity th at resu lts fro m y o u r tran sfo rm atio n o f/ (x) = sin x in Step 3. G rap h the fu n ctio n s to con firm y o u r identity. 2B. W rite the id en tity th at resu lts fro m y o u r alteratio n o f/ (x) = co s (x - tt ) in Step 3. G rap h the fu n ctio n s to con firm y o u r identity. 2C. MAKE A CONJECTURE W h a t d oes the reflectio n o f a g rap h su g g est fo r th e p u rp o se o f d ev elop in g an id en tity ? a tran slatio n ? Exercises Use the unit circle to write an identity relating the given expressions. Verify your identity by graphing. 1. cos (90° — x), sin x 2. cos — x j, sin x Insert the trigonometric function that completes each identity. 3. cos x = __________ (x - 4. co t x = ------------------- (x + 90°) 5. sec x = __________ (x — 180°) 6. csc x = ------------------- |x + y j Use transform ations to find the value of a for each expression. 7. sin ax = 2 sin x cos x 9. a s in 2 x = 1 — cos 2x cos 4 ax = c o s 2 x — s in 2 x 8. 10.1 + cos 6ax = 2 c o s 2 x connectED.m cgraw-hill.com | 345 Multiple-Angle and Product-to-Sum Identities : Then • You proved and used sum and difference identities. (Lesson 5-4) I Use double-angle, power-reducing, and half-angle identities to evaluate trigonometric expressions and solve trigonometric equations. The speed at which a plane travels can be described by a mach num ber, a ratio of the plane’s speed to the speed of sound. Exceeding the speed of sound produces a shock wave in the shape of a cone behind the plane. The angle 9 at the vertex of this cone is related to the mach number M describing the plane’s speed by the half-angle equation sin 2 Use product-to-sum identities to evaluate trigonometric expressions and solve trigonometric equations. I Use Multiple-Angle Identities B y letting a and (3 bo th equal 9 in each o f the angle sum identities you learned in the previous lesson, you can d erive the follow ing double-angle identities. KeyConcept Double-Angle Identities sin 29 = 2 sin 6 cos 9 cos 29 = cos2 6 - sin2 9 tan 2 0 = cos 2 0 = 2 cos2 ® “ 1 1 - tan2 0 cos 26 = 1- 2 sin2 0 P roof Double-Angle Identity for Sine sin 20 = sin(0 + 9 ) s» 29 = 9 + 0 = sin 0 cos 0 + cos 0 sin 0 Sine Sum Identity where a = 0 = 0 = 2 sin 0 cos 0 Simplify. J You will prove the double-angle identities for cosine and tangent in Exercises 6 3 -6 5 . H O S S H 3 0 I Evaluate Expressions Involving Double Angles If sin 9 = on the interval |tv, ^y-j, find sin 29, cos 29, and tan 20. S in ce sin 9 = ^ o n the in terv al y ^ j, on e p o in t o n the term in al sid e o f 9 h as y -co o rd in ate —7 and a d istan ce o f 25 u n its from the o rigin , as show n. T h e x -co o rd in ate o f this p o in t is th erefore —V 2 5 2 — 7 2 or —24. U sin g this p o in t, w e find that co s 0 = — or —~ 25 r and tan 9 = — or x 25 U se th ese v alu es and the d o u b le-an g le id e n tities for sin e and co sin e to fin d sin 20 and cos 29. T h en find tan 29 u sin g eith er the tan g en t d o u b le-an g le id en tity o r the d efin itio n o f tangent. sin 29 = 2 sin 9 cos 9 Method 1 tan 29 = cos 2 9 = 2 c o s 2 0 — 1 2 ta n ^ 1 - tan2 0 Method 2 tan 2 0 = - ^ ^ cos 20 2(1 \2 / 1 ► GuidedPractice 1. If cos 0 = o r 336 I 7 \2 527 U !4) on the in terv al |o, y j , find sin 20, cos 29, and tan 29. V..................................................................................................... ................. 346 | Lesson 5-5 _ 336 625 o r ^ 6 527 527 625 StudyTip More than One Identity Notice that there are three identities associated with cos 29. While there are other identities that could also be associated with sin 28 and tan 29, those associated with cos 29 are worth memorizing because they are more commonly used. H I B 3 ! E 3 Solve an Equation Using a Double-Angle Identity Solve sin 2 0 — sin 0 — 0 on the interval [0, 2tt]. U se the sin e d o u b le-an g le id en tity to rew rite the eq u atio n as a fu n ctio n o f a sin gle angle. Original equation sin 29 — sin d = 0 Sine Double-Angle identity 2 sin 6 cos 0 — sin 9 = 0 Factor. sin 9 (2 cos 9 — 1) = 0 sin 0 = 0 or Zero Product Property 2 cos 0 — 1 = 0 cos 9 = I = 0 o r tt 1 T h erefore, 0 = -?• o r - T h e solu tio n s o n th e in terv al [0, 2tt] are 0 = 0, y , ^ Gllided Practice tt , 3 ' o r -y-. Solve each equation on the interval [0, 2ir]. 