Graphing periodic functions - mathsbooks.netmathsbooks.net/Maths Quest 11B for Queensland/Ch 05... · Chapter 5 Graphing periodic functions 195 The length of time that it takes for

5syllabussyllabusrrefefererenceenceTopic:• Periodic functions and

applications

In thisIn this chachapterpter5A Period and amplitude of a

periodic function5B Radian measure5C Exact values5D Symmetry5E Trigonometric graphs5F Applications

Graphingperiodicfunctions

MQ Maths B Yr 11 Ch 05 Page 193 Thursday, October 25, 2001 3:50 PM

194 M a t h s Q u e s t M a t h s B Ye a r 1 1 f o r Q u e e n s l a n d

IntroductionAndrew is riding a ferris wheel20 metres in diameter. It takes2 minutes to complete one revol-ution. Andrew wants to graph hisheight above ground against timeand find a rule from which hecan calculate his height at anytime. The graph of height againsttime will be an example of aperiodic function because itrepeats itself every 2 minutes.

In earlier chapters we havelooked at several different typesof functions. In this chapter wewill look at periodic functions —functions that repeat themselvesover a period of time. Periodicfunctions usually involve thetrigonometric ratios that webegan looking at in chapter 4.

Period and amplitude of a periodic function

A function that repeats itself over a period of time is called a periodic function. Thefigure below, showing the amount of rainfall at a certain weather station each month, isan example of such a function, because the pattern of rainfall tends to repeat each year.

Note that all the previous function properties have been satisfied. The only newaspect is the periodicity.

The highlighted region on the graph shows the section of curve that repeats itself.There are many everyday occurrences — including the tides and daily temperatures —that can be closely represented by a periodic function.

Month

50

0 J J J A S O N D J F M AF M A M

100

150

Rai

nfal

l (m

m)


C h a p t e r 5 G r a p h i n g p e r i o d i c f u n c t i o n s 195

The length of time that it takes for a periodic function to begin repeating itself is called the period, T, of the function.

Amy is a keen snorkler who wants to know the best time to go snorkelling. She measures the depth of water in the inlet every 2 hours for several days and plots the results on the graph below.

Every 6 hours the graph repeats itself and so we can say that 6 hours is the period of this function. That is, T = 6 h.

From the graph it can be seen that the maximum depth of water in the inlet is 10 metres and the minimum depth is 4 metres.

The equilibrium position is the centre of the oscillation and so in this case theequilibrium position is a depth of 7 metres. The maximum depth of water is3 metres above this equilibrium and the minimum depth of water is 3 metres belowequilibrium.

The amplitude is the maximum variation in the function from its equilibrium. We cantherefore say that this periodic function has an amplitude, A, of 3 metres. That is,A = 3 m.

Generally the easiest way to calculate the amplitude is to halve the differencebetween the maximum and minimum values.

Temperature and tideTides

1 Use the newspaper or other sources to find the time and depth of each high and low tide in Moreton Bay over a period of 3 days.

2 Plot the points generated, showing time on the horizontal axis and depth on the vertical axis. Join the points with a smooth curve.

3 Over what period of time does there appear to be a repetition of the graph?

Temperature4 Use the newspaper or the Weather Bureau’s web site to find the temperature in

Brisbane at 2-hourly intervals over a period of 3 days.5 Plot the points generated, showing time on the horizontal axis and temperature

on the vertical axis. Join the points with a smooth curve.6 Over what period of time does there appear to be a repetition of the graph?

Time

2

04 a

m8 a

m

12 m

id

12 no

on4 p

m8 p

m4 a

m4 a

m8 a

m

12 m

id

12 no

on4 p

m8 p

m

4

6

8

10

Dep

th (

m)



By finding the point that lies halfway between the maximum and minimum values,we are able to find this equilibrium position. In the predecing case, we can see thathalfway between the maximum value of 10 and the minimum value of 4 is theequilibrium position of 7 metres.

Once we have calculated the period of a function we can calculate the frequency, f,

using the formula f = . The frequency tells us how many repetitions of the function

occur in one unit of time.

For the case of a tidal function, where the period of the function is 12 hours ( day),

the frequency would be f =

=

= 2.This means that the function repeats twice every day.

In many practical examples the function may not be exactly the same each period andso we make an estimate of both the period and amplitude.

x

y

1

1211109876543210

2

3

4For the function drawn at right state:a the periodb the amplitude.

THINK WRITE

a The function repeats itself every 4 units on the x-axis.

a T = 4

b The maximum value on the y-axis is 4. bThe minimum value on the y-axis is 2.

Halve the difference between 2 and 4. Amplitude =

Amplitude = 1

12

34 2–

2------------

1WORKEDExample

1T---

12---

1T---

1 12---

------

A periodic function has a period of 6.25 units. Calculate the frequency of the function.

THINK WRITE

Write the formula. f =

Substitute the value of T. f =

Calculate. f = 0.16

11T---

21

6.25----------

3

2WORKEDExample


C h a p t e r 5 G r a p h i n g p e r i o d i c f u n c t i o n s

197

The graph below shows the number of customers that pass through the checkouts of a 24-hour supermarket each day. State the period and amplitude of the function.

THINK WRITE

The pattern appears to repeat every 24 hours.

T = 24 hours

The maximum value is 600.

The minimum value is 40.

Halve the difference between 40 and 600. Amplitude =

Amplitude = 280

100

2000

1600

1200

0800

0400

0000

0000

2000

1600

1200

0800

0400

0000

200

300

400

500

600

Time of day

Num

ber

of c

usto

mer

s 800

700

1

2

3

4600 40–

2---------------------

3WORKEDExample

remember1. The period, T, of a function is the length of time (as shown on the horizontal

axis) that a function takes to begin to repeat itself.2. The amplitude, A, of a periodic function is the maximum distance (as shown on

the vertical axis) that the function differs from its equilibrium position.3. The amplitude is most easily found by halving the difference between the

maximum and minimum values of a periodic function.4. The equilibrium position can be determined by finding the average of the

maximum and minimum values.5. The frequency, f, tells us how many repetitions of the function occur in one unit

of time.

6. The frequency can be found using the formula f = .

7. The greater the period of the function, the smaller the frequency, and vice versa.

1T---

remember


198

M a t h s Q u e s t M a t h s B Ye a r 1 1 f o r Q u e e n s l a n d

Period and amplitude of a periodic function

1

For the function below, state:

a

the period

b

the amplitude.

2 For each of the functions drawn below, state the period and the amplitude.

3 A periodic function has a period of 4 units. Calculate the frequency of the function.

4 The graph at right shows the number of people crossing an intersection during a 12-minute period. For the function at right, state the period and amplitude.

a b c

d e f

g h

5AWWORKEDORKEDEExample

1

1

109876543210

2

3

4

5

6

y

x4π2π

–2

2

0

y

x

–1

1

0–4π 5 —

4π

y

x3π

–1.5

1.5

0

y

x

–4

4

0 3 —2π

y

x

–2

2

0 π–π

y

x

–3

3

0 π–3

π–

y

x

–2.5

2.5

0–2

π π–

y

x

–0.5

0.5

0 –6π –

2π

WWORKEDORKEDEExample

2


3

t

1

121086420

2

3

4

5

6


C h a p t e r 5 G r a p h i n g p e r i o d i c f u n c t i o n s 1995 The graph below shows the average humidity in

Brisbane each month of the year.

a Does the graph appear to be a periodic function?b If so, what appears to be the period of the function?c Which month appears to be the hottest month of

the year?d Which month appears to be the coldest month of

the year?e What is the amplitude of the function?

6

Which of the following graphs does NOT appear to represent a periodic function?A B

C D

E All are periodic functions.

