trigonometry graphs and equations

MATHEMATICS Learner’s Study and Revision

Guide for Grade 12

SOLUTION OF TRIGONOMETRIC EQUATIONS

AND INEQUALITIES

Revision Notes, Exercises and Solution Hints by

Roseinnes Phahle

Examination Questions by the Department of Basic Education

Preparation for the Mathematics examination brought to you by Kagiso Trust

Contents

Unit 17

Solving trigonometric equations and inequalities graphically 3

Solving trigonometric equations analytically – Type 1 5

Other inequalities that can be solved graphically 6



Answers to exercises 10

Refining techniques for sketching trigonometric functions 17

Examination questions with solution hints and answers 21

More questions from past examination papers 26

Answers 32

How to use this revision and study guide

1. Study the revision notes given at the beginning. The notes are interactive in that in some parts you are required to make a response based on your prior learning of the topic in class or from a textbook.

2. “Warm‐up” exercises follow the notes. Some exercises carry solution HINTS in the answer section. Do not read the answer or hints until you have tried to work out a question and are having difficulty.

3. The notes and exercises are followed by questions from past examination papers.

4. The examination questions are followed by blank spaces or boxes inside a table. Do the working out of the question inside these spaces or boxes.

5. Alongside the blank boxes are HINTS in case you have difficulty solving a part of the question. Do not read the hints until you have tried to work out the question and are having difficulty.

6. What follows next are more questions taken from past examination papers.

7. Answers to the extra past examination questions appear at the end. Some answers carry HINTS and notes to enrich your knowledge.

8. Finally, don’t be a loner. Work through this guide in a team with your classmates.

Solution of trigonometric equations and inequalities analytically and graphically

3

REVISION UNIT 17: TRIGONOMETRIC EQUATIONS AND INEQUALITIES

SOLVING TRIGONOMETRIC EQUATIONS AND INEQUALITIES GRAPHICALLY

Example 1: Below are shown the graphs of ( )2

cos xxf = and ( ) ( )o30sin −= xxg for

[ ]oo 180 ;180−∈x .

-180 -160 -140 -120 -100 -80 -60 -40 -20 20 40 60 80 100 120 140 160 180

-1

1

x

y

f g

A

B

1. Where the graphs meet is where ( ) ( )xgxf =

The graphs meet at the point A where o160−=x and at the point B where o80=x .

So the solution of the equation

( ) ( )xgxf =

or ( )o30sin2

cos −= xx

for values of [ ]oo 180 ;180−∈x is o160−=x and o80=x .

2. To find out the values of x for which ( ) ( )xgxf > , look at the graphs and see in what interval or

intervals the graph of f is above the graph of g .

Answer is oo 80160 <<− x .


Example 2: Below are shown again the graphs of ( )2

cos xxf = and ( ) ( )o30sin −= xxg for

[ ]oo 180 ;180−∈x .

-180 -160 -140 -120 -100 -80 -60 -40 -20 20 40 60 80 100 120 140 160 180

-1

1

x

y

f g

A

B

1. For what values of x is ( )( ) ?0<xgxf

The answer will be given by values of x for which ( )xf and ( )xg have opposite signs (one negative

and the other positive).

Looking at the graph the answer is given by ( )oo 30 ;150−∈x .

3. At what values of x is:

a) ( ) ( ) ?1=− xgxf Answer : =x

b) ( ) ( ) ?5,1=− xgxf Answer: =x

c) ( ) ( ) ?1−=− xgxf Answer: =x


5

SOLVING TRIGONOMETRIC EQUATIONS ANALYTICALLY

TYPE 1: REPLACING θtan by θθ

cossin whenever θtan appears in

the equation to be solved

Example 3: Solve xx tan212sin = for values of oo 18090 ≤≤− x .

Solution: xx tan212sin =

xxxx

cossin.

21cossin2 =

xxx sincossin4 2 =

0sincossin4 2 =− xxx

( ) 01cos4sin 2 =−xx

Either sin x = 0 or 4cos x2 ‐ 1 = 0

Cos x = 21

±

Therefore x = o0 or o180 or ooo 120or ;60 ;60−=x

Verifying the above analytical solutions by means of graphs

The graphs of ( ) xxf 2sin= and ( ) xxg tan21

= are drawn below on the same set of axes. The solution

to the equation is then given by the x ‐coordinates of the points of intersection of the two graphs indicated by the arrows in the diagram below.

