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Western Cape Education DepartmentTelematicsLearning Resource 2019MATHEMATICSGrade 12

Telematics Mathematics Grade 12 Resources2February to October 2019Dear Grade 12 LearnerIn 2019 there will be 8 Telematics sessions on grade 12 content and 6 Telematics sessions on grade11 content. In grade 12 in the June, September and end of year examination the grade 11 contentwill be assessed. It is thus important that you compile a study timetable which will consider therevision of the grade 11 content. The program in this book reflects the dates and times for all grade12 and grade 11 sessions. It is highly recommended that you attend both the grade 12 and 11Telematics sessions, this will support you with the revision of grade 11 work. This workbookhowever will only have the material for the grade 12 Telematics sessions. The grade 11 materialyou will be able to download from the Telematics website. Please make sure that you bring thisworkbook along to each and every Telematics session.In the grade 12 examination Trigonometry will be 50 marks and the Geometry 40 marks of the150 marks of Paper 2.Your teacher should indicate to you exactly which theorems you have to study for examinationpurposes. There are altogether 6 proofs of theorems you must know because it could be examined.These theorems are also marked with (**) in this Telematics workbook, 4 are grade 11 theoremsand 2 are grade 12 theorems. At school you should receive a book called “Grade 12 Tips forSuccess”. In it you will have a breakdown of the weighting of the various Topics in Mathematics.Ensure that you download a QR reader, this will enable you the scan the various QR codes.At the start of each lesson, the presenters will provide you with a summary of the importantconcepts and together with you will work though the activities. You are encouraged to comeprepared, have a pen and enough paper (ideally a hard cover exercise book) and your scientificcalculator with you.You are also encouraged to participate fully in each lesson by asking questions and working out theexercises, and where you are asked to do so, sms or e-mail your answers to the studio.Remember:” Success is not an event, it is the result of regular and consistent hard work”.GOODLUCK, Wishing you all the success you deserve!

Telematics Mathematics Grade 12 Resources3February to October 20192019 Mathematics Telematics ProgramDayDateTimeGradeSubjectTopicTerm 1: 9 Jan – 15 MarchTuesday12 February15:00 – 16:0012MathematicsTrigonometry RevisionWednesday13 February15:00 – 16:0012WiskundeTrigonometrieHersieningTERM 2: 2 April to 14 JuneMonday8 April15:00 – 16:0012MathematicsTrigonometryTuesday9 April15:00 – 16:0012WiskundeTrigonometrieWednesday15 May15:00 – 16:0011MathematicsGeometryThursday16 May15:00 – 16:0011WiskundeMeetkundeWednesday22 May15:00 – 16:0012MathematicsGeometryThursday23 May15:00 – 16:0012WiskundeMeetkundeTerm 3: 9 July – 20 SeptemberMonday29 July15:00 – 16:0012MathematicsDifferential CalculusTuesday30 July15:00 – 16:0012WiskundeDifferentiaalrekeningWednesday07 August15:00 – 16:0011MathematicsFunctionsMonday12 August15:00 – 16:0011WiskundeFunksiesTerm 4: 1 October – 4 DecemberTuesday15 October15:00 – 16:0011MathematicsPaper 1 RevisionWednesday16 October15:00 – 16:0011WiskundePaper 2 Revision

Telematics Mathematics Grade 12 Resources4February to October 2019Session 1: TrigonometryxDefinitions of trigonometric ratios:oIn a right-angled 'Sin eoppositSin TyrCosTxrTanTyxTadjacentoppositeadjacentTanTxo On a Cartesian Plane0 , 90 , 180 , 270 , 360 can beyx30 , 45 and 60 can beobtained from the following unit circleT.obtained from the following90qyr, the radius is1 since it is aunit circle(0 ; 1) 180q (-1 ; 0) (1 ; 0) xTTT270q3cos 45q 1T Tan is ve in the3rdquadrantSineSAllTan Cos180q TbecomesT 60q32sin 60q 2cos 60q 1 2tan 45q 13T180q- TS Sine is ve in the2nd quadrant1T 45qsin 45q 12145qT 30qsin 30q 1 2tan 30q 145q2360q1cos 30q The “CAST” rule enables you to obtain thesign of the trigonometric ratios in any of thefour quadrants.yThe trigonometric function of angles(180q T) or (360q T) or (-T)30q20q360q (0 ; -1)xrTxSpecial AnglesoTytan 60q 23A - ALL trigratios are ve inthe first quadrantxC Cos is ve inthe 4nd quadrant360q-T Trigonometric function of TThe sign is determined bythe “CAST” rule.(180q T )(180q T )(360q T )(360q T )( T )sin(180q T ) sin Tsin(180q T ) sin Tsin(360q T ) sin Tsin(360q T ) sin Tsin( T ) sin Tcos(180q T ) cos Tcos(180q T ) cos Tcos(360q T ) cos Tcos(360q T ) cos Tcos( T ) cos Ttan(180q T ) tan Ttan(180q T ) tan Ttan(360q T ) tan Ttan(360q T )tan( T ) tan Ttan T

