Solution Manual for Trigonometry, 5th Edition

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SOLUTIONSMANUALEDGARREYESSoutheastern Louisiana UniversityTRIGONOMETRYFIFTHEDITIONMark DugopolskiSoutheastern Louisiana University

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Table of ContentsChapter P .................................................................................................1Chapter 1................................................................................................32Chapter 2................................................................................................83Chapter 3..............................................................................................137Chapter 4..............................................................................................195Chapter 5..............................................................................................245Chapter 6..............................................................................................302

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P.1 The Cartesian Coordinate System1For Thought1.False, the point (2,−3) is in Quadrant IV.2.False, the point (4,0) does not belong to anyquadrant.3.False, since the distance is√(a−c)2+ (b−d)2.4.False, sinceAx+By=Cis a linear equation.5.True6.False, since√72+ 92=√130≈11.47.True8.True9.True10.False, since the radius is 3.P.1 Exercises1.ordered2.Cartesian3.x-axis4.origin5.Pythagorean theorem6.circle7.linear equation8.y-intercept9.(4,1), Quadrant I10.(−3,2), Quadrant II11.(1,0),x-axis12.(−1,−5), Quadrant III13.(5,−1), Quadrant IV14.(0,−3),y-axis15.(−4,−2), Quadrant III16.(−2,0),x-axis17.(−2,4), Quadrant II18.(1,5), Quadrant I19.c=√(√3)2+ 12=√4 = 220.Sincea2+a2=√22, we get 2a2= 2 ora2= 1.Thena= 1.21.Sinceb2+ 22= 32, we getb2+ 4 = 9 orb2= 5.Thenb=√5.22.Sinceb2+(12)2= 12, we getb2+ 14 = 1orb2= 34 . Thus,b=√32 .23.Sincea2+ 32= 52, we geta2+ 9 = 25ora2= 16. Thena= 4.24.c=√32+ 22=√9 + 4 =√1325.√4·7 = 2√726.√25·2 = 5√227.√5√9=√5328.√3√16=√3429.√2√3·√3√3=√6330.√3√5·√5√5=√15531.2√3√5·√5√5=2√15532.5√3·√3√3=5√3333.1√3·√3√3=√3334.3√2·√2√2=3√2235.√2√3·√3√3=√6336.√5√2·√2√2=√10237.Distance is√(4−1)2+ (7−3)2=√9 + 16 =√25 = 5, midpoint is (2.5,5)38.Distance is√144 + 25 = 13,midpoint is (3,0.5)39.Distance is√(−1−1)2+ (−2−0)2=√4 + 4 = 2√2, midpoint is (0,−1)

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2Chapter PAlgebraic Prerequisites40.Distance is√4 + 4 = 2√2, midpoint is (0,1)41.Distance is√√√√(√22−0)2+(√22−0)2=√24 + 24 =√1 = 1, midpoint is(√2/2 + 02,√2/2 + 02)=(√24,√24)42.Distance is√(√3−0)2+ (1−0)2=√3 + 1 = 2, midpoint is(√3 + 02,1 + 02)=(√32,12)43.Distance is√(√18− √8)2+(√12− √27)2=√(3√2−2√2)2+ (2√3−3√3)2=√(√2)2+ (−√3)2=√5,midpoint is(√18 +√82,√12 +√272)=(3√2 + 2√22,2√3 + 3√32)=(5√22,5√32)44.Distance is√(√72− √50)2+(√45− √20)2=√(6√2−5√2)2+ (3√5−2√5)2=√(√2)2+ (√5)2=√7,midpoint is(√72 +√502,√45 +√202)=(6√2 + 5√22,3√5 + 2√52)=(11√22,5√52)45.Distance is√(1.2 + 3.8)2+ (4.4 + 2.2)2=√25 + 49 =√74, midpoint is (−1.3,1.3)46.Distance is√49 + 81 =√130,midpoint is (1.2,−3)47.Distance is√π2+ 42, midpoint is(3π4,12)48.Distance is√π2+ 42, midpoint is(π4,12)49.Distance is√(2π−π)2+ (0−0)2=√π2=π,midpoint is(2π+π2,0 + 02)=(3π2,0)50.Distance is√(π−π2)2+ (1−1)2=√π24=π2 , midpoint is(π+π/22,1 + 12)=(3π4,1)51.Distanceis√(π2−π3)2+(−13−12)2=√π236 + 2536 =√π2+ 256,midpoint is⎛⎜⎜⎝π3 +π22,12−132⎞⎟⎟⎠=(5π12,112)52.Distanceis√(π−2π3)2+(−1 + 12)2=√π29+ 14 =√4π2+ 96,midpoint is⎛⎜⎜⎝2π3+π2,−12−12⎞⎟⎟⎠=(5π6,−34)53.Center(0,0), radius 435x35y54.Center (0,0), radius 12-2x2-2y