-80 -70 -60 -50 -40 -30 -20 -10 10 20 30 40 50 60 70 80 90 100 110 120 130 140 150 160 170

-1

1

x

y

f

g

↓↑ ↑

↓↑


ONCE MORE ON SOLVING TRIGONOMETRIC INEQUALITIES

As shown at the beginning of this unit, we can use graphs to solve not only trigonometric equations as in Example 1 but also trigonometric inequalities as in Example 2. Let’s use the above example to explain further how you would solve trigonometric inequalities.

For example, we can use the graphs to determine the values of x for which

( ) ( )xfxg >

or, in other words xx 2sintan21

>

All you need to do is to look at the graphs to find the range of values of x for which the y values of the

graph of ( )xg are greater than the y values of the graph of ( )xf .

Simply explained, this means that look for the range of x values for which the graph of ( )xg is above

the graph of ( )xf .

Answer: oooooo 180120or ;9060or 060 <<<<<<− xxx

OTHER INEQUALITIES THAT CAN BE SOLVED GRAPHICALLY

By looking at the graphs of say any ( )xf and any ( )xg drawn on the same set of axes, we can also

solve inequalities in the form:

( )( ) 0<xgxf

or ( ) ( ) 0. >xgxf

In the first case (which we looked at in Example 1), we would look for a range of x values for which ( )xf and ( )xg have opposite signs. That is, a range of x values in which the graphs of ( )xf and ( )xg

lie on opposite sides of the x ‐axis.

In the second case, we would look for a range of x for which both ( )xf and ( )xg are positive or

negative. That is, a range of x values in which the graphs of ( )xf and ( )xg lie on the same side of the

x ‐axis.

GENERAL SOLUTIONS

General solutions of trigonometric equations covered in Unit 16 are applied in the sections that follow. Commit them to memory or learn to derive them by repeatedly practising to write them down because they are not included on the information sheet that will be given to you in the examination. They are referred to in the next section.


7

TYPE 2: USING ( )θθ −= o90cossin or ( )θθ −= o90sincos Suppose you are solving the equation: αsinsin =x Obviously in this case the two sides could not be equal unless α=x Because sine is positive in the first and second quadrants we must in fact have α=x in the first quadrant and α−= o180x in the second quadrant Furthermore, sine has a period of o360 ; so these solutions repeat themselves every o360 . Hence the general solutions are:

o360.nx += α and ( ) oo 360.180 nx +−= α where Ζ∈n Does this solution ring a bell? It is in fact the general solution you met in Unit 16 with calc∠ replaced by α . So too application of the general solutions to: αcoscos =x will result in the solution: o360.nx +±= α and application of the general formula to: αtantan =x will result in the solution: o180.nax += We now show how the above together with the property of complementary angles in the box at the top of this page is used to solve equations.


Example 4: Solve the equation xx 2cossin = You can proceed as in either column 1 or column 2 below:

xx 2cossin =

( )xx 290sinsin −= o The general solution is thus:

( ) oo 360.290 nxx +−= and

( ) ooo 360.290180 nxx +−−=

xx 2cossin = xx sin2cos =

( )xx −= o90cos2cos The general solution is thus:

( ) oo 360.902 nxx +−= and

( )oo 360.902 nxx +−−=

Example 5: Solve the equation ( )o45cossin −= xx

You can proceed as in either column 1 or column 2 below:

( )o45cossin −= xx

( ){ }oo 4590sinsin −−= xx The general solution is thus:

( ){ } ooo 360.4590 nxx +−−= and

( ){ } oooo 360.4590180 nxx +−−−=

( )o45cossin −= xx

( ) xx sin45cos =− o

( ) ( )xx −=− oo 90cos45cos The general solution is thus:

( ) ( ) ooo 360.9045 nxx +−=− and

( ) ( ) ooo 360.9045 nxx +−−=−

Exercise 17.1: Use both answers in the above examples to find solutions for values of [ ]oo 180;270−∈x

Exercise 17.2: Do rough sketches to show how the above equations can be solved graphically. Where in the sketches do you find the solutions to the equations?

