Gr12 Mathematics JIT Term1 TeacherMaterial 2020

KZN DEPARTMENT OF EDUCATION MATHEMATICS JUST IN TIME MATERIAL GRADE 12 TERM 1 – 2020 TABLE OF CONTENTS TOPIC PAGE NO. EUCLIDEAN GEOMETRY 2 – 19 TRIGONOMETRY 20 – 33 This document has been compiled by the FET Mathematics Subject Advisors together with Lead Teachers. It seeks to unpack the contents and to give more guidance to teachers.

EUCLIDEAN GEOMETRY Weighting 50±3 Related concepts Axioms, Theorems, Corollaries, Riders, Converse Prior knowledge  Ratio  Lines – Parallel, Perpendicular, Transversal  Triangles – Types, congruency, exterior angle, Midpoint theorem, Pythagoras, Area  Similar figures  Properties of quadrilaterals  Circle geometry – centre, radius, chord, segments, tangents, quads, arc, sector Resources Calculator, set of instruments, tracing paper, coloured pencils/pens Teacher activity Let leaners explore with the aid of investigations using trace papers and paper folding Learner activity Application worksheet(s), explore theorem proofs and have knowledge of converses Methodology  Revise Euclidean Geometry from GET – 1 day  Revise grade 10 Euclidean Geometry – 2 days  Take note of acceptable reasons when writing proofs  Do overview of important theorems and corollaries  Important cases in in grade 12 similarity  How to go about solving a rider Misconceptions/ Errors  Assuming information that is not given  Incorrect naming of triangles in similarity  Ratio is usually confused with measurement in similarity and proportional intercept theorem.  Confusing “Required To Prove (RTP) and Given information.”  Confusing original reasons with converse reasons when giving concluding statements for proofs. FROM THE ATP: DATES CURRICULUM STATEMENT 1. Revise earlier work on the necessary and sufficient conditions for polygons to be similar. 2. Prove (accepting results established in earlier grades): 29/1 – 05/2 that a line drawn parallel to one side of a triangle divides the other two sides proportionally (6 days) (and the Midpoint Theorem as a special case of this theorem); Solve proportionality problems and prove riders. 3. Prove (accepting results established in earlier grades): 3.1 that equiangular triangles are similar; 6/2 – 14/2 3.2 that triangles with sides in proportion are similar; and (7 days) 3.3 the Pythagorean Theorem by similar triangles. Solve similarity problems and prove riders. 2

1. RATIO AND PROPORTION 1.1 Ratio  A comparison of two quantities with the same units.  A ratio is a fraction, and a fraction is a ratio. Example: There are 5 pupils: 2 are boys and 3 are girls 𝟐 The ratio of boys to girls is 2:3 or 𝟑 𝟑 The ratio of girls to boys is 3:2 or 𝟐 𝟐 The ratio of boys to all pupils is 2:5 or 𝟓 𝟑 The ratio of girls to all pupils is 3:5 or 𝟓  It is important to note that ratios do not necessarily represent the lengths of sides. This is the reason for adding a variable such as 𝒂, 𝒃, 𝒙, 𝒑, 𝒌 … Exercise 1: Consider the line segment PS. If PS = 60 cm and R divides PS in the ratio 6:4 then PR: RS = 6: 4 P 6 parts (6k) R 4 parts (4k) S a) Determine the value of k. b) Hence, or otherwise, determine the lengths of PR and RS. PR RS c) Determine the ratio of: (i) (ii) PS PS Exercise 2: D and E are points on AB and AC of ∆ABC so that DE BC. AB = 96 cm and AD:DB = 5:3. It is further given that AC = 128 cm. A D E C B a) Determine the length of AD b) Determine the length of DB c) Determine the ratio of AE:EC d) Determine the length of AE e) Determine the length of EC 3

1.2 Proportion Proportion is when two ratios (or fractions) are equal. Example: Sizes When shapes are "in proportion" their relative sizes are the same. Here we see that the ratios of head length to body length are the same in both drawings. So, they are proportional. Making the head too long or short would look bad! 1.3 Proportionality Theorem Prior Knowledge for proof of proportionality theorem  Parallel Lines, corresponding angles  Triangles with equal or common bases, lying between the same parallel lines are of equal area or have the same area. Area of triangles 1. In the diagram below, △ABC and △DEF have the same height (h) since both triangles are between the same parallel lines. Triangles with equal heights have areas which are proportional to their bases. 2. △WXY and △ZXY have the same base (XY) and the same height (h) since both triangles lie between the same parallel lines. Triangles with equal bases and between the same parallel lines are equal in area. 4

