Unit 15: Trigonometry

1

MATHEMATICS

Learner’s Study and

Revision Guide for

Grade 12

TRIGONOMETRY

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

2

Contents

Unit 16

Factorising trigonometric expressions 3 Compound angle formulae 3 Introducing trigonometric identities 4 Special angles 5 The sine and cosine of complementary angles 8 Rotation of a point about the origin 10 Examination questions with solution hints and answers 11 Reduction formulae angles greater than 90 13 Negative angles 13 Summary of results 15 Solving trigonometric ratios or triangles without using a calculator 18 Examination questions with solution hints and answers 19 General solutions in trigonometry 22 Proving trigonometric identities 23 Answers to all the exercises 28 Examination questions with solution hints and answers 24 More questions from past examination papers 34 Answers to past examination papers 40 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.

Trigonometry

REVISION UNIT 16: TRIGONOMETRY

PRIOR KNOWLEDGE ASSUMED

In this Unit it is assumed that you know that the trigonometric ratios are defined in a right angled triangle and that you can relate them to the hypotenuse, opposite and adjacent sides.

It is also assumed that you can use the inverse key on a calculator to work out the angle given the ratio.

FACTORISING TRIGONOMETRIC EXPRESSIONS

Do you know that you can also factorise some trigonometric expressions in the same way as you factorise algebraic expressions such as :

Differences of two squares: ( )( )yxyxyx +−=− 22

Some trinomials: ( )( )21232 ++=++ xxxx EXERCISE 16.1

Factorise the following (if you‘re at a loss as to what to do look up the answer for a HINT):

a. =− xx 22 sincos b. =+− xxxx 22 sinsincos2cos

c. =++ xxxx 22 sinsincos2cos d. =−+ 6sinsin 2 xx

COMPOUND ANGLE FORMULAE

Complete the following formulae:

sin (A+B) = sin (A-B) =

cos (A+B) = cos (A-B) =

=+ xx 22 sincos

=x2cos

=x2sin Factorise as a difference of two squares:

( )( ) cos1 2 =− x Factorise as a difference of two squares:

( )( ) sin1 2 =− x

sin 2 x =

=xtan

cos 2 x = or in terms of xsin we can write that cos 2 x = or in terms of xcos we can write that cos 2 x =


4

INTRODUCING TRIGONOMETRIC IDENTITIES

In proving identities, you have to show that the LHS (left hand side) reduces to the RHS (right hand side) or the RHS reduces to the LHS or that both LHS and RHS reduce to the same expression.

EXERCISE 16.2

Prove the following identities:

1. Use a right angled triangle to prove that the following:

a. 1cossin 22 =+ xx

b. xxx

cossintan =

2. Write x3sin in terms of xsin

3. ( ) ( ) θφθφθφ 22 sinsinsinsin −=−+

4. ( ) ( ) BABABA sinsin2coscos −=−−+

5. xxx 22 sin322coscos −=+

6. AAA 2tan

2cos12cos1

=+−

7. xxx 2cossincos 44 =−

8. AA

A tan2cos1

2sin=

+

9. θθθ

θθ tan2coscos1

2sinsin=

+++

Trigonometry

SPECIAL ANGLES: 90 and ;60 ;45 ;30 ;0

In order to work out the type of problems such as Questions 3.3 and Question 4 (see page 11, for example) you will need to know the value in surd form of the trigonometric ratios of special angles. These surd forms are not on the formula sheet you will be given in the examination. So you must memorise them or, if you have difficulty remembering them, then you can learn how to quickly derive them as shown below.

Derivations

Special angles Diagrams to be used

For angles 30 and 60 :

As shown in the diagram below, we draw an equilateral triangle We can make the sides any length but they must of course all be of equal length. For the sake of convenience we have chosen the length of each side to be 2 units.

We now drop a perpendicular from any corner of the equilateral triangle to the

side opposite the corner. This result s in angles of size 30 and 90 .


6


Calling the length of the perpendicular x , use Pythagoras Theorem to work out the value of x leaving your answer in surd form.

Complete the working out of x below: =+ 22 1x =2x = Therefore =x Leaving your answers in surd form, you can now refer to the last triangle to complete the table below:

=30sin

=60sin

=30cos

=60cos

=30tan

=60tan

Trigonometry


For angle 45

In this case a right angled isosceles triangle is drawn as shown:

We can make the equal sides any length we like. But for convenience we choose to make each equal to 1 unit in length.

Calling the hypotenuse x , we use Pythagoras Theorem to work out the value of x(again leaving the answer in surd form):

=2x

=

Therefore =x

Now you can use the triangle to complete the following (leaving your answers in surd form):

=45sin

=45cos

=45tan

Angles 0 and 90

You should know off by heart the values of the trigonometric ratios of the angles 0 and 90 . If you don’t then you can figure them out from the graphs of sine, cosine and tangent. Or, you could simply use your calculator to find them.