Exercise 17.3: Looking at the sketches, for what values of x is

a) xx 2cossin ≥

b) ( ) xx sin45cos >− o


9

TYPE 3: USING DOUBLE ANGLE FORMULAE Use of a double angle formula will apply if the equation includes a trigonometric ratio of a single angle

and another of a double angle. For example, x and x2 . The idea is to reduce all the ratios to one of a single angle. Factorization or use of the quadratic formula will usually follow substitution using a double angle formula.

Example 6: Solve xx 2sincos2 = by factorizing.

Solution: xx 2sincos2 =

xxx cossin2cos2 = xxx cossincos = :factorise nowcan you which 0cossincos =− xxx

( ) 0sin1cos =− xx

Either 0cos =x or 0sin1 =− x

Exercise 17.4: Complete the above example by finding the solutions for oo 180270 ≤≤− x .

Exercise 17.5: Do rough sketches to show how the equation in above example can be solved graphically.

Exercise 17.6: Looking at your rough sketches, for what values of x is xx 2sincos2 ≥ ?

Exercise 17.7: Looking at your rough sketches, for what values of x is 02sincos2 <xx ?

Example 7: Solve xx 2cossin = for oo 180270 ≤≤− x using the quadratic formula.

Solution Explanation Step 1: xx 2cossin = Step 2: xx 2sin21sin −=

Step 3: 01sinsin2 2 =−+ xx Thus

a

acbbx2

4sin2 −±−

=

The third step is a quadratic equation in xsin with ,2=a 1=b and 1−=c Substitute for cba and , or solve by factorization.

Exercise 17.8: Complete the above example by finding the solutions for oo 180270 ≤≤− x .

Exercise 17.9: Instead of using the quadratic formula, use factorization to solve the above problem.

Exercise 17.10: Verify the solutions of Exercises 17.8 and 17.9 graphically.


ANSWERS

EXERCISE 17.1 Solution to Example 6: The solutions to the first example are fully worked out in order to show you how to work them out.

Bear in mind that we are looking for solutions in the range ]180;270[ oo− . First answer in the first column:

( ) oo 360.290 nxx +−=

oo 360.902 nxx +=+

oo 360.903 nx +=

oo 120.30 nx +=

3−=n : ooo 33036030 −=−=x Reject this answer. Why?

2−=n : ooo 21024030 −=−=x

1−=n : ooo 9012030 −=−=x

0=n : o30=x

1=n : ooo 15012030 =+=x

2=n : ooo 27024030 =+=x Reject this answer. Why? Second answer in the first column:

( ) ooo 360.290180 nxx +−−=

Simplifies to oo 360.90 nx −−=

1−=n : oo 36090 +−=x No need to complete the answer. Why?

0=n : o90−=x 1=n : Is there need to complete this? Why? Combining the two sets of answers and writing them in ascending (or descending) order we have all the

solutions in the interval ]180;270[ oo− :

Answers: oooo 150 ;30 ;90 ;210 −−=x NOTE: Using the second column would yield the same answers. Give yourself practice by trying this. Solution to Example 7: Working out the general solutions for { };;;;2;1;0;1;2,, −−−−−−−∈n , the answers are given by:

oo 67,5 ;5,112−=x


11

EXERCISE 17.2

Sketch of Example 4

-270 -240 -210 -180 -150 -120 -90 -60 -30 30 60 90 120 150 180

-1

1

x

y

sin x

cos 2x

A

B

C↓ ↓ ↓ ↓D

Graphical solutions are given by the x ‐coordinates of the points A, B, C and D which are the points of intersection of the two graphs. The arrows indicate the solutions. Sketch of Example 5

-270 -240 -210 -180 -150 -120 -90 -60 -30 30 60 90 120 150 180

-1

1

x

y

sin x

cos (x-45)

A

B

↓ ↓

Similarly, the solutions which are indicated by the arrows are given by the x ‐coordinates of the points of intersection A and B of the two graphs give.


EXERCISE 17.3

Pay particular attention to whether the relationship in the question is ≥ , that is “greater than and equal to”; or simply >, that is “greater than”. The type of relation indicates how you write the solution as explained below. Solution to Example 4: Inspect the graphs for values of x for which the graph of xsin is above the graph of x2cos .