2A. cos 2a = —s in 2 a 2B. ta n 2/3 = 2 ta n /3 T h e d o u ble an g le id en tities can b e u sed to d eriv e the p o w er-red u cin g id en tities below . T h ese id en tities m ak e calcu lu s-related m an ip u latio n s o f fu n ctio n s lik e y = c o s 2 x m u ch easier. K eyC on cept Power-Reducing Identities ta n 2 0 = COS 2 0 = 1 + C ° S 2 * s in 2 e = 1 - C20S 2 e j ~ cos ^ 1 + COS 20 Proof Power-Reducing Identity for Sine 1 - cos 2 0 1 - (1 - 2 s in 2 6) 2 2 2 s in 2 2 = S in 2 Cosine Double-Angle Identity 0 Subtract. Simplify. 9 „ _ _ _ _ _ _ -/ _ You will prove the power-reducing identities for cosine and tangent in Exercises 82 and 83. Use an Identity to Reduce a Power Rewrite sin4 x in terms with no power greater than 1. Frangois Viete (1540- 1603) Born in a village in western France, Viete was called to Paris to decipher messages for King Henri III. Extremely skilled at manipulating equations, he used double-angle identities for sine and cosine to derive triple-, quadruple-, and quintuple-angle identities. s in 4 x = (sin 2 x )2 _ — (sin2 — cos2xj Sine 1 — 2 cos 2x + cos2 2x 4 1 — 2 cos 2x + x )2 = sin4 x Power-Reducing Identity ... , Multiply. 1 + cos 4x 2 4 2 — 4 cos 2x + 1 + cos 4x = -i (3 — 4 cos 2x + cos 4x) Cosine Power-Reducing Identity Common denominator Factor. Guided Practice Rewrite each expression in terms with no power greater than 1. 3A. c o s 4 x 3B. s in 3 0 ^j^connectE^mcgra^^^^^m1 347 B Q S 3 3 3 3 & Solve an Equation Using a Power-Reducing Identity Solve cos2 x — cos 2 x = 4 . 2 Solve Algebraically cos2 X ■ 1 + cos 2 x 2 cos 2x = Original equation 2 _ 1 Cosine Power-Reducing identity — cos — 2—x = 2 Multiply each side by 2. 1 + cos 2x — 2 cos 2 x = 1 Subtract 1 from each side. cos 2x — 2 cos 2x = 0 Subtract like terms. —cos 2x = 0 Multiply each side by —1. cos 2x = 0 TV 2 x = fo r f 4 4 Solutions for double angle in [0,2it] Divide each solution by 2. T h e grap h o f y = cos 2x h as a p erio d o f t t , s o the so lu tio n s are x = ^ 37V Yl'K O f — — b Yl'K, n 6 . Support Graphically T h e g rap h o f y = c o s 2 x — cos 2x — j h as zero s at ^ and o n the in terv al [0, 7r]. Ztro X=.7B£3SBifi Y = - i . i E - i 3 [0, p- GuidedPractice tv] s c l: ~ b y [ - 1 . 5 , 1 . 5 ] scl: 1 Solve each equation. 4A. cos4 a — sin 4 a = -j 4B. s in 2 3/3 = s in 2 /3 By replacing 6 w ith — in e ach o f the p o w er-red u cin g id en tities, y o u can d eriv e each o f the fo llo w in g h alf-an g le identities. T h e sig n o f e ach id e n tity th at in v o lv es the ± sy m b ol is d eterm in ed by ch eck in g the q u ad ran t in w h ich the term in al sid e o f — lies. WatchOut! Determining Signs To determine which sign is appropriate when using a half-angle identity, check KeyConcept Half-Angle Identities > - cos 6 + cos 0 tan sinf = ± ^ f * * -fL the quadrant in which - | lies, not the quadrant in which 0 lies. ta n ! = 1 — cos 9 sine tan 1 = 2 sine 1 + cose Proof Half-Angle Identity for Cosine /1 + cos 0 2 " I 11 + cos| _,_\l H ± V 2 + cos 2x ) Rewrite 0 as 2 • Substitute x - 6 = ±Vcos^ x Cosine Power-Reducing Identity = cos X Simplify. = cos| Substitute. You will prove the half-angle identities for sine and tangent in Exercises 66-68. 348 | Lesson 5-5 | M ultiple-Angle and Product-to-Sum Identities Evaluate an Expression Involving a Half Angle Find the exact value of cos 112.5°. N o tice th at 112.5° is h a lf o f 225°. T h erefore, ap p ly the h alf-an g le id e n tity fo r cosine, n o tin g th at sin ce 112.5° lies in Q u ad ran t II, its cosin e is n eg ativ e. 225° 112.5° = cos 112.5° = c o s ; + cos 225° - { 225° Cosine Half-Angle Identity (Quadrant II angle) k V2 cos 225° = 2-V2 -{ Subtract and then divide. V 2 -V 2 7=— V4 V 2 -V 2 o r Quotient Property of Square Roots 5---------- 2 CHECK U se a calcu lato r to su p p o rt y o u r assertio n th at co s 112.5° = — ^ cos 112.5° « -0 .3 8 2 6 8 3 4 3 2 4 and V 2 - V2 -0 .3 8 2 6 8 3 4 3 2 4 ✓ p GuidedPractice Find the exact value of each expression. StudyTip Tangent Half-Angle Identities When evaluating the tangent function for half-angle values, it is usually easiest to use the form of the tangent half-angle identity tan » = sinceits 2 sin 9 denominator has only one term. 5B. ta n ^ | 5A. sin 75° R ecall th at yo u can use su m and d ifference id en tities to so lv e equ ation s. H alf-an g le id en tities can also b e u sed to solv e equ ations. m m m Solve an Equation Using a Half-Angle Identity Solve sin2 x = 2 cos2 j on the interval [0, 2 ir]. Original equation s in 2 x = 2 cos + COS x s in 2 * • 2 "|/1 + COS sin x = 2 ^ j Cosine Half-Angle Identity Simplify. Multiply. sin2 x = 1 + cos x Pythagorean Identity 1 — cos2 x — 1 + cos x Subtract 1 from each side. —cos2 x — cos x = 0 Factor. cos z (—cos x — 1) = 0 cos x = 0 or Zero Product Property —cos x — 1 = 0 * = T2 o r ^2 cos x = —1; therefore, x = tt. T h e solu tio n s o n the in terv al [0, 2tv] are x = y , -y -, o r Solutions in [0 ,2 ir] tt. p GuidedPractice Solve each equation on the interval [0, 2 t v ] . 