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8x 52. cos 8x cos24x sin24x 65. 66. 67. 68. 69. 70. sin x sin 3x cos x cos 3x tan x sin x sin 3x cos x cos 3x cot 2x sin 4x sin 6x cos 4x cos 6x cot x cos 8x cos 4x cos 8x cos 4x cot 6x cot 2x sin x sin y cos x cos y cot x y 2 b a sin x sin y cos x cos y tan x y 2 b a 71. a. Prove that 1 cos x sin x sin x 1 cos x. b. ... |
Section 9.3 can also be used with the techniques shown in Section 8.3, where equations were rewritten into a basic form and then solved. Example 1 Use Double-Angle Identities Solve 5 cos x 3 cos 2x 3. Solution cos2x. 5 cos x 3 Use a double-angle identity to rewrite cos 2x in terms of 5 cos x 3 cos 2x 3 2 cos2x 1 3 5 c... |
1 For any integer k, 3x sin 3x p 2 x p 6 11 2pk 2pk 3 Example 5 Use Half-Angle Identities Find the solutions of sin x sin, where 0 x 2p. x 2 Solution CAUTION Squaring both sides of an equation may introduce extraneous solutions. Be sure to check all solutions in the original equation. 1 sin x sin x 2 1 cos x sin x ± B... |
x b 2a2 b2 cos x c c 2a2 b2 [2] Substitute the equivalent expressions from [1] into equation [2]. cos a sin x sin a cos x c 2a2 b2 Use the addition identity for sine to rewrite the left side of the equation. sin 1 x a 2 c 2a2 b2 The last equation can be solved by using the methods from Section 8.3. The steps of the pr... |
by the hypotenuse, 23 3 A B 2 1 1 2 2. 2 23 2 sin x 1 2 cos x 23 2 y 1 −1 0 −1 −2 3 α 2 x 1 ( 3, −1) Figure 9.4-3 Section 9.4 Using Trigonometric Identities 607 Step 5 Rewrite the equation by substituting cos 11p 6 for 23 2 and sin 11p 6 for 1 2 and then use the addition identity for sine. CAUTION When substituting si... |
5 x a cos x 1 1 sin 1 2 x a sin 11 p 2 x p 2 a 1.5708 0.9273 a cos 1 3 5 0.9273 0.6435 In one revolution, the maximum occurs at approximately 0.6435. Where the minimum value occurs is found in a similar manner. Exercises 9.4 In Exercises 1–27, find all solutions of the equation in the interval [0, 2P]. 1. sin2x 3 cos2... |
that c. Find all values of x in x maximum value of 0, 2p. can assume. that give the 36. 37. 38. 39. 40 23 sin x cos x sin x 23 cos x 2 sin x 2 cos x sin x cos x 4 sin x 3 cos Important Concepts Section 9.1 Basic trigonometry identities................ 574 Strategies for proving trigonometric identities... 574 Section ... |
Identities sin sin cos cos tan tan sin x cos y cos x sin y sin x cos y cos x sin y cos x cos y sin x sin y cos x cos y sin x sin y tan x tan y 1 tan x tan y tan x tan y 1 tan x tan y 610 Chapter Review 611 Cofunction Identities sin x cos p 2 a x b cos x sin p 2 a x b tan x cot sec x csc cot x tan a csc x sec a p 2 p 2... |
cos2x 1 1 tan2x 21 2 sin x 1 sin x 1 1 tan3x 21 1 a. c. e. b. 0 cot x 0 d. sec x b. sin x sin3x d. sin x 1 1 tan2x 2 Section 9.2 In Exercises 19–20, prove the given identity. 19. 20. cos x y 1 2 x y cos 2 cos x cos y 1 cos x y 2 1 cos2x sin2y 1 tan x tan y 21. Evaluate the following in exact form, where the angles and... |
2 cos x 2 cos3 x sin x sin 2x 34. sin 2x 1 tan x cot 2x 35. If tan x 5 12 and sin x 7 0, find sin 2x. 36. If cos x 15 17 and 0 6 x 6 p 2, find sin x 2. 37. If sin x 0, is it true that sin 2x 0? Justify your answer. 38. If cos x 0, is it true that cos 2x 0? Justify your answer. 39. Show 32 23 22 26 2 by computing cos p... |
simplified as follows. (See Example 4 of Section 9.2 for details of the simplification.) sin 1 x h 2 h sin x sin x cos h 1 h a b cos x sin h a h b The expression that represents the instantaneous rate of change is found by finding the limit of sin x a cos h 1 h cos x b sin h a h b as h approaches 0. lim hS0 a sin x co... |
parts are adequately supported. Problems like these can be modeled and solved by using vectors. See Exercise 50 in Section 10.6. 616 Chapter Outline 10.1 The Law of Cosines 10.2 The Law of Sines 10.3 The Complex Plane and Polar Form for Complex Numbers 10.4 DeMoivre’s Theorem and nth Roots of Complex Numbers 10.5 Vect... |