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P.1 The Cartesian Coordinate System355.Center (−6,0), radius 6-14-10x-66y56.Center (0,3), radius 33-3x37y57.Center (2,−2), radius 2√225x-2-5y58.Center (4,−2), radius 2√5410x-4-6y59.x2+y2= 760.x2+y2= 12 since (2√3)2= 1261.(x+ 2)2+ (y−5)2= 1/462.(x+ 1)2+ (y+ 6)2= 1/963.The distance between (3,5) and the originis√34 which is the radius. The standardequation is (x−3)2+ (y−5)2= 34.64.The distance between (−3,9) and the originis√90 which is the radius. The standardequation is (x+ 3)2+ (y−9)2= 90.65.Note, the distance between (√2/2,√2/2)and the origin is 1. Thus, the radius is 1.The standard equation isx2+y2= 1.66.Note, the distance between (√3/2,1/2)and the origin is 1. Thus, the radius is 1.The standard equation isx2+y2= 1.67.The radius is√(−1−0)2+ (2−0)2=√5.The standard equation is (x+1)2+(y−2)2= 5.68.Since the center is (0,0) and the radius is 2,the standard equation isx2+y2= 4.69.Note, the center is (1,3) and the radius is 2.The standard equation is (x−1)2+(y−3)2= 4.70.The radius is√(2−0)2+ (2−0)2=√8. Thestandard equation is (x−2)2+ (y−2)2= 8.71.We solve fora.a2+(35)2=1a2=1−925a2=1625a=±4572.We solve fora.a2+(−12)2=1a2=1−14a2=34a=±√3273.We solve fora.(−25)2+a2=1a2=1−425a2=2125a=±√215

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4Chapter PAlgebraic Prerequisites74.Solve fora:(23)2+a2=1a2=1−49a2=59a=±√5375.y= 3x−4 goes through (0,−4),(43,0).25x-3-4y76.y= 5x−5 goes through (0,−5),(1,0).24x-55y77.3x−y= 6 goes through (0,−6), (2,0).13x-63y78.5x−2y= 10 goes through (0,−5), (2,0).13x-55/2-4y79.x+y= 80 goes through (0,80), (80,0).8040x2080y80.2x+y=−100 goes through (0,−100),(−50,0).4020x10050y81.x= 3y−90 goes through (0,30),(−90,0).-90-45x3015y82.x= 80−2ygoes through (0,40),(80,0).8040x4020y

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P.1 The Cartesian Coordinate System583.12x−13y= 600 goes through (0,−1800),(1200,0).1200600x-1800-900y84.23y−12x= 400 goes through (0,600),(−800,0).-800-400x600300y85.Intercepts are (0,0.0025),(0.005,0).0.005x0.0025y86.Intercepts are (0,−0.3),(0.5,0).0.5-0.5x-0.30.3y87.x= 537x-55y88.y=−25-5x-1-3y89.y= 44-4x35y90.x=−3-2-4x-33y91.x=−4-3-5x-44y

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6Chapter PAlgebraic Prerequisites92.y= 5-55x46y93.Solving fory, we havey= 1.-11x2-1y94.Solving forx, we getx= 1 .-12x1-1y95.y=x−20 goes through (0,−20),(20,0).2020x2020y96.y= 999x−100 goes through (0,−100),(100/999,0).0.51 x100100y97.y= 3000−500xgoes through (0,3000),(6,0).64x30001000y98.y=1300 (200x−1) goes through (0,−1/300),(1/200,0).0.010.01x0.010.01y99.The hypotenuse is√62+ 82=√100 = 10.100.The other leg is√102−42=√84 = 2√21 ft.101.a) Letrbe the radius of the smaller circle.Consider the right triangle with verticesat the origin, another vertex at the centerof a smaller circle, and a third vertex atthe center of the circle of radius 1.Bythe Pythagorean Theorem, we obtain1 + (2−r)2=(1 +r)25−4r=1 + 2r