6A. 2 s in 2 ^ + cos x = 1 + sin x 6B. 8 tan -J + 8 cos x tan £ connectED.m cgraw-hill.com | 349 2 Use Product-to-Sum Identities Product-to-Sum Identities K eyC on cep t sin asm [3 = l[c o s (a - cos a cos (3 = f3) - cos (a + /3)] (a - -|-[cos sin a cos f3 = -|{sin ( a + /3) + sin ( a - /?)] p) + cos (a + f3)] cos a sin /3 = l[s in (a + /3) - sin ( a - /3)] P ro o f Product-to-Sum Identity for sin StudyTip Proofs Remember to work the more complicated side first when proving these identities. To w o rk w ith fu n ctio n s su ch as y = cos 5x sin 3x in calcu lu s, y o u w ill n eed to ap p ly one o f th e fo llo w in g p ro d u ct-to -su m identities. a cos f3 |[s in ( a + P ) + sin ( a - f3)] = l(s in a cos (3 + cos More complicated side of identity a sin (3 + sin a cos /3 - cos a sin /3) Sum and Difference Identities = -1(2 sin a cos (3) = sin a Combine like terms. cos /3 Multiply. J You will prove the remaining three product-to-sum identities in Exercises 8 4 -8 6 . ■ - i S B Use an Identity to Write a Product as a Sum or Difference Rewrite cos 5 x sin 3 x as a sum or difference. cos 5x sin 3x = -j[s in (5x + 3x) — sin (5x — 3x)] y = —(sin 8 x — sin 2 x ) Simplify. 1 1 = 2 sin 8 x - - sin 2x Distributive Property GuidedPractice Rewrite each product as a sum or difference. 7B. sin 7 x sin 6 x 7A. sin 4 0 cos 6 V T h ese p ro d u ct-to -su m id en tities h av e corresp on d in g su m -to -p rod u ct id entities. K ey C o n cep t Sum-to-Product Identities sin a + sin j3 = 2 sin cos | —2 sin a — sin f3 = 2 cos sin ( ' cos a + cos /3 = 2 cos | t t ^ ) cos | a g ~) cos a — cos 0 = - 2 sin j — P ro o f Sum-to-Product Identity for sin 2 sm( 2 1 cos( 2 I + sin (3 ...... a + [3 , Substitute x = — — 2 and jy = 2 s in x c o s y = 2 | | [ s i n (x + y) + sin (x - y )]j . ( a -f /3 = sm ( 2 + a a —(3\ = a —f3 — — 2 Product-to-Sum Identity . / a + /3 2 ) + s in ( si n | a ~ ^ | 2 a —f3\ 2 J = sin( ¥ ) + sin( f ) = sin a + sin (3 Substitute and simplify. Combine fractions. Simplify. J You will prove the remaining three sum -to-product identities in Exercises 8 7 -8 9 . 350 | Lesson 5-5 Multiple-Angle and Product-to-Sum Identities B E 2 E 0 3 0 Use a Product-to-Sum or Sum-to-Product Identity Find the exact value of sin ( 12 5tt 12 2 + sin 12 I 5tt IT 12 I COS 7T ' 12 \ 12 2 Sum-to-Product Identity , = 2 sin -?■ cos Simplify. -(#)(#) sinf = T - andcosf = T - V6 2 Simplify. ►GuidedPractice Find the exact value of each expression. 8A. 3 cos 37.5° cos 187.5° 8B. cos - cos Y J You can also use su m -to -p ro d u ct id e n tities to solv e so m e trig o n o m etric equ atio ns. PJ2EIEE13E3 Solve an Equation Using a Sum-to-Product Identity Solve cos 4x + cos 2x = 0. Solve Algebraically Original equation cos 4x 4- cos 2x = 0 2 cos ( 4 ^ ± 2 £ ) c o s ( i l ^ ) = 0 Cosine Sum -to-Product Identity Simplify. (2 cos 3x)(co s x) = 0 S et e ach facto r equ al to zero and fin d solu tio n s on the in te rv al [ 0 , 2ir]. First factor set equal to 0 2 cos 3x = 0 Divide each side by 2. cos 3x = 0 3x = f o r f WatchOut Periods for M ultiple Angle Trigonometric Functions Recall from Lesson 4 -4 that the periods of y = sin to and y = cos for are ~ k * =? “ ? > Second factor set equal to 0 cos x = 0 2 2 Solutions in [ 0 ,2 it] Multiple angle solutions in [0 ,2 it] Divide each solution by 3. T h e p erio d o f y = cos 3x is — , so the so lu tio n s are — + — n, 2 not 27^. 3 ■+ 2 t m , or ' 37T + 2 tm, n € Support Graphically /TT- 'TT C/TT* T h e grap h o f y = cos 4x + cos 2x h as zeros at —, 6 2 and 6 on the in terv al [0 , 27t]. ✓ Z tro X =.E 23E 9B 7B p GuidedPractice V=0 0, 2tt] s c l : b y [-3 , 3] scl: 1 Solve each equation. 9A. sin x + sin 5x = 0 9B. cos 3 x — cos 5x = 0 connected 351 Exercises = Step-by-Step Solutions begin on page R29. Find the values of sin 20, cos 26, and tan 26 for the given value and interval. (Example 1) 1. cos 6 = f , (270°, 360°) 2. tan 6 = 26. s in 2 9 — 1 = c o s 2 6 15 (90°, 180°) 4. sin 6 = 24. 1 — s in 2 6 — cos 29 = j 25. c o s2 6 — j cos 26 = 0 (180°, 270°) 3. cos 6 = Solve each equation. (Example 4) 27. c o s2 6 — sin 9 = 1 y - , 2tt) 28. 