Chapter 6 Additional Topics in Trigonometry

1111572836_0600.qxd 9/29/10 1:43 PM Page 403

Additional Topics 6 in Trigonometry

6.1 Law of Sines 6.2 Law of Cosines 6.3 Vectors in the Plane 6.4 Vectors and Dot Products 6.5 Trigonometric Form of a Complex Number Section 6.3, Example 10 Direction of an Airplane

Andresr 2010/used under license from Shutterstock.com 403

Copyright 2011 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 1111572836_0601.qxd 10/12/10 4:10 PM Page 404

404 Chapter 6 Additional Topics in Trigonometry

6.1 Law of Sines

Introduction What you should learn ● Use the Law of Sines to solve In Chapter 4, you looked at techniques for solving right triangles. In this section and the oblique triangles (AAS or ASA). next, you will solve oblique triangles—triangles that have no right angles. As standard ● Use the Law of Sines to solve notation, the angles of a triangle are labeled oblique triangles (SSA). A, , and CB ● Find areas of oblique triangles and use the Law of Sines to and their opposite sides are labeled model and solve real-life a, , and cb problems. as shown in Figure 6.1. Why you should learn it You can use the Law of Sines to solve C real-life problems involving oblique triangles. For instance, Exercise 46 b a on page 411 shows how the Law of Sines can be used to help determine A c B the distance from a boat to the shoreline. Figure 6.1 To solve an oblique triangle, you need to know the measure of at least one side and the measures of any two other parts of the triangle—two sides, two angles, or one angle and one side. This breaks down into the following four cases. 1. Two angles and any side (AAS or ASA) 2. Two sides and an angle opposite one of them (SSA) 3. Three sides (SSS) 4. Two sides and their included angle (SAS) The first two cases can be solved using the Law of Sines, whereas the last two cases can be solved using the Law of Cosines (see Section 6.2).

Law of Sines (See the proof on page 464.) If ABC is a triangle with sides a, b, and c, then a b c ϭ ϭ . sin A sin B sin C Oblique Triangles Study Tip C C Notice in Figure 6.2 a a that the height h of b h h b each triangle can be found using the formula

B h A c B A c ϭ sin A b A is acute. A is obtuse. Figure 6.2 or h ϭ b sin A.

The Law of Sines can also be written in the reciprocal form sin A sin B sin C ϭ ϭ . a b c lotusa 2010/used under license from Shutterstock.com Gabriela Trojanowska 2010/used under license from Shutterstock.com

Section 6.1 Law of Sines 405

Example 1 Given Two Angles and One Side—AAS For the triangle in Figure 6.3, C ϭ 102.3Њ, B ϭ 28.7Њ, andb ϭ 27.4 feet. Find the remaining angle and sides. C b = 27.4 ft Solution 102.3° a The third angle of the triangle is 28.7° ϭ ЊϪ Ϫ A 180 B C A c B ϭ 180ЊϪ28.7Њ Ϫ 102.3Њ Figure 6.3 ϭ 49.0Њ. Study Tip By the Law of Sines, you have When you are solving a b c ϭ ϭ . a triangle, a careful A B C sin sin sin sketch is useful as a Using b ϭ 27.4 produces quick test for the feasibility of an answer. Remember that b 27.4 ϭ ϭ Њ the longest side lies opposite a sin A Њ sin 49.0 43.06 feet sin B sin 28.7 the largest angle, and the and shortest side lies opposite the smallest angle. b 27.4 c ϭ sin C ϭ sin 102.3Њ 55.75 feet. sin B sin 28.7 Now try Exercise 9.

Example 2 Given Two Angles and One Side—ASA A pole tilts toward the sun at an 8Њ angle from the vertical, and it casts a 22-foot shadow. The angle of elevation from the tip of the shadow to the top of the pole is 43Њ. How tall is the pole? Solution In Figure 6.4,A ϭ 43Њ and C B ϭ 90Њϩ8Њϭ98Њ. So, the third angle is C ϭ ЊϪA Ϫ B b 180 a ° ϭ 180ЊϪ43Њ Ϫ 98Њ 8 ϭ 39Њ. ° By the Law of Sines, you have 43 B c = 22 ft A a c ϭ . sin A sin C Figure 6.4 Because c ϭ 22 feet, the length of the pole is c 22 a ϭ sin A ϭ sin 43Њ 23.84 feet. sin C sin 39Њ Now try Exercise 41.

For practice, try reworking Example 2 for a pole that tilts away from the sun under the same conditions.

406 Chapter 6 Additional Topics in Trigonometry The Ambiguous Case (SSA) In Examples 1 and 2, you saw that two angles and one side determine a unique triangle. However, if two sides and one opposite angle are given, then three possible situations can occur: (1) no such triangle exists, (2) one such triangle exists, or (3) two distinct triangles satisfy the conditions.

The Ambiguous Case (SSA) Consider a triangle in which you are given a, b, and A. (Notice that h ϭ b sin A. ) A is acute.A is acute.A is acute.A is acute.A is obtuse.A is obtuse. Sketch a a b b a b a h h a b h a h a b b

A A A A A A Necessary a < h a ϭ h a Ն b h < a < b a Յ b a > b condition Possible None One One Two None One triangles

Example 3 Single-Solution Case—SSA For the triangle in Figure 6.5, C a ϭ 22 inches,b ϭ 12 inches, and A ϭ 42Њ. b = 12 in. a = 22 in. 42° Find the remaining side and angles. A c B Solution Figure 6.5 One solution: a Ն b By the Law of Sines, you have sin B sin A ϭ Reciprocal form b a sin A sin B ϭ b Multiply each side by b. a sin 42Њ sin B ϭ 12 Substitute for A, a, and b. 22

B 21.41Њ. B is acute. Now you can determine that C 180ЊϪ42Њ Ϫ 21.41Њ ϭ 116.59Њ. Then the remaining side is given by c a ϭ Law of Sines sin C sin A a c ϭ sin C Multiply each side by sin C. sin A 22 c ϭ sin 116.59Њ Substitute for a, A, and C. sin 42Њ

c 29.40 inches. Simplify. Now try Exercise 19.

Section 6.1 Law of Sines 407

Example 4 No-Solution Case—SSA a = 15 Show that there is no triangle for which a ϭ 15, b ϭ 25, and A ϭ 85Њ.

Solution b = 25 h Begin by making the sketch shown in Figure 6.6. From this figure it appears that no triangle is formed. You can verify this by using the Law of Sines. 85° A sin B sin A ϭ Reciprocal form b a Figure 6.6 No solution: a < h sin A sin B ϭ b Multiply each side by a sin 85Њ sin B ϭ 25 1.6603 > 1 15 This contradicts the fact that sin B Յ 1. So, no triangle can be formed having sides a ϭ 15 and b ϭ 25 and an angle of A ϭ 85Њ. Now try Exercise 27.

Example 5 Two-Solution Case—SSA Find two triangles for which a ϭ 12 meters,b ϭ 31 meters, and A ϭ 20.5Њ. Study Tip Solution Because h ϭ b sin A ϭ 31sin 20.5Њ 10.86 meters, you can conclude that there are two When using the possible triangles (because h < a < b ). By the Law of Sines, you have Law of Sines, choose the sin B sin A form so that the ϭ Reciprocal form b a unknown variable is in the numerator. sin A sin 20.5Њ sin B ϭ b ϭ 31 0.9047. a 12 There are two angles Њ ЊϪ Њϭ Њ B1 64.8 and B2 180 64.8 115.2 Њ Њ Њ between 0 and 180 whose sine is 0.9047. For B1 64.8 , you obtain C 180ЊϪ20.5ЊϪ64.8Њϭ94.7Њ a 12 c ϭ sin C ϭ sin 94.7Њ 34.15 meters. sin A sin 20.5Њ Њ For B2 115.2 , you obtain C 180ЊϪ20.5ЊϪ115.2Њϭ44.3Њ a 12 c ϭ sin C ϭ sin 44.3Њ 23.93 meters. sin A sin 20.5Њ The resulting triangles are shown in Figure 6.7.

b = 31 m b = 31 m a = 12 m a = 12 m 115.2° 20.5° 64.8° 20.5° A B1 A B2 Figure 6.7 Two solutions: h < a < b Now try Exercise 29.

Andresr 2010/used under license from Shutterstock.com

408 Chapter 6 Additional Topics in Trigonometry Area of an Oblique Triangle The procedure used to prove the Law of Sines leads to a simple formula for the area of an oblique triangle. Referring to Figure 6.8, note that each triangle has a height of h ϭ b sin A. To see this when A is obtuse, substitute the reference angle 180ЊϪA for A. Now the height of the triangle is given by h ϭ b sin180ЊϪA. Using the difference formula for sine, the height is given by

h ϭ bsin 180Њ cos A Ϫ cos 180Њ sin A sinu Ϫ v ϭ sin u cos v Ϫ cos u sin v ϭ b0 и cos A Ϫ Ϫ1 и sin A ϭ b sin A. Consequently, the area of each triangle is given by 1 Area ϭ baseheight 2 1 ϭ cb sin A 2 1 ϭ bc sin A. 2 By similar arguments, you can develop the formulas 1 1 Area ϭ ab sin C ϭ ac sin B. 2 2

C C

a a b h h b

A B B c A c A is acute. A is obtuse. Figure 6.8

Area of an Oblique Triangle The area of any triangle is one-half the product of the lengths of two sides times the sine of their included angle. That is, 1 1 1 Area ϭ bc sin A ϭ ab sin C ϭ ac sin B. 2 2 2

Note that when angle A is 90Њ, the formula gives the area of a right triangle as 1 Area ϭ bc 2 1 ϭ baseheight. 2 Similar results are obtained for angles C and B equal to 90Њ.

Section 6.1 Law of Sines 409

Example 6 Finding the Area of an Oblique Triangle Find the area of a triangular lot having two sides of lengths 90 meters and 52 meters and an included angle of 102Њ. Solution Consider a ϭ 90 meters,b ϭ 52 meters, and C ϭ 102Њ, as shown in Figure 6.9. Then the area of the triangle is 1 Area ϭ ab sin C Formula for area 2 1 ϭ 9052sin 102Њ Substitute for a, b, and C. 2

2288.87 square meters. Simplify.

b = 52 m 102°

C a = 90 m Figure 6.9 Now try Exercise 35.

Example 7 An Application of the Law of Sines The course for a boat race starts at point A and proceeds in the direction S 52Њ W to N point B, then in the direction S 40Њ E to point C, and finally back to point A, as shown A in Figure 6.10. Point C lies 8 kilometers directly south of point A. Approximate the total W E distance of the race course. S 52° Solution B Because lines and ACBD are parallel, it follows that 8 km ° ЄBCA ЄDBC. 40 Consequently, triangle ABC has the measures shown in Figure 6.11. For angle B, you have ϭ ЊϪ ЊϪ Њϭ Њ B 180 52 40 88 . D C Using the Law of Sines Figure 6.10 a ϭ b ϭ c sin 52Њ sin 88Њ sin 40Њ A you can let b ϭ 8 and obtain c 8 52° a ϭ sin 52Њ 6.31 sin 88Њ B b = 8 km and a 40° 8 c ϭ sin 40Њ 5.15. sin 88Њ C The total length of the course is approximately Figure 6.11 Length 8 ϩ 6.31 ϩ 5.15 ϭ 19.46 kilometers. Now try Exercise 43.

410 Chapter 6 Additional Topics in Trigonometry

See www.CalcChat.com for worked-out solutions to odd-numbered exercises. 6.1 Exercises For instructions on how to use a graphing utility, see Appendix A.

Vocabulary and Concept Check In Exercises 1–4, fill in the blank(s). 1. An ______triangle is one that has no right angles. a c 2. Law of Sines:ϭ ______ϭ sin A sin C 3. To find the area of any triangle, use one of the following three formulas: Area ϭ ______, ______, or ______. 4. Two ______and one ______determine a unique triangle.

5. Which two cases can be solved using the Law of Sines? 6. Is the longest side of an oblique triangle always opposite the largest angle of the triangle? Procedures and Problem Solving Using the Law of Sines In Exercises 7–26, use the Law 22. B ϭ 40Њ, C ϭ 105Њ, c ϭ 20 of Sines to solve the triangle. 23. B ϭ 10Њ, C ϭ 135Њ, c ϭ 45 7. A 8. A 24. A ϭ 5Њ40Ј, B ϭ 8Њ15Ј, b ϭ 4.8 25. C ϭ 85Њ20Ј, a ϭ 35, c ϭ 50 35° c 26. B ϭ 2Њ45Ј, b ϭ 6.2, c ϭ 5.8 25° b c b 55° Using the Law of Sines In Exercises 27–30, use the Law C 18 mm B of Sines to solve the triangle. If two solutions exist, find 60° both. B 12 in. C 27. A ϭ 76Њ, a ϭ 18, b ϭ 20 9. B 10. C 28. A ϭ 110Њ, a ϭ 125, b ϭ 200 15° a 110° b 29. A ϭ 58Њ, a ϭ 11.4, b ϭ 12.8 40° a B A 30. A ϭ 58Њ, a ϭ 4.5, b ϭ 12.8 30 ft 20 cm 125° Using the Law of Sines In Exercises 31–34, find the C value(s) of b such that the triangle has (a) one solution, b (b) two solutions, and (c) no solution. A 31. A ϭ 36Њ, a ϭ 5 11. 12. 80°15′ A 88°35′ A ϭ Њ ϭ c 32. A 60 , a 10 c 2.8 km 50.2 yd 33. A ϭ 10Њ, a ϭ 10.8 B C B a C ° ′ a 25°30′ 22 45 34. A ϭ 88Њ, a ϭ 315.6

13. A ϭ 36Њ, a ϭ 8, b ϭ 5 Finding the Area of a Triangle In Exercises 35–40, find 14. A ϭ 76Њ, a ϭ 34, b ϭ 21 the area of the triangle having the indicated angle and 15. A ϭ 102.4Њ, C ϭ 16.7Њ, a ϭ 21.6 sides. 16. A ϭ 24.3Њ, C ϭ 54.6Њ, c ϭ 2.68 35. C ϭ 110Њ, a ϭ 6, b ϭ 10 17. A ϭ 110Њ 15Ј, a ϭ 48, b ϭ 16 36. B ϭ 130Њ, a ϭ 92, c ϭ 30 18. B ϭ 2Њ 45Ј, b ϭ 6.2, c ϭ 5.8 37. A ϭ 38Њ 45Ј, b ϭ 67, c ϭ 85 19. A ϭ 110Њ, a ϭ 125, b ϭ 100 38. A ϭ 5Њ 15Ј, b ϭ 4.5, c ϭ 22 ϭ Њ ϭ Њ ϭ 3 ϭ Њ Ј ϭ ϭ 20. A 55 , B 42 , c 4 39. B 75 15 , a 103, c 58 ϭ Њ ϭ Њ ϭ 5 ϭ Њ Ј ϭ ϭ 21. B 28 , C 104 , a 38 40. C 85 45 , a 16, b 20

Section 6.1 Law of Sines 411

41. Physics A flagpole at a right angle to the horizontal is 45. Environmental Science The bearing from the Pine located on a slope that makes an angle of 14Њ with the Knob fire tower to the Colt Station fire tower is horizontal. The flagpole casts a 16-meter shadow up the N 65Њ E, and the two towers are 30 kilometers apart. A slope when the angle of elevation from the tip of the fire spotted by rangers in each tower has a bearing of Њ shadow to the sun is 20 . N 80Њ E from Pine Knob and S 70Њ E from Colt Station. (a) Draw a triangle that represents the problem. Show Find the distance of the fire from each tower. the known quantities on the triangle and use a variable to indicate the height of the flagpole. N Colt Station (b) Write an equation involving the unknown quantity. W E S (c) Find the height of the flagpole. 70° 80° 30 km 42. Architecture A bridge is to be built across a small Fire 65° lake from a gazebo to a dock (see figure). The bearing Њ from the gazebo to the dock is S 41 W. From a tree 100 Pine Knob Not drawn to scale meters from the gazebo, the bearings to the gazebo and the dock are S 74Њ E and S 28Њ E, respectively. Find the 46. (p. 404) A boat is sailing distance from the gazebo to the dock. due east parallel to the shoreline at a speed of 10 miles per hour. At a given time the N Њ Tree bearing to a lighthouse is S 70 E, and ° 100 m W E Њ 74 B 15 minutes later the bearing is S 63 E (see S figure). The lighthouse is located at the 28° Gazebo shoreline. Find the distance d from the boat 41° to the shoreline.

N 63° ° W E Dock 70 d S 43. Aerodynamics A plane flies 500 kilometers with a bearing of 316Њ (clockwise from north) from Naples to Elgin (see figure). The plane then flies 720 kilometers from Elgin to Canton. (Canton is due west of Naples.) Find the bearing of the flight from Elgin to Canton. 47. Angle of Elevation A 10-meter telephone pole casts a 17-meter shadow directly down a slope when the angle N Elgin N of elevation of the sun is 42Њ (see figure). Find ␪, the W E angle of elevation of the ground. 720 km 500 km S A 44° ° 48 10 m Canton Naples Not drawn to scale 42° 42° − θ 44. Mechanical Engineering The circular arc of a C railroad curve has a chord of length 3000 feet and a B θ 17 m central angle of 40Њ. (a) Draw a diagram that visually represents the problem. ␪ ␾ Show the known quantities on the diagram and use 48. Aviation The angles of elevation and to an airplane the variables and sr to represent the radius of the are being continuously monitored at two observation arc and the length of the arc, respectively. points A and B, respectively, which are 2 miles apart, and the airplane is east of both points in the same (b) Find the radius r of the circular arc. vertical plane. (c) Find the length s of the circular arc. (a) Draw a diagram that illustrates the problem. lotusa 2010/used under license from Shutterstock.com Gabriela Trojanowska 2010/used under license from Shutterstock.com (b) Write an equation giving the distance d between the plane and point B in terms of ␪ and ␾.

