Trigonometric Functions...1.2.15 identify contexts suitable for modelling by trigonometric functions...

Year 11 ATAR Mathematics

Trigonometric Functions

Year 11 | Mathematics Methods | Trigonometric Functions | © Department of Education WA 2020

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Year 11 Mathematics Methods Trigonometric Functions

© Department of Education Western Australia 2020 1

Contents

Content page… ................................................................................................................... 1

Signposts… ....................................................................................................................... … 2

Overview ......................................................................................................................... 3

Lesson 1 – Graphs of sine, cosine and tangent……............................................................5

Lesson 2 – Transforming the trigonometric graphs, amplitude and period…………………..14

Lesson 3 – Change of period horizontal dilations and combination of transformations…27

Lesson 4 – Phase changes of 𝑦𝑦 = sin(𝑥𝑥 − 𝑐𝑐), 𝑦𝑦 = cos(𝑥𝑥 − 𝑐𝑐) and 𝑦𝑦 = tan (𝑥𝑥 − 𝑐𝑐)……………40

Lesson 5 – Angle sum and difference identities…………………………………………………………….50

Lesson 6 – Solving practical problems………………………………………………………………………….64

Lesson 7 – Solving equations involving trigonometric functions……………………………………74

Lesson 8 & 9 – Exam practice………………………………………………………………………………………..81

Glossary …………………………………………………………………….………………………………………………..82

Solutions……………………………………………………………….……………………………………………………..84



Signposts

Each symbol is a sign to help you.

Here is what each one means.

Important Information

Mark and Correct your work

You write an answer or response

Use your CAS calculator

A point of emphasis

Refer to a text book

Contact your teacher (if you can)

Check with your school about Assessment submission



Overview This Booklet contains approximately 3 weeks, or 12 hours, of work. Some students may find they need additional time.

To guide the pace at which you work through the booklet refer to the content page.

Space is provided for you to write your solutions in this PDF booklet. If you need more space, then attach a page to the page you are working on.

Answers are given to all questions: it is assumed you will use them responsibly, to maximise your learning. You should check your day to day lesson work.

Assessments All of your assessments are provided for you separately by your school.

Assessments will be either response or investigative. Weightings for assessments are provided by your school.

Calculator This course assumes the use of a CAS calculator. Screen displays will appear throughout the booklets to help you with your understanding of the lessons. Further support documents are available.

Textbook You are encouraged to use a text for this course. A text will further explain some topics and can provide you with extra practice questions.

Online Support

Search for a range of online support.



Content covered in this booklet

The syllabus content focused on in this booklet includes:

Trigonometric functions

1.2.7 understand the unit circle definition of sin𝜃𝜃, cos𝜃𝜃 and tan 𝜃𝜃 and periodicity using radians

1.2.8 recognise the exact values of sin𝜃𝜃, cos𝜃𝜃 and tan 𝜃𝜃 at integer multiples of 𝜋𝜋6

and 𝜋𝜋4

1.2.9 recognise the graphs of 𝑦𝑦 = sin𝑥𝑥,𝑦𝑦 = cos𝑥𝑥 , and 𝑦𝑦 = tan𝑥𝑥 on extended domains

1.2.10 examine amplitude changes and the graphs of 𝑦𝑦 = 𝑎𝑎 sin𝑥𝑥 and 𝑦𝑦 = 𝑎𝑎 cos𝑥𝑥

1.2.11 examine period changes and the graphs of 𝑦𝑦 = sin𝑏𝑏𝑥𝑥, 𝑦𝑦 = cos𝑏𝑏𝑥𝑥 and 𝑦𝑦 = tan 𝑏𝑏𝑥𝑥

1.2.12 examine phase changes and the graphs of 𝑦𝑦 = sin(𝑥𝑥 − 𝑐𝑐), 𝑦𝑦 = cos(𝑥𝑥 − 𝑐𝑐) and 𝑦𝑦 = tan (𝑥𝑥 − 𝑐𝑐)

1.2.13 examine the relationships sin �𝑥𝑥 + 𝜋𝜋2� = cos𝑥𝑥 and cos �𝑥𝑥 − 𝜋𝜋

2� = sin𝑥𝑥

1.2.14 prove and apply the angle sum and difference identities

1.2.15 identify contexts suitable for modelling by trigonometric functions and use them to solve practical problems

1.2.16 solve equations involving trigonometric functions using technology, and algebraically in simple cases

Click this link to access further information about the course syllabus:

https://senior-secondary.scsa.wa.edu.au/syllabus-and-support-materials/mathematics/mathematics-methods





Lesson 1 Graphs of sine, cosine and tangent functions

By the end of this lesson you should be able to:

• recall facts about the unit circle

• draw accurate graphs of sin x, cos x and tan x using common angles in radian formeg 𝜋𝜋

6, 𝜋𝜋4

, 𝜋𝜋3

, 𝜋𝜋2

, etc.

• Recognise the important points of the sine and cosine functions and be able to usethese points to draw quick sketches of their graphs.

• Sketch the tangent graph (tan x) using important points and features of thefunction.

Prior Learning - Sine functions

Recall some common angle measures in degrees and radians.

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Exercise 1.1

1. Using your calculator, complete the table of values below. Remember to work inradian mode. And give your answers to one decimal place.

Value of 𝑡𝑡 0 𝜋𝜋6

𝜋𝜋4

𝜋𝜋3

𝜋𝜋2

2𝜋𝜋3

3𝜋𝜋4

5𝜋𝜋6

𝜋𝜋

sin 𝑡𝑡 (to 2 d.p.)

Value of 𝑡𝑡 7𝜋𝜋6

5𝜋𝜋4

4𝜋𝜋3

3𝜋𝜋2

5𝜋𝜋3

7𝜋𝜋4

11𝜋𝜋6

2𝜋𝜋



9𝜋𝜋4

7𝜋𝜋3

5𝜋𝜋2

8𝜋𝜋3

11𝜋𝜋4

17𝜋𝜋6

3𝜋𝜋



13𝜋𝜋4

10𝜋𝜋3

7𝜋𝜋2

11𝜋𝜋3

15𝜋𝜋4

23𝜋𝜋6

4𝜋𝜋


2. Did you notice a pattern start to occur? Comment.



3. Plot the points from the table on the grid below and then join your points with a smoothcurve.



Graphing the sine curve – another way The next diagram can be used to graph the sine function.

𝑥𝑥 = 1.0,𝑦𝑦 = sin 1.0 ≈ 0.8 𝑥𝑥 = 5.5,𝑦𝑦 = sin 5.5 ≈ −0.7

Our wrapping function has been marked in the unit circle with graduations every 0.1 of a radian.

Do you remember? The sine of an angle is the 𝒚𝒚 coordinate of the point on the unit circle.

To graph sin 1.0 Locate 1.0 on the unit circle below.

Locate 1.0 on the horizontal axis of the grid to the right of the unit circle.

Transfer the 𝑦𝑦 coordinate from 1.0 on the unit circle across onto the grid, marking a point above 1.0 on the horizontal axis.

𝑥𝑥 = 1.0,𝑦𝑦 = sin 1.0 ≈ 0.8 𝑥𝑥 = 5.5,𝑦𝑦 = sin 5.5 ≈ −0.7



The working for sin 1.0 has been done for you and another point corresponding to sin 5.5 has also been marked on the grid.

It is not necessary to draw in all of the arrows as has been done in the diagram above.

All that is necessary to plot the points as has been done in this diagram.

You can see that we can discover the shape without plotting every 0.1. Complete the diagram by plotting every 0.5. That is, plot 𝑥𝑥 = 4.5, 5.0, 5.5 and 6.0.

4. What is the maximum value of sin (𝑡𝑡)?

5. What is the minimum value of sin (𝑡𝑡)?

6. At what values of 𝑡𝑡 does the graph of sin 𝑡𝑡 cross the horizontal axis? That is , what are thezeros of cos 𝑡𝑡

Cosine functions

Repeat for cos t. Remember to work in radian mode. And give your answers to one decimal place.


𝜋𝜋4

𝜋𝜋3

𝜋𝜋2

2𝜋𝜋3

3𝜋𝜋4

5𝜋𝜋6

𝜋𝜋

cos 𝑡𝑡 (to 2 d.p.)




5𝜋𝜋4

4𝜋𝜋3

3𝜋𝜋2

5𝜋𝜋3

7𝜋𝜋4

11𝜋𝜋6

2𝜋𝜋



9𝜋𝜋4

7𝜋𝜋3

5𝜋𝜋2

8𝜋𝜋3

11𝜋𝜋4

17𝜋𝜋6

3𝜋𝜋



13𝜋𝜋4

10𝜋𝜋3

7𝜋𝜋2

11𝜋𝜋3

15𝜋𝜋4

23𝜋𝜋6

4𝜋𝜋




The cosine curve again

Do you remember? The cosine of an angle is the 𝑥𝑥 coordinate of the point on the unit circle.

To use the same technique for cosine as we did for sine, above, we need to turn the grid around so that it runs down the page. Then we can transfer the 𝑥𝑥 −values down on to the grid. You can see that the positive values come from the first and fourth quadrants, and the negative values from the second and third quadrants.

The points have been marked in for cos 0, cos 0.5, cos 1.0,

cos 𝜋𝜋2

, cos 2.0, cos 2.5 , cos 3.0 , cos𝜋𝜋 , cos 3𝜋𝜋2

and cos 2𝜋𝜋.The

working needed to plot cos 2.0 is shown in detail.

Using the unit circle and a ruler, mark in a few more values of cos 𝑥𝑥 and complete the sketch up to 𝑥𝑥 = 2𝜋𝜋.

Does your graph look like you expect it should?

8. What is the maximum value of cos 𝑡𝑡?

9. What is the minimum value of cos 𝑡𝑡?

10. At what values of 𝑡𝑡 does the graph of cos 𝑡𝑡 cross the horizontal axis? That is , whatare the zeros of cos 𝑡𝑡?

11. Compare your graphs of sin 𝑡𝑡 and cos 𝑡𝑡. What is similar? What is different?



Graphing tan x

12. Complete the tables of values of tan below, using your calculator to help you. Remember towork in radian mode. And give your answers to one decimal place.

Value of 𝑥𝑥 -π −5𝜋𝜋6

−3𝜋𝜋4

−2𝜋𝜋3

−𝜋𝜋2

−𝜋𝜋3

−𝜋𝜋4

−𝜋𝜋6

0

tan 𝑥𝑥

Value of 𝑥𝑥 𝜋𝜋6

𝜋𝜋4

𝜋𝜋3

𝜋𝜋2

2𝜋𝜋3

3𝜋𝜋4

5𝜋𝜋6

π

tan 𝑥𝑥

Value of 𝑥𝑥 -1.4 -1.2 -1.0 -0.8 -0.6 -0.4 -0.2 0

tan 𝑥𝑥

Value of 𝑥𝑥 0 0.2 0.4 0.6 0.8 1.0 1.2 1.4

tan 𝑥𝑥

13. Use as many of the above points as necessary to graph tan 𝑥𝑥 accurately on this grid. (the 𝜋𝜋 isapproximate.)



Check your graph using your calculator to make sure you have it drawn correctly. Make any corrections before you move on.

14. Where are the zeros located?

15. As x gets close to 𝜋𝜋2 what seems to happen to tan 𝑥𝑥?

16. What happens to tan 𝑥𝑥 after 𝜋𝜋2 ?

17. What happens near −𝜋𝜋2

18. What seems to be the period of tan 𝑥𝑥?

19. What do you think happens to tan 𝑥𝑥 when 𝑥𝑥 > 𝜋𝜋?

20. What happens to the graph when 𝑥𝑥 < −𝜋𝜋?

21. What would be the maximum value of tan 𝑥𝑥?

22. What would be the minimum value of tan 𝑥𝑥?

23. Another value of x which is useful when sketching tan x is the value of 𝑥𝑥 when tan 𝑥𝑥 = 1.What is this value of 𝑥𝑥?



Lesson 2 Transforming the trigonometric graphs

By the end of this lesson you should be able to:

• Transform the graphs of sine, cosine and tangent by sliding up or down

• Transform the graphs of sine, cosine and tangent by dilating about the x-axis

• Transform the graphs of sine, cosine and tangent by reflecting about the x-axis.

• Understand the effect of changing ‘b’ in sin (bx) – horizontal dilations

• Determine the period of a sin (bx) from the value of b

• Determine the period of a sine function from its graph

Transformations Below are examples of the transformations obtained by sliding these functions up or down. Write their equations underneath.

Sine curve

Usual position translated up 1 unit translated down 1 unit

𝑦𝑦 = sin 𝑥𝑥 𝑦𝑦 =______________ 𝑦𝑦 =______________

Cosine curve


𝑦𝑦 = cos 𝑥𝑥 𝑦𝑦 =______________ 𝑦𝑦 =______________



Tangent curve


𝑦𝑦 = tan 𝑥𝑥 𝑦𝑦 =______________ 𝑦𝑦 =______________

Did you guess for the middle graphs that have been translated up one 𝑦𝑦 = sin 𝑥𝑥 + 1,𝑦𝑦 = cos 𝑥𝑥 + 1 𝑎𝑎𝑎𝑎𝑎𝑎 𝑦𝑦 = tan 𝑥𝑥 + 1?