8

SUMMARY OF THE RESULTS

Summarise all of the above results by completing the table below:

sin =0

cos =0

tan =0

sin 30 =

cos =30

tan =30

sin 45 =

cos =45

tan =45

sin 60 =

cos =60

tan =60

sin =090

cos =090

tan =090

THE SINE AND COSINE OF COMPLEMENTARY ANGLES

By observing the sine of any angle in the above table and the cosine of the complement of that angle, what do you notice?

Is your observation true in general?

Can you prove it by drawing a right angled triangle and naming one of the angles .θ So what is the size of the third angle in terms of ?θ Now use your triangle to prove the general result that

( )θθ −= 90cossin and

( )θθ −= 90sincos

We conclude that the sine of an angle is equal to the cosine of the complementary angle, and vice a versa.

You could also use the compound angle formulae to prove the above. Can you do that now?

Trigonometry

EXERCISE 16.3

Use compound angle formulae and the values of the trigonometric ratios of the special angles to prove the following identities:

1. ( ) θθ cos90sin =+

2. ( ) θθ sin90cos −=+

3. ( ) θθ sin270cos −=−

4. ( ) θθθ cossin45sin2 −=−

5. ( )BABABA

tantan1tantantan

−+

=+

6. A

AA 2tan1tan22tan

−=

7. ( )BABABA

tantan1tantantan

+−

=−

8. ( ) ( ) 145tan45tan =−+ DD

EXERCISE 16.4

By using the compound angle formulae as well as the surd forms of the trigonometric ratios of the special angles, evaluate the following:

1. Use the fact that 304515 −= to show that 221315sin −

= .

2. Use the fact that 304575 += to show that ( )2132175tan += .

EXERCISE 16.5

Without using a calculator, evaluate the following leaving you answers in surd form:

1. =15cos

2. =15sin

3. =15tan

4. =75cos

5. =75sin

6. =75tan


10

ROTATION OF A POINT ABOUT THE ORIGIN THROUGH AN ANGLE OF θ

Sketch x and y - axes in the space opposite. Mark a point A with coordinates ( )yx; in the first quadrant (but the point could be in any quadrant). Let the distance of A from the origin be r units. That is OA=r . Let OA make an angle of β with the positive direction of the x -axis. From A drop a perpendicular line to the x -axis. Looking at your diagram you have that:

cosrx

=β and sinry

=β

so that βcosrx = and βsinry =

Now A is rotated through an angle θ to a point ( )yx ′′′ ;A or the line OA to AO ′ . The direction of the

rotation could be clockwise or anti-clockwise. It does not matter which direction nor does the size of angle

θ but make it anti-clockwise and into the second quadrant.

In the space opposite, sketch both OA and AO ′on the

same set of axes showing the angles β and θ .

From A′ drop a perpendicular to the x -axis.

So you now will have

( )θβ +=′ cosrx and ( )θβ +=′ sinry

EXERCISE 16.6

Apply compound angle formulae to these last equations to show that

)sincos;sincos(');( θθθθ xyyxAyxA +−≡′′′

which gives the coordinates of A after rotation about the origin through an angle of θ .

Trigonometry

PAPER 2 QUESTION 3.3 DoE/ADDITIONAL EXEMPLAR 2008

PAPER 2 QUESTION 4 DoE/NOVEMBER 2008


12

PAPER 2 QUESTION 3.3 DoE/ADDITIONAL EXEMPLAR 2008

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

3.3.1 NOTE: This rule you must memorise because it has not always appeared on the information sheet given in the examination - see Exercise 16.6 on page 10.

Rotation about the origin through an angle θ :

)sincos;sincos(');( θθθθ xyyxAyxA +−→

θ is positive if the rotation is anti-clockwise; and negative if the rotation is clockwise.

NOTE: The rotations about the origin through 90

or 180 are obtained by substituting these angles for θ .

Answer:

+−

23

2;

23

2xyyx

3.3.2 Use the above result. Answer: ( )332;323 −−−

Paper 2 Question 4 DoE/November 2008

Number Hints and answers Work out the solutions in the box below

4 You know the rule now. Apply it taking into account that the rotation is clockwise and so has an implication for the sign of the angle of rotation.

Answer:

22;

225

or (3,54; 0,71)

Trigonometry

REDUCTION FORMULAE FOR ANGLES GREATER THAN 90

In the diagrams below showing the four quadrants, indicate and draw the size of angles in each

quadrant in terms of α , α−180 , α+180 and α−360 . Also indicate the trigonometric ratio that is positive in each quadrant.

