Solution Manual for Trigonometry, 2nd Edition

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SOLUTIONSMANUALDANIELS.MILLERNiagara County Community CollegeTRIGONOMETRYSECONDEDITIONRobert BlitzerMiami Dade College

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iiiTABLE OF CONTENTSTRIGONOMETRY2EChapter 1Angles and the Trigonometric Functions ...................................................................1Chapter 2Graphs of the Trigonometric Functions; Inverse Trigonometric Functions.............55Chapter 3Trigonometric Identities and Equations..................................................................175Chapter 4Laws of Sines and Cosines; Vectors.......................................................................297Chapter 5Complex Numbers, Polar Coordinates, and Parametric Equations ........................371

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Chapter 1Angles and the Trigonometric Functions1Section 1.1Check Point Exercises1.The radian measure of a central angle is the length ofthe intercepted arc,s,divided by the circle’s radius,r.The length of the intercepted arc is 42 feet:s= 42feet. The circle’s radius is 12 feet:r= 12 feet. Nowuse the formula for radian measure to find the radianmeasure of.θ42 feet3.512 feetsrθ ===Thus, the radian measure ofθis 3.52.a.radians606060radians180180radians3πππ° =° ⋅=°=b.radians270270270radians1801803radians2πππ° =° ⋅=°=c.radians300300300radians1801805radians3πππ−−° = −° ⋅=°= −3.a.oradians180radians4radiansπππ=⋅4o180o454==b.o44radians 180radiansπππ−= −⋅33o4 180o2403⋅= −= −c.o1806 radians6 radiansradiansπ=⋅o6 180o343.8π⋅=≈d.o1804.7 radians4.7 radiansradiansπ−= −⋅o4.7 180o269.3π−⋅=≈ −4.a.b.c.d.5.a.For a 400º angle, subtract 360º to find a positivecoterminal angle.400o– 360o= 40ob.For a –135º angle, add 360º to find a positivecoterminal angle.–135o+ 360o= 225o6.a.131310325555πππππ−=−=b.3029215151515πππππ−+= −+=

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Chapter 1Angles and the Trigonometric Functions27.a.8553602855720135° −° ⋅=° −° =°b.171722433ππππ−⋅=−17125333πππ=−=c.252523666ππππ−+⋅= −+253611666πππ= −+=d.()17.417.4 57991.8≈=For a991.8angle, we need to subtract twomultiples of360to find a positive coterminalangle less than360 .17.42217.444.8ππ−⋅=−≈8.The formulasrθ=can only be used whenθisexpressed in radians. Thus, we begin by converting45° to radians. Multiply byradians180π°.radians454545radians180180radians4πππ° =° ⋅=°=Now we can use the formulasrθ=to find thelength of the arc. The circle’s radius is 6 inches :r= 6 inches. The measure of the central angle inradians is:44ππθ =. The length of the arcintercepted by this central angle is(6 inches)46inches43inches24.71 inches.srθπππ====≈9.The formula212Arθ=can only be used whenθisexpressed in radians. Thus, we begin by converting150° to radians. Multiply byradians180π°.radians150150180150radians1805radians6πππ° =° ⋅°==The circle’s radius is 6 feet.Now use the formula212Arθ=to find the area ofthe sector.:2122125(6 feet)615square feet47.12 square feetArθππ===≈10.We are givenω, the angular speed.45ω =revolutions per minuteWe use the formularνω=to find,vthe linearspeed. Before applying the formula, we must expressωin radians per minute.45 revolutions2radians1 minute1 revolution90radians1 minuteπωπ=⋅=The angular speed of the propeller is90πradians perminute. The linear speed is901.5 inches 1 minute135inchesminuteinches135minuteinches424 minuterνωπππ==⋅==≈The linear speed is135πinches per minute, which isapproximately 424 inches per minute.