5. t a n 0 = - | , ( ^ 2 i r ) MACH NUMBER T h e an g le 9 at the v ertex o f the con e-sh ap ed sh o ck w av e p ro d u ced b y a p lan e break in g the sou n d b arrie r is related to the m ach n u m b er M d escrib in g th e p la n e 's sp eed b y the h alf-an g le eq u atio n 6. ta n 0 = V 3 , ( o , f ) f) 1 sin Tr = ~h- (Example 5) 2 M 7. s i n 0 . § , ( f , * ) 8. cos 0 = - A ( * , f ) .a a il La Solve each equation on the interval [0, 2 tt]. (Example 2) 9. sin 29 = cos 6 a. E xp ress the m ach n u m b er o f the p lan e in term s 10. cos 26 = cos 6 o f cosine. b. U se the exp ressio n fo u n d in p a rt a to find the m ach 11. cos 29 — sin 9 = 0 n u m b er o f a p lan e if cos 6 = 12.' t a n 29 — t a n 2 6 ta n 2 6 = 2 13. sin 29 csc 6 = 1 Find the exact value of each expression. 14. ^fcos 2 0 + 4 cos 6 = —3 ( 15) GOLF A g olf b all is h it w ith an in itial v elo city o f 88 feet p er second . T h e d istan ce the b all trav els is foun d by v02 sin 20 d= — > w here vQis the in itial velocity, 6 is the angle th at the p ath o f the b a ll m akes w ith the gro u n d , and 32 is in feet p er second squared . (Example 2) 29. sin 67.5° 30. 31. tan 157.5° 32. sin cos^j H tt 12 Solve each equation on the interval [0, 2ir]. (Example 6) 33. sin — + cos 0 = 1 34. tan — = sin — 35. 2 sin oc ■ 2 0— — cos 0— = 36. 11 — sin = sin 0 Rewrite each product as a sum or difference. (Example 7) a. If the b all travels 242 feet, w h at is 6 to the nearest degree? b. U se a d o u b le-an gle id en tity to rew rite the eq u atio n for d. Rewrite each expression in terms with no power greater than 1. (Example 3) 16. c o s3 6 17. ta n 3 9 18. s ec4 9 19. c o t3 6 20. cos •sin 22. s in 2 6 — c o s2 9 352 | Lesson 5-5 21. s in 2 6 c o s 3 23. sin4 0 cos2 0 37. cos 3 0 cos 0 38.cos 12.r sin 5x 39. sin 3x cos 2x 40.sin 8 0 sin 0 Find the exact value of each expression. (Example 8) 41. 2 sin 135° sin 75° 42. cos 43. | sin 172.5° sin 127.5° 44. sin 142.5° cos 352.5° 45. sin 75° + sin 195° 46. 2 cos 105° + 2 cos 195° at o ■ 17tC r\ • TT 47. 3 sm - j y - 3 sin — 48. cos ^ + cos + cos Solve each equation. (Example 9) 49. cos 0 — cos 3 0 = 0 50. 2 cos 4 0 + 2 cos 2 0 = 0 5 1 . sin 3 0 + sin 5 0 = 0 52. sin 2 0 — sin 0 = 0 53. 3 cos 60 — 3 cos 4 0 = 0 54. 4 sin 0 + 4 sin 30 = 0 M u ltip le -A n gle and Product-to-Sum Identities Rewrite each expression in terms of cosines of multiple angles with no power greater than 1. Simplify each expression. 55 . J 1 + c° s6* 56. 71. s in 6 9 72. s in 8 9 73. cos7 9 74. sin 4 9 c o s4 9 Write each expression as a sum or difference. 57. cos (a + b) cos (a —b) 59. sin (b -I- 9) cos ( b + tt) 58. sin (0 — 7r) sin (9 + 7t) 75. MULTIPLE REPRESENTATIONS In th is p ro blem , you w ill in v estig ate h o w g rap h s o f fu n ctio n s can b e u sed to 60. cos (a — b) sin (b — a) find identities. a. GRAPHICAL U se a g rap h in g calcu lato r to grap h 61. MAPS A M ercato r p ro jectio n is a flat p ro jectio n o f the f( x ) = 4|sin 9 cos ~ — co s 9 sin globe in w h ich the d istan ce b etw een the lin es o f latitu d e on the interval [—27V, 2tt]. increases w ith th eir d istan ce from the equator. b. ANALYTICAL W rite a sin e fu n ctio n h(x) th at m od els the g rap h o t f( x ) . T h e n v e rify th at/ (x) = h(x) algebraically. C. GRAPHICAL U se a g rap h in g calcu lato r to graph g(x) = c o s 2 I# — y j — s in 2 ( o — y j on the interval [—2tv, 2tt], d. ANALYTICAL W rite a co sin e fu n ctio n k(x) th at m od els the g rap h o f g(x). T h e n v e rify th at g(x) = k(x) algebraically. H.O.T. Problem s The calcu latio n o f a p o in t o n a M ercato r p ro jectio n Use Higher-Order Thinking Skills 76. CHALLENGE V erify the fo llo w in g identity, con tain s the exp ressio n tan ^45° + j^J, w here £ is the latitu d e o f the point. sin 29 co s 9 — co s 29 sin 9 = sin 9 a. W rite the exp ressio n in term s o f sin i and cos 1. REASONING Consider an angle in the unit circle. Determine w hat quadrant a double angle and half angle would lie in if the terminal side of the angle is in each quadrant. b. Find the v alu e o f this exp ressio n if i = 60°. 62. BEAN BAG TOSS Iv an con stru cted a b e a n b a g to ssin g gam e as sh ow n in the figu re below . 77. I 78. II 79. Ill CHALLENGE Verify each identity. 80. sin 4 0 = 4 sin 9 co s 0 — 8 s in 3 0 cos 9 (81^) cos 4 0 = 1 — 8 s in 2 0 c o s 2 0 PROOF Prove each identity. a. E xactly h o w far w ill th e b a ck ed g e o f th e b o ard b e b. from the ground ? 83. tan2 0 = j ~ cos^ 1 -I- cos 29 E xactly h o w lo n g is the entire setu p? PROOF Prove each identity. 63. cos 29 = c o s2 9 — s in 2 9 64. cos 2 9 = 2 c o s 2 9 — 1 65. tan 2 9 = 66 . sin - 2 tan 8. 