Solution Manual For Trigonometry: A Unit Circle Approach, 10th Edition

Preview (16 of 820 Pages)

100%

Solution Manual For Trigonometry: A Unit Circle Approach, 10th Edition - Page 1 preview image

Loading page ...

’SSOLUTIONSMANUALTIMBRITTJackson State Community CollegeTRIGONOMETRY:AUNITCIRCLEAPPROACHTENTHEDITIONMichael SullivanChicago State University

Solution Manual For Trigonometry: A Unit Circle Approach, 10th Edition - Page 2 preview image

Loading page ...

Solution Manual For Trigonometry: A Unit Circle Approach, 10th Edition - Page 3 preview image

Loading page ...

Table of ContentsPrefaceChapter 1Graphs and Functions1.1 The Distance and Midpoint Formulas......................................................................................... 11.2 Graphs of Equations in Two Variables; Circles........................................................................ 121.3 Functions and Their Graphs...................................................................................................... 331.4 Properties of Functions ............................................................................................................. 501.5 Library of Functions; Piecewise-defined Functions ................................................................. 661.6 Graphing Techniques: Transformations ................................................................................... 771.7 One-to-One Functions; Inverse Functions ................................................................................ 93Chapter Review.............................................................................................................................. 112Chapter Test................................................................................................................................... 121Chapter Projects............................................................................................................................. 125Chapter 2Trigonometric Functions2.1 Angles and Their Measure...................................................................................................... 1262.2 Trigonometric Functions: Unit Circle Approach .................................................................... 1342.3 Properties of the Trigonometric Functions ............................................................................. 1512.4 Graphs of the Sine and Cosine Functions ............................................................................... 1622.5 Graphs of the Tangent, Cotangent, Cosecant, and Secant Functions...................................... 1822.6 Phase Shift; Sinusoidal Curve Fitting ..................................................................................... 191Chapter Review.............................................................................................................................. 203Chapter Test................................................................................................................................... 211Cumulative Review........................................................................................................................ 214Chapter Projects............................................................................................................................. 216Chapter 3Analytic Trigonometry3.1 The Inverse Sine, Cosine, and Tangent Functions.................................................................. 2203.2 The Inverse Trigonometric Functions (Continued) ................................................................ 2313.3 Trigonometric Equations ........................................................................................................ 2433.4 Trigonometric Identities ......................................................................................................... 2623.5 Sum and Difference Formulas ................................................................................................ 2743.6 Double-angle and Half-angle Formulas.................................................................................. 2973.7 Product-to-Sum and Sum-to-Product Formulas...................................................................... 321Chapter Review.............................................................................................................................. 331Chapter Test................................................................................................................................... 345Cumulative Review........................................................................................................................ 350Chapter Projects............................................................................................................................. 353Chapter 4Applications of Trigonometric Functions4.1 Right Triangle Trigonometry; Applications ........................................................................... 3574.2 The Law of Sines .................................................................................................................... 3694.3 The Law of Cosines ................................................................................................................ 3834.4 Area of a Triangle ................................................................................................................... 3944.5 Simple Harmonic Motion; Damped Motion; Combining Waves ........................................... 402Chapter Review.............................................................................................................................. 411Chapter Test................................................................................................................................... 418Cumulative Review........................................................................................................................ 422Chapter Projects............................................................................................................................. 425

Solution Manual For Trigonometry: A Unit Circle Approach, 10th Edition - Page 4 preview image

Loading page ...

