Chapter4/5Part1-Trigonometryin

Radians

LessonPackage

MHF4U

Chapter4/5Part1OutlineUnitGoal:Bytheendofthisunit,youwillbeabletodemonstrateanunderstandingofmeaningandapplicationofradianmeasure.Youwillbeabletomakeconnectionsbetweentrigratiosandthegraphicalandalgebraicrepresentationsofthecorrespondingtrigfunctions.

Section Subject LearningGoals CurriculumExpectations

L1 RadianMeasure-recognizetheradianasanalternativeunittothedegreeforanglemeasurement-convertbetweendegreeandradianmeasures

B1.1,1.2

L2 TrigRatiosandSpecialAngles

-Determine,withouttechnology,theexactvaluesoftrigratiosforspecialangles B1.3,1.4

L3 GraphingTrigFunctions -sketchthegraphsofall6trigratiosandbeabletodescribekeyproperties B2.1,2.2,2.3

L4 TransformationsofTrigFunctions

-Givenequationstatepropertiesofsinusoidalfunction,andgraphitusingtransformations-Givengraph,writetheequationofthetransformedfunction

B2.4,2.5,2.6

L5 ApplicationsofTrigFunctions

-Solveproblemsarisingfromrealworldapplicationsinvolvingtrigfunctions B2.7

Assessments F/A/O MinistryCode P/O/C KTACNoteCompletion A P PracticeWorksheetCompletion F/A P Quiz–FindingTrigRatios F P PreTestReview F/A P Test–TriginRadians O B1.1,1.2,1.3,1.4

B2.1,2.2,2.3,2.4,2.5,2.6,2.7 P K(21%),T(34%),A(10%),C(34%)

L1–4.1RadianMeasureMHF4UJensenPart1:WhatisaRadian?Anglesarecommonlymeasuredindegrees.However,inmathematicsandphysics,therearemanyapplicationsinwhichexpressinganangleasapurenumber,withoutunits,ismoreconvenientthanusingdegrees.Whenmeasuringinradians,thesizeofanangleisexpressedintermsofthelengthofanarc,𝑎,thatsubtendstheangle,𝜃,where𝜃 = $

%.

1Radianisdefinedasthesizeofananglethatissubtendedbyanarcwithalengthequaltotheradiusofthecircle.Therefore,whenthearclengthandradiusareequal,𝜃 = $

%= %

%= 1radian

Howmanyradiansareinafullcircle?Orinotherwords,howmanytimescananarclengthequaltotheradiusfitaroundthecircumferenceofacircle?Remember:𝐶 = 2𝜋𝑟Soifweusethefullcircumferenceofthecircleforthearclength,𝜃 = $

%= +,%

%= 2𝜋radians

Part2:SwitchingBetweenDegreesandRadiansThekeyrelationshipyouneedtoknowinordertoswitchbetweendegreesandradiansis:

360° = 2𝜋radians

1degree= ,123

radiansToswitchfromdegreestoradians,multiplyby

𝜋180

1radian=123,degrees

Toswitchfromradianstodegrees,multiplyby

180𝜋

Example1:Startbyconvertingthefollowingcommondegreemeasurestoradians

a)180°

= 180°×𝜋180

= 𝜋radians

b)90°

= 90°×𝜋180

= ,

+radians

c)60°

= 60°×𝜋180

= ,

8radians

d)45°

= 45°×𝜋180

= ,

;radians

e)30°

= 30°×𝜋180

= ,

<radians

f)1°

= 1°×𝜋180

= ,

123radians

Example2:Converteachofthefollowingdegreemeasurestoradianmeasures

Example3:Converteachofthefollowingradianmeasurestodegreemeasures

Part3:ApplicationExample4:Suzettechoosesacameltorideonacarousel.Thecamelislocated9mfromthecenterofthecarousel.Itthecarouselturnsthroughanangleof=,

