MATHEMATICS Grade 12 - Western Cape

Western Cape Education Department

Telematics Learning Resource 2019

MATHEMATICS Grade 12

Telematics Mathematics Grade 12 Resources

2

February to October 2019

Dear Grade 12 Learner

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 will be assessed. It is thus important that you compile a study timetable which will consider the revision of the grade 11 content. The program in this book reflects the dates and times for all grade 12 and grade 11 sessions. It is highly recommended that you attend both the grade 12 and 11 Telematics sessions, this will support you with the revision of grade 11 work. This workbook however will only have the material for the grade 12 Telematics sessions. The grade 11 material you will be able to download from the Telematics website. Please make sure that you bring this workbook along to each and every Telematics session.

In the grade 12 examination Trigonometry will be + 50 marks and the Geometry + 40 marks of the 150 marks of Paper 2.

Your teacher should indicate to you exactly which theorems you have to study for examination purposes. 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 theorems and 2 are grade 12 theorems. At school you should receive a book called "Grade 12 Tips for Success". 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 important concepts and together with you will work though the activities. You are encouraged to come prepared, have a pen and enough paper (ideally a hard cover exercise book) and your scientific calculator with you.

You are also encouraged to participate fully in each lesson by asking questions and working out the exercises, 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 Resources

3

February to October 2019

2019 Mathematics Telematics Program

Day

Date

Term 1: 9 Jan ? 15 March

Tuesday

12 February

Wednesday 13 February

Time

15:00 ? 16:00 15:00 ? 16:00

Grade

12 12

TERM 2: 2 April to 14 June

Monday

8 April

15:00 ? 16:00 12

Tuesday

9 April

15:00 ? 16:00 12

Wednesday 15 May

15:00 ? 16:00 11

Thursday

16 May

15:00 ? 16:00 11

Wednesday 22 May

15:00 ? 16:00 12

Thursday

23 May

15:00 ? 16:00 12

Term 3: 9 July ? 20 September

Monday

29 July

15:00 ? 16:00 12

Tuesday

30 July

Wednesday 07 August

15:00 ? 16:00 12 15:00 ? 16:00 11

Monday

12 August

15:00 ? 16:00 11

Term 4: 1 October ? 4 December

Tuesday

15 October

15:00 ? 16:00 11

Wednesday 16 October

15:00 ? 16:00 11

Subject

Mathematics Wiskunde

Mathematics Wiskunde Mathematics Wiskunde Mathematics Wiskunde

Mathematics Wiskunde Mathematics Wiskunde

Mathematics Wiskunde

Topic

Trigonometry Revision Trigonometrie Hersiening

Trigonometry Trigonometrie Geometry Meetkunde Geometry Meetkunde

Differential Calculus Differentiaalrekening Functions Funksies

Paper 1 Revision Paper 2 Revision

Telematics Mathematics Grade 12 Resources

4

February to October 2019

Session 1: Trigonometry

x Definitions of trigonometric ratios:

o In a right-angled '

SinT opposite hypotenuse

CosT adjacent hypotenuse

TanT opposite adjacent

hypotenuse

opposit

T

adjacent

o On a Cartesian Plane

SinT y

y

r

CosT x r

TanT y x

ry

T

x

x

x Special Angles

o

0?, 90?, 180?, 270?, 360? can be

obtained from the following unit circle

.

T 90q

y

(0 ; 1)

T 180q (-1 ; 0)

T 0q

(1 ; 0) x T 360q

r, the radius is 1 since it is a unit circle

30?, 45? and 60? can be obtained from the following

30q

2

3

60q 1

45q

2

1

45q 1

(0 ; -1)

T 270q

x The "CAST" rule enables you to obtain the

sign of the trigonometric ratios in any of the

four quadrants.