Did you guess for the graphs on the right that have been moved down 𝑦𝑦 = sin 𝑥𝑥 − 1,𝑦𝑦 = cos 𝑥𝑥 − 1 𝑎𝑎𝑎𝑎𝑎𝑎 𝑦𝑦 = tan 𝑥𝑥 − 1?

Dilating about the x-axis Look at the following variations on the sine and cosine curves.

𝑦𝑦 = 12

sin 𝑥𝑥 𝑦𝑦 = 2 sin 𝑥𝑥 𝑦𝑦 = 3 cos 𝑥𝑥 𝑦𝑦 = 14

cos 𝑥𝑥

In each case, a constant has been placed in front of sin 𝑥𝑥 or cos 𝑥𝑥. We could just as easily do the same for tan 𝑥𝑥.

𝑦𝑦 = 12

tan 𝑥𝑥 𝑦𝑦 = 2 tan 𝑥𝑥

These functions are written in the general form by using the letter ‘a’ to represent any constant.

𝑦𝑦 = 𝑎𝑎 sin 𝑥𝑥 𝑦𝑦 = 𝑎𝑎 cos 𝑥𝑥 𝑦𝑦 = 𝑎𝑎 tan 𝑥𝑥

In the previous lesson you were asked to explore the effect of 𝑎𝑎 in 𝑦𝑦 = 𝑎𝑎 sin 𝑥𝑥.



You should have found that varying ‘𝑎𝑎’ is to dilate the graph in relation to the 𝑥𝑥-axis. You will see the effects in the graphs which follow.

The following are 𝑦𝑦 = cos 𝑥𝑥 ,𝑦𝑦 = 3 cos 𝑥𝑥 and 𝑦𝑦 = 14

cos 𝑥𝑥.

Try labelling them, then using your calculator, graph them and check.

The changes in ‘a’ changed the ‘height’ of the function. This is a change in the amplitude.

where a is 12

graph is contracted, all the y values are half the values of sin 𝑥𝑥

where a is 2

graph is enlarged, all the y values are double the values of sin 𝑥𝑥

where a = 1

graph is as shown above



The amplitude is the maximum displacement from the mean position of the function. It is always positive.

You can see that the above function has an amplitude equal to 2 units.

Exercise 2.1

Use the information on the graphs on the next page to work out its rule. You will need to decide whether it is sin, cos or tan and also a value of ‘𝑎𝑎’. Check your answers by drawing them on your calculator.

1. 2.



3. 4.

5. 6.

Reflecting

The top two graphs below show sin𝑥𝑥 and tan 𝑥𝑥 in their usual positions. If we reflect the graphs about the 𝑥𝑥-axis, the results are shown in the seconds row of diagrams.



The effect is to make every 𝑦𝑦 value opposite in sign. Their rules are 𝑦𝑦 = − sin 𝑥𝑥 and 𝑦𝑦 = − tan 𝑥𝑥.

Exercise 2.2 1. State the amplitude of the following

a) 𝑦𝑦 = 4 sin 𝑥𝑥

b) 𝑦𝑦 = −2 cos(𝑥𝑥 − 1)

c) 𝑦𝑦 = 2.5 sin 𝑥𝑥 + 3

2. State the amplitude of the following

a) b) c)



Change in period – sine graphs In the diagrams below we see the effect of horizontal dilations. The waves are squeezed up or spread out.



How long is a wave?

Obviously, if we decide where a wave starts and where it ends, we can measure how ‘long’ it is.

We call this distance the ‘wavelength’.

In most occasions where these waves occur, they are associated with time.

For example, we may wish to know, in an electrical current, how many wave or cycles occur every second.

This association of waves in a given period of time leads to another name or the wavelength, the ‘period’.

Horizontal dilations have the effect of changing the wavelength or period.

Horizontal dilations and sin 𝒃𝒃𝒃𝒃 We will now look at a number of sine graphs such as 𝑦𝑦 = sin 𝑥𝑥 or sin 2𝑥𝑥

Look at the following graphs which are named for you.

For one complete wave of sin 𝑥𝑥, we can see that the period is 2𝜋𝜋 (approximately 6.3 if the scale were numerical). Keep this fact in mind as you work though the following graphs.

For 𝑦𝑦 = sin 2𝑥𝑥, how many waves between 0 and 2𝜋𝜋?

What will the period be?

The period of sin x is 2π



You should have seen that for sin 2𝑥𝑥, there were 2 waves between 0 and 2𝜋𝜋. This would mean that the period would be 1

2 of 2𝜋𝜋.

That is, the period of sin 2𝑥𝑥 is 𝜋𝜋.

Exercise 2.3

Now examine the graphs below and from the previous page and record your findings in the table on the next page.

Using your CAS graph y = 3 - 2 sin (1/2 x). Look carefully at the CAS screen and see if you can draw a graph like these shown.



Now complete this table.

Curve Number of waves in 2𝜋𝜋 𝑏𝑏 Period 2𝜋𝜋 ÷ 𝑏𝑏

sin 𝑥𝑥

sin 2𝑥𝑥

sin12𝑥𝑥

sin 3𝑥𝑥

sin13𝑥𝑥

sin 4𝑥𝑥

sin14𝑥𝑥

3 - 2 sin (1/2 x)

Summary Let's put together a formal summary of the last few pages.

For 𝑦𝑦 = sin 𝑏𝑏𝑥𝑥,

a) b gives the number of waves in 2𝜋𝜋.

b) the period is 2𝜋𝜋 ÷ 𝑏𝑏 (or 2𝜋𝜋𝑏𝑏

)

c) the period is the length of one wave.

From the previous lessons we considered translation which med the graph up or down the vertical scale, by changing the value of '𝑎𝑎' in 𝑦𝑦 = sin 𝑥𝑥 + 𝑎𝑎.

We also look at vertical dilations which were dependent on the value of '𝑎𝑎' in 𝑦𝑦 = 𝑎𝑎 𝑠𝑠𝑠𝑠𝑎𝑎 𝑥𝑥.

This was called a change in 'amplitude'.

Let's put all of these transformations together.

The period of a sine function is calculated by 2𝜋𝜋𝑏𝑏

in radians or 360 𝑏𝑏

in degrees



Get it all together Adding '𝑏𝑏', and '𝑎𝑎' and '𝑎𝑎' into the equation gives us

Example

Find the period of this function: 𝑦𝑦 = 2 sin 23𝑥𝑥 − 1

Solution By now, we know what the graph of sin 𝑥𝑥 is like. We know its shape, its period, its amplitude, the maximum and minimum locations and the zeroes.

In this example, we have 𝑎𝑎 = 2, 𝑏𝑏 = 23, which means the period is found by dividing 2𝜋𝜋 by 2

3, and

we have 𝑎𝑎 = −1.

2𝜋𝜋 ÷ 23

= 2𝜋𝜋 × 32

= 3𝜋𝜋



Exercise 2.4 1. Explain the effect of changing ‘𝑏𝑏’ in sin 𝑏𝑏𝑥𝑥.

2. Write down the period of these functions

a) sin 𝑥𝑥 b) sin 14𝑥𝑥

c) 2 sin 𝑥𝑥 d) sin 6𝑥𝑥

e) sin 34𝑥𝑥 f) 3

2sin𝜋𝜋𝑥𝑥

g) 3 − sin 2𝑥𝑥

3. Write down the amplitude for each of the functions in question 2.

a) sin 𝑥𝑥 b) sin 14𝑥𝑥

c) 2 sin 𝑥𝑥 d) sin 6𝑥𝑥

e) sin 34𝑥𝑥 f) 3

2sin𝜋𝜋𝑥𝑥

g) 3 − sin 2𝑥𝑥

4. Explain the meaning of ‘period’ as it applies to 𝑦𝑦 = sin 𝑥𝑥.



5. Determine the value of 𝑏𝑏 from these graphs.

a) b)

6. Name these graphs of sin 𝑏𝑏𝑥𝑥.

a) b)

c) d)



Lesson 3 Graphs of trigonometric functions horizontal dilations

By the end of this lesson you should be able to

• extend your understanding of change in period to the cosine and tangent functions

• sketch the graphs of sin bx, cos bx and tan bx using graphic calculator

• draw accurate graphs of these functions when necessary. sin, cos, tan -what's thedifference?

• examine the changes which occur when the sine graph is moved to the left or right

• Understand the meaning of the term ‘phase shift’

• Determine the values of c in sin (x+c) from the graph

• sketch the graph of sin (x+c)

Sin, cos, tan -what's the difference? We have already seen similarities between sine and cosine graphs, and the basic period of cosine is the same as for sine. That is, cos 𝑥𝑥 repeats itself every 2𝜋𝜋 just the same as sin 𝑥𝑥.

Graphs of cos 𝒃𝒃𝒃𝒃 What happens when we change the value of 𝑏𝑏? Because of the close links with sin 𝑏𝑏𝑥𝑥, we can readily understand that changing 𝑏𝑏 will have the same effect on cos 𝑏𝑏𝑥𝑥 as it does on sin 𝑏𝑏𝑥𝑥.

Study the following graphs and complete the statements to the right of each graph. Remember to check your answers when you finish each one or two graphs. You can check by using the graphic calculator.



Exercise 3.1

1. a) This is the graph of 𝑦𝑦 = _______________

b) How many wavesbetween 0 and 2𝜋𝜋?______________

c) What is the period ofthis graph?____________















Summary For 𝑦𝑦 = cos 𝑏𝑏𝑥𝑥

• 𝑏𝑏 gives the number of waves in 2𝜋𝜋.

• The period is 2𝜋𝜋 ÷ 𝑏𝑏 or

• the period is the length of one wave.

If you think this looks familiar you are right. It is the same set of rules as for 𝑦𝑦 = sin 𝑥𝑥 + 𝑎𝑎.

Get it all together

Adding 'b', and 'a' and 'd' into the equation gives us



Tangent is a little different

We don't have waves with tangent!

However, we can still talk about wavelength or the period of tan 𝑥𝑥 repeats its basic shape every 𝜋𝜋 units along the horizontal axis. You won't win too many prizes for guessing the period of tan 𝑥𝑥, now!

The period of tan 𝑥𝑥 is 𝜋𝜋.

The period of sin 𝑥𝑥 is 2𝜋𝜋.

The period of cos 𝑥𝑥 is 2𝜋𝜋.

Sometimes, we need to use degrees, and so, in degrees, the periods are

sin 𝑥𝑥° and cos 𝑥𝑥° ∶ Period = 360°

tan 𝑥𝑥° : Period = 180°

Branches of 𝐭𝐭𝐭𝐭𝐭𝐭𝒃𝒃𝒃𝒃 The tangent graph doesn’t have the wave-like structure that sine and cosine have. It will suit our purpose to talk about the ‘branches’ of the tan graph, each separate curve being called a branch.



You will notice that there is a separate branch of tan 𝑥𝑥 at every interval of 𝜋𝜋 units on the horizontal axis.

Thus the period of tan 𝑥𝑥 is 𝜋𝜋 (or 180o). in other words, the basic shape is repeated every 𝜋𝜋 units.

As we say with sin 𝑏𝑏𝑥𝑥 and cos 𝑏𝑏𝑥𝑥, the number of branches of tan 𝑏𝑏𝑥𝑥 is controlled by ‘𝑏𝑏’.

Compare the number of branches with the value of 𝑏𝑏. (2 half branches count as 1 branch).

Exercise 3.2

a) (i) 𝑏𝑏 = 2

Branches in 𝜋𝜋 units =______

(ii) Period =______

b) (i) 𝑏𝑏 = 12


(ii) Period =______

c) (i) 𝑏𝑏 = 12


(ii) Period =______



Branching and ‘𝒃𝒃’ You would have noticed that the number of branches coincides with the value of 𝑏𝑏.

Did you discover the relationship between the period, 𝜋𝜋 and the value of ‘𝑏𝑏’? Having worked with sine and cosine, you shouldn’t have had much difficulty with this relationship.

Another nutshell For 𝑦𝑦 = tan 𝑏𝑏𝑥𝑥

• 𝑏𝑏 gives the number of branches in 𝜋𝜋

• The period is 𝜋𝜋 ÷ 𝑏𝑏 (or 𝜋𝜋𝑏𝑏

)

• The period is the distance between branches.

Extending tan 𝒃𝒃 We can of course add in other features which can change the nature of the tangent graph.

Using a CAS calculator Calculators have taken away much of the laborious work involved in graphing the different trigonometric functions. However, you will still need to be able to calculate the period and amplitude of a function. This will allow you to set the “window” quickly and efficiently.

Note: For sin 𝑏𝑏𝑥𝑥 and cos 𝑏𝑏𝑥𝑥 period = 2𝜋𝜋𝑏𝑏

For tan 𝑏𝑏𝑥𝑥, period = 𝜋𝜋𝑏𝑏



Exercise 3.3 1. Use your calculator to graph the following trigonometric functions over given domains.

• 𝑦𝑦 = 3 sin 4𝑥𝑥 over the domain 0 ≤ 𝑥𝑥 ≤ 360

• 𝑦𝑦 = cos 12𝑥𝑥 over the domain −2𝜋𝜋 ≤ 𝑥𝑥 ≤ 2𝜋𝜋

• 𝑦𝑦 = tan 3𝑥𝑥 over the domain −𝜋𝜋 ≤ 𝑥𝑥 ≤ 0

Make a sketch of each of the above three functions and in each case write down the following (where they exist)

(i) The period and amplitude

(ii) The 𝑥𝑥 −intercepts (zeroes)

(iii) The location of the maximum and minimum values.

