2. There are an infinite number of possible triangles, all similar, with three given angles whose sum is 180.
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1 SECTION CHAPTER 8 Setion 8 1. There re n infinite numer of possile tringles, ll similr, with three given ngles whose sum is If two ngles α nd β of tringle re known, the third ngle n e found immeditely s γ 180 α β. Then the given side will e inluded etween two of the three known ngles.. It is lled the miguous se euse there re two tringles determined y the given informtion. Note: Answers hve een rounded to the numer of signifint digits given in Tle 1; n sign hs een used rther thn. 8. γ 180 ( ) γ α 180 ( ) α β 180 ( ) β α 180 ( ) α 0 sin sin sin sin sin 41 1 sin10 sin 33 1 sin10 14 m 1 m sin sin 9 sin101 sin 3 55 mm sin sin 5 sin sin83 sin m sin sin 43 sin sin105 sin 3 sin 5 5 mi sin sin 77 sin 0 sin 9 sin101 4 mm sin 47 sin m sin 5 sin mi 1. 3 ft, ft, α 30 SSA: h sin α sin tringle; se () 0. ft, ft, α 30 SSA: h sin α sin 30 3 > 0 tringles; se () ft, ft, α 30 SSA: h sin α sin > 3 1 tringle; se (d). 5 ft, ft, α 30 SSA: h sin α sin < 5 < tringles; se ()
2 1 CHAPTER 8 ADDITIONAL TOPICS IN TRIGONOMETRY 4. β 7.5, γ 54.5, 9.7 inhes α 180 ( ) α 98 sin sin 9.7 sin 98 sin in sin sin sin sin in. α 1.7, β 34.4, 18.3 km γ 180 ( ) sin sin sin sin γ.9 sin sin 34.4 sin sin km 1. km 8. α.3, 14.7 inhes, 35. inhes sin sin 14.7 sin sin sin β > 1 No solution 30. h sin α 17 sin ; h < <, so two tringles possile sin sin sin 43.5 sin sin β β 59.1 or β 10.9 Tringle I β 59.1 α + β + γ γ 180 γ 77.4 Tringle II β' 10.9 α + β' + γ' γ' 180 γ' 15. sin sin sin sin sin 43.5 sin m 138 sin 43.5 sin m h sin β 44 sin ; 135 h < <, so two tringles possile sin sin sin sin sin α α 5.0 or α 14.0 Tringle I α 5.0 α + β + γ γ 180 γ 9.7 sin sin sin 7.3 sin m 135 Tringle II α' 14.0 α' + β + γ' γ' 180 γ' 8.7 sin sin sin 7.3 sin 8.7 ' 141 m 135
3 SECTION α 137.3, 13.9 m, 19.1 m sin sin 13.9 sin sin β 8.73 β + α 0.03 > 180 No solution 3. β 33 50', 73 m, 140 m sin sin sin sin sin 3350' sin sin 3350' sin18 30' α 17 40' 1740 m γ 180 (33 50' ') γ 18 30' 38. h sin β 9 sin 30 4 h, so there is one tringle sin sin sin sin sin α 1 α 90 α + β + γ γ 180 γ 0 sin sin sin 30 sin 0 80 in 4 4. From the lw of sines sin sin sin sin nd sin sin sin sin () 40. α 37.3, 4.8 m k 4.8 sin is k suh tht 0 < < k gives no solution; k gives one solution; k < < gives two solutions. (1) sin sin sin sin os sin os sin os sin sin os sin (3) dding (1) nd () ut sin sin 1 (180 γ) sin 90 os similrly, sutrting (1) nd () sin sin os sin sin os sin sin os sin sin sin os sin os sin (4) dividing (4) y (3) ut os os 1 (180 γ) os 90 sin