Telematics Mathematics Grade 12 ResourcesxFebruary to October 2019TRIGONOMETRIC IDENTITIEStan Tx5sin TcosT(cosT z 0)sin 2 T cos 2 TCo-functions or Co-ratiossin(90q T )cosTcos(90q T )sin Tsin 2 T1,1 cos 2 T ,rTxDetermine theReference angleEstablish inwhich twoquadrants θ is.Calculate θ inthe interval[0q; 360q]Write down thegeneral solution2.3.4.sin(90q T ) cos Tcos(90q T ) sin TxTrigonometric Equationssin T1.90q-T ycos 2 T 1 sin 2 Tcos T0,707tan T 0,866 1 1Reference sin (0,707) 45qReference cos (0,866) 30qReference tan 1 (1) 45q? θ 45qorθ 180q - 45q? θ 180q- 30q or θ 180q 30q? θ 180q - 45q? θ 45qorθ 135q? θ 150q orθ 210q? θ r150q? θ r150 k360º where k ? θ 135q? θ 45q k360º orθ 135q k360º where k 1? θ 135q k180º & k TRIGONOMETRIC GRAPHSSine FunctionCosine FunctionTangent FunctionEquationShapea 0a 0aAmplitudePeriodaNote:“c” refers to theside of the triangleopposite to angleC that is the sideSOLUTIONS OF TRIANGLESxArea RuleArea of 'ABC x111absin C acsin B bcsin A222Sine Rulesin A sin Bsin C abcxOrabc sin A sin Bsin CACosine Rulea2b2 c 2 2bc cos Ab c a2bc2orcos A2bc2BaC

Telematics Mathematics Grade 12 Resources6February to October 2019TRIGONOMETRY SUMMARYQuestiontypeSummary ofprocedure1. Calculatethe value of atrigexpressionwithout usinga calculator.Establish whether youneed a rough sketch orspecial triangles, ASTCrules or compoundangles.Example question3, D [0q; 270q] and E [0q; 180q] .4It is given that sin(D E ) sin D. cos E sin E . cos DDetermine, without using a calculator,a) sin Db) sin(D E ) .cos( 210 ). sin 2 405 . cos14 1.2 Calculate: a)tan 120 . sin 104 1.1 If 13 cos D5 and tan E b) sin 70q cos 40q cos 70q sin 40q2. If a trigratio is givenas a variableexpressanother trigratio in termsof the samevariable.3. Simplify atrigonometrical expression.4. Prove agiven identity.5. Solve atrig equation.Draw a rough sketch2. If sin 27q q , express each of the following in terms of q.with given angle anda) sin 117qb) cos( 27q)label 2 of the sides. The3rd side can then bedetermined usingPythagoras. Express eachof the angles in questionin terms of the angle inthe rough sketch.Use the ASTC rule to3. Simplify:simplify the givencos ( 720q x) . sin ( 360q x) . tan ( x 180q)a)expression if possible.sin ( x) . cos (90q x)See if any of thesin ( 90q x) . tan ( 360q x)identities can be used tob)sin (180q x) . cos (90q x) cos(540q x). cos( x)simplify it, if not see if itcan be factorized. Checksin 2 x cos x cos 3 xagain if any identity canc)cos xbe used. This includes2using the compound andsin x cos xd)double angle identities.1 cos 2 xSimplify the one side ofProve thatthe equation usingtan x . cos 3 x1sin xa)reduction formulae and221 sin x co s x 2identities until .b)Find the reference angleSolve for x [ 180q; 360q]by ignoring the “-“signand findinga) sin x 0,435b) cos 2 x 0,435sin 1 (0,435)1Write down the twoc) tan x 1 0,435solutions in the interval2x [0q; 360q] . Thenwrite down the generalsolution of this eq. Fromthe general solution youcan determine thesolution for the specifiedinterval by using variousvalues of k.