7 The graphs at right show the levelsof oestrogen and progesterone in a woman’s body over the 28 days of her menstrual cycle.a Which graph has the greater

amplitude?b What is the frequency for the

functions?

Months of the year

40

0 J J J A S O N DF M A M J JJ J A S O N DF M A M

50

60

70

80

90

Hum

idity

mmultiple choiceultiple choice

765421 3 1413121198 10 212019181615 17 282726252322 24

ProgesteroneOestrogen



Radian measureIn earlier chapters we saw that both the linear and quadratic functions could be definedalgebraically. A linear function can be written in the form y = mx + c, while a quadraticfunction takes the form y = ax2 + bx + c. A periodic function can also be related to analgebraic expression.

Before we investigate this expression, we need to explore other areas. One such area is the concept of a radian.

You should be familiar with measuring angles in degrees (°), and will recall that there are 360° in a full circle. An alternative unit for angle measurement is the radian.

Consider the circle drawn at right, which shows a sector that cuts off an arc on the circle equal to the radius of the circle.

This angle is defined to be equal to 1 radian, which is written 1c. In many examples nounit is given, in which case radians are assumed.The circumference of a circle = 2πr units.So the angle swept in one revolution is equivalent to 2π radians or 2πc.

Rhythm of lifeExamine your pulse rate 4 times a day for a week as you go about your regular routine. Try to get someone else to take your pulse at least once during the week while you sleep.1 Graph your pulse rate against the time of day.

a Does the graph appear to be a periodic function?b What is the period of the function?

2 Find the maximum and minimum values and, hence, the amplitude of the function.

Ferris wheelingAt the beginning of this chapter we looked at Andrew, who was riding the ferris wheel.1 Given that the ferris wheel takes 2 minutes to complete one revolution, state the

period of the function.2 What is the frequency of the function?

Finding a radian1 Draw a circle of radius 5 cm.2 Cut off a piece of string 5 cm long and place it on the circumference of the circle.3 Mark the end points of the arc formed by the strings and draw radii at these points.4 Use a protractor to measure the angle between these radii.

y

x

1 radius

1 radian


C h a p t e r 5 G r a p h i n g p e r i o d i c f u n c t i o n s 201Therefore, 2πc = 360°, so:

πc = 180°

1c =

1° = .

These formulas may be used to convert degrees to radians and vice versa. In mostcases an angle measured in radians will be given in terms of π. This is an exact radianmeasure.

180°

πc-----------

πc

180°-----------

Convert the following angles into exact radians.a 120∞b 70∞

THINK WRITE

a Multiply 120 by . a 120° = 120° ×

Simplify by dividing through by the highest common factor, 60.

120° =

b Multiply 70 by . b 70° = 70° ×

Simplify by dividing through by the highest common factor, 10.

70° =

1πc

180°----------- πc

180°-----------

22πc

3--------

1πc

180°----------- πc

180°-----------

27πc

18--------

4WORKEDExample

Convert the following angles to degrees.

a b 0.8πc

THINK WRITE

a Multiply by . a = ×

Cancel out the πc. = × 180°

Simplify. = 150°

b Multiply 0.8πc by . b 0.8πc = 0.8πc ×

Cancel out the π c. = 0.8 × 180°

Simplify. = 144°

5πc

6--------

15πc

6-------- 180°

πc----------- 5πc

6-------- 5πc

6-------- 180°

πc-----------

256---

3

1180°

πc----------- 180°

πc-----------

2

3

5WORKEDExample



When angles expressed in radians are not given in terms of π, they do not give awhole number when converted to degrees. In such cases the angle will need to beapproximated.

Radian measure

1 Convert the following angles into exact radians.

2 Convert the following angles to degrees.

a 30° b 45° c 60° d 20° e 50°f 90° g 270° h 360° i 150° j −225°k −420° l 300° m 48° n 72°

a b c d e 3.6π

f g h 0.375π

Use a calculator to make the following conversions.a 57.2∞ to radians, correct to 3 decimal placesb 2.75c to degrees, correct to the nearest degree

THINK WRITE

a Multiply 57.2 by . a 57.2° = 57.2° ×

Evaluate and round off to 3 decimal places.

= 0.998c

b Multiply 2.75 by . b 2.75c = 2.75c ×

Evaluate and round off to the nearest degree.

= 158°

1πc

180°----------- πc

180°-----------

2

1180°

πc----------- 180°

πc-----------

2

6WORKEDExample

remember1. An angular measure of 1 radian will cut off an arc equal to one radius on a

circle.2. πc = 180°3. To convert an angle in degrees to radians, we multiply by .

4. To convert an angle in radians to degrees, we multiply by .

5. If no unit is given for the angle then the angle is assumed to be in radians.

πc

180°-----------

180°πc

-----------

remember

5B

SkillSH

EET 5.1 WWORKEDORKEDEExample

4

SkillSH

EET 5.2 WWORKEDORKEDEExample

5 π5--- 2π

3------ 2π

9------ 11π

9---------

π6---– π

4---–



The smallest angle measurement listed below is:

4

The largest angle measurement listed below is:

5 Use a calculator to convert the following angles toradians, correct to 3 decimal places.

6 Use a calculator to convert each of the followingradian measures to degrees. Give each answer cor-rect to the nearest degree.

Exact valuesCalculated trigonometric values (sin, cos and tan) of most angles are rational approximations, correct to several decimal places.

However, for a few particular angles exact trigonometric values can be determined. Consider the isosceles triangle drawn below.

This triangle with equal sides of 1 unit has a hypotenuse of units. We can therefore calculate the exact trigonometric values for 45°.

So using SOH CAH TOA

sin 45° = cos 45° = tan 45° =

= × = × = 1

= =

A 85° B π C 1.92c D E .

A −3π B C 250°

D E 3.84c.

a 49° b 78° c −125°d 191° e 234° f 327°g 170° h 217°

a 0.35c b 1.47c c 3.142c

d 0.958c e 15.6c f −2.71g 4.06c h 8c


2π3

------ 2π5

------


4π3

------

5π4

------


6a


6b

1

145°

2

2

1

2------- 1

2------- 1

1---

1

2------- 2

2------- 1

2------- 2

2-------

22

------- 22

-------



To find the exact trigonometric values for 30° and 60° consider the equilateral tri-angle with each side length 2 units. The triangle has been bisected down the middle

giving a perpendicular height of units.

So using SOH CAH TOA

sin 30° = cos 30° = tan 30° =

= ×

=

sin 60° = cos 60° = tan 60° =

= These results should be memorised for both degrees and radians. The triangles using

radian measures are shown in the figures below.

The results are summarised in this table.

Angle(degrees)

Angle(radians) sin q cos q tan q

30°

45° 1

60°

3

30°

60°

2

1

3

12--- 3

2------- 1

3-------

1

3------- 3

3-------

33

-------

32

------- 12--- 3

1-------

3

1

1

2

π4—

2

1

3

π6—

π3—

π6--- 1

2--- 3

2------- 3

3-------

π4--- 2

2------- 2

2-------

π3--- 3

2-------

12--- 3



Without using a calculator, state the exact value of:a sin 45° b cos 60° c tan 30°.

THINK WRITE

a Read the value from the table, or determine using the special

triangle, sin θ = .

a sin 45° =

= ×

=

b Read the value from the table, or determine using the special

triangle, cos θ = .

b cos 60° =

c Read the value from the table, or determine using the special

triangle, tan θ = .

c tan 30° =

= ×

=

45°

45°2

1

1

opphyp---------

1

2-------

1

2------- 2

2-------

22

-------

30°

60°

23

1

adjhyp---------

12---

30°

60°

23

1

oppadj---------

1

3-------

1

3------- 3

3-------

33

-------

7WORKEDExample

Find the exact value of:

a sin b tan .