(iv) The location of any asymptotes.



Phase shift or phase change The term 'phase shift' or 'phase change' applies to changing or shifting the position of the waves or branches of a trigonometric graph to the left or right.

The rule used to define the function changes too, by varying the value of 'c' in

𝑦𝑦 = sin(𝑥𝑥 + 𝑐𝑐)

𝑦𝑦 = cos(𝑥𝑥 + 𝑐𝑐) 𝑦𝑦 = tan (𝑥𝑥 + 𝑐𝑐)



Before we move on

Where does the idea of a phase-shift have an application?

In physics, there is an application in the graphs which describe aspects of alternating currents. The characteristics of these currents depend on the phase differences between current and voltage.

The defining rules for current graph and the voltage graph would be different.

Your stereo speakers When installing speakers for a car stereo or home stereo, it is important to connect the wires the right way. The sound you hear is produced by the vibration of the speaker cones and these cones need to vibrate ‘in phase’, otherwise some of the sound quality is lost.

Sound waves can represented by sinusoidal curves, and if two similar instruments commenced playing a note as slightly different starting points, we might have the graphs looking like this.

The effect would be greater when the graphs looked like this.



Phase shift from the graphs

sin (𝑥𝑥 + 𝑐𝑐) Examine the shift that has occurred with these graphs.

Notice the direction and distance.

We can easily deduce the value of c from the graph.

The rules for these graphs can then be written down.



Phase shift with 𝐬𝐬𝐬𝐬𝐭𝐭(𝒃𝒃 + 𝒄𝒄) On the previous page you can see the changes which occur by sliding 𝑦𝑦 = sin 𝑥𝑥 to the left or right. This horizontal movement parallel to the 𝑥𝑥 axis is called a phase shift. Graphic calculators now make it very easy for us to see, not only to change in the original function 𝑦𝑦 = sin 𝑥𝑥, but the amount of the phase shift.

Exercise 3.4

Graphic Calculator Activity. 1. Graph 𝑦𝑦 = sin 𝑥𝑥 over the domain −2𝜋𝜋 ≤ 𝑥𝑥 ≤ 2𝜋𝜋.

2. Leave the graph of 𝑦𝑦 = sin 𝑥𝑥 on your screen and now graph 𝑦𝑦 = sin �𝑥𝑥 + 𝜋𝜋2� using the same

window settings. Write down the effect of adding 𝜋𝜋2. In this case 𝑐𝑐 = + 𝜋𝜋

2.

3. Leave the graph of 𝑦𝑦 = sin 𝑥𝑥 on your screen and now draw the graph of 𝑦𝑦 = sin �𝑥𝑥 − 𝜋𝜋2�.

Turn off 𝑦𝑦 = sin �𝑥𝑥 + 𝜋𝜋2�.

Write down the effect of subtracting 𝜋𝜋2. In this case = −𝜋𝜋

2

4. Graph the following sine curves and see if the effect of changing “c” fits in with your earlierfinding. Remember to compare each graph to the graph of 𝑦𝑦 = sin 𝑥𝑥

a) 𝑦𝑦 = sin �𝑥𝑥 + 𝜋𝜋4� Note 𝑐𝑐 = + 𝜋𝜋

4

b) 𝑦𝑦 = sin �𝑥𝑥 − 𝜋𝜋3� Note 𝑐𝑐 = −𝜋𝜋

3

c) 𝑦𝑦 = sin(𝑥𝑥 + 𝜋𝜋) Note 𝑐𝑐 = +𝜋𝜋

d) 𝑦𝑦 = sin �𝑥𝑥 − 𝜋𝜋4� Note 𝑐𝑐 = −𝜋𝜋

4



Summary Have you discovered what the pattern is?

You should have found that the phase shift is equal in size to ‘𝑐𝑐’ and that

• The graph moves to the left if ‘𝑐𝑐’ is positive.

• The graph moves to the right if ‘𝑐𝑐’ is negative.

Example Sketch the graph of 𝑦𝑦 = sin(𝑥𝑥 − 1.5)

In this case, 𝑐𝑐 = −1.5 which means the phase shift is 1.5 radians to the right. Alternatively we could say draw 𝑦𝑦 = sin 𝑥𝑥 and then move the vertical axis 1.5 units to the left.

Rather than use the ‘𝜋𝜋’ scale, it is better to use the simple numerical scale for this graph.



Exercise 3.5 1. Sketch the following over the domain −4 ≤ 𝑥𝑥 ≤ 4

a) 𝑦𝑦 = sin(𝜋𝜋𝑥𝑥),

b) 𝑦𝑦 = sin(𝜋𝜋𝑥𝑥 + 𝜋𝜋) ,

c) 𝑦𝑦 = − sin𝜋𝜋(𝑥𝑥 + 1),

d) 𝑦𝑦 = 1 − sin𝜋𝜋(𝑥𝑥 + 1)

Describe clearly the way in which graphs (b), (c) and (d) are related to graph (a). Comment on any phase shifts, including the size of the phase shift.

a) b)

c) d)



Lesson 4 Phase changes of trigonometric functions


• sketch the graph of 𝑐𝑐𝑐𝑐𝑠𝑠 (𝑥𝑥 + 𝑐𝑐)

• Sketch graphs in the form of 𝑡𝑡𝑎𝑎𝑎𝑎(𝑥𝑥 + 𝑐𝑐)

• Extend your ability to graph trigonometric functions in the form of 𝑠𝑠𝑠𝑠𝑎𝑎 𝑏𝑏(𝑥𝑥 + 𝑐𝑐),𝑐𝑐𝑐𝑐𝑠𝑠 𝑏𝑏(𝑥𝑥 + 𝑐𝑐) and 𝑡𝑡𝑎𝑎𝑎𝑎 𝑏𝑏(𝑥𝑥 + 𝑐𝑐)

Phase shift with cosine

Examine these graphs of sin(𝑥𝑥 + 𝑐𝑐) in which the value of c changes from 0 to 𝜋𝜋4 to 𝜋𝜋

2 to π.

The rules for these graphs are

1. 𝑦𝑦 = sin 𝑥𝑥 (𝑐𝑐 = 0)

2. 𝑦𝑦 = sin (𝑥𝑥 + 𝜋𝜋4

) �𝑐𝑐 = 𝜋𝜋4�

3. 𝑦𝑦 = sin (𝑥𝑥 + 𝜋𝜋2) �𝑐𝑐 = 𝜋𝜋

2�

4. 𝑦𝑦 = sin (𝑥𝑥 + 𝜋𝜋) (𝑐𝑐 = 𝜋𝜋)

When is the phase shift such that the graph of 𝑦𝑦 = sin(𝑥𝑥 + 𝑐𝑐) is the same as 𝑦𝑦 = cos 𝑥𝑥?

1. 2.

3. 4.

Graph 3 is the same as the graph of cos 𝑥𝑥. That is, sin �𝑥𝑥 + 𝜋𝜋2� = cos 𝑥𝑥.



Now, the next question for you to consider, is what phase shift is necessary so that cos(𝑥𝑥 + 𝑐𝑐) = sin 𝑥𝑥?

In other words, from graph 3, what shift is necessary to make the graph the same as graph 1?

The simplest shift would be to move the graph 𝜋𝜋2 units to the right. This would lead you to ‘c’

having the value of −𝜋𝜋2

.

The rule would be cos �𝑥𝑥 − 𝜋𝜋2� = sin 𝑥𝑥 .

The simplest shift A few lines back, the word ‘simplest’ was used when talking about a phase shift. Did you wonder why?

Clearly, because of the periodic nature of trigonometric functions, there could be other phase shifts which give rise to the same positioning of the waves.

What value would ‘𝑐𝑐’ have so that sin(𝑥𝑥 + 𝑐𝑐) = sin 𝑥𝑥 or cos(𝑥𝑥 + 𝑐𝑐) ?

What value would ‘𝑐𝑐’ have so that tan(𝑥𝑥 + 𝑐𝑐) = tan 𝑥𝑥 ?

Do you think ‘𝑐𝑐’ could have more than one value? How many values? What spacing in terms of 𝜋𝜋-units would these values of 𝑐𝑐 have?

Perhaps you knew the answers to these questions, perhaps there were some that you weren’t sure about.

Check with this summary:

• There in an infinite number of vales of 𝑐𝑐 which would make sin(𝑥𝑥 + 𝑐𝑐) = sin 𝑥𝑥,tan(𝑥𝑥 + 𝑐𝑐) = tan 𝑥𝑥 or cos(𝑥𝑥 + 𝑐𝑐) = cos 𝑥𝑥

• For sine and cosine, the values of 𝑐𝑐 would be spaced 2𝜋𝜋 units apart

• For tangent, the values of 𝑐𝑐 would be 𝜋𝜋 units apart.

• The simplest shifts would be

o sin(𝑥𝑥 ± 2𝜋𝜋) = sin 𝑥𝑥

o cos(𝑥𝑥 ± 2𝜋𝜋) = cos 𝑥𝑥

o tan(𝑥𝑥 ± 𝜋𝜋) = tan 𝑥𝑥

Other possible values of 𝑐𝑐 would be:

sin(𝑥𝑥 ± 4𝜋𝜋) , sin(𝑥𝑥 ± 6𝜋𝜋) sin(𝑥𝑥 ± 8𝜋𝜋)

cos(𝑥𝑥 ± 4𝜋𝜋) , cos(𝑥𝑥 ± 6𝜋𝜋) cos(𝑥𝑥 ± 8𝜋𝜋)



tan(𝑥𝑥 ± 2𝜋𝜋) tan(𝑥𝑥 ± 3𝜋𝜋) tan(𝑥𝑥 ± 4𝜋𝜋)

Phase shift and sketching 𝒚𝒚 = 𝐜𝐜𝐜𝐜𝐬𝐬(𝒃𝒃 + 𝒄𝒄) Previously we looked at the graphs of 𝑦𝑦 = sin(𝑥𝑥 + 𝑐𝑐) and how the value of ‘𝑐𝑐’ determines the phase shift. Graphing 𝑦𝑦 = cos (𝑥𝑥 + 𝑐𝑐) is done in the same way, since the sine and cosine graphs are the same shape but simply out of phase with each other.

Exercise 4.1 1. Sketch graphs of

a) 𝑦𝑦 = cos �𝑥𝑥 − 𝜋𝜋4�

b) 𝑦𝑦 = cos �𝑥𝑥 + 𝜋𝜋3�

c) Describe clearly how thegraphs (a) and (b) are relatedto the graph of

𝑦𝑦 = cos 𝑥𝑥

By now you should have discovered the following relationship. If the graph of 𝑦𝑦 = cos 𝑥𝑥 is moved 𝜋𝜋2 units right, parallel to the 𝑥𝑥 axis, it would then be the same as the graph of 𝑦𝑦 = sin 𝑥𝑥. We say

that sin 𝑥𝑥 and cos 𝑥𝑥 are 𝜋𝜋2 out of phase with each other.

It follows that

Use your calculator to check. The graphs should superimpose over each other.

cos 𝑥𝑥 = sin �𝑥𝑥 +𝜋𝜋2� and that sin 𝑥𝑥 = cos �𝑥𝑥 −

𝜋𝜋2�



Phase shift and sketching 𝒚𝒚 = 𝐭𝐭𝐭𝐭𝐭𝐭(𝒃𝒃 + 𝒄𝒄) Although the shape of the tangent graph is very different to the sine and cosine graph, the change of phase works in exactly the same way.

Exercise 4.2 Calculator Activity 1. Graph 𝑦𝑦 = tan 𝑥𝑥 over the domain − 2𝜋𝜋 ≤ 𝑥𝑥 ≤ 2𝜋𝜋.

2. Now in turn, graph 𝑦𝑦 = tan �𝑥𝑥 + 𝜋𝜋4� and 𝑦𝑦 = tan �𝑥𝑥 − 𝜋𝜋

2�.

3. Compare your screens to the ones right. You need to checkthem carefully because the screen is very “busy”.

4. What is the effect of changing the value of ‘𝑐𝑐’?

Summary

• The change in phase is equal in size to ‘𝑐𝑐’.

• The graph moves to the left if ‘𝑐𝑐’ is positive.

• The graph moves to the right if ‘𝑐𝑐’is negative.

You should have also noticed that we have done all of our graphs using radians. The rules for change of phase would be exactly the same if we had used degrees.



Exercise 4.3 1. Sketch the graphs of

a) 𝑦𝑦 = tan(𝑥𝑥 + 1) using a numericalscale −7 ≤ 𝑥𝑥 ≤ 7

b) 𝑦𝑦 = tan �𝑥𝑥 − 𝜋𝜋4� using a

‘𝜋𝜋 −units’ scale −2𝜋𝜋 ≤ 𝑥𝑥 ≤ 2𝜋𝜋

Life wasn’t meant to be easy

Have a look at the defining rules for these functions.

a) 𝑦𝑦 = sin 2 �𝑥𝑥 − 𝜋𝜋4� b) 𝑦𝑦 = sin �2𝑥𝑥 − 𝜋𝜋

2�

c) 𝑦𝑦 = sin 12

(𝑥𝑥 − 𝜋𝜋) d) 𝑦𝑦 = sin �𝑥𝑥2− 𝜋𝜋

2�

e) 𝑦𝑦 = tan 3 �𝑥𝑥 + 𝜋𝜋2� f) 𝑦𝑦 = tan �3𝑥𝑥 + 3𝜋𝜋

2�

Look at the angle part a) and b), c) and d), e) and f).

