Year 11 ATAR Mathematics
Trigonometric Functions
Year 11 | Mathematics Methods | Trigonometric Functions | Β© Department of Education WA 2020
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Contents
Content page⦠................................................................................................................... 1
Signposts⦠....................................................................................................................... ⦠2
Overview ......................................................................................................................... 3
Lesson 1 β Graphs of sine, cosine and tangentβ¦β¦............................................................5
Lesson 2 β Transforming the trigonometric graphs, amplitude and periodβ¦β¦β¦β¦β¦β¦β¦..14
Lesson 3 β Change of period horizontal dilations and combination of transformationsβ¦27
Lesson 4 β Phase changes of π¦π¦ = sin(π₯π₯ β ππ), π¦π¦ = cos(π₯π₯ β ππ) and π¦π¦ = tan (π₯π₯ β ππ)β¦β¦β¦β¦β¦40
Lesson 5 β Angle sum and difference identitiesβ¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦.50
Lesson 6 β Solving practical problemsβ¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦.64
Lesson 7 β Solving equations involving trigonometric functionsβ¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦74
Lesson 8 & 9 β Exam practiceβ¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦..81
Glossary β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦.β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦..82
Solutionsβ¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦.β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦..84
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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
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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.
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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 sinππ, cosππ and tan ππ and periodicity using radians
1.2.8 recognise the exact values of sinππ, cosππ and tan ππ at integer multiples of ππ6
and ππ4
1.2.9 recognise the graphs of π¦π¦ = sinπ₯π₯,π¦π¦ = cosπ₯π₯ , and π¦π¦ = tanπ₯π₯ on extended domains
1.2.10 examine amplitude changes and the graphs of π¦π¦ = ππ sinπ₯π₯ and π¦π¦ = ππ cosπ₯π₯
1.2.11 examine period changes and the graphs of π¦π¦ = sinπππ₯π₯, π¦π¦ = cosπππ₯π₯ and π¦π¦ = tan πππ₯π₯
1.2.12 examine phase changes and the graphs of π¦π¦ = sin(π₯π₯ β ππ), π¦π¦ = cos(π₯π₯ β ππ) and π¦π¦ = tan (π₯π₯ β ππ)
1.2.13 examine the relationships sin οΏ½π₯π₯ + ππ2οΏ½ = cosπ₯π₯ and cos οΏ½π₯π₯ β ππ
2οΏ½ = sinπ₯π₯
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
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Lesson 1 Graphs of sine, cosine and tangent functions
By the end of this lesson you should be able to:
β’ recall facts about the unit circle
β’ draw accurate graphs of sin x, cos x and tan x using common angles in radian formeg ππ
6, ππ4
, ππ3
, ππ2
, etc.
β’ Recognise the important points of the sine and cosine functions and be able to usethese points to draw quick sketches of their graphs.
β’ 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 π‘π‘ 0 ππ6
ππ4
ππ3
ππ2
2ππ3
3ππ4
5ππ6
ππ
sin π‘π‘ (to 2 d.p.)
Value of π‘π‘ 7ππ6
5ππ4
4ππ3
3ππ2
5ππ3
7ππ4
11ππ6
2ππ
sin π‘π‘ (to 2 d.p.)
Value of π‘π‘ 13ππ6
9ππ4
7ππ3
5ππ2
8ππ3
11ππ4
17ππ6
3ππ
sin π‘π‘ (to 2 d.p.)
Value of π‘π‘ 19ππ6
13ππ4
10ππ3
7ππ2
11ππ3
15ππ4
23ππ6
4ππ
sin π‘π‘ (to 2 d.p.)
2. Did you notice a pattern start to occur? Comment.
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3. Plot the points from the table on the grid below and then join your points with a smoothcurve.
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Graphing the sine curve β another way The next diagram can be used to graph the sine function.
π₯π₯ = 1.0,π¦π¦ = sin 1.0 β 0.8 π₯π₯ = 5.5,π¦π¦ = sin 5.5 β β0.7
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 ππ 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 π¦π¦ coordinate from 1.0 on the unit circle across onto the grid, marking a point above 1.0 on the horizontal axis.
π₯π₯ = 1.0,π¦π¦ = sin 1.0 β 0.8 π₯π₯ = 5.5,π¦π¦ = sin 5.5 β β0.7
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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 π₯π₯ = 4.5, 5.0, 5.5 and 6.0.
4. What is the maximum value of sin (π‘π‘)?
5. What is the minimum value of sin (π‘π‘)?
6. At what values of π‘π‘ does the graph of sin π‘π‘ cross the horizontal axis? That is , what are thezeros of cos π‘π‘
Cosine functions
Repeat for cos t. Remember to work in radian mode. And give your answers to one decimal place.
Value of π‘π‘ 0 ππ6
ππ4
ππ3
ππ2
2ππ3
3ππ4
5ππ6
ππ
cos π‘π‘ (to 2 d.p.)
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Value of π‘π‘ 7ππ6
5ππ4
4ππ3
3ππ2
5ππ3
7ππ4
11ππ6
2ππ
cos π‘π‘ (to 2 d.p.)
Value of π‘π‘ 13ππ6
9ππ4
7ππ3
5ππ2
8ππ3
11ππ4
17ππ6
3ππ
cos π‘π‘ (to 2 d.p.)
Value of π‘π‘ 19ππ6
13ππ4
10ππ3
7ππ2
11ππ3
15ππ4
23ππ6
4ππ
cos π‘π‘ (to 2 d.p.)
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The cosine curve again
Do you remember? The cosine of an angle is the π₯π₯ 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 π₯π₯ βvalues 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 π₯π₯ and complete the sketch up to π₯π₯ = 2ππ.
Does your graph look like you expect it should?
8. What is the maximum value of cos π‘π‘?
9. What is the minimum value of cos π‘π‘?
10. At what values of π‘π‘ does the graph of cos π‘π‘ cross the horizontal axis? That is , whatare the zeros of cos π‘π‘?
11. Compare your graphs of sin π‘π‘ and cos π‘π‘. What is similar? What is different?
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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 π₯π₯ -Ο β5ππ6
β3ππ4
β2ππ3
βππ2
βππ3
βππ4
βππ6
0
tan π₯π₯
Value of π₯π₯ ππ6
ππ4
ππ3
ππ2
2ππ3
3ππ4
5ππ6
Ο
tan π₯π₯
Value of π₯π₯ -1.4 -1.2 -1.0 -0.8 -0.6 -0.4 -0.2 0
tan π₯π₯
Value of π₯π₯ 0 0.2 0.4 0.6 0.8 1.0 1.2 1.4
tan π₯π₯
13. Use as many of the above points as necessary to graph tan π₯π₯ accurately on this grid. (the ππ isapproximate.)
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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 π₯π₯?
16. What happens to tan π₯π₯ after ππ2 ?
17. What happens near βππ2
18. What seems to be the period of tan π₯π₯?
19. What do you think happens to tan π₯π₯ when π₯π₯ > ππ?
20. What happens to the graph when π₯π₯ < βππ?
21. What would be the maximum value of tan π₯π₯?
22. What would be the minimum value of tan π₯π₯?
23. Another value of x which is useful when sketching tan x is the value of π₯π₯ when tan π₯π₯ = 1.What is this value of π₯π₯?
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Lesson 2 Transforming the trigonometric graphs
By the end of this lesson you should be able to:
β’ Transform the graphs of sine, cosine and tangent by sliding up or down
β’ Transform the graphs of sine, cosine and tangent by dilating about the x-axis
β’ Transform the graphs of sine, cosine and tangent by reflecting about the x-axis.
β’ Understand the effect of changing βbβ in sin (bx) β horizontal dilations
β’ Determine the period of a sin (bx) from the value of b
β’ 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
π¦π¦ = sin π₯π₯ π¦π¦ =______________ π¦π¦ =______________
Cosine curve
Usual position translated up 1 unit translated down 1 unit
π¦π¦ = cos π₯π₯ π¦π¦ =______________ π¦π¦ =______________
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Tangent curve
Usual position translated up 1 unit translated down 1 unit
π¦π¦ = tan π₯π₯ π¦π¦ =______________ π¦π¦ =______________
Did you guess for the middle graphs that have been translated up one π¦π¦ = sin π₯π₯ + 1,π¦π¦ = cos π₯π₯ + 1 ππππππ π¦π¦ = tan π₯π₯ + 1?
Did you guess for the graphs on the right that have been moved down π¦π¦ = sin π₯π₯ β 1,π¦π¦ = cos π₯π₯ β 1 ππππππ π¦π¦ = tan π₯π₯ β 1?