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P.1 The Cartesian Coordinate System74=6rr=23.The diameter of the smaller circleis 2r=43.b) Thesmallestcirclesarecenteredat(±r,0) or (±4/3,0). The equations of thecircles are(x−43)2+y2= 49and(x+ 43)2+y2= 49102.Draw a right triangle with vertices at the cen-ters of the circles, and another vertex at apoint of intersection of the two circles.Thelegs of the right triangle are 5 and 12.Bythe Pythagorean theorem, the hypotenuse is√52+ 122= 13.103.LetC(h, k) andrbe the center and radius ofthe smallest circle, respectively. Thenk=−r.We consider two right triangles each of whichhas a vertex atC.The right triangles have sides that are perpen-dicular to the coordinate axes. Also, one sideof each right triangle passes through the centerof a larger circle.Applying the Pythagorean Theorem, we list asystem of equations(r+ 1)2=h2+ (1−r)2(2−r)2=h2+r2.The solutions arer= 1/2,h=√2, andk=−r=−1/2.The equation of the smallest circle is(x− √2)2+ (y+ 1/2)2= 1/4.104.We apply symmetry to the centers of theremaining three circles.From the answer orequation in Exercise 103, the equations of theremaining circles are(x− √2)2+ (y−1/2)2=1/4(x+√2)2+ (y−1/2)2=1/4(x+√2)2+ (y+ 1/2)2=1/4.105.The midpoint of (0,20.8) and (48,27.4) is(0 + 482,20.8 + 27.42)= (24,24.1).In 1994 (= 1970 + 24), the median age at firstmarriage was 24.1 years.106.a) Ifh= 0, then 0 = 0.229n+ 5.203.Thenn=−5.203/0.229≈ −22.72.Then-intercept is(−22.72,0).There were no unmarried couples in 1977(≈2000−22.7). Nonsense.b) Ifn= 0, thenh= 0.229(0) + 5.203 =5.203.Theh-intercept is (0,5.203).In2000, there where 5,203,000 unmarried-couple households.107.The distance between (10,0) and (0,0) is 10.The distance between (1,3) and the origin is√10.If two points have integer coordinates,then the distance between them is of the form√s2+t2wheres2, t2∈ {0,1,22,32,42, ...}={0,1,4,9,16, ...}.Note, there are no numberss2andt2in{0,1,4,9,16, ...}satisfyings2+t2= 19.Thus, one cannot find two points with integercoordinates whose distance between them is√19.108.One can assume the vertices of the righttriangle areA(0,0),B(1,√3), andC(1,0).The midpoint of the hypotenuseABis(12,√32). The distance between the midpointandCis√(12−1)2+(√32−0)2= 1, whichis also the distance from the midpoint toA,and the distance from the midpoint toB.111.On day 1, break off a 1-dollar piece and paythe gardener.On day 2, break of a 2-dollar and pay the gar-dener.The gardener will give you back yourchange which is a 1-dollar piece.

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8Chapter PAlgebraic PrerequisitesOn day 3, you pay the gardener with the 1-dollar piece you received as change from theprevious day.On day 4, pay the gardener with the 4-dollarbar.The gardener will give you back yourchange which will consist of a 1-dollar pieceand a 2-dollar piece.On day 5, you pay the gardener with the 1-dollar piece you received as change from theprevious day.On day 6, pay the gardener with the 2-dollarpiece you received as change from day 4. Thegardener will give you back your change whichis a 1-dollar piece.On day 7, pay the gardener with the 1-dollarpiece you received as change from day 6.112.LetABCbe a right triangle with verticesatA(2,7),B(0,−3), andC(6,1). Notice, themidpoints of the sides ofABCare (3,−1),(4,4) and (1,2). The area ofABCis12AC×BC=12√42+ 62√62+ 42=12 (52)=26.P.1 Pop Quiz1.The distance is√16 + 4 =√20 = 2√5.2.Center (3,−5), radius 93.Completing the square, we find(x2+ 4x+ 4) + (y2−10y+ 25)=−28 + 4 + 25(x+ 2)2+ (y−5)2=1.The center is (−2,5) and the radius is 1.4.The distance between (3,4) and the origin is5, which is the radius.The circle is given by(x−3)2+ (y−4)2= 25.5.By settingx= 0 andy= 0 in 2x−3y= 12we find−3y= 12 and 2x= 12, respectively.Sincey=−4 andx= 6 are the solutions ofthe two equations, the intercepts are (0,−4)and (6,0).6.(5,−1)For Thought1True, since the number of gallons purchased is20 divided by the price per gallon.2.False, since a student’s exam grade is a functionof the student’s preparation. If two classmateshad the same IQ and only one prepared thenthe one who prepared will most likely achievea higher grade.3.False, since{(1,2),(1,3)}is not a function.4.True5.True6.True7.False, the domain is the set of all real numbers.8.True9.True, sincef(0) = 0−20 + 2 =−1.10.True, since ifa−5 = 0 thena= 5.P.2 Exercises1.function2.independent, dependent3.domain, range4.parabola5.function6.function7.Note,b= 2πais equivalent toa=b2π. Thenais a function ofb, andbis a function ofa.8.Note,b= 2(5 +a) is equivalent toa=b−102.Thenais a function ofb, andbis a functionofa.9.ais a function ofbsince a given denominationhas a unique length. Since a dollar bill and afive-dollar bill have the same length, thenbisnot a function ofa.