3. △PQR and △SQR have the same base (QR) and are equal in area. Triangles on the same side of the same base and equal in area, lie between parallel lines. 1.4 Formal proof of the theorem (Required for exam purposes) A line drawn parallel to one side of a triangle divides the other two sides proportionally. . In ∆APQ, draw height h1 relative to base AP; height h2 relative to AQ. Note: While the proof is complicated with three diagrams, you need only concentrate on the first simple diagram to do the proof, as long as you label the other two diagrams correctly, with P on AB/BA produced and Q on AC/CA produced. (When referring to ∆APQ etc. , we mean 𝐀𝐫𝐞𝐚∆APQ. ) Short acceptable reason: Line to one side of ∆ or preferably (Prop theorem, name parallel lines) Converse: If a line divides the two sides of a triangle proportionally, then the line is parallel to the third side. 5

1.5 Useful strategies in solving problems with proportion involving areas of triangles: Identify the type of problem using the following criteria: 1. If the two triangles in question have a common height 1 Use area formula: Area = 2 base × height and a common vertex. 2. If the two triangles in question have a common angle. 1 Use area rule: Area = 2 a b sin C 3. If none of the above. Identify a common triangle and relate the two triangles in question to it, then use any of the two methods mentioned above. Exercise 3: 3.1 ΔPQR has TS QP 3.2 NL PQ Write down all the different pairs of Write down all the different pairs of equal ratios. equal ratios. P L O S M R Q N T P 3.3 In ∆ PQR, ST QR, SU PR and TU PQ. P S T Q R U Complete the following proportions, and in each case, state the pair of parallel lines which make the proportions possible. PS TR (i) = = (ii) = = PQ PT 1.6 ACTIVITIES 1. In the accompanying figure MSQ R. Furthermore, PM = x cm, MQ = 2 cm, PS = (x+2) cm and SR =3 cm. 6

Determine, with reasons, the value of x. (3) 2. In ∆ABC, DE BC and AB = 6 cm. AD = x cm, AE = 5 cm and EC = 3 cm. Determine with reasons the value of x. (3) 3. In the accompanying figure, BF CG DH and DH EH. Furthermore, AB = 1 cm, BC = 3 cm and CD = 1cm. 3.1 Write down the values of the following ratios: AG (a) (1) GH AD (b) (1) DE 3.2 Determine the length of DE. (3) 4. In the accompanying figure, N is a point on PQ and M is a point on PR of ∆PQR such that NM QR. T is a point on QR such that NT PR. QT:TR = 3:2 7

PQ = 30 units and PM = 16 units Calculate, stating reasons, the lengths of: 4.1 PN (5) 4.2 MR (5) 5. In the diagram, E, H, D and G are points on the sides of ∆ABC. GH AC and ED BC. AD 5 BG 2 HG cuts ED at K. = and = DC 3 GC 1 If AB = 24 units, determine giving reasons the lengths of: 5.1 AE (3) 5.2 AH (3) 6. In the diagram below, ∆ AFC is drawn. B lies on AC with D and E on CF. BD AE and BE AF. CB:CA = 5:9 8

6.1 If the length of CA is 18 units, find the length of BA. (2) Area ∆BDC 6.2 Determine (3) Area ∆BED 6.3 If the length of CF is 45 units, find the length of DE. (4) 7. In the diagram below, two circles touch internally at A AB is the diameter of the larger circle and AL is the diameter of the smaller circle. S and L are the centres of the circles. D is a point on the smaller circle and C is a point on the larger circle ADC is a straight line. M is a point on LB so that MN LC 7.1 Prove that DL CB. (4) 7.2 Prove that 2SD = LC. (3) SL 7.3 Determine the value of . (2) AB SL 7 7.4 If AB = 30 units and = , then determine the length of LM. (3) AB 9 BQ 3 BR 1 8. In the figure AQ RT, = and = . QC 5 RA 2 9

8.1 If BT = k, calculate TQ in terms of k. (3) 8.2 Hence, or otherwise, calculate the numerical value of: CP 8.2.1 (3) PR Area ∆RCT 8.2.2 (4) Area ∆ABC 5 9. Given ∆ABC with PQ DC. BQ and DC intersect in E. DB = 9 AB and QC = 3 × AQ. Calculate: A P Q D E B C Area ∆QPA 9.1 (3) Area ∆QPB BE 9.2 (2) EQ 10. In ∆PQR , B lies on PR such that 2PR = BR. A lies on PQ such that PA:PQ = 3:8 BC is drawn parallel to AR. 10

Area ∆PRA 10.1 Write down the value of (2) Area ∆QRA BD 10.2 Calculate the value of the ratio . Show all your working to support your answer. (5) BQ 11. In the figure , PQRS is a quadrilateral. PQ = 24 units, QR = 8 units, RS = 12 units, SP = 32 units and QS = 16 units. P Q S R 11.1 Prove that QR PS (5) Area of triangle QRS 11.2 Determine the ratio (4) Area of quadrilateral PQRS 11.3 If PR and QS intersect at O, calculate the length of OQ. (5) 2. SIMILARITY  Similarity means two objects are of the same shape but not necessarily of the same size.  Congruency means that two figures or objects are identical in shape and size.  The order of naming triangles is important when working with similarity. 11