4 14 CHAPTER 8 ADDITIONAL TOPICS IN TRIGONOMETRY sin os os sin sin sin os os tn tn ut tn ot 1 tn tn tn tn (B) from 7, 41 α 73 0 β tn 738 tn Answers gree to two signifint digits, the ury of the given informtion. 44. sin 53 sin 830' 10 sin 9830' 8.08 miles from A 4.8 miles from B 4. tn 43 5' h x h x tn 43 5' h h tn h x 000 tn tn 38 + h tn 435' h tn 38 h 1 tn 43 5' 000 tn tn 38 h tn38 1 tn 435' tn 435' h The distne ove se level h 14,490 feet to 4 signifint digits sin 1.7 sin11 sin α 45 γ " α 135 γ " sin(sev) SEV
5 SECTION α α 48 β β β + γ γ 180 γ 31 h 157 sin sin h sin sin 48 h 109 ft, to the nerest foot sin 3. R sin 58.4 R 9.73 mm to 3 signifint digits s R mm to 3 signifint digits 5. Let x e the length of the side in the horizontl tringle tht is lso in the vertil tringle with ngle γ. In the horizontl tringle sin(180 (α + β)) sin 180 os(α + β) os 180 sin(α + β) x sin d 0 os(α + β) ( 1) sin(α + β) sin(180 ( )) sin(α + β) x d sin α s(α + β) 1 In the vertil tringle s( ) Setion 8 tn γ h x h x tn γ d sin α s(α + β) tn γ. Use the lw of osines to determine the lrgest ngle (lrgest euse it is opposite the longest side). Then use the lw of sines to determine seond ngle nd the ft tht the sum of the ngles is 180 to determine the third. 4. Sustituting into + os γ gives + os γ whih redues to os γ 1 or γ 0, onfirming tht in n equilterl tringle ll ngles mesure 0.. There is no SSA ongruene theorem, sine the given informtion does not determine tringle uniquely. There is n SSS ongruene theorem sine three sides tht stisfy the tringle inequlity determine the tringle uniquely. 8. A tringle n hve t most one otuse ngle. Sine α 93.5 is otuse oth γ nd β must e ute. [β + γ ; thus oth β nd γ re less thn 90.]
6 1 CHAPTER 8 ADDITIONAL TOPICS IN TRIGONOMETRY 10. β 57.3,.08 m, 5.5 m + os β (.08)(5.5)os m Solve for smllest ngle: 5.5 sin 5.48 sin sin 5.48 sin 57.3 γ 53.7 α 9.0 or α 180 ( ) 1. α ', 8.44 in, 0.3 in + os α (8.44)(0.3)os ' 7.0 in Solve for smllest ngle: sin13550' sin sin 7 sin13550' β 1 30' γ 31 40' or γ 180 (135 50' ') 14. If 1.5 m, 5.3 m, 10.7 m, then sides nd re not long enough to onstrut tringle: + < mi, 0.7 mi, 1. mi Solve for the lrgest ngle: + os β (10.5)(1.)os β os β β sin γ. sin α 180 ( ) m, 9.4 m, 33.7 m Solving for the lrgest ngle: + os γ (31.5)(9.4)os γ os γ γ sin sin α 59.4 β 180 ( ) β + γ > 180 no solution. β 7.3, 13.7 yds, 0.1 yd + os β (13.7)(0.1)os yd 10.1 sin sin α 38.5 γ 180 ( ) 114. (Answers will vry slightly depending on method used.) 4. β 13.4, γ 17.3, 7. km α 180 (β + γ) 180 ( ) 30.3 sin sin sin sin sin13.4 sin km; 7. km. γ.4, 5.5 m, 5.5 m γ β.4 α 180 (γ + β) 180 ((.4 )) sin.4 sin m m, 5.3 m, 8. m 30. α 4.7, 18.1 m,. m
