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π 2π
f(x)
1
–1
2
–2
x
3
–3
Year 13Mathematics
Contents
uLake Ltdu a e tduLake LtdInnovative Publisher of Mathematics Texts
Robert Lakeland & Carl NugentTrigonometry
• AchievementStandard .................................................. 2
• ReviewofEarlierTrigonometricWork ..................................... 3
• TrigonometricGraphs................................................... 5
• FindingaParticularSolutiontoaTrigonometricEquation .................... 11
• GeneralSolutionsofTrigonometricEquations ............................. 16
• ModellingPracticalSituationsusingTrigonometricFunctions................. 22
• ReciprocalTrigonometricFunctions ..................................... 26
• CompoundTrigonometricFormulae ..................................... 31
• DoubleAngleFormulae ................................................. 36
• TrigonometricEquationsExpressedasQuadratics .......................... 38
• SumandProductFormulae ............................................. 40
• ApplicationsofTrigonometry ............................................ 45
• PracticeInternalAssessment ............................................. 52
• Formulae .............................................................. 56
• Answers............................................................... 57
IAS 3.3
IAS 3.3 – Year 13 Mathematics and Statistics – Published by NuLake Ltd New Zealand © Robert Lakeland & Carl Nugent
2 IAS 3.3 – Trigonometry
Thisachievementstandardinvolvesapplyingtrigonometricmethodsinsolvingproblems.
◆ ThisachievementstandardisderivedfromLevel8ofTheNewZealandCurriculumandisrelatedto theachievementobjectives ❖ manipulatetrigonometricexpressions ❖ formanduseusetrigonometricequations intheMathematicsstrandoftheMathematicsandStatisticsLearningArea.
◆ Applytrigonometricmethodsinsolvingproblemsinvolves: ❖ selectingandusingmethods ❖ demonstratingknowledgeofconceptsandterms ❖ communicatingusingappropriaterepresentations.
◆ Relationalthinkinginvolvesoneormoreof: ❖ selectingandcarryingoutalogicalsequenceofsteps
❖ connectingdifferentconceptsorrepresentations ❖ demonstratingunderstandingofconcepts ❖ formingandusingamodel; andalsorelatingfindingstoacontext,orcommunicatingthinkingusingappropriatemathematical statements.
◆ Extendedabstractthinkinginvolvesoneormoreof: ❖ devisingastrategytoinvestigateorsolveaproblem
❖ identifyingrelevantconceptsincontext ❖ developingachainoflogicalreasoning,orproof ❖ formingageneralisation; andusingcorrectmathematicalstatements,orcommunicatingmathematicalinsight.
◆ Problemsaresituationsthatprovideopportunitiestoapplyknowledgeorunderstandingof mathematicalconceptsandmethods.Situationswillbesetinreal-lifeormathematicalcontexts.
◆ Methodsincludeaselectionfromthoserelatedto: ❖ trigonometricidentities ❖ reciprocaltrigonometricfunctions ❖ propertiesoftrigonometricfunctions ❖ solvingtrigonometricequations ❖ generalsolutions.
Achievement Achievement with Merit Achievement with Excellence• Applytrigonometricmethods
insolvingproblems.• Applytrigonometric
methods,usingrelationalthinking,insolvingproblems.
• Applytrigonometricmethods,usingextendedabstractthinking,insolvingproblems.
NCEA 3 Internal Achievement Standard 3.3 – Trigonometry
IAS 3.3 – Year 13 Mathematics and Statistics – Published by NuLake Ltd New Zealand © Robert Lakeland & Carl Nugent
8 IAS 3.3 – Trigonometry
Thegraphy=cosxstartsat(0,1)sowedrawour
graphstartingat π2,1
⎛⎝⎜
⎞⎠⎟.
Example
Graphf(x)=cos x – π2
⎛⎝⎜
⎞⎠⎟from0to2π.
Wedrawf(x)=cosxfromthetranslated
originatπ2,0
⎛⎝⎜
⎞⎠⎟ .
Note:Thefinalgraphoff(x)=cos x – π2
⎛⎝⎜
⎞⎠⎟isthe
graphy=sinx.