Chapter 5Polar Coordinates; Vectors5.1 Polar Coordinates.................................................................................................................... 4295.2 Polar Equations and Graphs.................................................................................................... 4365.3 The Complex Plane; De Moivre’s Theorem ........................................................................... 4655.4 Vectors.................................................................................................................................... 4765.5 The Dot Product...................................................................................................................... 4885.6 Vectors in Space ..................................................................................................................... 4945.7 The Cross Product................................................................................................................... 500Chapter Review.............................................................................................................................. 510Chapter Test................................................................................................................................... 519Cumulative Review........................................................................................................................ 523Chapter Projects............................................................................................................................. 525Chapter 6Analytic Geometry6.2 The Parabola ........................................................................................................................... 5296.3 The Ellipse .............................................................................................................................. 5436.4 The Hyperbola ........................................................................................................................ 5596.5 Rotation of Axes; General Form of a Conic ........................................................................... 5786.6 Polar Equations of Conics....................................................................................................... 5906.7 Plane Curves and Parametric Equations ................................................................................. 597Chapter Review.............................................................................................................................. 610Chapter Test................................................................................................................................... 619Cumulative Review........................................................................................................................ 624Chapter Projects............................................................................................................................. 625Chapter 7Exponential and Logarithmic Functions7.1 Exponential Functions ............................................................................................................ 6297.2 Logarithmic Functions............................................................................................................ 6487.3 Properties of Logarithms ........................................................................................................ 6687.4 Logarithmic and Exponential Equations................................................................................. 6777.5 Financial Models .................................................................................................................... 6967.6 Exponential Growth and Decay Models; Newton’s Law; Logistic Growthand Decay Models .................................................................................................................... 7037.7 Building Exponential, Logarithmic, and Logistic Models from Data..................................... 713Chapter Review.............................................................................................................................. 717Chapter Test................................................................................................................................... 726Cumulative Review........................................................................................................................ 729Chapter Projects............................................................................................................................. 731Appendix AReviewA.1 Algebra Essentials.................................................................................................................. 734A.2 Geometry Essentials............................................................................................................... 739A.3 Factoring Polynomials; Completing the Square..................................................................... 745A.4 Solving Equations .................................................................................................................. 748A.5 Complex Numbers; Quadratic Equations in the Complex Number System .......................... 762A.6 Interval Notation; Solving Inequalities .................................................................................. 768A.7nth Roots; Rational Exponents............................................................................................... 779A.8 Lines....................................................................................................................................... 788A.9 Building Linear Models from Data ........................................................................................ 804

Solution Manual For Trigonometry: A Unit Circle Approach, 10th Edition - Page 5 preview image

Loading page ...

Appendix BGraphing UtilitiesB.1 The Viewing Rectangle.......................................................................................................... 809B.2 Using a Graphing Utility to Graph Equations ........................................................................ 810B.3 Using a Graphing Utility to Locate Intercepts and Check for Symmetry .............................. 814B.5 Square Screens ....................................................................................................................... 816

Solution Manual For Trigonometry: A Unit Circle Approach, 10th Edition - Page 6 preview image

Loading page ...

1Chapter 1Graphs and FunctionsSection 1.11.02.()5388− −==3.2234255+==4.22211601213600372161+=+==Since the sum of the squares of two of the sidesof the triangle equals the square of the third side,the triangle is a right triangle.5.12bh6.true7.x-coordinate or abscissa;y-coordinate orordinate8.quadrants9.midpoint10.False; the distance between two points is nevernegative.11.False; points that lie in quadrant IV will have apositivex-coordinate and a negativey-coordinate.The point()1, 4−lies in quadrant II.12.True;1212,22xxyyM++=13.b14.a15.(a)quadrant II(b)x-axis(c)quadrant III(d)quadrant I(e)y-axis(f)Quadrant IV16.(a)quadrant I(b)quadrant III(c)quadrant II(d)quadrant I(e)y-axis(f)x-axis17.The points will be on a vertical line that is twounits to the right of they-axis.

Solution Manual For Trigonometry: A Unit Circle Approach, 10th Edition - Page 7 preview image

Loading page ...