<,determinethelengthofthearctravelledbythecamel,

tothenearesttenthofameter.𝜃 = $

%

𝑎 = 𝜃𝑟

𝑎 =5𝜋6 9

𝑎 ≅ 23.6m

a)225°

= 225°×𝜋180

= =,

;radians

b)80°

= 80°×𝜋180

= ;,

@radians

c)450°

= 450°×𝜋180

= =,

+radians

a)+,8radians

=2𝜋3 ×

180𝜋

= 120°

b)@,;radians

=9𝜋4 ×

180𝜋

= 405°

c)1radian

= 1×180𝜋

= 57.296°

1

1√2

2

1

√3

Part4:SpecialTrianglesUsingRadianMeasuresDrawthe45-45-90triangle Drawthe30-60-90triangle Also,youwillneedtoremembertheUNITCIRLCEwhichisacirclethathasaradiusof1.Usetheunitcircletowriteexpressionsforsin 𝜃,cos 𝜃,andtan 𝜃intermsof𝑥,𝑦,and𝑟.Ontheunitcircle,thesineandcosinefunctionstakeasimpleform:sin 𝜃 = 𝑦cos 𝜃 = 𝑥Thevalueofsin 𝜃isthe𝒚-coordinateofeachpointontheunitcircleThevalueofcos 𝜃isthe𝒙-coordinateofeachpointontheunitcircleTherefore,wecanusethepointswheretheterminalarmintersectstheunitcircletogetthesineandcosineratiosjustbylookingatthe𝑦and𝑥co-ordinatesofthepoints.

sin 𝜃 =𝑦𝑟 =

𝑦1 = 𝑦

cos 𝜃 =𝑥𝑟 =

𝑥1 = 𝑥

tan 𝜃 =𝑦𝑥

Example5:Filloutchartofratiosusingspecialtrianglesandtheunitcircle

0, 2𝜋𝜋6

𝜋4

𝜋3

𝜋2 𝜋 3𝜋

2

sin 𝑥 0 12

12 3

2 1 0 -1

cos 𝑥 1 32

12 1

20 -1 0

tan 𝑥 013 1 3 undefined 0 undefined

L2–4.2TrigRatiosandSpecialAnglesMHF4UJensenPart1:ReviewofLastYearTrigWhatisSOHCAHTOA?Ifweknowarightangletrianglehasanangleof𝜃,allotherrightangletriangleswithanangleof𝜃aresimilarandthereforehaveequivalentratiosofcorrespondingsides.Thethreeprimaryratiosareshowninthediagramtotheright.Whatisareferenceangle?Anyangleover90hasareferenceangle.Thereferenceangleisbetween0°and90°andhelpsusdeterminetheexacttrigratioswhenwearegivenanobtuseangle(angleover90degrees).Thereferenceangleistheanglebetweentheterminalarmandtheclosestx-axis(0/360or180).WhatistheCASTrule?Whenfindingthetrigratiosofpositiveangles,wearerotatingcounterclockwisefrom0degreestoward360.TheCASTrulehelpsusdeterminewhichtrigratiosarepositiveineachquadrantNote:Therearemultipleanglesthathavethesametrigratio.Youcanusereferenceanglesandthecastruletofindthem.

sinθ

cosθ

tanθ =

=

= opposite

opposite

hypotenuse

hypotenuse

adjacent

adjacent

1

1√2 2

1

√3

Part2:FindingExactTrigRatiosforSpecialAnglesStartbydrawingbothspecialtrianglesusingradianmeasuresExample1:Findtheexactvalueforeachgiventrigratio.

a)tan (()*= − (

√.

b)sin 1)*= (

2

c)cos 1)5= − (

√2 d)sec .)

5= −√2

Part3:ApplicationQuestionExample2:Findthevalueofall6trigratiosfor1)

.

Justinisflyingakiteattheendofa50-mstring.Thesunisdirectlyoverhead,andthestringmakesanangleof)*withtheground.Thewindspeedincreases,andthekiteflieshigheruntilthestringmakesanangleof)

.with

theground.Determineanexactexpressionforthehorizontaldistancebetweenthetwopositionsofthekitealongtheground.