412 Chapter 6 Additional Topics in Trigonometry

49. MODELING DATA Conclusions The Leaning Tower of Pisa in Italy leans because it True or False? In Exercises 51–53, determine whether was built on unstable soil—a mixture of clay, sand, and the statement is true or false. Justify your answer. water. The tower is approximately 58.36 meters tall from its foundation (see figure). The top of the tower 51. If any three sides or angles of an oblique triangle are leans about 5.45 meters off center. known, then the triangle can be solved. 52. If a triangle contains an obtuse angle, then it must be 5.45 m oblique. 53. Two angles and one side of a triangle do not necessarily β determine a unique triangle. 54. Writing Can the Law of Sines be used to solve a right triangle? If so, write a short paragraph explaining how α x 58.36 m to use the Law of Sines to solve the following triangle. Is there an easier way to solve the triangle? Explain. B ϭ 50Њ, C ϭ 90Њ, a ϭ 10 θ 55. Think About It Given A ϭ 36Њ and a ϭ 5, find values d Not drawn to scale of b such that the triangle has (a) one solution, (b) two solutions, and (c) no solution. (a) Find the angle of lean ␣ of the tower. (b) Write ␤ as a function of d and ␪, where ␪ is the 56. CAPSTONE In the figure, a triangle is to be formed angle of elevation to the sun. by drawing a line segment of length a from 4, 3 (c) Use the Law of Sines to write an equation for the to the positive x- axis. For what value(s) of a can length d of the shadow cast by the tower in terms you form (a) one triangle, (b) two triangles, and (c) no of ␪. triangles? Explain your reasoning. (d) Use a graphing utility to complete the table. y (4, 3) ␪ 10Њ 20Њ 30Њ 40Њ 50Њ 60Њ 3

d 2 a 1

50. Exploration In the figure,␣ and ␤ are positive angles. x (0, 0) 12345 γ 18 9 α β Cumulative Mixed Review c Evaluating Trigonometric Functions In Exercises 57 and (a) Write ␣ as a function of ␤. 58, use the given values to find (if possible) the values of ␪ (b) Use a graphing utility to graph the function. the remaining four trigonometric functions of . ␪ ϭ 5 ␪ ϭϪ12 Determine its domain and range. 57. cos 13, sin 13 (c) Use the result of part (b) to write c as a function 2 85 58. tan ␪ ϭ , csc ␪ ϭϪ of ␤. 9 2 (d) Use the graphing utility to graph the function in part (c). Determine its domain and range. Writing Products as Sums or Differences In Exercises (e) Use the graphing utility to complete the table. What 59–62, write the product as a sum or difference. can you conclude? 59. 6 sin 8␪ cos 3␪ 60. 2 cos 2␪ cos 5␪ ␤ 0.4 0.8 1.2 1.6 2.0 2.4 2.8 ␲ 5␲ 61. 3 cos sin ␣ 6 3 c 5 3␲ 5␲ 62. sin sin 2 4 6

Section 6.2 Law of Cosines 413

6.2 Law of Cosines

Introduction What you should learn ● Use the Law of Cosines to solve Two cases remain in the list of conditions needed to solve an oblique triangle—SSS and oblique triangles (SSS or SAS). SAS. To use the Law of Sines, you must know at least one side and its opposite angle. ● Use the Law of Cosines to model When you are given three sides (SSS), or two sides and their included angle (SAS), and solve real-life problems. none of the ratios in the Law of Sines would be complete. In such cases you can use the ● Use Heron’s Area Formula to find Law of Cosines. areas of triangles.

Law of Cosines (See the proof on page 465.) Why you should learn it You can use the Law of Cosines to Standard Form Alternative Form solve real-life problems involving b2 ϩ c 2 Ϫ a 2 oblique triangles. For instance, a2 ϭ b2 ϩ c 2 Ϫ 2bc cos A cos A ϭ 2bc Exercise 52 on page 418 shows you how the Law of Cosines can be used a 2 ϩ c 2 Ϫ b2 b2 ϭ a 2 ϩ c 2 Ϫ 2ac cos B cos B ϭ to determine the lengths of the guy 2ac wires that anchor a tower. a 2 ϩ b2 Ϫ c 2 c 2 ϭ a 2 ϩ b2 Ϫ 2ab cos C cos C ϭ 2ab

Example 1 Three Sides of a Triangle—SSS Find the three angles of the triangle shown in Figure 6.12.

a = 8 ft c = 14 ft

C b = 19 ft A Figure 6.12 Solution It is a good idea first to find the angle opposite the longest side—side b in this case. Using the alternative form of the Law of Cosines, you find that a 2 ϩ c 2 Ϫ b2 Explore the Concept cos B ϭ Alternative form 2ac What familiar formula 82 ϩ 142 Ϫ 192 do you obtain when ϭ Substitute for a, b, and c. you use the third form 2͑8͒͑14͒ of the Law of Cosines Ϸ Ϫ0.45089. Simplify. c2 ϭ a2 ϩ b2 Ϫ 2ab cos C Because cos B is negative, you know that B is an obtuse angle given by B Ϸ 116.80Њ. and you let C ϭ 90Њ? What is At this point it is simpler to use the Law of Sines to determine A. the relationship between the Law sin B sin 116.80Њ of Cosines and this formula? sin A ϭ a΂ ΃ Ϸ 8΂ ΃ Ϸ 0.37583 b 19 You know that A must be acute, because B is obtuse, and a triangle can have, at most, one obtuse angle. So, A Ϸ 22.08Њ and C Ϸ 180ЊϪ22.08ЊϪ116.80Њϭ41.12Њ Now try Exercise 7.

Dmitry Kalinovsky 2010/used under license from Shutterstock.com

414 Chapter 6 Additional Topics in Trigonometry

Do you see why it was wise to find the largest angle first in Example 1? Knowing the cosine of an angle, you can determine whether the angle is acute or obtuse. That is,

cos ␪ > 0 for 0Њ < ␪ < 90Њ Acute

cos ␪ < 0 for 90Њ < ␪ < 180Њ. Obtuse So, in Example 1, once you found that angle B was obtuse, you knew that angles A and C were both acute. Furthermore, if the largest angle is acute, then the remaining two angles are also acute.

Example 2 Two Sides and the Included Angle—SAS Find the remaining angles and side of the triangle shown in Figure 6.13.

b = 9 m a

25° A c = 12 m B Figure 6.13 Study Tip Solution Use the Law of Cosines to find the unknown side a in the figure. When solving an oblique triangle given a2 ϭ b2 ϩ c2 Ϫ 2bc cos A three sides, you use the a2 ϭ 92 ϩ 122 Ϫ 2͑9͒͑12͒ cos 25Њ alternative form of the Law of Cosines to solve for an angle. a2 Ϸ 29.2375 When solving an oblique a Ϸ 5.4072 triangle given two sides and their included angle, you use Because a Ϸ 5.4072 meters, you now know the ratio the standard form of the Law sin A of Cosines to solve for an a unknown side. and you can use the reciprocal form of the Law of Sines to solve for B. sin B sin A ϭ b a sin A sin B ϭ b΂ ΃ a sin 25Њ sin B Ϸ 9΂ ΃ 5.4072 sin B Ϸ 0.7034 There are two angles between 0Њ and 180Њ whose sine is 0.7034, Explore the Concept B Ϸ 44.7Њ and B Ϸ 180ЊϪ44.7Њϭ135.3Њ. 1 2 In Example 2, suppose Ϸ Њ For B1 44.7 , A ϭ 115Њ. After solving C Ϸ 180ЊϪ25ЊϪ44.7Њ ϭ 110.3Њ. for a, which angle 1 would you solve for next, B Ϸ Њ For B2 135.3 , or C? Are there two possible C Ϸ 180ЊϪ25ЊϪ135.3Њϭ19.7Њ. solutions for that angle? If so, 2 how can you determine which Because side c is the longest side of the triangle,C must be the largest angle of the angle is the correct measure? triangle. So,B Ϸ 44.7Њ and C Ϸ 110.3Њ. Now try Exercise 11.

Section 6.2 Law of Cosines 415 Applications

Example 3 An Application of the Law of Cosines The pitcher’s mound on a women’s softball field is 43 feet from home plate and the distance between the bases is 60 feet, as shown in Figure 6.14. (The pitcher’s mound is not halfway between home plate and second base.) How far is the pitcher’s mound from first base? 60 ft 60 ft Solution P h In triangle HPF, H ϭ 45Њ (line HP bisects the right angle at H ),f ϭ 43, and p ϭ 60. F Using the Law of Cosines for this SAS case, you have f = 43 ft h2 ϭ f 2 ϩ p2 Ϫ 2fp cos H Law of Cosines 60 ft 45° p = 60 ft ϭ 432 ϩ 602 Ϫ 2͑43͒͑60͒ cos 45º Substitute for H, f, and p. Ϸ 1800.33. Simplify. H So, the approximate distance from the pitcher’s mound to first base is Figure 6.14 h Ϸ Ί1800.33 Ϸ 42.43 feet. Now try Exercise 47.

Example 4 An Application of the Law of Cosines A ship travels 60 miles due east, then adjusts its course northward, as shown in Figure 6.15. After traveling 80 miles in the new direction, the ship is 139 miles from its point of departure. Describe the bearing from point B to point C.

N W E C S b = 139 mi

B a = 80 mi A Not drawn to scale c = 60 mi

Figure 6.15 Solution You have a ϭ 80, b ϭ 139, and c ϭ 60; so, using the alternative form of the Law of Cosines, you have a2 ϩ c2 Ϫ b2 cos B ϭ Alternative form 2ac 802 ϩ 602 Ϫ 1392 ϭ Substitute for a, b, and c. 2͑80͒͑60͒

Ϸ Ϫ0.97094. Simplify. So, B Ϸ arccos͑Ϫ0.97094͒ Ϸ 166.15Њ. Therefore, the bearing measured from due north from point to point CB is 166.15ЊϪ90Њϭ76.15Њ or N 76.15Њ E. Now try Exercise 49.

416 Chapter 6 Additional Topics in Trigonometry Heron’s Area Formula The Law of Cosines can be used to establish the following formula for the area of a triangle. This formula is called Heron’s Area Formula after the Greek mathematician Heron (ca. 100 B.C.).

Heron’s Area Formula (See the proof on page 466.) Given any triangle with sides of lengths a, b, and c, the area of the triangle is given by Area ϭ Ίs͑s Ϫ a͒͑s Ϫ b͒͑s Ϫ c͒ a ϩ b ϩ c where s ϭ . 2

Example 5 Using Heron’s Area Formula Find the area of the triangle shown in Figure 6.16.

C Explore the Concept b = 53 m a = 43 m Can the formulas at the bottom of the page be A c = 72 m B used to find the area of Figure 6.16 any type of triangle? Explain the advantages and disadvantages of Solution using one formula over another. Because a ϩ b ϩ c s ϭ 2 168 ϭ 2 ϭ 84 Heron’s Area Formula yields Area ϭ Ίs͑s Ϫ a͒͑s Ϫ b͒͑s Ϫ c͒ ϭ Ί84͑84 Ϫ 43͒͑84 Ϫ 53͒͑84 Ϫ 72͒ ϭ Ί84͑41͒͑31͒͑12͒ Ϸ 1131.89 square meters. Now try Exercise 55.

You have now studied three different formulas for the area of a triangle.

Formulas for Area of a Triangle ϭ 1 1. Standard Formula: Area 2bh ϭ 1 ϭ 1 ϭ 1 2. Oblique Triangle: Area 2bc sin A 2ab sin C 2ac sin B 3. Heron’s Area Formula: Area ϭ Ίs͑s Ϫ a͒͑s Ϫ b͒͑s Ϫ c͒

Amanda Rohde/iStockphoto.com

Section 6.2 Law of Cosines 417

See www.CalcChat.com for worked-out solutions to odd-numbered exercises. 6.2 Exercises For instructions on how to use a graphing utility, see Appendix A.

Vocabulary and Concept Check In Exercises 1 and 2, fill in the blank(s). a2 ϩ b2 Ϫ c2 1. The standard form of the Law of Cosines for cos C ϭ is ______. 2ab 2. Three different formulas for the area of a triangle are given by Areaϭ ______, ϭ 1 ϭ 1 ϭ 1 ϭ Area 2 bc sin A 2ab sin C 2ac sin B, and Area ______.

In Exercises 3– 6, one of the cases for the known measures of an oblique triangle is given. State whether the Law of Cosines can be used to solve the triangle. 3. ASA 4. SAS 5. SSS 6. AAS Procedures and Problem Solving Using the Law of Cosines In Exercises 7–24, use the Law Finding Measures in a Parallelogram In Exercises of Cosines to solve the triangle. 25–30, complete the table by solving the parallelogram 7. C 8. A shown in the figure. (The lengths of the diagonals are given by c and d. ) 16 in. 12 in. 12 cm 18 cm c AB18 in. φ B 8 cm C a θ d 9. C 10. C b 9.2 m 8.5 m 5.4 yd dcba ␪ ␾ 4.2 yd 25. 48 ᭿᭿ 30Њ ᭿ AB10.8 m A 2.1 yd B 26. 25 35 ᭿᭿᭿ 120Њ 27. 10 14 20 ᭿᭿᭿ 11. B 12. ° ′ B 10 mm 50 30 28. 40 60 80 ᭿᭿᭿ 20° C 12.5 ft 15 mm 10.4 ft 29. 15 25 20 ᭿᭿᭿ 30. ᭿ 25 50 35 ᭿᭿ A A C Solving a Triangle In Exercises 31–36, determine whether 13. a ϭ 6, b ϭ 8, c ϭ 12 the Law of Sines or the Law of Cosines can be used to find another measure of the triangle. Then solve the triangle. 14. a ϭ 9, b ϭ 3, c ϭ 11 ϭ ϭ ϭ Њ 15. A ϭ 50Њ, b ϭ 15, c ϭ 30 31. a 8, c 5, B 40 ϭ ϭ ϭ Њ 16. C ϭ 108Њ, a ϭ 10, b ϭ 7 32. a 10, b 12, C 70 ϭ Њ ϭ ϭ 17. a ϭ 9, b ϭ 12, c ϭ 15 33. A 24 , a 4, b 18 ϭ ϭ ϭ 18. a ϭ 45, b ϭ 30, c ϭ 72 34. a 11, b 13, c 7 ϭ Њ ϭ Њ ϭ 19. a ϭ 75.4, b ϭ 48, c ϭ 48 35. A 42 , B 35 , c 1.2 ϭ ϭ Њ ϭ Њ 20. a ϭ 1.42, b ϭ 0.75, c ϭ 1.25 36. a 160, B 12 , C 7 ϭ Њ Ј ϭ ϭ 21. B 8 15 , a 26, c 18 Using Heron’s Area Formula In Exercises 37–46, use 22. B ϭ 10Њ 35Ј, a ϭ 40, c ϭ 30 Heron’s Area Formula to find the area of the triangle. ϭ Њ Ј ϭ ϭ 23. B 75 20 , a 6.2, c 9.5 37. a ϭ 12, b ϭ 24, c ϭ 18 ϭ Њ Ј ϭ ϭ 24. C 15 15 , a 6.25, b 2.15 38. a ϭ 25, b ϭ 35, c ϭ 32

418 Chapter 6 Additional Topics in Trigonometry

39. a ϭ 5, b ϭ 8, c ϭ 10 50. Marine Transportation Two ships leave a port at 9 A.M. 40. a ϭ 13, b ϭ 17, c ϭ 8 One travels at a bearing of N 53Њ W at 12 miles per hour, and the other travels at a bearing of S 67Њ W at 16 miles 41. A 42. A per hour. Approximate how far apart the ships are at noon. 1.25 ft 2.75 cm 2.25 cm 51. Surveying A triangular parcel of ground has sides 2.45 ft B of lengths 725 feet, 650 feet, and 575 feet. Find the CB2.4 cm 1.24 ft measure of the largest angle. C 52. (p. 413) A 100-foot vertical 43. a ϭ 3.5, b ϭ 10.2, c ϭ 9 tower is to be erected on the side of a hill that makes a 6Њ angle with the horizontal (see 44. a ϭ 75.4, b ϭ 52, c ϭ 52 figure). Find the length of each of the two ϭ ϭ ϭ 45. a 10.59, b 6.65, c 12.31 guy wires that will be anchored 75 feet 46. a ϭ 4.45, b ϭ 1.85, c ϭ 3 uphill and downhill from the base of the tower. 47. Surveying To approximate the length of a marsh, a surveyor walks 380 meters from point A to point B. Then the surveyor turns 80Њ and walks 240 meters to point C (see figure). Approximate the length AC of the marsh. 100 ft B 80 ° 240 m 380 m

6 ° 75 ft C 75 ft A 53. Structural Engineering Q is the midpoint of the line 48. Geometry Determine the angle ␪ in the design of the segment PR in the truss rafter shown in the figure. What streetlight shown in the figure. are the lengths of the line segments PQ, QS, and RS? R

Q 3 10

S P θ 1 88 88 4 2 2 54. Architecture A retractable awning above a patio lowers at an angle of 50Њ from the exterior wall at a height of 10 feet above the ground (see figure). No direct sunlight is to enter the door when the angle of elevation of the sun 49. Geography On a map, Minneapolis is 165 millimeters is greater than 70Њ. What is the length x of the awning? due west of Albany, Phoenix is 216 millimeters from Minneapolis, and Phoenix is 368 millimeters from 50° Sunís rays Albany (see figure). x

Minneapolis 165 mm 10 ft Albany 216 mm 70° 368 mm

Phoenix 55. Architecture The Landau Building in Cambridge, Massachusetts has a triangular-shaped base. The lengths (a) Find the bearing of Minneapolis from Phoenix. of the sides of the triangular base are 145 feet, 257 feet, (b) Find the bearing of Albany from Phoenix. and 290 feet. Find the area of the base of the building. Dmitry Kalinovsky 2010/used under license from Shutterstock.com

Section 6.2 Law of Cosines 419

56. Geometry A parking lot has the shape of a parallelogram Conclusions (see figure). The lengths of two adjacent sides are 70 meters and 100 meters. The angle between the two True or False? In Exercises 59 and 60, determine whether sides is 70Њ. What is the area of the parking lot? the statement is true or false. Justify your answer. 59. Two sides and their included angle determine a unique triangle. 60. In Heron’s Area Formula,s is the average of the lengths 70 m of the three sides of the triangle.

Proof In Exercises 61 and 62, use the Law of Cosines to 70 ° prove the identity. 100 m 1 a ϩ b ϩ c Ϫa ϩ b ϩ c 61. bc͑1 ϩ cos A͒ ϭ ΂ ΃΂ ΃ 57. Mechanical Engineering An engine has a seven-inch 2 2 2 connecting rod fastened to a crank (see figure). 1 a Ϫ b ϩ c a ϩ b Ϫ c 62. bc͑1 Ϫ cos A͒ ϭ ΂ ΃΂ ΃ (a) Use the Law of Cosines to write an equation 2 2 2 giving the relationship between x and ␪. 63. Writing Describe how the Law of Cosines can be used (b) Write x as a function of ␪. (Select the sign that to solve the ambiguous case of the oblique triangle yields positive values of x. ) ABC, where a ϭ 12 feet,b ϭ 30 feet, and A ϭ 20Њ. Is (c) Use a graphing utility to graph the function in part (b). the result the same as when the Law of Sines is used to (d) Use the graph in part (c) to determine the total solve the triangle? Describe the advantages and the distance the piston moves in one cycle. disadvantages of each method. 3 in. 1.5 in . 7 in . 64. CAPSTONE Consider the cases SSS, AAS, ASA, SAS, and SSA. s θ d θ (a) For which of these cases are you unable to solve the triangle using only the Law of Sines? x 4 in. (b) For each case described in part (a), which form of 6 in. the Law of Cosines is most convenient to use?