-13 -12 -11 -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 1 2 3 4 5 6 7 8 9 10 11 12 13

-9

-8

-7

-6

-5

-4

-3

-2

-1

1

2

3

4

5

6

7

8

9

x

y

First Quadrant

-13 -12 -11 -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 1 2 3 4 5 6 7 8 9 10 11 12 13

-9

-8

-7

-6

-5

-4

-3

-2

-1

1

2

3

4

5

6

7

8

9

x

y

Second Quadrant

-13 -12 -11 -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 1 2 3 4 5 6 7 8 9 10 11 12 13

-9

-8

-7

-6

-5

-4

-3

-2

-1

1

2

3

4

5

6

7

8

9

x

y

ThirdQuadrant

-13 -12 -11 -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 1 2 3 4 5 6 7 8 9 10 11 12 13

-9

-8

-7

-6

-5

-4

-3

-2

-1

1

2

3

4

5

6

7

8

9

x

y

FourthQuadrant

NEGATIVE ANGLES

Negative angles are angles that are measured in a clockwise direction. Thus, if we measure the angle θ− from the origin of rectangular axes and taking the positive direction of the −x axis as a base line,

then the angle measured will fall in the 4th quadrant, assuming that θ is acute. In this quadrant cosine is positive and the other two trigonometric ratios are negative so that you can now complete the following:

( ) θθ sinsin −=− because sine is negative in the fourth quadrant

( ) =−θcos

( ) =θ-tan


14

In the above we have assumed that θ is acute. But θ can be any size. Example: )240sin( − = 240sin − Now reduce the angle 201 according to the reduction formulae. That means: ( )240sin − = 240sin−

( )

( ) quadrant 2nd in the is angle that theusing sin60--

60180sin

=

+−=

sin60 =

23 =

ANGLES = ( ) θ+90 We now show another way of reducing sines and cosines of ( )θ+90 .

Assume that θ+90 lies in the 2nd quadrant which is a reasonable assumption because we are adding to 90 . Then applying the reduction formula: ( ) ( )[ ] θθ +−+=+ 90180sin 90sin [ ] θ−−= 90180sin ( ) θ−= 90sin θcos = Can you complete the following? ( ) =+ 90cos θ = = =

Trigonometry

SUMMARY OF THE RESULTS Complete the following tables: Quadrant Angle size Reduction formula Reduction formula Reduction formula

II

( )α−180

sin ( ) =−α180

cos ( ) =−α180

tan ( ) =−α180

III

( )α+180

sin ( ) =+α180

cos ( ) =+α180

tan ( ) =+α180

IV

( )α−360

sin ( ) =−α360

cos ( ) =−α360

tan ( ) =−α360

sin ( ) =−α90 sin ( ) =+α90

cos ( ) =−α90 cos ( ) =+α90

tan ( )α

αtan

190 =−

tan ( ) =+α90

sin ( ) =− x

cos ( ) =− x

tan ( ) =− x

In the following, Ζ∈k :

( ) =±θ360.sin k

( )=±θ360.cos k

( ) =±θ360.tan k

EXERCISE 16.7 Write down the values of the following trigonometric ratios leaving your answer in surd form:

1. cos 150 =

2. tan 135 =

3. tan 330 =

4. cos 180 =

5. sin 240

6. cos 135

7. sin 510 =

8. cos 405 =

9. tan (-750 )

10. sin (-675 )

11. cos (-1140 )

12. tan 810


16

EXERCISE 16.8

Write each of the following as a trigonometric ratio of an acute angle, for example:

( ) 50cos130180cos130cos −=−−=

First, you must ask yourself: In what quadrant is angle 130 ? Then apply the appropriate reduction formula from the summary results shown above.

1. 175sin =

2. =313tan

3. 229cos

4. 328sin

5. 124tan

6. 196cos

7. 672sin

8. 500tan

9. 2108cos

10. ( )215sin −

11. ( )936tan −

12. ( )247cos −

EXERCISE 16.9

Without using a calculator, simplify each of the following (where possible leave answer in surd from):

1. =120sin

2. ( ) =− 120sin

3. =120tan

4. =225cos

5. ( )120cos −

6. =570sin

7. ( ) =− 135tan

8. =840sin

9. ( ) =− 120tan

10. =225cos2

11. ( ) =− 315sin2

12. =190sin

13. =100cos

14. =390tan

15. ( ) =− 258cos

16. =660cos

17. ( ) =− x90cos 18. ( ) =− x2180sin 19. ( ) =−− 180sin x 20. ( ) =−− 720cos x 21. ( ) =+ x90cos 22. ( ) =+ x540tan 23. ( ) =− 330tan2

Trigonometry

EXERCISE 16.10

Find the value of x between 0 and 90 in each of the following equations:

1. xsin60cos =

2. ( ) 40cos30sin =+ x

3. ( ) ( ) 45sin30cos2 +=+ xx

4. ( )60cos3cos += xx

5. ( )45sin2sin −= xx

EXERCISE 16.11

You are given that 2079,078cos = , calculate the following without using a calculator:

1. sin 12 =

2. cos 102 =

3. sin 102 =

4. sin 192 =


18

SOLVING TRIGONOMETRIC RATIOS WITHOUT USING A CALCULATOR

A calculator is not to be used for the solution of the problems below but drawing a sketch is essential in all cases.