Example
Graphy=sin x –π6
⎛⎝⎜
⎞⎠⎟ +1from0to2π.
Thishastwotranslations,across π6and
up1.
Wedrawy=sinxfromthetranslatedoriginatπ6,1
⎛⎝⎜
⎞⎠⎟ .
ExampleGraphf(x)=3cos(x–π)from0to2π.
Wedrawf(x)=cosxfromthetranslatedoriginat(π,0)thenextenditback.Thistranslatedgraphisthenstretchedthreetimesvertically.Justgotothekeypointsandmovethemthreetimesasfarfromthecentreline.
ExampleGraphy=2sin3xfrom0to2π.
Wedrawy=sinxbutwitha 13horizontalscale.
Wherewenormallywouldhaveamaximumatπ2,1
⎛⎝⎜
⎞⎠⎟wenowhaveamaximumat
π6,2⎛
⎝⎜⎞⎠⎟ .
Wecontinuethisgraphuntil2πcheckingtomakesurewehavethreecompletecycles.Nowwestretcheachpointverticallytwotimes.
π 2π
f(x)1
–1
x
f(x)=cos x – π2
⎛⎝⎜
⎞⎠⎟
π 2π
f(x)
1
–1
x
y=sin x –π6
⎛⎝⎜
⎞⎠⎟ +1
π6,1
⎛⎝⎜
⎞⎠⎟
π2,1
⎛⎝⎜
⎞⎠⎟
2
Afterthesinecurveisdrawnfrom( π6,1),the
graphisextendedback.
f(x)=3cos(x–π)
π 2π
f(x)
1
–1
x
2
–2
y=cos(x–π)
y=2sin3x
π 2π
y
1
–1
x
2
–2
y=sin3x
©uLake LtduLake Ltd Innovative Publisher of Mathematics Texts
IAS 3.3 – Year 13 Mathematics and Statistics – Published by NuLake Ltd New Zealand © Robert Lakeland & Carl Nugent
12 IAS 3.3 – Trigonometry
Example Findonesolutiontotheequation4sin x2
⎛⎝⎜
⎞⎠⎟=–1.234
Manipulationoftheequation.
4sin x2
⎛⎝⎜
⎞⎠⎟=–1.234
sin x2
⎛⎝⎜
⎞⎠⎟=–0.3085
x2=sin–1(–0.3085)
x=–0.31362 x 2 =–0.6272(4sf)
Example Findonesolutiontotheequation20–25cos x + π2
⎛⎝⎜
⎞⎠⎟=40
–25cos x + π2
⎛⎝⎜
⎞⎠⎟=20
cos x + π2
⎛⎝⎜
⎞⎠⎟=–0.8
x+ π2=cos–1(–0.8)
x=2.4981– π2
x=0.9273(4sf)
ThegraphicalapproachontheCasio9750GII.Drawthegraphofy=4sin x
2⎛⎝⎜
⎞⎠⎟andy=–1.234from
Thecalculatorwillfindthesolutionx=–0.6272.
0to2π.Wecanseefromthegraphthatthesedonotintersectfrom0to2πsointheviewwindowwechangethedomainto–πtoπthenselectandgraphsolve(G-Solv)andthenintersection(ISCT).
SHIFT F3 (-)VWindow Xmin
EXEmax
SHIFTSHIFT EXPπ
SHIFT F5G-Solv
F5ISCT
F6DRAW
ThegraphicalapproachontheTI-84Plus.Drawthegraphsof
y=40andy=20–25cos x + π2
⎛⎝⎜
⎞⎠⎟
fromx=0tox=2πandadjusttheyvaluestogofrom–50to50asthisgraphhasalargeamplitude.
Youwillbeaskedtoselectthecurvesthataretointersectandtoenteraguessneareachpointofintersection(orusethecursorkeystomovethepointclosetotheintersection).Thesolutionisx=0.9273.
(–) 05 ENTERWINDOW
05 ENTER GRAPH 2ND TRACECALC
5intersect
EXEEXPπ
EXE
UsingthesolverontheCasio9750GII.Checkyourcalculatorissettoradiansthenentertheequation4sin x
2⎛⎝⎜
⎞⎠⎟=–1.234.
