Trigonometry

MATHEMATICS

TRIGONOMETRY

Revision Notes, Exercises and Solution Hints by

Roseinnes Phahle Examination Questions by the Department of Basic of Basic Education

1

Preparation for the Mathematics examination brought to you by Kagiso Trust

Contents Unit 16 Factorising trigonometric expressions Compound angle formulae Introducing trigonometric identities Special angles The sine and cosine of complementary of complementary angles Rotation of a of a point about the origin Examination questions with solution hints and answers o

Reduction formulae angles greater than 90 Negative angles Summary of results of results Solving trigonometric ratios or triangles without using a calculator Examination questions with solution hints and answers General solutions in trigonometry Proving trigonometric identities Answers to all the exercises Examination questions with solution hints and answers More questions from past examination papers Answers to past examination papers

3 3 4 5 8 10 11 13 13 15 18 19 22 23 28 24 34 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 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 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 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.

2

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 :

− y 2 = ( x − y )( x + y ) 2 Some trinomials: x + 3 x + 2 = ( x + 1)(x + 2 )

Differences of two of two squares: x

2

EXERCISE 16.1 Factorise the following (if you‘re (if you‘re at a loss as to what to do look up the answer for a HINT): a.

cos 2 x − sin 2 x =

b.

cos 2 x − 2 cos x sin x + sin 2 x =

c.

cos 2 x + 2 cos x sin x + sin 2 x =

d.

sin 2 x + sin x − 6 =

COMPOUND ANGLE FORMULAE Complete the following formulae: sin (A+B) =

cos (A+B) =

sin (A‐B) =

cos (A‐B) =

cos x = 2

cos x + sin x = 2

2

sin 2 x = Factorise as a difference of two of two squares:

Factorise as a difference of two of two squares:

1 − cos x = (

1 − sin 2 x = (

2

)(

)

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

tan x =

)(

)


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. b.

sin 2 x + cos 2 x = 1 sin x tan x = cos x

2. Write sin 3 x in terms of sin x 3.

sin (φ + θ )sin (φ − θ ) = sin 2 φ − sin 2 θ

4.

cos( A + B) − cos( A − B) = −2 sin A sin B

5.

cos 2 x + cos 2 x = 2 − 3 sin 2 x

6.

7.

8.

9.

1 − cos 2 A 1 + cos 2 A

= tan 2 A

cos 4 x − sin 4 x = cos 2 x sin 2 A 1 + cos 2 A

= tan A

sin θ + sin 2θ 1 + cos θ + cos 2θ

= tan θ

4

Trigonometry

SPECIAL ANGLES: 0 o ; 30 o ; 45 o ; 60 o ; and 90 o In order to work out the type of problems such as Questions 3.3 and Question 4 (see page 112, 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

As shown in the diagram below, we draw an equilateral triangle We can make the

60 o and 30 o :

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 o

o

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


Special angles

Diagrams to be used 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:

x2 + 12 = 2 x =

= Therefore

x =

Leaving your answers in surd form, you can now refer to the last triangle to complete the table below:

sin 30o =

sin 60o =

cos 30o =

cos 60o =

tan 30o =

tan 60o =

6

Trigonometry

Special angles

o

For angle 45

Diagrams to be used

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):

x 2 =

= Therefore

x =

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

sin 45o =

cos 45o =

tan 45o =


Angles 0o and 90o o

o

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.

SUMMARY OF THE RESULTS Summarise all of the above results by completing the table below:

sin 0

=

o

cos 0

=

o

=

tan 30

o

=

tan 45

cos 60

o

=

tan 60

0

=

tan 90

cos 30

o

cos 45

o

sin 45 =

sin 60 =

sin 90

0

=

=

o

o

o

sin 30 =

tan 0

cos 90

o

=

o

=

o

=

0

=

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

sin θ = cos(90 o − θ ) and

cos θ = sin (90 o − θ )

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? 8

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.

2.

3.