Figure for 57 Figure for 58 65. Proof Use a half-angle formula and the Law of 58. MODELING DATA Cosines to show that, for any triangle, In a process with continuous paper, the paper passes C s͑s Ϫ c͒ cos΂ ΃ ϭ Ί across three rollers of radii 3 inches, 4 inches, and 2 ab 6 inches (see figure). The centers of the three-inch and where s ϭ 1͑a ϩ b ϩ c͒. six-inch rollers are d inches apart, and the length of the 2 arc in contact with the paper on the four-inch roller is 66. Proof Use a half-angle formula and the Law of s inches. Cosines to show that, for any triangle, C ͑s Ϫ a͒͑s Ϫ b͒ (a) Use the Law of Cosines to write an equation giving sin΂ ΃ ϭ Ί the relationship between d and ␪. 2 ab ϭ 1͑ ϩ ϩ ͒ (b) Write ␪ as a function of d. where s 2 a b c . (c) Write s as a function of ␪. Cumulative Mixed Review (d) Complete the table. Evaluating an Inverse Trigonometric Function In d (inches) 9 10 12 13 14 15 16 Exercises 67–70, evaluate the expression without using a calculator. ␪ (degrees) 67.arcsin͑Ϫ1͒ 68. cosϪ1 0 s (inches) Ί3 69.tanϪ1 Ί3 70. arcsin΂Ϫ ΃ 2

420 Chapter 6 Additional Topics in Trigonometry

6.3 Vectors in the Plane

Introduction What you should learn ● Represent vectors as directed Many quantities in geometry and physics, such as area, time, and temperature, can be line segments. represented by a single real number. Other quantities, such as force and velocity, ● Write the component forms of involve both magnitude and direction and cannot be completely characterized by a vectors. single real number. To represent such a quantity, you can use a directed line segment, \ ● Perform basic vector operations as shown in Figure 6.17. The directed line segment PQ has initial point P and terminal \ and represent vectors graphically. point Q. Its magnitude, or length, is denoted by PQ and can be found by using the ● Write vectors as linear Distance Formula. combinations of unit vectors. ● Find the direction angles of Q vectors. PQ Terminal ● point Use vectors to model and solve P real-life problems. Initia l point Why you should learn it Vectors are used to analyze numerous Figure 6.17 Figure 6.18 aspects of everyday life. Exercise 98 on page 431 shows you how vectors Two directed line segments that have the same magnitude and direction are equivalent. can be used to determine the tension For example, the directed line segments in Figure 6.18 are all equivalent. The set of \ in the cables of two cranes lifting an all directed line segments that are equivalent to a given directed line segment PQ is a object. vector v in the plane, written

\ v ϭ PQ. Vectors are denoted by lowercase, boldface letters such as , v, and wu .

Example 1 Equivalent Directed Line Segments Let u be represented by the directed line segment from y (4, 4) P0, 0 to Q3, 2 4 S v and let v be represented by the directed line segment from 3 (1, 2) (3, 2) R1, 2 to S4, 4 2 R Q u ϭ 1 as shown in Figure 6.19. Show that u v. (0, 0) x Solution P 1 234 From the Distance Formula, it follows that Figure 6.19 \ \ PQ and RS have the same magnitude.

\ PQ ϭ 3 Ϫ 02 ϩ 2 Ϫ 02 ϭ 13

\ RS ϭ 4 Ϫ 12 ϩ 4 Ϫ 22 ϭ 13 Moreover, both line segments have the same direction, because they are both directed toward the upper right on lines having the same slope.

\ 2 Ϫ 0 2 Slope of PQ ϭ ϭ 3 Ϫ 0 3

\ 4 Ϫ 2 2 Slope of RS ϭ ϭ 4 Ϫ 1 3

\ \ So,PQ and RS have the same magnitude and direction, and it follows that u ϭ v. Now try Exercise 11.

Patrick Hermans 2010/used under license from Shutterstock.com

Section 6.3 Vectors in the Plane 421 Component Form of a Vector The directed line segment whose initial point is the origin is often the most convenient representative of a set of equivalent directed line segments. This representative of the vector v is in standard position. A vector whose initial point is at the origin 0, 0 can be uniquely represented by the coordinates of its terminal point v1, v2 . This is the component form of a vector v , written as ϭ v v1, v2 .

The coordinates v1 and v2 are the components of v . If both the initial point and the terminal point lie at the origin, then v is the zero vector and is denoted by 0 ϭ 0, 0.

Component Form of a Vector The component form of the vector with initial point P p1, p2 and terminal point Q q1, q2 is given by \ ϭ Ϫ Ϫ ϭ ϭ PQ q1 p1, q2 p2 v1, v2 v. The magnitude (or length) of v is given by ϭ Ϫ 2 ϩ Ϫ 2 ϭ 2 ϩ 2 v q1 p1 q2 p2 v1 v2 . If v ϭ 1, then v is a unit vector. Moreover,v ϭ 0 if and only if v is the zero vector 0.

ϭ ϭ ϭ Two vectors u u1, u2 and v v1, v2 are equal if and only if u1 v1 and u ϭ v . For instance, in Example 1, the vector u from P0, 0 to Q3, 2 is 2 2 Technology Tip \ u ϭ PQ ϭ 3 Ϫ 0, 2 Ϫ 0 ϭ 3, 2 You can graph vectors and the vector v from R 1, 2 to S 4, 4 is with a graphing utility \ v ϭ RS ϭ 4 Ϫ 1, 4 Ϫ 2 ϭ 3, 2. by graphing directed line segments. Consult the user’s guide for your graphing Example 2 Finding the Component Form of a Vector utility for specific instructions. Find the component form and magnitude of the vector v that has initial point 4, Ϫ7 and terminal point Ϫ1, 5. Solution Let y P4, Ϫ7 ϭ p , p 6 1 2 Q(−1, 5) and Ϫ ϭ 2 Q 1, 5 q1, q2 x as shown in Figure 6.20. Then, the components −8 −6 −4 −2 2 4 6 ϭ − of v v1, v2 are 2 v ϭ Ϫ ϭ Ϫ Ϫ ϭϪ −4 v1 q1 p1 1 4 5 ϭ Ϫ ϭ Ϫ Ϫ ϭ −6 v2 q2 p2 5 7 12. − So, v ϭ Ϫ5, 12 and the magnitude of v is −8 P(4, 7) v ϭ Ϫ52 ϩ 122 ϭ 169 ϭ 13. Figure 6.20 Now try Exercise 15.

422 Chapter 6 Additional Topics in Trigonometry

1 2vvv −v − 3 v Vector Operations 2 2 The two basic vector operations are scalar multiplication and vector addition. Geometrically, the product of a vector v and a scalar k is the vector that is k times as long as v . If k is positive, then kv has the same direction as v , and if k is negative, then kv has the opposite direction of v , as shown in Figure 6.21. To add two vectors and vu geometrically, first position them (without changing their lengths or directions) so that the initial point of the second vector v coincides with the terminal point of the first vector u . The sum u ϩ v

is the vector formed by joining the initial point of the first vector u with the terminal Figure 6.21 point of the second vector v , as shown in Figure 6.22. This technique is called the parallelogram law for vector addition because the vector u ϩ v, often called the resultant of vector addition, is the diagonal of a parallelogram having and vu as its adjacent sides.

y y

u + v

u u

v x x

Figure 6.22

Definition of Vector Addition and Scalar Multiplication ϭ ϭ Let u u1, u2 and v v1, v2 be vectors and let k be a scalar (a real number). Then the sum of and vu is the vector ϩ ϭ ϩ ϩ u v u1 v1, u2 v2 Sum and the scalar multiple of k times u is the vector ϭ ϭ ku k u1, u2 ku1, ku2 . Scalar multiple

ϭ The negative of v v1, v2 is Ϫv ϭ Ϫ1v y ϭ Ϫ Ϫ v1, v2 Negative and the difference of and vu is

u Ϫ v ϭ u ϩ Ϫv Add Ϫv. See Figure 6.23.

ϭ u Ϫ v , u Ϫ v . Difference 1 1 2 2 − v u − v To represent u Ϫ v geometrically, you can use directed line segments with the same initial point. The difference u u Ϫ v is the vector from the terminal point of v to the terminal point of u , which is equal to v u + (−v) u ϩ Ϫv x as shown in Figure 6.23. Figure 6.23

Section 6.3 Vectors in the Plane 423

The component definitions of vector addition and scalar multiplication are illustrated y in Example 3. In this example, notice that each of the vector operations can be interpreted (−4, 10) 10 geometrically. 8 2v Example 3 Vector Operations 6 − Let v ϭ Ϫ2, 5 and w ϭ 3, 4, and find each of the following vectors. ( 2, 5)

a.2v b. w Ϫ v c. v ϩ 2w v

Solution x − − − − a. Because v ϭ Ϫ2, 5, you have 826 4 2 Figure 6.24 2 v ϭ 2Ϫ2, 5 y ϭ 2Ϫ2, 25 4 (3, 4) ϭ Ϫ4, 10. A sketch of 2v is shown in Figure 6.24. 3 w b. The difference of and vw is 2 −v w Ϫ v ϭ 3, 4 Ϫ Ϫ2, 5 1 ϭ 3 Ϫ Ϫ2, 4 Ϫ 5 x 345 w − v ϭ 5, Ϫ1. −1 (5, −1) Ϫ A sketch of w v is shown in Figure 6.25. Note that the figure shows the vector Figure 6.25 difference w Ϫ v as the sum w ϩ Ϫv. y c. The sum of and 2wv is 14 (4, 13) ϩ ϭ Ϫ ϩ v 2w 2, 5 2 3, 4 12 2w ϭ Ϫ2, 5 ϩ 23, 24 10 ϭ Ϫ2, 5 ϩ 6, 8 8 v + 2w ϭ Ϫ2 ϩ 6, 5 ϩ 8 (−2, 5)

ϭ 4, 13. v A sketch of v ϩ 2w is shown in Figure 6.26. x −626−4 −2 48 Now try Exercise 37. Figure 6.26

Vector addition and scalar multiplication share many of the properties of ordinary arithmetic.

Properties of Vector Addition and Scalar Multiplication Let , v, and wu be vectors and let and dc be scalars. Then the following properties are true. 1. u ϩ v ϭ v ϩ u 2. u ϩ v ϩ w ϭ u ϩ v ϩ w Study Tip 3. u ϩ 0 ϭ u 4. u ϩ Ϫu ϭ 0 Property 9 can be 5. cdu ϭ cdu 6. c ϩ du ϭ cu ϩ du stated as follows: The magnitude of the vector 7. cu ϩ v ϭ cu ϩ cv 8. 1u ϭ u, 0u ϭ 0 cv is the absolute value of c 9. cv ϭ c v times the magnitude of v .

424 Chapter 6 Additional Topics in Trigonometry Unit Vectors In many applications of vectors, it is useful to find a unit vector that has the same direction as a given nonzero vector v . To do this, you can divide v by its length to obtain v 1 u ϭ unit vector ϭ ϭ v. Unit vector in direction of v v v Note that u is a scalar multiple of v . The vector u has a magnitude of 1 and the same direction as v . The vector u is called a unit vector in the direction of v .

Example 4 Finding a Unit Vector Find a unit vector in the direction of v ϭ Ϫ2, 5 and verify that the result has a magnitude of 1. Solution The unit vector in the direction of v is v Ϫ2, 5 ϭ v Ϫ22 ϩ 52 1 ϭ Ϫ2, 5 4 ϩ 25 1 ϭ Ϫ2, 5 29 Ϫ2 5 ϭ , 29 29 Ϫ229 529 ϭ , . 29 29 This vector has a magnitude of 1 because Ϫ229 2 529 2 116 725 841 ϩ ϭ ϩ ϭ ϭ 1. 29 29 841 841 841 Now try Exercise 49.

The unit vectors 1, 0 and 0, 1 are called the standard unit vectors and are y denoted by

i ϭ 1, 0 and j ϭ 0, 1 2 as shown in Figure 6.27. (Note that the lowercase letter i is written in boldface to distinguish it from the imaginary number i ϭ Ϫ1. ) These vectors can be used to ϭ represent any vector v v1, v2 as follows. 1 j = 〈0, 1〉 ϭ v v1, v2 ϭ ϩ v1 1, 0 v2 0, 1 i = 〈1, 0〉 ϭ ϩ x v1i v2 j 12

The scalars v1 and v2 are called the horizontal and vertical components of v , respectively. The vector sum Figure 6.27 ϩ v1i v2 j is called a linear combination of the vectors and ji . Any vector in the plane can be written as a linear combination of the standard unit vectors and ji .

Section 6.3 Vectors in the Plane 425

Example 5 Writing a Linear Combination of Unit Vectors Let u be the vector with initial point 2, Ϫ5 and terminal point Ϫ1, 3. Write u as a linear combination of the standard unit vectors and ji . Solution Begin by writing the component form of the vector u . u ϭ Ϫ1 Ϫ 2, 3 Ϫ Ϫ5 ϭ Ϫ3, 8 ϭϪ3i ϩ 8j This result is shown graphically in Figure 6.28.

4 (−1, 3)

x −84−6 −4 −2 2 6 −2 u −4

−6 (2, −5)

Figure 6.28 Now try Exercise 63.

Example 6 Vector Operations Let u ϭϪ3i ϩ 8j and v ϭ 2i Ϫ j. Find 2u Ϫ 3v. Solution You could solve this problem by converting and vu to component form. This, however, is not necessary. It is just as easy to perform the operations in unit vector form. 2 u Ϫ 3v ϭ 2Ϫ3i ϩ 8j Ϫ 32i Ϫ j ϭϪ6i ϩ 16j Ϫ 6i ϩ 3j ϭϪ12i ϩ 19j Now try Exercise 69.

In Example 6, you could have performed the operations in component form. For instance, by writing u ϭϪ3i ϩ 8j ϭ Ϫ3, 8 and v ϭ 2i Ϫ j ϭ 2, Ϫ1 the difference of 2u and 3v is 2u Ϫ 3v ϭ 2Ϫ3, 8 Ϫ 32, Ϫ1 ϭ Ϫ6, 16 Ϫ 6, Ϫ3 ϭ Ϫ6 Ϫ 6, 16 Ϫ Ϫ3 ϭ Ϫ12, 19. Compare this result with the solution of Example 6.

426 Chapter 6 Additional Topics in Trigonometry Direction Angles y If u is a unit vector such that ␪ is the angle (measured counterclockwise) from the 1 positive x -axis to u , then the terminal point of u lies on the unit circle and you have (x, y) u ϭ x, y ϭ cos ␪, sin ␪ ϭ cos ␪i ϩ sin ␪j u y = sin θ ␪ θ as shown in Figure 6.29. The angle is the direction angle of the vector u . x Suppose that u is a unit vector with direction angle ␪. If v ϭ ai ϩ bj is any vector −1 x = cos θ 1 that makes an angle ␪ with the positive x -axis, then it has the same direction as u and you can write −1 v ϭ v cos ␪, sin ␪ ϭ v cos ␪i ϩ v sin ␪j. For instance, the vector v of length 3 that makes an angle of 30Њ with the positive x -axis is given by Figure 6.29 33 3 v ϭ 3cos 30Њi ϩ 3sin 30Њj ϭ i ϩ j 2 2 where v ϭ 3. Because v ϭ ai ϩ bj ϭ vcos ␪i ϩ vsin ␪j, it follows that the direction angle ␪ for v is determined from sin ␪ tan ␪ ϭ Quotient identity cos ␪ v sin ␪ ϭ Multiply numerator and denominator by v . v cos ␪ b ϭ . Simplify. a

Example 7 Finding Direction Angles of Vectors Find the direction angle of each vector. y (3, 3) a. u ϭ 3i ϩ 3j b. v ϭ 3i Ϫ 4j 3

Solution 2 u a. The direction angle is 1 b 3 θ = 45° tan ␪ ϭ ϭ ϭ 1. x a 3 1 2 3 So,␪ ϭ 45Њ, as shown in Figure 6.30. Figure 6.30 b. The direction angle is b Ϫ4 tan ␪ ϭ ϭ . a 3 y Moreover, because v ϭ 3i Ϫ 4j lies in Quadrant IV,␪ lies in Quadrant IV and its 1 θ = 306.87° reference angle is x 4 −13142 ␪Ј ϭ arctanϪ Ϫ53.13Њ ϭ 53.13Њ. −1 3 −2 v So, it follows that −3 ␪ 360ЊϪ53.13Њϭ306.87Њ −4 (3, −4) as shown in Figure 6.31. Figure 6.31 Now try Exercise 77.

Section 6.3 Vectors in the Plane 427 Applications Example 8 Finding the Component Form of a Vector Find the component form of the vector that represents the velocity of an airplane y descending at a speed of 100 miles per hour at an angle of 30Њ below the horizontal, as shown in Figure 6.32. 210° Solution x −100 −75 −50 The velocity vector v has a magnitude of 100 and a direction angle of ␪ ϭ 210Њ. v ϭ vcos ␪i ϩ vsin ␪j ϭ Њ ϩ Њ 100 100 cos 210 i 100 sin 210 j −50 3 1 ϭ 100Ϫ i ϩ 100Ϫ j 2 2 −75 ϭϪ503 i Ϫ 50j Figure 6.32 ϭ Ϫ503, Ϫ50 You can check that v has a magnitude of 100 as follows. v ϭ Ϫ5032 ϩ Ϫ502 ϭ 7500 ϩ 2500 ϭ 10,000

ϭ 100 Solution checks. ✓ Now try Exercise 97.

Example 9 Using Vectors to Determine Weight A force of 600 pounds is required to pull a boat and trailer up a ramp inclined at 15Њ from the horizontal. Find the combined weight of the boat and trailer. Solution Based on Figure 6.33, you can make the following observations.

\ BA ϭ force of gravity ϭ combined weight of boat and trailer

\ BC ϭ force against ramp

\ AC ϭ force required to move boat up ramp ϭ 600 pounds B W By construction,triangles BWD and ABC are similar. So,angle ABC is 15Њ. In triangle 15° D ABC you have 15° C \ A AC sin 15Њϭ \ BA 600 Figure 6.33 sin 15Њϭ \ BA

\ 600 BA ϭ sin 15Њ

\ BA 2318.