1 - tan2 9 67. ta n | = ± \ / | ^ i | 2 V 1 + cos 9 68. ta n | = 2 82. cos2 0 = 1 + c2os2e cos 9 sin 9 a 1 + cos e 84. cos a cos (3 = l[c o s (a —/3)+ cos (a + /3)] 85. sina 86. cos a sin /3 = -^-[sin (a + [3) — sin (a — (3)] 87. cos a cos /3 = -^-[sin ( a + f3)+ sin (a — f3)] + cos (3 = 2 cos cos 88. sin a — sin /3 = 2 cos j sin Verify each identity by using the pow er-reducing identities and then again by using the product-to-sum identities. 89. cos a — cos (3 = —2 sin |— 69. 2 c o s2 59 — 1 = cos 100 90. WRITING IN MATH D escrib e th e step s th at y o u w ou ld u se to 70. c o s2 29 — s in 2 29 = cos 49 sin j find th e e x a ct v alu e o f cos 8 0 if cos 0 = 1 353 Spiral Review Find the exact value of each trigonometric expression. (Lesson 5-4) 91 . cos y j 92-cos ^ 93. sin ^ 94. sin ^ 95.cos 96. sin 97. GARDENING E liza is w aitin g for the first d ay o f sp rin g in w h ich there w ill b e 1 4 h o u rs of d aylig h t to start a flow er garden. The n u m b er o f h o u rs o f d ay lig h t H in h e r to w n can be m o d eled b y H = 1 1 . 4 5 + 6 . 5 sin ( 0 . 0 1 6 8 d — 1 . 3 3 3 ) , w here d is the d ay o f the year, d = 1 rep resen ts Jan u ary 1 , d = 2 represents Ja n u a ry 2 , and so on. O n w h at d ay w ill E liza b egin gard ening ? (Lesson 5-3) Find the exact value of each expression. If undefined, write undefined. (Lesson 4-3) 98. csc 99. tan ( - j ) 210° 100. sin ^ 101. cos (-3 7 8 0 ° ) Graph and analyze each function. Describe the domain, range, intercepts, end behavior, continuity, and where the function is increasing or decreasing. (Lesson 2-1) 1 I i 102./(x) = - j r x 3 103. fi x ) = 4 x 4 1 0 4 ./(x) = —3 x 6 105./ (x) = 4 x 5 Skills Review for Standardized Tests 106. REVIEW Id en tify the eq u ation for the graph. 107. REVIEW F ro m a lo o k o u t p o in t o n a cliff abo v e a lake, the an g le o f d ep ressio n to a b o a t on the w ater is 12°. T h e b o a t is 3 k ilo m eters from the shore ju s t b elo w the cliff. W h a t is th e h e ig h t o f the cliff from the su rface o f the w ater to the lo o k o u t point? A y = 3 cos 29 1 B y = cos 29 3 C ¥ ¥ J C y = 3 cos 1 D y= 3 COS 3 ta n 12° 3 tan 12° FREE RESPONSE U se the grap h to an sw er each o f the follow ing. 108. a. W rite a fu n ctio n o f the fo rm f{ x ) = a cos {bx + c) + d that corresp on d s to the graph. b. R ew rite/ (x) as a sin e fun ction. C. R ew rite/ (x) as a cosin e fu n ctio n o f a sin gle angle. d. F in d all solu tio n s o f f( x ) = 0. e. H o w do the solu tio n s th at y ou fo u n d in p art d relate to the g rap h o f/ (x)? V. 354 | Lesson 5-5 | M u ltip le -A n gle and Product-to-Sum Identities Study Guide and Review Chapter Summary KeyConcepts Trigonometric Identities • • KeyVocabulary cofunction (Lesson 5- 1 ) (p. 337) Trigonom etric identities are identities th a t involve trigonom etric difference identity functions and can be used to find trigonom etric values. double-angle identity Trigonom etric expressions can be sim plified by w riting the expression half-angle identity in te rm s of one trigonom etric function or in te rm s of sine and cosine • p. 314) (p. 346) (p. 348) (p. 312) only. identity The m ost com m on trigonom etric odd-even identity identity is the Pythagorean power-reducing identity p. 347) product-to-sum identity (p. 350) identity sin2 9 + cos2 6 = 1. (p. 314) Pythagorean identity quotient identity Verifying Trigonometric Identities (p. 312) reduction identity (p. 340) (p. 337) trigonometric identity • Start with the more complicated side of the identity and work to transform it into the simpler side. • Use reciprocal, quotient, Pythagorean, and other basic trigonometric identities. • Use algebraic operations such as combining fractions, rewriting fractions as sums or differences, multiplying expressions, or factoring expressions. • Convert a denominator of the form 1 ± u or u ± 1 to a single term using its conjugate and a Pythagorean Identity. • Work each side separately to reach a common expression. Solving Trigonometric Equations (p. 312) reciprocal identity sum identity (Lesson 5- 2) (p. 313) verify an identity ip. 312) (P -320) VocabularyCheck Complete each identity by filling in the blank. Then name the identity. 1. s e c f l = ______________ (Lesson 5-3) • Algebraic techniques that can be used to solve trigonometric equations include isolating the trigonometric expression, taking the square root of each side, and factoring. • Trigonometric identities can be used to solve trigonometric equations by rewriting the equation using a single trigonometric function or by squaring each side to obtain an identity. 