Continued over page

THINK WRITE

a Read the exact value from the table or use a special triangle,

sin θ = .

a sin =

p3--- p

6---

32

1

–3π

–6π

opphyp---------

π3--- 3

2-------

8WORKEDExample



The unit circleTrigonometric ratios can be considered using a unit circle.

The coordinate axes divide the unit circle into four quadrants as shown in the diagram at right.

The position of point P can be described using the coordinates x and y and the angle, θ. The point, P, which is on the circle, is an angle of θ from the positive x-axis and can be expressed as P(θ ).

The x-coordinate of P is x = cosine θ.The y-coordinate of P is y = sine θ.These may be abbreviated: x = cos θ

y = sin θ

The point, P, can also be denoted as P (cos θ, sin θ).

The ‘tangent line’ on a unit circle is the line that is a tangent to the circle at the point (1, 0). Tangent θ (or tan θ for short) is the height at which a line along an angle hits the tangent line. The diagrams at right show tan θ for angles in the first and second quadrants.

Note: Tangent θ (or tan θ) may also be defined as the ratio . That is, tan θ = .

THINK WRITE

b Read the exact value from the table or use a special triangle,

tan θ = .

b tan = or

32

1

–6π

–3π

oppadj---------

π6--- 1

3------- 3

3-------

y

x

1stquadrant

4thquadrant

3rdquadrant

2ndquadrant

y

y

xx

P( )

O

–1

–1

1

1

1

θ

θ

Note: −1 ≤ sin θ ≤ 1, forall θ and −1 ≤ cos θ ≤ 1,for all θ.

y

x

tan

y

x

tan

The tangent line

θθ

θ

θ

yx-- sin θ

cos θ------------


C h a p t e r 5 G r a p h i n g p e r i o d i c f u n c t i o n s 207Therefore:

quadrant 1 is where 0° < θ < 90°quadrant 2 is where 90° < θ < 180°

quadrant 3 is where 180° < θ < 270°quadrant 4 is where 270° < θ < 360°

Since y = sin θ, sine is positive in quadrants 1 and 2 andsine is negative in quadrants 3 and 4.

Since x = cos θ, cosine is positive in quadrants 1 and 4 andcosine is negative in quadrants 2 and 3.

Since tan = , tangent is positive in quadrants 1 and 3and tangent is negative in quadrants 2and 4.

The diagram summarises which of the trigonometricfunctions is positive in each quadrant.

The word ‘CAST’ will assist in recalling this.You may also use a short phrase, such as ‘All Stations To Central’.

Note that for angles which are greater than 360° we can find the quadrant by makingmore than one revolution. The equivalent angle in the first revolution can be found byrepeatedly subtracting 360° until an angle between 0 and 360° is found. For example,an angle of 560° is in quadrant 3 (560° − 360° = 200°).

For angles that are negative, we move in the opposite (clockwise) direction. To find the equivalent positive angle, add 360° to the negative angle. For example, an angle of −210° will lie in quadrant 2 (−210° + 360° = 150°).

Before we can consider angles within quadrants 2, 3 and 4 we need to be able to calculate the trigonometric ratios for the angles on the boundary of each quadrant. Exact values of the boundary angles can be found using the following figure and the fact that sin θ = y,

cos θ = x and tan θ = .

yx--

y

x

Sinepositive

Allpositive

Tangentpositive

Cosinepositive

If a is any angle, find the quadrant where:a sin a > 0 and tan a < 0 b sin a < 0 and cos a > 0.

THINK WRITE

a Sine is positive and tangent is negative in quadrant 2.

a Quadrant 2

b Sine is negative and cosine is positive in quadrant 4.

b Quadrant 4

9WORKEDExample

0°/360°

270°

180°

90°

(–1,0)

(0,–1)

(0,1)

(0,1)

yx--



As with the exact values studied earlier, we should know these results in both degreesand radians. This figure can also be drawn in radians.

These results are summarised in the table below.

Angle(degrees)

Angle(radians) sin q cos q tan q

0° 0 0 1 0

90° 1 0 undefined

180° π 0 −1 0

270° −1 0 undefined

360° 2π 0 1 0

(–1,0)

(0,–1)

(0,1)

(0,1)0/2π

32—

π2—

π

π

π2---

3π2

------

Without using a calculator, find:a sin 90° b cos 180° c tan 270° d sin 360°.

THINK WRITE

a Sine corresponds to the y-coordinate on the unit circle.

a

The value of y at the point where θ = 90° is 1.

sin 90° = 1

b Cosine corresponds to the x-coordinate on the unit circle.

b

The value of x at the point where θ = 180° is −1.

cos 180° = −1

1

2

1

2

10WORKEDExample



THINK WRITE

c Tangent corresponds to the value of on the unit circle.

c

The value of y is −1 and the value of x is 0 at the point where θ = 270°.

tan 270° =

The value of is not defined. tan 270° is undefined

d Sine corresponds to the y-coordinate on the unit circle.

d

The value of y where θ = 360° is 0. sin 360° = 0

1yx--

21–

0------

31–

0------

1

2

rememberOn a unit circle:1. sin θ = y2. cos θ = x3. tan θ = height of point

where angle line meets tangent line at RHS of unit circle

4. tan θ = .

y

x

tan θ

θ

yx--

sin

cos

y

x

P( )

0, 360°, 2

, 180°

270°

90°

4th quadrant3rd quadrant

1st quadrant2nd quadrant

3 —2

–2

S

T

A

C

θθ

θ

θ

ππ

π

π

30°

60°

2

3

1π–3

π–6

45°

21

1π–4

Special triangles

Quadrants and angles

remember



Exact values

1 Without using a calculator, state the exact value of:

2 Without the use of a calculator, find the exact value of each of the following.

3 Without using a calculator, state whether the following values are positive (P) ornegative (N).

Verify your answers using a calculator.





6 If a is any angle, find the quadrant where:a sin a < 0 and tan a > 0 b sin a > 0 and cos a > 0.

7a If 0° < a < 360°, sin a < 0 and cos a < 0, then which one of the following is true?

b If 0° < a < 360°, sin a > 0 and tan a < 0, then which one of the following is true?

c Given that 0° < a < 360°, then the equation cos a = −1 has:

d If 0° < a < 360°, then the equation sin a = cos a has:

e If 0° ≤ a ≤ 180°, and sin a = cos a, then a is equal to:

8 Without using a calculator, find:

a sin 30° b cos 45° c tan 60° d cos 60° e sin 30°f tan 45° g cos 30° h tan 60°

a sin b cos c tan d sin e tan f cos g cos h tan

a sin 40° b sin 65° c sin 110° d sin 160° e sin 230°f sin 260° g sin 215° h sin 321° i sin 425° j sin (−36°)

a cos 27° b cos 68° c cos 115° d cos 200° e cos 250°f cos 295° g cos 402° h cos (−83°) i cos (−240°) j cos 157°

a tan 12° b tan 75° c tan 118° d tan 166° e tan 199°f tan 255° g tan 308° h tan 500° i tan (−45°) j tan (−137°)

A 0° < a < 90° B 90° < a < 180° C 180° < a < 270°D 270° < a < 360° E 0° < a < 180°

A 0° < a < 90° B cos a < 0 C 180° < a < 270°D cos a > 0 E 270° < a < 360°

A one solution B no solution C two solutionsD three solutions E four solutions

A no solutions B two solutions C three solutionsD one solution E four solutions

A 60° B 150° C 45° D 0° E 90°

a cos 180° b sin 270° c tan 360° d sin 180° e cos 270°f cos 360° g tan 270° h tan 180° i sin 630° j cos 720°.