of Cosines, given below. In any triangle ABC, with sides of lengths a, b, and c, as in Figure 10.1-1, cos A b2 c2 a2 2bc. The other two equations in the Law of Cosines can be similarly rewritten in an alternate form. In this form, the Law of Cosines provides a description of each angle of a triangle in terms of the th... |
of Cosines. cos A b2 c2 a2 cos A 152 8.32 202 15 2 21 1 cos A 0.4261 A 115.2° 2bc 8.3 2 2ac cos B a2 c2 b2 cos B 202 8.32 152 2 1 cos B 0.7346 B 42.7° 8.3 20 21 2 Use the sum of angle measures in a triangle to find the third angle. Thus, A 115.2°, B 42.7°, C 180° 115.2° 42.7° C 22.1°. 1 and 22.1° 2 ■ Example 3 The Dis... |
° 78° 1 As shown in Figure 10.1-5, adjacent angles ABC and CBE form a straight angle, which measures 180°. 2 mABC 180° 78° 102° Using the SAS information in triangle ABC, apply the Law of Cosines. b2 a2 c2 2ac cos B b2 952 1002 2 95 2 b2 19,025 19,000 cos 102° b 219,025 19,000 cos 102° b 151.58 100 21 1 cos 102° The le... |
a2 b2 cos2 A b2 sin2 A c2 2bc cos A a2 b2 c2 2bc cos A cos2 A sin2 A 2 a2 b2 c2 2bc cos A Rearrange terms. Square each side. Factor out b sin A Simplify. 2 2 b2. 1 2 2 2 1 1 2 Pythagorean identity Substitute for x and y. 622 Chapter 10 Trigonometric Applications This proves the first equation in the Law of Cosines. Si... |
Cleveland at the same time and flies southeast at 200 miles per hour. How far apart are the planes after 1 hour and 36 minutes? 21. The pitcher’s mound on a standard baseball diamond (which is actually a square) is 60.5 feet from home plate. How far is the pitcher’s mound from first base? 2nd base 90 ft 90 ft Pitcher'... |
Two people stand on opposite sides of the hill where the tunnel entrances are to be located. Both can see a stake located 530 meters from the first person and 755 meters from the second. The angle determined by the two people and the stake (the vertex) is How long must the tunnel be? 77°. 31. One diagonal of a paralle... |
11°. How 37. Assuming that the circles in the following figure are mutually tangent, find the lengths of the sides and the measures of the angles in triangle ABC. 8.23 C B 11.27 13 A 38. Assuming that the circles in the following figure are mutually tangent, find the lengths of the sides and the measures of the angles... |
sin C). A 0 and sin C 0. Divide a sin A c sin C 626 Chapter 10 Trigonometric Applications This proves one proportion in the Law of Sines. Similar arguments beginning with angles A or B in standard position prove the other proportions. The Law of Sines can be used to solve triangles in the following cases. 1. Two angles... |
. Place angle A in standard position with terminal side b. If angle A is less than then there are four possibilities for side a. 90°, Ambiguous Case: SSA Information with A 66 90 (i) a b, and side a is too short to reach the third side: no solution. (ii) a b, and side a just reaches the third side and is perpendicular ... |
6 b sin B 7 sin B sin 65° sin B 7 sin 65° 6 sin B 1.06 There is no angle B whose sine is greater than 1. Therefore, there is no triangle satisfying the given data. ■ Example 3 Solve a Triangle with SSA Information An airplane A takes off from carrier B and flies in a straight line for 12 kilometers. At that instant, a... |
° in each case, there C 12 7.5 7.5 35° A c 113.4° B c Figure 10.2-11 66.6° B Section 10.2 The Law of Sines 631 B 66.6° Case 1 C 180° 35° 66.6° 78.4° 2 1 B 113.4° Case 2 C 180° 35° 113.4° 31.6° 2 1 Use the Law of Sines. Use the Law of Sines. c sin C c sin 78.4° a sin A 7.5 sin 35° c 7.5 sin 78.4° c 12.8 sin 35° c sin C ... |
� ab sin C. y c B h a D C b Figure 10.2-13 x A Proof Place the vertex of angle C at the origin, with side b on the positive x-axis, as in Figure 10.2-13. Then b is the base and h is the height of the triangle. area of triangle ABC 1 2 base height 1 2 bh. The proof of the Law of Sines shows that area of triangle ABC 1 2... |
B 44°, C 48°, b 12 6. A 67°, C 28°, a 9 7. A 102.3°, B 36.2°, a 16 8. B 97.5°, C 42.5°, b 7 In Exercises 9–16, find the area of triangle ABC under the given conditions. 9. a 4, b 8, C 27° 10. b 10, c 14, A 36° 11. c 7, a 10, B 68° 12. a 9, b 13, C 75° 13. a 11, b 15, c 18 14. a 4, b 12, c 14 15. a 7, b 9, c 11 16. a 1... |