Did you see that 2 �𝑥𝑥 − 𝜋𝜋4� = �2𝑥𝑥 − 𝜋𝜋

2�?

Check the other pairs. Are they equal?

The answer is yes, they are equal. You might be asking why you need to be bothered with this algebraic manipulation. The ability to use this idea is very important when graphing functions that are similar to those in parts a) to f).



Let’s look at part b) 𝑦𝑦 = sin �2𝑥𝑥 − 𝜋𝜋2�, and ask the question, when is 𝑦𝑦 = 0?

Obviously, for sine functions, 𝑦𝑦 is equal to zero when the angle is equal to zero. That is, this case, when

Now look at part a) 𝑦𝑦 = sin 2 �𝑥𝑥 − 𝜋𝜋4�. By taking the ‘2’ out as a common factor, we have the

phase shift revealed.

As you can see, parts c) and e) are also in the form where the phase shift is obvious. We can see the value of ‘𝑐𝑐’ and also the value of ‘𝑏𝑏’. Remember the value of ‘𝑏𝑏’ is used to calculate the period of the function.

We started by suggesting that life wasn’t meant to be easy, but when we have the rule written in the form 𝑦𝑦 = sin 𝑏𝑏(𝑥𝑥 + 𝑐𝑐), then the working becomes very easy.

Exercise 4.4 Practise some manipulation.

1) Rewrite these functions in the form

𝑦𝑦 = sin 𝑏𝑏(𝑥𝑥 + 𝑐𝑐), 𝑦𝑦 = cos 𝑏𝑏(𝑥𝑥 + 𝑐𝑐) or 𝑦𝑦 = tan 𝑏𝑏(𝑥𝑥 + 𝑐𝑐)

a) 𝑦𝑦 = sin �2𝑥𝑥 + 𝜋𝜋2� =____________ b) 𝑦𝑦 = cos(2𝑥𝑥 − 𝜋𝜋)=_____________

c) 𝑦𝑦 = tan �3𝑥𝑥 + 𝜋𝜋2�=____________ d) 𝑦𝑦 = cos �𝑥𝑥

4− 𝜋𝜋

8�=______________

e) 𝑦𝑦 = tan �𝑥𝑥3− 𝜋𝜋

6�=_____________ f) 𝑦𝑦 = sin 1

2(3𝑥𝑥 + 𝜋𝜋)=____________

2) Write down the period and phase shift for each of these functions:

a) 𝑦𝑦 = cos 14�𝑥𝑥 − 𝜋𝜋

2�______________ b) 𝑦𝑦 = tan 3 �𝑥𝑥 + 𝜋𝜋

6�______________

c) 𝑦𝑦 = sin 32�𝑥𝑥 + 𝜋𝜋

3�______________ d) 𝑦𝑦 = cos 2 �𝑥𝑥 − 𝜋𝜋

2�______________

e) 𝑦𝑦 = sin 2 �𝑥𝑥 + 𝜋𝜋4�______________ f) 𝑦𝑦 = tan 1

3�𝑥𝑥 − 𝜋𝜋

2�______________

2𝑥𝑥 − 𝜋𝜋2

= 0

i.e. 2𝑥𝑥 = 𝜋𝜋2

i.e. 𝑥𝑥 = 𝜋𝜋4



Graphs involving multiple transformations You have already looked at graphs which involve some of the transformations, either one or two at a time. Now you will be sketching or drawing accurate graphs which could involve any or all of the transformations.

Recall: ‘𝑎𝑎’ controls vertical dilations – amplitude

‘𝑏𝑏’ controls horizontal dilations – �𝑝𝑝𝑝𝑝𝑝𝑝𝑠𝑠𝑐𝑐𝑎𝑎 = 2𝜋𝜋𝑏𝑏�

‘𝑐𝑐’ controls horizontal slides – phase shift

‘𝑎𝑎’ controls vertical slides.

Example

We want to sketch 𝑦𝑦 = 2 sin 2 �𝑥𝑥 − 𝜋𝜋2� − 1.

What transformations will we need to apply to the basic sine curve?

The following diagrams and explanations are these for you to see how this function could be graphed manually. You will of course do it on a CAS calculator.



Point A is the usual location for the origin. From the rule for the function, we can see that there is a phase shift 𝜋𝜋

2 units to the right. We need to work out the period so that we can determine the

horizontal scale. When we know the scale, we can draw in the vertical axis.

Because 𝑏𝑏 = 2, we know the period is 2𝜋𝜋2

or 𝜋𝜋. One full wave occupies 𝜋𝜋 radians on the horizontal scale.

Also, because ‘𝑎𝑎’ is 2, we know that the amplitude is 2. This enables us to mark unit divisions on the vertical axis.

Now, we notice that ‘𝑎𝑎’ is -1, indicating that the graph slides down one unit. In other words, the horizontal axis moves up one unit. We can then draw in the horizontal axis, and mark the scales on both axes.



Sketch it on your calculator over the domain −2𝜋𝜋 ≤ 𝑥𝑥 ≤ 2𝜋𝜋. Be careful when entering the function as it is important to use brackets correctly.

Graphical solutions of trigonometric equations Find all the example of the solutions to tan 𝑥𝑥 = 1.5 over the domain −𝜋𝜋 ≤ 𝑥𝑥 ≤ 2𝜋𝜋 and give your answer correct to one decimal place.

Here we have graphed 𝑦𝑦 = tan 𝑥𝑥 and 𝑦𝑦 = 1.5. the points of intersection represent the points where tan 𝑥𝑥 = 1.5.

What we need to give as the solutions, is the set of values of 𝑥𝑥 which make this equation true. We use a numerical scale to read off the solutions:

𝑥𝑥 ∈ {−2.1, 1.0, 4.1} (to 1 d.p.)

Check this on your calculator. Remember to enter a domain.

Solve this one yourself

Using a graph, find all the solutions for −5 ≤ 𝑥𝑥 ≤ 5 to the equation 2 sin 𝑥𝑥 = −1 to (1 d.p.)

Frist rewrite in the form sin 𝑥𝑥 = 𝑐𝑐

i.e. sin 𝑥𝑥 = _______

Now draw the curve 𝑦𝑦 = sin 𝑥𝑥, −5 ≤ 𝑥𝑥 ≤ 5



Did you get something like this?

Draw in your line 𝑦𝑦 = 𝑐𝑐.

Mark in the points of intersection.

Find the 𝒃𝒃 values to 1 decimal place.

Try to be as accurate as you can by using the diagram above.

State the solutions: 𝑥𝑥 ∈ { ______________________________________ }

Check your answers on your calculator, were you pretty close?



Lesson 5 Angle Sum and Difference


• Review the unit circle and in which quadrant sine, cosine and tangent are positive and negative

• Review exact values of sine, cosine and tangent for 30o, 45o and 60o and their correspondingvalue in radians.

• Prove and apply the angle sum and difference identities

• Use the identity tan 𝑥𝑥 = sin𝑥𝑥cos𝑥𝑥

to simplify suitable trigonometric expressions

• Establish and use the identity sin2 𝜃𝜃 + cos2 𝜃𝜃 = 1 to prove other identities

Review Look through the following pages to refresh your memory with concepts that should have been covered in year 10 or earlier in the year.

Trigonometric ratios in the unit circle For angle 𝜃𝜃,

sin𝜃𝜃 = 𝑦𝑦

cos𝜃𝜃 = 𝑥𝑥

tan𝜃𝜃 = 𝑦𝑦𝑥𝑥

Remember, tan𝜃𝜃 is not defined for values of 𝜃𝜃 which are odd multiples of 90o (or 𝜋𝜋2 radians).



Signs of the trigonometric ratios



Tangent in terms of sine and cosine In the right triangle ABC

The Pythagorean identity

We can use the Pythagorean theorem to show that sin2 𝜃𝜃 + cos2 𝜃𝜃 = 1

Let P be the point with co-ordinates (cos𝜃𝜃 , sin 𝜃𝜃).

We know that PA = sin𝜃𝜃 and OA = cos 𝜃𝜃.

In the ΔOAP, because ∠OAP is a right angle, then

PA2 + OA2 = OP2

i.e. sin2 𝜃𝜃 + cos2 𝜃𝜃 = 1 (OP is the radius of the circle)

sin 𝑥𝑥 = 𝑎𝑎𝑐𝑐 ,

rearranging to make 𝑎𝑎 the subject

𝑎𝑎 = 𝑐𝑐 sin 𝑥𝑥

cos 𝑥𝑥 = 𝑏𝑏𝑐𝑐

rearranging to make 𝑏𝑏 the subject

𝑏𝑏 = 𝑐𝑐 cos 𝑥𝑥

We know that tan 𝑥𝑥 = 𝑎𝑎𝑏𝑏

Substituting 𝑎𝑎 and 𝑏𝑏

tan 𝑥𝑥 = 𝑐𝑐 sin𝑥𝑥𝑐𝑐 cos𝑥𝑥

tan 𝑥𝑥 = sin𝑥𝑥cos𝑥𝑥

(𝑠𝑠𝑠𝑠𝑎𝑎 𝜃𝜃)2 is usually written as 𝑠𝑠𝑠𝑠𝑎𝑎2 𝜃𝜃

So as not to be confused by 𝑠𝑠𝑠𝑠𝑎𝑎(𝜃𝜃2)



Let’s write the identity here again.

sin2 𝜃𝜃 + cos2 𝜃𝜃 = 1

Rearranging we get

sin2 𝜃𝜃 = 1 − cos2 𝜃𝜃 or cos2 𝜃𝜃 = 1 − sin2 𝜃𝜃

These may then be factorised to

sin2 𝜃𝜃 = (1 − cos 𝜃𝜃)(1 + cos 𝜃𝜃) or cos2 𝜃𝜃 = (1 − sin𝜃𝜃)(1 + sin𝜃𝜃)

Test yourself

Try and write down the different forms of the Pythagorean identity that we have just generated, without looking back.

1. sin2 𝑥𝑥 + cos2 𝑥𝑥 = 1

2. cos2 𝜃𝜃 = =

3. sin2 𝜃𝜃 = =

Sine and cosine of �𝝅𝝅𝟐𝟐− 𝜽𝜽� or (𝟗𝟗𝟗𝟗°− 𝜽𝜽)

Consider the diagram below which is part of a unit circle diagram.



Let ∠ POS = ∠ROQ = 𝜃𝜃.

Then ∠QOS = 𝜋𝜋2− 𝜃𝜃.

Draw QB and PA perpendicular to OS.

We then have two right triangles, QBD and OAP.

Because QB is parallel to RO, and OQ meets them, we have

∠ ROQ = ∠ OQB = 𝜃𝜃 (alternate ∠𝑠𝑠).

In ∆𝑠𝑠 QBO, OAP,

∠OQB = ∠POA (= 𝜃𝜃)

∠OBQ = ∠POA (both right angles)

and OQ = PO (radii of circle)

∴ ∆𝑠𝑠 QBO and OAP are congruent

i.e. QB = OA = 𝑎𝑎 P(𝑎𝑎, 𝑏𝑏)

and OB = 𝑃𝑃𝑃𝑃 = 𝑏𝑏 Q(𝑏𝑏,𝑎𝑎)

Now,

sin �𝜋𝜋2− 𝜃𝜃� = sin∠SOQ = QB = OA = 𝑎𝑎 = cos𝜃𝜃

and

cos �𝜋𝜋2− 𝜃𝜃� = cos∠SOQ = OB = PA = 𝑏𝑏 = sin 𝜃𝜃.

sin �𝜋𝜋2− 𝜃𝜃� = cos 𝜃𝜃 cos �

𝜋𝜋2− 𝜃𝜃� = sin𝜃𝜃



Sine and cosine of �𝝅𝝅𝟐𝟐

+ 𝜽𝜽� or (𝟗𝟗𝟗𝟗° + 𝜽𝜽)

If we use the following diagram, and assume that we can prove the

triangles OAP and QBO are congruent as on the previous page, we can

arrive at the identities for sin �𝜋𝜋2

+ 𝜃𝜃� and cos �𝜋𝜋2

+ 𝜃𝜃� .

Again, we have QB = OA = 𝑎𝑎 P(𝑎𝑎, 𝑏𝑏)

and OB = PA = 𝑏𝑏 Q(−𝑏𝑏,𝑎𝑎)

Then sin �𝜋𝜋2

+ 𝜃𝜃� = ∠SOQ

= QB

= OA

= 𝑎𝑎 (𝑦𝑦 − coordinate of Q)

= cos 𝜃𝜃 (𝑥𝑥 − coordinate of P)

Also, cos �𝜋𝜋2

+ 𝜃𝜃� = cos∠SOQ

= -OB

= -PA

= -𝑏𝑏 (𝑥𝑥 − coordinate of Q)

= − sin𝜃𝜃 (−ve 𝑦𝑦 − coordiante of P)

sin �𝜋𝜋2

+ 𝜃𝜃� = cos 𝜃𝜃 cos �𝜋𝜋2

+ 𝜃𝜃� = −sin𝜃𝜃



The distributive property does not apply This lesson is concerned with developing and using the formula for

sin(𝑃𝑃 ± 𝐵𝐵) and

cos(𝑃𝑃 ± 𝐵𝐵).