5CWWORKEDORKEDEExample

7


8 π4--- π

3--- π

3--- π

6--- π

4--- π

6--- π

4--- π

6---

EXCEL

Spreadsheet

The unit circle

The unit circle

GCpro

gram

The unit circle


9


WorkS

HEET 5.1WWORKEDORKEDEExample

10



SymmetryThe unit circle can be divided into symmetrical sections, as shown in the diagram on the right.

Relationships between the circular functions — sine, cosine and tangent — can be established, based on these symmetrical properties.

For simplicity, assume θ is an acute angle,although the following properties hold for any θ.

Quadrant 1As already seen:

sin θ = ycos θ = x

tan θ =

Quadrant 2By symmetry:

sin (π − θ) = y = sin θcos (π − θ) = −x = −cos θ

tan (π − θ) = = −tan θ


sin (π + θ) = −y = −sin θcos (π + θ) = −x = −cos θ

tan (π + θ) = − = tan θ


sin (2π − θ) = −y = −sin θcos (2π − θ) = x = cos θ

tan (2π − θ) = = −tan θ

Notes1. These relationships also apply if degrees are used in

place of radians — that is, if π is replaced by 180° or 2π is replaced by 360°.

2. An angle measurement is assumed to be in radiansunless the degree symbol is given.

Provided an angle is expressed as π ± or 2π ± ,the trig function (sin or cos) remains the same, only the sign (+ or −) may change.Use this diagram to determine the sign.

–1

–1

1

10 or 2 c

c–2

cx

–y

y

–x

P(2 – )

P( ) = (cos , sin )= (x, y)

θ

θP( + )θ

P( – )θ θ θ θ

3 c —2π

π

π

π

π

π π

Mathcad

The unitcircle

yx---

EXCEL Spreadsheet

The unitcircle

yx–

------

Cabri Geometry

The unitcircle

y–x

------

Mathcad

Unitcircle

symmetry

y–x

------

θ θ

S A

T C



a If sin θ ° = 0.93, find sin (180 + θ)°. b If cos θ ° = 0.44, find cos (360 − θ)°.c If tan θ = 1.72, find tan (π − θ ). d If cos θ = 0.83, find cos (π + θ ).

THINK WRITE

a Sketch the angle on a unit circle and relate it to the first quadrant.

a

By symmetry sin (180 + θ)° = −sin θ. sin (180 + θ)° = −sin θ °Replace sin θ ° with 0.93. = −0.93

b Sketch the angle on a unit circle and relate it to the first quadrant.

b

By symmetry cos (360 − θ)° = cos θ. cos (360 − θ)° = cos θ °Replace cos θ ° with 0.44. = 0.44

c Sketch the angle on a unit circle and relate it to the first quadrant.

c

By symmetry tan (π − θ) = −tan θ. tan (π − θ) = −tan θReplace tan θ with 1.72. = −1.72

d Sketch the angle on a unit circle and relate it to the first quadrant.

d

By symmetry cos (π + θ) = −cos θ. cos (π + θ) = −cos θReplace cos θ with 0.83. cos (π + θ) = −0.83

1

sin (180 + )°180° sin

(180 + )°x

y

S

T

A

Cθ

θθ θ

θ

23

1 y

x

(360 – )°

cos

cos (360 – )°

S A

T Cθ

θθ

θ

θ

23

1 y

x

S A

T C

– tan

tan ( – )

θ

θ

θ

θ

θππ

π

23

1

cos

cos ( )

+x

y

S

T

A

Cπ θ

+π θ

θθθ

23

11WORKEDExample



12WORKEDExampleFind, without using a calculator, the exact value of each of the following.

a tan 150° b sin 330° c cos d tan

Continued over page

THINK WRITE

a Express tan 150° as tan (180 − 30)°. a tan 150° = tan (180 − 30)°Sketch the angle on a unit circle and relate it to the first quadrant.

By symmetry tan (180 − 30)° = −tan 30°. tan 150° = −tan 30°

Replace tan 30° with its exact value, or . = − or −

b Express sin 330° as sin (360 − 30)°. b sin 330° = sin (360 − 30)°Sketch the angle on a unit circle and relate it to the first quadrant.

By symmetry sin (360 − 30)° = −sin 30°. sin 330° = −sin 30°Replace sin 30° with its exact value, . = −

c Express cos as cos . c cos = cos

Sketch the angle on a unit circle and relate it to the first quadrant.

3p4

------7p6

------

12 y

x30°

tan 30°

tan (180 – 30)°

S A

CT

(180 – 30)°

3

41

3------- 3

3-------

1

3------- 3

3-------

12 y

x30°

sin (360 – 30)°

S A

CT

sin 30°

(360 – 30)°

3

412---

12---

13π4

------ π π4---–

3π4

------ π π4---–

2 y

x

S A

CT

cos –4

–4

cos ( )–4 –

–4

–

π

π

π

π

π

π



THINK WRITE

By symmetry cos = −cos . cos = −cos

Replace cos with its exact value of or

.

= − or −

d Express tan as tan . d tan = tan

Sketch the angle on a unit circle and relate it to the first quadrant.

By symmetry tan = tan . tan = tan

Replace tan with its exact value or . = or

3 π π4---–

π4--- 3π

4------ π

4---

4π4--- 1

2-------

22

-------

1

2------- 2

2-------

17π6

------ π π6---+

7π6

------ π π6---+

2 y

x

S A

CT

–6

–6 +

–6tan =

–6+tan ( )π

π

π

π

π

π

3 π π6---+

π6--- 7π

6------ π

6---

4π6---

1

3------- 3

3-------

1

3------- 3

3-------

If sin = 0.924, evaluate each of the following (without using a calculator).

a sin b sin .

THINK WRITE

a Express sin as sin = sin . a sin = sin

3p8

------

5p8

------11p

8---------

15π8

------ 8π8

------ – 3π8

------ π –

3π8

------ 5π

8------ π –

3π8

------

y

x

S A

CT

3 —8 – ππ 3 —

8π

13WORKEDExample



Symmetry

1 If sin θ° = 0.63, find:

2 If cos θ = 0.25, find:

3 If tan θ = 2.1, find:

a sin (180 − θ )° b sin (180 + θ )° c sin (−θ )° d sin (360 − θ )°.

a cos (π − θ ) b cos (π + θ ) c cos (2π − θ ) d cos (−θ ).

a tan (2π − θ ) b tan (−θ ) c tan (π + θ ) d tan (π −θ ).

THINK WRITE

Using symmetry express sin as sin . = sin

Replace sin with 0.924. = 0.924

b Express sin as sin = sin . b sin = sin

By symmetry express sin as −sin . = −sin

Replace sin with 0.924. = −0.924

2 π – 3π8

------ 3π

8------

3π8

------

33π8

------

111π

8--------- 8π

8------ 3π

8------+

π 3π8

------+ 11π

8--------- π 3π

8------+

y

x

S A

CT

3 —8 + ππ

3 —8π

2 π 3π8

------+ 3π

8------

3π8

------

33π8

------

remember1. A unit circle sketch of the given angle related back

to the first quadrant is often helpful.2. Provided an angle is expressed as π ± θ or 2π ± θ,

the trig function (sin or cos) remains the same, only the sign (+ or −) may change.

3. Use this diagram to determine the sign.

4. π may be written as , , , and so on.3π3

------ 4π4

------ 6π6

------ 8π8

------

remember

S A

T C

5D

EXCEL Spreadsheet

The unitcircle

Cabri Geometry

The unitcircle


11

MQ Maths B Yr 11 Ch 05 Page 215 Monday, October 29, 2001 6:55 AM


4 Given that sin a° = 0.3, cos b° = 0.7 and tan c° = 0.9, write down the value of each ofthe following.a sin (180 + a)° b cos (180 − b)° c tan (360 − c)°d sin (−a)° e sin (180 − a)° f cos (−b)°g cos (360 − b)° h tan (180 − c)° i tan (180 + c)°

5 Find, without using a calculator, the exact value of each of the following.

6 Find, without using a calculator, the exact value of each of the following.

7 If sin = 0.383, cos = 0.924 and tan = 0.414, evaluate each of the following

(without using a calculator).

