year 13 mathematics ias 3

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



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



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




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




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)


51IAS 3.3 – Trigonometry


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.




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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