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P.2 Functions910.Since different U.S. coins have differentdiameters, thenais a function ofb, andbis a function ofa.11.Since an item has only one price,bis a functionofa. Since two items may have the same price,ais not a function ofb.12.ais not a function ofbsince there may betwo students with the same semester gradesbut different final exams scores.bis not afunction ofasince there may be identical finalexam scores with different semester grades.13.ais not a function ofbsince it is possible thattwo different students can obtain the samefinal exam score but the times spent onstudying are different.bis not a function ofasince it is possible thattwo different students can spend the same timestudying but obtain different final exam scores.14.ais not a function ofbsince it is possible thattwo adult males can have the same shoe sizebut have different ages.bis not a function ofasince it is possible fortwo adults with the same age to have differentshoe sizes.15.Since 1 in≈2.54 cm,ais a function ofbandbis a function ofa.16.Since there is only one cost for mailing a firstclass letter, thenais a function ofb.Sincetwo letters with different weights each under1/2-ounce cost 47 cents to mail first class,bisnot a function ofa.17.Sinceb=a3anda=3√b, we get thatbis a function ofa, andais a function ofb.18.Sinceb=a4anda=±4√b, we get thatbis a function ofa, butais not a function ofb.19.Sinceb=|a|, we getbis a function ofa. Sincea=±b, we findais not a function ofb.20.Note,b=√asincea≥0, anda=b2. Thus,bis a function ofa, andais a function ofb.21.A=s222.s=√A23.s=√2d224.d=s√225.P= 4s26.s=P/427.A=P2/1628.d=√2A29.y= 2x−1 has domain (−∞,∞) and range(−∞,∞), some points are (0,−1) and (1,1)11x23y30.y=−x+ 3 has domain (−∞,∞) and range(−∞,∞), some points are (0,3) and (3,0)23x32y31.y= 5 has domain (−∞,∞) and range{5},some points are (0,5) and (1,5)22 x46y32.y=−4 has domain (−∞,∞) and range{−4},some points are (0,−4) and (1,−4)33 x35y

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10Chapter PAlgebraic Prerequisites33.y=x2−20 has domain (−∞,∞) and range[−20,∞), some points are (0,−20) and (6,16)66x301030y34.y=x2+ 50 has domain (−∞,∞) and range[50,∞), some points are (0,50) and (5,75)510x100150y35.y= 40−x2has domain (−∞,∞) and range(−∞,40], some points are (0,40) and (6,4)66x3050y36.y=−10−x2has domain (−∞,∞) and range(−∞,−10], some points are (0,−10)and (4,−26)44x203010y37.y=x3hasdomain(−∞,∞)andrange(−∞,∞), some points are (0,0) and (2,8)12 x88y38.y=−x3has domain (−∞,∞) and range(−∞,∞), some points are (0,0) and (0,−8)12 x88y39.y=√x−10 has domain [10,∞) and range[0,∞), some points are (10,0) and (14,2)101040x24y40.y=√x+ 30 has domain [−30,∞) and range[0,∞), some points are (−30,0) and (−26,2)3010x26y