EXERCISE 16.12

1. You are given that 53cos =α and that α is an acute angle, find the values of:

a. αsin

b. αtan

2. You are given that 257sin =φ and that φ is an obtuse angle, find the values of:

a. φcos

b. φtan

3. You are now given that 135cos =θ and that 360270 <≤ θ , find the values of:

a. θsin

b. θtan

4. If tx =sin where x is an acute angle, find the following:

a. xcos

b. xtan

5. If 135sin =α and

257cos =β and that the angles α and β are acute, find:

a. ( )βα +sin b. ( )βα −sin

c. ( )βα +cos d. ( )βα −cos

6. If 7248,055,43cos = , find using compound angle formulae the values of :

a. 55,223cos b. 45,136cos

Trigonometry

PAPER 2 QUESTION 4 DoE/ADDITIONAL EXEMPLAR 2008



20


Number Hints and answers Work out the solutions in the boxes below 4.1.1 Use the reduction formulae and surd

forms, for example,

( ) 30cos30360cos330cos =−=

2330cos =

cos (angle) = sin (complement of angle)

etc.

to simplify each and every term as far as possible.

Answer:

21

−

4.1.2 Use the reduction formulae, for example,

( ) xx 2sin2180sin =−

and also compound angle formula,

xxx cossin22sin =

and from and from magnitudes over 360

such as 720 take away multiples of 360

to reduce to a magnitude less than 360 .

to simplify each and every term as far as possible.

Answer: 1

4.2 Invoke special angles, for example,

( ) 3045sin15sin −=

then apply a compound angle formula and thereafter the surd forms.

Trigonometry


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

5.1.1 Use the reduction formulae and surd forms.

Note that 120480 ≡ . Why?

Answer:

23

5.1.2 Find a way of invoking the special

angles 30 , 45 and/or 60 .

Thereafter, use a compound angle formula.

Answer: ( )

4132 −

5.2 Use reduction formulae.


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GENERAL SOLUTIONS IN TRIGONOMETRY

In the formulae below, calc∠ stands for the angle given by the calculator.

A. If sin k=θ then 360calc ⋅+∠= nθ or ( ) Ζ∈⋅+∠−= nn ,360calc180 θ

B. If cos k=θ then 360calc ⋅+∠= nθ or ( ) Ζ∈⋅+∠−= nn ,360calc θ

C. If tan k=θ then Ζ∈⋅+∠= nn ,180calc θ

Example: Solve 174,02cos −=x for 180180 ≤≤− x .

Solution: calc 10002,100 ≅=∠

Therefore, 360.1002 nx += and 360.1002 nx +−=

180.50 nx +=

180.50 nx +−=

Solve these equations for =n . . . -2, -1, 0, 1, 2, . . and select values that lie in [ ] 360 ;360− .

Answer: 230 ;130 ;50 ;130 ;230 ;310 −−−=x

EXERCISE 16.13

Find

a) the general solutions and

b) the solutions that lie between 270− and 180

of the following equations:

1. Cos 21

=θ

2. tan θ = -1

3. sinθ =23

−

4. 2 cos θ2 - 1,2 = 0

5. tan 3

12 −=θ

6. cos θ sin =θ2 cos θ

7. 2 sin θ2 - sin θ = 1

8. 6 cos θ2 - cos θ = 2

Trigonometry

PROVING TRIGONOMETRIC IDENTITIES

We mentioned at the beginning of this unit that in proving identities, you have to show that the LHS (left hand side) reduces to the RHS (right hand side) or the RHS reduces to the LHS or that both LHS and RHS reduce to the same expression. EXERCISE 16.14

1. Prove that ( ) ( ) xxx o cos22145cos45cos o =−⋅+

2. 2.1 ( ) ( ) 2cossincossin 22 =−++ αααα 2.2 φφφ 244 cos21sincos =+− 3. 3.1 Prove that ( ) 145sin22sin 2 =−°+ xx