8 Given that sin 75° = 0.966, cos 75° = 0.259 and tan 75° = 3.732, find the value of eachof the following (without using a calculator).

9 If sin 0.7 = 0.644, cos 0.7 = 0.765 and tan 0.7 = 0.842, find the value of each of thefollowing, without using a calculator. (Hint: π = 3.142, approximately.)

Trigonometric graphsThe trigonometric functions are the most basic of all periodic functions. Now that we can find the exact values of the most important angles we are able to graph the trigonometric functions.

Convention has it that trigonometric functions are usually graphed with radians on the x-axis. Because we are now working with the standard coordinate axes and the standard form of a function is y = f(x) we use x rather than θ as our variable.

a sin 150° b cos 135° c tan 240° d tan 330°e sin 240° f cos 210° g tan 120° h sin 300°i cos (−60)° j sin (−135)° k tan 180° l sin 270°

a cos b sin c tan d sin

e cos f tan g sin h cos

i tan j sin k cos π l sin

a sin b cos c tan

d cos e sin f tan

a sin 105° b cos 255° c tan 285°d sin 255° e cos 435° f tan (−75)°

a sin 2.442 b cos 3.842 c tan 5.584 d sin (−0.7)

Mathca

d

Unit circle symmetry


12a, b

Mathca

d

The unit circle


12c, d 5π4

------ 5π6

------ 7π4

------ 2π3

------

7π3

------ π6---–

4π3

------ 7π6

------

11π6

--------- 5π5

------– π

2---–


13

π8--- π

8--- π

8---

9π8

------ 7π8

------ 17π8

---------

π8---–

7π8

------ 15π8

---------

MQ Maths B Yr 11 Ch 05 Page 216 Friday, October 26, 2001 9:43 AM


The graph of y = sin xTo get an idea of what the graph of y = sin x looks like, we may first construct a tableof values.

Next, these values are plotted on a set of coordinate axes and a smooth curve isdrawn to join the points. The range (the set of possible y-values for the functions) of y= sin x is {y: −1 ≤ y ≤ 1}.

Verify the shape of this curve using a graphics calculator. It can be observed that thecurve repeats itself in cycles after an interval of 2π units. Due to this repetition it iscalled a periodic function and the period is the interval between repetitions. The periodof y = sin x is 2π radians (or 360°).

When dealing with graphs of periodic functions such as sin x, unless otherwise stated, we assume that the units for x are radians. This is because on the unit circle from which all trigonometric ratios are developed, the arc length corresponds to the angle at the centre of the circle in radians.

The mean position of the curve is y = 0 and the maximum and minimum values are 1 and −1 respectively. The distance from the mean position to the maximum (or minimum) position is called the amplitude of the periodic function. The amplitude of y = sin x is 1 unit.

The graph of y = cos xAs for the graph of y = sin x, the graph of y = cos x can be established by firstcompleting a table of values.

x −π − − − 0 π 2π

y = sin x 0 −0.71 −1 −0.71 0 0.71 1 0.71 0 −0.71 −1 −0.71 0

x −π − − − 0 π 2π

y = cos x −1 −0.71 0 0.71 1 0.71 0 −0.71 −1 −0.71 0 0.71 0

3π4

------ π2--- π

4--- π

4--- π

2--- 3π

4------ 5π

4------ 3π

2------ 7π

4------

y

x2

(2 , 0)

–4π–

4 π–

2 π –

2π 3 —

2π

( , –1)3 —2π

( , –0.71)7 —4π( , –0.71)5 —

4π( , –0.71)3 —

4 π

( , 0.71)3 —4π

( , 0.71) –4π

–2π( , 1)

π π–π

π( , 0)π

y = sin x

–1

1

1–2

1–2

0(0, 0)–π( , 0)

–2

π( , –1)–

– –

– –

1c

1 unity

x

Period

Period

Amplitude

Mean position

3π4

------ π2--- π

4--- π

4--- π

2--- 3π

4------ 5π

4------ 3π

2------ 7π

4------



When these points are plotted on a set of axes and joined with a smooth curve, the graph looks like this:

Verify the shape of this curve using a graphics calculator.

It can be seen that the basic shape of the curve y = cos x is the same as that of y = sin x. It is also a periodic function with:

period = 2πamplitude = 1.

Note: The graph of y = cos x is exactly the same as that of y = sin x translated radians or 90° to the left.

The graph of y = tan xConsider the unit circle with an angle of θ at the centre.

From this circle we can see that sin θ = y, cos θ = x and

tan θ = . From this we are able to conclude that tan θ = .

We can use this information to generate the table of values below, or we can use our knowledge of special angles.

(Note: Multiples of could be used to give more points to plot and a clearer indication

of the shape of the graph.)Note the presence of some undefined y-values. This is because tan x = and

cos x = 0 at these values.These undefined values are shown as vertical asymptotes through the given value of

x for which they occur. (An asymptote is a line that a graph approaches, but never quitereaches. In the case of y = tan x, y approaches (but never actually reaches) −∞ and +∞for particular x-values.)

The graph of y = tan x is shown in the following figure.

Verify this graph using a graphics calculator.

The features of the graph of y = tan x are:1. It has vertical asymptotes though

x = . . . − , , , . . .

2. It has no amplitude.3. It has a period of π.4. It has a range of R (the set of all Real

numbers).

x −π − − − 0 π 2π

y = tan x 0 1 undefined −1 0 1 undefined −1 0 1 undefined −1 0

( , –0.71)5 —4π

( , 0.71)7 —4π

( , –0.71)3 —4π

( , 0)3 —2π

( , –0.71)3 —4 π

( , 0.71) –4π( , 0.71) –

4 π

π –2π( , 0)–

2( , 0)

(2 , 1)π

( , –1)π

y = cos x

2π

y

x–4π–

4 π–

2 π –

2π 3 —

2ππ–π

–1

1

1–2

1–2

0

–π( , –1)

(0, 1)–

–

– –––

π2---

θ–1

–1

1

y

x 1

yx-- sin θ

cos θ------------

3π4

------ π2--- π

4--- π

4--- π

2--- 3π

4------ 5π

4------ 3π

2------ 7π

4------

π8---

sin xcos x------------

y

x

–1

1

0(0, 0)

π–π( , 0)–π

ππ

π–2

π3 —2

π–4

π–4

π–2

2π7 —4

( , –1)

π5 —4

( , 1)π –4( , 1)

π3 —4

( , –1)

π 3 —4

( , 1)

(2 , 0)π ( , 0)

y = tan x

Vertical asymptotes

–

––

π2--- π

2--- 3π

2------



Further trigonometric graphsFrom the basic sketch of y = sin x and y = cos x we need to be able to sketch a graphthat is in the form y = a sin bx.

From the above investigation we can see that the effect of a on the graph ofy = a sin bx is to multiply the amplitude of y = sin x by a. The period of the function isnot affected by the value of a; that is, the period of y = a sin x remains 2π.

The amplitude of any periodic function is taken to be positive. We can say that theamplitude of y = a sin bx is equal to .

The effect of b is to divide the period of y = sin x by b. The amplitude of the functionis not affected by the value of b; that is, the amplitude of y = sin bx remains 1.

We can therefore sketch the graph of y = a sin bx by using y = sin x as our basic

model, but with an amplitude of a and a period of .