These values are [ ]o210 ;2700 −−∈x and [ ]oo 150 ;30∈x We use the square brackets [ …. ] because of ≥ in the question. This means that x takes the value that is against the square bracket [ . Using set notation the two sets can be combined by using the symbol ∪ which stands for “union”:

[ ] [ ]oooo 150 ;30210 ;270 ∪−−∈x Solution to Example 5:

We must inspect the graphs looking out for values of x for which the graph of ( )o45cos −x lies above the graph of xsin .

These values are ( )15 ;30o∈x . Note that in this case we do not use the square brackets because of > in the questions. The bracket ( against a value means that x does not take that value.


13

EXERCISE 17.4

0cos =x

Solution on calculator or calc o90=∠ Applying the general solution:

o360.calc nx +∠±=

oo 360.90 nx +±=

oo 360190 :1 ×−±=−= xn

oooo 36090-or 36090 −−+=x

o270−=x √ or o450− Reject. Why? o90 :0 ±== xn √ that is accept. Why?

0sin1 =− x 1sin =x

Solution on calculator or calc o90=∠ Applying the general solution:

o360.calc nx +∠=

or oo 360.)calc180( nx +∠−=

oo 360.90 nx +=

or ooo 360.)90180( nx +−= which both reduce to

oo 360.90 nx +=

o270 :1 −=−= xn √ Why? o90 :0 == xn √ Why?

Putting together all the answers we thus get: ∗−−= 90 ;90 ;270 oox

EXERCISE 17.5

-270 -240 -210 -180 -150 -120 -90 -60 -30 30 60 90 120 150 180

-2

-1

1

2

x

y

2cos x

sin 2x

↓ ↓ ↓

The solutions are indicated by the arrows and confirm those found in Exercise 17.4 above.


EXERCISE 17.6

-240 -210 -180 -150 -120 -90 -60 -30 30 60 90 120 150 180

-2

-1

1

2

x

y

2cos x

sin 2x

← →The ordinates of 2cos x are greaterthan those of sin 2x in this range

We must inspect the graphs and look for where the graph of xcos2 lies above the graph of x2sin . In other words, where are the ordinates or the y coordinates of xcos2 greater than or equal to those of x2sin ?

The answer is indicated by the arrows, that is [ ]oo 90 ;90−∈x Why do we use “close” brackets in this case? EXERCISE 17.7

-240 -210 -180 -150 -120 -90 -60 -30 30 60 90 120 150 180

-2

-1

1

2

x

y

2cos x

sin 2x

← →The ordinates have opposite signs in this range

In this case we inspect the graphs to observe a range in which the graphs or their ordinates have opposite signs. This range is indicated by the arrows in the above diagram.

The answer is thus: ( )oo 0 ;180−∈x Why do we use “open” brackets in this case?


15

EXERCISE 17.8

Finding the solutions by using the quadratic formula:

42or

44

431

4811sin −

=±−

=+±−

=x

0,5or 1sin −=x

Calculator answers are: o90−=x or o30=x Inserting these solutions into the two general solutions:

First, o90−=x : oo 360.90 nx +−= or ( ){ } ooooo 360.270360.90180 nnx +=+−−=

Determine the solutions in the interval oo 180270 ≤≤− x :

oo 270360190 :1 =×+−== xn Reject. Why?

o90 :0 −== xn Accept. Why?

( ) oo 360190 :1 ×−+−=−= xn Reject. Why?

oo 3601270 ×+=x Reject. Why?

o270=x Reject. Why?

( ) ooo 903601270 −=×−+=x Accept. Why?

You can extend the integer values of n in both negative and positive directions to see if you will uncover

more solution in the interval oo 180270 ≤≤− x . Do try.

Second, o30=x : oo 360.30 nx += or ( ) ooooo 360.150360.30180 nnx +=+−=

:1−=n ( ) oo 360130 ×−+=x Reject. Why?

:0=n o30=x Accept. Why?

ooo 210360150 −=−=x Accept. Why?

o150=x Accept. Why? Again, you can try more integer values of n to see if there are any more solutions in the given interval. Combining all the solutions and writing them in ascending order of magnitude, the answer is:

{ } 150 ;30 ;90 ;210 oooo −−∈x Compare these solutions with the graphical solutions illustrated on previous page.