region. Hint: Divide the region into triangles. 39. 120 55 89° 96° 103° 68.4 72° 40. 40 135 80° 20 120° 135° 23 1 3 75° 130° 30 30 41. A surveyor marks points A and B 200 meters apart on one bank of a river. She sights a point C on the opposite bank and determines the angles shown in the figure below. What is the dist... |
garden area. The 40-foot fence makes a building. What is the area of the garden? angle with the 58° 57° 49. Two straight roads meet at an angle of 40° in Harville, one leading to Eastview and the other to Wellston (see the figure on the next page). Eastview is 18 kilometers from Harville and 20 kilometers from Wellsto... |
It is 85°? 81° 67° 800 ft 8.6° 1 mile 52. A plane flies in a direction of 105° from airport A. 55. A hinged crane makes an angle of 50° with the ground. A malfunction causes the lock on the hinge to fail and the top part of the crane swings down (see the figure). How far from the base of the crane does the top hit the... |
a 1 2 s b 2 21. Combine the facts to prove Heron’s Formula. 10.3 The Complex Plane and Polar Form for Complex Numbers* Objectives • Graph a complex number in the complex plane • Find the absolute value of a complex number • Express a complex number in polar form • Perform polar multiplication The real number system is... |
a. 3 2i b. 4 5i Solution 3 2i a. 0 232 22 213 0 b. 0 4 5i 0 242 5 1 2 241 ■ 2 Absolute values and trigonometry lead to a useful way of representing be a nonzero complex number and denote complex numbers. Let a bi by r. Then r is the length of the line segment joining (a, b) and 0 (0, 0) in the plane. Let be the angle ... |
3-3. Therefore, must be a second- 13, 1 A u B quadrant angle. So, conditions. u 5p 6 satisfies these – 3 + i i 1 2 3 5π 6 real Thus, 23 i 2 cos a 5p 6 i sin 5p 6 b. Figure 10.3-3 ■ 640 Chapter 10 Trigonometric Applications Example 3 Find Polar Form Express 2 5i in polar form. Solution In this case, a 2 and b 5. r 2a2 b... |
polar form can be written in rectangular form by evaluating each term and simplifying. For example, Solution In this case, r1 2, u1 5p 6, r2 3, and u2 z1z2 r1r23 2 3 1 21 cos u1 1 cos 2 c 2 cos a 5p 6 i sin 2 23 2 a 23 i 5p cos cos 10p a 12 31p 12 u1 1 i sin. 7p 4 u22 4 5p 6 10p 12 a a i sin a i sin 7p 4 b u22 5p 6 21... |
1 Exercises 10.3 In Exercises 1–8, plot the point in the complex plane that corresponds to each number. 23. 1. 3 2i 2. 7 6i 4. 22 7i 7. 2i 3 5 a 2 i b 5. 8. 1 i 1 i 2 21 1 4 3 i 1 6 3i 2 3. 6. 8 3 5 3 i 2 i 1 21 1 2i 2 In Exercises 9–14, find each absolute value. 9. 12. 5 12i 0 2 3i 0 0 0 10. 13. 2i 0 12i 0 0 0 11. 14... |
34. 3 cos a p 12 p 8 i sin p 12b 2 cos a 7p 12 i sin 7p 12 b i sin p 8 b 12 cos a 3p 8 i sin 3p 8 b 7 2 a cos p 4 i sin p 4 b 35. 12 cos a 11p 12 i sin 11p 12 b 36. 37. 8 cos a 5p 18 cos 4 a p 9 6 cos a 7p 20 4 cos a p 10 i sin i sin i sin i sin 5p 18 b p 9 b 7p 20 b p 10b Section 10.3 The Complex Plane and Polar Form... |
and c di. b. Find the slope of the line N from 0 to c. Find the equation of the line L, through a bi and parallel to line N of part b. Hint: The point (a, b) is on L; find the slope of L by using part b and facts about the slope of parallel lines. d. Find the equation of the line M, through c di and parallel to line K... |
. and b. If, 1 2 2 2 644 Chapter 10 Trigonometric Applications c. If cos b cos u b u angles and same terminal side. Hint: cos u, sin u 1 2 sin b sin u, and show that in standard position have the 1 are points on the unit circle. 2 cos b, sin b and d. Use parts a–c to prove this equality rule for polar form: cos b i sin... |
25p 6 p 6 23 1 2 2 1623 16i cos a 32 a i sin p 6 b polar form i b rectangular form ■ NOTE tan 1 1 p 4 Example 2 Find Powers of Complex Numbers Evaluate 1 i 10. 2 1 Solution Express the complex number 1 i in polar form. 1 i 22 cos a p 4 i sin p 4 b Apply DeMoivre’s Theorem. 1 i 2 1 10 22 10p 4 i sin 10p 4 b 10 B cos a ... |