7 SECTION 8-17 Angle opposite longest side: + os α (5.3)(8.)os α α sin sin β 9.7 γ 180 (α + β) 180 ( ) γ γ 47.9, 35. in, 5.5 in sin sin 5.5 sin sin sin β No solution 34. h sin β 98.5 sin h < < There re two tringles. sin sin sin 5.1 sin sin γ γ 51.1 or 18.9 Tringle I γ 51.1 α + β + γ α α sin sin sin 5.1 sin m sin sin 18.1 sin 4.7. β 5.3 or β' sin β 5.3 γ 180 ( ) 8.0 β' γ' 180 ( ) sin 4.7 sin 8.0 ' sin m ' 7.93 m Tringle I: β 5.3, γ 8.0, 3.1 m Tringle II: β' 114.7, γ' 18., ' 7.93 m Tringle II γ' 18.9 α' + β + γ' α' α'.0 sin sin sin 5.1 sin ' 55.5 m 3. h A ,100sq.yds o h o sin 49.3 h 10.1sin sin 49.3 o A 701 sq. m. o h sin 41 h 4sin 41 4 o
8 18 CHAPTER 8 ADDITIONAL TOPICS IN TRIGONOMETRY A sin 41,900 o sq. ft. 4. o o o o o o o sin sin sin 41 o 9.5 sin o h 9.5sin 41 sin 7 h sin 7 o o 9.5sin 41 sin o sin sin41 o o A sin 7 sq. m. o sin o o s 109 A sq. yds. 4. Given + : + os γ lw of osines + + os γ sustitution os γ 0 os γ 0 γ x + y : os β x x os β : os γ y y os γ x y x os β + os γ os β x + x os β os β x os β ( os γ) os β + os γ os(180 γ) x os γ x x os γ 50. It is wiser to strt y using the lw of osines to find the lrgest ngle, in this se β. Then os 11 so β 10. Then sine, α must equl γ, hene from α + γ we get α γ 30. By finding α first, the inorret onlusion β 30 ws drwn from the eqution sin β we just sw, β ; tully, s 5. AB (85)(73) os m (8.)(8.)os θ (8)(3)os(144 50) 10. ft
9 SECTION θ After hours, Plne A hs trveled 800 miles, Plne B 1000 miles. The ngle etween them is (1000)(800) os mi 0. The ngle t the enter is An isoseles tringle is formed, so the other two ngles re Let x e the hord formed in the irle. Using the lw of sines to solve, 7.09 sin 70 x sin 40 x The perimeter, to 3 signifint digits, is 9x 43. m.. OA OB AB AB OA + OB (OA)(OB)os θ 5 1 ( 5 ) 5 + ( ) (5)( )os θ (4 5) (3 1) 5 θ 0.44 rdin 4. The sides of the tringle hve lengths of 5 + 7, , nd Find ngle γ first (ngle opposite longest side): (10)(7)os γ γ ' 13 7 Angle α: sin 9810 sin α ' Angle β: β 180 (α + γ) 180 (98 10' ') 49 40'. AB AC BC (AB) (AC) + (CB) (AC)(BC)os(ACB) os(acb) Angle ACB (CS) R + (ST) (R)(ST) os 1.4 (CS) (394)(1034) os 1.4 CS 40 height CS R miles Setion 8 3. Answers will vry. 4. A vetor tht hs mgnitude 1 is lled unit vetor.. The differene is the vetor tht represents the wind or urrent! Tht is, the pprent veloity is the veloity reltive to the ir, while the tul veloity is the resultnt of the pprent veloity nd the wind veloity, tht is, the veloity reltive to the ground. 8. The oordintes of P re given y (x p, y p ) (x x, y y ) (3, 15 7) (1, 8) Hene OP 1, 8 1. The oordintes of P re given y (x p, y p ) (x x, y y ) (0 9, 0 ( 7)) ( 9, 7) Hene OP 9,7 10. The oordintes of P re given y (x p, y p ) (x x, y y ) (8 ( 5), ( 1) ) (13, 3) Hene OP 13, OP AB 7,1 1. v v ( 4) 3 5 5