Thehorizontaldistancebetweenthetwokites= 𝑥( − 𝑥2 = 25 3 − 25 = 25 3 − 1 meters

cos𝜋6 =

𝑥(50

√32 =

𝑥(50

𝑥( =50√32

𝑥( = 25√3

cos𝜋3 =

𝑥250

12 =

𝑥250

𝑥2 =502

𝑥2 = 25

L3–5.1/5.2GraphingTrigFunctionsMHF4UJensenPart1:RemembertheUnitCircleTheunitcircleisacircleisacirclethatiscenteredattheoriginandhasaradiusof1unit.Ontheunitcircle,thesineandcosinefunctionstakeasimpleform:

sin 𝜃 =𝑦𝑟 =

𝑦1 = 𝑦

cos 𝜃 =𝑥𝑟 =

𝑥1 = 𝑥

Thevalueofsin 𝜃isthe𝒚-coordinateofeachpointontheunitcircleThevalueofcos 𝜃isthe𝒙-coordinateofeachpointontheunitcircle

Part2:GraphingSineandCosine

Tographsineandcosine,wewillbeusingaCartesianplanethathasanglesfor𝑥values.

Example1:Completethefollowingtableofvaluesforthefunction𝑓 𝑥 = sin(𝑥).Usespecialtriangles,theunitcircle,oracalculatortofindvaluesforthefunctionat30° = 5

6radianintervals.

Example2:Completethefollowingtableofvaluesforthefunction𝑓 𝑥 = cos(𝑥).Usespecialtriangles,theunitcircle,oracalculatortofindvaluesforthefunctionat30° = 5

6radianintervals.

𝑥 𝑓(𝑥)0 056 7

8

856= 5

9 9

8~0.87

956= 5

8 1

>56= 85

9 9

8~0.87

?56 7

8

656= 𝜋 0A56 − 7

8

C56= >5

9 − 9

8~ − 0.87

D56= 95

8 −1

7E56= ?5

9 − 9

8~ − 0.87

7756 − 7

8

7856= 2𝜋 0

𝑥 𝑓(𝑥)0 156 9

8~0.87

856= 5

9 7

8

956= 5

8 0

>56= 85

9 − 7

8

?56 − 9

8~ − 0.87

656= 𝜋 −1A56 − 9

8~ − 0.87

C56= >5

9 − 7

8

D56= 95

8 0

7E56= ?5

9 7

8

7756 9

8~0.87

7856= 2𝜋 1

PropertiesofbothSineandCosineFunctionsDomain:{𝑋 ∈ ℝ}Range:{𝑌 ∈ ℝ| − 1 ≤ 𝑦 ≤ 1}Period:2𝜋radians

Amplitude:OPQROST8

= 7R(R7)8

= 1PERIOD:thehorizontallengthofonecycleonagraph.AMPLITUDE:halfthedistancebetweenthemaximumandminimumvaluesofaperiodicfunction.Part3:GraphingtheTangentFunction

Recall:tan 𝜃 = WXYZ[\WZ

Note:Sincecos 𝜃isinthedenominator,anytimecos 𝜃 = 0,tan 𝜃willbeundefinedwhichwillleadtoaverticalasymptote.

Sincesin 𝜃isinthenumerator,anytimesin 𝜃 = 0,tan 𝜃willequal0whichwillbean𝑥-intercept.

Example3:Completethefollowingtableofvaluesforthefunction𝑓 𝑥 = tan(𝑥).Usethequotientidentitytofind𝑦-values.

PropertiesoftheTangentFunctionDomain: 𝑋 ∈ ℝ 𝑥 ≠ 𝑘𝜋

2𝑤ℎ𝑒𝑛𝑘𝑖𝑠𝑜𝑑𝑑} Range:{𝑌 ∈ ℝ}

Period:𝜋radians Amplitude:none(nomaxormin)

𝑥 𝑓(𝑥)0 E

7= 0

56 7

9~0.58

856= 5

9 3~1.73

956= 5

8 7

E=und.

>56= 85

9 − 3~ − 1.73

?56 − 7

9~ − 0.58

656= 𝜋 E

R7= 0

A56

79~0.58

C56= >5

9 3~1.73

D56= 95

8 R7

E=und.

7E56= ?5

9 − 3~ − 1.73

7756 − 7

9~ − 0.58

7856= 2𝜋 E

7= 0

Part4:GraphingReciprocalTrigFunctionsThegraphofareciprocaltrigfunctionisrelatedtothegraphofitscorrespondingprimarytrigfunctioninthefollowingways:

• Reciprocalhasaverticalasymptoteateachzeroofitsprimarytrigfunction• Reciprocalhasazeroateachverticalasymptoteofitsprimarytrigfunction• Hasthesamepositive/negativeintervalsbutintervalsofincreasing/decreasingarereversed• 𝑦-valuesof1and-1donotchangeandthereforethisiswherethereciprocalandprimaryintersect• Localminpointsoftheprimarybecomelocalmaxofthereciprocalandviceversa.