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Section 1.1Angles and Trigonometric Utilities3Concept and Vocabulary Check 1.11.origin;x-axis2.counterclockwise; clockwise3.acute; right; obtuse; straight4.sr5.180π°6.180π°7.coterminal;360°;2π8.rθ9.212rθ10.false11.rω; angularExercise Set 1.11.obtuse2.obtuse3.acute4.acute5.straight6.right7.40 inches4 radians10 inchessrθ===8.30 feet6 radians5 feetsrθ===9.8 yards4 radians6 yards3srθ===10.18 yards2.25 radians8 yardssrθ===11.400 centimeters4 radians100 centimeterssrθ===12.600 centimeters6 radians100 centimeterssrθ===13.radians454518045radians180radians4πππ° =° ⋅°==14.radians181818018radians180radians10πππ° =° ⋅°==15.radians135135180135radians1803radians4πππ° =° ⋅°==16.radians150150180150radians1805radians6πππ° =° ⋅°==17.radians300300180300radians1805radians3πππ° =° ⋅°==18.radians330330180330radians18011radians6πππ° =° ⋅°==19.radians225225180225radians1805radians4πππ−° = −° ⋅°= −= −

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Chapter 1Angles and the Trigonometric Functions420.radians270270180270radians1803radians2πππ−° = −° ⋅°= −= −21.oradians180radians22radiansπππ=⋅o1802o90==22.oradians180radians99radiansπππ=⋅o180o209==23.o22radians180radians33radiansπππ=⋅o2 1803o120⋅==24.0o3radians1803 180o1354radians4ππ⋅⋅==25.o77radians180radians66radiansπππ=⋅o7 1806o210⋅==26.oo11radians18011 180o3306radians6ππ⋅⋅==27.o1803radians3radiansradiansπππ−= −⋅o3 180o540= − ⋅= −28.o180oo4radians4 180720radiansππ−⋅= −⋅= −29.radians181818018radians1800.31 radiansππ° =° ⋅°=≈30.radians767618076radians1801.33 radiansππ° =° ⋅°=≈31.radians404018040radians1800.70 radiansππ−° = −° ⋅°= −≈ −32.radians505018050radians1800.87 radiansππ−° = −° ⋅°= −≈ −33.radians200200180200radians1803.49 radiansππ° =° ⋅°=≈34.radians250250180250radians1804.36 radiansππ° =° ⋅°=≈35.o1802 radians2 radiansradiansπ=⋅o2 180o114.59π⋅=≈36.oo1803 180o3 radians171.89radiansππ⋅⋅=≈37.oradians180radians1313radiansπππ=⋅o18013 o13.85=≈

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Section 1.1Angles and Trigonometric Utilities538.oo180180oradians10.5917radians17ππ⋅=≈39.o1804.8 radians4.8 radiansradiansπ−= −⋅o4.8 180o275.02π−⋅=≈ −40.oo1805.2 1805.2 radiansradiansππ−⋅−⋅=o297.94≈ −41.42.43.44.45.46.47.48.49.50.51.52.

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Chapter 1Angles and the Trigonometric Functions653.54.55.56.57.39536035° −° =°58.41536055° −° =°59.150360210−° +° =°60.160360200−° +° =°61.76536037651080315−° +° ⋅= −° +° =°62.76036037601080320−° +° ⋅= −° +° =°63.191912726666πππππ−=−=64.171710725555πππππ−=−=65.23232320322455555πππππππ−⋅=−=−=66.2525252422466666πππππππ−⋅=−=−=67.10099250505050πππππ−+= −+=68.8079240404040πππππ−+= −+=69.313123677ππππ−+⋅= −+314211777πππ= −+=70.383823699ππππ−+⋅= −+385416999πππ= −+=71.12 inches,45rθ==°Begin by converting 45° to radians, in order to usethe formulasrθ=.radians4545radians1804ππ° =°⋅=°Now use the formulasrθ=.123inches9.42 inches4srπθπ==⋅=≈72.16 inches,60rθ==°Begin by converting 60° to radians, in order to usethe formulasrθ=.radians6060radians1803ππ° =° ⋅=°Now use the formulasrθ=.1616inches16.76 inches33srππθ==⋅=≈73.8 feet,225rθ==°Begin by converting 225° to radians, in order to usethe formulasrθ=.radians5225225radians1804ππ° =°⋅=°Now use the formulasrθ=.5810feet31.42 feet4srπθπ==⋅=≈74.9 yards,315rθ==°Begin by converting 315° to radians, in order to usethe formulasrθ=.radians7315315radians1804ππ° =° ⋅=°Now use the formulasrθ=.7639yards49.48 yards44srππθ==⋅=≈