Note: If two triangles are congruent, then they are also similar, but two similar triangles are not necessarily congruent. 2.1 Triangles are special polygons  If two triangles are equiangular, then their sides will always be in the same proportion, so the triangles are similar.  Equiangular means the corresponding angles are equal.  If the sides of two triangles are in the same proportion, then the triangles will be equiangular, so the triangles are similar. Example 1: A 40 D 40 115 115 C F B E In ∆ABC and ∆DEF 1. ̂A=̂ D (Given) 2. ̂B=̂ E (Given) 3. Ĉ = ̂F (sum of  ‘s of a ∆) ∴ ∆ABC ∣∣∣ ∆DEF (  ) or (AAA) or (Equiangular ∆′s) Example 2: B E 10 cm 5 cm A C D F 3 cm 6 cm AC 6 2 AB 10 2 = = and = = DF 3 1 DE 5 1 BC 2 Provided is also in the ratio , then all three pairs of sides will be in proportion, and so the EF 1 triangles are similar. 2.2 Similarity Theorem If two triangles are equiangular then their corresponding sides are in proportion and the triangles are similar. Formal proof of the theorem (Required for exam purposes) 12

Converse: If the corresponding sides of two triangles are in proportion, then the triangles are similar. 2.3 How to prove two triangles similar There are three easy ways to prove similarity: (i) AAA(Angle-Angle-Angle) If two pairs of corresponding angles in a pair of triangles are equal, then the triangles are similar. We know this because if two angle pairs are the same, then the third pair must also be equal. 13

When using this method, structure your answer as in example 1 above. (ii) SSS (side-side-side) If the measures of corresponding sides are known, then their proportionality can be calculated. If all three pairs are in proportion, then the triangles are similar. (iii) SAS (Side-Angle-Side) If two pairs of corresponding sides are in proportion, and the included angle of each pair is equal, then the two triangles they form are similar. 2.4 Ways in which similarity can be asked 1. Prove that ∆ABC ∣∣∣ ∆DEF. AB BC 2. Prove that = . First prove: ∆ABC ∣∣∣ ∆DEF and then deduce the proportion of the sides. DE EF 3. Prove that: KN. PX = NR. YP. Find two triangles in which KN, PX, NR and YP (or sides equal to these), and thus prove that: ∆KNR ∣∣∣ ∆YPX , then deduce what you were asked to prove. 2.5 Useful strategies in locating the triangles when proving that they similar 1. Try taking letters from the top and bottom and see if you can locate similar triangles Use the diagram to check whether ABC and ABD form triangles If yes, it is ok. If not, you go for option 2. 2. Try taking letters from the left and right and see if you can locate similar triangles 14

then refer to the diagram to check whether ABC and ACD form triangles, if yes, it is ok if not you go for option 3. 3.1 Replace lengths with equal other lengths and then try to locate triangles. 3.2 Use information from previous parts of the question to assist you. 3.3 Look for other pairs of triangles which might be similar and have a bearing on what you are trying to prove. 2.6 ACTIVITIES 1. State with reason whether the following triangles are similar: (a) (b) 2. Given that the following triangles are similar, determine the values of 𝑥 and 𝑦. 3. ABCD is a cyclic quadrilateral with BC = CD. The tangent through C meets AB produced at S. 15

Let 𝐶̂1 = 𝑥. Prove that: 3.1 𝐶̂1 = 𝐴̂2 3.2 𝛥BCS|||ΔDAC 3.3 BC2 = DA.BS NSC Mar 2006 (SG) 4. PQ is a tangent to circle QRBAT. RT is produced to meet tangent QP at P. TA and RB are produced to meet at S. SP BA. Let 𝑆̂1 = 𝑥. 4.1 Names, with reasons, TWO other angles each equal to x. 4.2 Prove that 𝛥PTS|||ΔPSR. 4.3 4.3.1 Prove that 𝛥PQT|||ΔPRQ. 4.3.2 Hence show that PQ2 = PR.PT. 4.4 Hence show that PQ = PS. NSC Nov 2006 SG 5. FA ia a tangent to circle ABCD at A. AD and FBC produced meet at E. 5.1 Prove that 𝛥DEC|||ΔBEA. 5.2 Prove that 𝛥FAB|||ΔFCA. 5.3 Hence show that FA.CA= FC.AB. NSC Nov 2007 SG 16