4.

sin (90 o + θ ) = cos θ

(

tan ( A + B ) =

6.

tan 2 A =

7.

tan ( A − B ) =

8.

tan 45 + D tan 45 − D = 1

)

cos 90 + θ = − sin θ o

(

cos θ − 270

o

(

1 − tan A tan B

2 tan A 1 − tan 2 A

) = − sin θ

2 sin (θ − 45 o ) = sin θ − cos θ

tan A + tan B

5.

tan A − tan B 1 + tan A tan B

) (

o

o

)

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 15

o

2. Use the fact that 75

o

= 45o − 30o to show that sin 15o =

= 45 o + 30 o to show that tan 75 o =

3 −1 2 2 1

( 2

.

)

2

3 +1 .

EXERCISE 16.5 Without using a calculator, evaluate the following leaving you answers in surd form:

1.

cos 15 o =

4.

cos 75 o =

2.

sin 15 o =

5.

sin 75 o =

3.

tan 15 o =

6.

tan 75 o =


ROTATION OF A POINT ABOUT THE ORIGIN THROUGH AN ANGLE OF θ o Sketch x and y ‐ axes in the space opposite. Mark a point A with coordinates ( x; y ) 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 . o

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: cos β =

x r

y

and

sin β =

and

y = r sin β

r

so that

x = r cos β

o

Now A is rotated through an angle θ to a point

A′( x′; y ′) or the line OA to OA′ . The direction of the rotation could be clockwise or anti‐clockwise. It does not matter which direction nor does the size of angle

θ o but make it anti‐clockwise and into the second quadrant. In the space opposite, sketch both OA and OA ′ on the o

o

same set of axes showing the angles β and θ . From A ′ drop a perpendicular to the x ‐axis. So you now will have

x ′ = r cos( β + θ ) and y ′ = r sin ( β + θ )

EXERCISE 16.6 Apply compound angle formulae to these last equations to show that

A′( x ′; y ′) ≡ A' ( x cos θ − y sin θ ; y cos θ + x sin θ ) o

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

10

Trigonometry

PAPER 2 QUESTION 3.3

PAPER 2 QUESTION 4

DoE/ADDITIONAL EXEMPLAR 2008

DoE/NOVEMBER 2008


PAPER 2 QUESTION 3.3 Number

DoE/ADDITIONAL EXEMPLAR 2008 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 Example 16.5 on page 113. Rotation about the origin through an angle θ :

A( x; y ) → A' ( x cos θ − y sin θ ; y cos θ + x sin θ )

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

NOTE: The rotations about the origin through 90

o

o

or 180 are obtained by substituting these angles for θ . Answer:

⎛ x 3 3 ⎞ ⎜ − y ; y + x ⎟ ⎜2 2 2 2 ⎠⎟ ⎝ 3.3.2

Use the above result.

Answer:

− 3 − 2 3 ;2 − 3 3 Paper 2 Question 4 Number

DoE/November 2008 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:

⎛ 5 2 2 ⎞ ⎜ ⎟ ⎜ 2 ; 2 ⎟ or (3,54; 0,71) ⎝ ⎠

12

Trigonometry

REDUCTION FORMULAE FOR ANGLES GREATER THAN 90 o In the diagrams below showing the four quadrants, indicate and draw the size of angles in each quadrant in terms of α , 180

− α , 180 o + α and 360o − α . Also indicate the trigonometric ratio that

o

is positive in each quadrant.