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Section 1.1Angles and Trigonometric Utilities775.Begin by converting the angle to radians.radians1818radians18010ππ° =°⋅=°Now use the formula212Arθ=to find the area ofthe sector.:212212(10 meters)105square meters15.71 square metersArθππ===≈76.Begin by converting the angle to radians.radians1515180radians12ππ° =° ⋅°=Now use the formula212Arθ=to find the area ofthe sector.:212212(6 yards)123square yards24.71 square yardsArθππ===≈77.Begin by converting the angle to radians.radians2402401804radians3ππ° =° ⋅°=Now use the formula212Arθ=to find the area ofthe sector.:2122124(4 inches)332square inches333.51 square inchesArθππ===≈78.Begin by converting the angle to radians.radians33033018011radians6ππ° =° ⋅°=Now use the formula212Arθ=to find the area ofthe sector.:21221211(3 inches)633square inches425.92 square inchesArθππ===≈79.6 revolutions per second6 revolutions2radians12radians1 second1 revolutions1 seconds12radians per secondπππ=⋅==80.20 revolutions per second20 revolutions2radians40radians1 second1 revolution1 second40radians per secondπππ=⋅==81.42and33ππ−82.75and66ππ−83.35and44ππ−84.7and44ππ−85.3and22ππ−86.−πandπ87.55112606ππ⋅=88.3572606ππ⋅=

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Chapter 1Angles and the Trigonometric Functions889.3 minutes and 40 seconds equals 220 seconds.220222603ππ⋅=90.4 minutes and 25 seconds equals 265 seconds.265532606ππ⋅=91.212Arθ=21221225 square feet(10 feet)25 square feet(10 feet)25501 radians2θθθθ====92.21221221236 square yards(6 yards)36 square yards(6 yards)36182 radiansArθθθθθ=====93.First, convert to degrees.11360revolutionrevolution661 revolution1 360606°=⋅=⋅° =°Now, convert 60° to radians.radians606060radians180180radians3πππ° =° ⋅=°=Therefore,16revolution is equivalent to 60° or3πradians.94.First, convert to degrees.o11360revolutionsrevolutions331 revolution1oo3601203=⋅=⋅=Now, convert 120° to radians.radians1202120120radiansradians1801803πππ° =° ⋅==°Therefore,13revolution is equivalent to 120° or23πradians.95.The distance that the tip of the minute hand moves isgiven by its arc length,s. Sincesrθ=, we begin byfindingrandθ. We are given thatr= 8 inches. Theminute hand moves from 12 to 2 o'clock, or 16 of acomplete revolution. The formulasrθ=can only beused whenθis expressed in radians. We mustconvert16revolution to radians.112radiansrevolutionrevolution661 revolutionradians3ππ=⋅=The distance the tip of the minute hand moves is8(8 inches)inches8.38 inches.33srππθ===≈96.The distance that the tip of the minute hand moves isgiven by its arc length,s. Sincesrθ=, we begin byfindingrandθ. We are given thatr= 6 inches. The minute hand moves from 12 to 4o’clock, or13of a complete revolution. The formulasrθ=can only be used whenθis expressed inradians. We must convert13revolution to radians.112radiansrevolutionrevolution331 revolution2radians3ππ=⋅=The distance the tip of the minute hand moves is212(6 inches)inches334inches12.57 inches.srππθπ====≈