6. BC and CAE are tangents to circle DAB. BDE is a straight line. BD= BA. 6.1 Prove that: 6.1.1 ̂2 = A D ̂2 + A ̂3 6.1.2 DA BC ED EA 6.2 Hence deduce that AB = AC. 6.3 Calculate the length of AB if it is Further given that EC:EA = 5: 2 and ED= 18 units. 6.4 Prove that ΔEDA|||ΔEAB. 6.5 Hence deduce that EA2 = ED.EB. 6.6 Name another triangle which is similar to ΔEDA. NSC May 2008 SG 7. DRS is a tangent to circle ARTM. Chords AT and MR intersect at P. ̂1 = T T ̂2 . 7.1 Prove, stating reasons, that: 7.1.1 ̂R3 = ̂ R4 7.1.2 ΔAPR|||ΔMPT 7.2 Hence complete: AP.PT =. .. NSC May 2009 SG 8. Circle KRST is drawn. Chords RK and ST produced meet at J. JT = RS. Prove that: 8.1 𝛥JRS|||ΔJTK 8.2 JR.TK = RS2 NSC May 2010 SG 17

9. Two circles, with centres at A and B, intersect at X and Y. The radius of the larger circle, with centre A, is R and the radius of the smaller circle, with centre B, is r. CX is a tangent to circle B at X and DX is a tangent to circle A at X. Prove that: 9.1 XY2 = DY.YC 9.2 𝐴̂1 = 𝐵̂1 9.3 𝛥CAY|||ΔYBX NSC Mar 2002 HG 10. EF, EG and EH are chords of the circle such that 𝐸̂1 = 𝐸̂2 . KG is a tangent to the circle at G. EH is produced to meet KG at K. HG and GF are drawn. Prove that: 10.1 GK2 = EK.HK EK GK2 10.2 EF = HG2 NSC Nov 2002 HG 11. ST is a tangent to circle PRT, diameter PT. K is a point on PT such that PK:KT = 1: 2. PR = √18 units, PQ = √2 units. 11.1 Prove that: 11.1.1 TR KQ 11.1.2 TKQS is a cyclic Quadrilateral 11.1.3 𝛥QRT|||ΔKTS 11.1.4 𝛥RTS|||ΔTPS 11.2 If PS = √32 units, calculate, with reasons and without the use of a calculator, the length of: 18

11.2.1 ST 11.2.2 KT NSC Nov 2005 HG 12. AD is a diameter of circle ABCD. AD is produced to meet tangent NCP at P. AV ⊥ NQ. 12.1 Prove that NQ CD. 12.2 Prove that ANCQ is a cyclic quadrilateral. 12.3 12.3.1 Prove that 𝛥PCD|||ΔPAC. 12.3.2 Complete: PC2 =. .. 12.4 Prove that BC2 = CD.NB. 12.5 If it is further given that PC= MC, BM2 AP.DP prove that 1 − BC2 = CD.NB NSC Nov 2006 HG 13. RP is the diameter of circle RSP with centre O. NP and NS are tangents to the circle at P and S. E is the midpoint of PS. Let 𝑆1 = 𝑥. 13.1 Find, with reasons, THREE other angles equal to x. 13.2 Prove that: 13.2.1 𝛥SEN|||ΔOPN. 13.2.2 PN2 = EN.ON 13.3 Prove that PE2 = OE.EN 13.4 Based on the fact that 𝛥PEO|||ΔPSR, determine the ratio OE:RS. PN2 2ON 13.5 Hence prove that = PE2 RS NSC Nov 2007 HG 19

TRIGONOMETRY Weighting 40 ±3 marks in the final NSC examination Related concepts Use of quotient identities Square identities Reduction formulae Special angles Negative angles Right angled triangle General solution Area rule, sine rule, cosine rule Trig graphs Prior knowledge Definition of trig ratios Pythagoras theorem Definition of quadrants Factorisation Angle of elevation Angle of depression Amplitude Period Resources Calculator Textbooks Activities Baseline assessment Classwork Methodology Revise grade 11 trigonometry Use the activities below to deduce that : cos  A  B   cos A-cosB but cos  A  B   cosAcosB+sinAsinB Guiding activity:  Given (i) ˆA=90 and ˆB=30 ˆ (ii) A=120 ˆ  and B=30 ˆ (iii) A=225 ˆ  and B=135 Use a calculator to calculate the following:  cos  A  B 20