y

y

9

9

8

8

7

7

Second Quadrant

First Quadrant

6

5

4

6

5

4

3

3

2

2

1

1

x -13

-12

-11

-10

- 9

-8

-7

-6

-5

-4

- 3

-2

-1

1

2

3

4

5

6

7

8

9

10

11

1 2

x

13

-13

-12

-11

-10

-9

- 8

-7

-6

-5

-4

- 3

-2

-1

1

-1

-1

-2

-2

-3

-3

-4

-4

-5

-5

-6

-6

-7

-7

-8

-8

-9

-9

y

2

3

4

5

6

7

8

9

7

8

9

1 0

10

11

12

y

9

9

8

8

7

7

6

6

5

5

4

4

3

3

2

2

1

1

x

x -13

-12

-11

-10

-9

-8

-7

- 6

-5

-4

- 3

-2

-1

Third Quadrant

13

1

2

3

4

5

6

7

8

9

10

11

12

13

- 13

- 12

-11

-10

-9

- 8

-7

- 6

-5

- 4

-3

- 2

-1

1

-1

-1

-2

-2

-3

-3

-4

-4

-5

-5

-6

-6

-7

-7

-8

-8

-9

-9

2

3

4

5

6

11

12

13

Fourth Quadrant

NEGATIVE ANGLES Negative angles are angles that are measured in a clockwise direction. Thus, if we measure the angle

− θ o from the origin of rectangular axes and taking the positive direction of the x − axis as a base line, o

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:

sin (− θ ) = − sin θ because sine is negative in the fourth quadrant cos(− θ ) = tan(- θ ) =


In the above we have assumed that θ is acute. But θ can be any size.

Example:

sin( −240 o ) =

− sin 240 o o

Now reduce the angle 201 according to the reduction formulae. That means:

(

sin − 240 o

)

− sin 240 o

=

= − sin (180 o + 60 o )⎫⎪ ⎬ using that the angle is in the 2nd quadrant ⎪⎭ = - (- sin60 o ) = sin60 o

=

3 2

ANGLES = (90 o + θ o )

(

We now show another way of reducing sines and cosines of 90 Assume that 90

o

o

+ θ ) .

+ θ o lies in the 2nd quadrant which is a reasonable assumption because we are adding

o

to 90 . Then applying the reduction formula:

sin (90 o + θ o ) = + sin 180 o − (90 o + θ o )

= sin 180 o − 90 o − θ o = sin (90 o − θ o ) = cosθ o Can you complete the following?

(

)

cos 90 o + θ o = = = =

14

Trigonometry

SUMMARY OF THE RESULTS Complete the following tables:

Quadrant

Angle size

II

(180

o

− α )

sin 180

(

o

− α ) =

cos 180

III

(180

o

+ α )

sin 180

(

o

+ α ) =

cos 180

IV

(360

− α ) sin (360 o − α ) =

cos 360

(

o

(

o

sin 90 sin 90

o

− α ) =

Reduction formula

+ α ) =

(

(

o

(

o

cos 90

− α ) =

Reduction formula

− α ) =

tan 180

(

o

− α ) =

o

+ α ) =

tan 180

(

o

+ α ) =

o

− α ) =

tan 360

(

o

− α ) =

o

(

(

cos 90

Reduction formula

(

o

(

o

tan 90

+ α ) =

tan 90

sin

(− x ) =

cos

(− x ) =

tan

1

− α ) =

tanα

+ α ) =

(− x ) =

In the following, k ∈ Ζ :

(

)

sin k .360 o ± θ =

(

cos k .360 o ± θ =

)

tan (k .360 o ± θ ) =

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

o

5. sin 240

o

6. cos 135

o

7. sin 510 =

o

8. cos 405 =

1. cos 150 = 2. tan 135 = 3. tan 330 = 4. cos 180 =

o o

o

o

o

9. tan (‐750 ) o

10. sin (‐675 ) o

11. cos (‐1140 ) 12. tan 810

o


EXERCISE 16.8 Write each of the following as a trigonometric ratio of an acute angle, for example:

cos130 o = − cos(180 o − 130 o ) = − cos 50 o o

First, you must ask yourself: In what quadrant is angle 130 ? Then apply the appropriate reduction formula from the summary results shown above. 1.

sin 175 o =

7.

sin 672 o

2.

tan 313 o =

8.

tan 500 o

3.

cos 229 o

9.

cos 2108 o

4.

sin 328 o

10. sin − 215

o

5.

tan 124 o

11.

o

6.

cos 196 o

12.