z4 16 cos a cos b i sin b 2 1 2p 3 i sin 2p 3 b, find s and b such that is a solution. In other words, find s and b such that cos b i sin b s 1 3 4 16 2 4 cos a 2p 3 i sin 2p 3 b. Use DeMoivre’s Theorem to rewrite the left side. cos 4b i sin 4b s4 1 16 2 cos a 2p 3 i sin 2p 3 b The equality rules for complex numbers (... |
sin nb 1 n r r 2 cos u i sin u cos u i sin u 2 2 1 1 s 3 sn 1 Therefore, sn r s 2n r and nb u 2kp b u 2kp n where k is any integer. Letting angles b. This is stated in the following formula for nth roots. k 0, 1, 2, p, n 1 produces n distinct Formula for nth Roots For each positive integer n, the nonzero complex numbe... |
■ The n distinct nth roots of 1 (the solutions of roots of unity. Because number 1 is u 0 produces a formula for roots of unity. cos 0 i sin 0. sin 0 0, Applying the root formula with ) are called the nth the polar form of the and cos 0 1 r 1 and zn 1 Formula for Roots of Unity For each positive integer n, there are n... |
ometric Applications Therefore, in the complex plane, every nth root of unity is exactly 1 unit from the origin. That is, the nth roots of unity all lie on the unit circle in the complex plane. Example 6 Find nth Roots of Unity Find the fifth roots of unity. Solution The fifth roots of unity have the following form. co... |
along the graph. The cursor will jump from vertex to vertex, that is, from one fifth root of unity to the next. Example 7 Graph Roots of Unity Graph the tenth roots of unity, and estimate the two tenth roots of unity in the first quadrant. Solution With a graphing calculator in parametric mode, set the range values as... |
5 29. x4 1 i 23 30. x4 8 8i 23 In Exercises 31–35, represent the roots of unity graphto obtain ically. Then use feature approximations of the form for each root (round to four places). trace a bi the 31. seventh roots of unity 32. fifth roots of unity 33. eighth roots of unity 34. twelfth roots of unity 35. ninth root... |
sin x cos u i sin u to show that the nonzero complex number r has two square roots and that 1 these square roots are negatives of each other. 2 Section 10.5 Vectors in the Plane 653 10.5 Vectors in the Plane Objectives • Find the components and magnitude of a vector • Perform scalar multiplication of vectors, vector a... |
1, y2 – y1) x Figure 10.5-5 The distance formula shows that and! OR have the same length.! PQ 4 2 2 0! 2 PQ! OR 2 5 1 1 4 0 1! PQ 2 2 and 2 2! OR 1 1 2 242 22 220 2 242 22 220 2 2 have the same slope: The lines containing › 4 2 5 1! OR and! OR ‹ PQ slope! PQ! PQ Because slope, and 2 4 1 2 ‹ slope OR › 2 0 4 0 2 4 1 2 b... |
which is also called its norm. H Magnitude The magnitude (or norm) of the vector 2a2 b2. v 7 7 v a, b I H is CAUTION Example 2 Find Components and Magnitude of a Vector The order in which the coordinates of the initial point and terminal point are subtracted to a, b obtain I H cant. For the points x2, y22 Q P and : x1,... |
〉 v 〈9, 3〉 3v x −2v 〈−6, −2〉 Figure 10.5-6 Also note that 2v 7 7 Therefore, v 7 7 6, 2 H I 3, 1 I H 2 232 12 210 6 1 2 2 22 240 2210 3v Similarly, it can be verified that 7 an illustration of the following facts. 7 2v 7 7 2210. Figure 10.5-6 is Geometric Interpretation of Scalar Multiplication The magnitude of the vect... |
a vector c, d c, d tor H is defined using the negative of a vector as follows. v and is denoted c, d H v 1 1, I 2H I I 2 1 1 is defined to be the vecv. Vector subtraction Vector Subtraction If u a, b II HH and v, c, d II HH u (v) then u v a, b II HH a c, b d HH II is the vector c, d II HH Example 5 Perform Vector Subt... |
0v 0 and r0 0 associative for addition commutative additive identity additive inverse distributive distributive associative for multiplication multiplicative identity multiplication by 0 u v v u Proof of u a, b Let and I H therefore, v c, d H Addition of real numbers is commutative;. I H u v c, d a, b I I H a c, b d H... |
, d II, v a, b II u HH 26. v 0 v 0 v 28. 30. r 1 u v 2 v r rs 2 1 ru rv sv 2 1 s rv 2 1 27. 29. 31. v v 0 2 1 r s v rv sv 2 1 1v v and 0v 0 32. Let v be the vector with initial point x1, y12 and let k be any real 1 and 1 x2, y22 terminal point number. a. Find the component form of v and kv. b. Calculate and c. Use the ... |