Chapter 1:Graphs and Functions218.The points will be on a horizontal line that isthree units above thex-axis.19.221222(,)(20)(10)21415d P P=−+−=+=+=20.221222(,)( 20)(10)( 2)1415d P P=−−+−=−+=+=21.221222(,)( 21)(21)( 3)19110d P P=−−+−=−+=+=22.()221222(,)2( 1)(21)319110d P P=− −+−=+=+=23.()()( )221222(,)(53)4428464682 17d P P=−+− −=+=+==24.()()()( )221222(,)214034916255d P P=− −+−=+=+==25.()221222(,)6( 3)(02)9(2)81485d P P=− −+−=+ −=+=26.()()221222(,)422( 3)2542529d P P=−+− −=+=+=27.()221222(,)(64)4( 3)2744953d P P=−+− −=+=+=28.()()221222(,)6(4)2( 3)1051002512555d P P=− −+− −=+=+==29.22122222(,)(0)(0)()()d P Pababab=−+−=−+ −=+30.221222222(,)(0)(0)()()22d P Paaaaaaaa=−+−=−+ −=+==31.( 2,5),(1,3),( 1, 0)ABC= −== −()()()222222222222(,)1( 2)(35)3( 2)9413(,)11(03)( 2)( 3)4913(,)1( 2)(05)1( 5)12526d A Bd B Cd A C=− −+−=+ −=+==− −+−=−+ −=+==− − −+−=+ −=+=Verifying that ∆ ABC is a right triangle by thePythagorean Theorem:[][][]()()()222222(,)(,)(,)1313261313262626d A Bd B Cd A C+=+=+==The area of a triangle is12Abh=⋅. In thisproblem,

Solution Manual For Trigonometry: A Unit Circle Approach, 10th Edition - Page 8 preview image

Loading page ...

Section 1.1:The Distance and Midpoint Formulas3[] []1(,)(,)2111313132213 square units2Ad A Bd B C=⋅⋅=⋅=⋅⋅=32.( 2, 5),(12, 3),(10,11)ABC= −==−()()()222222222222(,)12( 2)(35)14( 2)1964200102(,)1012( 113)( 2)( 14)4196200102(,)10( 2)( 115)12(16)14425640020d A Bd B Cd A C=− −+−=+ −=+===−+ −−=−+ −=+===− −+ −−=+ −=+==Verifying that ∆ ABC is a right triangle by thePythagorean Theorem:[][][]()()()222222(,)(,)(,)10210220200200400400400d A Bd B Cd A C+=+=+==The area of a triangle is12Abh=. In thisproblem,[] []1(,)(,)21 102 10221 100 2100 square units2Ad A Bd B C=⋅⋅=⋅⋅=⋅⋅=33.(5,3),(6, 0),(5,5)ABC= −==()()()222222222222(,)6(5)(03)11(3)1219130(,)56(50)(1)512526(,)5(5)(53)1021004104226d A Bd B Cd A C=− −+−=+ −=+==−+−=−+=+==− −+−=+=+==Verifying that ∆ ABC is a right triangle by thePythagorean Theorem:[][][]()()()222222(,)(,)(,)1042613010426130130130d A Cd B Cd A B+=+=+==The area of a triangle is12Abh=. In this

Solution Manual For Trigonometry: A Unit Circle Approach, 10th Edition - Page 9 preview image

Loading page ...

Chapter 1:Graphs and Functions4problem,[] []1(,)(,)211042621 2262621 2 26226 square unitsAd A Cd B C=⋅⋅=⋅⋅=⋅⋅=⋅⋅=34.( 6, 3),(3,5),( 1, 5)ABC= −=−= −()()()222222222222(,)3( 6)( 53)9( 8)8164145(,)13(5( 5))( 4)1016100116229(,)1(6)(53)5225429d A Bd B Cd A C=− −+ −−=+ −=+==− −+− −=−+=+===− − −+−=+=+=Verifying that ∆ ABC is a right triangle by thePythagorean Theorem:[][][]()()()222222(,)(,)(,)29229145294 2914529116145145145d A Cd B Cd A B+=+=+⋅=+==The area of a triangle is12Abh=. In thisproblem,[] []1(,)(,)2129 22921 2 29229 square unitsAd A Cd B C=⋅⋅=⋅⋅=⋅⋅=35.(4,3),(0,3),(4, 2)ABC=−=−=()()()()222222222222(,)(04)3( 3)(4)0160164(,)402( 3)45162541(,)(44)2( 3)05025255d A Bd B Cd A C=−+ −− −=−+=+===−+− −=+=+==−+− −=+=+==Verifying that ∆ ABC is a right triangle by thePythagorean Theorem:[][][]()222222(,)(,)(,)45411625414141d A Bd A Cd B C+=+=+==The area of a triangle is12Abh=. In this

Solution Manual For Trigonometry: A Unit Circle Approach, 10th Edition - Page 10 preview image

Loading page ...