The same results apply for graphs of the form y = a cos bx.Next year we will extend this further to look at graphs of the form y = a sin (bx + c).

The effect of 2The following investigation is best completed using a graphics calculator but can also be done on paper.

1 On the same set of axes sketch the graphs of y = sin x and y = 2 sin x.• Press .• For Y1 = press , and .

In the display you should seeY1 = sin(X).• For Y2 = press , , and .

In the display you should seeY2 = 2sin(X).• Press and enter the following

settings:Xmin = −6.28Xmax = 6.28Xscl = 1Ymin = −2Ymax = 2Yscl = 0.5Xres = 1.• Press . The calculator should then

display both graphs.

2 What has been the effect of the 2 on the graph of y = 2sin x?

3 Now, on the same axes draw the graphs of y = sin x and y = sin 2x.• The graphics calculator instructions are the same except for Y2. You will need

to press , , and .

4 What has been the effect of the 2 on the graph of y = sin 2x?CA

SIO

Effectof 2

Y=SIN X,T,q,n )

2 SIN X,T,q,n )

WINDOW

GRAPH

SIN 2 X,T,q,n )

a

2πb

------



Once we have determined both the period and amplitude of the periodic function, weare able to sketch the function using either y = sin x or y = cos x as a model.

The shape of each function remains unchanged with only the amplitude and periodbeing adjusted. The pattern of the periodicfunction must be continued over the domainfor which we are asked to sketch the function.

Note that if a is negative, the shape of thegraph is reflected in the x-axis.

The Maths Quest CD-ROM contains anumber of technology files which can be usedto show the graphs of sine and cosine func-tions. A sample screen shows a graph of asine function.

State i the period and ii the amplitude of each of the following functions.

a b y = −1.5 sin 4x c y = 6 cos

THINK WRITE

a From the graph the cycle repeats after 3π units. a ii Period = 3πFrom the graph, the distance from the mean position to the maximum position is 4 units.

ii Amplitude = 4

b State the function being considered. b y = −1.5 sin 4x

Write the formula for the period of y = a sin bx.

ii Period = where b = 4

Substitute b = 4. ii Period =

Simplify. ii Period =

By rule, the amplitude is | a | or | −1.5 |.

ii Amplitude = | −1.5 |ii Amplitude = 1.5

c State the function being considered. c y = 4 cos

Write the formula for the period. ii Period = where b =

Substitute b = . ii Period =

Simplify. ii Period = 6πBy rule, the amplitude is 6. ii Amplitude = 6

y

x3 —2π 3π

–4

4

0

x3---

12

1

22πb

------

32π4

------

4π2---

5

1x3---

22πb

------ 13---

313---

2π13---

------

4

5

14WORKEDExample

EXCEL

Spreadsheet

Sine graphs



Sketch the graphs of the following functions and state i the period and ii the amplitude of each.

a y = 4 cos 0 ≤ x ≤ 12π b y = − sin 4x 0 ≤ x ≤ 2π

THINK WRITE

a State the function being considered. a y = 4 cos

Write the formula for the period. ii Period =

Substitute b = . =

Simplify the value of the period. = 6πThe amplitude is the value in front of cos, written as a positive value.

ii Amplitude = 4

Draw a set of axes.Since the amplitude is 4, mark or imagine horizontal guidelines at y = −4 and 4.Sketch one cycle of the graph every period (every 6π) along the x-axis, for 0 < x < 12π, showing key x-values.

b State the function being considered. b y = − sin 4x

Write the formula for the period. ii Period =

Substitute b = 4. ii Period =

Simplify the value of the period. ii Period =

The amplitude is the value in front of sin, written as a positive value.

ii Amplitude =

Draw a set of axes.Since the amplitude is , mark or imagine horizontal guidelines at y = − and .Sketch one cycle of the graph every period(every ) along the x-axis, for 0 < x < 2π.

Determine the x-intercepts and mark these on the graph.

x3--- 2

3---

1x3---

22πb

------

313---

2π13---

------

4

5

6

7

8 y

x

–4

4

0 3 63 —2π 9 —

2π 15 —

2π 21 —

2ππ π 9π 12π

123---

22πb

------

32π4

------

4π2---

523---

6

723---

23--- 2

3---

y

x2–4π –

2π 3 —

2π5 —

4π3 —

4π 7 —

4ππ π0

2 —2π(= )

2–3

2–3–

8π2---

9

15WORKEDExample



The graph of y = a tan bx can be drawnusing a similar technique. The period

will be and although it has no

amplitude, the value of a will cause thegraph to increase more rapidly. Experi-ment with the graph of y = a tan bx ona graphics calculator, although at thisstage you do not need to sketch graphsof this type yourself.

There are a number of technologyfiles on the Maths Quest CD-ROM thatcan be used to verify your graph. TheMathcad screen shows the graph (blueline) for worked example 16. As acomparison, the red line indicates thegraph of y = cos x.

Sketch the graph of f(x) = 2 cos x in the domain -2 ≤ x ≤ 4.

Verify that this graph is correct by using a graphics calculator.

THINK WRITE

State the function being considered. f(x) = 2 cos x

The period of the function is ,

where b = .

Period =

Simplify the value of the period. = 4The amplitude is 2. Amplitude = 2Draw a set of axes.Using an interval of 4 (= period), mark the x-axis from −2 to 4 (the specified domain).Show −2 and 2 on the y-axis, since the amplitude is 2.

Visualise, or check, the general shape of the graph of y = a cos bx.Starting from the point (0, 2) complete one cycle of the cosine function forward to (4, 2).Complete half of a cycle back from (0, 2) to (−2, −2).The x-intercepts are halfway between the maximum and minimum points.

p2---

1π2---

22πb

------π2---

2ππ2---

------

345 y

x

–2

2

0 1–2 –1 2 3 4

y = 2 cos x — 2 π

6

7

8

9

10

11

16WORKEDExample

Mathca

d

Cosine graphs

πb---



Trigonometric graphs

1 State i the period and ii the amplitude for each of the following functions:a b c

d e f

g h

remember1. Basic graph types

(a) y = a sin bx y = a cos bx

(b) Period = , amplitude = | a |

2. Sketching (a) Recall basic graph type.(b) Find period and amplitude.(c) Sketch in sections of one period.(d) Find x-intercepts between other known intercepts or minimum and

maximum points.(e) If a < 0 the shape of the graph is reflected in the x-axis.

2πb

------

y

x

–a

a

0 2 —bπ3 —

2bπ —

2bπ –

bπ

y = a cos bx

y

x

–a

a

0 2 —bπ –

bπ

y = a sin bx

= period

remember

5EWWORKEDORKEDEExamplexample

14a

Cabri Geometry

Sine andcosinegraphs

GC program

Trigonometricgraphs

EXCEL Spreadsheet

Sinegraphs

EXCEL Spreadsheet

Cosinegraphs

Mathcad

Sinegraphs

Mathcad

Cosinegraphs

y

x4π2π

–2

2

0

y

x

–1

1

0–4π 5 —

4π

y

x3π

–1.5

1.5

0

y

x

–4

4

0 3 —2π

y

x

–2

2

0 π–π

y

x

–3

3

0 π–3

π–

y

x

–2.5

2.5

0–2

π π–

y

x

–0.5

0.5

0 –6π –

2π


224

M a t h s Q u e s t M a t h s B Ye a r 1 1 f o r Q u e e n s l a n d

2

State

i

the period and

ii

the amplitude of each of the following functions.

3

Sketch the graph of the following functions for 0

≤

x

≤

2

π

and state

i

the period and

ii

the amplitude of each.

Check your answers using a graphics calculator.