Example4:Completethefollowingtableofvaluesforthefunction𝑓 𝑥 = csc(𝑥).Usethereciprocalidentitytofind𝑦-values.

PropertiesoftheCosecantFunctionDomain: 𝑋 ∈ ℝ 𝑥 ≠ 𝑘𝜋}where𝑘 ∈ ℕ Range:{𝑌 ∈ ℝ}Period:2𝜋radians Amplitude:none(nomaxormin)

ReciprocalIdentities

𝒄𝒔𝒄𝜽 =𝟏

𝒔𝒊𝒏𝜽 𝒔𝒆𝒄𝜽 =𝟏

𝒄𝒐𝒔𝜽 𝒄𝒐𝒕𝜽 =𝟏

𝒕𝒂𝒏𝜽

𝑥 𝑓(𝑥)0 und.56 2

856= 5

9

89~1.15

956= 5

8 1

>56= 85

9

89~1.15

?56 2

656= 𝜋 0A56 −2

C56= >5

9 − 8

9~ − 1.15

D56= 95

8 −1

7E56= ?5

9 − 8

9~ − 1.15

7756 −2

7856= 2𝜋 und.

Example5:Completethefollowingtableofvaluesforthefunction𝑓 𝑥 = sec(𝑥).Usethereciprocalidentitytofind𝑦-values.

PropertiesoftheSecantFunctionDomain: 𝑋 ∈ ℝ 𝑥 ≠ 𝑘𝜋

2𝑤ℎ𝑒𝑛𝑘𝑖𝑠𝑜𝑑𝑑} Range:{𝑌 ∈ ℝ}

Period:2𝜋radians Amplitude:none(nomaxormin)

𝑥 𝑓(𝑥)0 156 8

9~1.15

856= 5

9 2

956= 5

8 und.

>56= 85

9 −2

?56 − 8

9~ − 1.15

656= 𝜋 −1A56 − 8

9~ − 1.15

C56= >5

9 −2

D56= 95

8 und.

7E56= ?5

9 2

7756

89~1.15

7856= 2𝜋 1

Example6:Completethefollowingtableofvaluesforthefunction𝑓 𝑥 = tan(𝑥).Usethereciprocalidentitytofind𝑦-values.

PropertiesoftheSecantFunctionDomain: 𝑋 ∈ ℝ 𝑥 ≠ 𝑘𝜋}where𝑘 ∈ ℕ Range:{𝑌 ∈ ℝ}Period:𝜋radians Amplitude:none(nomaxormin)

𝑥 𝑓(𝑥)0 und.56 3~1.73

856= 5

9

79~0.58

956= 5

8 0

>56= 85

9 − 7

9~ − 0.58

?56 − 3~ − 1.73

656= 𝜋 und.A56 3~1.73

C56= >5

9

79~0.58

D56= 95

8 0

7E56= ?5

9 − 7

9~ − 0.58

7756 − 3~ − 1.73

7856= 2𝜋 und.

L4–5.3TransformationsofTrigFunctionsMHF4UJensenPart1:TransformationProperties𝑦 = 𝑎 sin 𝑘 𝑥 − 𝑑 + 𝑐 DesmosDemonstration

𝑎 𝑘 𝑑 𝑐

Verticalstretchorcompressionbyafactorof𝑎.Verticalreflectionif𝑎 < 0𝑎 = 𝑎𝑚𝑝𝑙𝑖𝑡𝑢𝑑𝑒

Horizontalstretchorcompressionbyafactorof67.Horizontalreflectionif𝑘 <0.2𝜋|𝑘|

= 𝑝𝑒𝑟𝑖𝑜𝑑

Phaseshift𝑑 > 0; 𝑠ℎ𝑖𝑓𝑡𝑟𝑖𝑔ℎ𝑡𝑑 < 0; 𝑠ℎ𝑖𝑓𝑡𝑙𝑒𝑓𝑡

Verticalshift𝑐 > 0; 𝑠ℎ𝑖𝑓𝑡𝑢𝑝𝑐 < 0; 𝑠ℎ𝑖𝑓𝑡𝑑𝑜𝑤𝑛

Example1:Forthefunction𝑦 = 3 sin 6

G𝜃 + 𝜋

I− 1,statethe…

Amplitude:

𝑎𝑚𝑝 = 𝑎 = 3

Period:

𝑝𝑒𝑟𝑖𝑜𝑑 =2𝜋|𝑘|

=2𝜋12

= 4𝜋

Phaseshift:

𝑑 = − 𝜋I;shiftleft𝜋

I

Verticalshift:

𝑐 = −1;shiftdown1

Max:

𝑚𝑎𝑥 = 𝑐 + 𝑎𝑚𝑝 = −1 + 3 = 2

Min:

𝑚𝑖𝑛 = 𝑐 − 𝑎𝑚𝑝 = −1 − 3 = −4

Part2:GivenEquationàGraphFunctionExample2:Graph𝑦 = 2 sin 2 𝑥 − 𝜋

I+ 1usingtransformations.Thenstatetheamplitudeandperiodof

thefunction. Amplitude:LMNOLPQ

G= IO O6

G= 2 Period:𝜋radians

𝑦 = sin 𝑥

𝒙 𝒚

0 0𝜋2 1

𝜋 03𝜋2 −1

2𝜋 0

𝑦 = 2 sin 2 𝑥 −𝜋3 + 1

𝒙𝟐 +

𝝅𝟑 𝟐𝒚 + 𝟏

𝜋3 =

2𝜋6 1

7𝜋12 =

3.5𝜋6 3

5𝜋6 1

13𝜋12 =

6.5𝜋6 -1

4𝜋3 =

8𝜋6 1

Part3:GiventheGraphàWritetheEquation𝑦 = 𝑎 sin 𝑘 𝑥 − 𝑑 + 𝑐

𝑎 𝑘 𝑑 𝑐

Findtheamplitudeofthefunction:

𝑎 =𝑚𝑎𝑥 −𝑚𝑖𝑛

2

Findtheperiod(inradians)ofthefunctionusingastartingpointandendingpointofafullcycle.

𝑘 =2𝜋

𝑝𝑒𝑟𝑖𝑜𝑑

for𝐬𝐢𝐧 𝒙:tracealongthecenterlineandfindthedistancebetweenthe𝑦-axisandthebottomleftoftheclosestrisingmidline.for𝐜𝐨𝐬 𝒙:thedistancebetweenthe𝑦-axisandtheclosestmaximumpoint

𝑑bPQ = 𝑑cdb −𝜋2𝑘

𝑑cdb = 𝑑bPQ +𝜋2𝑘

Findtheverticalshift𝑐 = 𝑚𝑎𝑥 − 𝑎𝑚𝑝𝑙𝑖𝑡𝑢𝑑𝑒

OR

𝑐 =𝑚𝑎𝑥 +𝑚𝑖𝑛

2 (thisfindsthe‘middle’ofthefunction)

Example3:Determinetheequationofasineandcosinefunctionthatdescribesthefollowinggraph

Example4:DeterminetheequationofasineandcosinefunctionthatdescribesthefollowinggraphExample5:

a)Createasinefunctionwithanamplitudeof7,aperiodof𝜋,aphaseshiftof𝜋eright,andavertical

displacementof-3.𝑎 = 7

𝑘 =2𝜋

𝑝𝑒𝑟𝑖𝑜𝑑 =2𝜋𝜋= 2

𝑐 = −3𝑑 =

𝜋4

b)Whatwouldbetheequationofacosinefunctionthatrepresentsthesamegraphasthesinefunctionabove?