Section 1.1:The Distance and Midpoint Formulas5problem,[] []1(,)(,)21 4 5210 square unitsAd A Bd A C=⋅⋅=⋅⋅=36.(4,3),(4, 1),(2, 1)ABC=−==()()()()222222222222(,)(44)1( 3)04016164(,)2411( 2)04042(,)(24)1( 3)( 2)44162025d A Bd B Cd A C=−+− −=+=+===−+−=−+=+===−+− −=−+=+==Verifying that ∆ ABC is a right triangle by thePythagorean Theorem:[][][]()222222(,)(,)(,)4225164202020d A Bd B Cd A C+=+=+==The area of a triangle is12Abh=. In this problem,[] []1(,)(,)21 4 224 square unitsAd A Bd B C=⋅⋅=⋅⋅=37.The coordinates of the midpoint are:1212( ,),224435 ,2280,22(4, 0)xxyyx y++=−++===38.The coordinates of the midpoint are:()1212( ,),222204,2204,220, 2xxyyx y++=−++===39.The coordinates of the midpoint are:1212( ,),223620,2232,223 ,12xxyyx y++=−++===40.The coordinates of the midpoint are:1212( ,),222432,2261,2213,2xxyyx y++=+−+=−==−

Solution Manual For Trigonometry: A Unit Circle Approach, 10th Edition - Page 11 preview image

Loading page ...

Chapter 1:Graphs and Functions641.The coordinates of the midpoint are:1212( ,),224631,22210 ,22(5,1)xxyyx y++=+−+=−==−42.The coordinates of the midpoint are:1212( ,),224232,2221,2211,2xxyyx y++=−+−+=−−==−−43.The coordinates of the midpoint are:1212( ,),2200,22,22xxyyx yabab++=++==44.The coordinates of the midpoint are:1212( ,),2200,22,22xxyyx yaaaa++=++==45.The x coordinate would be235+=and the ycoordinate would be523−=. Thus the newpoint would be()5,3.46.The new x coordinate would be123− −= −andthe new y coordinate would be6410+=. Thusthe new point would be()3,10−47.a.If we use a right triangle to solve theproblem, we know the hypotenuse is 13 units inlength. One of the legs of the triangle will be2+3=5. Thus the other leg will be:222225132516914412bbbb+=+===Thus the coordinates will have an y value of11213− −= −and11211− +=. So the pointsare()3,11and()3,13−.b.Consider points of the form()3,ythat are adistance of 13 units from the point()2,1−−.()()()()( )()2221212222223( 2)1512512226dxxyyyyyyyy=−+−=− −+ − −=+ − −=+++=++()()()22222213226132261692260214301113yyyyyyyyyy=++=++=++=+−=−+11011yy−==or13013yy+== −Thus, the points()3,11and()3,13−are adistance of 13 units from the point()2,1−−.48.a.If we use a right triangle to solve theproblem, we know the hypotenuse is 17 units inlength. One of the legs of the triangle will be2+6=8. Thus the other leg will be:222228176428922515bbbb+=+===Thus the coordinates will have an x value of11514−= −and11516+=. So the points are()14,6−−and()16,6−.

Solution Manual For Trigonometry: A Unit Circle Approach, 10th Edition - Page 12 preview image

Loading page ...