Dilating about the x-axis Look at the following variations on the sine and cosine curves.
π¦π¦ = 12
sin π₯π₯ π¦π¦ = 2 sin π₯π₯ π¦π¦ = 3 cos π₯π₯ π¦π¦ = 14
cos π₯π₯
In each case, a constant has been placed in front of sin π₯π₯ or cos π₯π₯. We could just as easily do the same for tan π₯π₯.
π¦π¦ = 12
tan π₯π₯ π¦π¦ = 2 tan π₯π₯
These functions are written in the general form by using the letter βaβ to represent any constant.
π¦π¦ = ππ sin π₯π₯ π¦π¦ = ππ cos π₯π₯ π¦π¦ = ππ tan π₯π₯
In the previous lesson you were asked to explore the effect of ππ in π¦π¦ = ππ sin π₯π₯.
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You should have found that varying βππβ is to dilate the graph in relation to the π₯π₯-axis. You will see the effects in the graphs which follow.
The following are π¦π¦ = cos π₯π₯ ,π¦π¦ = 3 cos π₯π₯ and π¦π¦ = 14
cos π₯π₯.
Try labelling them, then using your calculator, graph them and check.
The changes in βaβ changed the βheightβ 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 π₯π₯
where a is 2
graph is enlarged, all the y values are double the values of sin π₯π₯
where a = 1
graph is as shown above
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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 βππβ. Check your answers by drawing them on your calculator.
1. 2.
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3. 4.
5. 6.
Reflecting
The top two graphs below show sinπ₯π₯ and tan π₯π₯ in their usual positions. If we reflect the graphs about the π₯π₯-axis, the results are shown in the seconds row of diagrams.
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The effect is to make every π¦π¦ value opposite in sign. Their rules are π¦π¦ = β sin π₯π₯ and π¦π¦ = β tan π₯π₯.
Exercise 2.2 1. State the amplitude of the following
a) π¦π¦ = 4 sin π₯π₯
b) π¦π¦ = β2 cos(π₯π₯ β 1)
c) π¦π¦ = 2.5 sin π₯π₯ + 3
2. State the amplitude of the following
a) b) c)
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Change in period β sine graphs In the diagrams below we see the effect of horizontal dilations. The waves are squeezed up or spread out.
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How long is a wave?
Obviously, if we decide where a wave starts and where it ends, we can measure how βlongβ it is.
We call this distance the βwavelengthβ.
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 βperiodβ.
Horizontal dilations have the effect of changing the wavelength or period.
Horizontal dilations and sin ππππ We will now look at a number of sine graphs such as π¦π¦ = sin π₯π₯ or sin 2π₯π₯
Look at the following graphs which are named for you.
For one complete wave of sin π₯π₯, 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 π¦π¦ = sin 2π₯π₯, how many waves between 0 and 2ππ?
What will the period be?
The period of sin x is 2Ο
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You should have seen that for sin 2π₯π₯, 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 2π₯π₯ 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.
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Now complete this table.
Curve Number of waves in 2ππ ππ Period 2ππ Γ· ππ
sin π₯π₯
sin 2π₯π₯
sin12π₯π₯
sin 3π₯π₯
sin13π₯π₯
sin 4π₯π₯
sin14π₯π₯
3 - 2 sin (1/2 x)
Summary Let's put together a formal summary of the last few pages.
For π¦π¦ = sin πππ₯π₯,
a) b gives the number of waves in 2ππ.
b) the period is 2ππ Γ· ππ (or 2ππππ
)
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 'ππ' in π¦π¦ = sin π₯π₯ + ππ.
We also look at vertical dilations which were dependent on the value of 'ππ' in π¦π¦ = ππ π π π π ππ π₯π₯.
This was called a change in 'amplitude'.
Let's put all of these transformations together.
The period of a sine function is calculated by 2ππππ
in radians or 360 ππ
in degrees
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Get it all together Adding 'ππ', and 'ππ' and 'ππ' into the equation gives us
Example
Find the period of this function: π¦π¦ = 2 sin 23π₯π₯ β 1
Solution By now, we know what the graph of sin π₯π₯ is like. We know its shape, its period, its amplitude, the maximum and minimum locations and the zeroes.
In this example, we have ππ = 2, ππ = 23, which means the period is found by dividing 2ππ by 2
3, and
we have ππ = β1.
2ππ Γ· 23
= 2ππ Γ 32
= 3ππ
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Exercise 2.4 1. Explain the effect of changing βππβ in sin πππ₯π₯.
2. Write down the period of these functions
a) sin π₯π₯ b) sin 14π₯π₯
c) 2 sin π₯π₯ d) sin 6π₯π₯
e) sin 34π₯π₯ f) 3
2sinπππ₯π₯
g) 3 β sin 2π₯π₯
3. Write down the amplitude for each of the functions in question 2.
a) sin π₯π₯ b) sin 14π₯π₯
c) 2 sin π₯π₯ d) sin 6π₯π₯
e) sin 34π₯π₯ f) 3
2sinπππ₯π₯
g) 3 β sin 2π₯π₯
4. Explain the meaning of βperiodβ as it applies to π¦π¦ = sin π₯π₯.
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5. Determine the value of ππ from these graphs.
a) b)
6. Name these graphs of sin πππ₯π₯.
a) b)
c) d)
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Lesson 3 Graphs of trigonometric functions horizontal dilations
By the end of this lesson you should be able to
β’ extend your understanding of change in period to the cosine and tangent functions
β’ sketch the graphs of sin bx, cos bx and tan bx using graphic calculator
β’ draw accurate graphs of these functions when necessary. sin, cos, tan -what's thedifference?
β’ examine the changes which occur when the sine graph is moved to the left or right
β’ Understand the meaning of the term βphase shiftβ
β’ Determine the values of c in sin (x+c) from the graph
β’ 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 π₯π₯ repeats itself every 2ππ just the same as sin π₯π₯.
Graphs of cos ππππ What happens when we change the value of ππ? Because of the close links with sin πππ₯π₯, we can readily understand that changing ππ will have the same effect on cos πππ₯π₯ as it does on sin πππ₯π₯.
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.
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Exercise 3.1
1. a) This is the graph of π¦π¦ = _______________
b) How many wavesbetween 0 and 2ππ?______________
c) What is the period ofthis graph?____________
2. a) This is the graph of π¦π¦ = _______________
b) How many wavesbetween 0 and 2ππ?______________
c) What is the period ofthis graph?____________
3. a) This is the graph of π¦π¦ = _______________
b) How many wavesbetween 0 and 2ππ?______________
c) What is the period ofthis graph?____________
4. a) This is the graph of π¦π¦ = _______________
b) How many wavesbetween 0 and 2ππ?______________
c) What is the period ofthis graph?____________
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5. a) This is the graph of π¦π¦ = _______________
b) How many wavesbetween 0 and 2ππ?______________
c) What is the period ofthis graph?____________
Summary For π¦π¦ = cos πππ₯π₯
β’ ππ gives the number of waves in 2ππ.
β’ The period is 2ππ Γ· ππ or
β’ 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 π¦π¦ = sin π₯π₯ + ππ.
Get it all together
Adding 'b', and 'a' and 'd' into the equation gives us
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Tangent is a little different
We don't have waves with tangent!
However, we can still talk about wavelength or the period of tan π₯π₯ repeats its basic shape every ππ units along the horizontal axis. You won't win too many prizes for guessing the period of tan π₯π₯, now!
The period of tan π₯π₯ is ππ.
The period of sin π₯π₯ is 2ππ.
The period of cos π₯π₯ is 2ππ.
Sometimes, we need to use degrees, and so, in degrees, the periods are
sin π₯π₯Β° and cos π₯π₯Β° βΆ Period = 360Β°
tan π₯π₯Β° : Period = 180Β°
Branches of ππππππππππ The tangent graph doesnβt have the wave-like structure that sine and cosine have. It will suit our purpose to talk about the βbranchesβ of the tan graph, each separate curve being called a branch.
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You will notice that there is a separate branch of tan π₯π₯ at every interval of ππ units on the horizontal axis.
Thus the period of tan π₯π₯ is ππ (or 180o). in other words, the basic shape is repeated every ππ units.
As we say with sin πππ₯π₯ and cos πππ₯π₯, the number of branches of tan πππ₯π₯ is controlled by βππβ.
Compare the number of branches with the value of ππ. (2 half branches count as 1 branch).