( ) tan ( − 936 ) cos( − 247 ) o

EXERCISE 16.9 Without using a calculator, simplify each of the following (where possible leave answer in surd from):

(

)

1.

sin 120o =

9.

2.

sin (− 120o ) =

10.

cos 2 225o =

3.

tan 120o =

11.

sin 2 (− 315o ) =

tan − 120o =

(

(

cos 225o =

(

5.

cos − 120o

6.

sin 570 =

7.

8.

12.

)

(− x − 180 ) = o

sin 190o = 20. cos

(− x − 720 ) = o

13.

cos100o =

14.

tan 390 =

tan − 135o =

15.

cos(− 258o ) =

22. tan 540

sin 840o =

16.

cos 660o =

23. tan

o

(

)

− 2x) =

o

18. sin 180

19. sin 4.

− x ) =

o

17. cos 90

(

21. cos 90

o

16

( 2

o

+ x ) = o

+ x ) =

(− 330 ) = o

Trigonometry

EXERCISE 16.10 o

o

Find the value of x between 0 and 90 in each of the following equations: 1.

cos 60 o = sin x

2.

sin 30 o + x = cos 40 o

3.

2 cos( x + 30 o ) = sin (x + 45o )

4.

cos x = 3 cos x + 60 o

5.

sin x = 2 sin x − 45o

(

)

(

)

(

)

EXERCISE 16.11 You are given that cos 78

o

o

= 0,2079 , calculate the following without using a calculator:

o

1. sin 12 =

3. sin 102 =

o

4. sin 192 =

2. cos 102 =

o


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 cos α = a.

sin α

b.

tan α

2. You are given that sin φ = a.

cos φ

b.

tan φ

3 5

7 25

and that α is an acute angle, find the values of:

and that φ is an obtuse angle, find the values of:

3. You are now given that cos θ = a.

sin θ

b.

tan θ

4. If sin x =

13

and that 270

o

≤ θ < 360 o , find the values of:

t where x is an acute angle, find the following:

a.

cos x

b.

tan x

5. If sin α =

5

5 13

and cos β =

7 25

and that the angles α and β are acute, find:

a. sin (α + β )

b. sin (α − β )

c. cos(α + β )

d. cos(α − β )

6. If cos 43,55

o

= 0,7248 , find using compound angle formulae the values of :

a. cos 223,55

o

b. cos 136,45

18

o

Trigonometry

PAPER 2 QUESTION 4


PAPER 2 QUESTION 5

DoE/NOVEMBER 2008


PAPER 2 QUESTION 4 Number 4.1.1


Hints and answers Use the reduction formulae and surd


forms, for example,

cos 330 o = cos(360 o − 30 o ) = cos 30 o cos 30 o =

3 2

cos (angle) = sin (complement of angle) etc. to simplify each and every term as far as possible.

Answer:

− 4.1.2

1 2

Use the reduction formulae, for example,

(

)

sin 180 o − 2 x = sin 2 x and also compound angle formula,

sin 2 x = 2 sin x cos x and from and from magnitudes over 360 o

such as 720 take away multiples of 360

o o

o

to reduce to a magnitude less than 360 . to simplify each and every term as far as possible.

4.2

Answer: 1 Invoke special angles, for example,

(

sin 15 o = sin 45 o − 30 o

)

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

20

Trigonometry


DoE/NOVEMBER 2008

Hints and answers Use the reduction formulae and surd forms. Note that 480

≡ 120 o . Why?

o

Answer:

3 2 5.1.2

Find a way of invoking the special o

o

o

angles 30 , 45 and/or 60 . Thereafter, use a compound angle formula.

Answer:

2

(

)

3 −1 4

5.2

Use reduction formulae.