operations with linear combinations of vectors • Determine the direction angle of a vector • Determine resultant forces in physical applications In the previous section, vectors were introduced and vector arithmetic was defined. In this section, vectors are applied to real-world situations. Unit Vectors A vector with ... |
to write the rules for vector addition and scalar multiplication in terms of i and j. ai bj. I I ai bj 1 2 1 ci dj and cai cbj ai bj c 1 2 Example 2 Perform Operations with Linear Combinations If u 2i 6j and v 5i 2j, find 3u 2v. Solution 3u 2v 3 2i 6j 2 5i 2j 1 2 6i 18j 10i 4j 16i 22j 1 2 ■ Direction Angles ai bj v H ... |
a. The direction angle of u satisfies u tan u b a 13 5 2.6. y u θ 16 12 8 4 〈5, 13〉 13 4 5 8 a Figure 10.6-3 664 Chapter 10 Trigonometric Applications TAN1 Using the vector u is shown in Figure 10.6-3. b. The direction angle of v satisfies u key on a calculator indicates that u 68.96°. The y tan u b a 7 10 0.7. 〈–10, ... |
100 sin 70° j 2! OR 2175.162 145.272 227.56. is! OR 7 7 The direction angle of the resultant force satisfies u tan u 145.27 175.16 0.8294 A calculator in degree mode shows that u 39.67°. ■ Example 6 Resultant Force A 200-pound box lies on a ramp that makes an angle of with the horizontal. A rope is tied to the box fro... |
c, and its ground speed is the magnitude of 240 120 60° w –60 Figure 10.6-8 The direction angle of p (the angle it makes with the positive x-axis) is 90° 50° 40°. The angle that w makes with the positive y-axis is Section 10.6 Applications of Vectors in the Plane 667 so the direction angle of w, as measured from the po... |
6. u 2w w 8. 10. 1 2 1 4 7. 1 2 1 3v w 2 9. 2u 3v 20. 22. 24. v 4, 4 H v I 8, 0 H v 6j 21. v 5, 523 H I I 23. 25. v 4, 5 H v 4i 8j I 8u 4v w 2 1 11. 3 1 u 2v 2 6w 26. v 2i 8j 27. v 15i 10j 668 Chapter 10 Trigonometric Applications In Exercises 28–31, find a unit vector that has the same direction as the given vector. ... |
-6! is the component of the of Example 6, where TP 7 weight of the object parallel to the plane and is the component of the weight perpendicular to the plane.! TQ 7 7 7 38. An object weighing 50 pounds lies on an inclined angle with the horizontal. plane that makes a Find the components of the weight parallel and perpe... |
a point on the opposite shore directly north of her starting point. She can swim at 2.8 miles per hour, and there is a 1-mile-per-hour current in the river. In what direction should she swim in order to travel directly north (that is, what angle should the swimmer make with the south bank of the river)? 49. A river fl... |
d. I H I H w Section 10.5, 7 figure with Exercise 34 of Section 10.5). Show that u v is equivalent to the vector w with initial point (c, d) and terminal point (a, b) by now showing that and w have the same direction. u v 7 7 670 Chapter 10 Trigonometric Applications 10.6.A Excursion: The Dot Product Objectives • Find... |
Therefore, a b b a2 b2 2 2, I 1 then 1 c, d H ac bd ca db 2a2 b2 a 2 b 2. u 7 7 c, d H I a, b I H v u. v θ u Figure 10.6.A-1 The proofs of the last three statements are asked for in the exercises. Angles Between Vectors u v I u c, d H a, b I H are any nonzero vectors, then the angle between If and u and v is the small... |
a triangle, as shown in Figure 10.6.A-2. and the angle 0 and 0. c, d H u If, I v u 7 7 7 The lengths of two sides of the triangle are 2c2 d2. v 7 side (opposite angle Cosines produces the following result. The distance formula shows that the length of the third b d Therefore, the Law of a c u ) is 2 2 2. u 1 1 2 2 7 7... |
v Vectors u and v are said to be orthogonal (or perpendicular) if the angle between them is p 2 fact about orthogonal vectors follows. radians 90°, 1 2 or if at least one of them is 0. The key Orthogonal Vectors Let u and v be vectors. Then u and v are orthogonal exactly when uv 0. u v 0. Proof If u or v is 0, then ve... |
initial point at the origin and the same length and direction as as in the two cases shown in Figure 10.6.A-5.! QP, P u v Q projvu w O P u w v Q O projvu Figure 10.6.A-5 Because w is parallel to 10.6.A-5, the dot product: u projvu w kv w. it is orthogonal to v. As shown in Figure Consequently, by the properties of! QP... |