Gettingasoln.x=–0.6272.
MENU 4 sin8EQUA
(
2
X ÷
( – )) SHIFT .=
21 .
3 4 EXE
TousethesolverontheTI-84calculatortheequationmustbeintheform0=equation.
0=20+25cos x + π2
⎛⎝⎜
⎞⎠⎟
For these Examples, three different approaches are presented for finding a particular solution. Students should concentrate on the approach they find easiest to understand. Approach 1 (in blue) is manipulation of the equation, approach 2 (in pink) is by graphing the problem and approach 3 (in yellow) is using the solver built into the graphics calculator.
Manipulationoftheequation.
F3
F6
MATH CLEAR
+X ^ ÷2NDπ
)2 ENTER Enteraguess(e.g.0)and
togetx=0.9273.
Solver 2 +
2 5
0
COS
ALPHA ENTERSOLVE
ENTER
©uLake LtduLake Ltd Innovative Publisher of Mathematics Texts
IAS 3.3 – Year 13 Mathematics and Statistics – Published by NuLake Ltd New Zealand © Robert Lakeland & Carl Nugent
22 IAS 3.3 – Trigonometry
Modelling Practical Situations using Trigonometric Functions
Modelling Practical Situations using Trigonometric Functions
Ifyouhaveinformationaboutaproblemthatisbestmodelledbyatrigonometricfunction,thenwecanuseourunderstandingofamplitude,frequencyandperiodtowriteanequationandsubsequentlysolvethisequation.Aperiodicfunctionthatfollowsasinecurvewillhaveanequationf(x)=AsinB(x+C)+D.TheconstantsA,B,CandDweredefinedpreviously.Whenwearemodellingapracticalsituationwearelikelytoknowthemaximum,minimumvaluesalongwiththeperiodandhorizontalshift.WeusethesedatavaluestofindtheconstantsA,B,CandD.TheamplitudeAisgivenby
A=(maximum–minimum)2
ThehorizontalstretchBisgivenby
period=2πB
B=2π
period
wheretheperiodisthexvalue(oftentime)beforethegraphstartstorepeatitself.
Thehorizontaltranslationorshift–Cisthedistancealongthexaxisfromx=0(ort=0)untilthesinecurvestarts.Alternativelyitisthetimetothemaximumpointminusaquarteroftheperiod.
C=–shift
=–(timetomaximum– 14period)
OR =–timetomaximum+ 14period
TheverticaltranslationDisdefinedastheaverageheight.
D =(maximum+minimum)2
Ifwestudythegraphshownherewecanseethestandardsinecurvestartsatthepoint(4,3).Ithasamaximumvalueof5andaminimumvalueof1withaperiodof6.Theshiftfromthestartis+4.
ThereforewecanfindtheequationbyfindingeachconstantA,B,CandD.
A =(maximum–minimum)2
= 5 –12
=2
B =2π
period
= 2π6
π3
⎛⎝⎜
⎞⎠⎟
C =–shift
=–(timetomaximum– 14period)
OR =–timetomaximum+ 14period
=–4
D =(maximum+minimum)2
=(5+1)2
=3Theequationisthereforey=2sin 2π
6(x – 4) +3
1
2
3
4
5
x
2 4 6 8 10 12
shiftStartofsinegraph
A
Period
yshift=timeuntilmax– 1
4period
Summary: If T = period then A = max -min2
B = 2!T
C = –time to max + 14
period OR –time to max + T4
D = max+min2
©uLake LtduLake Ltd Innovative Publisher of Mathematics Texts
IAS 3.3 – Year 13 Mathematics and Statistics – Published by NuLake Ltd New Zealand © Robert Lakeland & Carl Nugent
46 IAS 3.3 – Trigonometry
0.1140.185
600 mTown Centre
Acommunicationtowerisontopofanislandinariver.Onflatgroundyounotethetopofthetowerisatanangleofelevationof0.114radians.
Afterwalking600moverlevelgrounddirectlytowardsthetowertheangleofelevationisincreasedto0.185radians.
Whatistheverticalheighttothetopofthetower?