EXERCISE 17.9

Finding the solutions by means of factorization: xx 2cossin =

xx 2sin21sin −=

01sinsin2 2 =−+ xx ( )( ) 01sin1sin2 =+− xx So that either 01sin2 =−x or 01sin =+x

1sinor 5,021sin −=== xx

Leading as we have seen in the previous Exercise to the solutions already found, namely:

{ } 150 ;30 ;90 ;210 oooo −−∈x EXERCISE 17.10

Finding the solutions graphically Below is a graphical illustration of the solution of sin x = cos 2x which confirms the solutions obtained by the alternative methods used in Exercises 17.8 and 17.9:

-270 -240 -210 -180 -150 -120 -90 -60 -30 30 60 90 120 150 180

-1

1

x

y

sin x

cos 2x

A

B

C↓ ↓ ↓ ↓D

The solutions are seen to be { } 150 ;30 ;90 ;210 oooo −−∈x


17

REFINING TECHNIQUES FOR SKETCHING TRIGONOMETRIC FUNCTIONS

We sketched trigonometric functions in UNIT 7 and in the preceding section. Let’s use a few more examples to show in greater detail what to do in order to obtain sketches of the curves of trigonometric functions.

Example 8: Sketch the graph of ( ) xxf 3sin= for the domain [ ]oo 180 ;90−∈x .

Characteristics Parent function xsin ( ) xxfy 3sin== Coordinates

Period o360 oo

1203

360=

−x intercepts at which 0=y

oo 180.0 kx += x is replaced by x3 oo 180.03 kx +=

oo 60.0 kx +=

o180 ,3 −=−= xk o120 ,2 −=−= xk o60 ,1 −=−= xk

o0 ,0 == xk o60 ,1 == xk o120 ,2 == xk o180 ,3 == xk o240 ,4 == xk

Coordinates that fall within the domain:

( )oo 0 ;120−

( )oo 0 ;60−

( )oo 0 ;0

( )oo 0 ;60

( )oo 0 ;120

( )oo 0 ;180

=y intercepts at

which o0=x 10sin == oy 00sin03sin ==×= ooy (0 o ; 0)

Maximum value at 1=y

oo 360.90 kx += x is replaced by x3 oo 360.903 kx +=

o90 ,1 −=−= xk

o30 ,0 == xk o150 ,1 == xk o270 ,2 == xk


( )1 ;90o−

( )1 ;30o

( )1 ;150o

Minimum value 1−=y

oo 360.90 kx +−= x is replaced by x3 oo 360.903 kx +−=

o150 ,1 −=−= xk

o30 ,0 −== xk o90 ,1 == xk o210 ,2 == xk


( )1- ;30o−

( )1 - ;90o


What a lot of working! It had to be at this stage because we are making explanations. But with much practice on your part you will not have to do so much working out and writing down. On many occasions you are required to state or indicate on your sketch the coordinates of the intercepts and turning points so you must know how to derive them. In practice though these can be more easily derived from your knowledge of the parent function instead of getting bogged down by using the general formulae.

What you next do is to plot the coordinates on your axes:

-120 -105 -90 -75 -60 -45 -30 -15 15 30 45 60 75 90 105 120 135 150 165 180

-1

1

x

y

Now join the points you see with a smooth curve (not straight lines!):

-120 -105 -90 -75 -60 -45 -30 -15 15 30 45 60 75 90 105 120 135 150 165 180

-1

1

x

y

The first sketch above is for purposes of showing that first you must plot the points. Otherwise, all you need to show when solving this question is the last sketch.

What is the amplitude of ( ) xxf 3sin= ?


19

Example 9: Sketch the graph of ( ) ( )o30cos += xxf for the domain [ ]oo 240 ;180−∈x .

Characteristics Parent function xcos ( ) ( )o30cos +== xxfy Coordinates

Period o360 o360


( ) o90.12 += kx x is replaced by o30+x

( ) oo 90.1230 +=+ kx oo 6090.2 += kx

o120 ,1 −=−= xk

o60 ,0 == xk

Coordinates that fall within the

domain:

( )oo 0 ;120−

( )oo 0 ;60

=y intercepts at

which o0=x

( )o30cos += xy

( )oo 300cos +=

o30cos=

87,023==

(0; 0,87)

Maximum value at 1=y

1cos =x oo 360.0 kx +=

x is replaced by o30+x ooo 360.030 kx +=+

oo 360.30 kx +−=

o30 ,0 −== xk o330 ,1 == xk


domain:

( )1 ;30o−

Minimum value 1−=y

1cos −=x oo 360.180 kx +=

x is replaced by o30+x ooo 360.18030 kx +=+

oo 360.150 kx +=

o150 ,0 == xk


domain:

( )1- ;150o

Plot the points and join them with a smooth curve to get:

-180 -165 -150 -135 -120 -105 -90 -75 -60 -45 -30 -15 15 30 45 60 75 90 105 120 135 150 165 180 195 210 225 240

-1

1

x

y

What is the amplitude of ( ) ( )o30cos += xxf ?