u, is the angle between u and v, u where compvu u v v u v cos u v u cos u This result is stated formally as follows. Projections and Components If u and v are nonzero vectors, and them, then u is the angle between and compvu u v v u cos U projvu compvu. 00 00 Example 6 Find Component Vectors If u 2i 3j and v 5i 2j, fi... |
foot-pounds. work done is 3850 110 35 1 2 When a force F moves an object in the direction of a vector d rather than in the direction of F, as shown in Figure 10.6.A-9, then the motion of the which is a force object can be considered as the result of the vector in the same direction as d. projdF, projvF 15° y v 75° F R... |
dot product when u w and, II 2, 5 HH u, v 4, 3 HH v w II 2 v w 1 u v 2 3u v 1 1 2 1 2w 7. 9. 11. 8. 10. 12. 2 2, 1 HH u. II v w 1 u v 2 u 4v 1 1 1 1 2 2 u v 2 2u w 2 In Exercises 13–18, find the angle between vectors u and v. 13. 14. 15. H u, v 4, 3 u 1, 2 H 0, 5 H I, v I u 2i 3j, v i 2, 4 H I I 16. u 2j, v 4i j 17. u... |
, s II HH c, d II, v a, b II u HH HH. 37. 38. u v w u v u w 1 ku v k 2 u v 1 u kv 2 39. 0 u 0 40. Suppose u parallel vectors. c 0, a. If a, b I H and v c, d H I are nonzero show that u and v lie on the same nonvertical straight line through the origin. v c show that a 0, (that is, v is a scalar b. If a u multiple of u)... |
, 2 2 1, Q 1 4, 3 2 53. A lawn mower handle makes an angle of with the ground. A woman pushes on the handle with a force of 30 pounds. How much work is done as she moves the lawn mower a distance of 75 feet on level ground? 60° 54. A child pulls a wagon along a level sidewalk by exerting a force of 18 pounds on the wag... |
................ 638 Real axis............................... 638 Imaginary axis........................... 638 Absolute value of a complex number......... 638 Argument.............................. 639 Modulus............................... 639 Polar form.............................. 639 Polar multiplication and divi... |
...... 662 Direction angle of a vector................. 663 Section 10.6.A Dot product............................. 670 Angle between vectors.................... 671 Parallel vectors.......................... 671 Angle theorem.......................... 672 Schwarz inequality....................... 673 Orthogonal vector... |
1 u22 i sin u1 1 u22 4 cos u i sin u r 3 The distinct nth roots of 1 r 3 2 4 n rn cos u i sin u u 2kp n b d a 1 i sin 2n r cos a c u 2kp n b cos nu i sin nu 4 are: 2 k 0, 1, 2, p, n 1 1 2 The distinct nth roots of unity are: cos 2kp n i sin 2kp n 1 k 0, 1, 2, p, n 1 Q x2, y22 1, then PQ › x1, y2 x2 H 2 y1I. and x1, y1... |
10.1 In Exercises 1–6, use the Law of Cosines to solve triangle ABC. 1. a 12, b 10, c 15 2. a 7.5, b 3.2, c 6.4 3. a 10, c 14, B 130° 4. a 7, b 8.6, C 72.4° 5. a 5, c 8, B 76° 6. a 90, b 70, c 40 7. Two trains depart simultaneously from the same station. The angle between their two tracks is 45 miles per hour and the ... |
34-foot shadow on level ground. The angle of elevation from the end of the 64°. shadow to the top of the pole is How long is the pole? Chapter Review 685 23. Two surveyors, Joe and Alice, are 240 meters apart on a riverbank. Each sights a flagpole on the opposite bank. The angle from the pole to Joe (vertex) to Alice ... |
i sin p 6 b 3p 8 b 686 Chapter Review 12 39. cos a i sin 3 cos a i sin 7p 12 5p 12 7p 12 b 5p 12 b Section 10.4 40. cos a p 12 i sin 18 p 12b 41. 13 3 c cos a 5p 36 i sin 12 5p 36b d In Exercises 42–46, solve the given equation in the complex number system, and express your answers in polar form. 42. x3 i 45. x4 i 43.... |
? Chapter Review 687 In Exercises 67 and 68, let ing: u 4i 3j and v 2i j. Find each of the follow- 67. projvu 68. compuv 69. If u and v have the same magnitude, show that u v and u v are orthogonal. 70. If u and v are nonzero vectors, show that the vector u kv is orthogonal to v, where k u v v2. 71. A 3500-pound automo... |