A few minutes with your calculator will demonstrate to you, quite conclusively, that sin(𝑃𝑃 + 𝐵𝐵) is not equal to sin𝑃𝑃 + sin𝐵𝐵.

e.g. sin(20° + 10°)

We know that sin(20° + 10°) = 30°.

We also know that sin 30° = 12

or 0.5.

Try adding sin 20° and sin 10°. What do you get? _____________.

Since this is not equal to 0.5, we have made the point.

If you haven’t been convinced by this example, here are a few more to try.

Exercise 5.1

1. cos(30° + 60°) = cos 90° = __________

cos 30° + cos 60° = _________

2. tan(45° + 45°) = tan 90° = __________

tan 45° + tan 45° = ________

3. sin(60° + 60°) = sin 120° = __________

sin 60° + 60° = ________

We could try examples with sin(𝑃𝑃 − 𝐵𝐵) or with cos(𝑃𝑃 − 𝐵𝐵), etc, but it seems obvious that we will achieve the same results.



We can state the results, and then demonstrate how they are derived.

[Note where the signs do or do not correspond.]

The addition formulae

If we take the angle POQ in fig 4.1., and rotate it in the clockwise direction, through an angle equal in size to ∠𝐵𝐵, we now have:

Q coincides with Q’ on the on the 𝑥𝑥 axis,

P coincides with P′ in fig 4.2.

and the ∠P′OQ ≡ ∠POQ = A − B.

Now, let’s extend our diagrams as follows.

sin (𝑃𝑃 ± 𝐵𝐵) = sin𝑃𝑃 cos𝐵𝐵 ± cos𝑃𝑃 sin𝐵𝐵

cos (𝑃𝑃 ± 𝐵𝐵) = cos𝑃𝑃 cos𝐵𝐵 ∓ sin𝑃𝑃 sin𝐵𝐵

tan (𝑃𝑃 ± 𝐵𝐵) = tan𝑃𝑃 ± tan𝐵𝐵

1 ∓ tan𝑃𝑃 tan𝐵𝐵



PQ is the same length as P’Q’ because they are chords formed by angles of equal size.

PD, QE and P’G are drawn perpendicular to the 𝑥𝑥 axis.

Now, (PQ) = (P′Q′)²

i.e. * (FQ)2 + (PF)² = (GQ’)² + (P’G)² [Pythagoras]

But, FQ = OE − OD

cos𝐵𝐵 − cos𝑃𝑃 in Fig 4.3

PF = PD – QE

= sin𝑃𝑃 − sin𝐵𝐵

GQ’ = OQ’ − OG

= 1 − cos(𝑃𝑃 − 𝐵𝐵) in Fig 4.4

and P’G = sin(𝑃𝑃 − 𝐵𝐵)

Now, substituting these in the line marked with the asterisk on the previous page, we have:

(FQ)2 + (PF)2 = (GQ′)2 + (P′G)2

i.e. (cos𝐵𝐵 − cos𝑃𝑃)² + (sin𝑃𝑃 − sin𝐵𝐵)² = [1 − cos(𝑃𝑃 − 𝐵𝐵)]² + [sin(𝑃𝑃 − 𝐵𝐵)]²

Multiplying out

co𝑠𝑠2𝐵𝐵 − 2 cos𝑃𝑃 cos𝐵𝐵 + cos2𝑃𝑃 + sin2𝑃𝑃 − 2 sin𝑃𝑃 sin𝐵𝐵 + sin²𝐵𝐵

= 1 − 2 cos(𝑃𝑃 − 𝐵𝐵) + cos2(𝑃𝑃 − 𝐵𝐵) + sin2(𝑃𝑃 − 𝐵𝐵)

Simplifying using the identity sin2𝜃𝜃 + cos2𝜃𝜃 = 1

2 − 2 cos𝑃𝑃 cos𝐵𝐵 − 2 sin𝑃𝑃 sin𝐵𝐵 = 2 − 2 cos(𝑃𝑃 − 𝐵𝐵)

Rearranging

cos(𝑃𝑃 − 𝐵𝐵) = cos𝑃𝑃 cos𝐵𝐵 + sin𝑃𝑃 sin𝐵𝐵



Exercise 5.2 Having derived one formula, we can operate on it in various ways to obtain the others.

Part of the working is left for you to complete.

1. cos (𝑃𝑃 + 𝐵𝐵) = cos𝑃𝑃 cos𝐵𝐵 − sin𝑃𝑃 sin𝐵𝐵

Replace 𝐵𝐵 by (−𝐵𝐵).

cos (𝑃𝑃 − 𝐵𝐵) = cos𝑃𝑃 cos(−𝐵𝐵) − sin𝑃𝑃 sin(−𝐵𝐵)

i.e. cos (A−𝐵𝐵) = __________________________

2. sin (𝑃𝑃 + 𝐵𝐵) = cos �𝜋𝜋2− (𝑃𝑃 + 𝐵𝐵)�

= cos �𝜋𝜋2− 𝑃𝑃_________�

= cos ��𝜋𝜋2− 𝑃𝑃� _________] 𝑇𝑇𝑝𝑝𝑝𝑝𝑎𝑎𝑡𝑡 𝑎𝑎𝑠𝑠 cos(𝑃𝑃 − 𝐵𝐵)

= cos �𝜋𝜋2− 𝑃𝑃� cos(−𝐵𝐵) − sin �𝜋𝜋

2− 𝑃𝑃� sin(−𝐵𝐵)

= ________ cos𝐵𝐵 ____________sin𝐵𝐵

3. sin (𝑃𝑃 − 𝐵𝐵) = sin[𝑃𝑃 + (−𝐵𝐵)]

= sin𝑃𝑃 cos(−𝐵𝐵) + cos𝑃𝑃 sin(−𝐵𝐵)

= _________________________

Check in your textbook to confirm your working.



We use the fact that tan𝑃𝑃 = sin𝐴𝐴cos𝐴𝐴

to derive the tan(𝑃𝑃 + 𝐵𝐵) and tan(𝑃𝑃 − 𝐵𝐵) formulae.

4. tan (𝑃𝑃 + 𝐵𝐵) =sin(𝐴𝐴+𝐵𝐵)cos(𝐴𝐴+𝐵𝐵)

= sin𝐴𝐴 cos𝐵𝐵+cos𝐴𝐴 sin𝐵𝐵cos𝐴𝐴 cos𝐵𝐵−sin𝐴𝐴 sin𝐵𝐵

We use a mathematician’s trick here and divide numerator and denominator by cos𝑃𝑃 cos𝐵𝐵. The resulting expression is a complex fraction, but as long as we keep our wits about us, that shouldn’t be a problem.

sin𝐴𝐴cos𝐵𝐵cos𝐴𝐴cos𝐵𝐵+

cos𝐴𝐴sin𝐵𝐵𝑐𝑐𝑐𝑐𝑐𝑐 𝐴𝐴cos𝐵𝐵

cos𝐴𝐴cos𝐵𝐵cos𝐴𝐴cos𝐵𝐵−

sin𝐴𝐴sin𝐵𝐵cos𝐴𝐴cos𝐵𝐵

= tan𝐴𝐴+tan𝐵𝐵1−tan𝐴𝐴 tan𝐵𝐵

5. Fill in the steps for tan (𝑃𝑃 − 𝐵𝐵) for yourself

tan (𝑃𝑃 − 𝐵𝐵) =

=

=

= _____________________



Alternative method:

tan(𝑃𝑃 − 𝐵𝐵) = tan[𝑃𝑃 + (−𝐵𝐵)] Note: use tan(−𝐵𝐵) = − tan(𝐵𝐵)

= tan𝐴𝐴+tan(−𝐵𝐵)1−tan𝐴𝐴 tan(−𝐵𝐵)

= _____________________

Now we have established all of the additional formulae, we will put them to use.

Example Find the exact formula of cos 15

Solution 1

cos 15

= cos(45 − 30 )

= cos 45 cos 30 + sin 45 sin 30

= 1√2

× √32

+ 1√2

× 12

= √32√2

+ 12√2

= √3+12√2

Solution 2

cos 15

= cos(60 − 45 )

= cos 60 cos 45 + sin 60 sin 45

= 12

× 1√2

+ √32

× 1√2

= 12√2

+ √32√2

= √3+12√2



Try this one yourself 6. Use the addition formulae to find the exact value of tan 165 .

tan 165 = tan(_______ + _______)

= tan _______+ tan _______ 1−tan _______ tan _______

= − tan _______ +tan _______

1−(−tan _______ tan _______)

= 1− tan______ 1+ tan_______

= 1− ______ 1+ _______



Lesson 6

Graphs of trigonometric functions horizontal dilations


• identify contexts suitable for modelling by trigonometric functions and use them so solvepractical problems.

Periodic functions There are many types of functions which are 'periodic' in nature, but we only need to consider the trigonometric functions at the moment.

Image by Oleg Alexandrov - Own work, Public Domain, https://commons.wikimedia.org/w/index.php?curid=2683866

Image by Krishnavedala - Own work, CC BY-SA 3.0, https://commons.wikimedia.org/w/index.php?curid=15289177

It is easy to see which functions are periodic by looking at their graphs.

Where the basic shape of the graph is repeated time and time again, we can call it periodic.

The distance over which the function has its basic shape is called the period of the function. For a sine curve, the period is the length of one wave.

Image by Geek3 - Own work, CC BY 3.0, https://commons.wikimedia.org/w/index.php?curid=9531683

https://commons.wikimedia.org/w/index.php?curid=2683866



https://www.google.com.au/url?sa=i&url=https%3A%2F%2Far.m.wikipedia.org%2Fwiki%2F%25D9%2585%25D9%2584%25D9%2581%3APeriodic_function_illustration.svg&psig=AOvVaw0N0LNis88UFOllnaxSTuRF&ust=1585712171743000&source=images&cd=vfe&ved=0CAIQjRxqFwoTCLDY-q_kw-gCFQAAAAAdAAAAABAD

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Where do we find periodic functions? Many practical situations in the sciences are applications of periodic functions. The functions concerned relate to some cyclic or periodic process unfolding in time.

In the biological sciences, there are many phenomena that show cyclic behaviour. These are called biological rhythms. The pattern on an electrocardiogram, which shows the length of time between heartbeats, is an example. Many animals show periodic, seasonal variations in their behaviour. Certain diseases also show cyclic seasonal variations.


The following diagram shows a non-trigonometric periodic function

In the physical sciences, the incidences of periodic behaviour are such that they are characteristically very precise. The swing of a pendulum in a clock duplicates the previous swing exactly. This is of course how the clock manages to keep accurate time. A stone or a single drop of water falling into a pond produces a wave pattern which is periodic.

There are many other examples, some of which you may have encountered:

• sound waves

• oscillating springs

• strings on a musical instrument

• the vibration of air in a pipe.


https://www.google.com.au/url?sa=i&url=https%3A%2F%2Fcommons.wikimedia.org%2Fwiki%2FFile%3ABiological_clock_human.svg&psig=AOvVaw3CjsOU_eTYte8DHjGwv-dX&ust=1585712318471000&source=images&cd=vfe&ved=0CAIQjRxqFwoTCLiV_vHkw-gCFQAAAAAdAAAAABAD



Example Entry to a deep water harbour is through a shallow entrance over a reef. At low tide the entrance is 5 m deep and at high tide the entrance is 10 m deep. The water depth at the entrance can be modelled by a cosine function where depth is measured in metres and time in hours.

A fully loaded ship drawing 7.5 metres (ie 7.5 m of the vessel is below the waterline) approaches the harbour entrance at 10.00 am on a particular day. If high tide on that day was 8.00 am:

a) when will the ship be able to first enter the harbour?

b) when will the ship be able to leave the harbour if it takes 10 hours to be ready to leave(dock, unload, fully reload and be back at the entrance)?

(You must give evidence of how you have calculated these times.)



Solution a) The ship needs to wait until the water is above 7.5 m. this is approximately 4.5 hours

after 8 am, therefore not until 12.30 p.m.

b) 10 + 4.5 is 14.5 hours. By then the tide is too low, they will need to wait untilapproximately 16.5 hours after which is approximately 12.30 am the followingmorning.



Exercise 6.1 Biorhythms are cycles that are thought to track a person’s intellectual, physical and emotional well-being.