Example 10: Sketch the graph of ( ) ( )o60tan −= xxf for the domain [ ]oo 180 ;135−∈x .

Characteristics Parent function xtan ( ) ( )o60tan −== xxfy Coordinates

Period o180 o180


o180.kx = x is replaced by o60−x oo 180.60 kx =− oo 180.60 kx +=

o120 ,1 −=−= xk

o60 ,0 == xk


domain:

( )0 ;120o−

( )0 ;60o

=y intercepts

at which o0=x 00tan == oy ( ) 73,160tan600tan ==+ ooo (0; 1,73)

Asymptotes ( ) o90.12 += kx x is replaced by o60−x

( ) oo 90.1260 +=− kx

( ) oo 6090.12 ++= kx o30 ,1 −=−= xk o150 ,0 == xk

Asymptotes at

o30 −=x

o150 =x

Plot the points and join them with a smooth curve to get:

-130 -120 -110 -100 -90 -80 -70 -60 -50 -40 -30 -20 -10 10 20 30 40 50 60 70 80 90 100 110 120 130 140 150 160 170 180

-9

-8

-7

-6

-5

-4

-3

-2

-1

1

2

3

4

5

6

7

8

9

x

y

We will now apply all of the above to solving trigonometric equations and inequalities.


21

PAPER 2 QUESTION 6 DoE/ADDITIONAL EXEMPLAR 2008

PAPER 2 QUESTION8 DoE/NOVEMBER 2008

PAPER 2 QUESTION 7 DoE/PREPARATORY EXAMINATION 2008

Given the following functions: ( ) ( )o30sin += xxf and ( ) xxg 2cos=

7.1 Sketch the functions for [ ]oo 180;90−∈x on DIAGRAM SHEET 1.

Clearly show the intercepts with axes as well as the turning points. (8)

7.2 Calculate values of x for ( ) ( )xgxf = for [ ]oo 90;90−∈x . (8)

7.3 Use your graph to determine the value(s) of x for which ( ) ( )xgxf ≥ for [ ]oo 180;0∈x (2)

7.4 State the value(s) of x for which ( )xg is negative and x is increasing for [ ]oo 180;90−∈x .

(2)[20]


PAPER 2 QUESTION 6 DoE/ADDITIONAL EXEMPLAR 2008

Number Hints and answers Work out the solutions in the boxes below 6.1

Look at each function. State its amplitude. State its period, and its x ‐ and y ‐ intercepts. State its turning points. Sketch each graphs. Check your answer with a friend.

DIAGRAM SHEET 1

6.2 Some of the trigonometric equations you will

have met in Grades 11 & 12 can be solved like algebraic equations in one unknown. To get a single unknown we must transform the equation so that it only has sines or cosines, and furthermore, simplified to sines or cosines of a single angle x rather than a double angle x2 .

xsin as a sine of a single x is already in simplified form. So tackle x2cos reducing this to an expression in terms of a sine of single x by using an appropriate formula. Do this and you will end up with a quadratic equation in xsin . Solve for xsin by using the quadratic formula. Or, first let kx =sin .

Answer:

oo 5,158or 5,21 == xx

6.3 Look at your two graphs drawn on the same axes and find out where along the x ‐axis the difference between the two graphs is equal to 3 units. If you could not see where the graphs differ by 3 units, use the answer to see this and this will show you how to work out such a question in future. Answer: o90=x