both sides of the identity p 2 i2 e 2 1 p e 2 i e ii p 2 i e A B p A calculator computes the value of Figure 10.C-1. So ii 0.2079. p 2 i i e 1 2 e to the power i. p 2 to be 0.2078795764, as shown in Figure 10.C-1 Euler’s formula can be used to define complex powers of e, that is, xiy e x e cos y i sin y The equation t... |
ependence of Sections ⎫ 11.1 ⎪ ⎬ 11.2 ⎪ ⎭ 11.3 11.5 11.7 > 11.4 >> 11.6 Sections 3.1, 3.2, and 3.4 are prerequisites for this chapter. Except for the discussion of standard equations for conics in Sections 11.1–11.4, Chapter 6 (Trigonometry) is also a prerequisite. When a right circular cone is cut by a plane, the inte... |
of all points X such that the sum of the distance from X to P and the distance from X to Q is k. Written algebraically with X representing a point XP XQ k x, y 1 2 on the ellipse, To draw the ellipse, pin the ends of a string of length k at points P and Q, as shown in Figure 11.1-1. Place a pencil against the string, ... |
] 2˛x˛ b˛ 2 a˛ 2˛y˛ 2 a˛ 2 2˛b˛ Dividing both sides by the ellipse satisfy the equation 2 2b˛ a˛ shows that the coordinates of every point on 2 x˛ 2 a˛ 2 y˛ 2 b˛ 1 Standard form of an ellipse Conversely, it can be shown that every point whose coordinates satisfy 0, c the equation is on the ellipse. The equation for an ... |
� (a, 0) and 2 and (c, 0), (c, 0) 2 b˛ (a, 0) where ±±b ±±a x-intercepts: y-intercepts: major axis is on the y-axis vertices foci c 2a˛ (0, a) and 2 and (0, c), (0, c) 2 b˛ (0, a) where Notice the following facts in both cases. • the foci are within the ellipse • the major axis always contains the foci and is determine... |
at the vertices. and 2 1 0, 5 0, 5, 2 The minor axis lies on the x-axis, with endpoints 4, 0 1 and 4, 0. 2 1 2 ■ Example 2 Graph an Ellipse on a Calculator Graph the ellipse with equation 4x˛ 2 9y˛ 2 36 on a calculator. Solution Solve the equation for y. 3.1 9y˛ 2 36 4x˛ 2 2 36 4x˛ 2 y˛ −4.7 4.7 y ± B 9 36 4x2 9 ± 2 3... |
with a parabolic dish at each focus. (Parabolas are discussed in Section 11.3.) The shape of the room and two parabolic dishes carry the quietest sound from one focus to the other. The width of the ellipse is 13 feet 6 inches and the length of the ellipse is 47 feet 4 inches. Assume that the ellipse is centered at the... |
are the minimum and maximum distances from Earth to the Sun? Solution Choose a coordinate system with the center of the ellipse at the origin and the Sun at the point to get a diagram of the orbit. See Figure 11.1-8a. c, 0 2 1 Earth Sun Figure 11.1-8a Figure 11.1-8b 698 Chapter 11 Analytic Geometry The length of the m... |
18 x˛ 1 8. y 4 2 0 −4 −2 −2 −4 x 2 4 9. y 4 2 −4 −2 0 −2 −4 4 2 −4 0 −2 −2 −4 4 2 0 −4 −2 −2 −4 y y y 10. 11. 124 −2 −2 −4 2 4 Section 11.1 Ellipses 699 Calculus can be used to show that the area of the ellipse 2 with equation y˛ 2 b˛ the area of each ellipse in Exercises 17 – 22. Pab. 2 x˛ 2 a˛ 1 is Use this fact to ... |
that the last equation in part a may be further simplified as a2 1 x c 2 y˛ 2 ˛ 2 a˛ 2 cx. 2 x˛ 25 2 y˛ 4 1 14. 2 x˛ 6 2 y˛ 16 1 4x˛ 2 3y˛ 2 12 16. 9x˛ 2 4y˛ 2 72 13. 15. 2 c˛ 2 a˛ 1 x˛ 2 2 a˛ 2 y˛ 2 a˛ 2 c˛ 2 a˛ 2 1. 2 27. Sketch the graph of b 8, b 12, and 2 2 y˛ 4 b 20. 1 x˛ 2 b˛ What happens to the b 2, b 4, for 7... |
of a Hyperbola Let P and Q be points in the plane and k be a positive number. The hyperbola with foci P and Q is the set of all points X such that the absolute value of the difference of the distance from X to P and the distance from X to Q is k. That is, 0 XP XQ 0 k, where X represents the point (x, y) and k is calle... |
a • the center is the midpoint between the foci and the midpoint between the vertices 2 a˛ c˛ 2 b˛ 2 • When the equation is in standard form with the x term positive and y term negative, the hyperbola intersects the x-axis and opens left and right. When the x term is negative and the y term is positive, the hyperbola i... |
2 x. 2 First plot the vertices and sketch 1 2 1 the auxiliary rectangle determined by the vertical lines and y ± b ± 3. the horizontal lines The asymptotes go through the origin and the corners of this rectangle, as shown on the left in Figure 11.2-2. It is then easy to sketch the hyperbola by drawing curves that are ... |