1. Below is Vince’s chart for his emotional cycle, where the horizontal scale shows days andd = 0 represents today. The vertical scale shows his emotional state as a percentage.

a) Write a description of his emotional changes over the last two weeks and the next twoweeks.

b) Which graph of the standard trigonometric functions: 𝑦𝑦 = sin 𝑥𝑥, 𝑦𝑦 = cos 𝑥𝑥 or𝑦𝑦 = tan 𝑥𝑥, for 𝑥𝑥 from −𝜋𝜋 ≤ 𝑥𝑥 ≤ 𝜋𝜋, best matches the emotional cycle in Vince’s chart?(Check these functions on your calculator if you are not sure.) remember to useradians

c) If today is 23rd May, what dates will Vince’s emotional cycle be at a 100% low duringthe next three months?

d) If the days crossing from positive to negative are considered critical, identify the nexttwo days, after today, that need to be watched out for as emotionally critical.



e) Use the function given to complete the following table of approximate values for thefunction, rounding to two decimal places when required. 𝑓𝑓(𝑎𝑎) = 100 cos �𝜋𝜋𝜋𝜋

14�

remember to use radians

f) What is the period it takes for the graph to repeat itself?

g) Vince’s physical cycle peaked, at 100%, two days before his emotional cycle reached itshighest value. His physical cycle has the same shape and period length as his emotionalcycle. When would you expect him to be at his lowest (-100%) physical value and onwhich day would it change from positive values to negative values?

h) Show that 𝑓𝑓(𝑎𝑎) = 100 cos �𝜋𝜋(𝜋𝜋+2)14

�models Vince’s physical cycle and give the approximate physical percentages for 𝑎𝑎 = 0, 2, 4, 6, 8, 10, 12, 14.

i) How is the two-day difference between Vince’s physical and emotional cycles reflectedin the defining rule of the function?

j) Vince’s intellectual cycle peaked, at 100%, six days before his emotional cycle reachedits highest value. His intellectual cycle has the same shape and period length as hisother two cycles. Find a cosine function that models Vince’s intellectual cycle and givethe approximate intellectual percentages for 𝑎𝑎 = 0, 2, 4, 6, 8, 10, 12, 14.

d 0 2 4 6 7 8 10 12 14

𝑓𝑓(𝑎𝑎) 2dp

d 0 2 4 6 8 10 12 14


d 0 2 4 6 7 8 10 12 14




There are also secondary cycles for Vince, for mastery, passion and wisdom which are modelled with the same period by the following functions:

Mastery ( 4)( ) 90cos14df d + =

π

Passion: ( 3)( ) 85cos14df d + =

π

Wisdom: ( 1)( ) 80cos14df d + =

π

k) Sketch the graphs of the functions for these three secondary cycles for Vince’s last twoweeks and future two weeks and write a description to compare them to each otherand to the original three cycles in the earlier tasks. Use their main features, such ashigh and low values and intercepts, to compare them. How do the high and low valuesaffect the function rule?

l) The amplitude for each of the emotional, physical and intellectual cycles is 100. Givethe amplitude for mastery, passion and wisdom.



m) Consider another secondary cycle for balance, modelled with the same period by thefollowing function:

Balance: ( 7)( ) 95cos14df d − =

π

Give the high and low values and amplitude for this cycle and explain what effect the –7 has.

n) Create defining rules for functions to model these cycles if the following information isknown:

Highest value

Lowest value

Peak compared to emotional cycle

𝑓𝑓(𝑎𝑎)

Cycle 1 72 -72 3 days earlier

Cycle 2 55 -55 2 days later

Cycle 3 68 -68 5 days later

o) Vince realises that his biorhythms can be modelled by sine functions as well as cosine

functions because he knows that: cos sin2

x x = −

π .

He rewrites the function for the emotional cycle as (7 )( ) 100sin14

df d − =

π

Show how he obtains this sine function.

p) Rewrite the functions for her physical and intellectual cycles as sine functions.



2. A prison cell block, B, has a security spotlight, on the roof, with back-to-back beams thatrotate in an anticlockwise direction to light up a wall joining two cell blocks, A and C andcontaining a gate, as shown in the diagram right:

The time for a full rotation, of one beam, is 8 seconds and the starting point is shiningdirectly in front at the middle of the gate.

a) Sketch the position of the centre of the beams of light on the wall, 3 m above andbelow the gate, for the first 16 seconds, using the axes below:

b) Compare your sketch in a) to the graph of 𝑦𝑦 = tan 𝑥𝑥. How many complete cycles areshown?

c) For what values of time is the function that models the sketch undefined? Relate thesetimes to the angle of rotation (in radians) of the first beam.

d) What is the period of the function in seconds and the corresponding positive angle inradians?

Cell block A

Cell block C

Gate Spotlight

Cell block B

Exercise yard

time4 8 12 16

position on wall above gate

-3

3

(seconds)



e) Compare the graphs of the following functions and choose one to model the two beam spotlight rotations.

𝑦𝑦 = 3tan �𝜋𝜋𝑥𝑥2� ; 𝑦𝑦 = tan �𝜋𝜋𝑥𝑥

2� ,𝑦𝑦 = 3tan �𝜋𝜋𝑥𝑥

4� ,𝑦𝑦 = tan �𝜋𝜋𝑥𝑥

4�

f) One of the lights is not working. Redraw your sketch over the same 16 as seconds if only one beam is operational. Does it make a difference if the back light rather than the front light is not working?

time4 8 12 16


-3

3

(seconds)

time4 8 12 16


-3

3

(seconds)



Lesson 7

Solving trigonometric equations By the end of this lesson you should be able to

• solve equations involving trigonometric functions using technology

• solve equations involving trigonometric functions algebraically

Solving trigonometric equations As an example, suppose you were asked to solve sin 𝜃𝜃 = 0.8. Your calculator would give

𝜃𝜃 = 0.927… in radian mode or 𝜃𝜃 = 53.13°… in degree mode.

However you have seen that sine is a periodic function, so there can be an infinite number of solutions unless we specify the domain of 𝜃𝜃.

A look at the graph of sin𝜃𝜃 demonstrates this fact. There are eight solutions for −2𝜋𝜋 ≤ 𝜃𝜃 ≤ 6𝜋𝜋

A unit circle diagram is helpful in working out the solutions too. Adding or subtracting 360 or (2𝜋𝜋) will give further solutions.



If we worked in radians, for the given equation, sin𝜃𝜃 = 0.8, we could have the following solutions for −2𝜋𝜋 ≤ 𝜃𝜃 ≤ 4𝜋𝜋.

The reference angle 𝜃𝜃 is 0.927 radians.

The second quadrant solution for −2𝜋𝜋 ≤ 𝜃𝜃 ≤ 4𝜋𝜋 is (𝜋𝜋 − 𝜃𝜃) or 2.215

Solutions for (0 ≤ 𝜃𝜃 ≤ 2𝜋𝜋) are 0.927 and 2.215

Adding 2𝜋𝜋 for (2𝜋𝜋 ≤ 𝜃𝜃 ≤ 4𝜋𝜋) gives 7.210 and 8.498

Subtracting 2𝜋𝜋 for (−2𝜋𝜋 ≤ 𝜃𝜃 ≤ 0) gives -5.356 and -4.068

The working for sine and cosine equations is similar, except that cosine is positive in the first and fourth quadrants, while tangent is positive in the first and third quadrants.

A unit circle sketch is always helpful.

Study the next few diagrams to refamiliarise yourself with these concepts.

Ratios for (−𝜽𝜽) = −𝐬𝐬𝐬𝐬𝐭𝐭𝜽𝜽), �𝝅𝝅𝟐𝟐

± 𝜽𝜽� , (𝝅𝝅 ± 𝜽𝜽)

Ratios for (−𝜽𝜽)

sin(−𝜃𝜃) = − sin𝜃𝜃 tan(−𝜃𝜃) = − tan𝜃𝜃 cos(−𝜃𝜃) = cos 𝜃𝜃



Ratios for �𝝅𝝅𝟐𝟐

± 𝜽𝜽�

sin �𝜋𝜋2

+ 𝜃𝜃� = cos 𝜃𝜃 sin �𝜋𝜋2− 𝜃𝜃� = cos 𝜃𝜃

cos �𝜋𝜋2

+ 𝜃𝜃� = −sin𝜃𝜃 cos �𝜋𝜋2− 𝜃𝜃� = sin𝜃𝜃

tan �𝜋𝜋2

+ 𝜃𝜃� = − 1 tan 𝜃𝜃

tan �𝜋𝜋2− 𝜃𝜃� = 1

tan 𝜃𝜃

Ratios for (𝝅𝝅 ± 𝜽𝜽)

sin(𝜋𝜋 + 𝜃𝜃) = −sin𝜃𝜃 sin �𝜋𝜋2− 𝜃𝜃� = cos 𝜃𝜃

cos(𝜋𝜋 + 𝜃𝜃) = − cos𝜃𝜃 cos �𝜋𝜋2− 𝜃𝜃� = sin𝜃𝜃

tan(𝜋𝜋 + 𝜃𝜃) = tan𝜃𝜃 tan (𝜋𝜋 − 𝜃𝜃) = tan𝜃𝜃

Note: 𝝅𝝅 𝒄𝒄𝒄𝒄𝒄𝒄 𝒃𝒃𝒃𝒃 𝒓𝒓𝒃𝒃𝒓𝒓𝒓𝒓𝒄𝒄𝒄𝒄𝒃𝒃𝒓𝒓 𝒃𝒃𝒚𝒚 𝟏𝟏𝟏𝟏𝟗𝟗 𝒄𝒄𝒄𝒄𝒓𝒓𝝅𝝅𝟐𝟐𝒄𝒄𝒄𝒄𝒄𝒄 𝒃𝒃𝒃𝒃 𝒓𝒓𝒃𝒃𝒓𝒓𝒓𝒓𝒄𝒄𝒄𝒄𝒃𝒃𝒓𝒓 𝒃𝒃𝒚𝒚 𝟗𝟗𝟗𝟗



Example Follow the steps below to solve cos𝜃𝜃 = 0.6, 0 ≤ 𝜃𝜃 ≤ 2𝜋𝜋.

Solution Use the domain to find the number of solutions.

cos𝜃𝜃 = 0.6, 0 ≤ 𝜃𝜃 ≤ 2𝜋𝜋 has two solutions, one in each of the first and fourth quadrants where cosine is positive.

The domain is in radians so the solutions must be in radians in exact form or rounded to a given number of decimal places.

The first quadrant solution is cos−1 0.6 = 0.93 (2 d.p.)

The fourth quadrant solution is 2π − 0.9273 = 5.36 (2 dp).

Note also that sinθ = (−0.5), for (0 ≤ θ ≤ 2π) has two exact solutions, one in each of the third and fourth quadrants with a reference angle of π

6. Solutions are 7π

6 and 11π

6.

Also, when solving trigonometric equations, you must be wary of how you use the formulae. Consider this example:

Solve sin(2𝑥𝑥) = cos(𝑥𝑥). 0 ≤ θ ≤ 2π

You can't do this:

You’ve lost solution(s) because you haven’t considered the case where cos(𝑥𝑥) = 0

sin(𝑥𝑥 + 𝑥𝑥) = cos(𝑥𝑥)

sin(𝑥𝑥)cos(𝑥𝑥) + cos(𝑥𝑥) sin(𝑥𝑥) = cos(𝑥𝑥)

2 sin(𝑥𝑥) cos(𝑥𝑥) = cos(𝑥𝑥)

2 sin(𝑥𝑥) = 1

sin(𝑥𝑥) = 12

DO NOT DO THIS



So you MUST factorise instead:

Now we have 2 equations to solve for 𝑥𝑥. We need to look at the domain. 0 ≤ θ ≤ 2π.

For cos 𝑥𝑥 = 0

𝑥𝑥 = 𝜋𝜋2

or 3𝜋𝜋2

For sin 𝑥𝑥 = 12

𝑥𝑥 = 𝜋𝜋6 or 5𝜋𝜋

6

Therefore for Solve sin(2𝑥𝑥) = cos(𝑥𝑥). 0 ≤ θ ≤ 2π

𝑥𝑥 = 𝜋𝜋2

, 𝜋𝜋6

, 3𝜋𝜋2

, 5𝜋𝜋6

sin(2𝑥𝑥) = cos(𝑥𝑥)

2 sin(𝑥𝑥) cos(𝑥𝑥) = cos(𝑥𝑥)

2 sin(𝑥𝑥) cos(𝑥𝑥) − cos(𝑥𝑥) = 0

cos(𝑥𝑥)[2 sin(𝑥𝑥) − 1] = 0

So either cos(𝑥𝑥) = 0 or 2 sin(𝑥𝑥) = 1

or sin(𝑥𝑥) = 1 sin(𝑥𝑥) = 1

2

DO THIS

Remember to use the horizontal axis for cos.

Remember to use the vertical axis for sin.



Exercise 7.1 1) Solve the following equations for the given domain, leaving answers exact where you can.

Check your answers by using your calculator. Remember to include the correct domain.

a) sin 0.5θ = , 2 2π θ π− ≤ ≤

b) 1sin , 0 22

θ θ= − ≤ ≤ π

c) 4 sin �𝑥𝑥 + 𝜋𝜋2� = 2 for 0 ≤ 𝑥𝑥 ≤ 2𝜋𝜋

d) tan 2𝑥𝑥 = 1 for −180 ≤ 𝑥𝑥 ≤ 180



e) sin(2𝑥𝑥) = cos 𝑥𝑥 , for − 𝜋𝜋 ≤ 𝜃𝜃 ≤ 𝜋𝜋,

f) cos 𝑥𝑥 + 2 cos2 𝑥𝑥 = 0 for − 𝜋𝜋 ≤ 𝜃𝜃 ≤ 𝜋𝜋

g) cos �𝑥𝑥 + 𝜋𝜋4� = 1

√2 for −2𝜋𝜋 ≤ 𝑥𝑥 ≤ 0



Lesson 8 and 9

Exam practice


• refer to SCSA past exams

• Complete text practice as needed

The image below shows a screen shot from the SCSA website. In this location you can access past examinations and other support materials. Click the link below to be taken to the website:






Glossary https://senior-secondary.scsa.wa.edu.au/syllabus-and-support-materials/mathematics/mathematics-methods

Trigonometric functions

Angle sum and difference identites

The angle sum and difference identites for sine and cosine are given by

sin(𝑃𝑃 ± 𝐵𝐵) = sin𝑃𝑃 cos𝐵𝐵 ± cos𝑃𝑃 sin𝐵𝐵

cos(𝑃𝑃 ± 𝐵𝐵) = cos𝑃𝑃 cos𝐵𝐵 ∓ sin𝑃𝑃 sin𝐵𝐵

Area of a sector The area of a sector of a circle is given by 𝑃𝑃 = 12𝑝𝑝2𝜃𝜃, where 𝑃𝑃 is the sector area,

𝑝𝑝 is the radius and 𝜃𝜃 is the angle subtended at the centre, measured in radians.