Telematics Mathematics Grade 12 Resources7February to October 2019Question typeSummary of procedureExample question6. Sketch a trig graph.1st sketch the trig graph without thevertical or horizontal transformation.Then shift the graph in this case 1 unitup.Sketch7.Find the area of atriangle.8. Finding an unknownside or angle in atriangle.b)y2 cos 3 x 1 for x [ 90q; 120q]c)y sin( x 60)for x [ 240q; 120q]'ABC, with B 104,5q , ABIf it is a right-angled triangle then1areabase u height , otherwise use2the area rule1Area of 'ABC ab sin C2Draw a rough sketch with the giveninformation. If it is not a right-angledtriangle you will use either the sine orcosine rule.BC6cm and9cm . Calculate, correct to one decimal placearea 'ABCa)b)'ABC, with B 104,5q , AB 6cm andBC 9cm . Calculate the length of AC.'ABC, with C 43,2q , AB 4,5cm andBC 5,7cm . Calculate the size of A .SKETCHING TRIG GRAPHCalculatethe periodWrite downtheamplitude ifit is a sine orcosinegraph.Identify the shape of the graph and drawa sine, cosine or tan graph withdetermined period and amplitude. Labelthe other x-intercepts. Repeat this patternover the specified domain.SKETCHNow do the vertical or horizontaltransformation if required.2 cos 3 x 1 for x [ 90q; 120q]yyyPeriod Amplitude 2360q120q32312x-90-60-3030-1-260190x-60-3030-16090

Telematics Mathematics Grade 12 Resources8February to October 2019QUESTION .1In the figure below, the point P(–5 ; b) is plotted on the Cartesian plane.OP 13 units and RÔP D .yP(–5 ; b)x 13xROxWithout using a calculator, determine the value of the following: .2 .1.1cos D(1) .1.2tan(180q D )(3)Consider:sin(T 360q) sin(90q T ) tan( T )cos(90q T )sin(T 360q) sin(90q T ) tan( T )to a single trigonometric ratio.cos(90q T ) .2.1Simplify .2.2Hence, or otherwise, without using a calculator, solve for T if0q d T d 360q :sin(T 360q) sin(90q T ) tan( T )cos(90q T ) .3 .484 2sin A 1 cos A0,54.1 cos A .3.1Prove that .3.2For which value(s) of A in the interval 0q d A d 360q is the identity inQUESTION 5.3.1 undefined?Determine the general solution of 8 cos 2 x 2 cos x 1 0 .(5)(3)(5)(3)(6)[26]

Telematics Mathematics Grade 12 Resources9February to October 2019QUESTION In the diagram below, the graphs of f (x) cos(x p) and g(x) q sin x are shown for theinterval 180q d x d 180q .yyg1Af0,5xx-180 -135 -90q-45q0q45q90q180q135q- 0,5B-1 .1Determine the values of p and q.(2) .2The graphs intersect at A(–22,5 ; 0,38) and B. Determine the coordinates of B.(2) .3Determine the value(s) of x in the intervalf ( x) g ( x) 0 . .4 180q d x d 180qfor which(2)The graph f is shifted 30 to the left to obtain a new graph h. .4.1Write down the equation of h in its simplest form. .4.2Write down the value of x for which h has a minimum in the interval 180q d x d 180q .(2)(1)[9]

Telematics Mathematics Grade 12 Resources10February to October 2019QUESTION 1 2sin AaProve that in any acute-angled sin C.c(5) P̂ 132q, PQ 27,2 cm and QR 73,2 cm.P132 27,2 cmRQ .373,2 cm .2.1Calculate the size of R̂ .(3) .2.2Calculate the area of .(3)In the figure below, SP̂Q a , PQ̂S b and PQ h. PQ and SR are perpendicularto RQ.PaShbRQ .3.1Determine the distance SQ in terms of a, b and h. .3.2Hence show that RSh sin a cos b.sin( a b)(3)(3)[17]