2. sin 9 cos 9 3. 1 = sec2 9 4. cos (9 0 ° - 9) ■ 5. tan ( — 0 ) = ___ Sum and Difference Identities (Lesson 5-4) 6 . s in ( o + /3 ) = s i n a . • Sum and difference identities can be used to find exact values of trigonometric functions of uncommon angles. 7. • Sum and difference identities can also be used to rewrite a trigonometric expression as an algebraic expression. 8. Multiple-Angle and Product-Sum Identities • (Lesson 5-5) Trigonometric identities can be used to find the values of expressions that otherwise could not be evaluated. 9. 10. cos a . . = cos2 a - sin 2 a + cos 9 1 - cos 29 . = ^-[cos ( a - (3) + cos (a + /3 )j .. . y /H connectED .m cgraw -hill.com | 355 'V Study Guide and Review Continued Lesson-by-Lesson Review Trigonometric Identities (pp. 312-319) Find the value of each expression using the given information. Example 1 11. sec 0 and cos 0; tan 0 = 3, cos 9 > 0 If sec 0 = —3 and sin 0 > 0, find sin 0. 1 12. cot 9 and sin 0; cos i tan 9 < 0 13. csc 0 and tan 0; cos 9 = -f, sin 9 < 0 0 14. cot 9 and cos 9; tan 9 = cos 0 = 1 3 sec 9 < 0 Now you can use the Pythagorean identity that includes sin 9 and cos 9 to find sin 9. cos‘ 17. sin2 ( - x ) + cos2 ( - x ) 21. Pythagorean Identity = 1 sin20 + ( _ l ) 2 = = 1 Simplify each expression. 18. sin2 x + cos2 x + cot2 * sec2 x - tan2 x COS sec 0 = - 3 3 sin‘ 19. Reciprocal Identity sec 9 csc 9 > 0 15. sec 9 and sin 0; cot 0 = - 2 , csc 0 < 0 16. cos 9 and sin 0; cot 9 = Since sin 0 > 0 and sec 9 < 0 , 9 must be in Quadrant II. To find sin O, first find cos 0 using the Reciprocal Identity for sec 9 and cos 9. 20 . (-X ) 1 22 . 1 - sin x 1 = 1 Multiply. sin2 9 = ^ Subtract. sin2 9 tan2x + 1 COS 0 = . 0„ = _V8 sm cosx 1 + sec x 2V2 o r _ 1 Simplify. Verifying Trigonometric Identities (pp. 3 2 0 - 3 2 6 ) Example 2 Verify each identity. sin 9 + i ±cose = 2csc0. 1 + cos 0 sin 0 23. ■+ . sin - „ = 2 csc i 1 - cos 9 1 + cos ( Verify that 24. cos 0 , sin 9■= 1 sec 9 csc 9 The left-hand side of this identity is more complicated, so start with that expression. 25. cot 9 , 1 + csc 9 1 + csc 9 cot 9 26. cos 9 1 - sin 9 1 + sin 9 cos 27. cot2 9 1 + csc 9 csc 0 — 1 28. S +Hf=SK(,+csc' 29. sec 9 + csc 1 1 + tan 9 - 2 sec 9 CSC I 30. cot 9 csc 9 + sec 9 = csc2 9 sec 9 31. sin 9 sin 9 + cos 9 32. COS4 tan 9 1 + tan 9 sin4 9 = 1 - tan2 9 sec‘ 1 + cos 9 _ sin2 0 + (1 + cos t sin 9 •+ sin 0 sin 0(1 + cos 9) 1 + cos _ sin2 0 + 1 + 2 cos 9 + cos2 9 sin 0(1 + cos 0) _ sin2 0 + cos2 0 + 1 + 2 cos 9 sin 0(1 + cos 0) _ 1 + 1 + 2 cos 9 ~ sin 0(1 + cos 0) _ 2 + 2 COS 0 ~ sin 0(1 + cos 0) 2(1 + cos 0) ~ sin 0(1 + cos 0) 2 sin 0 = 2 csc 0 356 C h a p te r 5 I Study G uide and Review Lesson-by-Lesson Review Solving Trigonometric Equations ............................................. (pp. 3 2 7 - 3 3 3 ) Find all solutions of each equation on the interval [ 0 ,2 - * ] . Example 3 Solve the equation sin 9 = 1 — cos 6 for all values of 6. 33. 2 s i n x = V 2 34. 4 c o s 2 x = 3 35. ta n 2 x - 3 = 0 36. 9 + c o t2 x = 1 2 37. 2 s in 2 x = sin x 38. 3 cos a-+ 3 = s in 2 * sin 6 = 1 - cos 6 Original equation. s in 2 9 = (1 - cos 9 )2 Square each side. s in 2 9 = 1 - 2 cos 9 + c o s 2 9 Expand. 1 - co s 2 0 = 1 - Solve each equation for all values of x. 2 cos 0 = 2 co s 2 39. sin 2 x - sin x = 0 9 + cos2 9 Pythagorean Identity 9 - 2 cos 9 Subtract. 0 = 2 cos 0 (c o s 0 - 1 ) 40. ta n 2 x = tan x Factor. Solve fo r x on [ 0 , 27r], 41. 3 cos x = cos x - 1 cos 0 = 0 or cos 0 = 1 42. sin 2 x = sin x + 2 0 = cos-1 0 9 = cos'“ 1 1 43. sin2 x = 1 - cos x 0 = | : Or ^ 0 = o Q'TTA c h eck show s th a t ~ is an extraneous solution. So the solutions 44. sin x = cos x + 1 are 0 = y + 2 itk or 0 = 0 + 2/77T. J v_____ ------------------------------- --------------- bum ana uitterence identities (pp. 3 3 6 - 3 4 3 ) Find the exact value of each trigonometric expression. Exam ple 4 45. cos 1 5 ° 46. sin 3 4 5 ° 47. t a n ^ y Find the exact value of tan 48. s i n ^ | 49. c o s - ^ 50. t a n - ^ | . tan 1 7 " = ' = " ( t tan ^ Simplify each expression. + 4 tan Sum Identity 3 1 - V3 52. cos 2 4 ° cos 3 6 ° - sin 2 4 ° sin 3 6 ° 1 - (-V 3 ) 53. sin 9 5 ° cos 5 0 ° - cos 9 5 ° sin 5 0 ° 1 - V3 5 4 . cos I j f cos J L + sin f sin % Verify each identity. 55. cos (0 + 3 0 ° ) - sin (0 + 6 0 °) = - s i n 0 ^ (cos 0 - sin 0 ) Evaluate for tangent. Simplify. 