Area of a segment The area of a segment of a circle is given by 𝑃𝑃 = 12𝑝𝑝2(𝜃𝜃 − sin𝜃𝜃), where 𝑃𝑃 is the

segment area, 𝑝𝑝 is the radius and 𝜃𝜃 is the angle subtended at the centre, measured in radians.

Circular measure Circular measure is the measurement of angle size in radians.

Length of an arc The length of an arc in a circle is given by ℓ = 𝑝𝑝𝜃𝜃, where ℓ is the arc length, 𝑝𝑝 is the radius and 𝜃𝜃 is the angle subtended at the centre, measured in radians. This is simply a rearrangement of the formula defining the radian measure of an angle.

Length of a chord The length of a chord in a circle is given by ℓ = 2𝑝𝑝 sin 12𝜃𝜃, where ℓ is the chord

length, 𝑝𝑝 is the radius and 𝜃𝜃 is the angle subtended at the centre, measured in radians.

Period of a function The period of a function 𝑓𝑓(𝑥𝑥) is the smallest positive number 𝑝𝑝 with the property that 𝑓𝑓(𝑥𝑥 + 𝑝𝑝) = 𝑓𝑓(𝑥𝑥) for all 𝑥𝑥. The functions sin 𝑥𝑥 and cos 𝑥𝑥 both have period 2𝜋𝜋 and tan 𝑥𝑥 has period 𝜋𝜋.

Radian measure The radian measure 𝜃𝜃 of an angle in a sector of a circle is defined by 𝜃𝜃 = ℓ𝑟𝑟

, where 𝑝𝑝 is the radius and ℓ is the arc length. Thus, an angle whose degree measure is 180 has radian measure 𝜋𝜋.

Sine rule and cosine rule The lengths of the sides of a triangle are related to the sine of its angles by the equations 𝑎𝑎

sin 𝐴𝐴= 𝑏𝑏

sin𝐵𝐵= 𝑐𝑐

sin 𝐶𝐶 .

This is known as the sine rule.





The lengths of the sides of a triangle are related to the cosine of one of its angles by the equation 𝑐𝑐2 = 𝑎𝑎2 + 𝑏𝑏2 − 2𝑎𝑎𝑏𝑏 cos𝐶𝐶.

This is known as the cosine rule.

Sine, cosine and tangent functions

Since each angle 𝜃𝜃 measured anticlockwise from the positive 𝑥𝑥-axis determines a point 𝑃𝑃 on the unit circle, we will define

the cosine of 𝜃𝜃 to be the 𝑥𝑥-coordinate of the point 𝑃𝑃

the sine of 𝜃𝜃 to be the 𝑦𝑦-coordinate of the point 𝑃𝑃

the tangent of 𝜃𝜃 is the gradient of the line segment 𝑂𝑂𝑃𝑃.



Solutions

Exercise 1.1

1.


𝜋𝜋4

𝜋𝜋3

𝜋𝜋2

2𝜋𝜋3

3𝜋𝜋4

5𝜋𝜋6

𝜋𝜋

sin 𝑡𝑡 (to 2 d.p.) 0 0.5 0.7 0.9 1 0.9 0.7 0.5 0


5𝜋𝜋4

4𝜋𝜋3

3𝜋𝜋2

5𝜋𝜋3

7𝜋𝜋4

11𝜋𝜋6

2𝜋𝜋

sin 𝑡𝑡 (to 2 d.p.) -0.5 -0.7 -0.9 -1 -0.9 -0.7 -0.5 0


9𝜋𝜋4

7𝜋𝜋3

5𝜋𝜋2

8𝜋𝜋3

11𝜋𝜋4

17𝜋𝜋6

3𝜋𝜋

sin 𝑡𝑡 (to 2 d.p.) 0.5 0.7 0.9 1 0.9 0.7 0.5 0


13𝜋𝜋4

10𝜋𝜋3

7𝜋𝜋2

11𝜋𝜋3

15𝜋𝜋4

23𝜋𝜋6

4𝜋𝜋

sin 𝑡𝑡 (to 2 d.p.) -0.5 -0.7 -0.9 -1 -0.9 -0.7 -0.5 0

2. the numbers started to repeat themselves after 2π

3.

4. What is the maximum value of sin (𝑡𝑡)?

4. -1

5. 0, π, 2π, 3π, 4π

6. Repeat for cos t. Remember to work in radian mode. And give your answers to one decimal place.



Value of 𝒕𝒕 0 𝝅𝝅𝟔𝟔

𝝅𝝅𝟒𝟒

𝝅𝝅𝟑𝟑

𝝅𝝅𝟐𝟐

𝟐𝟐𝝅𝝅𝟑𝟑

𝟑𝟑𝝅𝝅𝟒𝟒

𝟓𝟓𝝅𝝅𝟔𝟔

𝝅𝝅

cos 𝑡𝑡 (to 2 d.p.) 1 0.9 0.7 0.5 0 -0.5 -0.7 -0.9 -1

Value of 𝒕𝒕 𝟕𝟕𝝅𝝅𝟔𝟔

𝟓𝟓𝝅𝝅𝟒𝟒

𝟒𝟒𝝅𝝅𝟑𝟑

𝟑𝟑𝝅𝝅𝟐𝟐

𝟓𝟓𝝅𝝅𝟑𝟑

𝟕𝟕𝝅𝝅𝟒𝟒

𝟏𝟏𝟏𝟏𝝅𝝅𝟔𝟔

𝟐𝟐𝝅𝝅

cos 𝑡𝑡 (to 2 d.p.) -0.9 -0.7 -0.5 0 0.5 0.7 0.9 1

Value of 𝒕𝒕 𝟏𝟏𝟑𝟑𝝅𝝅𝟔𝟔

𝟗𝟗𝝅𝝅𝟒𝟒

𝟕𝟕𝝅𝝅𝟑𝟑

𝟓𝟓𝝅𝝅𝟐𝟐

𝟏𝟏𝝅𝝅𝟑𝟑

𝟏𝟏𝟏𝟏𝝅𝝅𝟒𝟒

𝟏𝟏𝟕𝟕𝝅𝝅𝟔𝟔

𝟑𝟑𝝅𝝅

cos 𝑡𝑡 (to 2 d.p.) 0.9 0.7 0.5 0 -0.5 -0.7 -0.9 -1

Value of 𝒕𝒕 𝟏𝟏𝟗𝟗𝝅𝝅𝟔𝟔

𝟏𝟏𝟑𝟑𝝅𝝅𝟒𝟒

𝟏𝟏𝟗𝟗𝝅𝝅𝟑𝟑

𝟕𝟕𝝅𝝅𝟐𝟐

𝟏𝟏𝟏𝟏𝝅𝝅𝟑𝟑

𝟏𝟏𝟓𝟓𝝅𝝅𝟒𝟒

𝟐𝟐𝟑𝟑𝝅𝝅𝟔𝟔

𝟒𝟒𝝅𝝅

cos 𝑡𝑡 (to 2 d.p.) -0.9 -0.7 -0.5 0 0.5 0.7 0.9 1

14. 1

15. 2

16. 𝜋𝜋2

, 3𝜋𝜋2

, 5𝜋𝜋2

, 7𝜋𝜋2

17. Both graphs have the same wave shape and repeat every 2π. Sin starts at (0,0)whilst cos starts at (0, 1)



18. Complete the tables of values of tan below, using your calculator to help you. Remember towork in radian mode. And give your answers to one decimal place.

Value of 𝒃𝒃 -π −𝟓𝟓𝝅𝝅𝟔𝟔

−𝟑𝟑𝝅𝝅𝟒𝟒

−𝟐𝟐𝝅𝝅𝟑𝟑

−𝝅𝝅𝟐𝟐

−𝝅𝝅𝟑𝟑 −

𝝅𝝅𝟒𝟒

−𝝅𝝅𝟔𝟔

0

tan 𝑥𝑥 0 0.6 1 1.7 Und -1.7 -1 -0.6 0

Value of 𝒃𝒃 0 𝝅𝝅𝟔𝟔

𝝅𝝅𝟒𝟒

𝝅𝝅𝟑𝟑

𝝅𝝅𝟐𝟐

𝟐𝟐𝝅𝝅𝟑𝟑

𝟑𝟑𝝅𝝅𝟒𝟒

𝟓𝟓𝝅𝝅𝟔𝟔

π

tan 𝑥𝑥 0 0.6 1 1.7 Und -1.7 -1 -0.6 0

Value of 𝒃𝒃 -1.4 -1.2 -1.0 -0.8 -0.6 -0.4 -0.2 0

tan 𝑥𝑥 -5.8 -2.6 -1.6 -1.0 -0.7 -0.4 -0.2 0

Value of 𝒃𝒃 0 0.2 0.4 0.6 0.8 1.0 1.2 1.4

tan 𝑥𝑥 0 0.2 0.4 0.7 1.0 1.6 2.6 6.0

19. Use as many of the above points as necessary to graph tan 𝑥𝑥 accurately on this grid. (the 𝜋𝜋 isapproximate.)

14. -π, 0, π

15. It gets very large

16. It becomes very small

17. On the left it is very large, on the right very small.

18. 𝜋𝜋

19. When 𝑥𝑥 > 𝜋𝜋, tan 𝑥𝑥 will become positive and increasingly larger as 𝑥𝑥 gets closer to −3𝜋𝜋2

20. When 𝑥𝑥 < −𝜋𝜋, tan 𝑥𝑥 will become negative and will continue to become smaller as larger as𝑥𝑥 gets closer to −3𝜋𝜋

2.

21. No maximum. Function is infinite

22. No minimum



23. 𝑥𝑥 = 𝜋𝜋4, tan 𝜋𝜋

4= 1

Exercise 2.1 Use the information on the graphs on the next page to work out its rule. You will need to decide whether it is sin, cos or tan and also a value of ‘𝑎𝑎’. Check your answers by drawing them on your calculator.

7. 𝑦𝑦 = 3 sin 𝑥𝑥 8. 𝑦𝑦 = 2 cos 𝑥𝑥

9. 𝑦𝑦 = 2 tan 𝑥𝑥 10. 𝑦𝑦 = 14

sin 𝑥𝑥

11. 𝑦𝑦 = 4 cos 𝑥𝑥 12. 𝑦𝑦 = 12

tan 𝑥𝑥

Exercise 2.2 3.

a) 4

b) 2

c) 2.5

4. State the amplitude of the following

d) 3 e) 1.5 f) 2

Exercise 2.3

Curve Number of waves in 2𝜋𝜋 𝑏𝑏 Period 2𝜋𝜋 ÷ 𝑏𝑏

sin 𝑥𝑥 1 1 2𝜋𝜋 2𝜋𝜋

sin 2𝑥𝑥 2 2 𝜋𝜋 𝜋𝜋

sin12𝑥𝑥

12

12

4𝜋𝜋 4𝜋𝜋

sin 3𝑥𝑥 2𝜋𝜋3

2𝜋𝜋3

sin13𝑥𝑥

13

13

6𝜋𝜋 6𝜋𝜋

sin 4𝑥𝑥 4 4 𝜋𝜋2

𝜋𝜋2

sin14𝑥𝑥

14

14

8𝜋𝜋 8𝜋𝜋

Exercise 2.4 7. Changes the number of waves in 2π, ie the period

8. Write down the period of these functions

a) 𝑏𝑏 = 1 b) 𝑏𝑏 = 14



Period = 2𝜋𝜋1

= 2𝜋𝜋 Period = 2𝜋𝜋14

= 8𝜋𝜋

c) 𝑏𝑏 = 1

Period = 2𝜋𝜋1

= 2𝜋𝜋

d) 𝑏𝑏 = 6

Period = 2𝜋𝜋6

= 𝜋𝜋3

e) 𝑏𝑏 = 34

Period = 2𝜋𝜋34

= 8𝜋𝜋3

f) 𝑏𝑏 = 𝜋𝜋

Period = 2𝜋𝜋𝜋𝜋

= 1

g) 𝑏𝑏 = 2

Period = 2𝜋𝜋2

= 𝜋𝜋

9. Write down the amplitude for each of the functions in question 2.

a) 𝑎𝑎 = 1 b) 𝑎𝑎 = 1

c) 𝑎𝑎 = 2 d) 𝑎𝑎 = 1

e) 𝑎𝑎 = 1 f) 𝑎𝑎 = 32

g) 𝑎𝑎 = −1 amplitude is 1

10. For 𝑦𝑦 = sin 𝑥𝑥 the period is the length of one horizontal wave, ie 2𝜋𝜋 .

11. Determine the value of 𝑏𝑏 from these graphs.

a) 2𝜋𝜋𝑏𝑏

= 𝜋𝜋2

𝑏𝑏 = (2×2𝜋𝜋)𝜋𝜋

𝑏𝑏 = 4

b) 2𝜋𝜋𝑏𝑏

= 𝜋𝜋

𝑏𝑏 = (2𝜋𝜋)𝜋𝜋

𝑏𝑏 = 2

12.

a) 𝑦𝑦 = sin 3𝑥𝑥 b) 𝑦𝑦 = sin 14𝑥𝑥

c) 𝑦𝑦 = sin 4𝑥𝑥 d) 𝑦𝑦 = sin 13𝑥𝑥

e) 𝑦𝑦 = sin 12𝑥𝑥

Exercise 3.1

6. a) 𝑦𝑦 = cos 𝑥𝑥

b) 1

c) 2𝜋𝜋1

= 2𝜋𝜋



7. a) 𝑦𝑦 = cos 2𝑥𝑥 _

b) 2

c) 2𝜋𝜋2

8. a) 𝑦𝑦 = cos 12𝑥𝑥

b) 12

c) 2𝜋𝜋12

= 4𝜋𝜋


b) 3

c) 2𝜋𝜋3


b) 14

c) 2𝜋𝜋14

= 8𝜋𝜋

Exercise 3.2

d) (i) 2

(ii) Period = 𝜋𝜋2

e) (i) 12

(ii) Period =𝜋𝜋12

= 2𝜋𝜋

f) (i) 12

(ii) Period =𝜋𝜋2



Exercise 3.3 2.