 cos A  cos B  cos AcosB  sinAsinB What deduction can you make in 1 above?  cos  A  B  cosAcosB  sinAsinB will then be used to derive the other compound and double angles.  Then demonstrate with some examples the application of the identities from any relevant textbook.  Revise grade 11 solution of triangles  Cutting and fitting to recognize the visualisation of 3D figures  Application of the method o Analyse the given information o Look for the right angled triangle (Pythagoras theorem) o If there is no right angled triangle, then use cosine and or the sine rule o Break up diagram into its constituents parts o Look at the triangle that has more information than the others  Look for a common side Misconceptions/  sin x  tan x  Errors  cos x 2 tan   x  3/ y  2 3 cos 2 x  sin 2 x  1 sin tx  t sin x cos x cos  A±B   cos A  cos B FROM THE ATP: DATES TOPIC CURRICULUM STATEMENT Compound angle identities: 1. cos(𝛼 ± 𝛽) = 𝑐𝑜𝑠𝛼𝑐𝑜𝑠𝛽 ∓ 𝑠𝑖𝑛𝛼𝑠𝑖𝑛𝛽 17/2 – 03/3 TRIGONOMETRY: 2. sin(𝛼 ± 𝛽) = 𝑠𝑖𝑛𝛼𝑐𝑜𝑠𝛽 ± 𝑐𝑜𝑠𝛼𝑠𝑖𝑛𝛽 (12 days) COMPOUND ANGLES 3. 𝑠𝑖𝑛2𝛼 = 2𝑠𝑖𝑛𝛼𝑐𝑜𝑠𝛼 4. 𝑐𝑜𝑠2𝛼 = 𝑐𝑜𝑠 2 𝛼 − 𝑠𝑖𝑛2 𝛼 5. 𝑐𝑜𝑠2𝛼 = 2𝑐𝑜𝑠 2 𝛼 − 1 6. 𝑐𝑜𝑠2𝛼 = 1 − 2𝑠𝑖𝑛2 𝛼 04/3 – 10/3 TRIGONOMETRY: 2D/3D Solve problems in two and three dimensions. (5 days) 21

BASELINE ASSESSMENT: Revision of Grade 11 Trigonometry  Definitions and signs of ratios  Reduction formulae  Special angles  Identities  General solutions Definitions and signs of ratios 1. If 7sinθ + 4 = 0 and cosθ > 0 Find, without a calculator, the value of : 1.1 cosθ sinθ 1.2 cosθ 1.3 1 – 49cos2θ 1.4 cosθ. tanθ Reduction formulae cos(180°+θ).tan(−θ) 2. Simplify: sin(360°−θ).tan(720°+θ) cos(180°−β).cos(90°−β) 3. Prove that = −1 sin(90°+β).sin(−β−180°) Special angles 4. Simplify: sin210° − tan120°. cos330° Identities 2 𝑠𝑖𝑛 2 𝑥 𝑐𝑜𝑠 𝑥 5. Prove that:2 𝑡𝑎𝑛 𝑥−2 𝑠𝑖𝑛 𝑥 𝑐𝑜𝑠 𝑥 = 𝑠𝑖𝑛 𝑥 General solution 6. Determine the general solution of: 6.1 2cos2 ∝ −1 + cos ∝= 0 6.2 sin3θ = sin(2θ + 10°) 6.3. cos(θ + 30°) = sin2θ 22

COMPOUND ANGLES Given α=60o and β= 30o Evaluate: (a) cos (α – β) (b) cosα – cosβ (c) cosαcosβ + sinαsinβ Compare the answers of (a) and (b) Compare the answers of (a) and (c) Compare the answers of (b) and (c) What can you conclude? Prove that 𝐜𝐨𝐬( 𝐀 − 𝐁) = 𝐜𝐨𝐬 𝐀 . 𝐜𝐨𝐬 𝐁 + 𝐬𝐢𝐧 𝐀 . 𝐬𝐢𝐧 𝐁 PROOF Let the line PR = d 1 R(cosA;sinA) Construct a circle with centre 0 and a radius of 1. P(cosB;sinB) This is known as a unit circle. A B 0 From the distance formula d2 = (sin A − sin B)2 + (cos A − cos B)2 ………..d2 = (y2 − y1 )2 + (𝑥2 − 𝑥1 )2 = sin2 A − 2 sin A sin B + sin2 B + cos2 A − 2 cos A cos B + cos2 B = 1 + 1 − 2 sin A sin B − 2 cos A cos B = 2 − 2 sin A sin B − 2 cos A cos B = 2 − 2(sin A sin B + cos A cos B) …………..(1) 23

Using cosine rule d2 = OP2 + OR2 − 2(OP)(OR) cos( A − B) = 12 + 12 − 2(1)(1) cos( A − B) = 2 − 2 cos( A − B). . . . . . . . . . . . . . . . . . . . . (2) (2) = (1): 2 − 2 cos( a − B) = 2 − 2(sin A sin B + cos A cos B) −2 cos( A − B) = −2(sin A sin B + cos A cos B) cos( A − B) = sin A sin B + cos A cos B The angle (α – β) is called a compound angle Other compound angles done in grade 11 which we proved by using the reduction formulae are:  cos(-B) = cos B  sin (90 – A) = cos A  cos (90 – B) = sin B Using the compound angle formula cos(α – β) = cosα.cosβ + sinα.sinβ, show that: cos(α + β) = cosα.cosβ – sinαsinβ Hint: cos(α + β) = cos[(α – ( - β)] sin(α + β) = sinα.cosβ + cosαsinβ Hint: sin(α + β) = cos[90 o - (α + β)] sin(α – β) = sinα.cosβ – cosαsinβ compound angle formulae cos(α – β) = cosα.cosβ + sinα.sinβ cos(α + β) = cosα.cosβ – sinα.sinβ sin(α + β) = sinα.cosβ + cosα.sinβ sin(α – β) = sinα.cosβ – cosα.sinβ ACTIVITIES 1. Expand each of the following 1.1 sin(θ + 50°) 1.2 cos(3𝑥 − 40°) Solutions 1.1 sinθcos50° + cosθsin50° 1.2 cos3𝑥cos40° + sin3𝑥sin40° 2. Simply the following into a single trigonometric ratio 2.1 sin𝑥cosy − cos𝑥siny 2.2 cos70°cos10° − sin70°sin10° 24