\ So, the combined weight is approximately 2318 pounds. (In Figure 6.33, note that AC is parallel to the ramp.) Now try Exercise 99.

428 Chapter 6 Additional Topics in Trigonometry

Example 10 Using Vectors to Find Speed and Direction An airplane is traveling at a speed of 500 miles per hour with a bearing of 330Њ at a fixed Study Tip altitude with a negligible wind velocity, as shown in Figure 6.34(a). As the airplane reaches a certain point, it encounters a wind blowing with a velocity of 70 miles per Recall from Section 4.8 hour in the direction N 45Њ E, as shown in Figure 6.34(b). What are the resultant speed that in air navigation, and direction of the airplane? bearings can be measured in degrees clockwise y y from north. In Figure 6.34, north is in the positive y -direction.

Wind v v1 v1 120° θ x x (a) (b) Figure 6.34 Solution Using Figure 6.34, the velocity of the airplane (alone) is ϭ Њ Њ v1 500 cos 120 , sin 120 ϭ Ϫ250, 2503 and the velocity of the wind is ϭ Њ Њ v2 70 cos 45 , sin 45 ϭ 352, 352. So, the velocity of the airplane (in the wind) is ϭ ϩ v v1 v2 ϭ Ϫ250 ϩ 352, 2503 ϩ 352 Ϫ200.5, 482.5 and the resultant speed of the airplane is vϪ200.52 ϩ 482.52 522.5 miles per hour. Finally, given that ␪ is the direction angle of the flight path and 482.5 tan ␪ Ϫ2.4065 Ϫ200.5 Airplane Pilot you have ␪ 180ЊϩarctanϪ2.4065 180ЊϪ67.4Њ ϭ 112.6Њ. You can use a graphing utility in degree mode to check this calculation, as shown in Figure 6.35. So, the true direction of the airplane is approximately 337.4Њ.

Figure 6.35 Now try Exercise 105.

Sean Prior 2010/used under license from Shutterstock.com

Section 6.3 Vectors in the Plane 429

See www.CalcChat.com for worked-out solutions to odd-numbered exercises. 6.3 Exercises For instructions on how to use a graphing utility, see Appendix A.

Vocabulary and Concept Check In Exercises 1–8, fill in the blank(s). 1. A ______can be used to represent a quantity that involves both magnitude and direction. \ 2. The directed line segment PQ has ______point P and ______point Q. \ \ 3. The ______of the directed line segment PQ is denoted by PQ . 4. The set of all directed line segments that are equivalent to a given directed line \ segment PQ is a ______v in the plane. 5. The directed line segment whose initial point is the origin is said to be in ______. 6. The two basic vector operations are scalar ______and vector ______. 7. The vector u ϩ v is called the ______of vector addition. ϩ 8. The vector sum v1i v2 j is called a ______of the vectors and ji , and the scalars v1 and v2 are called the ______and ______components of v , respectively. 9. What two characteristics determine whether two directed line segments are equivalent? 10. What do you call a vector that has a magnitude of 1? Procedures and Problem Solving Equivalent Directed Line Segments In Exercises 11 and 17. y 18. y v. 4 (3, 3) 4 ؍ show that u ,12 3 3 y y 2 11. 12. 2 (0, 4) 1 4 1 v x − 6 (6, 5) x 5 v 4 u (3, 3) −2−1 1 2 4 5 (−4, −1)−2 (3, −1) u − 4 v − 3 (2, 4) x 2 − (3, −2) 4 2 (4, 1) −4 2 4 −3 −5 (0, 0) v −2 x Initial Point Terminal Point −2 2 4 6 −4 − − − − 2 ( 3, 4) (0, 5) 19. Ϫ3, Ϫ5 5, 1 20. Ϫ3, 11 9, 40 Finding the Component Form of a Vector In Exercises 2 2 13–24, find the component form and the magnitude of 21. 5, 1 1, 5 7 Ϫ7 the vector v . 22. 2, 0 0, 2 Ϫ2 Ϫ 1 4 13. y 14. y 23. 3, 1 2, 5 24. 5, Ϫ2 1, 2 4 1 2 5 (1, 3) 3 x 2 3 4 Sketching the Graph of a Vector In Exercises 25–30, use 2 −1 the figure to sketch a graph of the specified vector. To v v 1 −2 print an enlarged copy of the graph, go to the website (4, −2) www.mathgraphs.com. x −3 −121 3 y 15. y 16. y − 5 6 (3, 5) ( 1, 4) 5 v 4 v 3 3 v u 2 2 x 1 (2, 2) 1 x x − − − − 2 1 2 3 4 5 3 2 1 1 2 3 (−1, −1) 25. Ϫv 26. 3u 27. u ϩ v 28. u Ϫ v ϩ Ϫ 1 29. u 2v 30. v 2u

430 Chapter 6 Additional Topics in Trigonometry

Sketching the Graph of a Vector In Exercises 31–36, use Finding a Vector In Exercises 57–62, find the vector v the figure to sketch a graph of the specified vector. To with the given magnitude and the same direction as u. print an enlarged copy of the graph, go to the website Magnitude Direction www.mathgraphs.com. 57. v ϭ 8 u ϭ 5, 6 y 58. v ϭ 3 u ϭ 4, Ϫ4 ϭ ϭ ϩ v 59. v 7 u 3i 4j ϭ ϭ Ϫ x 60. v 10 u 2i 3j u 61. v ϭ 8 u ϭϪ2i 62. v ϭ 4 u ϭ 5j

Writing a Linear Combination of Unit Vectors In Ϫ 31. 2u 32. 3v Exercises 63–66, the initial and terminal points of a vector ϩ 1 33. u 2v 34. 2v are given. Write the vector as a linear combination of the Ϫ 1 ϩ standard unit vectors i and j. 35. v 2u 36. 2u 3v Initial Point Terminal Point ,Vector Operations In Exercises 37–42, find (a) u ؉ v b) u ؊ v, (c) 2u ؊ 3v, and (d) v ؉ 4u. Then sketch each 63. Ϫ3, 1 4, 5) resultant vector. 64. 0, Ϫ2 3, 6 37. u ϭ 4, 2, v ϭ 7, 1 38. u ϭ 5, 3, v ϭ Ϫ4, 0 65. Ϫ1, Ϫ5 2, 3 39. u ϭ Ϫ6, Ϫ8, v ϭ 2, 4 66. Ϫ6, 4 0, 1 40. u ϭ 0, Ϫ5, v ϭ Ϫ3, 9 Vector Operations In Exercises 67–72, find the ϭ ϩ ϭ Ϫ ϭ Ϫ ϭϪ ϩ 41. u i j, v 2i 3j 42. u 2i j, v i j component form of v and sketch the specified vector 2i ؊ j and ؍ operations geometrically, where u .i ؉ 2j ؍ Writing a Vector In Exercises 43–46, use the figure and w write the vector in terms of the other two vectors. 67. v ϭ 3u 68. v ϭ 2 w y 2 3 69. v ϭ u ϩ 2w 70. v ϭϪu ϩ w ϭ 1 ϩ ϭ Ϫ 71. v 2 3u w 72. v 2u 2w u v w Finding Direction Angles of Vectors In Exercises 73–78, x find the magnitude and direction angle of the vector v. 73. v ϭ 5cos 30Њi ϩ sin 30Њj 43. w 44. v 74. v ϭ 8cos 135Њi ϩ sin 135Њj 45. u 46. 2v 75. v ϭ 6i Ϫ 6j 76. v ϭϪ4i Ϫ 7j 77. v ϭϪ2i ϩ 5j 78. v ϭ 12i ϩ 15j Finding a Unit Vector In Exercises 47–56, find a unit vector in the direction of the given vector. Verify that the Finding the Component Form of a Vector In Exercises result has a magnitude of 1. 79–86, find the component form of v given its magnitude 47. u ϭ 6, 0 and the angle it makes with the positive x-axis. Sketch v. ϭ Ϫ 48. u 0, 2 Magnitude Angle 49. v ϭ Ϫ 1, 1 79. v ϭ 3 ␪ ϭ 0Њ 50. v ϭ 3, Ϫ4 80. v ϭ 1 ␪ ϭ 45Њ 51. v ϭ Ϫ24, Ϫ7 7 81. v ϭ ␪ ϭ 150Њ ϭ Ϫ 2 52. v 8, 20 ϭ 5 ␪ ϭ Њ 82. v 2 45 53. v ϭ 4i Ϫ 3j 83. v ϭ 32 ␪ ϭ 150Њ 54. w ϭ i Ϫ 2j 84. v ϭ 43 ␪ ϭ 90Њ 55. w ϭ 2j 85. v ϭ 2 v in the direction i ϩ 3j 56. w ϭϪ3i 86. v ϭ 3 v in the direction 3i ϩ 4j

Section 6.3 Vectors in the Plane 431

Finding the Component Form of a Vector In Exercises 98. (p. 420) The cranes shown 87–90, find the component form of the sum of u and v in the figure are lifting an object that weighs ␪ ␪ with direction angles u and v. 20,240 pounds. Find the tension in the cable of each crane. Magnitude Angle ϭ ␪ ϭ Њ 87. u 5 u 60 ϭ ␪ ϭ Њ v 5 v 90 ϭ ␪ ϭ Њ 88. u 4 u 60 ϭ ␪ ϭ Њ v 4 v 90 ϭ ␪ ϭ Њ ° 89. u 20 u 45 44.5 ϭ ␪ ϭ Њ ° v 50 v 150 24.3 ϭ ␪ ϭ Њ 90. u 35 u 25 ϭ ␪ ϭ Њ v 50 v 120

Using the Law of Cosines In Exercises 91 and 92, use the Law of Cosines to find the angle ␣ between the vectors. Physics In Exercises 99 and 100, use the figure to .Assume 0؇ Յ ␣ Յ 180؇. determine the tension in each cable supporting the load 91. v ϭ i ϩ j, w ϭ 2i Ϫ j 99. 10 in. 20 in . 100. A 50° 30° B ϭ ϩ ϭ Ϫ 92. v 3i j, w 2i j B A C Graphing Vectors In Exercises 93 and 94, graph the 24 in . 2000 lb vectors and the resultant of the vectors. Find the magnitude and direction of the resultant. C 5000 lb 93. y 94. y 101. MODELING DATA 300 300 A loaded barge is being towed by two tugboats, and the ° 400 135° magnitude of the resultant is 6000 pounds directed 70 400 ° ° along the axis of the barge (see figure). Each tow line 25 x 25 x makes an angle of ␪ degrees with the axis of the barge.

Resultant Force In Exercises 95 and 96, find the angle between the forces given the magnitude of their resultant. θ (Hint: Write force 1 as a vector in the direction of the positive x -axis and force 2 as a vector at an angle ␪ with θ the positive x -axis.) Force 1 Force 2 Resultant Force (a) Write the resultant tension T of each tow line as a ␪ 95. 45 pounds 60 pounds 90 pounds function of . Determine the domain of the function. 96. 3000 pounds 1000 pounds 3750 pounds (b) Use a graphing utility to complete the table. 97. Physical Education A ball is thrown with an initial velocity of 70 feet per second, at an angle of 40Њ ␪ 10Њ 20Њ 30Њ 40Њ 50Њ 60Њ with the horizontal (see figure). Find the vertical and T horizontal components of the velocity.

ft (c) Use the graphing utility to graph the tension function. 70 sec (d) Explain why the tension increases as ␪ increases. 40°

Patrick Hermans 2010/used under license from Shutterstock.com

432 Chapter 6 Additional Topics in Trigonometry

102. MODELING DATA 104. MODELING DATA To carry a 100-pound cylindrical weight, two people A commercial jet is flying from Miami to Seattle. The lift on the ends of short ropes that are tied to an eyelet jet’s velocity with respect to the air is 580 miles per on the top center of the cylinder. Each rope makes an hour, and its bearing is 332Њ. The wind, at the altitude angle of ␪ degrees with the vertical (see figure). of the plane, is blowing from the southwest with a velocity of 60 miles per hour. (a) Draw a figure that gives a visual representation of θ θ the problem. (b) Write the velocity of the wind as a vector in component form.

100 lb (c) Write the velocity of the jet relative to the air as a vector in component form. (d) What is the speed of the jet with respect to the (a) Write the tension T of each rope as a function of ␪. ground? Determine the domain of the function. (e) What is the true direction of the jet? (b) Use a graphing utility to complete the table. 105. Aviation An airplane is flying in the direction ␪ Њ Њ Њ Њ Њ Њ 10 20 30 40 50 60 148Њ with an airspeed of 860 kilometers per hour. T Because of the wind, its groundspeed and direction are 800 kilometers per hour and 140Њ, respectively. Find (c) Use the graphing utility to graph the tension function. the direction and speed of the wind. (d) Explain why the tension increases as ␪ increases. y N 148° 140° W E 103. MODELING DATA x S Forces with magnitudes of 150 newtons and 220 newtons act on a hook (see figure).

y 800 kilometers per hour 150 newtons Wind 860 kilometers per hour

106. MODELING DATA θ 220 newtons A tetherball weighing 1 pound is pulled outward from x the pole by a horizontal force u until the rope makes an angle of ␪ degrees with the pole (see figure).

Tension (a) Find the direction and magnitude of the resultant of the forces when ␪ ϭ 30Њ. θ (b) Write the magnitude M of the resultant and the u direction ␣ of the resultant as functions of ␪, 1 lb where 0Њ Յ ␪ Յ 180Њ. (c) Use a graphing utility to complete the table.

␪ 0Њ 30Њ 60Њ 90Њ 120Њ 150Њ 180Њ (a) Write the tension T in the rope and the magnitude of u as functions of ␪. Determine the domains of M the functions. ␣ (b) Use a graphing utility to graph the two functions for 0Њ Յ ␪ Յ 60Њ. (d) Use the graphing utility to graph the two functions. (c) Compare T and u as ␪ increases. (e) Explain why one function decreases for increasing ␪, whereas the other does not.

Section 6.3 Vectors in the Plane 433

Conclusions 121. Proof Prove that cos ␪i ϩ sin ␪j is a unit vector for any value of ␪. True or False? In Exercises 107–110, determine whether the statement is true or false. Justify your answer. 122. Writing Write a program for your graphing utility that graphs two vectors and their difference given the 107. If and vu have the same magnitude and direction, then vectors in component form. u ϭ v. 123. Writing Give geometric descriptions of (a) vector 108. If u is a unit vector in the direction of v , then v ϭ vu. addition and (b) scalar multiplication. 109. If v ϭ a i ϩ bj ϭ 0, then a ϭϪb. 110. If u ϭ a i ϩ bj is a unit vector, then a 2 ϩ b2 ϭ 1. 124. CAPSTONE The initial and terminal points of vector v are 3, Ϫ4 and 9, 1, respectively. True or False? In Exercises 111–118, use the figure to (a) Write v in component form. determine whether the statement is true or false. Justify (b) Write v as the linear combination of the standard your answer. unit vectors i and j. (c) Sketch v with its initial point at the origin. b t (d) Find the magnitude of v. a d w c s Finding the Difference of Two Vectors In Exercises 125 u v and 126, use the program in Exercise 122 to find the difference of the vectors shown in the graph. 125.y 126. y ϭϪ 111. a d 8 112. c ϭ s (1, 6) 100 (80, 80) 6 (4, 5) (−20, 70) ϩ ϭ 113. a u c 4 (10, 60 ) ϩ ϭϪ (9, 4) 114. v w s 2 (5, 2) x − 115. a ϩ w ϭϪ2d x ( 100, 0) 50 2468 −50 116. a ϩ d ϭ 0 117. u Ϫ v ϭϪ2b ϩ t Cumulative Mixed Review 118. t Ϫ w ϭ b Ϫ a Simplifying an Expression In Exercises 127–132, 119. Think About It Consider two forces of equal simplify the expression. magnitude acting on a point. 6x 4 127. 14xϪ1y5 (a) If the magnitude of the resultant is the sum of the 7yϪ2 magnitudes of the two forces, make a conjecture 3sϪ2 about the angle between the forces. 128. 5s5tϪ5 50tϪ1 (b) If the resultant of the forces is 0, make a conjecture 129. 18x04xy23xϪ1 about the angle between the forces. 130. 5ab2aϪ3b02a0bϪ2 (c) Can the magnitude of the resultant be greater than 9 Ϫ4 the sum of the magnitudes of the two forces? 131. 2.1 ϫ 10 3.4 ϫ 10 Explain. 132. 6.5 ϫ 1063.8 ϫ 104 120. Exploration Consider two forces Solving an Equation In Exercises 133–136, solve the ϭ ϭ ␪ ␪ F1 10, 0 and F2 5 cos , sin . equation. ϩ ␪ (a) Find F1 F2 as a function of . 133. cos xcos x ϩ 1 ϭ 0 (b) Use a graphing utility to graph the function for 134. sin x2 sin x ϩ 2 ϭ 0 0 Յ ␪ < 2␲. 135. 3 sec x ϩ 4 ϭ 10 (c) Use the graph in part (b) to determine the range of 136. cos x cot x Ϫ cos x ϭ 0 the function. What is its maximum, and for what value of ␪ does it occur? What is its minimum, and for what value of ␪ does it occur? (d) Explain why the magnitude of the resultant is never 0.

434 Chapter 6 Additional Topics in Trigonometry

6.4 Vectors and Dot Products

The Dot Product of Two Vectors What you should learn ● Find the dot product of two So far you have studied two vector operations—vector addition and multiplication by a vectors and use the properties scalar—each of which yields another vector. In this section, you will study a third of the dot product. vector operation, the dot product. This product yields a scalar, rather than a vector. ● Find the angle between two vectors and determine whether Definition of Dot Product two vectors are orthogonal. ● Write vectors as the sums of The dot product of u u , u and v v , v is given by 1 2 1 2 two vector components. ● u v u1v1 u2v2. Use vectors to find the work done by a force. Why you should learn it Properties of the Dot Product (See the proofs on page 467.) You can use the dot product of two Let u, , and wv be vectors in the plane or in space and let c be a scalar. vectors to solve real-life problems involving two vector quantities. For 1. u v v u 2. 0 v 0 instance, Exercise 73 on page 441 3. u v w u v u w 4. v v v 2 shows you how the dot product can be used to find the force necessary 5. c u v cu v u cv to keep a truck from rolling down a hill.

Example 1 Finding Dot Products

a. 4, 5 2, 3 42 53 8 15 23 b. 2, 1 1, 2 21 12 2 2 0 c. 0, 3 4, 2 04 32 0 6 6 Now try Exercise 7.

In Example 1, be sure you see that the dot product of two vectors is a scalar (a real number), not a vector. Moreover, notice that the dot product can be positive, zero, or negative.