4 −2 0 2 x 4 −2 −4 (0, − 10) Figure 11.2-4 Writing the Equation of a Hyperbola Example 3 Find the Equation of a Hyperbola Find the equation of the hyperbola that has vertices at and passes through the point A asymptotes, and label the foci. 0, 1 1 2 Then sketch its graph by using the 3, 12 and 0, 1 B. 1 2 Solution The ... |
before sailors on Ship B. The speed of sound in air is approximately 1100 feet per second. Describe the possible locations of the explosion. Solution In 1 2 second, the sound traveled 1 2 1 1100 2, or 550 feet. Therefore, the explo- sion occurred at a point 550 feet closer to ship A than to ship B. That is, the differ... |
closer to the passenger ship at A, as illustrated in Figure 11.2-6c. y x2 75,625 − y2 1,666,775 = 1 A B x (−1320, 0) (−275, 0) (1320, 0) (275, 0) Figure 11.2-6c To find the exact location of the explosion, the sound must be detected at a third location that is the focus of another hyperbola that shares one of the two ... |
˛ 2 y˛ 2 16 14. 16. x2 4 y 2 1 3y˛ 2 5x˛ 2 15 18y˛ 2 8x˛ 2 2 0 18. x˛ 2 2y˛ 2 1 13. 15. 17. 708 Chapter 11 Analytic Geometry In Exercises 19–24, graph each equation using a graphing calculator. 30. y2 8 x2 36 1 at 6, 4 1 2 19. 4x2 y 2 16 21. 2 2y 2 1 x 20. 22. x2 4 x2 6 y2 1 y2 16 1 23. 3y 2 5x 2 15 24. 18y 2 8x 2 2 0 ... |
a point on the hyperbola with origin as follows. a. Let x, y P 2 1 c, 0 F11 foci F1 P 7 F2 P. the distance formula, and Exercise 31, and 2 By the definition of a hyperbola, Assume that F21 c, 0. 2 F1 P F2 2a 2 Show that the last equation simplifies as shown. cx a2 a2 x c 2 1 2 y 2 b. Show that the last equation in par... |
If x, y to the x, y is the length of the vertical segment from to is any point on the parabola, then the distance from as shown in Figure 11.3-2. y p 0 By the definition of a parabola, y (x, y) (0, p) (0, −p) x (x, −p) Figure 11.3-2 2 1 distance from distance from x, y 1 x, y 1 x 0 to to 2 2 2 0, p 2 1 0, p 2 1 y p 2 ... |
is 2p • the distance from the vertex to the focus and the distance from the vertex to the directrix is p • the vertex is the midpoint of the line segment joining the focus and the directrix Characteristics of Parabolas Section 11.3 Parabolas 711 For a nonzero real number p, parabolas have the following characteristics... |
, and equation of the parabola that passes through and has its focus on the x-axis. 0, 0 the point 1 Sketch the graph of the parabola, and label its focus and directrix. has vertex 1, 112 A B,, 2 Solution Because the focus is on the x-axis, the equation has the form Since is on the graph, 1, 112 A B 112 A 1 2 2 4p B 1 ... |
0 −20 Figure 11.3-8b 20 (41, 12) x 40 All radio waves hitting the dish are reflected into the focus, so the receiver should be located there. To find the focus, draw a cross section of the dish, with the vertex at the origin, as shown in Figure 11.3-8b. The equation of Because the point (41, 12) is on the this parabol... |
ola with vertex at the origin passing through the given points. 2 2 8x y˛ 19. 21. 1 1 1, 2 and 1, 2 1 2 2 20. 3,3 2 b a and 3 2 b 3, a 1, 2 and 1 2 1, 2 2 22. 1 2, 5 and 2, 5 2 1 23. Find the point on the graph of that is closest to the focus of the parabola. 24. Find the point on the graph of 2 3y x˛ that is closest t... |
ions, a similar result applies. in the rule of shifts the graph of the function 5 units to the right, 5 shifts the graph 5 units y f then shifts the graph 4 units upward, because For equations that are not func- that is, y 4 y f 4 and y with in an equation shifts the graph of the equation: • Let h and k be constant. Re... |
: 16 and 25 were not added to the left side of the equation. Actually were added, when the left side is multiplied out. Therefore, to leave the original equation unchanged, 64 and 225 must be added to the right side. 9 25 225 and x 4 1 2 8x 16 2 4 4 1 253 64 225 2 2 36 36 x 4 9 2 10y 25 36 NOTE Review the technique of ... |
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