Solutions 2.1 sin(𝑥 − y) 2.2 cos(70 + 10) =cos80° 3. Prove the following: 3.1 cos(180° + θ) = −cosθ √2−√6 3.2 cos105° = 4 Solutions 3.1 cos180°cosθ − sin180°sinθ LHS = cos105° =(−1)cosθ − (0)sinθ = cos(60° + 45°) =−cosθ = cos60°cos45° − sin60°sin45° 1 √2 √3 √2 =(2) ( 2 ) − ( 2 ) ( 2 ) √2 √6 =4 − 4 √2−√6 = = RHS 4 ACTIVITIES 1. Expand each of the following: 1.1 sin(𝑥 − 10°) 1.2 cos(∝ +70°) 1.3 sin(3θ + 2β) 1.4 sin(𝑥 − 4y) 2. Simplify to a single trigonometric ratio: 2.1 cos3θ. cosθ + sin3θsinθ 2.2 sin70°cos10° − cos80°cos70° 2.3 cos70°cos10° − cos20°cos80° 2.4 sin8ycos2y + cos8ysin2y 3. Evaluate (no calculator): 3.1 sin15° 3.2 sin195° 3.3 tan15° 4. Prove without the use of a calculator that: 4.1 cos78° + cos42° = cos18° 4.2 sin5𝑥 − 2cos3𝑥sin2𝑥 = cos(90° − 𝑥) 1 4.3 cos(45° + 𝑥) cos(45° − 𝑥) = 2 cos2𝑥 25

DOUBLE ANGLES Double angles involve trigonometric functions of double angles, i.e. sin 2A, cos 2A and tan2A sin 2A = sin(A + A) Use the compound angle formula to work out double angle formulae sin(A + A) = sinAcosA + cosAsinA = 2sinAcosA Therefore 𝐬𝐢𝐧𝟐𝐀 = 𝟐𝐬𝐢𝐧𝐀𝐜𝐨𝐬𝐀 Similarly cos2A = cos(A + A) = cosAcosA – sinAsinA = cos2 A – sin2 A sin2α = 2sinαcosα cos2α = cos2α – sin2α sin2α + cos2α = 1 = 2 cos2α – 1 = 1 – 2sin2α Since there are 3 formulae for cos2A it is important to know which formula to use to avoid complications. For example 1 - cos2A cos2A + 1 =1 - (1 - 2sin2A) = (2 cos2 A– 1) + 1 = 2sin2A = 2 cos2 A EDUCATORS’ ACTIVITIES 1. Expand the following using double angle formulae: 1.1 sin6θ 1.2 2cos22° 1.3 4sin44° Solutions 26

1.1 sin6θ = sin 2(3θ) = 2sin3θ. cos3θ 1.4 2cos22° = 2cos2(11°) 1.5 4sin44° = 4sin2(22°) = 2[cos211° − sin211°] = 4(2sin22°. cos22°) = 2 cos211° − 2sin211° = 8sin22°. cos22° OR = 2(1 – 2sin2 11°) = 2 – 4sin2 11° OR = 2(2cos 2 11 – 1) = 4cos2 11° – 2 2. Write the following as a single trigonometric function: 2.1 2sin3A. cos3A 2.2 2sin30°. cos30° 2.3 2cos 2 15° – 1 2.4 sin𝑥. cos𝑥 Solutions 2.1 2sin3A.cos3A = sin2(3A) 2.2 2sin30o.cos30o = sin2(30o) = sin6A =sin60o √3 =2 2.3 2cos215 – 1 = cos2(15o) 1 2.4 sin 𝑥 cos 𝑥 = . 2 sin 𝑥 cos 𝑥 2 = cos30o 1 √3 = 2 sin 2 𝑥 = 2 LEARNERS’ ACTIVITIES 1. Expand the following using double angle formulae: (a) cos 4 𝑥 (b) sin 7 0° (c) − cos 8 6° 2. Simplify (a) 2 cos 2 2 2,5° − 1 (b) 8 sin 𝑥 cos 𝑥 27