S Sine is +ve in the 2nd quadrant

y

180q- T

T =30q

sin 30q

=

1 2

cos 30q =

3 2

tan 30q = 1

3

T

T Tan is +ve in the3rd quadrant

SinSe

All

x

Tan

Cos

+

180q+T

360q-T

T = 45q sin 45q = 1

2

cos 45q = 1

2

tan 45q = 1

T = 60q

sin 60q =

3 2

cos 60q =

1 2

tan 60q = 3

A - ALL trig ratios are +ve in the first quadrant

C Cos is +ve in the 4nd quadrant

The trigonometric function of angles (180q?T) or (360q?T) or (-T)

becomes

?

Trigonometric function of T

The sign is determined by the "CAST" rule.

(180q T ) sin(180q T) sinT cos(180q T ) cosT tan(180q T) tanT

(180q T ) sin(180q T ) sinT cos(180q T ) cosT tan(180q T) tanT

(360q T ) sin(360q T ) sinT cos(360q T ) cosT tan(360q T ) tanT

(360q T ) sin(360q T) sinT cos(360q T ) cosT tan(360q T ) tanT

(T ) sin(T ) sinT cos(T ) cosT tan(T ) tanT

Telematics Mathematics Grade 12 Resources

5

February to October 2019

x TRIGONOMETRIC IDENTITIES

tan T

sin T cosT

(cosT z 0)

sin 2 T cos2 T 1 ,

sin2 T 1 cos2 T ,

cos2 T 1 sin2 T

x Co-functions or Co-ratios

sin(90q T ) cosT cos(90q T ) sin T

r 90q-T y

sin(90q T) cosT cos(90q T) sinT

x Trigonometric Equations

1. Determine the Reference angle

2. Establish in which two quadrants is.

3. Calculate in the interval

[0q; 360q]

4. Write down the general solution

sinT 0,707 Reference = sin1(0,707) = 45q

? = 45q or = 180q - 45q

? = 45q or = 135q

? = 45q+ k360? or = 135q + k360? where k =

T x

cosT 0,866 Reference = cos1(0,866) = 30q

tanT 1 Reference = tan 1(1) = 45q

? = 180q- 30q or = 180q + 30q ? = 180q - 45q

? = 150q or = 210q ? = r150q ? = r150+ k360? where k =

? = 135q ? = 135q+ k180? & k =

TRIGONOMETRIC GRAPHS

Equation

Shape a > 0

Sine Function

Cosine Function

Tangent Function

a < 0

Amplitude

a

a

Period

SOLUTIONS OF TRIANGLES

x Area Rule

Area of 'ABC = 1 absin C = 1 acsin B = 1 bcsin A

2

2

2

x Sine Rule

sin A = sin B = sin C Or a = b = c

a

b

c

sin A sin B sin C

x Cosine Rule

a2 b2 c2 2bc cos A

or

cos A b 2 c 2 a 2

2bc

Note: "c" refers to the side of the triangle opposite to angle C that is the side

A

c

b

B

a

C

Telematics Mathematics Grade 12 Resources

6

February to October 2019

TRIGONOMETRY SUMMARY

Question Summary of

type

procedure

Example question

1. Calculate the value of a trig expression without using a calculator.

Establish whether you need a rough sketch or special triangles, ASTC rules or compound angles.

1.1 If 13cosD

5 and tan E

3 , 4

D [0q;

270q]

and

E [0q;

180q] .

It is given that sin(D E) sinD.cos E sin E.cosD

Determine, without using a calculator,

a) sinD

b) sin(D E ) .

1.2

Calculate: a)

cos(210$ ).sin2 405$.cos14$ tan 120 $. sin104 $

b) sin 70q cos 40q cos 70qsin 40q

2. If a trig ratio is given as a variable express another trig ratio in terms of the same variable. 3. Simplify a trigonometric al expression.

4. Prove a given identity.

5. Solve a trig equation.

Draw a rough sketch with given angle and label 2 of the sides. The 3rd side can then be determined using Pythagoras. Express each of the angles in question in terms of the angle in the rough sketch. Use the ASTC rule to simplify the given expression if possible. See if any of the identities can be used to simplify it, if not see if it can be factorized. Check again if any identity can be used. This includes using the compound and double angle identities.