Exercise 3.2
a) (i) ππ = 2
Branches in ππ units =______
(ii) Period =______
b) (i) ππ = 12
Branches in ππ units =______
(ii) Period =______
c) (i) ππ = 12
Branches in ππ units =______
(ii) Period =______
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Branching and βππβ You would have noticed that the number of branches coincides with the value of ππ.
Did you discover the relationship between the period, ππ and the value of βππβ? Having worked with sine and cosine, you shouldnβt have had much difficulty with this relationship.
Another nutshell For π¦π¦ = tan πππ₯π₯
β’ ππ gives the number of branches in ππ
β’ The period is ππ Γ· ππ (or ππππ
)
β’ The period is the distance between branches.
Extending tan ππ 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 βwindowβ quickly and efficiently.
Note: For sin πππ₯π₯ and cos πππ₯π₯ period = 2ππππ
For tan πππ₯π₯, period = ππππ
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Exercise 3.3 1. Use your calculator to graph the following trigonometric functions over given domains.
β’ π¦π¦ = 3 sin 4π₯π₯ over the domain 0 β€ π₯π₯ β€ 360
β’ π¦π¦ = cos 12π₯π₯ over the domain β2ππ β€ π₯π₯ β€ 2ππ
β’ π¦π¦ = tan 3π₯π₯ over the domain βππ β€ π₯π₯ β€ 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 π₯π₯ βintercepts (zeroes)
(iii) The location of the maximum and minimum values.
(iv) The location of any asymptotes.
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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
π¦π¦ = sin(π₯π₯ + ππ)
π¦π¦ = cos(π₯π₯ + ππ) π¦π¦ = tan (π₯π₯ + ππ)
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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 βin phaseβ, 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.
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Phase shift from the graphs
sin (π₯π₯ + ππ) 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.
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Phase shift with π¬π¬π¬π¬ππ(ππ + ππ) On the previous page you can see the changes which occur by sliding π¦π¦ = sin π₯π₯ to the left or right. This horizontal movement parallel to the π₯π₯ 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 π¦π¦ = sin π₯π₯, but the amount of the phase shift.
Exercise 3.4
Graphic Calculator Activity. 1. Graph π¦π¦ = sin π₯π₯ over the domain β2ππ β€ π₯π₯ β€ 2ππ.
2. Leave the graph of π¦π¦ = sin π₯π₯ on your screen and now graph π¦π¦ = sin οΏ½π₯π₯ + ππ2οΏ½ using the same
window settings. Write down the effect of adding ππ2. In this case ππ = + ππ
2.
3. Leave the graph of π¦π¦ = sin π₯π₯ on your screen and now draw the graph of π¦π¦ = sin οΏ½π₯π₯ β ππ2οΏ½.
Turn off π¦π¦ = sin οΏ½π₯π₯ + ππ2οΏ½.
Write down the effect of subtracting ππ2. In this case = βππ
2
4. Graph the following sine curves and see if the effect of changing βcβ fits in with your earlierfinding. Remember to compare each graph to the graph of π¦π¦ = sin π₯π₯
a) π¦π¦ = sin οΏ½π₯π₯ + ππ4οΏ½ Note ππ = + ππ
4
b) π¦π¦ = sin οΏ½π₯π₯ β ππ3οΏ½ Note ππ = βππ
3
c) π¦π¦ = sin(π₯π₯ + ππ) Note ππ = +ππ
d) π¦π¦ = sin οΏ½π₯π₯ β ππ4οΏ½ Note ππ = βππ
4
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Summary Have you discovered what the pattern is?
You should have found that the phase shift is equal in size to βππβ and that
β’ The graph moves to the left if βππβ is positive.
β’ The graph moves to the right if βππβ is negative.
Example Sketch the graph of π¦π¦ = sin(π₯π₯ β 1.5)
In this case, ππ = β1.5 which means the phase shift is 1.5 radians to the right. Alternatively we could say draw π¦π¦ = sin π₯π₯ and then move the vertical axis 1.5 units to the left.
Rather than use the βππβ scale, it is better to use the simple numerical scale for this graph.
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Exercise 3.5 1. Sketch the following over the domain β4 β€ π₯π₯ β€ 4
a) π¦π¦ = sin(πππ₯π₯),
b) π¦π¦ = sin(πππ₯π₯ + ππ) ,
c) π¦π¦ = β sinππ(π₯π₯ + 1),
d) π¦π¦ = 1 β sinππ(π₯π₯ + 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)
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Lesson 4 Phase changes of trigonometric functions
By the end of this lesson you should be able to
β’ sketch the graph of πππππ π (π₯π₯ + ππ)
β’ Sketch graphs in the form of π‘π‘ππππ(π₯π₯ + ππ)
β’ Extend your ability to graph trigonometric functions in the form of π π π π ππ ππ(π₯π₯ + ππ),πππππ π ππ(π₯π₯ + ππ) and π‘π‘ππππ ππ(π₯π₯ + ππ)
Phase shift with cosine
Examine these graphs of sin(π₯π₯ + ππ) in which the value of c changes from 0 to ππ4 to ππ
2 to Ο.
The rules for these graphs are
1. π¦π¦ = sin π₯π₯ (ππ = 0)
2. π¦π¦ = sin (π₯π₯ + ππ4
) οΏ½ππ = ππ4οΏ½
3. π¦π¦ = sin (π₯π₯ + ππ2) οΏ½ππ = ππ
2οΏ½
4. π¦π¦ = sin (π₯π₯ + ππ) (ππ = ππ)
When is the phase shift such that the graph of π¦π¦ = sin(π₯π₯ + ππ) is the same as π¦π¦ = cos π₯π₯?
1. 2.
3. 4.
Graph 3 is the same as the graph of cos π₯π₯. That is, sin οΏ½π₯π₯ + ππ2οΏ½ = cos π₯π₯.
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Now, the next question for you to consider, is what phase shift is necessary so that cos(π₯π₯ + ππ) = sin π₯π₯?
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 βcβ
having the value of βππ2
.
The rule would be cos οΏ½π₯π₯ β ππ2οΏ½ = sin π₯π₯ .
The simplest shift A few lines back, the word βsimplestβ 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 βππβ have so that sin(π₯π₯ + ππ) = sin π₯π₯ or cos(π₯π₯ + ππ) ?
What value would βππβ have so that tan(π₯π₯ + ππ) = tan π₯π₯ ?
Do you think βππβ could have more than one value? How many values? What spacing in terms of ππ-units would these values of ππ have?
Perhaps you knew the answers to these questions, perhaps there were some that you werenβt sure about.
Check with this summary:
β’ There in an infinite number of vales of ππ which would make sin(π₯π₯ + ππ) = sin π₯π₯,tan(π₯π₯ + ππ) = tan π₯π₯ or cos(π₯π₯ + ππ) = cos π₯π₯
β’ For sine and cosine, the values of ππ would be spaced 2ππ units apart
β’ For tangent, the values of ππ would be ππ units apart.
β’ The simplest shifts would be
o sin(π₯π₯ Β± 2ππ) = sin π₯π₯
o cos(π₯π₯ Β± 2ππ) = cos π₯π₯
o tan(π₯π₯ Β± ππ) = tan π₯π₯
Other possible values of ππ would be:
sin(π₯π₯ Β± 4ππ) , sin(π₯π₯ Β± 6ππ) sin(π₯π₯ Β± 8ππ)
cos(π₯π₯ Β± 4ππ) , cos(π₯π₯ Β± 6ππ) cos(π₯π₯ Β± 8ππ)
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tan(π₯π₯ Β± 2ππ) tan(π₯π₯ Β± 3ππ) tan(π₯π₯ Β± 4ππ)
Phase shift and sketching ππ = πππππ¬π¬(ππ + ππ) Previously we looked at the graphs of π¦π¦ = sin(π₯π₯ + ππ) and how the value of βππβ determines the phase shift. Graphing π¦π¦ = cos (π₯π₯ + ππ) 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) π¦π¦ = cos οΏ½π₯π₯ β ππ4οΏ½
b) π¦π¦ = cos οΏ½π₯π₯ + ππ3οΏ½
c) Describe clearly how thegraphs (a) and (b) are relatedto the graph of
π¦π¦ = cos π₯π₯
By now you should have discovered the following relationship. If the graph of π¦π¦ = cos π₯π₯ is moved ππ2 units right, parallel to the π₯π₯ axis, it would then be the same as the graph of π¦π¦ = sin π₯π₯. We say
that sin π₯π₯ and cos π₯π₯ are ππ2 out of phase with each other.