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Section 1.1Angles and Trigonometric Utilities997.The length of each arc is given bysrθ=. We aregiven thatr= 24 inches and90 .θ=°The formulasrθ=can only be used whenθis expressed in radians.radians909090radians180180=radians2πππ° =° ⋅=°The length of each arc is(24 inches)12inches237.70 inches.srπθπ===≈98.The distance that the wheel moves is given bysrθ=. We are given thatr= 80 centimeters andθ= 60°. The formulasrθ=can only be used whenθis expressed in radians.radians606060radians180180=radians3πππ° =° ⋅=°The length that the wheel moves is80(80 centimeters)centimeters3383.78 centimeters.srππθ===≈99.Begin by converting the angle to radians.radians1351351803radians4ππ° =° ⋅°=Now use the formula212Arθ=to find the area ofthe sector.:2122123(40 feet)4600square feet1884.96 square feetArθππ===≈100.Begin by converting the angle to radians.radians1201201802radians3ππ° =° ⋅°=Now use the formula212Arθ=to find the area ofthe sector.:2122122(30 feet)3300square feet942.48 square feetArθππ===≈101.Recall thatsrθ=. We are given thats= 8000 miles andr= 4000 miles.8000 miles2 radians4000 milessrθ===Now, convert 2 radians to degrees.o180o2 radians2 radians114.59radiansπ=⋅≈102.Recall thatsrθ=. We are given thats= 10,000 miles andr= 4000 miles.10, 000 miles2.5 radians4000 milessrθ ===Now, convert 2.5 radians to degrees.o180o2.5 radians143.242radiansπ⋅≈103.Recall thatsrθ=. We are given thatr= 4000 miles and30θ=°. The formulasrθ=canonly be used whenθis expressed in radians.radians303030radians180180radians6==(4000 miles)2094 miles6srππππθ° =° ⋅=°=≈To the nearest mile, the distance fromAtoBis2094 miles.104.Recall thatsrθ=. We are given thatr= 4000 miles and10θ=°. We can only use theformulasrθ=whenθis expressed in radians.radians101010radians180180radians18==(4000 miles)698 miles18srππππθ° =° ⋅=°=≈To the nearest mile, the distance fromAtoBis698 miles.

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Chapter 1Angles and the Trigonometric Functions10105.Linear speed is given byrνω=. We are given that12πω=radians per hour andr= 4000 miles. Therefore,(4000 miles) 124000miles per hour121047 miles per hourrπνωπ===≈The linear speed is about 1047 miles per hour.106.Linear speed is given byrνω=. We are given thatr= 25 feet and the wheel rotates at 2 revolutions perminute. We need to convert 2 revolutions per minuteto radians per minute.2 revolutions per minute2radians2 revolutions per minute 1 revolution4radians per minuteππ=⋅=(25 feet)(4)314 feet per minuterνωπ==≈The linear speed of the Ferris wheel is about314 feet per minute.107.Linear speed is given byrνω=. We are given thatr= 12 feet and the wheel rotates at 20 revolutions perminute.20 revolutions per minute2radians20 revolutions per minute 1 revolution40radians per minuteππ=⋅=(12 feet)(40)1508 feet per minuterνωπ==≈The linear speed of the wheel is about1508 feet per minute.108.Begin by converting 2.5 revolutions per minute toradians per minute.2.5 revolutions per minute2radians2.5 revolutions per minute 1 revolution5radians per minuteππ=⋅=The linear speed of the animals in the outer rows is(20 feet)(5)100 feet per minuterνωπ==≈The linear speed of the animals in the inner rows is(10 feet)(5)50 feet per minuterνωπ==≈The difference is1005050πππ−=feet per minuteor about 157 feet per minute.109. – 120.Answers may vary.121.30 15'10 ''30.25=122.65 45'20 ''65.76=123.30.4230 25'12 ''=124.50.4250 25'12 ''=125.does not make sense; Explanations will vary.Sample explanation: Angles greater thanπwillexceed a straight angle.126.does not make sense; Explanations will vary.Sample explanation: It is possible forπto be usedin an angle measured using degrees.127.makes sense128.does not make sense; Explanations will vary.Sample explanation: That will not be possible if theangle is a multiple of2.π129.A right angle measures 90° and902π° =radians≈1.57 radians.If3 radians1.5 radians,2θθ==is smaller than aright angle.130.srθ=Begin by changing20θ=°to radians.2020radians1809100= 9900286 milesrrππππ° =° ⋅=°=≈To the nearest mile, a radius of 286 miles should beused.131.212Arθ=212212700 square feet(30 feet)700 square feet(30 feet)70045014 radians9θθθθ====Covvert radians to degrees:1414 180radians8999π°=⋅≈°The sprinkler should rotate about89 .°