sin 2θ (c) 1+cos 2θ sin 2𝑥−cos 2𝑥 (d) 1−cos 2𝑥−sin𝑥 MIXED EXERCISES P(-6;5) 1. Use the diagram alongside to find the following: 1.1 the length of OP 1.2 sin θ 1.3 cos 2 θ 1.4 tan( θ − 180°) θ 1 0 1.5 tan 2θ 2. If OM is 13 units in the Cartesian plane and the reflex angle of XỘM = α. Calculate the following without the use of a calculator: 2.1 cos α 2.2 1 − sin2 α cos(180°−α).cos(90°−α) 0 2.3 sin(90°+α).sin(−α−180°) 2.4 tan 2 θ 13 33 M(-5 ; t) 2√6 3. If cos A = 5 and A ∈ [90°; 360°] calculate without using a calculator and with the aid of a diagram the value of 5tanA.cosA 3 4. If tan θ = 4 and θ ∈ (90°; 270°) calculate without the use of a calculator and with the aid of a diagram the value of: 4.1 sin 2 θ 4.2 cos 2 θ 5. If 13 sin A + 12 = 0 for 0° ≤ A ≤ 270°, and 17 cos B = 15 for 90° ≤ B ≤ 360°. Determine the value of cos( A + B). 2 6. If 2 sin α = −1 with α ∈ (90°; 270°)and tan β = 12 with β ∈ (90°; 270) calculate without the use √ of a calculator and with aid of a diagram the values of : 6.1 cos( α − β) 28

6.2 cos 2 α − cos 2 β 7. If cos 5 2° = m, write the following in terms of m 7.1 sin 5 2° 7.2 sin 3 8° 7.3 cos( − 128°) 7.4 cos 8 ° 7.5 cos 1 04° 7.6 cos 2 2 6° 8. If sin 1 2° = k determine the following in terms of k: 8.1 cos 2 4° 8.2 sin 2 4° 8.3 sin 7 8° 8.4 sin 4 2° 8.5 tan 1 2° sin 25° cos 25° 9. If sin 5 0° = t, determine the value of in terms of t. 4−8 cos2 25 10. If cos 4 ° = m,determine the value of: 10.1 cos 3 8° cos 3 4° + sin 3 4° sin 3 8° 10.2 cos 2 1 78° − cos 2 2 72° 11. Simplify the following to a single trigonometric ratio: 1−cos 2𝑥 (cos2 𝑥−sin2 𝑥)2 11.1 11.2 sin 2𝑥 cos4 𝑥−sin4 𝑥 sin 2𝑥−cos𝑥 1−cos2 𝑥−sin(−2𝑥) 11.3 1−cos 2𝑥−sin 𝑥 11.4 sin 𝑥+2 cos 𝑥 12. Simplify the following expressions without using a calculator: cos θ.tan(180°−θ).sin θ 12.1 sin(540°+θ).cos(90°−θ) sin(𝑥−180°).tan 𝑥 cos 690° 12.2 cos2 (x−90°) 12.3 − sin2 ( 90° − 𝑥) − tan 𝑥 . cos( − 𝑥). sin( − 𝑥 − 360°) sin 𝑥.sin(90+y)−cos 𝑥.sin(180°+y) 12.4 cos 𝑥 cos(y−360°)+sin(−𝑥) sin y 13. Prove that: 13.1 sin 3 𝑥 = 3 sin 𝑥 − 4 sin3 𝑥 cos 2θ+cos θ+1 1 13.2 = tan θ sin θ+sin 2θ 13.3 cos 4 𝑥 − sin4 𝑥 = 2 cos 2 𝑥 − 1 13.4 cos 7 A + cos A = 2 cos 4 A cos 3 A 29