Example 2 Using Properties of Dot Products Let u 1, 3, v 2, 4, and w 1, 2. Use the vectors and the properties of the dot product to find the indicated quantity. Study Tip a. u vw b. u 2v c. u In Example 2, notice that the product in part Solution (a) is a vector, whereas Begin by finding the dot product of and vu and the dot product of and uu . the product in part (b) is a scalar. Can you see why? u v 1, 3 2, 4 12 34 14 u u 1, 3 1, 3 11 33 10 a. u vw 141, 2 14, 28 b. u 2v 2u v 214 28 c. Because u2 u u 10, it follows that u u u 10. Now try Exercise 15.

Franck Boston 2010/used under license from Shutterstock.com

Section 6.4 Vectors and Dot Products 435 The Angle Between Two Vectors The angle between two nonzero vectors is the angle , 0 , between their respective standard position vectors, as shown in Figure 6.36. This angle can be found using the dot product. (Note that the angle between the zero vector and another vector is not defined.)

v − u

u θ v Origin Figure 6.36

Angle Between Two Vectors (See the proof on page 467.) If is the angle between two nonzero vectors and vu , then u v cos . uv

Example 3 Finding the Angle Between Two Vectors Find the angle between u 4, 3 and v 3, 5. Solution

u v cos u v 4, 3 3, 5 4, 3 3, 5 27 534 This implies that the angle between the two vectors is 27 arccos 22.2 534 as shown in Figure 6.37.

6 5 v = 〈3, 5〉 4 3 22.2° 2 θ u = 〈4, 3〉 1 x 1234 56 Figure 6.37 Now try Exercise 23.

436 Chapter 6 Additional Topics in Trigonometry

Rewriting the expression for the angle between two vectors in the form

u v u v cos Alternative form of dot product produces an alternative way to calculate the dot product. From this form, you can see that because u and v are always positive, u v and cos will always have the same sign. Figure 6.38 shows the five possible orientations of two vectors.

u u u θ θ θ θ v uv v v v u ؍ ␲ ␪ ␲ ␪ ؍ ␲ ␲ ␪ ␲ ␪ ؍ ␪ 2 < < 2 0 < < 2 0 1 ؍ cos ␪ < 1 cos ␪ > 0 0 ؍ ؊1 ؊1 < cos ␪ < 0 cos ␪؍ cos ␪ Opposite direction Obtuse angle 90° angle Acute angle Same direction Figure 6.38

Definition of Orthogonal Vectors Technology Tip The vectors and vu are orthogonal when u v 0. The graphing utility program Finding the The terms orthogonal and perpendicular mean essentially the same thing—meeting Angle Between Two at right angles. Even though the angle between the zero vector and another vector is not Vectors, found at this textbook’s defined, it is convenient to extend the definition of orthogonality to include the zero Companion Website, graphs vector. In other words, the zero vector is orthogonal to every vector u because 0 u 0. two vectors u a, b and v c, d in standard position Example 4 Determining Orthogonal Vectors and finds the measure of the angle between them. Use the Are the vectors program to verify Examples 3 u 2, 3 and v 6, 4 and 4. orthogonal? Solution Begin by finding the dot product of the two vectors. u v 2, 3 6, 4 26 34 0 Because the dot product is 0, the two vectors are orthogonal, as shown in Figure 6.39.

4 v = 〈6, 4〉 3 2 1 x 1234567 −1 〈 − 〉 −2 u = 2, 3 −3

Figure 6.39 Now try Exercise 43.

studio_chki 2010/used under license from Shutterstock.com

Section 6.4 Vectors and Dot Products 437 Finding Vector Components You have already seen applications in which two vectors are added to produce a resultant vector. Many applications in physics and engineering pose the reverse problem—decomposing a given vector into the sum of two vector components. Consider a boat on an inclined ramp, as shown in Figure 6.40. The force F due to gravity pulls the boat down the ramp and against the ramp. These two orthogonal

forces,w1 and w2, are vector components of F . That is, w1 F w1 w2. Vector components of F w2 The negative of component w1 represents the force needed to keep the boat from rolling F down the ramp, and w represents the force that the tires must withstand against the 2 Figure 6.40 ramp. A procedure for finding w1 and w2 is shown below.

Definition of Vector Components Let and vu be nonzero vectors such that u w2 u w1 w2

where w1 and w2 are orthogonal and w1 is parallel to (or a scalar multiple of) v , as shown in Figure 6.41. The vectors w1 and w2 are called vector components θ v of u. The vector w1 is the projection of onto vu and is denoted by w1 w1 projvu. ␪ is acute.

The vector w2 is given by w2 u w1. u w2 From the definition of vector components, you can see that it is easy to find the

component w2 once you have found the projection of onto vu . To find the projection, θ you can use the dot product, as follows. v

w1 u w1 w2 ␪ is obtuse. w v u cv w2 1 is a scalar multiple of . Figure 6.41 u v cv w2 v Take dot product of each side with v . u v cv v w2 v 2 u v c v 0 2 and vw are orthogonal. So, u v c v 2 and u v w proj u cv v. 1 v v 2

Projection of onto vu Let and vu be nonzero vectors. The projection of onto vu is given by u v projvu v. v 2

438 Chapter 6 Additional Topics in Trigonometry

Example 5 Decomposing a Vector into Components Find the projection of y v = 〈6, 2〉 u 3, 5 onto v 6, 2. 2 Then write u as the sum of two orthogonal vectors, one of which is proj u. 1 v w1 x Solution −1 1 2 3 4 5 6 −1 The projection of onto vu is −2 u v 8 6 2 w2 w proj u v 6, 2 , 1 v v2 40 5 5 −3 −4 as shown in Figure 6.42. The other component,w2, is −5 u = 〈3, −5〉 6 2 9 27 w2 u w1 3, 5 , , . 5 5 5 5 Figure 6.42 So, u w1 w2 6 2 9 27 , , 5 5 5 5 3, 5.

Now try Exercise 57.

Example 6 Finding a Force A 200-pound cart sits on a ramp inclined at 30, as shown in Figure 6.43. What force is required to keep the cart from rolling down the ramp? Solution Because the force due to gravity is vertical and downward, you can represent the gravitational force by the vector

F 200j. Force due to gravity v w1 To find the force required to keep the cart from rolling down the ramp, project F onto 30° a unit vector v in the direction of the ramp, as follows. F 3 1 Figure 6.43 v cos 30i sin 30j i j Unit vector along ramp 2 2 Therefore, the projection of onto vF is F v w1 projvF v v2 F vv 1 200 v 2 3 1 100 i j. 2 2 The magnitude of this force is 100, and therefore a force of 100 pounds is required to keep the cart from rolling down the ramp. Now try Exercise 73.

Section 6.4 Vectors and Dot Products 439 Work The work W done by a constant force F acting along the line of motion of an object is given by

\ W magnitude of forcedistance F PQ as shown in Figure 6.44. If the constant force F is not directed along the line of motion (see Figure 6.45), then the work W done by the force is given by

\ \ W proj PQ F PQ Projection form for work

\ \ cos F PQ proj PQ F cos F

\ F PQ. Dot product form for work

F F θ

proj PQ F P Q P Q Force acts along the line of motion. Force acts at angle ␪ with the line of motion. Figure 6.44 Figure 6.45 This notion of work is summarized in the following definition.

Definition of Work The work W done by a constant force F as its point of application moves along \ the vector PQ is given by either of the following.

\ \ 1. W projPQ F PQ Projection form

\ 2. W F PQ Dot product form

Example 7 Finding Work To close a barn’s sliding door, a person pulls on a rope with a constant force of 50 pounds at a constant angle of 60, as shown in Figure 6.46. Find the work done in moving the door 12 feet to its closed position. 12 ft Solution proj F P PQ Q Using a projection, you can calculate the work 60° as follows. \ F \ W projPQ F PQ \ cos 60 F PQ 12 ft 1 Figure 6.46 5012 2 300 foot-pounds So, the work done is 300 foot-pounds. You can verify this result by finding the vectors \ F and PQ and calculating their dot product. Now try Exercise 75.

440 Chapter 6 Additional Topics in Trigonometry

See www.CalcChat.com for worked-out solutions to odd-numbered exercises. 6.4 Exercises For instructions on how to use a graphing utility, see Appendix A.

Vocabulary and Concept Check 1. For two vectors u and v, does u v v u? 2. What is the dot product of two orthogonal vectors? 3. Is the dot product of two vectors an angle, a vector, or a scalar?

In Exercises 4–6, fill in the blank(s). 4. If is the angle between two nonzero vectors u and v, then cos ______. 5. The projection of onto vu is given by projvu ______. 6. The work W done by a constant force F as its point of application moves along \ the vector PQ is given by either W ______or W ______. Procedures and Problem Solving Finding a Dot Product In Exercises 7–10, find the dot 30. u cos i sin j product of and v.u 4 4 7. u 6, 3 8. u 4, 1 2 2 v cos i sin j v 2, 4 v 2, 3 3 3 9. u 5i j 10. u 3i 2j Finding the Angle Between Two Vectors In Exercises v 3i j v 2i j 31–34, graph the vectors and find the degree measure of the angle between the vectors. Using Properties of Dot Products In Exercises 11–16, ؊3, 4>, and 31. u 2i 4j 32. u 6i 3j> ؍ v ,<2 ,2> ؍ use the vectors u ؊4> to find the indicated quantity. State v 3i 5j v 8i 4j ,1> ؍ w whether the result is a vector or a scalar. 33. u 6i 2j 34. u 2i 3j 11. u u 12. v w v 8i 5j v 4i 3j 13. u 2v 14. 4u v 15. 3w vu 16. u 2vw Finding the Angles in a Triangle In Exercises 35–38, use vectors to find the interior angles of the triangle with the Finding the Magnitude of a Vector In Exercises 17–22, given vertices. use the dot product to find the magnitude of u. 35. 1, 2, 3, 4, 2, 5 36. 3, 4, 1, 7, 8, 2 17. u 5, 12 18. u 2, 4 37. 3, 0, 2, 2, 0, 6) 38. 3, 5, 1, 9, 7, 9 19. u 20i 25j 20. u 6i 10j Using the Angle Between Two Vectors In Exercises 21. u 4j 22. u 9i 39–42, find u v , where ␪ is the angle between and v.u Finding the Angle Between Two Vectors In Exercises 3 39. u 9, v 36, 23–30, find the angle ␪ between the vectors. 4 23. u 1, 0 24. u 4, 4 40. u 4, v 12, v 0, 2 v 2, 0 3 2 25. u 3i 4j 26. u 2i 3j 41. u 4, v 10, 3 v 2i 3j v i 2j 27. u 2i 28. u 4j 42. u 100, v 250, 6 v 3j v 3i 29. u cos i sin j Determining Orthogonal Vectors In Exercises 43–46, 3 3 determine whether and vu are orthogonal. 3 3 v cos i sin j 43. u 10, 6 44. u 12, 4 4 4 1 1 v 9, 15 v 4, 3

Section 6.4 Vectors and Dot Products 441

45. u j 46. u 2i 2j Finding Orthogonal Vectors In Exercises 65–68, find v i j v i j two vectors in opposite directions that are orthogonal to the vector u. (There are many correct answers.) A Relationship of Two Vectors In Exercises 47–50, 65. u 2, 6 66. u 7, 5 determine whether and vu are orthogonal, parallel, or 67. u 1i 3 j 68. u 5i 3j neither. 2 4 2 47. u 10, 20 48. u 15, 9 Finding Work In Exercises 69 and 70, find the work v 5, 10 v 5, 3 done in moving a particle from to QP if the magnitude 3 7 9 and direction of the force are given by v . 49. u 5 i 10j 50. u 10 i 3j 3 69. P 0, 0, Q 4, 7, v 1, 4 v 12i 14j v 5i 2 j 70. P 1, 3, Q 3, 5, v 2i 3j Finding an Unknown Vector Component In Exercises 51–56, find the value of k such that the vectors and vu 71. Business The vector u 1225, 2445 gives the are orthogonal. numbers of hours worked by employees of a temp agency at two pay levels. The vector v 12.20, 8.50 51. u 2i kj 52. u 3i 2j gives the hourly wage (in dollars) paid at each level, v 3i 2j v 2i kj respectively. (a) Find the dot product u v and explain 53. u i 4j 54. u 3ki 5j its meaning in the context of the problem. (b) Identify the vector operation used to increase wages by 2 percent. v 2ki 5j v 2i 4j 72. Business The vector u 3240, 2450 gives the 55. u 3ki 2j 56. u 4i 4kj numbers of hamburgers and hot dogs, respectively, sold v 6i v 3j at a fast food stand in one week. The vector v 1.75, 1.25 gives the prices in dollars of the food Decomposing a Vector into Components In Exercises items. (a) Find the dot product u v and explain its 57–60, find the projection of ontovu . Then write u as the meaning in the context of the problem. (b) Identify the sum of two orthogonal vectors, one of which is proj u. 1 v vector operation used to increase prices by 22 percent. 57. u 3, 4 58. u 4, 2 73. (p. 434) A truck with a v 8, 2 v 1, 2 gross weight of 30,000 pounds is parked on a slope of d (see figure). Assume that the only 59. u 0, 3 60. u 5, 1 force to overcome is the force of gravity. v 2, 15 v 1, 1

Finding the Projection of u onto v Mentally In Exercises 61–64, use the graph to determine mentally the projection of ontovu . (The coordinates of the terminal points of the vectors in standard position are given.) Use the formula for the projection of ontovu to verify your result. d° 61. y 62. y Weight = 30,000 lb 5 (6, 4) (6, 4) 4 4 (a) Find the force required to keep the truck from 2 v 3 (3, 2) v rolling down the hill in terms of the slope d. 2 u x (b) Use a graphing utility to complete the table. 1 u 246 x (−3, −2) −1 135246 d 0 1 2 3 4 5

63. y 64. y Force 6 (6, 4) (6, 4) 4 d 6 7 8 9 10 (−2, 3) v 2 v u Force x x −2 264 −2 2 46 −2 −2 u (2, −3) (c) Find the force perpendicular to the hill when d 5. − 4 Franck Boston 2010/used under license from Shutterstock.com

442 Chapter 6 Additional Topics in Trigonometry

74. Physics A sport utility vehicle with a gross weight of 80. The work W done by a constant force F acting along the 5400 pounds is parked on a slope of 10. Assume that line of motion of an object is represented by a vector. the only force to overcome is the force of gravity. Find the force required to keep the vehicle from rolling down 81. If u cos , sin and v sin , cos , are u the hill. Find the force perpendicular to the hill. and v orthogonal, parallel, or neither? Explain. 82. Error Analysis Describe the error. 75. MODELING DATA 5, 8 2, 7 10, 56 One of the events in a local strongman contest is to drag 83. Think About It Let u be a unit vector. What is the a cement block. One competitor drags the block with a value of u u? Explain. constant force of 250 pounds at a constant angle of 30 with the horizontal (see figure). 84. CAPSTONE What is known about , the angle between two nonzero vectors u and v, under each condition (see figure)? 30°

θ u v (a) Find the work done in terms of the distance d. Origin (b) Use a graphing utility to complete the table. (a) u v 0 (b) u v > 0 (c) u v < 0 d 25 50 100 85. Think About It What can be said about the vectors u Work and v under each condition? (a) The projection of onto equals uvu . 76. Public Safety A ski patroller pulls a rescue toboggan (b) The projection of onto vu equals 0. across a flat snow surface by exerting a constant force 86. Proof Use vectors to prove that the diagonals of a of 35 pounds on a handle that makes a constant angle of rhombus are perpendicular. 22 with the horizontal. Find the work done in pulling 87. Proof Prove the following. the toboggan 200 feet. u v2 u2 v2 2u v 88. Proof Prove that if u is orthogonal to and , then uwv is orthogonal to cv dw for any scalars c and d. 89. Proof Prove that if u is a unit vector and is the angle between u and i, then u cos i sin j. 22° 90. Proof Prove that if u is a unit vector and is the angle between u and j, then u cos i sin j. 77. Finding Work A tractor pulls a log 800 meters, and 2 2 the tension in the cable connecting the tractor and the log is approximately 15,691 newtons. The direction of Cumulative Mixed Review the force is 35 above the horizontal. Approximate the Transformation of a Graph In Exercises 91–94, describe work done in pulling the log. how the graph of g is related to the graph of f. 78. Finding Work A mover exerts a horizontal force of 25 pounds on a crate as it is pushed up a ramp that is 91. gx f(x 4 92. gx fx 12 feet long and inclined at an angle of 20 above the 93. gx f(x 6 94. gx f2x horizontal. Find the work done in pushing the crate up the ramp. Operations with Complex Numbers In Exercises 95–100, perform the operation and write the result in Conclusions standard form. True or False? In Exercises 79 and 80, determine whether 95. 3i4 5i 96. 2i1 6i the statement is true or false. Justify your answer. 97. 1 3i1 3i 98. 7 4i7 4i 79. The vectors u 0, 0 and v 12, 6 are orthogonal. 3 2 6 3 99. 100. 1 i 2 3i 4 i 1 i

Section 6.5 Trigonometric Form of a Complex Number 443

6.5 Trigonometric Form of a Complex Number

The Complex Plane What you should learn ● Plot complex numbers in the Just as real numbers can be represented by points on the real number line, you can complex plane and find absolute represent a complex number z a bi as the point ͑a, b͒ in a coordinate plane (the values of complex numbers. complex plane). The horizontal axis is called the real axis and the vertical axis is called ● Write trigonometric forms of the imaginary axis, as shown in Figure 6.47. complex numbers. ● Multiply and divide complex Imaginary axis numbers written in trigonometric form. 3 (−1, 3) ● Use DeMoivre’s Theorem to find or 2 (3, 2) powers of complex numbers. −1 + 3i or ● Find n th roots of complex 1 3 + 2i numbers. Real − axis 23(−2, −1) 1 2 Why you should learn it or You can use the trigonometric form −2 − i of a complex number to perform operations with complex numbers. Figure 6.47 For instance, in Exercises 153–160 on pages 454 and 455, you can use The absolute value of a complex number a bi is defined as the distance the trigonometric form of a complex ͑ ͒ ͑ ͒ between the origin 0, 0 and the point a, b . number to help you solve polynomial equations. Definition of the Absolute Value of a Complex Number The absolute value of the complex number z a bi is given by Խa biԽ Ίa2 b2.

When the complex number a bi is a real number (that is, when b 0 ), this definition agrees with that given for the absolute value of a real number Խa 0iԽ Ίa2 02 ԽaԽ.

Example 1 Finding the Absolute Value of a Complex Number Plot z 2 5i and find its absolute value.