10 0 CHAPTER 8 ADDITIONAL TOPICS IN TRIGONOMETRY 0. v 10 ( 10) v ( 1) 5 4. u + v u v u v os(180 ). u + v u v u v os(180 ) (10)(84)os(18044 ) 190 gm 190 sin13 84 sin α (8.0)(.0) os(1804 ) 9.1 knots.0 sin 9.1 sin11 α sin137 v sin17 u v 14 kg; sin 33 sin137 u 1 kg sin19.5 u sin 3.7 v sin17.8 u 30 mph nd v 173 mph 3. (A) u + v 1, 3, 1 3, ( ), 0 (B) u v 1, 3, 1 3, ( ) 4, 4 (C) u v + 3w 1, 3, 3 0, 30,4 5,0 34. (A) u + v 3,, 3 ( ), 5,4 (B) u v 3,, 3 ( ), 1,0 (C) u v + 3w 3,, 3 3,0 9,40 13, 3., 5,0 0, 5 1,0 5 0,1 i 5 j 38. 0, 7 7 0,1 7 j uuur 40. AB 0 ( ), ( 1),3,0 0,3 1,0 3 0,1 i 3j 4. u v 3i j (i + 4j) 3i j i 4j i j 44. 3u + v 3(3i j) + (i + 4j) 9i j + 4i + 8j 13i + j 4. u 3v + w 3i j 3(i + 4j) + (i) 3i j i 1j + 4i i 14j 48. v u 1 v v ,1 5 1, v ( 3) 13 u 1 v v 1 13, 3 3, v 0 ( 17) u 1 v v , 17 0, v ( ) ( 7) 9 3 u 1 v v 1 ( i 7 j) 3 5. True. The stndrd vetor is n ext representtion of the vetor. 58. Flse. i is nother unit vetor, nd there re n infinite numer. 3 i 7 3 j 7, 3 3
11 SECTION True. v + v v hs the sme diretion s v nd twie the mgnitude.. Flse. i + j is not unit vetor. 4. u + v, d,, d. u + ( u),,,,, d d,, v + u ( ), ( ) 0, m(u + v) m,, d m, d 70. 1u 1, 1,1, u 7. v m m, m md m, m m, md m, m, d mu + mv (15)(3.9) os (5 45 ) 14 mph 3.9 sin 14 sin(70 ) θ 15 heding mph t R (300)(900) os sin sin α The third ngle: 180 ( ) 17 diretion v sin θ 4 55 θ mph t 349. ( ) 78. (A) prllel fore 500 sin l (B) fore perpendiulr 500 os l 80. Left: 41 sin 31 1; Right: 31 sin > 0 slide left 8. Let the left tension e represented y T L nd the right tension y T R. Then T L sin 4. + T R sin T L os 4. T R os 5.3 Solving the seond eqution for T L : T L T os5.3 R os 4. TR (os5.3 )(sin 4. ) Sustituting: + T os 4. R sin x(os5.3 )(sin 4. ) Grph y 1 + x sin 5.3 nd y 11. os 4. T R 77 l, T L 7 l. 84. Let the left tension e represented y T L nd the right tension y T R. Then T L sin 45 + T R sin T L os 45 T R os 0 Solving the seond eqution for T : T L L TR os 0 os 45 TR (os 0 )(sin 45 ) Sustituting: + T os 45 R sin x(os 0 )(sin 45 ) Grph y 1 + x sin 0 nd y 500. os 45 T R 390 l, T L 518 l
12 CHAPTER 8 ADDITIONAL TOPICS IN TRIGONOMETRY 8. Angle ABC 30 BC sin BC 000 kg, tension BC os 30 AB AB 000 os kg, ompression 88. Angle ABC: os(abc) 5 ABC 33. (Answers my vry due to rounding.) AB sin AB 9040 kg, ompression BC AB os os kg, tension Setion 8 4. If r is positive, θ represents the ngle etween the polr xis nd the ry onneting the point to the pole; θ my e ny of the oterminl suh ngles. If r is negtive θ represents the ngle etween the polr xis nd the ry opposite to the ry onneting the point to the pole; gin, θ my e ny of the oterminl suh ngles. 4. One set of polr oordintes is given y r x y, tn θ y x. If x 0, then θ n e given s if y is positive nd y. Answers will vry. or 3 if y is negtive
13 SECTION (, 10 ): The polr xis is rotted 10 lokwise (negtive diretion) nd the point is loted units from the pole long the negtive polr xis. (, 150 ): The polr xis is rotted 150 ounterlokwise (positive diretion) nd the point is loted units from the pole long the negtive polr xis. (, 330 ): The polr xis is rotted 330 ounterlokwise (positive diretion) nd the point is loted units long the positive polr xis os r 5. θ 4.. ( 3.500,.0) (0.8,.95)