GENERAL SOLUTIONS IN TRIGONOMETRY In the formulae below, calc ∠ stands for the angle given by the calculator. A. If sin θ = k then θ = calc∠ + n ⋅ 360

o

B. If cos θ = k then θ = calc∠ + n ⋅ 360

o

(

)

θ = 180 o − calc∠ + n ⋅ 360 o , n ∈ Ζ

or

θ = −(calc∠) + n ⋅ 360 o , n ∈ Ζ

or

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

Example: Solve cos 2 x = −0,174 for − 180 ≤ x ≤ 180 . o

Solution: calc ∠

= 100,02 o ≅ 100 o

2 x = 100 o + n.360 o

Therefore,

o

2 x = −100 o + n.360 o

and

x = 50 o + n.180 o

Solve these equations for n

Answer: x

=.

x = −50 o + n.180 o

. . ‐2, ‐1, 0, 1, 2, . . and select values that lie in

= −310 o ; − 230 o ; − 130 o ; 50 o ; 130 o ; 230 o

EXERCISE 16.13 Find a) the general solutions and b) the solutions that lie between

− 270 o and 180 o

of the following equations:

1. Cos θ =

1

5. tan 2θ =

2

1 3

6. cos θ sin 2θ = cos θ

2. tan θ = ‐1 3. sin θ = −

−

3

2 4. 2 cos 2θ ‐ 1,2 = 0

2

7. 2 sin θ ‐ sin θ = 1 2

8. 6 cos θ ‐ cos θ = 2

22

− 360 o ; 360 o .

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

(

o

+ x ) ⋅ cos(45 o − x ) =

1

1.

Prove that cos 45

2.

2.1

(sin α + cos α )2 + (sin α − cos α )2 = 2

2.2

cos 4 φ − sin 4 φ + 1 = 2 cos 2 φ

3.1

Prove that sin 2 x + 2 sin (45° − x ) = 1

3.2

Hence deduce, without the use of a calculator, that sin 15°

3.

4.

Given:

4.1

2

cos2 x

2

2

sin 3θ 1 + 2 cos 2θ

=

2− 3 4

= sin θ

Prove the above identity. o

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

5.

sin x + sin 2 x

= tan x

5.1

Prove that

5.2

For what values of  is the identity not valid if   0°; 180°.

6.

Prove that

7.

Prove that

sin 2 x 1 + cos 2 x 1 − cos 2θ sin 2θ

1 + cos x + cos 2 x

= tan x = tan θ


ANSWERS TO ALL THE EXERCISES EXERCISE 16.1 1. HINT: Recall the factorization of algebraic expressions:

− β 2 2 2 b. α − 2αβ + β 2 2 c. a + 2αβ + β 2 d. α + α − 6 2

a. α

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.

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.

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.

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 α = cos x and β = sin x and

The third problem in this exercise is fully worked out below in order to show you how to set out proving identities.

you will see the similarity.

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!

No answers are given because to do so would rob you of the opportunity to do the factorizations yourself.

LHS = sin (φ + θ ) sin(φ − θ ) =

EXERCISE 16.2 1. HINT: If you cannot prove these formulae you can look them up in your textbook.

[sin φ cosθ + sin θ cosφ ][sin φ cosθ − sin θ cosφ ] = sin φ cos θ − sin θ cos φ 2

2

2

(

) − sin θ (1 − sin φ )

= sin φ 1 − sin θ 2

2. HINT: Split 3 x into ( x + 2 x ) and thereafter

2

2

2

2

=

apply the compound angle formula for

sin 2 φ − sin 2 φ sin 2 θ − sin 2 θ + sin 2 θ sin 2 φ

sin ( A + B ) .

= sin φ − sin θ 2

Answer: sin 3 x = 3 sin x − 4 sin x 3

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.

2

= RHS

∴ sin (φ + θ )sin (φ − θ ) = sin 2 φ − sin 2 θ 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 24

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.

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: 15 o = 45 o − 30 o 75 o = 45 o + 30 o EXERCISE 16.5 1. 2. 3. 4.

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 tan(angle ) ratio replace this ratio by

sin (angle ) cos(angle ) sin ( A + B ) cos( A + B )

. So replace tan( A + B ) by .