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Section 1.2Right Triangle Trigonometry11132.srθ=Begin by changing26θ=°to radians.132626radians1809013=4000901815 milessπππ° =° ⋅=°⋅≈To the nearest mile, Miami, Florida is 1815 milesnorth of the equator.133.First find the hypotenuse.222222225122514416913cabcccc=+=+=+==Next write the ratio.513ac=134.First find the hypotenuse.22222222111122cabcccc=+=+=+==Next write the ratio and simplify.12122222ac==⋅=135.222222222ababccccabc+=++=Since222,cab=+continue simplifying bysubstituting2cfor22.ab+Section 1.2Check Point Exercises1.Use the Pythagorean Theorem,222cab=+, to findc.222223,43491625255abcabc===+=+=+===Referring to these lengths as opposite, adjacent, andhypotenuse, we haveopposite3sinhypotenuse5adjacent4coshypotenuse5opposite3tanadjacent4θθθ======hypotenuse5cscopposite3hypotenuse5secadjacent4adjacent4cotopposite3θθθ======2.Use the Pythagorean Theorem,222cab=+, tofindb.2222222215125242426abcbbbb+=+=+====Note that sideais oppositeθand sidebis adjacenttoθ.opposite1sinhypotenuse5adjacent26coshypotenuse5opposite16tanadjacent1226hypotenuse5csc5opposite1hypotenuse556secadjacent1226adjacent26cot26opposite1θθθθθθ================

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Chapter 1Angles and the Trigonometric Functions123.Apply the definitions of these three trigonometric functions.length of hypotenusecsc 45length of side opposite 45221length of hypotenusesec 45length of side adjacent to 45221length of side adjacent to 45cot 45length of side opposite 45111° =°==° =°==°° =°==4.length of side opposite 60tan 60length of side adjacent to 60331length of side opposite 30tan 30length of side adjacent to 3011333333°° =°==°° =°==⋅=5.2sin3tancos53θθθ==2323552525555=⋅==⋅=113csc2sin23113seccos5533535555115cot2tan25θθθθθθ=======⋅====

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Section 1.2Right Triangle Trigonometry136.We can find the value ofcosθby using the Pythagorean identity.22sincos1212cos1212cos1412cos1432cos4 33cos42θθθθθθθ+=+=+==−===7.a.oooosin 46cos(9046)cos 44=−=b.cottan122126tan12125tan 12ππππππ=−=−=8.Many Scientific CalculatorsFunctionModeKeystrokesDisplay(rounded to four places)a.sin 72.8°Degree72.8 SIN0.9553b.csc 1.5Radian1.5 SIN 1/x1.0025Many Graphing CalculatorsFunctionModeKeystrokesDisplay(rounded to four places)a.sin 72.8°DegreeSIN 72.8 ENTER0.9553b.csc 1.5Radian(SIN 1.5 )x-1ENTER1.00259.Because we have a known angle, an unknown opposite side, and a known adjacent side, we select thetangent function.otan 24750o750 tan 24750(0.4452)333.9aaa==≈≈The distance across the lake is approximately 333.9 yards.

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