3.2 Hence deduce, without the use of a calculator, that 4

3215sin2 −=°

4. Given: θθ

θ sin2cos21

3sin=

+

4.1 Prove the above identity.

4.2 Without using a calculator, prove that the identity is not valid for .60=θ

5. 5.1 Prove that xxx

xx tan2coscos1

2sinsin=

+++

5.2 For what values of 𝑥 is the identity not valid if 𝑥 ∈ [0°; 180°].

6. Prove that xx

x tan2cos1

2sin=

+

7. Prove that θθθ tan

2sin2cos1

=−


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ANSWERS TO ALL THE EXERCISES

EXERCISE 16.1 1. HINT: Recall the factorization of algebraic expressions: a. 22 βα −

b. 22 2 βαβα +−

c. 22 2 βαβ ++a

d. 62 −+αα Satisfy yourself that you can factorise each of the above expressions by factorizing them . Then check if you have factorised correctly by multiplying the factors to see if you get back the expressions. Each of the trigonometric expressions in this exercise are of the same form as the algebraic expressions above. They thus factorise in the same way. Just let xcos=α and xsin=β and you will see the similarity. No answers are given because to do so would rob you of the opportunity to do the factorizations yourself. EXERCISE 16.2 1. HINT: If you cannot prove these formulae you can look them up in your textbook. 2. HINT: Split x3 into ( )xx 2+ and thereafter apply the compound angle formula for

( )BA +sin .

Answer: xxx 3sin4sin33sin −= HINT: A general HINT to solving identities: Identities have two sides: a left hand side (LHS) and a right hand side (RHS). Proving identities means showing that the two sides are identical even though they do not look alike. What you must do is to show that one side reduces to an expression that has the same look as the other side.

One side of an identity may already be so simple that it cannot be simplified further. You don’t choose this side to work on because there is nothing you can do to make it look more simple. You must choose the side that does not look simple, the side that looks complicated because this is the side you can unravel step by step until it looks like the side that is simple. However, there are identities in which neither side looks simple. In these cases, you must take each side in turn, manipulate it step by step until you cannot simplify it further. The two sides will reduce to expressions that look alike. The third problem in this exercise is fully worked out below in order to show you how to set out proving identities. 3. HINT: Choose to work on the LHS because you can expand it as shown below – it is the side to which you can do something! LHS = ( ) ( )θφθφ −+ sinsin = [ ][ ]φθθφφθθφ cossincossincossincossin −+

= φθθφ 2222 cossincossin −

= ( ) ( )φθθφ 2222 sin1sinsin1sin −−− =

φθθθφφ 222222 sinsinsinsinsinsin +−−

= θφ 22 sinsin − = RHS ( ) ( ) θφθφθφ 22 sinsinsinsin −=−+∴

As you will have noticed from the above, the answer to proving an identity is not a simple statement or a value. But the answer consists in working out in a logical step by step process to show that one side of an identity can be reduced to look like the other side; or both sides can be worked upon to show that they reduce to the

Trigonometry

same expression. For this reason, no answers are provided to the rest of the identity questions. Only hints are given. 4. HINT: On the RHS is a single term which is not only simple enough but cannot be simplified any further. You must therefore choose to work on the LHS to reduce it step by step till it is the same expression as on the RHS. 5. HINT: Neither side looks simpler than the other. So choose either side to work on and show it reduces to the other.

6. HINT: The RHS consists of a single simple term. Work on the LHS. The angle on the LHS is a double angle 2A whereas the one on the RHS is a single angle A. This suggests that cos2A must be replaced by its formula in terms of a single A and 1 will also have to be replaced. By what? 7. HINT: Again you have no option but to work on the LHS because the RHS is made up of a single term that is simple enough. Factorise the LHS as a difference of two squares. One of the factors will be yet another difference of two squares which you must factorise.

8. HINT: Apply some of the hints given above.

9. HINT: Apply some of the hints given above. EXERCISE 16.3:

1. HINTS: Generally as above 2. HINTS: Generally as above 3. HINTS: Generally as above 4. HINTS: Generally as above 5. HINT: As a general rule in simplifying

an expression that contains the ( )angletan ratio replace this ratio by ( )( )angleangle

cossin . So replace ( )BA +tan by

( )( )BA

BA++

cossin .

6. HINT: Write 2A as (A+A) and then replace B in Question 5 by A.

7. HINT: You can proceed as in Question 5 above, or you can look at A-B as A+(-B) and use the formula you derived in Question 5.

8. HINT: It’s time to leave you to your own devices to prove this identity!

EXERCISE 16.4 HINTS: To invoke the special angles you must make the following conversions: 304515 −= 304575 += EXERCISE 16.5

1. ( ) ( )4

132or 1322

1 ++

2. ( ) ( )4

132or 1322

1 −−

3. 1313

−+

4. ( ) ( )4

132or 1322

1 −−

5. ( ) ( )4

132or 1322

1 ++

6. 1313

−+

EXERCISE 16.6 HINT: A hint is given in the question telling you what to apply. After applying the compound angle formulae, you must replace βcosr and βsinr by x and y as obtained in the notes introducing the exercise (on page 103).