14 4 CHAPTER 8 ADDITIONAL TOPICS IN TRIGONOMETRY ( 7.115, 5.557) (8.3, 34 ) (31, 59 ) (9.37, 15.8 ) 3. os 4os vries from vries from vries from 0 to 1 to 0 4 to 0 to 0 to 1 0 to 4 to3 1 to 0 4 to 0 3 to 0 to 1 0 to os 8os vries from vries from vries from vries from 0 to 4 0 to 1 to 0 8 to 0 4 to to 0 to 1 0 to 8 to3 4 to 3 1 to 0 8 to to 3 to 0 to 1 0 to 8 to 5 4 to 5 1 to 0 8 to to3 5 to 3 0 to 1 0 to 8 3 to7 4 3 to 7 1 to 0 8 to to 7 to 4 0 to 1 0 to sin3 sin3 vries from vries from vries from vries from 0 to 0 to 0 to 1 0 to to 3 to 1 to 0 to 0 3 to to 3 0 to 1 0 to to 3 3 to 1 to 0 to 0 3 to 5 to 5 0 to 1 0 to 5 to 5 to 3 1 to 0 to 0 to7 3 to 7 0 to 1 0 to 7 to to 5 1 to 0 to 0 M M M M
15 SECTION os 3os 3 3os vries from vries from vries from vries from 0 to 1 to 0 3 to 0 to 3 to 0 to 1 0 to 3 3 to 0 to3 1 to 0 3 to 0 0 to 3 3 to 0 to 1 0 to 3 3 to 44. os 4os 4os vries from vries from vries from vries from 0 to 1 to 0 4 to 0 to to 0 to 1 0 to 4 to to3 1 to 0 4 to 0 to 3 to 0 to 1 0 to 4 to 4. r + os θ r 4 + os θ r + 4 os θ (A) r 4 os θ r 4 os 3θ r 4 os 5θ (B) 7 leves in r 4 os 7θ (C) n leves in r os(nθ) ( > 0 nd n odd) -
16 CHAPTER 8 ADDITIONAL TOPICS IN TRIGONOMETRY 50. (A) r 4 os θ r 4 os 4θ r 4 os θ (B) 1 leves in r 4 os 8θ (C) n leves in r os nθ ( > 0 nd n even) x x y x x + y r os θ r r os θ 54. x + y 9 r 9 or r ±3 5. xy 1 (r os θ r sin θ) 1 r ( sin θ os θ) 1 r sin θ 1 r 1 sin r s θ 58. r( os θ + sin θ) 4 (r os θ) + r sin θ 4 x + y 4 0. r 8 os θ r 8r os θ x + y 8x. r 4 r 1 x + y 1 4. n r 1 + os (nθ) 1 1 smll petl inside 1 lrge petl smll petls etween lrge petls 3 3 smll petls inside 3 lrge petls 4 4 smll petls etween 4 lrge petls r 1 + os(nθ) will hve n lrge nd n smll petls. For n odd the smll petls re within the lrge petls. For n even the smll petls re etween the lrge petls.. r os θ (1) r sin θ () 0 θ π Divide () y (1): 1 sin os tn θ θ r, 4 4 [Note: (0, 0) is not solution to this system even though the grphs ross t the origin.]
17 SECTION (1) r sin θ () r os θ 0 θ 30 sin θ os θ 1 sin θ sin θ + sin θ 1 0 ( sin θ 1)(sin θ + 1) 0 sin θ 1 0 sin θ sin θ 1 sin θ 1 θ 30,150 θ 70 r 4, 4 r 8 (4, 30 ), (4, 150 ), ( 8, 70 ) [Note: (0, 0) is not solution to this system even though the grphs ross t the origin.] 70. P 1 (, 30 ) nd P (3, 0 ) d ( r) ( r ) rr os( ) 3 ()(3) os(030 ) 13 1os t 45 : 9k, t 90 : 14k, t 10 : 13k, t 150 : 11k 74. (A) e 0.: 11.5 (B) e 1: ellipse prol (C) e : hyperol Setion If z re iθ then the ngle θ is lled the rgument of z, usully hosen so tht π < θ π or 180 < θ 180. i 1 4. If z 1 re i nd z re z1 i( 1) re the two numers, then their quotient is e whih lies on the unit z irle with rgument θ 1 θ.