It follows that
Use your calculator to check. The graphs should superimpose over each other.
cos π₯π₯ = sin οΏ½π₯π₯ +ππ2οΏ½ and that sin π₯π₯ = cos οΏ½π₯π₯ β
ππ2οΏ½
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Phase shift and sketching ππ = ππππππ(ππ + ππ) 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 π¦π¦ = tan π₯π₯ over the domain β 2ππ β€ π₯π₯ β€ 2ππ.
2. Now in turn, graph π¦π¦ = tan οΏ½π₯π₯ + ππ4οΏ½ and π¦π¦ = tan οΏ½π₯π₯ β ππ
2οΏ½.
3. Compare your screens to the ones right. You need to checkthem carefully because the screen is very βbusyβ.
4. What is the effect of changing the value of βππβ?
Summary
β’ The change in phase is equal in size to βππβ.
β’ The graph moves to the left if βππβ is positive.
β’ The graph moves to the right if βππβis 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.
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Exercise 4.3 1. Sketch the graphs of
a) π¦π¦ = tan(π₯π₯ + 1) using a numericalscale β7 β€ π₯π₯ β€ 7
b) π¦π¦ = tan οΏ½π₯π₯ β ππ4οΏ½ using a
βππ βunitsβ scale β2ππ β€ π₯π₯ β€ 2ππ
Life wasnβt meant to be easy
Have a look at the defining rules for these functions.
a) π¦π¦ = sin 2 οΏ½π₯π₯ β ππ4οΏ½ b) π¦π¦ = sin οΏ½2π₯π₯ β ππ
2οΏ½
c) π¦π¦ = sin 12
(π₯π₯ β ππ) d) π¦π¦ = sin οΏ½π₯π₯2β ππ
2οΏ½
e) π¦π¦ = tan 3 οΏ½π₯π₯ + ππ2οΏ½ f) π¦π¦ = tan οΏ½3π₯π₯ + 3ππ
2οΏ½
Look at the angle part a) and b), c) and d), e) and f).
Did you see that 2 οΏ½π₯π₯ β ππ4οΏ½ = οΏ½2π₯π₯ β ππ
2οΏ½?
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).
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Letβs look at part b) π¦π¦ = sin οΏ½2π₯π₯ β ππ2οΏ½, and ask the question, when is π¦π¦ = 0?
Obviously, for sine functions, π¦π¦ is equal to zero when the angle is equal to zero. That is, this case, when
Now look at part a) π¦π¦ = sin 2 οΏ½π₯π₯ β ππ4οΏ½. By taking the β2β 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 βππβ and also the value of βππβ. Remember the value of βππβ is used to calculate the period of the function.
We started by suggesting that life wasnβt meant to be easy, but when we have the rule written in the form π¦π¦ = sin ππ(π₯π₯ + ππ), then the working becomes very easy.
Exercise 4.4 Practise some manipulation.
1) Rewrite these functions in the form
π¦π¦ = sin ππ(π₯π₯ + ππ), π¦π¦ = cos ππ(π₯π₯ + ππ) or π¦π¦ = tan ππ(π₯π₯ + ππ)
a) π¦π¦ = sin οΏ½2π₯π₯ + ππ2οΏ½ =____________ b) π¦π¦ = cos(2π₯π₯ β ππ)=_____________
c) π¦π¦ = tan οΏ½3π₯π₯ + ππ2οΏ½=____________ d) π¦π¦ = cos οΏ½π₯π₯
4β ππ
8οΏ½=______________
e) π¦π¦ = tan οΏ½π₯π₯3β ππ
6οΏ½=_____________ f) π¦π¦ = sin 1
2(3π₯π₯ + ππ)=____________
2) Write down the period and phase shift for each of these functions:
a) π¦π¦ = cos 14οΏ½π₯π₯ β ππ
2οΏ½______________ b) π¦π¦ = tan 3 οΏ½π₯π₯ + ππ
6οΏ½______________
c) π¦π¦ = sin 32οΏ½π₯π₯ + ππ
3οΏ½______________ d) π¦π¦ = cos 2 οΏ½π₯π₯ β ππ
2οΏ½______________
e) π¦π¦ = sin 2 οΏ½π₯π₯ + ππ4οΏ½______________ f) π¦π¦ = tan 1
3οΏ½π₯π₯ β ππ
2οΏ½______________
2π₯π₯ β ππ2
= 0
i.e. 2π₯π₯ = ππ2
i.e. π₯π₯ = ππ4
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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: βππβ controls vertical dilations β amplitude
βππβ controls horizontal dilations β οΏ½πππππππ π ππππ = 2πππποΏ½
βππβ controls horizontal slides β phase shift
βππβ controls vertical slides.
Example
We want to sketch π¦π¦ = 2 sin 2 οΏ½π₯π₯ β ππ2οΏ½ β 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.
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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 ππ = 2, we know the period is 2ππ2
or ππ. One full wave occupies ππ radians on the horizontal scale.
Also, because βππβ is 2, we know that the amplitude is 2. This enables us to mark unit divisions on the vertical axis.
Now, we notice that βππβ 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.
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Sketch it on your calculator over the domain β2ππ β€ π₯π₯ β€ 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 π₯π₯ = 1.5 over the domain βππ β€ π₯π₯ β€ 2ππ and give your answer correct to one decimal place.
Here we have graphed π¦π¦ = tan π₯π₯ and π¦π¦ = 1.5. the points of intersection represent the points where tan π₯π₯ = 1.5.
What we need to give as the solutions, is the set of values of π₯π₯ which make this equation true. We use a numerical scale to read off the solutions:
π₯π₯ β {β2.1, 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 β5 β€ π₯π₯ β€ 5 to the equation 2 sin π₯π₯ = β1 to (1 d.p.)
Frist rewrite in the form sin π₯π₯ = ππ
i.e. sin π₯π₯ = _______
Now draw the curve π¦π¦ = sin π₯π₯, β5 β€ π₯π₯ β€ 5
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Did you get something like this?
Draw in your line π¦π¦ = ππ.
Mark in the points of intersection.
Find the ππ values to 1 decimal place.
Try to be as accurate as you can by using the diagram above.
State the solutions: π₯π₯ β { ______________________________________ }
Check your answers on your calculator, were you pretty close?
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Lesson 5 Angle Sum and Difference
By the end of this lesson you should be able to
β’ Review the unit circle and in which quadrant sine, cosine and tangent are positive and negative
β’ Review exact values of sine, cosine and tangent for 30o, 45o and 60o and their correspondingvalue in radians.
β’ Prove and apply the angle sum and difference identities
β’ Use the identity tan π₯π₯ = sinπ₯π₯cosπ₯π₯
to simplify suitable trigonometric expressions
β’ Establish and use the identity sin2 ππ + cos2 ππ = 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 ππ,
sinππ = π¦π¦
cosππ = π₯π₯
tanππ = π¦π¦π₯π₯
Remember, tanππ is not defined for values of ππ which are odd multiples of 90o (or ππ2 radians).
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Signs of the trigonometric ratios
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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 ππ + cos2 ππ = 1
Let P be the point with co-ordinates (cosππ , sin ππ).
We know that PA = sinππ and OA = cos ππ.
In the ΞOAP, because β OAP is a right angle, then
PA2 + OA2 = OP2
i.e. sin2 ππ + cos2 ππ = 1 (OP is the radius of the circle)
sin π₯π₯ = ππππ ,
rearranging to make ππ the subject
ππ = ππ sin π₯π₯
cos π₯π₯ = ππππ
rearranging to make ππ the subject
ππ = ππ cos π₯π₯
We know that tan π₯π₯ = ππππ
Substituting ππ and ππ
tan π₯π₯ = ππ sinπ₯π₯ππ cosπ₯π₯
tan π₯π₯ = sinπ₯π₯cosπ₯π₯
(π π π π ππ ππ)2 is usually written as π π π π ππ2 ππ
So as not to be confused by π π π π ππ(ππ2)
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Letβs write the identity here again.
sin2 ππ + cos2 ππ = 1
Rearranging we get
sin2 ππ = 1 β cos2 ππ or cos2 ππ = 1 β sin2 ππ
These may then be factorised to
sin2 ππ = (1 β cos ππ)(1 + cos ππ) or cos2 ππ = (1 β sinππ)(1 + sinππ)
Test yourself
Try and write down the different forms of the Pythagorean identity that we have just generated, without looking back.
1. sin2 π₯π₯ + cos2 π₯π₯ = 1
2. cos2 ππ = =
3. sin2 ππ = =
Sine and cosine of οΏ½π π ππβ π½π½οΏ½ or (ππππΒ°β π½π½)
Consider the diagram below which is part of a unit circle diagram.