(i) Period = 360 4

= 90 , amplitude = 3

(ii) (0 , 0)(45 , 0) (90 , 0) (135 , 0) (180 , 0)(225 , 0)(270 , 0) (315 , 0)

(iii) max(22.5 , 3)(112.5 , 3)(202.5 , 3)(292.5 , 3)min (67.5 ,−3)(157.5 ,−3)(247.5,−3)(337.5,−3)

(iv) nil.

(i) Period =2𝜋𝜋12

= 4𝜋𝜋 amplitude = 1

(ii) (−𝜋𝜋, 0) (𝜋𝜋, 0)(iii) Max (0,1)

min(−2𝜋𝜋,−1)(2𝜋𝜋,−1)(iv) nil

(i) Period = 𝜋𝜋3

= 4𝜋𝜋 amplitude doesn’t exist

(ii) (−𝜋𝜋, 0) �− 2𝜋𝜋3

, 0� (−𝜋𝜋, 3)(0,0)

�𝜋𝜋3

, 0� �2𝜋𝜋3

, 0� (𝜋𝜋, 0) (iii) No max or min.(iv) 𝑥𝑥 = −5𝜋𝜋

6,−𝜋𝜋

2,−𝜋𝜋

6, 𝜋𝜋6

, 𝜋𝜋2

, 5𝜋𝜋6

,



Exercise 3.4

Graphic Calculator Activity. 5.

6.

moves 𝜋𝜋2 to the left

7.


8. a)


b)

moves 𝜋𝜋3 to the right

c)

moves 𝜋𝜋 to the left, (or reflected in the x axis)

d)

.moves 𝜋𝜋4

to the right



Exercise 3.5 2. Sketch the following over the domain −4 ≤ 𝑥𝑥 ≤ 4

a) 𝑦𝑦 = sin(𝜋𝜋𝑥𝑥),

b) 𝑦𝑦 = sin(𝜋𝜋𝑥𝑥 + 𝜋𝜋) ,

c) 𝑦𝑦 = − sin𝜋𝜋(𝑥𝑥 + 1),

d) 𝑦𝑦 = 1 − sin𝜋𝜋(𝑥𝑥 + 1)

Describe clearly the way in which graphs (b), (c) and (d) are related to graph (a). Comment on any phase shifts, including the size of the phase shift.

e) f) is the same shape as 𝑦𝑦 = sin(𝜋𝜋𝑥𝑥) but translated left 𝜋𝜋 units

g) is the same as 𝑦𝑦 = sin(𝜋𝜋𝑥𝑥) butranslated left 𝜋𝜋 units

h) is the same shape as 𝑦𝑦 = sin(𝜋𝜋𝑥𝑥) but translated up 1 unit



Exercise 4.1 2.

a)

b)

c) a) Is the same shape as 𝑦𝑦 = co𝑠𝑠(𝑥𝑥) but has been translated right 𝜋𝜋4, b) is the same

shape as 𝑦𝑦 = cos(𝑥𝑥) but has been translated left 𝜋𝜋3

Exercise 4.2 Calculator Activity 5. 𝑐𝑐 moves 𝑦𝑦 = tan 𝑥𝑥 𝑐𝑐 units to the right

Exercise 4.3 2) Sketch the graphs of

a)

b)



Exercise 4.4 3)

g) 𝑦𝑦 = sin 2 �𝑥𝑥 + 𝜋𝜋4� h) 𝑦𝑦 = cos 2 �2 − 𝜋𝜋

4�

i) 𝑦𝑦 = tan 3 �𝑥𝑥 + 𝜋𝜋6� j) 𝑦𝑦 = cos 1

4�𝑥𝑥 − 𝜋𝜋

2�

k) 𝑦𝑦 = tan 13�𝑥𝑥 − 𝜋𝜋

2� l) 𝑦𝑦 = sin 3

2�𝑥𝑥 + 𝜋𝜋

3�

4)

g) Period =2𝜋𝜋14

=8

Phase change 𝜋𝜋2 right

h) Period = 2𝜋𝜋3

Phase change = 𝜋𝜋6 left

i) Period =2𝜋𝜋32

= 4𝜋𝜋3

Phase change 𝜋𝜋3 left

j) Period = 2𝜋𝜋2

= 𝜋𝜋Phase change = 𝜋𝜋

2 right

Exercise 5.1 4. cos(30° + 60°) = cos 90° =0

cos 30° + cos 60° = 1.37 5. tan(45° + 45°) = tan 90° undefined

tan 45° + tan 45° = 2 6. sin(60° + 60°) = sin 120° =0.87

sin 60° + 60° = 1.73 Exercise 5.2 6. cos (𝑃𝑃 + 𝐵𝐵) = cos𝑃𝑃 cos𝐵𝐵 − sin𝑃𝑃 sin𝐵𝐵

Replace 𝐵𝐵 by (−𝐵𝐵).cos (𝑃𝑃 − 𝐵𝐵) = cos𝑃𝑃 cos(−𝐵𝐵) − sin𝑃𝑃 sin(−𝐵𝐵)

i.e. cos (A−𝐵𝐵) = cos𝑃𝑃 cos𝐵𝐵 + sin𝑃𝑃 sin𝐵𝐵



7. sin (𝑃𝑃 + 𝐵𝐵) = cos �𝜋𝜋2− (𝑃𝑃 + 𝐵𝐵)�

= cos �𝜋𝜋2− 𝑃𝑃 − 𝐵𝐵�

= cos ��𝜋𝜋2− 𝑃𝑃� − 𝐵𝐵 � 𝑇𝑇𝑝𝑝𝑝𝑝𝑎𝑎𝑡𝑡 𝑎𝑎𝑠𝑠 cos(𝑃𝑃 − 𝐵𝐵)

= cos �𝜋𝜋2− 𝑃𝑃� cos(−𝐵𝐵) − sin �𝜋𝜋

2− 𝑃𝑃� sin(−𝐵𝐵)

= sin𝑃𝑃 cos𝐵𝐵 + cos𝑃𝑃 sin𝐵𝐵

8. sin (𝑃𝑃 − 𝐵𝐵) = sin[𝑃𝑃 + (−𝐵𝐵)]= sin𝑃𝑃 cos(−𝐵𝐵) + cos𝑃𝑃 sin(−𝐵𝐵) = sin𝑃𝑃 cos𝐵𝐵 − cos𝑃𝑃 sin𝐵𝐵

5. Fill in the steps for tan (𝑃𝑃 − 𝐵𝐵) for yourselftan (𝑃𝑃 − 𝐵𝐵) =

sin(𝐴𝐴−𝐵𝐵)cos(𝐴𝐴−𝐵𝐵)

= sin𝐴𝐴 cos𝐵𝐵−cos𝐴𝐴 sin𝐵𝐵cos𝐴𝐴 cos𝐵𝐵+sin𝐴𝐴 sin𝐵𝐵

= sin𝐴𝐴cos𝐵𝐵cos𝐴𝐴cos𝐵𝐵+

cos𝐴𝐴sin𝐵𝐵cos𝐴𝐴 cos𝐵𝐵

cos𝐴𝐴cos𝐵𝐵cos𝐴𝐴cos𝐵𝐵−

sin𝐴𝐴sin𝐵𝐵cos𝐴𝐴cos𝐵𝐵

= tan𝐴𝐴−tan𝐵𝐵1−tan𝐴𝐴 tan𝐵𝐵

6. tan 165 = tan(135 + 30 )

= tan135 +tan30 1−tan135 tan30

= −1+ 1

√3

1−(−1)� 1√3�

= −1+ 1

√3

1+ 1√3

rationalise the denominator by multiplying by 1− 1

√3

1− 1√3

= �−1+ 1

√3��1− 1

√3�

�1− 1√3��1− 1

√3�

= −43+

2√33

1−13

= −4+2√32

= −2 + √3



Exercise 6.1

1. a) 2 weeks ago Vince was at his lowest in his emotional cycle. Today he at his highest.

In 2 weeks he will be at the lowest level again.

b) 𝑦𝑦 = cos 𝑥𝑥

c) 23rd May + 2 weeks = 6 June.

6 June +4 weeks = 4 July

4 July + 1 August

d) 23rd May + 7 days = 30 May.

6 June +4 weeks = 4 July

4 July + 4 weeks = 1 August

e) Use the function given to complete the following table of approximate values for thefunction, rounding to two decimal places when required. 𝑓𝑓(𝑎𝑎) = 100 cos �𝜋𝜋𝜋𝜋

14�

f) 28 days

g) 12 days, 5 days from today, 28 May

h) It is his emotional cycle translated 2 units left

i) +2 translates 2 units left

j) 𝑓𝑓(𝑎𝑎) = 100 cos �𝜋𝜋(𝜋𝜋+6)14

�

d 0 2 4 6 7 8 10 12 14

𝑓𝑓(𝑎𝑎) 2dp 100 90 62 22 0 -22 -62 -90 -100

d 0 2 4 6 8 10 12 14

𝑓𝑓(𝑎𝑎) 2dp 90.10 62.35 22.25 -22.25 -62.35 -90.10 -100 -90.10

d 0 2 4 6 8 10 12 14

𝑓𝑓(𝑎𝑎) 2dp 22 -22 -62 -90 -100 -90 -62 -22



k)

l) Mastery 90%, passion 85%, wisdom 80%

m) Amplitude of 95%, peaks again in 7 days.

n)

Cycle 1 72 cos �𝜋𝜋(𝜋𝜋+3)14

�

Cycle 2 55 cos �𝜋𝜋(𝜋𝜋−2)14

�

Cycle 3 68 cos �𝜋𝜋(𝜋𝜋−5)14

�

o) 100 cos �𝜋𝜋𝜋𝜋14� = 100 sin �𝜋𝜋

2− 𝜋𝜋𝜋𝜋

14�

= 100 sin �7𝜋𝜋−𝜋𝜋𝜋𝜋14

�

= 100 sin �𝜋𝜋(7−𝜋𝜋)14

�

p) Physical 100 sin �𝜋𝜋(5−𝜋𝜋)14

� , intellectual 100 sin �𝜋𝜋(1−𝜋𝜋)14

�



2. a) Sketch the position of the centre of the beams of light on the wall, 3 m above and

below the gate, for the first 16 seconds, using the axes below:

b) The graph is the same shape as tany x= with a period of 4 seconds and a restrictedrange (-3) 3y≤ ≤ . Four complete cycles are shown

c) Undefined at 2 seconds, 6 seconds, 10 seconds and 14 seconds. These relate to when the

first beam is rotated through an angle of. 3 5 7, , ,2 2 2 2π π π π

d) 4 seconds, π radians

e) tan4xy =

π

f)

time4 8 12 16


-3

3

(seconds)

time4 8 12 16


-3

3

(seconds)

time4 8 12 16


-3

3

(seconds)



Exercise 7.1 1) Solve the following equations for the given domain, leaving answers exact where you can.

Check your answers by using your calculator. Remember to include the correct domain.

a) −11𝜋𝜋6

,−7𝜋𝜋6

, 𝜋𝜋6

, 5𝜋𝜋6

b) 5𝜋𝜋4

, 7𝜋𝜋4

c) 𝜋𝜋3

, 5𝜋𝜋3

d) -157.5, -67.5, 22.5

e) −𝜋𝜋2

, 𝜋𝜋2

, 𝜋𝜋6

, 5𝜋𝜋6

f) −2𝜋𝜋3

,−𝜋𝜋2

, 𝜋𝜋2

, 2𝜋𝜋3

g) 0,−2𝜋𝜋,−𝜋𝜋2

Trigonometric Functions...1.2.15 identify contexts suitable for modelling by trigonometric functions...

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Transcript of Trigonometric Functions...1.2.15 identify contexts suitable for modelling by trigonometric functions...