1 13.5 sin( 45° − 𝑥) sin( 45° − 𝑥) = 2 − sin2 𝑥 14. Determine the general solution of: 14.1 2 sin2 α + sin α = 3 14.2 cos 4 θ. cos 4 0° + sin 4 θ. sin 4 0° = −1 14.3 tan2 y − 2 tan y − 3 = 0 14.4 cos 2 α + cos α = 0 14.5 sin( 𝑥 − 45°) = cos 2 𝑥 14.6 cos( 4𝑥 − 36°) = sin 2 𝑥 sin 2x cos 2𝑥 14.7 Given the equation sin 60° + sin 30° = 2, show that the equation can be written as √3 sin( 2𝑥 + 600 ) = 2 . Then find the general solution for x. 14.8 Solve for α if 1 + sin 2 α − 4 sin2 α = 0 and α ∈ [−1800 ; 900 ] 𝑠𝑖𝑛 2𝑥−𝑐𝑜𝑠 𝑥 𝑐𝑜𝑠 𝑥 14.9 For which values of x is the identity 1−𝑐𝑜𝑠 2𝑥−𝑠𝑖𝑛 𝑥 = 𝑠𝑖𝑛 𝑥 undefined? Do not prove the identity. 1 14.10 2 cos( 𝑥 + 300 ) = cos( 𝑥 − 300 ), deduce that tan 𝑥 = 3,hence write down the general √ solution of the first equation. 1 1 14.11 If 2 sin 2 𝑥 = a − ; −900 ≤ 𝑥 ≤ 900 , and 𝑎2 + 2 = 3, calculate the value of x without a 𝑎 using a calculator. sin 𝑥+cos 𝑥 14.12 Show that = sin( 45° + 𝑥), and hence find the general solution for √2 sin 𝑥+√3 cos 𝑥 = 0,5 2 15. On the same the same set of axes, sketch the graph of y = tan 𝑥 − 1 and y = 2 tan 𝑥 for 𝑥 ∈ (−90°; 180°]. 16. Given 𝑓(𝑥) = sin 2 𝑥 and 𝑔(𝑥) = cos 𝑥 , 𝑥 ∈ [−90°; 90°] 16.1 Sketch the graphs of 𝑓(𝑥) and 𝑔(𝑥) on the same set of axes. 16.2 Solve for 𝑓(𝑥) = 𝑔(𝑥) 16.3 For which values of 𝑥 is 𝑓(𝑥 ) < 𝑔(𝑥 ), 𝑥 ∈ [−90°; 90]? 16.4 For which values of 𝑥 is 𝑓(𝑥) increasing? 𝑥 17. Sketched below are 𝑓(𝑥) = sin( 𝑥 − 45°)and 𝑔(𝑥 ) = cos 2 for 𝑥 ∈ [−180°; 180]. 30

The curves intersect at points P and Q. 17.1 Determine the coordinates of the points P and Q, correct to 2 decimal places. 17.2 For which value(s) of 𝑥 is……. (i) 𝑓 < 𝑔? (ii) 𝑔 decreasing? (iii) 𝑓 increasing and negative? 18. The graph of 𝑓(𝑥) = 𝑐𝑜𝑠 𝑥 + 𝑏 and 𝑔(𝑥) = 𝑐𝑜𝑠 𝑘 𝑥 are sketched below for 𝑥 ∈ [−90°; 90°]: g f 18.1 Determine: (i) the value of 𝑏 if 𝑓 touches the 𝑥 -axis at the origin (ii) the amplitude of 𝑓 (iii) the value of 𝑘 if the period of 𝑔 is half the period of 𝑓 (iv) the coordinates of the 𝑥 -intercepts of 𝑔 (v) the maximum value of f . 18.2 Use the graphs to determine the values of 𝑥 in the interval 𝑥 ∈ [−90°; 90°] for which: (i) 𝑓(𝑥). 𝑔(𝑥) ≤ 0 𝑔(𝑥) (ii) 𝑓(𝑥) is undefined. 31

𝑔(𝑥) (iii) < 1 given that 𝑓 and 𝑔 intesect at 𝑥 = ±60°and 𝑥 = ±90°. 𝑓(𝑥) 19. In the figure Q is the foot of a vertical tower PQ, while R and S are two points in P the same horizontal plane as Q. The angle of elevation of P, as measured from R, is 𝑥. R𝑄̂S= 𝑦, QS= 𝑎 metres and the area of triangle RQS=A 𝑚2 . 2𝐴 𝑡𝑎𝑛 𝑥 Q Prove that 𝑃𝑄 = 𝑎 𝑠𝑖𝑛 𝑦 . a y x S R 20. In the figure below A, B and C are in the P same horizontal plane. P is a point vertically above A and B and C are 20 units apart. The angle of elevation from B to P is 𝛼. ̂ C = θ and AĈB = β. AB A α 20.1 Prove that AP = 20 sin β tan α B θ sin(θ+β) 20 tan α β C 20.2 Given that AB = AC , show that AP = cos β 21. PQ is a flagpole of height h metres, PS is a stay wire P (a stay has a similar purpose to a guy rope) inclined to the horizontal at an angle of 𝛽. R is equidistant from Q and S. RQ = 𝑥 and QŜR = α β 21.1 Show that h = 𝑥√2(1 + cos 2 α). tan β. α S 21.2 If 𝑥 = 2,58m, α = 58° and β = 64°, find h. x R 22. G, K and J are three points on the same horizontal plane. HG and IJ are the two vertical poles. I Wires are strung from K to the tops of the H poles. The angle of elevation of I from K is β. The length of wire KI is ̂ J = 𝑥. p metres. GK G θ J 22.1 Show that KJ = p cos β α β x p.cos β.sin(𝑥+θ) 22.2 Show that GK = K sin θ 32

p.cos β.sin(𝑥+θ).tan α 22.3 Prove that: HG = sin θ REFERENCES: Clever Mathematics Mind Action Series Study and Master Mathematics The Answer Series St Stithians Maths prelims Maths Handbook and Study Guide Maritzburg College textbook Maths is fun.com Classroom Mathematics Platinum Mathematics 33