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Let β POS = β ROQ = ππ.
Then β QOS = ππ2β ππ.
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
β ROQ = β OQB = ππ (alternate β π π ).
In βπ π QBO, OAP,
β OQB = β POA (= ππ)
β OBQ = β POA (both right angles)
and OQ = PO (radii of circle)
β΄ βπ π QBO and OAP are congruent
i.e. QB = OA = ππ P(ππ, ππ)
and OB = ππππ = ππ Q(ππ,ππ)
Now,
sin οΏ½ππ2β πποΏ½ = sinβ SOQ = QB = OA = ππ = cosππ
and
cos οΏ½ππ2β πποΏ½ = cosβ SOQ = OB = PA = ππ = sin ππ.
sin οΏ½ππ2β πποΏ½ = cos ππ cos οΏ½
ππ2β πποΏ½ = sinππ
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Sine and cosine of οΏ½π π ππ
+ π½π½οΏ½ or (ππππΒ° + π½π½)
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 οΏ½ππ2
+ πποΏ½ and cos οΏ½ππ2
+ πποΏ½ .
Again, we have QB = OA = ππ P(ππ, ππ)
and OB = PA = ππ Q(βππ,ππ)
Then sin οΏ½ππ2
+ πποΏ½ = β SOQ
= QB
= OA
= ππ (π¦π¦ β coordinate of Q)
= cos ππ (π₯π₯ β coordinate of P)
Also, cos οΏ½ππ2
+ πποΏ½ = cosβ SOQ
= -OB
= -PA
= -ππ (π₯π₯ β coordinate of Q)
= β sinππ (βve π¦π¦ β coordiante of P)
sin οΏ½ππ2
+ πποΏ½ = cos ππ cos οΏ½ππ2
+ πποΏ½ = βsinππ
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The distributive property does not apply This lesson is concerned with developing and using the formula for
sin(ππ Β± π΅π΅) and
cos(ππ Β± π΅π΅).
A few minutes with your calculator will demonstrate to you, quite conclusively, that sin(ππ + π΅π΅) is not equal to sinππ + sinπ΅π΅.
e.g. sin(20Β° + 10Β°)
We know that sin(20Β° + 10Β°) = 30Β°.
We also know that sin 30Β° = 12
or 0.5.
Try adding sin 20Β° and sin 10Β°. What do you get? _____________.
Since this is not equal to 0.5, we have made the point.
If you havenβt been convinced by this example, here are a few more to try.
Exercise 5.1
1. cos(30Β° + 60Β°) = cos 90Β° = __________
cos 30Β° + cos 60Β° = _________
2. tan(45Β° + 45Β°) = tan 90Β° = __________
tan 45Β° + tan 45Β° = ________
3. sin(60Β° + 60Β°) = sin 120Β° = __________
sin 60Β° + 60Β° = ________
We could try examples with sin(ππ β π΅π΅) or with cos(ππ β π΅π΅), etc, but it seems obvious that we will achieve the same results.
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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 β π΅π΅, we now have:
Q coincides with Qβ on the on the π₯π₯ axis,
P coincides with Pβ² in fig 4.2.
and the β Pβ²OQ β‘ β POQ = A β B.
Now, letβs extend our diagrams as follows.
sin (ππ Β± π΅π΅) = sinππ cosπ΅π΅ Β± cosππ sinπ΅π΅
cos (ππ Β± π΅π΅) = cosππ cosπ΅π΅ β sinππ sinπ΅π΅
tan (ππ Β± π΅π΅) = tanππ Β± tanπ΅π΅
1 β tanππ tanπ΅π΅
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PQ is the same length as PβQβ because they are chords formed by angles of equal size.
PD, QE and PβG are drawn perpendicular to the π₯π₯ axis.
Now, (PQ) = (Pβ²Qβ²)Β²
i.e. * (FQ)2 + (PF)Β² = (GQβ)Β² + (PβG)Β² [Pythagoras]
But, FQ = OE β OD
cosπ΅π΅ β cosππ in Fig 4.3
PF = PD β QE
= sinππ β sinπ΅π΅
GQβ = OQβ β OG
= 1 β cos(ππ β π΅π΅) in Fig 4.4
and PβG = sin(ππ β π΅π΅)
Now, substituting these in the line marked with the asterisk on the previous page, we have:
(FQ)2 + (PF)2 = (GQβ²)2 + (Pβ²G)2
i.e. (cosπ΅π΅ β cosππ)Β² + (sinππ β sinπ΅π΅)Β² = [1 β cos(ππ β π΅π΅)]Β² + [sin(ππ β π΅π΅)]Β²
Multiplying out
coπ π 2π΅π΅ β 2 cosππ cosπ΅π΅ + cos2ππ + sin2ππ β 2 sinππ sinπ΅π΅ + sinΒ²π΅π΅
= 1 β 2 cos(ππ β π΅π΅) + cos2(ππ β π΅π΅) + sin2(ππ β π΅π΅)
Simplifying using the identity sin2ππ + cos2ππ = 1
2 β 2 cosππ cosπ΅π΅ β 2 sinππ sinπ΅π΅ = 2 β 2 cos(ππ β π΅π΅)
Rearranging
cos(ππ β π΅π΅) = cosππ cosπ΅π΅ + sinππ sinπ΅π΅
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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 (ππ + π΅π΅) = cosππ cosπ΅π΅ β sinππ sinπ΅π΅
Replace π΅π΅ by (βπ΅π΅).
cos (ππ β π΅π΅) = cosππ cos(βπ΅π΅) β sinππ sin(βπ΅π΅)
i.e. cos (Aβπ΅π΅) = __________________________
2. sin (ππ + π΅π΅) = cos οΏ½ππ2β (ππ + π΅π΅)οΏ½
= cos οΏ½ππ2β ππ_________οΏ½
= cos οΏ½οΏ½ππ2β πποΏ½ _________] πππππππππ‘π‘ πππ π cos(ππ β π΅π΅)
= cos οΏ½ππ2β πποΏ½ cos(βπ΅π΅) β sin οΏ½ππ
2β πποΏ½ sin(βπ΅π΅)
= ________ cosπ΅π΅ ____________sinπ΅π΅
3. sin (ππ β π΅π΅) = sin[ππ + (βπ΅π΅)]
= sinππ cos(βπ΅π΅) + cosππ sin(βπ΅π΅)
= _________________________
Check in your textbook to confirm your working.
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We use the fact that tanππ = sinπ΄π΄cosπ΄π΄
to derive the tan(ππ + π΅π΅) and tan(ππ β π΅π΅) formulae.
4. tan (ππ + π΅π΅) =sin(π΄π΄+π΅π΅)cos(π΄π΄+π΅π΅)
= sinπ΄π΄ cosπ΅π΅+cosπ΄π΄ sinπ΅π΅cosπ΄π΄ cosπ΅π΅βsinπ΄π΄ sinπ΅π΅
We use a mathematicianβs trick here and divide numerator and denominator by cosππ cosπ΅π΅. The resulting expression is a complex fraction, but as long as we keep our wits about us, that shouldnβt be a problem.
sinπ΄π΄cosπ΅π΅cosπ΄π΄cosπ΅π΅+
cosπ΄π΄sinπ΅π΅ππππππ π΄π΄cosπ΅π΅
cosπ΄π΄cosπ΅π΅cosπ΄π΄cosπ΅π΅β
sinπ΄π΄sinπ΅π΅cosπ΄π΄cosπ΅π΅
= tanπ΄π΄+tanπ΅π΅1βtanπ΄π΄ tanπ΅π΅
5. Fill in the steps for tan (ππ β π΅π΅) for yourself
tan (ππ β π΅π΅) =
=
=
= _____________________
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Alternative method:
tan(ππ β π΅π΅) = tan[ππ + (βπ΅π΅)] Note: use tan(βπ΅π΅) = β tan(π΅π΅)
= tanπ΄π΄+tan(βπ΅π΅)1βtanπ΄π΄ tan(βπ΅π΅)
= _____________________
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 β 30 )
= cos 45 cos 30 + sin 45 sin 30
= 1β2
Γ β32
+ 1β2
Γ 12
= β32β2
+ 12β2
= β3+12β2
Solution 2
cos 15
= cos(60 β 45 )
= cos 60 cos 45 + sin 60 sin 45
= 12
Γ 1β2
+ β32
Γ 1β2
= 12β2
+ β32β2
= β3+12β2
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Try this one yourself 6. Use the addition formulae to find the exact value of tan 165 .
tan 165 = tan(_______ + _______)
= tan _______+ tan _______ 1βtan _______ tan _______
= β tan _______ +tan _______
1β(βtan _______ tan _______)
= 1β tan______ 1+ tan_______
= 1β ______ 1+ _______
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Lesson 6
Graphs of trigonometric functions horizontal dilations
By the end of this lesson you should be able to
β’ 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
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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.
https://commons.wikimedia.org/w/index.php?curid=3017148
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:
β’ sound waves
β’ oscillating springs
β’ strings on a musical instrument
β’ the vibration of air in a pipe.