Telematics Mathematics Grade 12 Resources11February to October 2019Session 2: TRIGONOMETRY(rr 50/150 Marks)Compound and Double AnglesIn order to master this section it is best to learn the identities given below. These identities will also be givenon the formulae sheet in the Examination paper.xxCompound Angle Identities:(a) cos( A B)cos A cos B sin Asin Bcos( A B)cos A cos B sin Asin B(b) sin( A B)sin A cos B sin B cos Asin( A B)sin A cos B sin B cos AWhen two anglesare added orsubtracted to forma new angle, then acompound or adouble angle isformed.Double Angle Identities(c) sin 2 A2 sin A cos A(d) cos 2 Acos2 A sin 2 AReferred toas doubleangleformulae 1 2 sin 2 A 2 cos2 A 1What should you ensure you can do at the end of this section for examination purposes:A. Accepting the Compound Angle formulae cos( A B)cos A cos B sin Asin B use it to deriveThe following formulae:cos( A B)cos A cos B sin Asin Bsin( A B)sin A cos B sin B cos Asin( A B)sin A cos B sin B cos Acos 2 Acos2 A sin 2 ACo-functions or Coratiossin(90q T )cosTNegative Anglescos 2 A 1 2 sin A2sin( T )cos 2 A 2 cos2 A 1 sin Tcos( T ) cosTtan( T ) tan Tsin 2 A 2 sin A cos AYou must rememberB. Use compound angle and double angle identities to:1. Evaluate an expression without using a calculatorsin2 T cos2 T1sin2 T1 cos2 Tcos2 T1 sin2 T2. Simplifying trigonometric expressions3. Prove identities4. Solve trigonometric equations (both specific and general solutions)

Telematics Mathematics Grade 12 Resources12February to October 2019The sketches below gives a visual of compound and double angles.Sketch 1Sketch 3Sketch 2Sketch 1: The compound angle AB̂C is equal to the sum of D and β. eg. 75q45q 30qSketch 2: The compound angle EĜH is equal to the difference between D and β. eg. 15q 60q 45q or15q 45q 30qSketch 3: The double angle PT̂R is equal to the sum of D and D. eg. 45q 22.5q 22.5qGiven any special angles D and β, we can find the values of the sine and cosine ratios of the anglesα β , α β and 2α .Are you clear on the difference betweena compound and double angle?Please note:0q ; 30q ; 45q ; 60q and 90q are special angles, you are able to evaluate any trigonometric function ofthese angles without using a calculator.Exercises: Do not use a calculator.A. Derive each of the compound and double angle formulae in the box on the previous page.B.1.1.1 Evaluate each of the following without using a calculator.b) cos15qc) cos105qd) sin 165qa) sin 75qe) sin 36q.cos 54q cos 36q sin 54qf) cos 42q. cos18q sin 42q sin18qg) sin 85q. sin 25q cos 85q cos 25qh) sin 70q. cos 40q cos 70q sin 40q2 sin 40q. cos 40qi) 2 sin 30q. cos 30qj)cos10q21.2 If sin α, tan β2 and D and β are acute angles determine the value of sin(α β ) .31.3 Iftan A23and90q A 360q , determine without using a calculator2.Simplify the following expression to a single trigonometric function:3.Prove that4 cos( x). cos(90q x)sin(30q x). cos x cos(30q x). sin xa) cos 75q2 ( 3 1)4b) cos(90q 2 x). tan(180q x) sin 2 (360q x)c) (tan x 1)(sin 2 x 2 cos 2 x)4.cos 2 A .3 sin 2 x2(1 2 sin x cos x)Determine the general solution for x in the following:a) sin 2x. cos10q cos 2x. sin10q cos 3xb)cos2 x3 sin 2 xc)2 sin xsin( x 30q)Scan the QR code for revision from examinationpapers on this section with solutions.

Telematics Mathematics Grade 12 Resources13February to October 2019ADDITIONAL QUESTIONS1.8Given sin α; where 900 d α d 270q17With the aid of a sketch and without the use of a calculator, calculate:a) tan αb) sin(90q α)c) cos 2α2.a) Using the expansions for sin( A B) and cos( A B) , prove the identity of:sin( A B)cos( A B)tan A tan B1 tan A. tan B(3)sin( A B), prove in any ΔABC thatcos( A B)tan A. tan B. tan C tan A tan B tan Cb) If tan( A B)3.4.If sin 36q cos12qa) sin 48qb) sin 24qc) cos 24qp and6.cos 36q sin12q q ,determine in terms of p and q :32and 80q 60q 20qShow that sin 2 20q sin 2 40q sin 2 80q(7)Given: f ( x) 1 sin x and g ( x) cos 2 xCalculate the points of intersection of the graphs f and g for x [180q ; 360q]Given that sin T1, calculate the numerical value of sin 3T , WITHOUT using a3calculator.7.(4)(3)(3)(3)(HINT: 40q 60q 20q5.(3 2 3)(7)(5)Prove that, for any angle A:4 sin A cos A cos 2 A sin 15qsin 2 A(tan 225q 2 sin 2 A)8.Solve for x if 2 cos x9.If cos β6 22(6)tan 2 x and x [ 90q ; 90q] . Show ALL working details.(8)p; where p 0 and β [00 ; 900 ] , determine, using a diagram, an5expression in terms of p for:a) tan βb)cos 2 β10.1 If sin 28 a and cos 32 b, determine the following in terms of a and/or b :a) cos 28qb) cos 64qc) sin 4 10.2 Prove without the use of a calculator, that if sin 28 a and cos 32 b, thenb 1 a2 a 1 b21.2(4)(3)(2 3 4)(4)