26

EXERCISE 16.7 1.HINT: Question 1 is fully worked out below and gives a hint using the reduction formulae as to how to solve Questions 2 to 6.

( )2330cos30180cos150cos −=−=−=

2. -1

3. 3

1−

4. -1

5. 23

−

6. 2

1−

7. HINT: First, you must reduce the angle by taking away multiples of 360 :

( ) 150sin3601510sin510sin =×−= Then use the reduction formulae:

( )2130sin30180sin150sin =+=−=

8. 2

1 9. 3

1− 10.

21

11. 21 12. Undefined.

EXERCISE 16.8 HINT: This has been given at the beginning of the exercise. Hints in the previous exercise also come into play. Work out each question in a logical step by step way. 1. 5sin 2. 47tan− 3. 49cos− 4. 32sin− 5. 56tan−

6. 16cos− 7. 48sin− 8. 40tan− 9. 52cos 10. 35sin 11. 36tan− 12. 67cos− EXERCISE 16.9

1. 23

2. 23

−

3. 3−

4. 2

1−

5. 21

−

6. 21

−

7. 1

8. 23

9. 3

10. 21

11. 21

12. 10sin− 13. 80cos−

14. 3

1

15. 78cos

16. 21

17. xsin 18. x2sin 19. xsin 20. xcos

Trigonometry

21. xsin− 22. xtan

23. 31

EXERCISE 16.10 The HINTS are given first and the answers follow below. 1. HINT: Use the sine and cosine of complementary angles. 2. HINT: Use the sine and cosine of complementary angles. 3. HINT: Use the compound angle formulae, substitute in the surd forms and simplify. 4. HINT: Use the compound angle formulae, substitute in the surd forms and simplify. 5. HINT: Use the compound angle formulae, substitute in the surd forms and simplify. ANSWERS: 1. 30=x 2. 20=x

3. 3198,302116tan 1 ≈=

+−

= −x

4. 89,1033

1tan 1 =

= −x

5. 675,7312

2tan 1 =

−= −x

EXERCISE 16.11 HINT: Deploy all the techniques you have learnt in the above exercises.

1. 0,2079 2. -0.2079 3. 0.2079 4. -0,2079

EXERCISE 16.12

1. HINT: By definition hypoyenuse

adjacent=αcos .

Draw a rough right angled triangle. Mark one of the angles as α .

We are given that hypotenuseadjacent

==53cosα .

So mark the side adjacent to angle α as 3 units in length and the hypotenuse as 5 units in length. Next, use Pythagoras’ Theorem to work out the length of the side opposite to angle α . Using the triangle you should get the following answers:

a) 54sin =α

b) 34tan =α

2. HINT: The angle φ in this case is obtuse and so you need to do a small rough sketch of rectangular axes and situate the angle so it overlaps into the second quadrant. Then do as above but using the definition of sine. You should get:

a) 2524cos −

=φ

b) 257tan −=φ

3. HINT: First determine in which quadrant to situate the angle θ and the right angled triangle. Then proceed as above to get the following answers:

a) 1312sin −=θ

b) 5

12tan −=θ


28

4. HINT: Refer to the definition:

hypotenuse

oppositex =sin

Write t in the form 1t

so the side opposite angle

x is t units in length and the hypotenuse is 1 unit in length. The answers become:

a) 1cos −= tx

b) 1

tan−

=t

tx

5. HINT: First, apply the appropriate compound angle formula. In this formula you know the values of αsin and

βcos because these are given. What we do not know are the values of βsin and

αcos . We must calculate these. To do so, roughly construct two right angles triangles.

In one you use hypotenuse

opposite==

135sinα

to label the sides and angle α appropriately. From this triangle, use Pythagoras’ Theorem to work out the adjacent side to angle α . You can now write down the value of αcos . In the other triangle, you use

hypotenuse

adjacent==

257cosβ

From the triangle, work out the side opposite to angle

β . You can now write down the value of βsin . You are now in a position to make substitutions into the compound formula of ( )βα +sin and calculate the value. Answers:

a. 325323

b. 325253

−

c. 32536

−

d. 325204

e. 36323

−

f. 204253

−

6. Answers a. 7248,055,223cos −=

b. 7248,045,136cos −=

EXERCISE 16.13 NOTE: The general solutions are not complete if you do not write Ζ∈n . 1 (a) 36060 ⋅+±= nθ , Ζ∈n (b) 60±=θ 2 (a) 18045 ⋅+−= nθ , Ζ∈n (b) 135 ;45 ;225 −−=θ 3 (a)