1 + \/3 1 - V3 1 56. cos ( 0 + 2 3 lt _ 5 tt , 2 it 12 4 3 ) t + tan ^ 1 - tan ta n f + ta n - f 51 ■ -------- --------------- 5 1 -te n -ta n L . OQ-rr + V3 1 - V3 Rationalize the denominator. 1 - V3 4-2V 3 1 -3 Multiply. z W 3 Simplify. 57. cos ( 0 - y j + cos ( 0 + y j = cos 0 58. tan ( 0 + 3? ) = Jan e ~ ] \ 4 1 tan 6 + 1 v j connectED.m cgraw-hill.com | 357 H Study Guide and Review l l ■ a Continued ■ Lesson-by-Lesson Review Multiple-Angle and Product-Sum Identities (pp. 346-354) Example 5 Find the values of sin 29, cos 26, and tan 26 for the given value and interval. Find the values of sin 29, cos 26, and tan 26 if 6 is in the 59. cos 6 = 1 (0°, 90°) 60.tan 6 = 2, (180°, 270°) fourth quadrant and tan 6 = — - y . 61. sin 6 = j , ( y , if ) 62.sec 0 = 9 is in the fourth quadrant, so cos 9 2nj sin 2 9 = 2 sin 9 cos 9 Find the exact value of each expression. 64. cos 65. tan 67.5° 66. c o s -^ 67. s i n ^ 68. tan cos 20 = 2 cos2 9 - 1 11 TT 63. sin 75° ^ a n d s in 0 = _ | . = 2(i) 12 tan 29 = 13tt ■ (-*) 2 tan e 1 - tan2 _ / _ 2 4 \2 I 7I 12 48 ' 7 or 336 527 _ 527 49 Applications and Problem Solving 69. CONSTRUCTION Find the tangent of the angle that the ramp makes with the building if sin 0 = (Lesson 5-1) V t4 5 and cos i 145 12VT45 145 ' 72. PROJECTILE MOTION A ball thrown with an initial speed v0 at an angle 9 that travels a horizontal distance d w ill remain in the air t d Suppose a ball is thrown with an seconds, where t v0 cos 9' initial speed of 50 feet per second, travels 100 feet, and is in the air for 4 seconds. Find the angle at which the ball was thrown. (Lesson 5-3) 73. BROADCASTING Interference occurs when two waves pass 70. LIGHT The intensity of light that emerges from a system of two /n polarizing lenses can be calculated by / = / „ - • 0 CSC2 -, where l0 is 9’ the intensity of light entering the system and 9 is the angle of the axis of the second lens with the first lens. Write the equation for the light intensity using only tan 9. (Lesson 5-1 71. MAP PROJECTIONS Stenographic projection is used to project the contours of a three-dimensional sphere onto a two-dimensional map. Points on the sphere are related to points on the map using sin a . Verify that r = 1 + c o s a . (Lesson 5-2) r= sin a 1 - cos a through the same space at the same time. It is destructive if the amplitude of the sum of the waves is less than the amplitudes of the individual waves. Determine whether the interference is destructive when signals modeled by y = 20 sin (31 + 45°) and y = 20 sin (3f + 225°) are combined. (Lesson 5-4) 74. TRIANGULATION Triangulation is the process of measuring a distance d using the angles a and (3 and the distance I using £ = d , d (Lesson 5-5) tan a + tan /3 d Aa I a. Solve the formula for d. t sin a sin (3 sin a cos (3 + cos a sin f3' b. Verify that d = — c. Verify that d = e sin a sin (3 sin (a + f3) ' d. Show that if a = (3, then d = 0.5£ tan a . 358 C h a p te r s Study G uide and Review ill! Practice Test Find the value of each expression using the given information. Find the exact value of each trigonometric expression. 1. sin 9 and cos 9, csc 6 = - 4 , cos 9 < 0 19. tan 165° 2. csc 9 and sec 9 , tan 9 = 20 . cos 5 csc 9 < 0 12 21. sin 75° Simplify each expression. 3. 22. cos 465° - cos 15° sin (90° - x) tan (90° - x) 23. 6 sin 675° - 6 sin 45° 4. sec2 x - 1 tan2 x + 1 5. 24. MULTIPLE CHOICE Which identity is true? sin 9 (1 + cot2 9) F cos (9 + 1 tt sec‘ 9 H sin ^0 - - y - j = cos 9 = 1 J sin cos 9 , 1 - sin9 2 cos9 1 + sin 9 cos < 1 + sin 9 1 1 + cos 9 = - s in G cos (7x - 9) = cos 9 Verify each identity. csc‘ csc‘ 9 tt) 1 1 - cos( : 2 CSC 9. - s e c 2 0 sin2 9 = cos2 6 ~ 1 cos2 9 A tan (-9 ) = - ta n 9 B tan (—0) = 1 cot (- 9 ) C tan ( - 9 ) ■ sin (-9) cos (-9) + 9) = sin 9 Simplify each expression. 25. cos ^ cos 26. 10. sin4 x - cos4 x = 2 sin2 x - 1 11. MULTIPLE CHOICE Which expression is not true? (tt - sin ~ sin tan 135° - tan 15° 1 + tan 135° tan 15° 27. PHYSICS A soccer ball is kicked from ground level with an initial speed of v at an angle of elevation 9. A.0 D tan ( -9 ) + 1 = sec (-9 ) -A Find all solutions of each equation on the interval [ 0 , 2n], a. The horizontal distance dthe ball will travel can be determined i/2 cin O/i using d = ^ , where g is the acceleration due to gravity. 12. V 2 sin 0 + 1 = 0 13. sec2 0 = | Verify that this expression is the same as | v2 (tan 9 - tan 9 sin2 9). Solve each equation for all values of 6. b. The maximum height h the object will reach can be determined using h = v ™ 14. tan2 9 - tan 9 = 0 15. 16. 