18 8 CHAPTER 8 ADDITIONAL TOPICS IN TRIGONOMETRY. If z e iθ i 3 is the numer, then w1 e i( ) 3, w e i( 4 ) 3, nd w3 e re the ue roots. They lie on the 4 unit irle with rguments,, nd, tht is, 10 prt (A) 1 + i 3 A sketh shows tht 1 + i 3 is ssoited with θ 10, r 1 + i 3 (os 10 + i sin 10 ) e 10 i (B) 3i A sketh shows tht θ 90, r 3 3i 3(os( 90 ) + i sin( 90 )) 3e ( 90 )i (C) 7 4i γ ( 7) ( 4) θ tn i 8.0(os( 150. ) + i sin( 150. )) 8.0e ( 150. )i 18. (A) 3 i A sketh shows tht θ, r 3 i os + i sin e (π/)i (B) + i A sketh shows tht θ 3, r i 3 3 os i sin 4 4 e(3π/4)i (C) 5i r ( 5) θ tn 1 5 5i 7.81(os( 0.9) + i sin( 0.9)) 7.81e 0.9i 0. (A) e 30 i (os 30 + i sin 30 ) i 3 + i
19 SECTION (B) e ( 3π/4)i 3 3 os i sin 4 4 i 1 i (C) 5.71e ( 0.48)i 5.71(os( 0.48) + i sin( 0.48)) 5.0.4i. (A) 3 e ( π/)i 3 os i sin 3 (0 i) i 3 (B) e 135 i (os i sin 135 ) i 1 + i (C).83e ( )i.83(os( ) + i sin( )).83( i( )).0.4i z 1 z r 1 e iθ 1 r e iθ r 1 r e i(θ 1 + θ ) 1 z1 re 1 r1 i z re z 1 z e 13 i 3e 93 i 18e 5 i 13i z1 e e 39 i 93 i z 3e. z 1 z r 1 e i θ1 r e i θ r 1 r e i(θ 1 + θ ) i1 z1 re 1 r1 i z re z 1 z 3e 7 i e 97 i e 14 i 1 7i z 3e 97 i z e 1.5e ( 30 )i 8. z 1 z r 1 e i θ1 r e i θ r 1 r e i(θ 1 + θ ) z1 z z 1 z 7.11e 0.79i.e 1.07i 18.91e 1.8i 18.91e 1.8i z 1 z i i1 re 1 r1 i re 7.11e.e 0.79i 1.07i r ei(θ 1 θ ) r ei(θ 1 θ ) r ei(θ 1 θ ).7e ( 0.8)i 30. z n r n e nθi (5e 15 i ) e (3 15 )i 15e 45 i 3. z n r n e nθi ( e 15 i ) 8 ( ) 8 e (8 15 )i 1e 10 i
20 30 CHAPTER 8 ADDITIONAL TOPICS IN TRIGONOMETRY 34. ( 3 + i): r, θ 30 e 30 i z n r n e nθi ( 3 + i) 8 (e 30 i ) 8 8 e (8 30 )i 5e 40 i 3. ( 1 + i): r, θ 135 e 135 i z n r n e nθi ( e 135 i ) 4 ( ) 4 e (4 135 )i 4e 540 i 4(os i sin 180 ) 4( 1 + 0i) ( 3 + i): r, θ 150 e 150 i z n r n e nθi (e 150 i ) 5 5 e (5 150 )i 3e 750 i 3e 30 i 3(os 30 + i sin 30 ) i i : r 1, θ 40 1e 40 i z n r n e nθi (1e 40 i ) e (3 40 )i 1e 70 i 1e 0 i os 0 + i sin z 1/n r 1/n e [(θ/n) + ((30 k)/n) ]i (8e 45 i ) 1/3 8 1/3 e [(45 /3) + ((30 k)/3)]i e w 1 e 15 i w e ( )i e 135 i w 3 e ( )i e 55 i ( k)i 4. ( 1 + i): r, θ 135 e 135 i z 1/n r 1/n e [(θ/n) + ((30 k)/n) ]i ( e 135 i ) 1/3 ( 1/ ) 1/3 e [(135 /3) + ((30 k)/3)]i 1/ e w 1 1/ e 45 i w 1/ e 15 i w 3 1/ e 85 i 48. z 1 1e 0 i z 1/n r 1/n e [(θ/n) + ((30 k/n)]i (1e 0 i ) 1/4 1 1/4 e [0 + ((30 k)/4)]i 1e 90 ki w 1 1e 0 i 1 + 0i w 1e 90 i 0 + i w 3 1e 180 i 1 + 0i w 4 1e 70 i 0 i 50. z 8 8e 180 i z 1/n r 1/n e [(θ/n) + ((30 k)/n]i (8e 180 i ) 1/3 8 1/3 e [(180 /3) + ((30 k)/3)]i e ( k)i w 1 e 0 i (os 0 + i sin 0 ) i w e 180 i (os i sin 180 ) w 3 e 300 i (os i sin 300 ) 1 3 i ( k)i 44. z 1/n r 1/n e [(θ/n) + ((30 k)/n)]i (1e 90 i ) 1/4 1 1/4 e [(90 /4) + ((30 k)/4)]i e w 1 e.5 i w e 11.5 i w 3 e 0.5 i w 4 e 9.5 i w 3-1 w y y w w 4 w 1 w 3 w 1 1 x x ( k)i 5. z i e 90 i z 1/n r 1/n e [(θ/n) + ((30 k)/n)]i (1e 90 ) 1/5 1 1/5 e [( 90 /5) + ((30 k)/5]i 1e w 1 1e ( 18 )i w 1e 54 i w 3 1e 1 i w 4 1e 198 i w 5 1e 70 i ( k)i -1 w 4 w 3 y 1-1 w 5 w 1 w 1 x