Solution Imaginary axis The complex number z = −2 + 5i 5 z 2 5i 4 is plotted in Figure 6.48. The absolute value of z is 3 29 ԽzԽ Ί͑2͒2 52 Real Ί 29. −43−3 −2 −1 1 24axis Now try Exercise 11. Figure 6.48

Supri Suharjoto 2010/used under license from Shutterstock.com

444 Chapter 6 Additional Topics in Trigonometry Trigonometric Form of a Complex Number In Section 2.4, you learned how to add, subtract, Imaginary multiply, and divide complex numbers. To work axis effectively with powers and roots of complex numbers, it is helpful to write complex numbers in trigonometric form. In Figure 6.49, consider (a, b) the nonzero complex number a bi. By letting be the angle from the positive real axis r b (measured counterclockwise) to the line segment θ connecting the origin and the point ͑a, b͒, you Real can write a axis a r cos and b r sin

where Figure 6.49 r Ίa2 b2. Consequently, you have a bi ͑r cos ͒ ͑r sin ͒i from which you can obtain the trigonometric form of a complex number.

Trigonometric Form of a Complex Number The trigonometric form of the complex number z a bi is given by z r͑cos i sin ͒ where a r cos , b r sin , r Ίa2 b2, and tan b͞a. The number r is the modulus of z, and is called an argument of z.

The trigonometric form of a complex number is also called the polar form. Because there are infinitely many choices for , the trigonometric form of a complex number is not unique. Normally, is restricted to the interval 0 < 2, although on occasion it is convenient to use < 0.

Example 2 Writing a Complex Number in Trigonometric Form Write the complex number Imaginary axis z 2i Real − − axis in trigonometric form. 2 1 3π 12 2 Solution ⎢⎢z = 2

The absolute value of z is −2 z = −2i r Խ2iԽ Ί02 ͑2͒2 Ί4 2. −3 With a 0, you cannot use tan b͞a to find . Because z 2i lies on the negative −4 imaginary axis (see Figure 6.50), choose 3͞2. So, the trigonometric form is Figure 6.50 z r͑cos i sin ͒ 3 3 2΂cos i sin ΃. 2 2 Now try Exercise 23.

Section 6.5 Trigonometric Form of a Complex Number 445

Example 3 Writing a Complex Number in Trigonometric Form Write the complex number z 2 2Ί3i in trigonometric form. What’s Wrong? Solution You use a graphing utility to check the answer to Example 3, The absolute value of z is as shown in the figure. You r Խ2 2Ί3iԽ Ί͑2͒2 ͑2Ί3͒2 Ί16 4 determine that r 4 and Ϸ 1.047 Ϸ ͞3. Your value and the angle is given by for does not agree with the b 2Ί3 value found in the example. tan Ί3. a 2 What’s wrong? Because tan͑͞3͒ Ί3 and z 2 2Ί3i lies in Quadrant III, choose to be ͞3 4͞3. So, the trigonometric form is z r͑cos i sin ͒ 4 4 4΂cos i sin ΃. 3 3 See Figure 6.51.

Imaginary Technology Tip axis

Real A graphing utility axis can be used to convert −3 −2 4π 1 3 a complex number in trigonometric form to standard form. For instance, ⎢⎢z = 4 − 2 enter the complex number Ί2͑cos ͞4 i sin ͞4͒ in − 3 your graphing utility and press ENTER . You should obtain the z − − i − = 2 2 3 4 standard form 1 i, as shown below. Figure 6.51 Now try Exercise 29.

Example 4 Writing a Complex Number in Standard Form Write the complex number in standard form a bi. z Ί8΄cos΂ ΃ i sin΂ ΃΅ 3 3 Solution Because cos͑͞3͒ 1͞2 and sin͑͞3͒ Ί3͞2, you can write z Ί8΄cos΂ ΃ i sin΂ ΃΅ 3 3 1 Ί3 Ί8΄ i΅ 2 2 1 Ί3 2Ί2΄ i΅ 2 2 Ί2 Ί6i. Now try Exercise 47.

iofoto 2010/used under license from Shutterstock.com

446 Chapter 6 Additional Topics in Trigonometry Multiplication and Division of Complex Numbers The trigonometric form adapts nicely to multiplication and division of complex numbers. Suppose you are given two complex numbers ͑ ͒ z1 r1 cos 1 i sin 1 and ͑ ͒ z2 r2 cos 2 i sin 2 .

The product of z1 and z 2 is ͑ ͒͑ ͒ z1z2 r1r2 cos 1 i sin 1 cos 2 i sin 2 ͓͑ ͒ ͑ ͔͒ r1r2 cos 1 cos 2 sin 1 sin 2 i sin 1 cos 2 cos 1 sin 2 ͓ ͑ ͒ ͑ ͔͒ r1r2 cos 1 2 i sin 1 2 . Sum and difference formulas This establishes the first part of the following rule. The second part is left for you to verify (see Exercise 171).

Product and Quotient of Two Complex Numbers ͑ ͒ ͑ ͒ Let z1 r1 cos 1 i sin 1 and z2 r2 cos 2 i sin 2 be complex numbers. ͓ ͑ ͒ ͑ ͔͒ z1z2 r1r2 cos 1 2 i sin 1 2 Product z r 1 1 ͓ ͑ ͒ ͑ ͔͒ cos 1 2 i sin 1 2 , z 2 0 Quotient z2 r2

Note that this rule says that to multiply two complex numbers you multiply moduli and add arguments, whereas to divide two complex numbers you divide moduli and subtract arguments.

Example 5 Multiplying Complex Numbers in Trigonometric Form

Find the product z1z2 of the complex numbers. z 3΂cos i sin ΃ 1 4 4 3 3 z 2΂cos i sin ΃ 2 4 4 Solution 3 3 z z 3΂cos i sin ΃ 2΂cos i sin ΃ 1 2 4 4 4 4 3 3 Imaginary 6΄cos΂ ΃ i sin΂ ΃΅ axis 4 4 4 4 4 6͑cos i sin ͒ 3 2 z1 6͓1 i͑0͔͒ z2 z z 1 1 2 Real 6 −6−5 −4 −3 −2 −1 123 axis −1 The numbers z1, z2, and z1z2 are plotted in Figure 6.52. Figure 6.52 Now try Exercise 65.

Section 6.5 Trigonometric Form of a Complex Number 447

Example 6 Multiplying Complex Numbers in Trigonometric Form

Find the product z1z2 of the complex numbers. 2 2 11 11 Technology Tip z 2΂cos i sin ΃ z 8΂cos i sin ΃ 1 3 3 2 6 6 Some graphing utilities Solution can multiply and divide 2 2 11 11 complex numbers in z z 2΂cos i sin ΃ 8΂cos i sin ΃ trigonometric form. If you have 1 2 3 3 6 6 access to such a graphing utility, 2 11 2 11 use it to find z z and z ͞z in 16΄cos΂ ΃ i sin΂ ΃΅ 1 2 1 2 3 6 3 6 Examples 6 and 7. 5 5 16΂cos i sin ΃ 2 2 16΂cos i sin ΃ 2 2 16͓0 i͑1͔͒ 16i You can check this result by first converting to the standard forms Ί Ί z1 1 3i and z2 4 3 4i and then multiplying algebraically, as in Section 2.4. ͑ Ί ͒͑ Ί ͒ z1z2 1 3i 4 3 4i 4Ί3 4i 12i 4Ί3 16i Now try Exercise 67.

Example 7 Dividing Complex Numbers in Trigonometric Form Find the quotient

z2 of the complex numbers. ͑ ͒ ͑ ͒ z1 24 cos 300 i sin 300 z 2 8 cos 75 i sin 75 Solution z 24͑cos 300i sin 300͒ 1 ͑ ͒ z2 8 cos 75 i sin 75 24 ͓cos͑30075͒ i sin͑30075͔͒ 8 3͑cos 225i sin 225͒ Ί2 Ί2 3΄΂ ΃ i΂ ΃΅ 2 2 3Ί2 3Ί2 i 2 2 Now try Exercise 73.

Hasan Kursad Ergan/iStockphoto.com

448 Chapter 6 Additional Topics in Trigonometry Powers of Complex Numbers The trigonometric form of a complex number is used to raise a complex number to a power. To accomplish this, consider repeated use of the multiplication rule. z r͑cos i sin ͒ z 2 r͑cos i sin ͒r͑cos i sin ͒ r 2͑cos 2 i sin 2͒ z3 r 2͑cos 2 i sin 2͒r͑cos i sin ͒ r 3͑cos 3 i sin 3͒ z4 r 4͑cos 4 i sin 4͒ z5 r5͑cos 5 i sin 5͒ . . This pattern leads to DeMoivre’s Theorem, which is named after the French mathematician Abraham DeMoivre (1667–1754).

DeMoivre’s Theorem If z r͑cos i sin ͒ is a complex number and n is a positive integer, then Explore the Concept n zn ͓r͑cos i sin ͔͒ Plot the numbers i, i 2, 3 4 5 r n ͑cos n i sin n͒. i , i , and i in the complex plane. Write each number in trigonometric form and describe what Example 8 Finding a Power of a Complex Number happens to the angle as you Use DeMoivre’s Theorem to find form higher powers of i n. 12 ͑1 Ί3i͒ . Solution First convert the complex number to trigonometric form using r Ί͑1͒2 ͑Ί3͒2 2 and Ί3 arctan . 1 3 So, the trigonometric form is 1 Ί3i 2΂cos i sin ΃. 3 3 Then, by DeMoivre’s Theorem, you have 12 ͑1 Ί3i͒12 ΄2΂cos i sin ΃΅ 3 3 12 12 212΂cos i sin ΃ 3 3 4096͑cos 4 i sin 4͒ 4096͑1 0͒ 4096. Now try Exercise 107.

Section 6.5 Trigonometric Form of a Complex Number 449 Roots of Complex Numbers Recall that a consequence of the Fundamental Theorem of Algebra is that a polynomial equation of degree has nn solutions in the complex number system. So, an equation such as x6 1 has six solutions, and in this particular case you can find the six solutions by factoring and using the Quadratic Formula. x 6 1 0 ͑ x3 1͒͑x3 1͒ 0 ͑ x 1͒͑x2 x 1͒͑x 1͒͑x2 x 1͒ 0 Consequently, the solutions are 1 ± Ί3i 1 ± Ί3i x ±1, x , and x . 2 2 Each of these numbers is a sixth root of 1. In general, the nth root of a complex number is defined as follows.

Definition of an nth Root of a Complex Number The complex number u a bi is an nth root of the complex number z when z un ͑a bi͒n.

To find a formula for an n th root of a complex number, let be an nu th root of z, where u s͑cos i sin ͒ and z r͑cos i sin ͒. By DeMoivre’s Theorem and the fact that un z, you have sn ͑cos n i sin n͒ r͑cos i sin ͒. Taking the absolute value of each side of this equation, it follows that sn r. Substituting back into the previous equation and dividing by r, you get cos n i sin n cos i sin . So, it follows that cos n cos and sin n sin . Because both sine and cosine have a period of 2, these last two equations have solutions if and only if the angles differ by a multiple of 2. Consequently, there must exist an integer k such that n 2k 2k . n By substituting this value of into the trigonometric form of u, you get the result stated in the theorem on the next page.

Explore the Concept

The n th roots of a complex number are useful for solving some polynomial equations. For instance, explain how you can use DeMoivre’s Theorem to solve the polynomial equation x4 16 0. [Hint: Write 16 as 16͑cos i sin ͒.]

450 Chapter 6 Additional Topics in Trigonometry

nth Roots of a Complex Number For a positive integer n, the complex number z r͑cos i sin ͒ has exactly n distinct n th roots given by 2k 2k Ίn r΂cos i sin ΃ n n where k 0, 1, 2, . . . , n 1.

When k > n 1, the roots begin to repeat. For instance, when k n, the angle Imaginary axis 2n 2 n n is coterminal with ͞n, which is also obtained when k 0. 2π n The formula for the n th roots of a complex number z has a nice geometrical n r 2π interpretation, as shown in Figure 6.53. Note that because the n th roots of z all have the n Real same magnitude Ίn r, they all lie on a circle of radius Ίn r with center at the origin. axis Furthermore, because successive n th roots have arguments that differ by 2 n the n th roots are equally spaced around the circle. Figure 6.53 You have already found the sixth roots of 1 by factoring and by using the Quadratic Formula. Example 9 shows how you can solve the same problem with the formula for nth roots.

Example 9 Finding the nth Roots of a Real Number Find all the sixth roots of 1. Solution First write 1 in the trigonometric form 1 1͑cos 0 i sin 0͒. Then, by the n th root formula with n 6 and r 1, the roots have the form Imaginary 0 2k 0 2k k k axis Ί6 1΂cos i sin ΃ cos i sin . 6 6 3 3 13 1 3 − + i + i 22 2 2 So, for k 0, 1, 2, 3, 4, and 5, the sixth roots are as follows. (See Figure 6.54.) −1 + 0i 1 + 0 i cos 0 i sin 0 1 Ί Real 1 3 2 2 cos i sin i Incremented by axis 3 3 2 2 n 6 3 2 2 Ί 1 3 cos i sin i 1133 3 3 2 2 − − i − i 2222 cos i sin 1 Figure 6.54 4 4 1 Ί3 cos i sin i 3 3 2 2 5 5 1 Ί3 cos i sin i 3 3 2 2 Now try Exercise 147.

Section 6.5 Trigonometric Form of a Complex Number 451

In Figure 6.54, notice that the roots obtained in Example 9 all have a magnitude of 1 and are equally spaced around the unit circle. Also notice that the complex roots occur in conjugate pairs, as discussed in Section 2.5. The distinct nn th roots of 1 are called the nth roots of unity.

Example 10 Finding the nth Roots of a Complex Number Find the three cube roots of z 2 2i. Solution The absolute value of z is r Խ2 2iԽ Ί͑2͒2 22 Ί8 and the angle is given by b 2 tan 1. a 2 Because z lies in Quadrant II, the trigonometric form of z is z 2 2i Ί8 ͑cos 135i sin 135͒. By the formula for n th roots, the cube roots have the form 135º 360k 135º 360k Ί6 8΂cos i sin ΃. 3 3 Finally, for k 0, 1, and 2, you obtain the roots 135360͑0͒ 135360͑0͒ Ί6 8 ΂cos i sin ΃ 3 3 Ί2͑cos 45i sin 45͒ 1 i 135360͑1͒ 135360͑1͒ Ί6 8 ΂cos i sin ΃ Imaginary 3 3 axis Ί2͑cos 165i sin 165͒ −1.3660 + 0.3660i 1 + i Ϸ 1.3660 0.3660i 1 135360͑2͒ 135360͑2͒ Real Ί6 8 ΂cos i sin ΃ 3 3 −2 1 2 axis −1 Ί2͑cos 285i sin 285͒ 0.3660 − 1.3660i −2 Ϸ 0.3660 1.3660i. See Figure 6.55. Figure 6.55 Now try Exercise 151.

Explore the Concept

Use a graphing utility set in parametric and radian modes to display the graphs of X1T cos T and Y1T sin T. Set the viewing window so that 1.5 X 1.5 and 1 Y 1. Then, using 0 T 2, set the “Tstep” to 2͞n for various values of n. Explain how the graphing utility can be used to obtain the n th roots of unity.

452 Chapter 6 Additional Topics in Trigonometry

See www.CalcChat.com for worked-out solutions to odd-numbered exercises. 6.5 Exercises For instructions on how to use a graphing utility, see Appendix A.

Vocabulary and Concept Check In Exercises 1–3, fill in the blank. 1. The ______of a complex number a bi is the distance between the origin ͑0, 0͒ and the point ͑a, b͒. 2. ______Theorem states that if z r͑cos i sin ͒ is a complex number and n is a positive integer, then zn r n͑cos n i sin n͒. 3. The complex number u a bi is an ______of the complex number z when z un ͑a bi͒n.

4. What is the trigonometric form of the complex number z a bi? 5. When a complex number is written in trigonometric form, what does r represent? 6. When a complex number is written in trigonometric form, what does represent?