4

Parts

a

to

c

refer to the graph.

a

The amplitude of the function is:

b

The period of the function is:

c

The equation of the function could be:

a

y

=

sin

x

b

y

=

3 sin

x

c

y

=

2 sin 2

x

d

y

= −

4 sin 3

x

e

y

=

sin

f

y

=

2 cos 4

x

g

y

=

0.4 cos

h

y

= −

3 cos 5

x

i

y

=

2.5 cos

j

y

= −

sin 6

x

k

y

=

sin

π

x

l

y

=

a

y

=

sin 2

x

b

y

=

2 cos

x

c

y

=

3 sin

d

y

=

4 cos 2

x

e

y

=

sin 3

x

f

y

=

cos

g

y

=

5 sin

h

y

= −

4 cos

i

y

=

2 cos 4

x

j

y

= −

3 sin 2

x

A B

−

3

C

π

D

3

E

6

A

2

π

B C

π

D

6

E

3

A

y

=

3 sin 2

x

B

y

=

3 sin

x

C

y

=

3 cos 2

x

D

y

= −

3 sin 2

x

E

y

=

3 sin

WWORKEDORKEDEExamplexample

14b, c

12---

x3---

x3--- πx

4------

15---

cos πx2

------

4----------------

WWORKEDORKEDEExamplexample

15

x2---

12--- 2

3---

x2---

x3--- x

2---


y

x

–3

3

0 –2π

π2---

π2---

x2---

MQ Maths B Yr 11 Ch 05 Page 224 Monday, June 24, 2002 12:24 PM


The equation of this curve could be:

A y = 2 cos x B y = 2 sin C y = −2 cos x

D y = cos E y = −2 cos

6 State the equation of each of the functions graphed below.

7 Sketch the graph of each of the following functions.

Check your graphs using a graphics calculator.

a b

c d

e f

a f(x) = cos 2x for −2π ≤ x ≤ 2π b f(x) = 3 cos for −4π ≤ x ≤ 4π

c f(x) = 2 sin 3x for 0 ≤ x ≤ 2π d f(x) = −3 cos 2x for 0 ≤ x ≤ 2π

e f(x) = 1.5 sin for −6π ≤ x ≤ 6π f f(x) = −4 sin 4x for 0 ≤ x ≤ π

mmultiple choiceultiple choicey

x

–2

2

0 π π3π2

x2---

x2--- x

2---

y

x

–1.5

1.5

0 π3

y

x

–2

2

0 π3—4

π–4

π–4–

y

x

–5

5

0 π4π2π–2

y

x

–4

4

0 π3—2

y

x

–1

1

0 π2—3

π4—3

y

x

–3

3

0 π–6

π–6

–

x2---

x3---



8 Sketch the graphs of each of the following functions.Check your graphs using a graphics calculator.

9 For each of the functions graphed below, state the rule using full function notation.

10 Sketch the graph of y = tan x for the domain 0 ≤ x ≤ 2π.

a f(x) = sin for −π ≤ x ≤ 2π b f(x) = 1.8 cos for −6 ≤ x ≤ 6

c f(x) = −1.4 sin for 0 ≤ x ≤ 6 d f(x) = −3 cos for 0 ≤ x ≤ 8

a b c

d e f

How high?We are now ready to try and draw the graph of Andrew’s height above ground level as he rides the ferris wheel.1 What is the maximum and minimum height of the ferris wheel?2 Draw a set of axes allowing us to follow the path of the ferris wheel for 3

revolutions (6 minutes) on the x-axis and allowing for the minimum and maximum height on the y-axis.

3 Assuming that the ferris wheel revolves at a constant rate, complete the table of values below, then plot the points generated on your axes.

4 If the wheel is revolving at a constant rate, what will be the height of the wheel after minute. Plot the point generated on your graph. Similarly, plot the points generated after 1 minutes, 2 minutes, 3 minutes, 4 minutes and 5 minutes.


1632---

2x3

------ πx3

------

πx2

------ 3πx4

---------

WorkS

HEET 5.2

y

x

–3

3

80 π—6

f(x)

y

x50 π

f(x)5–2

5–2

–

y

x

–2

–1 1

2

0f(x)

y

x

–1.8

–1 3

1.8

0f(x)

y

x

–3

3

3

0f(x)

y

x

–2.4

1

2.4

0

f (x)

1–3–

Time (min) 0 1 2 3 4 5 6

Height (m)

12---

12--- 1

2--- 1

2--- 1

2--- 1

2---



ApplicationsMany situations arise in science and nature where relationships between two variablesexhibit periodic behaviour. Tide heights, sound waves, biorhythms and ovulation cyclesare examples.

In these situations trigonometric functions can be used to model the behaviour of thevariables. The independent variable, x, is often a measurement such as time. When mod-elling with trigonometric functions you should work in radians unless otherwise instructed.

5 We now need to consider some of the intermediate points. After 15 seconds the

wheel will have passed through an angle of radians. The height of the ferris

wheel can be found by finding h in the figure below. (Note : h = 10 − x.) Plot this point on your axes.

6 Draw a diagram and use the same method to find the height of the wheel after 10 seconds and after 20 seconds. Plot the points generated on your graph.

7 You should now be able to plot the height of the wheel after 40, 45, 50, 70, 75, 80, 100, 105 and 110 seconds to complete the first revolution.

8 Join the points plotted with a smooth curve and repeat the function for the second and third revolutions of the ferris wheel.

9 Use your knowledge of trigonometric graphs to state the function of height against time.

π4---

10 – x

10 mx

h

A river flows through Tony’s property. Tony measures the water level by measuring the distance above or below a marker in the water. Tony lets d be the level of water above or below the marker and t is the number of hours since midnight. Tony’s results are shown in the graph at right.Express d as a function of t.

Continued over page

THINK WRITE

The graph is in the basic shape of a sine graph.

The graph is of the form y = a sin bx.

The amplitude of the graph is 2. Amp = 2: y = 2 sin bx

t

1

–1

–2

16141210840

2

3

4

2 6

5d

1

2

17WORKEDExample



Having found the equation, we are able to then find other information about theparticular function.

THINK WRITE

The period of the function is 4. Put this equal to the period and solve for b.

= 4

Write the equation. 2π = 4b

b =

=

y = 2 sin x

32πb

------

4

2π4

------

π2---

π2---

E. coli is a type of bacterium. Its concentration, P parts per million (ppm), at a particular beach over a 12-hour period t hours after 6 am, is described by the function:

P = 0.05 sin + 0.1.

a Find the i maximum and ii minimum E. coli levels at this beach.

b What is the level at 3 pm?

THINK WRITE

a Write the function. a P = 0.05 sin + 0.1

ii The maximum value of the sine function is 1.

ii The maximum P occurs when sin = 1.

Substitute sin = 1 into the

equation for P and evaluate.

Max. P = 0.05(1) + 0.1 = 0.15

State the solution. The maximum E. coli level is 0.15 ppm.ii The minimum value of the sine

function is −1.ii The minimum P occurs when

sin = −1.

Substitute sin = −1 into the

equation for P and evaluate.

Min. P = 0.05(−1) + 0.1Min. P = 0.05

State the solution. The minimum E. coli level is 0.05 ppm.

pt12------

πt12------

1πt12------

2πt12------

3

1πt12------

2πt12------

3

WORKEDExample 18



Applications

1 The weight of a rabbit over a period oftime is modelled by the graph.a State i the amplitude and ii the period.b Express W as a function of t.

2 The diagram shows the heart rate of anathlete during a particular hour of aworkout.a Find the initial heart rate.b State i the amplitude and ii the period.c Express H as a function of t.

THINK WRITE

b At 3 pm it is 9 hours since 6 am. b At 3 pm, t = 9.

Substitute t = 9 into the equation for P, and evaluate.