23

PAPER 2 QUESTION 8 DoE/NOVEMBER 2008

Number Hints and answers Work out the solutions in the boxes below 8.1 Have either a cosine or a sine on both

sides of the equation by using the cosine of an angle is the same as the sine of the complement of the angle, and vis a verse. First find the general solutions and then

find the solutions within [ ]oo 180;90− Answer:

ooo 5,112;5,22;5,67−=x

or oo 135;45−=x

8.2 Sketch strictly within the given domain; otherwise you will lose a mark or two. Use a different colour pen for each curve so you can see each clearly when you sketch them. Use broken horizontal lines to indicate the maximum and minimum of each function. The sine curve is easy to sketch in because its period is o360 . Mark the points where it crosses the x ‐axis, attains its maximum and minimum values. Do likewise for the cosine curve dividing by 3 the points where its parent curve cos x crosses the x ‐axis and attains its maximum and minimum values within the domain [ ]oo 180;90− . DIAGRAM SHEET 2

8.3 Look at the graphs and see for which values of x within [ ]oo 180;90− the graph of ( )xf is below the graph of ( )xg .

Answer:


PAPER 2 QUESTION 7 DoE/PREPARATORY EXAMINATION 2008

Number Hints and answers Work out the solutions in the boxes below 7.1

First do sketches on rough paper if this approach suits you better and enables you to see the shape of its curve together with the location of its turning points and intercepts on the axes.

On rough paper make a quick sketch of sin x for [ ]oo 180;90−∈x . Then shift the graph by o30 to the left. Mark its intercepts with the axes and its turning points.

Similarly on rough paper make a quick sketch of cos x for [ ]oo 360;180−∈x taking into consideration that the period is to be halved when sketching cos 2 x . Now halve the markings on the x ‐axis to give the x ‐intercepts of cos 2 x for

[ ]oo 180;90−∈x . On the diagram given to you, now sketch the graphs as neatly and correctly as possible, superimposing one on the other: DIAGRAM SHEET 1

Answers:

( )xf ( )xg

x ‐intercepts ( )0;30o−

( )0;150o

( )0;45o−

( )0;45o

( )0;135o

y ‐ intercepts ( )5.0;0o ( )1;0o


25

Number Hints and answers Work out the solutions in the boxes below Turning points ( )1;60o ( )1;0o

( )1;90 −− o

( )1;90 −o

( )1;180o

7.2 You are solving the equation

( ) xx cos230sin =+ o in the interval

[ ]oo 90;90−∈x . Use the fact that the sin of an angle is equal to the cosine of the complement of the angle, or vice a versa. This way you will have either a sin or a cosine on both sides of the equation enabling you to equate the angles. Remember to apply the general solutions and only thereafter restrict the solutions to

[ ]oo 90;90−∈x . Answers:

oo 20;60−=x

7.3 Inspect your sketches in 7.1 to see where the graph of ( )xf is above the

graph of ( )xg . Note: answer is sought in interval

[ ]oo 180;0∈x . For what range of values is ( )xf

above ( )xg ? This will give you the answer to the question. Use your solutions in 7.2 to mark the end points of the range of values of x for which gf ≥ . Answer:

[ ]oo 140;20∈x

7.4 Answer: You write the answer.


MORE QUESTIONS FROM PAST EXAMINATION PAPERS

Exemplar 2008 Paper 2


27

Feb – March 2008 Paper 2

DIAGRAM SHEET 2



DIAGRAM SHEET 1


29

November 2009 (Unused paper) Paper 2

DIAGRAM SHEET 1

This sheet is the one shown above.


November 2009(1) Paper 2

DIAGRAM SHEET 4


31


DIAGRAM SHEET 3


ANSWERS

Exemplar 2008

8.1 )174,0;160( o−A

( )766,0;80oB

8.2 oo 80160 <<− x Feb/March 2009 8.1 Sketch:

8.2 o0=x or o180 or

ooo 120or 60 ;60 −=x

8.3 9060or 060 oooo <<<<− xx or oo 180120 << x

9 Maximum value = 31

Minimum value = 41

November 2009 (Unused paper) 8.1 2=a 8.2 Sketch:

8.3 2

8.4 Reading from the graph, o5,14 lies to the right of the point of intersection.

So o5,14<θ .

November 2009(1) 12.1 Sketch:

12.2 Points A and B shown on the graph. 12.3 oo 30or 90 −== xx 12.4 oo 210 and 30 == xx

12.5 [ ) ( ]oooo U 270;12060;90 −−∈x Feb/March 2010

11.1 { }oooo 360;330;210;180∈x 11.2 Sketch:

11.3 oo 210180 ≤≤ x or oo 360330 ≤≤ x

trigonometry graphs and equations

Documents