5.

6.

1 2 2 1 2 2

)

2

)

2

)

2

)

2

(

3 + 1 or

(

3 − 1 or

(

)

3 +1 4

(

)

3 −1 4

3 +1 3 −1 1 2 2 1 2 2

(

3 − 1 or

(

3 + 1 or

(

)

3 −1 4

(

)

3 +1 4

3 +1 3 −1

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 r cos β and r sin β by

x and y as obtained in the notes introducing the exercise (on page 103).


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.

(

)

cos 150 o = cos 180 o − 30 o = − cos 30 o = −

3 2

2. ‐1

6. −

o

− cos 67 o

1.

3 2

2.

−

2

3.

− 3

4.

−

o

(

)

Then use the reduction formulae:

(

)

sin 150 o = sin 180 o − 30 o = + sin 30 o =

1

9. −

2 1

1 3

10.

1 2

1

5.

−

6.

−

7. 8. 9.

2

12. Undefined.

2

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.

10. 11.

− tan 47 o 3. − cos 49 o 4. − sin 32 o 5. − tan 56 o

2

1 1 2 1 2

1

3 2

3 1 2 1 2

− sin 10 o o 13. − cos 80 12.

14.

1

3 o 15. cos 78 1 16.

o

3

2

sin 510 o = sin 510 o − 1 × 360 o = sin 150 o

2.

11. − tan 36

2 1

1. sin 5

o

3

away multiples of 360 :

11.

10. sin 35

EXERCISE 16.9

7. HINT: First, you must reduce the angle by taking

8.

o

3

4. ‐1 5. −

9. cos 52

12.

1

3. −

− cos 16 o o 7. − sin 48 o 8. − tan 40 6.

17. 18. 19. 20.

26

2 sin x sin 2 x sin x cos x

Trigonometry

21. − sin x 22. tan x

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

1

23.

We are given that cosα =

3

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:

5

=

adjacent hypotenuse

.

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)

sin α =

b)

tan α =

4 5 4 3

2. HINT: The angle φ in this case is obtuse and so

= 30 o o 2. x = 20 1. x

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:

⎛ 6 − 1 ⎞ ⎟ = 30,98 o ≈ 31o = tan −1 ⎜⎜ ⎟ ⎝ 1 + 2 ⎠ −1 ⎛ 1 ⎞ 4. x = tan ⎜⎜ ⎟⎟ = 10,89 o ⎝ 3 3 ⎠ ⎛ 2 ⎞ −1 o ⎟ 5. x = tan ⎜ ⎜ 2 − 1 ⎟ = 73,675 ⎝ ⎠ 3. x

− 24

a)

cos φ =

b)

tan φ = −

25

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 adjacent hypoyenuse

b) tan θ = − .

7 25

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) sin θ = −

1. HINT: By definition cos α =

3

12 13 12 5


4. HINT: Refer to the definition:

sin x = Write

x is

β .

opposite

You can now write down the value of sin β .

hypotenuse

t in the form

t

1

You are now in a position to make substitutions so the side opposite angle

t units in length and the hypotenuse is 1 unit

in length.

b)

calculate the value.

Answers: a.

The answers become: a)

into the compound formula of sin (α + β ) and

cos x = t − 1 tan x =

5. HINT: First, apply the appropriate compound angle formula. In this formula you know the values of sin α and because these are given. cos β What we do not know are the values of sin β and

−

c.

−

13

=

325 204

325 323

e.

−

f.

−

36 253 204

= −0,7248 o b. cos 136,45 = −0,7248 a. cos 223,55

To do so, roughly construct two right angles triangles.

5

d.

325 36

6. Answers

cos α . We must calculate these.

In one you use sinα =

325 253

b.

t t − 1

323

opposite

o

EXERCISE 16.13

hypotenuse

NOTE: The general solutions are not complete if you do not write n ∈ Ζ .

to label the sides and angle α appropriately. 1 (a) θ = From this triangle, use Pythagoras’ Theorem to work out the adjacent side to angle α .