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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.)
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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.
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Exercise 6.1 Biorhythms are cycles that are thought to track a personβs intellectual, physical and emotional well-being.
1. Below is Vinceβs 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: π¦π¦ = sin π₯π₯, π¦π¦ = cos π₯π₯ orπ¦π¦ = tan π₯π₯, for π₯π₯ from βππ β€ π₯π₯ β€ ππ, best matches the emotional cycle in Vinceβs chart?(Check these functions on your calculator if you are not sure.) remember to useradians
c) If today is 23rd May, what dates will Vinceβs 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.
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e) Use the function given to complete the following table of approximate values for thefunction, rounding to two decimal places when required. ππ(ππ) = 100 cos οΏ½ππππ
14οΏ½
remember to use radians
f) What is the period it takes for the graph to repeat itself?
g) Vinceβs 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 ππ(ππ) = 100 cos οΏ½ππ(ππ+2)14
οΏ½models Vinceβs physical cycle and give the approximate physical percentages for ππ = 0, 2, 4, 6, 8, 10, 12, 14.
i) How is the two-day difference between Vinceβs physical and emotional cycles reflectedin the defining rule of the function?
j) Vinceβs 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 Vinceβs intellectual cycle and givethe approximate intellectual percentages for ππ = 0, 2, 4, 6, 8, 10, 12, 14.
d 0 2 4 6 7 8 10 12 14
ππ(ππ) 2dp
d 0 2 4 6 8 10 12 14
ππ(ππ) 2dp
d 0 2 4 6 7 8 10 12 14
ππ(ππ) 2dp
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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 + =
Ο
Passion: ( 3)( ) 85cos14df d + =
Ο
Wisdom: ( 1)( ) 80cos14df d + =
Ο
k) Sketch the graphs of the functions for these three secondary cycles for Vinceβs 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.
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m) Consider another secondary cycle for balance, modelled with the same period by thefollowing function:
Balance: ( 7)( ) 95cos14df d β =
Ο
Give the high and low values and amplitude for this cycle and explain what effect the β7 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
ππ(ππ)
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 = β
Ο .
He rewrites the function for the emotional cycle as (7 )( ) 100sin14
df d β =
Ο
Show how he obtains this sine function.
p) Rewrite the functions for her physical and intellectual cycles as sine functions.
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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 π¦π¦ = tan π₯π₯. 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)
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e) Compare the graphs of the following functions and choose one to model the two beam spotlight rotations.
π¦π¦ = 3tan οΏ½πππ₯π₯2οΏ½ ; π¦π¦ = tan οΏ½πππ₯π₯
2οΏ½ ,π¦π¦ = 3tan οΏ½πππ₯π₯
4οΏ½ ,π¦π¦ = tan οΏ½πππ₯π₯
4οΏ½
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
position on wall above gate
-3
3
(seconds)
time4 8 12 16
position on wall above gate
-3
3
(seconds)
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Lesson 7
Solving trigonometric equations By the end of this lesson you should be able to
β’ solve equations involving trigonometric functions using technology
β’ solve equations involving trigonometric functions algebraically
Solving trigonometric equations As an example, suppose you were asked to solve sin ππ = 0.8. Your calculator would give
ππ = 0.927β¦ in radian mode or ππ = 53.13Β°β¦ 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 ππ.
A look at the graph of sinππ demonstrates this fact. There are eight solutions for β2ππ β€ ππ β€ 6ππ
A unit circle diagram is helpful in working out the solutions too. Adding or subtracting 360 or (2ππ) will give further solutions.
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If we worked in radians, for the given equation, sinππ = 0.8, we could have the following solutions for β2ππ β€ ππ β€ 4ππ.
The reference angle ππ is 0.927 radians.
The second quadrant solution for β2ππ β€ ππ β€ 4ππ is (ππ β ππ) or 2.215
Solutions for (0 β€ ππ β€ 2ππ) are 0.927 and 2.215
Adding 2ππ for (2ππ β€ ππ β€ 4ππ) gives 7.210 and 8.498
Subtracting 2ππ for (β2ππ β€ ππ β€ 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 (βπ½π½) = βπ¬π¬π¬π¬πππ½π½), οΏ½π π ππ
Β± π½π½οΏ½ , (π π Β± π½π½)
Ratios for (βπ½π½)
sin(βππ) = β sinππ tan(βππ) = β tanππ cos(βππ) = cos ππ
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Ratios for οΏ½π π ππ
Β± π½π½οΏ½
sin οΏ½ππ2
+ πποΏ½ = cos ππ sin οΏ½ππ2β πποΏ½ = cos ππ
cos οΏ½ππ2
+ πποΏ½ = βsinππ cos οΏ½ππ2β πποΏ½ = sinππ
tan οΏ½ππ2
+ πποΏ½ = β 1 tan ππ
tan οΏ½ππ2β πποΏ½ = 1
tan ππ
Ratios for (π π Β± π½π½)
sin(ππ + ππ) = βsinππ sin οΏ½ππ2β πποΏ½ = cos ππ
cos(ππ + ππ) = β cosππ cos οΏ½ππ2β πποΏ½ = sinππ
tan(ππ + ππ) = tanππ tan (ππ β ππ) = tanππ
Note: π π ππππππ ππππ ππππππππππππππππ ππππ ππππππ πππππππ π ππππππππ ππππ ππππππππππππππππ ππππ ππππ
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Example Follow the steps below to solve cosππ = 0.6, 0 β€ ππ β€ 2ππ.
Solution Use the domain to find the number of solutions.
cosππ = 0.6, 0 β€ ππ β€ 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 cosβ1 0.6 = 0.93 (2 d.p.)
The fourth quadrant solution is 2Ο β 0.9273 = 5.36 (2 dp).
Note also that sinΞΈ = (β0.5), for (0 β€ ΞΈ β€ 2Ο) has two exact solutions, one in each of the third and fourth quadrants with a reference angle of Ο
6. Solutions are 7Ο
6 and 11Ο
6.
Also, when solving trigonometric equations, you must be wary of how you use the formulae. Consider this example:
Solve sin(2π₯π₯) = cos(π₯π₯). 0 β€ ΞΈ β€ 2Ο
You can't do this:
Youβve lost solution(s) because you havenβt considered the case where cos(π₯π₯) = 0
sin(π₯π₯ + π₯π₯) = cos(π₯π₯)
sin(π₯π₯)cos(π₯π₯) + cos(π₯π₯) sin(π₯π₯) = cos(π₯π₯)
2 sin(π₯π₯) cos(π₯π₯) = cos(π₯π₯)
2 sin(π₯π₯) = 1
sin(π₯π₯) = 12
DO NOT DO THIS
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So you MUST factorise instead:
Now we have 2 equations to solve for π₯π₯. We need to look at the domain. 0 β€ ΞΈ β€ 2Ο.
For cos π₯π₯ = 0
π₯π₯ = ππ2
or 3ππ2
For sin π₯π₯ = 12
π₯π₯ = ππ6 or 5ππ
6
Therefore for Solve sin(2π₯π₯) = cos(π₯π₯). 0 β€ ΞΈ β€ 2Ο
π₯π₯ = ππ2
, ππ6
, 3ππ2
, 5ππ6
sin(2π₯π₯) = cos(π₯π₯)
2 sin(π₯π₯) cos(π₯π₯) = cos(π₯π₯)
2 sin(π₯π₯) cos(π₯π₯) β cos(π₯π₯) = 0
cos(π₯π₯)[2 sin(π₯π₯) β 1] = 0
So either cos(π₯π₯) = 0 or 2 sin(π₯π₯) = 1
or sin(π₯π₯) = 1 sin(π₯π₯) = 1
2
DO THIS
Remember to use the horizontal axis for cos.
Remember to use the vertical axis for sin.