Telematics Mathematics Grade 12 Resources14February to October 2019Revision: Grade 11 Geometry Theorems and ConversesThe proofs of the theorems marked with (**) must be studied because it could be examined. The part in boldin bracket is the abbreviation for the theorem, which we use as reasons when writing up geometry solutions.123Theorem**The line drawn from the centre of a circle perpendicular to a chord bisects the chord;(line from centre ٣ to chord)ConverseThe line from the centre of a circle to the midpoint of a chord is perpendicular to the chord.(line from centre to midpt of chord)The perpendicular bisector of a chord passes through the centre ofthe circle; (perp bisector of chord)Theorem** The angle subtended by an arc at the centre of a circle is double the size of the anglesubtended by the same arc at the circle (on the same side of the chord as the centre);( ס at centre 2 ס at circumference)Corollary1. Angle in a semi-circle is 900 ( ס s in semi circle)2. Angles subtended by a chord of the circle, on the same side of the chord, are equal( ס s in the same seg)3. Equal chords subtend equal angles at the circumference (equal chords; equal ס s)4. Equal chords subtend equal angles at the centre (equal chords; equal ס s)5. Equal chords in equal circles subtend equal angles at the circumference of the circles.(equal circles; equal chords; equal ס s)Corollary1. If the angle subtended by a chord at the circumference of the circleis 90q, then the chord is a diameter. (converse ס s in semi circle)Converse2. If a line segment joining two points subtends equal angles at two points on the same sideof the line segment, then the four points are concyclic.Theorem** The opposite angles of a cyclic quadrilateral are supplementary; (opp ס s of cyclic verse5Theorem6Theorem**ConverseIf the opposite angles of a quadrilateral are supplementary then the quadrilateral is a cyclicquadrilateral. (opp ס s quad sup OR converse opp ס s of cyclic quad)The exterior angle of a cyclic quadrilateral is equal to the interior opposite angle of thequadrilateral. (ext ס of cyclic quad)If the exterior angle of a quadrilateral is equal to the interior opposite angle of thequadrilateral, then the quadrilateral is cyclic.(ext ס int opp ס OR converse ext ס of cyclic quad)The tangent to a circle is perpendicular to the radius/diameter of thecircle at the point of contact.(tan ٣ radius)If a line is drawn perpendicular to a radius/diameter at the point where the radius/diametermeets the circle, then the line is a tangent to the circle. (line ٣ radius)Two tangents drawn to a circle from the same point outside the circle are equal in length.(Tans from common pt OR Tans from same pt)The angle between the tangent to a circle and the chord drawn from the point of contact isequal to the angle in the alternate segment. (tan chord theorem)If a line is drawn through the end-point of a chord, making with the chord an angle equal toan angle in the alternate segment, then the line is a tangent to the circle.(converse tan chord theorem OR ס between line and chord)Scan the QR code for grade 11geometry revision with solutions.

Telematics Mathematics Grade 12 Resources15February to October 2019Session 3 Grade 12 GeometryThe Grade 11 geometry entails the circle geometry theorems dealing with angles in a circle, cyclicquadrilaterals and tangents. The Grade 12 geometry is based on ratio and proportion as wellas similar triangles. Grade 11 geometry is especially important in order to do the grade 12Geometry hence this work must be thoroughly understood and regularly practiced to acquire thenecessary skills. The grade 11 geometry is summarized on the previous page.Below are Grade 12 Theorems, Converse Theorems and their Corollaries which you must know. Theproofs of the theorems marked with (**) must be studied because i

In 2019 there will be 8 Telematics sessions on grade 12 content and 6 Telematics sessions on grade 11 content. In grade 12 in the June, September and end of year examination the grade 11 content ... 16:00 11 Wiskunde Paper 2 Revision Telematics Mathematics Grade 12 Resources 3 February to October 2019. ... Question type Summary of procedure ...