2 (a) θ = −45

+ n ⋅ 180 o , n ∈ Ζ o o o (b) θ = −225 ; − 45 ; 135

You can now write down the value of cos α .

7 25

=

o

3 (a)

In the other triangle, you use

cos β =

±60 o + n ⋅ 360 o , n ∈ Ζ o (b) θ = ±60

θ ∈ {−60 o + n ⋅ 360 o } U {240 o + n ⋅ 360 o } , n∈Ζ

adjacent hypotenuse

From the triangle, work out the side opposite to angle

28

(b) θ = −120 ; o

− 60 o

Trigonometry

NOTE how the above answer in (a) has been written in terms of the symbols U and ∈ which come from set theory. U 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 − 60 + n ⋅ 360 for n ∈ Ζ and in the case of the second bracket it means all the angles o

given by 240

o

Writing θ ∈ { . . } U { . . } for n ∈ Ζ means that θ takes the values of all the angles in each set for which n∈Ζ.

+ n ⋅ 180 o , n ∈ Ζ θ = −206,57 o ;−153,43o ;−26,57 o ;26,57 o ;153,43o o

n∈Ζ

o

(b) θ = −240 ; o

− 120 o ; 0 o ; 120 o

6 (a) θ =

−15 o + n ⋅ 90 o , n ∈ Ζ o o o o o (b) θ = −195 ; − 105 ; − 15 ; 75 ; 165

7 (a) θ =

n∈Ζ

±90 o + n ⋅ 360 o or θ = 45 o + n ⋅ 180 o ,

(b) θ = −135 ;

− 90 o ; 45o ; 90 o

o

8 (a) for all n ∈ Ζ

θ = −30 o + n ⋅ 360 o or 150 o + 360 o or 90 o + n ⋅ 360 o o o o o (b) θ = −210 ; − 30 ; 90 ; 150 9 (a) for all n ∈ Ζ

(

) (

θ ∈ ± 120 o + n ⋅ 360 o U ± 48,19 + n ⋅ 360 o

)

(b) θ = −240 ; − 120 ; − 48,19 ; 48,19 ; 120 o

o

o

o

o

10 (a) for all n ∈ Ζ

θ = n ⋅ 360 o ; 180 o + n ⋅ 360 o ; or ± 75,52 o + n ⋅ 360 o (b) θ = −180 ; − 75,52 ; 0 ; 75,52 ; 180 o

o

o

4.2 The identity is undefined if the denominator Is equal to zero. That is,

1 + 2 cos 2θ = 0

In each case the angles form a set.

5 (a) θ = n ⋅ 120 ,

Only HINTS given in this exercise are for:

o

+ n ⋅ 360 o for n ∈ Ζ .

4 (a) θ = ±26,57

EXERCISE 16.14 Identities were dealt with in Exercise 16.2 and a HINT given there as to how to solve identities.

o

o

Solve this equation and see what you get. 5.1 x = 90 because tan x is undefined at this value of x . o


PAPER 2 QUESTIONS 5


PAPER 2 QUESTIONS 6

DoE/NOVEMBER 2008

PAPER 2 QUESTIONS 5 Given the following identity:

DoE/PREPARATORY EXAMINATION 2008

1 cos 2 A

+

sin 2 A cos 2 A

=

sin A + cos A cos A − SinA

5.1

State the values of A for which the above identity is undefined.

5.2

Hence prove the given identity.

(2) (6)[8]

PAPER 2 QUESTIONS 6


6.1

Express cos 2 x in terms of sin x .

6.2

Without using a calculator, determine the general solution of cos 2 x + sin x

(1)

30

=0

(6)[7]

Trigonometry



Hints and answers Draw a right‐angled triangle. By definition

cos x = t =

t

1

=

adjacent hypotenuse

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

Answer: tan x 5.1.2

1 − t

=

t

Use compound angle formula on sin 2 x .

cos x is given as

t . Use the triangle in 5.1.1 above to work out sin x . Substitute for cos x and sin x in the formula for sin 2 x .