Section 1.1:The Distance and Midpoint Formulas7b.Consider points of the form(),6x−that area distance of 17 units from the point()1, 2.()()()()()( )2221212222221262182164265dxxyyxxxxxxx=−+−=−+− −=−++=−++=−+()()()22222217265172652892650222401416xxxxxxxxxx=−+=−+=−+=−−=+−14014xx+== −or16016xx−==Thus, the points()14,6−−and()16,6−are adistance of 13 units from the point()1, 2.49.Points on thex-axis have ay-coordinate of 0. Thus,we consider points of the form(), 0xthat are adistance of 6 units from the point()4,3−.()()()()()22212122222243016831689825dxxyyxxxxxxx=−+−=−+ −−=−++ −=−++=−+()222222268256825368250811( 8)( 8)4(1)( 11)2(1)864448108228634332xxxxxxxxx=−+=−+=−+=−−− −±−−−=±+±==±==±433x=+or433x=−Thus, the points()433, 0+and()433, 0−areon thex-axis and a distance of 6 units from thepoint()4,3−.50.Points on they-axis have anx-coordinate of 0.Thus, we consider points of the form()0,ythatare a distance of 6 units from the point()4,3−.()()()()2221212222224034961696625dxxyyyyyyyyy=−+−=−+ −−=+++=+++=++()222222266256625366250611( 6)(6)4(1)( 11)2(1)63644680226453252yyyyyyyyy=++=++=++=+−−±−−=−±+−±==−±== −±325y= −+or325y= −−Thus, the points()0,325−+and()0,325−−are on they-axis and a distance of 6 units from thepoint()4,3−.51.a.To shift 3 units left and 4 units down, wesubtract 3 from thex-coordinate and subtract4 from they-coordinate.()()23,541,1−−=−b.To shift left 2 units and up 8 units, wesubtract 2 from thex-coordinate and add 8 tothey-coordinate.()()22,580,13−+=

Solution Manual For Trigonometry: A Unit Circle Approach, 10th Edition - Page 13 preview image

Loading page ...

Chapter 1:Graphs and Functions852.Let the coordinates of pointBbe(),xy. Usingthe midpoint formula, we can write()182,3,22xy− ++=.This leads to two equations we can solve.122145xxx− +=− +==832862yyy+=+== −PointBhas coordinates()5,2−.53.()1212,,22xxyyMx y++==.()111,( 3, 6)Pxy== −and( ,)( 1, 4)x y= −, so122222312231xxxxxx+=−+−=−= −+=and122222642862yyyyyy+=+==+=Thus,2(1, 2)P=.54.()1212,,22xxyyMx y++==.()222,(7,2)Pxy==−and( ,)(5,4)x y=−, so1211127521073xxxxxx+=+==+=and121112( 2)428( 2)6yyyyyy+=+ −−=−=+ −−=Thus,1(3,6)P=−.55.The midpoint of AB is:()0600,223, 0D++==The midpoint of AC is:()0404,222, 2E++==The midpoint of BC is:()6404,225, 2F++==()2222(,)04(34)(4)(1)16117d C D=−+−=−+ −=+=()2222(,)26(20)(4)21642025d B E=−+−=−+=+==2222(,)(20)(50)2542529d A F=−+−=+=+=56.Let12(0, 0),(0, 4),( ,)PPPx y===()()()221222122222222222,(00)(40)164,(0)(0)416,(0)(4)(4)4(4)16dPPdPPxyxyxydPPxyxyxy=−+−===−+−=+=→+==−+−=+−=→+−=Therefore,()222248168162yyyyyyy=−=−+==which gives2222161223xxx+=== ±Two triangles are possible. The third vertex is()()23, 2or23, 2−.

Solution Manual For Trigonometry: A Unit Circle Approach, 10th Edition - Page 14 preview image

Loading page ...

Section 1.1:The Distance and Midpoint Formulas957.Let()10, 0P=,()20,Ps=,()3, 0Ps=, and()4,Ps s=.yx(0,)s(0, 0)( , 0)s( ,)s sThe points1Pand4Pare endpoints of onediagonal and the points2Pand3Pare theendpoints of the other diagonal.1,400,,2222ssssM++==2,300,,2222ssssM++==The midpoints of the diagonals are the same.Therefore, the diagonals of a square intersect attheir midpoints.58.Let()10, 0P=,()2, 0Pa=, and33,22aaP=. To show that these verticesform an equilateral triangle, we need to showthat the distance between any pair of points is thesame constant value.()()()()()22122121222,000dP Pxxyyaaa=−+−=−+−==()()()22232121222222,302234444dPPxxyyaaaaaaaa=−+−=−+−=+===()()()22132121222222,3002234444dP Pxxyyaaaaaaa=−+−=−+−=+===Since all three distances have the same constantvalue, the triangle is an equilateral triangle.Now find the midpoints:1 22 31 3456000,, 02223330,22,4422300322,,2244P PP PP PaaPMaaaaaPMaaaaPM++===++===++===()2245222233,0424344316162aaadPPaaaaa=−+−=+=+=()224622223,0424344316162aaadPPaaaaa=−+−=−+=+=()2256222333,44440242aaaadPPaaa=−+−=+==Since the sides are the same length, the triangleis equilateral.