}360240{}36060{ ⋅+⋅+−∈ nnθ , Ζ∈n (b) 60 ;120 −−=θ

Trigonometry

NOTE how the above answer in (a) has been written in terms of the symbols and ∈ which come from set theory. stands for “union” and ∈ for “member of”. The brackets { . . } enclose a set. In the case of the first brackets, it means all the angles given by 36060 ⋅+− n for Ζ∈n and in the case of the second bracket it means all the angles given by 360240 ⋅+ n for Ζ∈n . In each case the angles form a set. Writing ∈θ { . . } { . . } for Ζ∈n means that θ takes the values of all the angles in each set for which

Ζ∈n . 4 (a) 18057,26 ⋅+±= nθ , Ζ∈n

43,153;57,26;57,26;43,153;57,206 −−−=θ 5 (a) 120⋅= nθ , Ζ∈n (b) 120 ;0 ;120 ;240 −−=θ 6 (a) 9015 ⋅+−= nθ , Ζ∈n (b) 165 ;75 ;15 ;105 ;195 −−−=θ 7 (a) 36090 ⋅+±= nθ or 18045 ⋅+= nθ ,

Ζ∈n (b) 90 ;45 ;90 ;135 −−=θ 8 (a) for all Ζ∈n

36090or 360150or 36030 ⋅++⋅+−= nnθ (b) 150 ;90 ;30 ;210 −−=θ 9 (a) for all Ζ∈n

( ) ( ) 36019,48360120 ⋅+±⋅+±∈ nnθ

(b) 120 ;19,48 ;19,48 ;120 ;240 −−−=θ 10 (a) for all Ζ∈n

36052,75or ;360180 ;360 ⋅+±⋅+⋅= nnnθ

(b) 180 ;52,75 ;0 ;52,75 ;180 −−=θ

EXERCISE 16.14 Identities were dealt with in Exercise 16.2 and a HINT given there as to how to solve identities. Only HINTS given in this exercise are for: 4.2 The identity is undefined if the denominator Is equal to zero. That is, 02cos21 =+ θ Solve this equation and see what you get. 5.1 90=x because xtan is undefined at this value of x .


30

PAPER 2 QUESTIONS 5 DoE/ADDITIONAL EXEMPLAR 2008

PAPER 2 QUESTIONS 6 DoE/NOVEMBER 2008

PAPER 2 QUESTIONS 5 DoE/PREPARATORY EXAMINATION 2008

Given the following identity: SinAA

AAAA

A −+

=+cos

cossin2cos2sin

2cos1

5.1 State the values of A for which the above identity is undefined. (2)

5.2 Hence prove the given identity. (6)[8]


6.1 Express cos x2 in terms of sin x . (1)

6.2 Without using a calculator, determine the general solution of 0sin2cos =+ xx (6)[7]

Trigonometry


Number Hints and answers Work out the solutions in the boxes below 5.1.1 Draw a right-angled triangle.

By definition

hypotenuseadjacent

1cos ===

ttx

Use the above definition of cosine and Pythagoras theorem to work the units in each length of the triangle.

Answer: t

tx −=

1tan

5.1.2 Use compound angle formula on x2sin .

xcos is given as t . Use the triangle in 5.1.1 above to work out xsin . Substitute for xcos and xsin in the formula for x2sin .

Answer: 22 tt −

5.2.1 When solving identities always choose the more complicated side to work on. The side that is simple cannot be simplified further.

So put xx

xxLHS 22 cossin1cossin+−

=

And simplify as far as possible using appropriate compound angle and reduction formulae.

5.2.2 Compare with 5.2.1 above and you will see that what in effect you have is

0tan21

=x

which is a simple equation. Answer: Ζ∈+= kkx ;180.0


32


Number Hints and answers Work out the solutions in the boxes below 6.1.1 Neither the LHS nor the RHS of the identity

looks simple. So choose any side and work on it to show that it is identical with the other side. Or, work in turn on each side and show that the sides simplify to the same expression. When proving identities, always replace tan x

by xx

cossin

.

You may need to replace 1 by xx 22 cossin + , or vice versa. Factorise, multiply out brackets, sum up fractions, do any or all of these procedures, whichever is appropriate, to simplify in order to prove identities.

6.1.2 Isolate tan x , that is do a transposition so that you have tan x standing alone on the LHS. Solve for x using the inverse tan key on your calculator. Express the answer given by the calculator as a general solution. Answer:

180.3,281 kx += or 180.3,101 kx +=

6.2.1 In order to relate β to its correct quadrant, your diagram must take into account the sign of cos β which will come from the given sign of p and also the given range of β .