Procedures and Problem Solving Finding the Absolute Value of a Complex Number In 21. Imaginary 22. Imaginary Exercises 7–14, plot the complex number and find its axis axis absolute value. 4 3 − 2 z = 1 + 3i 7. 4i 8. 2i Real 2 axis 9. 5 10. 8 −2 z = 3 − i Real −4 − − − axis 11. 4 4i 3 2 1 12. 5 12i Writing a Complex Number in Trigonometric Form In 13. 9 7i Exercises 23–46, represent the complex number 14. 10 3i graphically, and find the trigonometric form of the number. Writing a Complex Number in Trigonometric Form 23. 8i 24. 4i In Exercises 15–22, write the complex number in trigonometric form without using a calculator. 25. 5i 26. 12i 27. 5 5i 28. 2 2i 15. Imaginary 16. Imaginary axis axis 29. Ί3 i 30. 1 Ί3i 4 4 31. 1 i 3 2 z = 5 2 z = 2i Real 32. 4 4i 1 642 axis Ί Real −2 33. 2͑1 3i͒ − − axis −4 5͑Ί ͒ 2 1 1 2 34. 2 3 i 17. Imaginary 18. Imaginary 35. 7 4i axis axis 36. 5 i 4 4 37. 3 z = −4 2 2 Real Real 38. 6 − − axis − axis 6 4 −2 2 z = i 2 39. 3 Ί3i − − 4 4 40. 2Ί2 i 19. Imaginary 20. Imaginary 41. 1 2i axis axis 42. 1 3i Real 6 − − − axis + 43. 5 2i 4 3 2 4 z = 3 3i −2 2 44. 3 i Real −3 Ί − − axis 45. 3 2 7i z = 3 3i −2 2 46 46. 8 5Ί3i

Section 6.5 Trigonometric Form of a Complex Number 453

Writing a Complex Number in Standard Form In ͓5͑ ͔͒ ͓ 2͑ ͔͒ 69. 3 cos 140 i sin 140 3 cos 60 i sin 60 Exercises 47–58, represent the complex number ͓1͑ ͔͒ ͓ 4͑ ͔͒ graphically, and find the standard form of the number. 70. 2 cos 115 i sin 115 5 cos 300 i sin 300 ͓11͑ ͔͒ ͓ 2͑ ͔͒ 47. 2͑cos 120i sin 120͒ 48. 5͑cos 135i sin 135͒ 71. 20 cos 290 i sin 290 5 cos 200 i sin 200 3͑ ͒ 3͑ ͒ 72. ͑cos 5i sin 5͒͑cos 20i sin 20͒ 49. 2 cos 330 i sin 330 50. 4 cos 315 i sin 315 3 3 cos 50 i sin 50 51. 3.75΂cos i sin ΃ 73. 4 4 cos 20 i sin 20 5͑cos 4.3 i sin 4.3͒ 52. 1.5΂cos i sin ΃ 74. 2 2 4͑cos 2.1 i sin 2.1͒ 2͑cos 120i sin 120͒ 53. 6΂cos i sin ΃ 75. ͑ ͒ 3 3 4 cos 40 i sin 40 5 5 7 7 54. 8΂cos i sin ΃ cos΂ ΃ i sin΂ ΃ 6 6 4 4 76. 3 3 cos i sin 55. 4΂cos i sin ΃ 2 2 18͑cos 54i sin 54͒ 77. ͑ ͒ 56. 7͑cos 0 i sin 0͒ 3 cos 102 i sin 102 ͓ ͑ ͒ ͑ ͔͒ 9͑cos 20i sin 20͒ 57. 3 cos 18 45 i sin 18 45 78. 5͑cos 75i sin 75͒ 58. 6͓cos͑230º 30͒ i sin͑230º 30͔͒

Writing a Complex Number in Standard Form In Operations with Complex Numbers in Trigonometric Exercises 59–62, use a graphing utility to represent the Form In Exercises 79–94, (a) write the trigonometric complex number in standard form. forms of the complex numbers, (b) perform the indicated operation using the trigonometric forms, and (c) perform the indicated operation using the standard forms and 59. 5΂cos i sin ΃ 9 9 check your result with that of part (b). 3 3 ͑ ͒͑ ͒ 60. 12΂cos i sin ΃ 79. 2 2i 1 i 5 5 80. ͑3 3i͒͑1 i͒ ͑ ͒ 61. 9 cos 58º i sin 58º 81. ͑2 2i͒͑1 i͒ ͑ ͒ 62. 2 cos 155 i sin 155 82. ͑Ί3 i͒͑1 i͒ ͑ ͒ Representing a Power In Exercises 63 and 64, represent 83. 2i 1 i the powers z, z2, z3, and z4 graphically. Describe the 84. 3i͑1 Ί2i͒ pattern. 85. 2i͑Ί3 i͒ Ί2 86. i͑1 Ί3i͒ 63. z ͑1 i͒ 2 87. 2͑1 i͒ 1 88. 4͑1 i͒ 64. z ͑1 Ί3i͒ 2 3 4i 89. 1 Ί3i Multiplying or Dividing Complex Numbers In Exercises 2 2i 65–78, perform the operation and leave the result in 90. Ί trigonometric form. 1 3i 5 3 3 91. 65. ΄2΂cos i sin ΃΅΄5΂cos i sin ΃΅ 2 2i 2 2 2 2 2 92. 66. ΄3΂cos i sin ΃΅΄4΂cos i sin ΃΅ Ί3 i 3 3 6 6 4i 2 4 4 5 5 93. 67. ΄ ΂cos i sin ΃΅΄9΂cos i sin ΃΅ 1 i 3 3 3 3 3 2i 3 94. 68. ΄ ΂cos i sin ΃΅΄6΂cos i sin ΃΅ 1 Ί3i 2 6 6 4 4

454 Chapter 6 Additional Topics in Trigonometry

Sketching the Graph of Complex Numbers In Exercises 2 125. ΄3΂cos i sin ΃΅ 95–106, sketch the graph of all complex numbers z 8 8 satisfying the given condition. 5 126. ΄2΂cos i sin ΃΅ 95. ԽzԽ 2 10 10 96. ԽzԽ 5 127. Show that 1͑1 Ί3 i͒ is a sixth root of 1. 97. ԽzԽ 4 2 128. Show that 21͞4͑1 i͒ is a fourth root of 2. 98. ԽzԽ 6 99. ԽzԽ 7 Finding Square Roots of a Complex Number In 100. ԽzԽ 8 Exercises 129–136, find the square roots of the complex number. 101. 6 129. 2i 130. 5i 102. 4 131. 3i 132. 6i 103. 3 133. 2 2i 5 104. 134. 2 2i 6 135. 1 Ί3i 2 105. 136. 1 Ί3i 3 3 Finding the n th Roots of a Complex Number In 106. 4 Exercises 137–152, (a) use the theorem on page 450 to find the indicated roots of the complex number, Finding a Power of a Complex Number In Exercises (b) represent each of the roots graphically, and (c) write 107–126, use DeMoivre’s Theorem to find the indicated each of the roots in standard form. power of the complex number. Write the result in 137. Square roots of 5͑cos 120i sin 120͒ standard form. 138. Square roots of 16͑cos 60i sin 60͒ ͑ ͒3 107. 1 i 2 2 6 139. Fourth roots of 8΂cos i sin ΃ 108. ͑2 2i͒ 3 3 109. ͑1 i͒6 5 5 140. Fifth roots of 32΂cos i sin ΃ 110. ͑3 2i͒8 6 6 111. 2͑Ί3 i͒5 141. Cube roots of 25i 3 112. 4͑1 Ί3i͒ 142. Fourth roots of 625i ͓ ͑ ͔͒3 125͑ Ί ͒ 113. 5 cos 20 i sin 20 143. Cube roots of 2 1 3i 114. ͓3͑cos 150i sin 150͔͒4 144. Cube roots of 4Ί2͑1 i͒ 5 5 10 145. Cube roots of 64i 115. ΂cos i sin ΃ 4 4 146. Fourth roots of i 12 116. ΄2΂cos i sin ΃΅ 147. Fifth roots of 1 2 2 148. Cube roots of 1000 ͓ ͑ ͒ 4 117. 2 cos 1.25 i sin 1.25 ] 149. Cube roots of 125 ͓ ͑ ͔͒5 118. 4 cos 2.8 i sin 2.8 150. Fourth roots of 4 ͓ ͑ ͔͒8 119. 2 cos i sin 151. Fifth roots of 128͑1 i͒ ͑ ͒20 120. cos 0 i sin 0 152. Sixth roots of 729i 121. ͑3 2i͒5 122. ͑Ί5 4i͒4 ( p. 443) In Exercises 153–160, 123. ͓4(cos 10i sin 10͔͒6 use the theorem on page 450 to find all the solutions of the equation, and represent the solutions graphically. 124. ͓3͑cos 15i sin 15͔͒4 153. x4 i 0 154. x3 1 0

Section 6.5 Trigonometric Form of a Complex Number 455

155. x5 243 0 173. Use trigonometric forms of z and z in Exercise 172 to 156. x3 27 0 find the following. 157. x4 16i 0 (a) zz z 158. x6 64i 0 z (b) z , 0 159. x3 ͑1 i͒ 0 174. Show that the negative of z r͑cos i sin ͒ is 160. x4 ͑1 i͒ 0 z r͓cos͑ ͒ i sin͑ ͔͒. Electrical Engineering In Exercises 161–166, use the 175. Writing The famous formula formula to find the missing quantity for the given conditions. The formula eabi ea͑cos b i sin b͒ I Z is called Euler’s Formula, after the Swiss mathematician ؍ E Leonhard Euler (1707–1783). This formula gives rise where E represents voltage,I represents current, and Z to the equation represents impedance (a measure of opposition to a sinusoidal electric current), is used in electrical ei 1 0. engineering. Each variable is a complex number. This equation relates the five most famous numbers in 161. I 10 2i mathematics—0, 1,, , and ie —in a single equation. Z 4 3i Show how Euler’s Formula can be used to derive this equation. Write a short paragraph summarizing your 162. I 12 2i work. Z 3 5i 163. I 2 4i 176. CAPSTONE Use the graph of the roots of a E 5 5i complex number. 164. I 10 2i (a) Write each of the roots in trigonometric form. E 4 5i (b) Identify the complex number whose roots are 165. E 12 24i given. Use a graphing utility to verify your results. Z 12 20i (i) Imaginary (ii) Imaginary axis 166. E 15 12i axis Z 25 24i 33 22 ° ° 30° 30° Real 45 45 Real Conclusions axis axis 2 1 45° 45° True or False? In Exercises 167–170, determine whether −1 33 the statement is true or false. Justify your answer. 1͑ Ί ͒ 167. 2 1 3i is a ninth root of 1. 168. Ί3 i is a solution of the equation x2 8i 0. Cumulative Mixed Review 169. The product of two complex numbers is 0 only when Harmonic Motion In Exercises 177–180, for the simple the modulus of one (or both) of the complex numbers harmonic motion described by the trigonometric function, is 0. find the maximum displacement from equilibrium and .0 ؍ Geometrically, the n th roots of any complex number z the lowest possible positive value of t for which d .170 are all equally spaced around the unit circle centered at the origin. 177. d 16 cos t 4 z r ͑ i ͒ 171. Given two complex numbers 1 1 cos 1 sin 1 1 5 ͑ ͒ 178. d sin t and z2 r2 cos 2 i sin 2 , z2 0, show that 16 4 z r 179. d 1 cos 12t 1 1 ͓cos͑ ͒ i sin͑ ͔͒. 8 1 2 1 2 1 z 2 r2 180. d 12 sin 60 t 172. Show that z r͓cos͑͒ i sin͔͑͒ is the complex conjugate of z r͑cos i sin ͒.

456 Chapter 6 Additional Topics in Trigonometry

6 Chapter Summary

What did you learn? Explanation and Examples Review Exercises

Use the Law of Sines to solve Law of Sines oblique triangles (AAS or ASA) If ABC is a triangle with sides a, b, and c, then (p. 404). a b c . sin A sin B sin C C C 1–6

b a a h h b

ABc ABc A is acute. A is obtuse. 6.1 Use the Law of Sines to solve If two sides and one opposite angle are given, then three oblique triangles (SSA) (p. 406). possible situations can occur: (1) no such triangle exists (see Example 4), (2) one such triangle exists (see Example 7–10 3), or (3) two distinct triangles satisfy the conditions (see Example 5).

Find areas of oblique triangles The area of any triangle is one-half the product of the (p. 408), and use the Law of Sines lengths of two sides times the sine of their included angle. 1 1 1 to model and solve real-life That is, Area 2 bc sin A 2 ab sin C 2 ac sin B. 11–16 problems (p. 409). The Law of Sines can be used to approximate the total distance of a boat race course. (See Example 7.)

Use the Law of Cosines to solve Law of Cosines oblique triangles (SSS or SAS) Standard Form Alternative Form (p. 413). b2 c2 a2 a2 b2 c2 2bc cos A cos A 2bc 17–24 a2 c2 b2 b2 a2 c2 2ac cos B cos B 2ac a2 b2 c2 c2 a2 b2 2ab cos C cos C 2ab 6.2 Use the Law of Cosines to model The Law of Cosines can be used to find the distance and solve real-life problems between the pitcher’s mound and first base on a women’s 25, 26 (p. 415). softball field. (See Example 3.)

Use Heron’s Area Formula to find Heron’s Area Formula: Given any triangle with areas of triangles (p. 416). sides of lengths a, b, and c, the area of the triangle is Area ss as bs c, where 27–30 s a b c2.

Q Represent vectors as directed line PQ P Terminal point segments (p. 420). Initial point 31, 32

Write the component forms of The component form of the vector with initial point vectors (p. 421). P p1, p2 and terminal point Q q1, q2 is given by 31–34 6.3 \ PQ q1 p1, q2 p2 v1, v2 v.

Perform basic vector operations Let u u1, u2 and v v1, v2 be vectors and let k be a and represent vectors graphically scalar (a real number). 35–44 (p. 422). u v u1 v1, u2 v2 ku ku1, ku2 v v1, v2 u v u1 v1, u2 v2

Chapter Summary 457

What did you learn? Explanation and Examples Review Exercises Write vectors as linear v v1, v2 v1 1, 0 v2 0, 1 v1i v2 j combinations of unit vectors The scalars v and v are the horizontal and vertical (p. 424). 1 2 45–50 components of v, respectively. The vector sum v1i v2 j is the linear combination of the vectors i and j.

6.3 Find the direction angles of If u 2i 2j, then the direction angle is vectors (p. 426). tan 22 1. 51–56 So, 45.

Use vectors to model and solve Vectors can be used to find the resultant speed and 57–60 real-life problems (p. 427). direction of an airplane. (See Example 10.)

Find the dot product of two The dot product of u u1, u2 and v v1, v2 is vectors and use the properties u v u1v1 u2v2. 61–68 of the dot product (p. 434).

Find the angle between two If is the angle between two nonzero vectors u and v, then vectors and determine whether u v cos . two vectors are orthogonal u v 69–84 (p. 435). Vectors and vu are orthogonal when u v 0.

6.4 Write vectors as the sums of two Many applications in physics and engineering require the vector components (p. 437). decomposition of a given vector into the sum of two vector 85–88 components. (See Example 6.)

Use vectors to find the work done The work W done by a constant force F as its point of \ by a force (p. 439). application moves along the vector PQ is given by either of the following.

\ 89, 90 \ 1. W projPQ F PQ Projection form

\ 2. W F PQ Dot product form

Plot complex numbers in the A complex number z a bi can be represented as the complex plane and find absolute point a, b in the complex plane. The horizontal axis is values of complex numbers the real axis and the vertical axis is the imaginary axis. 91–94 (p. 443). The absolute value of z a bi is a bi a2 b2.

Write trigonometric forms of The trigonometric form of the complex number z a bi complex numbers (p. 444). is z rcos i sin where a r cos , b r sin , 95–98 r a2 b2, and tan ba.

Multiply and divide complex Let and 6.5 z1 r1 cos 1 i sin 1 z2 r2 cos 2 i sin 2 . numbers written in trigonometric z1z2 r1r2 cos 1 2 i sin 1 2 form (p. 446). 99–106 z r 1 1 cos 1 2 i sin 1 2 , z2 0 z2 r2

Use DeMoivre’s Theorem to find DeMoivre’s Theorem: If z rcos i sin is a powers of complex numbers complex number and n is a positive integer, then 107–110 (p. 448). zn rcos i sin n rncos n i sin n.

Find n th roots of complex The complex number u a bi is an n th root of the numbers (p. 449). complex number z when z un a bin. 111–122

458 Chapter 6 Additional Topics in Trigonometry

See www.CalcChat.com for worked-out solutions to odd-numbered exercises. 6 Review Exercises For instructions on how to use a graphing utility, see Appendix A.

6.1 18. C

Using the Law of Sines In Exercises 1–10, use the Law b = 4 100° a = 7 of Sines to solve the triangle. If two solutions exist, find both. ABc 1. A 32, B 50, a 16 19. a 9, b 12, c 20 2. A 38, B 58, a 12 20. a 7, b 15, c 19 3. B 25, C 105, c 25 21. a 6.5, b 10.2, c 16 4. B 20, C 115, c 30 22. a 6.2, b 6.4, c 2.1 5. A 60 15, B 45 30, b 4.8 23. C 65, a 25, b 12 6. A 82 45, B 28 45, b 40.2 24. B 48, a 18, c 12 7. A 75 , a 2.5, b 16.5 25. Geometry The lengths of the diagonals of a parallel- 8. A 15, a 5, b 10 ogram are 10 feet and 16 feet. Find the lengths of the 9. B 115, a 9, b 14.5 sides of the parallelogram if the diagonals intersect at an angle of 28. 10. B 150, a 10, b 3 26. Surveying To approximate the length of a marsh, a Finding the Area of a Triangle In Exercises 11–14, find surveyor walks 425 meters from point A to point B. The the area of the triangle having the indicated angle and surveyor then turns 65 and walks 300 meters to point C sides. (see figure). Approximate the length AC of the marsh. 11. A 33 , b 7, c 10 B 12. B 80, a 4, c 8 65° 13. C 122 , b 18, a 29 300 m 425 m 14. C 100, a 120, b 74

15. Landscaping A tree stands on a hillside of slope 28 C A from the horizontal. From a point 75 feet down the hill, the angle of elevation to the top of the tree is 45 (see figure). Find the height of the tree. Using Heron’s Area Theorem In Exercises 27–30, use Heron’s Area Formula to find the area of the triangle. 27. a 3, b 6, c 8 75 ft 28. a 15, b 8, c 10 29. a 64.8, b 49.2, c 24.1 45° 30. a 8.55, b 5.14, c 12.73

28° 6.3

16. Surveying A surveyor finds that a tree on the opposite Finding the Component Form of a Vector In Exercises bank of a river has a bearing of N 22 30 E from a 31–34, find the component form and the magnitude of certain point and a bearing of N 15 W from a point the vector v. 400 feet downstream. Find the width of the river. 31. y 32. y 6.2 6 6 (−5, 4) 6, 7 4 4 ( 2 ( Using the Law of Cosines In Exercises 17–24, use the v 2 2 Law of Cosines to solve the triangle. v (0, 1) x x 17. C −4 −2 246 (2, −1) b = 14 a = 8 33. Initial point:0, 10; terminal point: 7, 3 A c = 17 B 34. Initial point:1, 5 ; terminal point: 15, 9

Review Exercises 459

Vector Operations In Exercises 35–40, find (a) u ؉ v, 59. Physics In a manufacturing process, an electric hoist b) u ؊ v, (c) 3u, and (d) 2v ؉ 5u. Then sketch each lifts 200-pound ingots. Find the tension in the) resultant vector. supporting cables (see figure). 35. u 1, 3, v 3, 6 36. u 4, 5, v 0, 1 37. u 5, 2, v 4, 4 38. u 1, 8, v 3, 2 39. u 2i j, v 5i 3j 40. u 6j, v i j 60° 60° 200 lb 24 in. Vector Operations In Exercises 41–44, find the component form of w and sketch the specified vector operations 60. Aviation An airplane has an airspeed of 430 miles per ؉ ؍ ؊ ؍ geometrically, where u 6i 5j and v 10i 3j. hour at a bearing of 135. The wind velocity is 1 41.w 3v 42. w 2v 35 miles per hour in the direction N 30 E. Find the 43.w 4u 5v 44. w 3v 2u resultant speed and direction of the plane. 6.4 Finding a Unit Vector In Exercises 45–48, find a unit vector in the direction of the given vector. Verify that the Finding a Dot Product In Exercises 61–64, find the dot result has a magnitude of 1. product of and v.u 45.u 0, 6 46. v 12, 5 61.u 0, 2 62. u 7, 12 47.v 5i 2j 48. w 7i v 1, 10 v 4, 14 63.u 6i j 64. u 8i 7j Writing a Linear Combination In Exercises 49 and 50, v 2i 5j v 3i 4j the initial and terminal points of a vector are given. Write the vector as a linear combination of the standard Using Properties of Dot Products In Exercises 65–68, to find the <1 ,2> ؍ ؊3, ؊4> and v> ؍ unit vectors i and j. use the vectors u 49. Initial point: 8, 3 indicated quantity. Terminal point: 1, 5 65.u u 66. v 3 50. Initial point: 2, 3.2 67.4u v 68. u vu Terminal point: 6.4, 10.8 Finding the Angle Between Two Vectors In Exercises ␪ Finding the Magnitude and Direction Angle of a Vector 69–72, find the angle between the vectors. In Exercises 51–56, find the magnitude and the direction 69. u 22, 4, v 2, 1 angle of the vector v. 70. u 3, 1, v 4, 5 51. v 7cos 60i sin 60j 7 7 5 5 71. u cos i sin j, v cos i sin j 52. v 3cos 150i sin 150j 4 4 6 6 53.v 5i 4j 54. v 4i 7j 72. u cos 45i sin 45j 55.v 3i 3j 56. v 8i j v cos 300i sin 300j