1 - sin i COS I 1 sec 9 - 1 attained to the horizontal distance traveled. : COS I 1 = 2 sec 9 + 1 17. s e c 0 - 2 t a n 0 = O Find the values of sin 20, cos 26, and tan 26 for the given value and interval. 28. 18. CURRENT The current produced by an alternator is given by / = 40 sin 135-jrf, where I is the current in amperes and t is the time in seconds. At what time t does the current first reach 20 amperes? Round to the nearest ten-thousandths. Find the ratio of the maximum height tan 9 = - 3 , 2 tt| 29. cos 9 = i , (0°, 90°) 5 30. cos 9 = Connect to AP Calculus Rates of Change for Sine and Cosine' L,.*................... Objective Approximate rates of change for sine and cosine functions using the difference quotient. ** In Chapter 4, you learned that many real-world situations exhibit periodic behavior over time and thus, can be modeled by sinusoidal functions. Using transformations of the parent functions sin xand cos x, trigonometric models can be used to represent data, analyze trends, and predict future values. Hours of Daylight 1 16 i* * __ > 12 V) k. 3 X 8 i __ * ( __ __ < * i» n 4 0 J M M J Month S N While you are able to model real-world situations using graphs of sine and cosine, differential calculus can be used to determine the rate that the model is changing at any point in time. Your knowledge of the difference quotient, the sum identities for sine and cosine, and the evaluation of limits now makes it possible to discover the rates of change for these functions at any point in time. Activity 1 Approximate Rate of Change Approxim ate the rate of change of fix ) = sin x at several points. Substitute/(x) = sin x into the difference quotient. m ■ FfflTO f{x + h) - f( x ) h sin (x + h) —sin x h Approximate the rate of change of f i x ) at x = Repeat Steps 1 and 2 for x = 0 and for x = f TT Let h = 0.1,0.01, 0.001, and 0.0001. tt . A n a ly ze the Results 1. Use tangent lines and the graph of f i x ) = sin x to interpret the values found in Steps 2 and 3. 2. What will happen to the rate of change of f i x ) as x increases? V ....... ... ............... Unlike the natural base exponential function g(x) = ex and the natural logarithmic function h{x) = In x, an expression to represent the rate of change of fix) = sin x a t any point is not as apparent. However, we can substitute f(x) into the difference quotient and then simplify the expression. fix + h) - fix) m Difference quotient h sin {x + h) - sin x h (sin xcos h + cos / sin h) - sin x sin xcos h - sin x = sinx ( “ 360 C h a p te r 5 i ^ l ) + c o s x (^ ) fix) = sin x Sine Sum Identity Group terms with sin xand cos x. Factor sin x and cos x. We now have two expressions that involve h, sin x — -j and cos x obtainan accurateapproximation of the rate of change of f(x) at a point, we want h to be as close to 0 as possible. Recall that in Chapter 1, we were able to substitute h = 0 into an expression to find the exact slope of a function at a point. However, both of the fractional expressions are undefined at h = 0 . sin x ^ 1j expressions cosx Original undefined undefined We can approximate values for the two expressions by finding the limit of each as h approaches 0 using techniques discussed in Lesson 1 -3. Activity 2 Calculate Rate of Change Find an expression for the rate of change of f(x ) = sin x. cos h —1 Use a graphing calculator to estimate lim ■ ° h-> 0 h 'K Verify the value found in Step 1 by using the T A B L E function of your calculator. E5HEI sin h Repeat Steps 1 and 2 to estimate lim — h->0 h FIHfH Substitute the values found in Step 2 and Step 3 into the slope equation K=.(i66H4 [ - I T , tt] Vi -.0 0 3 0 -.0 0 2 0 B rans 0.0000 .00100 .00200 HHfH V = - .0 3 3 H 1 scl:-?- by [-1.5,1.5] scl: 1 .00300 Simplify the expression in Step 4. .00150 i.O E-3 E.0E-H ERROR "EE "H - I E "3 -.001E X= -.0 0 1 p A n a ly z e the Results 3. Find the rate of change of/(x) = sin x at x = 2, and 4. Make a conjecture as to why the rates of change for all trigonometric functions must be modeled by other trigonometric functions. M odel and Apply 5. In this problem, you will find an expression for the rate of change o f f (x) = cos x at any point x. a. Substitute/(x) = cos x into the difference quotient. b. Simplify the expression from part a. C. Use a graphing calculator to find the limit of the two fractional expressions as h approaches 0. d. Substitute the values found in part c into the slope equation found in part b. e. Simplify the slope equation in part d. f. Find the rate of change of f(x ) = cos x at x = 0, tt , 3 tt and 2 ' 361