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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 0.5ΞΈ = , 2 2Ο ΞΈ Οβ β€ β€
b) 1sin , 0 22
ΞΈ ΞΈ= β β€ β€ Ο
c) 4 sin οΏ½π₯π₯ + ππ2οΏ½ = 2 for 0 β€ π₯π₯ β€ 2ππ
d) tan 2π₯π₯ = 1 for β180 β€ π₯π₯ β€ 180
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e) sin(2π₯π₯) = cos π₯π₯ , for β ππ β€ ππ β€ ππ,
f) cos π₯π₯ + 2 cos2 π₯π₯ = 0 for β ππ β€ ππ β€ ππ
g) cos οΏ½π₯π₯ + ππ4οΏ½ = 1
β2 for β2ππ β€ π₯π₯ β€ 0
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Lesson 8 and 9
Exam practice
By the end of this lesson you should be able to
β’ refer to SCSA past exams
β’ 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:
https://senior-secondary.scsa.wa.edu.au/syllabus-and-support-materials/mathematics/mathematics-methods
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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(ππ Β± π΅π΅) = sinππ cosπ΅π΅ Β± cosππ sinπ΅π΅
cos(ππ Β± π΅π΅) = cosππ cosπ΅π΅ β sinππ sinπ΅π΅
Area of a sector The area of a sector of a circle is given by ππ = 12ππ2ππ, where ππ is the sector area,
ππ is the radius and ππ 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 ππ = 12ππ2(ππ β sinππ), where ππ is the
segment area, ππ is the radius and ππ 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 β = ππππ, where β is the arc length, ππ is the radius and ππ 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 β = 2ππ sin 12ππ, where β is the chord
length, ππ is the radius and ππ is the angle subtended at the centre, measured in radians.
Period of a function The period of a function ππ(π₯π₯) is the smallest positive number ππ with the property that ππ(π₯π₯ + ππ) = ππ(π₯π₯) for all π₯π₯. The functions sin π₯π₯ and cos π₯π₯ both have period 2ππ and tan π₯π₯ has period ππ.
Radian measure The radian measure ππ of an angle in a sector of a circle is defined by ππ = βππ
, where ππ is the radius and β 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 ππ
sin π΄π΄= ππ
sinπ΅π΅= ππ
sin πΆπΆ .
This is known as the sine rule.
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The lengths of the sides of a triangle are related to the cosine of one of its angles by the equation ππ2 = ππ2 + ππ2 β 2ππππ cosπΆπΆ.
This is known as the cosine rule.
Sine, cosine and tangent functions
Since each angle ππ measured anticlockwise from the positive π₯π₯-axis determines a point ππ on the unit circle, we will define
the cosine of ππ to be the π₯π₯-coordinate of the point ππ
the sine of ππ to be the π¦π¦-coordinate of the point ππ
the tangent of ππ is the gradient of the line segment ππππ.
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Solutions
Exercise 1.1
1.
Value of π‘π‘ 0 ππ6
ππ4
ππ3
ππ2
2ππ3
3ππ4
5ππ6
ππ
sin π‘π‘ (to 2 d.p.) 0 0.5 0.7 0.9 1 0.9 0.7 0.5 0
Value of π‘π‘ 7ππ6
5ππ4
4ππ3
3ππ2
5ππ3
7ππ4
11ππ6
2ππ
sin π‘π‘ (to 2 d.p.) -0.5 -0.7 -0.9 -1 -0.9 -0.7 -0.5 0
Value of π‘π‘ 13ππ6
9ππ4
7ππ3
5ππ2
8ππ3
11ππ4
17ππ6
3ππ
sin π‘π‘ (to 2 d.p.) 0.5 0.7 0.9 1 0.9 0.7 0.5 0
Value of π‘π‘ 19ππ6
13ππ4
10ππ3
7ππ2
11ππ3
15ππ4
23ππ6
4ππ
sin π‘π‘ (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 2Ο
3.
4. What is the maximum value of sin (π‘π‘)?
4. -1
5. 0, Ο, 2Ο, 3Ο, 4Ο
6. Repeat for cos t. Remember to work in radian mode. And give your answers to one decimal place.
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Value of ππ 0 π π ππ
π π ππ
π π ππ
π π ππ
πππ π ππ
πππ π ππ
πππ π ππ
π π
cos π‘π‘ (to 2 d.p.) 1 0.9 0.7 0.5 0 -0.5 -0.7 -0.9 -1
Value of ππ πππ π ππ
πππ π ππ
πππ π ππ
πππ π ππ
πππ π ππ
πππ π ππ
πππππ π ππ
πππ π
cos π‘π‘ (to 2 d.p.) -0.9 -0.7 -0.5 0 0.5 0.7 0.9 1
Value of ππ πππππ π ππ
πππ π ππ
πππ π ππ
πππ π ππ
πππ π ππ
πππππ π ππ
πππππ π ππ
πππ π
cos π‘π‘ (to 2 d.p.) 0.9 0.7 0.5 0 -0.5 -0.7 -0.9 -1
Value of ππ πππππ π ππ
πππππ π ππ
πππππ π ππ
πππ π ππ
πππππ π ππ
πππππ π ππ
πππππ π ππ
πππ π
cos π‘π‘ (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 2Ο. Sin starts at (0,0)whilst cos starts at (0, 1)
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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 ππ -Ο βπππ π ππ
βπππ π ππ
βπππ π ππ
βπ π ππ
βπ π ππ β
π π ππ
βπ π ππ
0
tan π₯π₯ 0 0.6 1 1.7 Und -1.7 -1 -0.6 0
Value of ππ 0 π π ππ
π π ππ
π π ππ
π π ππ
πππ π ππ
πππ π ππ
πππ π ππ
Ο
tan π₯π₯ 0 0.6 1 1.7 Und -1.7 -1 -0.6 0
Value of ππ -1.4 -1.2 -1.0 -0.8 -0.6 -0.4 -0.2 0
tan π₯π₯ -5.8 -2.6 -1.6 -1.0 -0.7 -0.4 -0.2 0
Value of ππ 0 0.2 0.4 0.6 0.8 1.0 1.2 1.4
tan π₯π₯ 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 π₯π₯ accurately on this grid. (the ππ isapproximate.)
14. -Ο, 0, Ο
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 π₯π₯ > ππ, tan π₯π₯ will become positive and increasingly larger as π₯π₯ gets closer to β3ππ2
20. When π₯π₯ < βππ, tan π₯π₯ will become negative and will continue to become smaller as larger asπ₯π₯ gets closer to β3ππ
2.
21. No maximum. Function is infinite
22. No minimum
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23. π₯π₯ = 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 βππβ. Check your answers by drawing them on your calculator.
7. π¦π¦ = 3 sin π₯π₯ 8. π¦π¦ = 2 cos π₯π₯
9. π¦π¦ = 2 tan π₯π₯ 10. π¦π¦ = 14
sin π₯π₯
11. π¦π¦ = 4 cos π₯π₯ 12. π¦π¦ = 12
tan π₯π₯
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 ππ Period 2ππ Γ· ππ
sin π₯π₯ 1 1 2ππ 2ππ
sin 2π₯π₯ 2 2 ππ ππ
sin12π₯π₯
12
12
4ππ 4ππ
sin 3π₯π₯ 2ππ3
2ππ3
sin13π₯π₯
13
13
6ππ 6ππ
sin 4π₯π₯ 4 4 ππ2
ππ2
sin14π₯π₯
14
14
8ππ 8ππ
Exercise 2.4 7. Changes the number of waves in 2Ο, ie the period
8. Write down the period of these functions
a) ππ = 1 b) ππ = 14
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Period = 2ππ1
= 2ππ Period = 2ππ14
= 8ππ
c) ππ = 1
Period = 2ππ1
= 2ππ
d) ππ = 6
Period = 2ππ6
= ππ3
e) ππ = 34
Period = 2ππ34
= 8ππ3
f) ππ = ππ
Period = 2ππππ
= 1
g) ππ = 2
Period = 2ππ2
= ππ
9. Write down the amplitude for each of the functions in question 2.
a) ππ = 1 b) ππ = 1
c) ππ = 2 d) ππ = 1
e) ππ = 1 f) ππ = 32
g) ππ = β1 amplitude is 1
10. For π¦π¦ = sin π₯π₯ the period is the length of one horizontal wave, ie 2ππ .
11. Determine the value of ππ from these graphs.
a) 2ππππ
= ππ2
ππ = (2Γ2ππ)ππ
ππ = 4
b) 2ππππ
= ππ
ππ = (2ππ)ππ
ππ = 2
12.
a) π¦π¦ = sin 3π₯π₯ b) π¦π¦ = sin 14π₯π₯
c) π¦π¦ = sin 4π₯π₯ d) π¦π¦ = sin 13π₯π₯
e) π¦π¦ = sin 12π₯π₯
Exercise 3.1
6. a) π¦π¦ = cos π₯π₯
b) 1
c) 2ππ1
= 2ππ
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7. a) π¦π¦ = cos 2π₯π₯ _
b) 2
c) 2ππ2
8. a) π¦π¦ = cos 12π₯π₯
b) 12
c) 2ππ12
= 4ππ
9. a) π¦π¦ = cos 3π₯π₯
b) 3
c) 2ππ3
10. a) π¦π¦ = cos 14π₯π₯
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
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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 ,β3)(157.5 ,β3)(247.5,β3)(337.5,β3)
(iv) nil.