𝑑cdb = 𝑑bPQ +𝜋2𝑘

=𝜋4 +

𝜋2(2)

=𝜋4 +

𝜋4 =

2𝜋4 =

𝜋2

𝑦 = 7 sin h2 i𝑥 −𝜋4jk − 3

𝑦 = 7 cos h2 i𝑥 −𝜋2jk − 3

Part4:EvenandOddFunctions

EvenFunctions OddFunctionsEVENFUNCTIONif:

Linesymmetryoverthe𝑦-axis

ODDFUNCTIONif:

Pointsymmetryabouttheorigin(0,0)

Rule:𝑓 −𝑥 = 𝑓(𝑥)

Rule:−𝑓 𝑥 = 𝑓(−𝑥)

Example:

𝑦 = cos 𝑥

𝑓 𝜋 = −1

𝑓 −𝜋 = −1Therefore,

𝑓 𝜋 = 𝑓 −𝜋

Example:

𝑦 = sin 𝑥

𝑓𝜋2 = 1

𝑓 −

𝜋2 = −1

Therefore,

−𝑓𝜋2 = 𝑓 −

𝜋2

𝑦 = tan 𝑥isalsoanoddfunction

L5–5.3TrigApplicationsMHF4UJensenExample1:AFerriswheelhasadiameterof15mandis6mabovegroundlevelatitslowestpoint.Ittakestherider30sfromtheirminimumheightabovethegroundtoreachthemaximumheightoftheferriswheel.Assumetheriderstartstherideattheminpoint.a)Modeltheverticaldisplacementoftheridervs.timeusingasinefunction.

𝑎 =𝑚𝑎𝑥 −𝑚𝑖𝑛

2 =21 − 62 = 7.5

𝑘 = /𝜋

123456= /𝜋

78= 𝜋

98

𝑐 = 𝑚𝑎𝑥 − 𝑎 = 21 − 7.5 = 13.5𝑑=4> = 15ℎ(𝑡) = 7.5 sin

𝜋30 𝑡 − 15 + 13.5

b)Sketchagraphofthisfunction

Example2:Mr.Ponsenhastheheatingsystem,inthisroom,turnonwhentheroomreachesaminof66oF,itheatstheroomtoamaximumtemperatureof76oFandthenturnsoffuntilitreturnstotheminimumtemperatureof66oF.Thiscyclerepeatsevery4hours.Mr.Ponsenensuresthatthisroomisatitsmaxtemperatureat9am.a)WriteacosinefunctionthatgivesthetemperatureatKing's,T,inoF,asafunctionofhhoursaftermidnight.

𝑎 =𝑚𝑎𝑥 −𝑚𝑖𝑛

2 =76 − 66

2 = 5𝑘 = /𝜋

123456= /𝜋

H= 𝜋

/

𝑐 = 𝑚𝑎𝑥 − 𝑎 = 76 − 5 = 71𝑑I5= = 9𝑇(ℎ) = 5 cos

𝜋2 ℎ − 9 + 71

b)Sketchagraphofthefunction.

Example3:ThetidesatCapeCapstan,NewBrunswick,changethedepthofthewaterintheharbor.OnonedayinOctober,thetideshaveahighpointofapproximately10metersat2pmandalowpointof1.2metersat8:15pm.Aparticularsailboathasadraftof2meters.Thismeansitcanonlymoveinwaterthatisatleast2metersdeep.Thecaptainofthesailboatplanstoexittheharborat6:30pm.Assuming𝑡 = 0isnoon,findtheheightofthetideat6:30pm.Isitsafeforthecaptaintoexittheharboratthistime?

𝑎 =𝑚𝑎𝑥 −𝑚𝑖𝑛

2 =10 − 1.2

2 = 4.4𝑘 = /𝜋

123456= /𝜋

O/.P= 4𝜋

/P

𝑐 = 𝑚𝑎𝑥 − 𝑎 = 10 − 4.4 = 5.6𝑑I5= = 2

ℎ(𝑡) = 4.4 cos4𝜋25 𝑡 − 2 + 5.6

ℎ(6.5) = 4.4 cos4𝜋25 6.5 − 2 + 5.6

ℎ(6.5) = 4.4 cos4𝜋25 4.5 + 5.6

ℎ(6.5) = 4.4 cos 18𝜋

/P+ 5.6

ℎ(6.5) ≅ 2.8metersSincethedepthofthewaterisgreaterthan2meters,thesailboatcansafelyexittheharbor.

Note:Horizontaldistancebetweenmaxandminpointsrepresenthalfacycle.Sincemaxandmintideare6.25hoursapart,onecyclemustbe12.5hours.