Answer: 2 t − t When solving identities always choose the more complicated side to work on. The side that is simple cannot be simplified further. 2

5.2.1

So put LHS =

5.2.2

sin x cos x 1 − sin 2 x + cos 2 x

And simplify as far as possible using appropriate compound angle and reduction formulae. Compare with 5.2.1 above and you will see that what in effect you have is

1 2

tan x = 0

which is a simple equation.

Answer: x

= 0 o + k .180 o ; k ∈ Ζ




DoE/NOVEMBER 2008

Hints and answers Neither the LHS nor the RHS of theidentity looks simple. So choose any side and work on it to show that it is identical with the other 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

sin x cos x

.

You may need to replace 1 by sin x + cos x , 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. 2

6.1.2

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: o o o o x = 281,3 + k .180 or x = 101,3 + k .180

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: tan β = 6.2.2

- 5 - p 2 p

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

Answer: cos2 β =

2 p2 − 5 5 32


Trigonometry

PAPER 2 QUESTIONS 5 Number 5.1


Hints and answers Division by zero is not permitted. Thus the identity is undefined if the denominators on both sides are 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 = 45 5.2

o

+ k .90 o , k ∈ Ζ

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

PAPER 2 QUESTIONS 6 Number 6.1


Hints and answers Reducing the 2 x 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. 2

6.2

Answer: cos 2 x = 1 – 2sin x Use the answer to 6.1 to replace cos 2 x . 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: x = 210 + k .360 o

o

o o o o x = 90 + k .360 and x = 330 + k .360



MORE QUESTIONS FROM PAST EXAMINATION PAPERS Exemplar 2008

Preparatory Examination 2008

34

Trigonometry

QUESTIONS 4 Answer all questions without using a calculator. Show ALL calculations. 4.1

If sin β = a and 90

o

< β < 180 o , determine using a sketch the value of tan β in terms of a . (4)

4.2

Simplify the following:

(

sin 15o + 2 cos − 135o

)

sin 300 o 4.3

(7)

Simplify the following expression to one trigonometric ratio of θ :

sin(− θ ). sin (180o − θ ) + cos(90o + θ )

− sin (360o − θ ) − tan 315o

(6)[17]


Feb – March 2009

36

Trigonometry

November 2009 (Unused paper)


November 2009(1)

38

Trigonometry

Feb – March 2010


ANSWERS Exemplar 2008 4.1 4.2 4.3

−

2 3

1 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 5.1

θ = ±60 o + k ⋅ 360 o ;

k ∈ Ζ

She used an incorrect expansion of

12

6.2.1

cos α = −

6.2.2

cos(α + β ) =

6.3

13

6.1 6.2 6.3

2 2

(t +

1 − t 2

)

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

o

5

5.1.1

tan θ = −

5.1.2

cos θ sin θ = −

5.2 5.3

tan x

5.4

12

4.2 4.3

3 2+ 6

−2 3

60 169

1

x = 209,08 + k ⋅ 360 o

o

x = 330,92 + k ⋅ 360 o

o

x = −150,92 + k ⋅ 360 o

o

or

x = −29,08 + k ⋅ 360 k ∈ Ζ o o x = 60 or x = 240 o

Preparatory Examination 2008

−

o

November 2009 (Unused papers)

C (a;−a )

4.1 tan β =

65

x = 29,54 or 330,46

sin ( A + B ) 5.2

33

a

5.5

o

1− a2 or

(

2 3+ 3

−2 3

− sin θ

Feb/March 2009 5.1 − sin x 5.2 tan x 6.1.1

cos113o = − p

6.1.2

cos 23 o = 1 − p 2

6.1.3

sin 46 o = 2 p 1 − p 2

)

6.1.1 6.1.2

6.2

40

cos 24 o = 1 − p 2 3 p 2 Proof required. Provide a proof and check with the teacher if it is correct

Trigonometry

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