57. Resultant Force Forces with magnitudes of 85 pounds Finding the Angle Between Two Vectors In Exercises and 50 pounds act on a single point. The angle between 73–76, graph the vectors and find the degree measure of the forces is 15. Describe the resultant force. the angle between the vectors. 58. Physics A 180-pound weight is supported by two 73.u 4i j 74. u 6i 2j ropes, as shown in the figure. Find the tension in each v i 4j v 3i j rope. 75. u 7i 5j 30° 30° v 10i 3j 76. u 5.3i 2.8j v 8.1i 4j 180 lb

460 Chapter 6 Additional Topics in Trigonometry

A Relationship of Two Vectors In Exercises 77–80, 20cos 320i sin 320 101. determine whether and vu are orthogonal, parallel, or 5cos 80i sin 80 neither. 3cos 230i sin 230 102. 77.u 39, 12 78. u 8, 4 9cos 95i sin 95 v 26, 8 v 5, 10 79.u 8, 5 80. u 15, 51 Operations with Complex Numbers in Trigonometric Form In Exercises 103–106, (a) write the trigonometric v 2, 4 v 20, 68 forms of the complex numbers, (b) perform the indicated operation using the trigonometric forms, and (c) perform Finding an Unknown Vector Component In Exercises the indicated operation using the standard forms and 81–84, find the value of k such that the vectors and vu check your result with that of part (b). are orthogonal. 103.2 2i3 3i 104. 4 4i1 i 81.u i kj 82. u 2i j 3 3i 1 i 105. 106. v i 2j v i kj 2 2i 2 2i 83.u ki j 84. u ki 2j v 2i 2j v i 4j Finding a Power In Exercises 107–110, use DeMoivre’s Theorem to find the indicated power of the complex Decomposing a Vector into Components In Exercises number. Write the result in standard form. 85–88, find the projection of onto v.u Then write u as the 4 sum of two orthogonal vectors, one of which is proj u. 107. 5cos i sin v 12 12 85. u 4, 3 , v 8, 2 4 4 5 108. 2cos i sin 86. u 5, 6, v 10, 0 15 15 87. u 2, 7, v 1, 1 109.2 3i6 110. 1 i8 88. u 3, 5, v 5, 2 Finding Square Roots In Exercises 111–114, find the 89. Finding Work Determine the work done by a crane square roots of the complex number. lifting an 18,000-pound truck 48 inches. 111.3 i 112. 3 i 90. Physics A 500-pound motorcycle is stopped on a hill inclined at 12. What force is required to keep the 113.2i 114. 5i motorcycle from rolling back down the hill? Finding Roots In Exercises 115–118, (a) use the 6.5 theorem on page 450 to find the indicated roots of the Finding the Absolute Value of a Complex Number In complex number, (b) represent each of the roots graphi- Exercises 91–94, plot the complex number and find its cally, and (c) write each of the roots in standard form. absolute value. 115. Sixth roots of 729i 116. Fourth roots of 256i 91.7i 92. 6i 117. Cube roots of 8 118. Fifth roots of 1024 93.5 3i 94. 10 4i Solving an Equation In Exercises 119–122, use the Writing a Complex Number in Trigonometric Form theorem on page 450 to find all solutions of the equation, In Exercises 95–98, write the complex number in and represent the solutions graphically. trigonometric form without using a calculator. 119.x4 256 0 120. x5 32i 0 95.2 2i 96. 2 2i 121.x3 8i 0 122. x4 81 0 97.3 i 98. 3 i Conclusions Multiplying or Dividing Complex Numbers In Exercises True or False? In Exercises 123 and 124, determine 99–102, perform the operation and leave the result in whether the statement is true or false. Justify your trigonometric form. answer. 5 99. cos i sin 4cos i sin 123. The Law of Sines is true if one of the angles in the 2 2 2 4 4 triangle is a right angle. 2 2 124. When the Law of Sines is used, the solution is always 100. 2cos i sin 3cos i sin 3 3 6 6 unique.

Chapter Test 461

See www.CalcChat.com for worked-out solutions to odd-numbered exercises. 6 Chapter Test For instructions on how to use a graphing utility, see Appendix A.

Take this test as you would take a test in class. After you are finished, check your work against the answers in the back of the book. In Exercises 1– 6, use the given information to solve the triangle. If two solutions exist, find both solutions. A 1. A 36, B 98, c 16 2. a 2, b 4, c 5 80° 3. A 35, b 8, c 12 565 ft 480 ft 4. A 25, b 28, a 18 5. B 130, c 10.1, b 5.2 6. A 150, b 4.8, a 9.4 B C

7. Find the length of the pond shown at the right. Figure for 7 8. A triangular parcel of land has borders of lengths 55 meters, 85 meters, and 100 meters. Find the area of the parcel of land. 9. Find the component form and magnitude of the vector w that has initial point 8, 12 and terminal point 4, 1.

.In Exercises 10–13, find (a) 2v ؉ u, (b) u ؊ 3v, and (c) 5u ؊ v 10. u 0, 4, v 4, 6 11. u 5, 2, v 1, 10 12. u i j, v 6i 9j 13. u 2i 3j, v i 2j

14. Find a unit vector in the direction of v 6i 4j. 15. Find the component form of the vector v with v 12, in the same direction as u 3, 5. 16. Forces with magnitudes of 250 pounds and 130 pounds act on an object at angles of 45 and 60, respectively, with the positive x -axis. Find the direction and magnitude of the resultant of these forces. 17. Find the dot product of u 9, 4 and v 1, 2. 18. Find the angle between the vectors u 7i 2j and v 4j. 19. Are the vectors u 9, 12 and v 4, 3 orthogonal? Explain. 20. Find the projection of u 6, 7 onto v 5, 1. Then write u as the sum of

two orthogonal vectors, one of which is projv u. 21. Write the complex number z 6 6i in trigonometric form. 22. Write the complex number 100cos 240i sin 240 in standard form.

In Exercises 23 and 24, use DeMoivre’s Theorem to find the indicated power of the complex number. Write the result in standard form. 5 5 8 23. 3cos i sin 6 6 24. 3 3i6

25. Find the fourth roots of 1281 3i. 26. Find all solutions of the equation x4 625i 0 and represent the solutions graphically.

462 Chapter 6 Additional Topics in Trigonometry

See www.CalcChat.com for worked-out solutions to odd-numbered exercises. 4–6 Cumulative Test For instructions on how to use a graphing utility, see Appendix A.

Take this test to review the material in Chapters 4–6. After you are finished, check your work against the answers in the back of the book. 1. Consider the angle 150. (a) Sketch the angle in standard position. (b) Determine a coterminal angle in the interval 0, 360. (c) Convert the angle to radian measure. (d) Find the reference angle . (e) Find the exact values of the six trigonometric functions of . 2. Convert the angle 2.55 radians to degrees. Round your answer to one decimal place. 12 3. Find cos when tan and sin > 0. 5 4 In Exercises 4–6, sketch the graph of the function by hand. (Include two full

periods.) Use a graphing utility to verify your graph. −6 6 1 4. f x 3 2 sin x 5. f x tan 3x 6. f x 2 sec x

7. Find positive values of a, , and cb such that the graph of the function −4 hx a cosbx c matches the graph in the figure at the right. Figure for 7

In Exercises 8 and 9, find the exact value of the expression without using a calculator. 3 8. sin arctan 4 1 9. tan arcsin 2 10. Write an algebraic expression equivalent to sinarctan 2x. sin 1 cos 11. Subtract and simplify: . cos sin 1

In Exercises 12–14, verify the identity. 12. cot2 sec2 1 1 13. sinx y sinx y sin2 x sin2 y 2 2 1 14. sin x cos x 8 1 cos 4x

In Exercises 15 and 16, solve the equation. 15. sin2 x 2 sin x 1 0 16. 3 tan cot 0

17. Approximate the solutions to the equation cos2 x 5 cos x 1 0 in the interval 0, 2.

In Exercises 18 and 19, use a graphing utility to graph the function and approximate its zeros in the interval [0, 2␲. If possible, find the exact values of the zeros algebraically. 1 sin x cos x 18. y 4 cos x 1 sin x 19. y tan3 x tan2 x 3 tan x 3

Cumulative Test for Chapters 4–6 463

12 3 20. Given that sin u 13, cos v 5, and angles and vu are both in Quadrant I, find tanu v. 1 21. If tan , find the exact value of tan 2, 0 < < . 2 2 4 3 22. If tan , find the exact value of sin , < < . 3 2 2 23. Write cos 8x cos 4x as a product.

In Exercises 24–27, verify the identity. 2 1 24. tan x 1 sin x 2 sin 2x 1 25. sin 3 sin 2 cos 2 cos 4 1 26. sin 3x cos 2x 2 sin 5x sin x 2 cos 3x 27. csc x sin 4x sin 2x C b In Exercises 28–31, use the information to solve the triangle shown at the right. a 28.A 46, a 14, b 5 29. A 32, b 8, c 10 A c B 30.A 24, C 101, a 10 31. a 24, b 30, c 47 Figure for 28–31

32. Two sides of a triangle have lengths 14 inches and 19 inches. Their included angle measures 82. Find the area of the triangle. 33. Find the area of a triangle with sides of lengths 12 inches, 16 inches, and 18 inches. 34. Write the vector u 3, 5 as a linear combination of the standard unit vectors i and j. Imaginary axis 35. Find a unit vector in the direction of v i 2j. z = −3 + 3i 36. Find u v for u 3i 4j and v i 2j. 3 2 37. Find k such that u i 2kj and v 2i j are orthogonal. Real 38. Find the projection of u 8, 2 onto v 1, 5. Then write u as the sum of two −3 −2 −11axis orthogonal vectors, one of which is projv u. 39. Find the trigonometric form of the complex number plotted at the right. Figure for 39 5 5 40. Write the complex number 63cos i sin in standard form. 6 6 41. Find the product 4cos 30i sin 306cos 120i sin 120. Write the answer in standard form. 42. Find the square roots of 2 i. 43. Find the three cube roots of 1. 44. Write all the solutions of the equation x4 625 0. 45. From a point 200 feet from a flagpole, the angles of elevation to the bottom and top of the flag are 16 45 and 18, respectively. Approximate the height of the flag to the nearest foot. 46. Write a model for a particle in simple harmonic motion with a maximum displacement of 7 inches and a period of 8 seconds. 47. An airplane’s velocity with respect to the air is 500 kilometers per hour, with a bearing of 30. The airplane is in a steady wind blowing from the northwest with a velocity of 50 kilometers per hour. What is the true direction of the airplane? What is its speed relative to the ground? 48. Forces of 60 pounds and 100 pounds have a resultant force of 125 pounds. Find the angle between the two forces.

464 Chapter 6 Additional Topics in Trigonometry

Proofs in Mathematics

Law of Sines (p. 404) Law of Tangents If ABC is a triangle with sides a, b, and c, then Besides the Law of Sines and a b c the Law of Cosines, there is . sin A sin B sin C also a Law of Tangents, which was developed by Francois Oblique Triangles Vi`ete (1540–1603). The Law of C C Tangents follows from the Law of Sines and the sum-to-product formulas for sine and is defined b a a h h b as follows. a b tan A B 2 A cB A cB a b tanA B2 A is acute. A is obtuse. The Law of Tangents can be used to solve a triangle when two sides and the included Proof angle are given (SAS). Before Let h be the altitude of either triangle shown in the figure above. Then you have calculators were invented, the Law of Tangents was used to h sin A or h b sin A solve the SAS case instead of b the Law of Cosines, because h computation with a table of sin B or h a sin B. a tangent values was easier. Equating these two values of h, you have a b a sin B b sin A or . sin A sin B C Note that sin A 0 and sin B 0 because no angle of a triangle can have a measure of or 180.0 In a similar manner, construct an altitude from vertex B to side AC b a (extended in the obtuse triangle), as shown at the right. Then you have h h sin A or h c sin A A cB c A is acute. h sin C or h a sin C. a Equating these two values of h, you have C a c a sin C c sin A or . a sin A sin C b By the Transitive Property of Equality, you know that A c B a b c . sin A sin B sin C h So, the Law of Sines is established. A is obtuse.

Proofs in Mathematics 465

Law of Cosines (p. 413) Standard Form Alternative Form b2 c 2 a 2 a2 b2 c 2 2bc cos A cos A 2bc a 2 c 2 b2 b2 a 2 c 2 2ac cos B cos B 2ac a 2 b2 c 2 c 2 a 2 b2 2ab cos C cos C 2ab

Proof To prove the first formula, consider the top triangle at the right, which has three acute y angles. Note that vertex B has coordinates c, 0. Furthermore,C has coordinates x, y, C = (x, y)) where x b cos A and y b sin A.

Because a is the distance from vertex C to vertex B, it follows that b y a a x c2 y 02 Distance Formula 2 2 2 a x c y 0 Square each side. x a2 b cos A c2 b sin A2 x Substitute for x and y. A c B = (c, 0) a2 b2 cos2 A 2bc cos A c2 b2 sin2 A Expand. 2 2 2 2 2 a b sin A cos A c 2bc cos A Factor out b2.

a2 b2 c2 2bc cos A. sin2 A cos2 A 1 To prove the second formula, consider the bottom triangle at the right, which also y has three acute angles. Note that vertex A has coordinates c, 0. Furthermore,C has C = (x, y)) coordinates x, y, where x a cos B and y a sin B.

Because b is the distance from vertex C to vertex A, it follows that a y b b x c2 y 02 Distance Formula 2 2 2 b x c y 0 Square each side. x b2 a cos B c2 a sin B2 x Substitute for x and y. B c A = (c, 0) b2 a2 cos2 B 2ac cos B c2 a2 sin2 B Expand. 2 2 2 2 2 b a sin B cos B c 2ac cos B Factor out a2.

b2 a2 c2 2ac cos B. sin2 B cos2 B 1 A similar argument is used to establish the third formula.

466 Chapter 6 Additional Topics in Trigonometry

Heron’s Area Formula (p. 416) Given any triangle with sides of lengths a, b, and c, the area of the triangle is given by Area ss as bs c a b c where s . 2

Proof From Section 6.1, you know that

1 Formula for the area Area bc sin A 2 of an oblique triangle 1 Area2 b2c2 sin2 A Square each side. 4 1 Area b2c2 sin2 A Take the square root 4 of each side. 1 b2c21 cos2 A Pythagorean Identity 4 1 1 bc1 cos A bc1 cos A. Factor. 2 2 Using the Law of Cosines, you can show that 1 a b c a b c bc1 cos A 2 2 2 and 1 a b c a b c bc1 cos A . 2 2 2 Letting a b c s 2 these two equations can be rewritten as 1 bc1 cos A ss a 2 and 1 bc1 cos A s bs c. 2 By substituting into the last formula for area, you can conclude that Area ss as bs c.

Proofs in Mathematics 467

Properties of the Dot Product (p. 434) Let u, , and wv be vectors in the plane or in space and let c be a scalar. 1. u v v u 2. 0 v 0 3. u v w u v u w 4. v v v 2 5. cu v cu v u cv

Proof Let u u1, u2 , v v1, v2 , w w1, w2 , 0 0, 0 , and let c be a scalar. 1. u v u1v1 u2v2 v1u1 v2u2 v u 2. 0 v 0 v1 0 v2 0 3. u v w u v1 w1, v2 w2 u1 v1 w1 u2 v2 w2 u1v1 u1w1 u2v2 u2w2 u1v1 u2v2 u1w1 u2w2 u v u w 2 2 2 22 2 4. v v v1 v2 v1 v2 v 5. c u v c u1, u2 v1, v2 c u1v1 u2v2 cu1 v1 cu2 v2 cu1, cu2 v1, v2 cu v

Angle Between Two Vectors (p. 435) If is the angle between two nonzero vectors and vu , then u v cos . uv

Proof Consider the triangle determined by vectors u, v, and v u, as shown in the figure. By v − u the Law of Cosines, you can write u θ v u2 u2 v2 2u v cos v v u v u u2 v2 2u v cos Origin v u v v u u u2 v2 2u v cos v v u v v u u u u2 v2 2u v cos v2 2u v u2 u2 v2 2u v cos u v cos . u v

468 Chapter 6 Additional Topics in Trigonometry Progressive Summary (Chapters 1–6) This chart outlines the topics that have been covered so far in this text. Progressive Summary charts appear after Chapters 2, 3, 6, and 9. In each Progressive Summary, new topics encountered for the first time appear in red.

ALGEBRAIC FUNCTIONS TRANSCENDENTAL FUNCTIONS OTHER TOPICS Polynomial, Rational, Radical Exponential, Logarithmic Trigonometric, Inverse Trigonometric ᭿ Rewriting ᭿ Rewriting ᭿ Rewriting ↔ Polynomial form Factored form Exponential form ↔ Logarithmic form Operations with polynomials Condense/expand logarithmic Rationalize denominators expressions Simplify rational expressions Simplify trigonometric expressions Operations with complex numbers Prove trigonometric identities Use conversion formulas Operations with vectors Powers and roots of complex numbers

᭿ Solving ᭿ Solving ᭿ Solving Equation Strategy Equation Strategy Linear ...... Isolate variable Exponential ...... Take logarithm of Quadratic ...... Factor, set to zero each side Extract square roots Logarithmic ...... Exponentiate Complete the square each side Quadratic Formula Trigonometric . . . . . Isolate function Polynomial ...... Factor, set to zero Factor, use inverse Rational Zero Test function Rational ...... Multiply by LCD Multiple angle . . . . . Use trigonometric Radical ...... Isolate, raise to power or high powers identities Absolute value . . . . . Isolate, form two equations ᭿ Analyzing ᭿ Analyzing ᭿ Analyzing Graphically Algebraically Graphically Algebraically Intercepts Domain, Range Intercepts Domain, Range Symmetry Transformations Asymptotes Transformations Slope Composition Minimum values Composition Asymptotes Standard forms Maximum values Inverse Properties End behavior of equations Amplitude, period Minimum values Leading Coefficient Reference angles Maximum values Test Numerically Synthetic division Table of values Descartes’s Rule of Signs Numerically Table of values