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P.2 Functions1141.y=√x+ 30 has domain [0,∞) and range[30,∞), some points are (0,30) and (400,50)200600x3060y42.y=√x−50hasdomain[0,∞)andrange [−50,∞), some points are (0,−50) and(900,−20)10002000x5010y43.y=|x| −40 has domain (−∞,∞) and range[−40,∞), some points are (0,−40) and (40,0)2060x4010y44.y=2|x|hasdomain(−∞,∞)andrange[0,∞), some points are (0,0) and (1,2)11x224y45.y=|x−20|has domain (−∞,∞) and range[0,∞), some points are (0,20) and (20,0)102050x2040y46.y=|x+ 30|has domain (−∞,∞) and range[0,∞), some points are (0,30) and (−30,0)3060x3010y47.3·4−2 = 1048.3(16) + 4 = 5249.−4−2 =−650.−8−2 =−1051.|8|= 852.| −1|= 153.4 + (−6) =−254.24·6 = 14455.80−2 = 7856.2/2 = 157.3a2−a58.4b−259.f(−x) = 3(−x)2−(−x) = 3x2+x60.g(−x) = 4(−x)−2 =−4x−261.Factoring, we getx(3x−1) + 0. Sox= 0,1/3.62.Since 4x−2 = 3, we getx= 5/4.63.Since|a+ 3|= 4 is equivalent toa+ 3 = 4ora+ 3 =−4, we havea= 1,−7.64.Since 3t2−t−10 = (t−2)(3t+ 5) = 0,we findt= 2,−5/3.65.C= 353n66.P= 580n

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12Chapter PAlgebraic Prerequisites67.C= 35n+ 5068.C= 2.50 + 0.50n69.We findC=4B3√D= 4(12 + 11/12)3√22,800≈1.822and a sketch of the graph ofC=4B3√22,800is given below.51020B123C70.Solving forB, we get4B3√22,800<2B<3√22,8002B<14 ft,2 in.Then the maximum displacement is 14 ft, 2 in.WithDfixed, we get thatCbecomes larger asthe beamBbecomes larger.Thus, a boat ismore likely to capsize as the beam gets larger.71.LetN= 2,B= 3.498, andS= 4.250. ThenD=π4B2·S·N=81.686 in.3ThenD≈81.7 in.3.72.LetN= 2,B= 3.518, andS= 4.250. ThenD=π4B2·S·N=82.622 in.3Using the unrounded answer to Exercise 71,the difference in the displacement is82.622−81.686≈0.94 in.3.73.Solving forB,D=π4B2·S·N4DπS·N=B2B=2√DπS·N .74.Solving forV,CR=1 +πB2·S4VCR−1=πB2·S4V1CR−1=4VπB2·SV=πB2·S4(CR−1).75.Pythagorean, legs, hypotenuse76.circle, radius, center77.√(2 + 3)2+ (−4 + 6)2=√2978.(4−62,−8 + 162)= (−1,4)79.If we replacex= 0 in 4x−6y= 40, then−6y= 40 ory=−20/3.They-intercept is(0,−203).If we replacey= 0 in 4x−6y= 40, then4x= 40 orx= 10. Thex-intercept is (10,0).80.The diagonal is√32+ 72=√58 ft.81.Rewriting the equation, we find133·3100·134·32x=13·3x133·3100·134·32x=3x−132x+93=3x−12x+ 93=x−1x=−94.

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P.3 Families of Functions, Transformations, and Symmetry1382.First, 9 = (a+b)2= (a2+b2) + 2ab= 89+ 2ab.Thenab=−40. Thus,a3+b3=(a+b)(a2+b2−ab)=3(89 + 40)=387.P.2 Pop Quiz1.Yes, sincer≥0 andr=√A/π2.SinceA=s2ands≥0, we obtains=√A.3.No, sinceb=±a.4.[1,∞)5.[2,∞)6.f(3) = 3(3) + 6 = 157.We find2a−4=102a=14a=7.For Thought1.False, it is a reflection in the y-axis.2.False, the graph ofy=x2−4 is shifted down 4units from the graph ofy=x2.3.False, rather it is a left translation.4.True5.True6.False, the down shift should come after thereflection.7.True8.False, since the domains are different.9.True10.True, sincef(−x) =−f(x) wheref(x) =x3.P.3 Exercises1.rigid2.nonrigid3.reflection4.upward translation, downward translation5.right, left6.stretching, shrinking7.odd8.even9.transformation10.family11.f(x) =√x,g(x) =−√x24x2-2y12.f(x) =x2+ 1,g(x) =−x2−12-2x4-4y13.y=x,y=−x4-4x4-4y

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