Simplify the one side of the equation using reduction formulae and identities until .

Find the reference angle by ignoring the "-"sign and finding sin 1 (0,435)

Write down the two solutions in the interval

2. If sin 27q q , express each of the following in terms of q. a) sin117q b) cos(27q)

3. Simplify: cos ( 720q x) . sin ( 360q x) . tan ( x 180q)

a) sin ( x) . cos (90q x)

b)

sin ( 90q x) . tan ( 360q x)

sin (180q x) . cos (90q x) cos(540q x).cos(x)

sin 2 x cos x cos3 x c)

cos x

sin2 x cos x d)

1 cos2 x

Prove that

a)

tan x . cos3 x

1 sin x

1 sin2 x cos2 x 2

b)

Solve for x [180q; 360q] a) sin x 0,435 b) cos 2x 0,435 c) tan 1 x 1 0,435 2

x [0q; 360q] . Then

write down the general solution of this eq. From the general solution you can determine the solution for the specified interval by using various values of k.

Telematics Mathematics Grade 12 Resources

7

February to October 2019

Question type Summary of procedure

Example question

6. Sketch a trig graph.

1st sketch the trig graph without the vertical or horizontal transformation. Then shift the graph in this case 1 unit up.

7.Find the area of a triangle.

8. Finding an unknown side or angle in a triangle.

If it is a right-angled triangle then

area 1 baseu height , otherwise use 2

the area rule Area of 'ABC = 1 ab sin C

2 Draw a rough sketch with the given information. If it is not a right-angled triangle you will use either the sine or cosine rule.

Sketch b) y 2 cos 3x 1 for x [90q; 120q] c) y sin(x 60) for x [240q; 120q]

'ABC, with B 104,5q , AB 6cm and BC 9cm . Calculate, correct to one decimal place area 'ABC

a) 'ABC, with B 104,5q , AB 6cm and BC 9cm . Calculate the length of AC.

b) 'ABC, with C 43,2q , AB 4,5cm and BC 5,7cm . Calculate the size of A .

Calculate the period

Write down the amplitude if it is a sine or cosine graph.

SKETCHING TRIG GRAPH

Identify the shape of the graph and draw Now do the vertical or horizontal

a sine, cosine or tan graph with

transformation if required.

determined period and amplitude. Label

the other x-intercepts. Repeat this pattern

over the specified domain.

SKETCH y 2 cos 3x 1 for x [90q; 120q]

Period = Amplitude =

360q

2 120q

3

y 2

1

-90 -60 -30 -1

x 30 60 90

-2

y 3

2

1

-60

-30

-1

x

30

60

90

Telematics Mathematics Grade 12 Resources

8

February to October 2019

QUESTION

.1

In the figure below, the point P(?5 ; b) is plotted on the Cartesian plane.

OP = 13 units and RO^ P D .

y

P(?5x; b)

13

O

xx R

Without using a calculator, determine the value of the following:

.1.1

cosD

(1)

.1.2

tan(180q D )

(3)

.2

Consider:

sin(T 360q) sin(90q T ) tan(T ) cos(90q T )

.2.1

Simplify

sin(T

360q) sin(90q T ) tan(T ) cos(90q T )

to

a

single

trigonometric

ratio.

(5)

.2.2

Hence, or otherwise, without using a calculator, solve for T if 0q d T d 360q :

sin(T 360q) sin(90q T ) tan(T ) 0,5

(3)

cos(90q T )

.3

.3.1

Prove that 8 4

4.

sin 2 A 1 cos A 1 cos A

(5)

.3.2

For which value(s) of A in the interval 0q d A d 360q is the identity in

QUESTION 5.3.1 undefined?

(3)

.4

Determine the general solution of 8cos2 x 2 cos x 1 0 .

(6)

[26]

 ................
................

In order to avoid copyright disputes, this page is only a partial summary.

Google Online Preview   Download