21 SECTION (A) x ( ) , is root of x x is degree 3 so there re two more roots. (B) y is the sping etween roots 3 w 1 From prolem 50, w i, w 1 3 i - w x w 3 (C) (1 + 3 i) [1 + 3( 3 i) + 3( 3 i) + ( 3 i) 3 ] + 8 [ i i] (D) In the sme mnner, (1 3 i) x x 3 4 x 4 1/3 (4e 0 i ) 1/3 z 1/n r 1/n e [(θ/n) + ((30 k)/n)]i (4e 0 i ) 1/3 4 1/3 e [(0 /3) + ((30 k)/3)]i 4e 10 ki x 1 4e 0 i 4 x 4e 10 i 1 3 4(os 10 + i sin 10 ) 4 i + 3 i x 3 4e 40 i 1 3 4(os 40 + i sin 40 ) 4 i 3 i 58. x x 3 7 x ( 7) 1/3 7e 180 i z 1/n r 1/n e [(θ/n) + ((30 k)/n)]i (7e 180 i ) 1/3 7 1/3 e [(180 /3) + ((30 k)/n)]i 3e ( k)i x 1 3e 0 i 3(os 0 + i sin 0 ) i i x 3e 180 i 3(os i sin 180 ) 3( 1 + 0i) 3 x 3 3e 300 i 3(os i sin 300 ) i i 0. Flse. If z 1 r 1 e i90 nd z r e i90, z1 r1 then z r ei0.. True. The squre roots of re i0 re r e i0 nd r e i180, whih re the rel numers r nd r. 4. Flse. e i0 is sixth root of 1, ut (e i0 ) e i10 is not 1. z1 r. 1 (os 1 isin 1 ) r 1 z r(os isin ) r os 1isin 1 os isin os isin os isin r 1 r os1os ios 1sin ios sin 1i sin 1sin os i sin r1 r (os1os sin 1sin ) i(ossin 1os 1sin ) r1 1 r [os(θ 1 θ ) + i sin(θ 1 θ )] r 1 r ei(θ 1 θ )
22 3 CHAPTER 8 ADDITIONAL TOPICS IN TRIGONOMETRY 8. For k 0, r 1/n e (θ/n + (k 30 )/n)i r 1/n e (θ/n)i For k n, r 1/n e (θ/n + (k 30 )/n)i r 1/n e (θ/n + 30 )i r 1/n e (θ/n)i 70. x x 1 x ( 1) 1/ 1e 180 i z 1/n r 1/n e [(θ/n) + ((30 k)/n)]i (1e 180 i ) 1/ 1 1/ e [(180 /) + ((30 k)/)]i 1e x 1 1e 30 i x 1e 90 i x 3 1e 150 i x 4 1e 10 i x 5 1e 70 i x 1e 330 i ( k)i 7. x 3 i 0 x 3 i x i 1/3 1e 90 i z 1/n r 1/n e [(θ /n) + ((30 k)/n)]i (1e 90 i ) 1/3 1 1/3 e [(90 /3) + ((30 k)/3)]i 1e x 1 1e 30 i x 1e 150 i x 3 1e 70 i 74. P(x) x 1; find x 1 1/ (1e 0 i ) 1/ nd write s ftors. z 1/n r 1/n e [(θ/n) + ((30 k)/n) ]i (1e 0 i ) 1/ 1 1/ e [(0 /) + ((30 k)/)]i 1e 0 ki x 1 e 0 i 1 x e 0 i i x 3 e 10 i i x 4 e 180 i 1 x 5 e 40 i 1 3 i x e 300 i 1 3 i P(x) x (x 1)(x + 1) x i x i x i x i Chpter 8 Group Ativity ( k)i Prolem 1: In the figure, the distne from P to F is r. Sine P hs retngulr oordintes (r os θ, r sin θ) nd Q hs retngulr oordintes ( p, r sin θ), the distne from P to d is r os θ ( p) r os θ + p. Then using PF e PD s stted: r e(r os θ + p) r er os θ + ep r(1 e os θ) ep ep r 1 e os Prolems 4 re left to the student. If 0 < e < 1, the oni setion is n ellipse, If e 1, the oni setion is prol. If e > 1, the oni setion is hyperol.
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