Answer: p

p 2-5-tan =β

6.2.2 Quickest to use the formula of cos 2 β in terms of cos β .

Answer: 5

52cos22 −

=pβ

Trigonometry


Number Hints and answers Work out the solutions in the boxes below 5.1 Division by zero is not permitted. Thus the

identity is undefined if the denominator on either side is equal to zero. The values of A for which the identity is undefined are given by solutions to the two equations:

1. cos 2A =0 2. cos A – Sin A =0 or tan A = 1 (why?)

Obtain general solutions to these equations. Answer: the two general solutions boil down to one general solution (how come?) which is A = Ζ∈+ kk ,90.45

5.2 Neither side looks simple. So see if you can work on each of the sides to reduce them to the same expression.


Number Hints and answers Work out the solutions in the boxes below 6.1 Reducing the x2 on the LHS to a single x

that is on the RHS will require use of a compound angle formula. Do just that and then finish off with an expression that is in terms of sin x only. Answer: cos x2 = 1 – 2sin x2

6.2 Use the answer to 6.1 to replace cos x2 . What you get is a quadratic equation in sin x . If you cannot see this then put w in place of sin x and you will see a quadratic equation in w . Solve this equation. Answers: 360.210 kx +=

360.90 kx += and 360.330 kx +=


34

MORE QUESTIONS FROM PAST EXAMINATION PAPERS

Exemplar 2008

Trigonometry

Preparatory Examination 2008

QUESTIONS 4

Answer all questions without using a calculator. Show ALL calculations. 4.1 If a=βsin and 18090 << β , determine using a sketch the value of βtan in terms of a . (4) 4.2 Simplify the following:

( )

300sin135cos215sin −+

(7)

4.3 Simplify the following expression to one trigonometric ratio of θ :

( ) ( ) ( )

( )

315tan360sin90cos180sin.sin

−−−++−−

θθθθ

(6)[17]


36

Feb – March 2009

Trigonometry

November 2009 (Unused paper)


38

November 2009(1)

Trigonometry

Feb – March 2010


40

ANSWERS

Exemplar 2008

4.1 3

2−

4.2 1 4.3 Proof required. Provide a proof and check with the teacher if it is correct. 4.4.1 Proof required. Provide a proof and check with the teacher if it is correct. 4.4.2 Ζ∈⋅+±= kk ;36060 θ 5.1 She used an incorrect expansion of ( )BA +sin

5.2 ( )2122 tt −+

6.1 Proof required. Provide a proof and check with the teacher if it is correct. 6.2 Proof required. Provide a proof and check with the teacher if it is correct. 6.3 ( )aaC −; Preparatory Examination 2008

4.1 21

tana

a−

−=β

4.2 32

623−

+ or

( )32

332−

+

4.3 θsin− Feb/March 2009 5.1 xsin− 5.2 xtan 6.1.1 p−=113cos

6.1.2 2123cos p−=

6.1.3 21246sin pp −=

6.2.1 1312cos −=α

6.2.2 ( )6533cos =+ βα

6.3 54,29=x or 46,330 November 2009 (Unused papers)

5.1.1 125tan −=θ

5.1.2 16960sincos −=θθ

5.2 xtan 5.3 1 5.4 36008,209 ⋅+= kx

36092,330 ⋅+= kx

36092,150 ⋅+−= kx or 36008,29 ⋅+−= kx Ζ∈k 5.5 60=x or 240=x

6.1.1 2124cos p−=

6.1.2 2

3p

6.2 Proof required. Provide a proof and check with the teacher if it is correct

Trigonometry

November 2009(1)

8.1 158tan −=α

8.2 ( )171590sin −=+α

8.3 2891612cos =α

9.1 -1

9.2 3

1−

9.3 36030 ⋅+= kx ; Ζ∈k There are other possible forms of the answer. 10.1 Proof required. Provide a proof and check with the teacher if it is correct. 10.2 Proof required. Provide a proof and check with the teacher if it is correct.

Feb/March 2010

9.1.1 57cossin −=+ θθ

9.1.2 7242tan =θ

9.2.1 Proof required. Provide a proof and check with the teacher if it is correct.

9.2.2 3

2

10.1.1 qp +=48sin

10.1.2 qp −=24sin

10.1.3 ( )qpqp−+

=2

24cos

or

( )

+−+= 211

2124cos qp

or

( )2124cos qp −−= 10.2 Proof required. Provide a proof and check with the teacher if it is correct. 10.3.1 Proof required. Provide a proof and check with the teacher if it is correct. 10.3.2 Undefined for 360180 ⋅+= kx , Ζ∈k

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Unit 15: Trigonometry

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