Solution Manual For Trigonometry: A Unit Circle Approach, 10th Edition - Page 15 preview image

Loading page ...

Chapter 1:Graphs and Functions1059.221222(,)(42)(11)(6)0366d P P=−−+−=−+==()222322(,)4(4)( 31)0(4)164d PP=−− −+ −−=+ −==221322(,)(42)( 31)(6)(4)3616522 13d P P=−−+ −−=−+ −=+==Since[][][]222122313(,)(,)(,)d P Pd PPd P P+=,the triangle is a right triangle.60.()221222(,)6( 1)(24)7(2)49453d P P=− −+−=+ −=+=()222322(,)46( 52)(2)(7)44953d PP=−+ −−=−+ −=+=()221322(,)4( 1)( 54)5(9)2581106d P P=− −+ −−=+ −=+=Since[][][]222122313(,)(,)(,)d P Pd PPd P P+=,the triangle is a right triangle.Since()()1223,,dPPdPP=, the triangle isisosceles.Therefore, the triangle is an isosceles righttriangle.61.()()221222(,)0(2)7( 1)28464682 17d P P=− −+− −=+=+==()222322(,)30(27)3(5)92534d PP=−+−=+ −=+=()()221322(,)3( 2)2( 1)5325934d P P=− −+− −=+=+=Since2313(,)(,)d PPd P P=, the triangle isisosceles.Since[][][]222132312(,)(,)(,)d P Pd PPd P P+=,the triangle is also a right triangle.Therefore, the triangle is an isosceles righttriangle.62.()()221222(,)4702( 11)(2)121412555d P P=−−+−=−+ −=+==()222322(,)4(4)(60)86643610010d PP=− −+−=+=+==()()221322(,)4762( 3)4916255d P P=−+−=−+=+==Since[][][]222132312(,)(,)(,)d P Pd PPd P P+=,the triangle is a right triangle.

Solution Manual For Trigonometry: A Unit Circle Approach, 10th Edition - Page 16 preview image

Loading page ...

Section 1.1:The Distance and Midpoint Formulas1163.Using the Pythagorean Theorem:222229090810081001620016200902127.28 feetdddd+=+====≈90909090d64.Using the Pythagorean Theorem:222226060360036007200720060284.85 feetdddd+=+=→===≈60606060d65.a.First: (90, 0), Second: (90, 90),Third: (0, 90)(0,0)(0,90)(90,0)(90,90)XYb.Using the distance formula:2222(31090)(1590)220( 75)5402552161232.43 feetd=−+−=+ −==≈c.Using the distance formula:2222(3000)(30090)30021013410030 149366.20 feetd=−+−=+==≈66.a.First: (60, 0), Second: (60, 60)Third: (0, 60)(0,0)(0,60)(60,0)(60,60)xyb.Using the distance formula:2222(18060)(2060)120(40)1600040 10126.49 feetd=−+−=+ −==≈c.Using the distance formula:2222(2200)(22060)2201607400020 185272.03 feetd=−+−=+==≈67.The Focus heading east moves a distance30tafterthours. The truck heading south moves adistance40tafterthours. Their distance apartafterthours is:22222(30 )(40 )9001600250050milesdtttttt=+=+==d40t30t68.15 miles5280 ft1 hr22 ft/sec1 hr1 mile3600 sec⋅⋅=()2221002210000484feetdtt=+=+

Preview Mode

This document has 820 pages. Sign in to access the full document!

Report

Study Now!

Document Details

Related Documents

Solution Manual for Trigonometry, 11th Edition

Precalculus - Polynomial and Rational Functions

Precalculus - Functions

Precalculus - Exponential and Logarithmic Functions

Math Word Problems - Translating Expressions

Math Word Problems - Equations

Algebra II – Word Problems

Algebra II – Sequences and Series

Algebra II - Segments Lines and Inequalities

Algebra II - Relations and Functions

Company

Explore

Study Tools