welcome to math analysis 2 honors! - bergen · welcome to math analysis 2 honors! i hope you have...

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WelcometoMathAnalysis2Honors!Ihopeyouhaveenjoyedyoursummerbreaksofar,butnowitistimetogetbackinto“school-mode”andreviewsomethingsfromMA1.WestartMA2withtheendoftheTrigonometryUnityoustartedinMA1;soitisveryimportantthatthosetopicsarerefreshedbeforeyoureturninSeptember.

BeforeyoubeginMathAnalysis2,youwillbeexpectedtohavemasteredthefollowingtopicswhichwerecoveredinMA1:• RightTriangleTrigonometry• SolvingTrigonometricEquations• Simplifying&ProvingTrigonometricIdentities• LawofSinesandLawofCosines• NavigationProblems

TechnologyneededforMA2:TI-84(orhigher)andaScientificCalculator.**THEREWILLBEATESTTHEFIRSTFEWDAYSOFSCHOOLONTHISMATERIAL**Thisyearwillbechallengingandalotofworkforyou.ButifyouhaveagoodattitudeandworkethicitcanalsobeafunandrewardingcoursetosetyouoncoursefortheAPCalculustrackhereatBergenTech.Emailmeifyouhaveanyquestions.Seeyousoon!Mr.Mellina(nicmel@bergen.org)&Dr.Ge(peige@bergen.org)MathAnalysis2Teachers

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RightTriangleTrigFindtheexactvalueofeachtrigonometricfunction.Theseshouldbedonewithoutacalculator.1. tan $%

& 2. sec− &%

+ 3. cos --%

. 4. cot /%

&

5. csc− %

$ 6. tan− &%

$ 7. sec 0 8. tan− 1%

+

9. sec 𝜃 10. sin 𝜃11. cos 𝜃 12. sin 𝜃13. cos 𝜃 14. cos 𝜃

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SolvingTrigEquationsSolvethefollowingequationsfor0 ≤ 𝑥 < 360°tothenearestdegree.15. cos 𝑥 + 1 = 0 16. sin$ 𝑥 = 017. 2 cos 𝑥 − 3 = 0 18. 2 sin 𝑥 + 3 = 019. 2 + sec 𝑥 = 0 20. tan 𝑥 csc 𝑥 + 2 = 021. cos 𝑥 tan 𝑥 − 3 = 0 22. cot$ 𝑥 + cot 𝑥 = 023. csc$ 𝑥 + 2 csc 𝑥 = 0 24. tan$ 𝑥 − 3 = 0

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25. 1 − cot$ 𝑥 = 0 26. 2 cos 𝑥 csc 𝑥 = 3 csc 𝑥27. 2 sin 𝑥 sec 𝑥 = sec 𝑥 28. 2 cos$ 𝑥 − cos 𝑥 = 129. 1 + cos 𝑥 − 2 sin$ 𝑥 = 0 30. cot$ 𝑥 − csc 𝑥 = 131. tan$ 𝑥 + 3 sec 𝑥 + 3 = 0 32. sin 𝑥 + 4 csc 𝑥 + 5 = 033. 3 sec 𝑥 − cos 𝑥 − 2 = 0 34. 4 sin+ 𝑥 + 3 sin$ 𝑥 − 1 = 0

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SimplifyingandProvingTrigIdentitiesSimplifyeachexpressiontoasingletrigfunction.35. sec 𝑥 sin 𝑥 36. cos 𝑥 tan 𝑥 37. csc 𝑥 tan 𝑥38. tan$ 𝑥 − sec$ 𝑥 39. 1 − cos$ 𝑥 40. 1 − cos 𝑥 1 + cos 𝑥 41. sec 𝑥 − 1 sec 𝑥 + 1 42. -

@ABC D− -

EFBC D 43. 1 − @ABC D

EFBC D

44. -

GH@C D− -

GHEC D 45. cos 𝑥 sec 𝑥 − cos 𝑥 46. cos$ 𝑥 sec$ 𝑥 − 1

47. 1 − cos 𝑥 1 + sec 𝑥 cos 𝑥 48. @AB D GH@ D

-IGH@C D 49. EFBC D

@JG DK-+ 1

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SolvingTriangles50. Twotowersareconstructedonaplotofland.Theangleofdepressionofthetopofthe

towerAwhenseenfromthetopofthetowerBis50°.IftheheightofthetowerBis120feetandtheheightofthetowerAis68feet,thenfindthehorizontaldistancebetweenthetwotowers.

51. Fromapointonthegroundawayfromthebaseofabuilding,theangleofelevationto

thetopofthebuildingis51°.Fromapoint4metersclosertothebuilding,theangleofelevationtothetopofthebuildingincreasesto67°.Howtallisthebuilding?(Roundyouranswerstothenearesttenth)

SolvetriangleABCwiththegivenspecifications.Roundallsidelengthstothenearesttenthandtheanglestothenearestangle.52. A=28°,C=90°,b=15 53. C=90°,c=9,a=4

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54. A=50°,C=65°,b=60 55. C=50°,a=12,b=1456. a=16,b=23,c=17 57. A=34°,a=6,b=7

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58. B=80°,b=7,c=5 59. A=65°,a=3.8,b=4.560. A=36°,a=9,andb=12

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Navigation61. Ashipproceedsonacourseof300°for2hoursataspeedof15knots(1knot=1

nauticalmileperhour).Thenitchangescourseto230°,continuingat15knotsfor3morehours.Atthattime,howfarandonwhatcourseistheshipfromitsstartingpoint?

62. TownTis8kmnortheastofvillageV.CityCis4kmfromTonabearingof150°fromT.

WhatisthebearinganddistanceofCfromV.63. Aplaneflies600kmonacourseof300°.Itthenfliessouthforawhileandfinallyflies

ona40°coursetoreturntoitsstartingpoint.Findthetotaldistancetraveledandthecoursefromitsstartingpointtoitslastturningpoint.