Example
Redrawthediagramlettingtheheightbehandthedistancetothebase(notrequired)bex.
Usingthesineruleweusetheanglepropertiesofatriangletofindtheangleatthetopandthencalculatelengthc.
angleattop=0.185–0.114
=0.071
Externalangleofatriangle=sum2oppositeinternalangles.
asinA
= csinC
600sin0.071
= csin0.114
c=600 x sin0.114sin0.071
c=962.103
Nowwiththesmalltrianglewecancalculateh.
sin0.185=opphyp
sin0.185= h962.103
h=962.103sin0.185
h=177m(3sf)
Thetwoapproachesexplainedherearetousetangentsortousethesinerule(bothwork).
0.114 0.185
600 m
h
x
c
0.071Usingtangents tanθ=opp
adj tan0.114= height
horizontal distanceSubstituteheight=handhorizontaldist=600+x
= hx + 600
Similarlywiththesmalltriangle
tan0.185= heighthorizontal distance
Substituteheight=handhorizontaldist=x
tan0.185=hx
Aswedonotwantxwerearrangeoneequationsoxbecomesthesubjectandsubstituteitintotheother.
xtan0.185=h
x= htan0.185
Substituteinthefirstequationandsolve
tan0.114=h
htan0.185
+ 600
h=( htan0.185
+600)tan0.114
8.734h=5.344h+600 h=177m(3sf)
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51IAS 3.3 – Trigonometry
IAS 3.3 – Year 13 Mathematics and Statistics – Published by NuLake Ltd New Zealand © Robert Lakeland & Carl Nugent
143. AplaneissightedatpointAfromaradarstationatO.Theplaneisatanangleofelevationof44˚andataheightof6800metres.SometimelaterthesameplaneissightedatpointBfromtheradarstationatanangleofelevationof27˚andataheightof4900metres.CandDarepointsonthegroundverticallybelowpointsAandBandangleCODis48˚.
a) Drawadiagramtorepresentthesituationdescribedabove.
Remembertolabelallapplicablepoints,anglesandlengths.
b) CalculatethedistancebetweenthepointsCandDtothreesignificantfigures.
Drawatwo-dimensionaldiagramhere.
144. Alargetreestandsattheendofaroad.Fromoneparticularpointontheroad,thetopofthetreeisatangleAdegrees.Afterwalkingdirectlytowardsthetreealongalevelroadfor23metrestheangleofelevationisincreasedtoBdegrees.
a) Showthath,theheightofthetreecan berepresentedbyh=
23cotA – cotB
b) IfangleAis27°andangleBis85° findtheheightofthetreetothe nearestmetre.
Drawatwo-dimensionaldiagramhere.
©uLake LtduLake Ltd Innovative Publisher of Mathematics Texts
IAS 3.3 – Year 13 Mathematics and Statistics – Published by NuLake Ltd New Zealand © Robert Lakeland & Carl Nugent
60 IAS 3.3 – Trigonometry
Page 37 cont...