(i) Period =2ππ12
= 4ππ amplitude = 1
(ii) (βππ, 0) (ππ, 0)(iii) Max (0,1)
min(β2ππ,β1)(2ππ,β1)(iv) nil
(i) Period = ππ3
= 4ππ amplitude doesnβt exist
(ii) (βππ, 0) οΏ½β 2ππ3
, 0οΏ½ (βππ, 3)(0,0)
οΏ½ππ3
, 0οΏ½ οΏ½2ππ3
, 0οΏ½ (ππ, 0) (iii) No max or min.(iv) π₯π₯ = β5ππ
6,βππ
2,βππ
6, ππ6
, ππ2
, 5ππ6
,
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Exercise 3.4
Graphic Calculator Activity. 5.
6.
moves ππ2 to the left
7.
moves ππ2 to the left
8. a)
moves ππ4 to the left
b)
moves ππ3 to the right
c)
moves ππ to the left, (or reflected in the x axis)
d)
.moves ππ4
to the right
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Exercise 3.5 2. Sketch the following over the domain β4 β€ π₯π₯ β€ 4
a) π¦π¦ = sin(πππ₯π₯),
b) π¦π¦ = sin(πππ₯π₯ + ππ) ,
c) π¦π¦ = β sinππ(π₯π₯ + 1),
d) π¦π¦ = 1 β sinππ(π₯π₯ + 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 π¦π¦ = sin(πππ₯π₯) but translated left ππ units
g) is the same as π¦π¦ = sin(πππ₯π₯) butranslated left ππ units
h) is the same shape as π¦π¦ = sin(πππ₯π₯) but translated up 1 unit
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Exercise 4.1 2.
a)
b)
c) a) Is the same shape as π¦π¦ = coπ π (π₯π₯) but has been translated right ππ4, b) is the same
shape as π¦π¦ = cos(π₯π₯) but has been translated left ππ3
Exercise 4.2 Calculator Activity 5. ππ moves π¦π¦ = tan π₯π₯ ππ units to the right
Exercise 4.3 2) Sketch the graphs of
a)
b)
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Exercise 4.4 3)
g) π¦π¦ = sin 2 οΏ½π₯π₯ + ππ4οΏ½ h) π¦π¦ = cos 2 οΏ½2 β ππ
4οΏ½
i) π¦π¦ = tan 3 οΏ½π₯π₯ + ππ6οΏ½ j) π¦π¦ = cos 1
4οΏ½π₯π₯ β ππ
2οΏ½
k) π¦π¦ = tan 13οΏ½π₯π₯ β ππ
2οΏ½ l) π¦π¦ = sin 3
2οΏ½π₯π₯ + ππ
3οΏ½
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(30Β° + 60Β°) = cos 90Β° =0
cos 30Β° + cos 60Β° = 1.37 5. tan(45Β° + 45Β°) = tan 90Β° undefined
tan 45Β° + tan 45Β° = 2 6. sin(60Β° + 60Β°) = sin 120Β° =0.87
sin 60Β° + 60Β° = 1.73 Exercise 5.2 6. cos (ππ + π΅π΅) = cosππ cosπ΅π΅ β sinππ sinπ΅π΅
Replace π΅π΅ by (βπ΅π΅).cos (ππ β π΅π΅) = cosππ cos(βπ΅π΅) β sinππ sin(βπ΅π΅)
i.e. cos (Aβπ΅π΅) = cosππ cosπ΅π΅ + sinππ sinπ΅π΅
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7. sin (ππ + π΅π΅) = cos οΏ½ππ2β (ππ + π΅π΅)οΏ½
= cos οΏ½ππ2β ππ β π΅π΅οΏ½
= cos οΏ½οΏ½ππ2β πποΏ½ β π΅π΅ οΏ½ πππππππππ‘π‘ πππ π cos(ππ β π΅π΅)
= cos οΏ½ππ2β πποΏ½ cos(βπ΅π΅) β sin οΏ½ππ
2β πποΏ½ sin(βπ΅π΅)
= sinππ cosπ΅π΅ + cosππ sinπ΅π΅
8. sin (ππ β π΅π΅) = sin[ππ + (βπ΅π΅)]= sinππ cos(βπ΅π΅) + cosππ sin(βπ΅π΅) = sinππ cosπ΅π΅ β cosππ sinπ΅π΅
5. Fill in the steps for tan (ππ β π΅π΅) for yourselftan (ππ β π΅π΅) =
sin(π΄π΄βπ΅π΅)cos(π΄π΄βπ΅π΅)
= sinπ΄π΄ cosπ΅π΅βcosπ΄π΄ sinπ΅π΅cosπ΄π΄ cosπ΅π΅+sinπ΄π΄ sinπ΅π΅
= sinπ΄π΄cosπ΅π΅cosπ΄π΄cosπ΅π΅+
cosπ΄π΄sinπ΅π΅cosπ΄π΄ cosπ΅π΅
cosπ΄π΄cosπ΅π΅cosπ΄π΄cosπ΅π΅β
sinπ΄π΄sinπ΅π΅cosπ΄π΄cosπ΅π΅
= tanπ΄π΄βtanπ΅π΅1βtanπ΄π΄ tanπ΅π΅
6. tan 165 = tan(135 + 30 )
= tan135 +tan30 1βtan135 tan30
= β1+ 1
β3
1β(β1)οΏ½ 1β3οΏ½
= β1+ 1
β3
1+ 1β3
rationalise the denominator by multiplying by 1β 1
β3
1β 1β3
= οΏ½β1+ 1
β3οΏ½οΏ½1β 1
β3οΏ½
οΏ½1β 1β3οΏ½οΏ½1β 1
β3οΏ½
= β43+
2β33
1β13
= β4+2β32
= β2 + β3
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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) π¦π¦ = cos π₯π₯
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. ππ(ππ) = 100 cos οΏ½ππππ
14οΏ½
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) ππ(ππ) = 100 cos οΏ½ππ(ππ+6)14
οΏ½
d 0 2 4 6 7 8 10 12 14
ππ(ππ) 2dp 100 90 62 22 0 -22 -62 -90 -100
d 0 2 4 6 8 10 12 14
ππ(ππ) 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
ππ(ππ) 2dp 22 -22 -62 -90 -100 -90 -62 -22
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k)
l) Mastery 90%, passion 85%, wisdom 80%
m) Amplitude of 95%, peaks again in 7 days.
n)
Cycle 1 72 cos οΏ½ππ(ππ+3)14
οΏ½
Cycle 2 55 cos οΏ½ππ(ππβ2)14
οΏ½
Cycle 3 68 cos οΏ½ππ(ππβ5)14
οΏ½
o) 100 cos οΏ½ππππ14οΏ½ = 100 sin οΏ½ππ
2β ππππ
14οΏ½
= 100 sin οΏ½7ππβππππ14
οΏ½
= 100 sin οΏ½ππ(7βππ)14
οΏ½
p) Physical 100 sin οΏ½ππ(5βππ)14
οΏ½ , intellectual 100 sin οΏ½ππ(1βππ)14
οΏ½
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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) 3yβ€ β€ . 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 2Ο Ο Ο Ο
d) 4 seconds, Ο radians
e) tan4xy =
Ο
f)
time4 8 12 16
position on wall above gate
-3
3
(seconds)
time4 8 12 16
position on wall above gate
-3
3
(seconds)
time4 8 12 16
position on wall above gate
-3
3
(seconds)
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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) β11ππ6
,β7ππ6
, ππ6
, 5ππ6
b) 5ππ4
, 7ππ4
c) ππ3
, 5ππ3
d) -157.5, -67.5, 22.5
e) βππ2
, ππ2
, ππ6
, 5ππ6
f) β2ππ3
,βππ2
, ππ2
, 2ππ3
g) 0,β2ππ,βππ2