103. RHS= 2cos2xsin 2x
= 2(cos2 x – sin2 x)
2sinxcosx
= cos2 x – sin2 xsinxcosx
= cosxsinx
– sinxcosx
=cotx–tanx
=LHS
104. sin3A=sin(A+2A) =sinAcos2A+cosAsin2A =sinA(1–2sin2A)+2sinAcos2A =sinA–2sin3A+2sinA(1–sin2A) =3sinA–4sin3A
105. cos3A=cos(A+2A) =cosAcos2A–sinAsin2A =cos3A–cosAsin2A–2sin2AcosA =cos3A–3cosAsin2A =4cos3A–3cosA
Page 39
106. x =2nπ±2.0944 or x =nπ+(–1)n0.3398 x = 0.3398,2.0944,2.8018,4.1888
107. x =nπ+(–1)n0.5236 or x =nπ–(–1)n0.7297 x = 0.5236,2.6180,3.8713,5.5535
108. x =2nπ±0.8411 or x =2nπ±2.4189
x = 0.8411,2.4189,3.8643,5.4421
109. x =nπ+1.2490 or x =nπ–1.1071
x = 1.2490,2.0344,4.3906,5.1760
110. (2sinx–1)(sinx+2)=0 x=nπ+(–1)n0.5236only
x = 0.5236,2.6180
111. x=2nπ±0.8411 or x=2nπ±3.1416 x = 0.8411,3.1416,5.4421
112. tanx=0 x = 0,π,2π
113. x=nπ+(–1)n0.3398 or x=nπ–(–1)n0.2526 x = 0.3398,2.8018,3.3943,6.0305
Page 41
114. P=4sin9θ+4sinθ
115. Q = 12(cos4A+cos(2A+2B))
116. R=5cos10θ+5cos2θ
117. S=2(cos(3A+2B)–cos5A)
118. T=3(cos2x–cos4x)
119. U=cos 2x + π2
⎛⎝⎜
⎞⎠⎟+cosπ
2
=cos 2x + π2
⎛⎝⎜
⎞⎠⎟
120. V=sin4x+sinπ2
=sin4x+1121. W=0.5cosπ–0.5cos2x = –0.5–0.5cos2x122. X=3(cos2π+cos2x) =3+3cos2x
123.Y=5(cos–2x–cosπ4+ π4
⎛⎝⎜
⎞⎠⎟ )
=5cos2x
Page 42
124. 4 cos π3
⎛⎝⎜
⎞⎠⎟– cos(2x)⎛
⎝⎜⎞⎠⎟ = 4
2x=2nπ±2.0944 x=nπ±1.0472 x=1.0472,2.0944,4.1888,5.2359
or x = π3, 2π3, 4π3, 5π3
125.cos2x=1 2x=2nπ x=nπ x=0,π,2π
126. sin(2x–0.5)+sin(0.5)=2 x 0.4567 2x=nπ+(–1)n0.4489+0.5 x=0.5nπ+(–1)n0.2244+0.25 x=0.4744,1.596,3.616,4.738
127. 2x=nπ+(–1)n0.4058–0.5 x=0.5nπ+(–1)n0.2029–0.25 x=1.118,3.095,4.259,6.236
128. 2(cos(4x+π)+cosπ))=–3 4x+π=2nπ±2.0944 x=0.5nπ±0.5236–0.7854 x = 0.2618,1.3090,1.8326,2.8798,
3.4034,4.4506,4.9742,6.0214
129. (2x–0.5267)=2nπ±1.8140 x=nπ±0.9070+0.2634 x=1.1703,2.4980,4.3120,5.6396
Page 35 cont...97. LHS=sin(x+y).sin(x–y) =(sinxcosy+sinycosx)(sinxcosy–sinycosx) =sin2x.cos2y–sin2y.cos2 x = sin2x.(1–sin2y)–sin2y.(1–sin2 x) = sin2x–sin2x.sin2y–sin2y+sin2y.sin2 x =sin2x–sin2y =RHS
98. LHS=tan A– π4
⎛⎝⎜
⎞⎠⎟tan A + π
4⎛⎝⎜
⎞⎠⎟
= tanA − tan π
4⎛⎝⎜
⎞⎠⎟tanA + tan π
4⎛⎝⎜
⎞⎠⎟
1+ tanAtan π4
⎛⎝⎜
⎞⎠⎟1− tanAtan π
4⎛⎝⎜
⎞⎠⎟
= tanA −1( ) tanA +1( )1+ tanA( ) 1− tanA( )
= − 1− tanA( )1− tanA( )
= –1 =RHS
99. LHS=cot(A+B)
= 1tan A + B( )
= 1− tanAtanBtanA + tanB
Dividetopandbottomby tanA.tanB
= 1
tanAtanB– tanAtanBtanAtanB
tanAtanAtanB
+ tanBtanAtanB
= cotAcotB−1cotB+ cotA
=RHS
Page 37
100. cos24x–0.5=0.5(2cos2 4x – 1) =0.5cos(8x)
101. sin x2
⎛⎝⎜
⎞⎠⎟cos x
2⎛⎝⎜
⎞⎠⎟=0.5sinx
102. LHS=(sinA–cosA)2 =sin2A–2sinAcosA+cos2A =1–2sinAcosA =1–sin2A =RHS
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