When t = 9, P = 0.05 sin + 0.1

= 0.05 sin + 0.1

= 0.05 + 0.1

= 0.035 + 0.1= 0.135

State the solution. The E. coli level at 3 pm is approximately 0.135 ppm.

1

29π12------

3π4

------

1

2-------

3

remember1. General equations: y = a sin bx + c, y = a cos bx + c

2. Period = , amplitude = | a |.

3. To find maximum value of a function, replace sin x or cos x with +1.4. To find minimum value of a function, replace sin x or cos x with −1.5. Initial values occur at t = 0.6. Sketch the graph for greater understanding.

2πb

------

remember

5F

EXCEL Spreadsheet

Trigonometricequations

Mathcad

Solvingcosine

equations


17

Mathcad

Solvingsine

equations

SkillSH

EET 5.3

W

t

2

1

4

3

03 6 (days)

(kg)

60

160

110

015 30 45 60 t (min)

H (beats/min)



3 The height above the ground, h metres,of a child on a swing at any time t seconds after being released is:

h = 1 + 0.6 cos .Find:a the maximum height of the swingb the height after

i 3 seconds and ii seconds.

4 The temperature, T (°C), inside a building on a given day is given by the function:

T = 8 sin

where t is the number of hours after 8 am.a What is the maximum temperature in the building and the time at which it first occurs?b Find the temperature at i 8 pm, ii 6 pm, iii 12 am (midnight).

5 The displacement, x (in mm), of a harp string t seconds after it is initially plucked ismodelled by the function:

x(t) = 12 sin 20πt.a What is the i amplitude and

ii period of this function?b How many vibrations (that is, cycles) will it complete in one second?c Find the displacement after 0.08 seconds.

6 The height of a bungee jumper, h metres, above a pool of water at any time t secondsafter jumping is described by the function:

h(t) = 20 cos 0.8t + 20.a What is the initial height of the bungee jumper?b When, if at all, does the bungee jumper first touch the water?c Assuming the cord is perfectly elastic, how long it is until the bungee jumper

returns to the lowest position?


18

πt2-----

43---

πt12------ 18+


C h a p t e r 5 G r a p h i n g p e r i o d i c f u n c t i o n s 2317 A cyclist rides one lap of a circular

track at a constant speed so that her distance, d metres, from her starting point at any time, t seconds, after starting is:

d = 50 − 50 cos

What is:a the time taken to complete one

lap?b the radius of the track?c the maximum distance from the

start?d the length of the track?e the distance from her starting

point after: ii 15 seconds and ii 40 seconds?

8 The depth of water, d metres, at a port entrance is given by the function

d(t) = 4.5 + 1.5 sin

where t is in hours.a Find i the maximum and ii the minimum depth at the port entrance.b A certain ship needs the depth at the port entrance to be more than 5 metres. The

ship can be loaded and unloaded, and in and out of the port, in 9 hours. Assumingthat the ship enters the port just as the depth at the entrance passes 5 metres, will theship be able to exit 9 hours later? How long will it have to spare, or by how manyminutes will it miss out?

πt30------

πt12------



Definition of a periodic function• A periodic function is one that repeats itself over a period of time.• The period of the function is the amount of time taken for the function to repeat

itself.• The amplitude of the function is the maximum displacement of the function from

its equilibrium.

Radian measure• An angular measure of 1 radian will cut off an arc of one unit on the unit circle.• πc = 180°.

• To convert an angle in degrees to radians we multiply by .

• To convert an angle in radians to degrees we multiply by .

Exact values• For exact values — use special triangles:

SymmetryA unit circle sketch of the given angle related back to the first quadrant is often helpful. Provided an angle is expressed as π ± θ or 2π ± θ, the trig function (sin or cos) remains the same, only the sign (+ or −) may change. So π may be written as

, , , etc.

Angle (θ ) sin θ cos θ tan θ0 0 1 0

or 30° =

or 45° = = 1

or 60°

or 90° 1 0 undefined

summary

πc

180°-----------

180°πc

-----------

45°

2 3

1

1–4

60°

30°

2

1–6

–3

π π

π

π6--- 1

2--- 3

2------- 1

3------- 3

3-------

π4--- 1

2------- 2

2------- 1

2------- 2

2-------

π3--- 3

2------- 1

2--- 3

π2---

3π3

------ 4π4

------ 6π6

------ 8π8

------



Trigonometric graphs• Basic graph types:

1. y = a sin bx y = a cos bx

2. Period = , amplitude = | a |

• Sketching:1. Recall basic graph type.2. Find period and amplitude.3. Sketch in sections of one period.4. Find x-intercepts between other known intercepts or minimum and maximum

points.

• Tangent graphs:1. y = a tan bx

2. Period = , no amplitude

3. Asymptotes at x = ± , and every period to the left and right of these.

4. The formula for asymptotes applies:

xasymptote = ± , where n = 0, 1, 2, . . .

Applications of sine and cosine functions• To find the maximum value of a function, replace sin x or cos x with +1.• To find the minimum value of a function, replace sin x or cos x with −1.• Initial values occur at t = 0.• A sketch graph may provide greater understanding.

• sin (π − θ) = sin θ• cos (π − θ) = −cos θ• tan (π − θ) = −tan θ

• sin (π + θ) = −sin θ• cos (π + θ) = −cos θ• tan (π + θ) = tan θ

• sin (2π − θ) = −sin θ• cos (2π − θ) = cos θ• tan (2π − θ) = −tan θ

2πb

------

y

x

–a

a

0 2 —bπ3 —

2bπ —

2bπ –

bπ

y = a cos bx

y

x

–a

a

0 2 —bπ –

bπ

y = a sin bx

= period

πb---

π2b------

2n 1+( )π2b

------------------------

y

x0 π3 —2b

π 3 —2b

π–b

π–b

π—2b

π—2b– – –



1 State the period and the amplitude of the function drawn below.

2 A periodic function has a period of 6 units. Find the frequency of the function.

3 Describe the features of a periodic function.

4 Convert the following angles to radians.a 45° b 150° c 50° d 260°

5 Convert the following angles to degrees.

a b 1.2πc c d 3c

6

In which quadrants is tan x positive?A 1 and 2 B 1 and 4 C 2 and 4 D 2 and 3 E 1 and 3

7

The value of cos 320° is:A positive, as 320° is in the 1st quadrant B negative, as 320° is in the 2nd quadrantC negative, as 320° is in the 3rd quadrant D negative, as 320° is in the 4th quadrantE positive, as 320° is in the 4th quadrant.

8

If tan a° < 0, sin a° < 0 and 0° < a° < 360°, then which one of the following is correct?

9

If tan θ = , then sin θ could be equal to:

10

The angle 150° is equivalent to:

A 0° < a° < 90° B 180° < a° < 270° C 90° < a° < 180°D 0° < a° < 180° E 270° < a° < 360°

A B C −1 D E 0

A B C D E

CHAPTERreview

5A

x

y

–1

1

–2

765420

2

1 3

5A5A5B

5Bπc

6----- 5πc

3--------


5C

5C mmultiple choiceultiple choice


5C


5C 312--- 3

2-------

1

2-------


5C 2π3

------ 7π6

------ 5π9

------ 5π6

------ 4π3

------



The angle which is equivalent to is:

12

The value of cos is:

13 Without using a calculator, state the exact value of:a sin 30° b cos 60° c tan 45°.

14 Without the use of a calculator, state the exact value of:

a b c .

15 If cos θ = 0.69 find:a sin (π − θ) b cos (2π − θ) c tan (π + θ).

Questions 16 to 18 refer to the function: f(x) = −2 sin 3x

16

The amplitude of f(x) is equal to:

17

The period of f(x) is equal to:

18 The range of f(x) is:

19

The rule for this graph is:

A 405° B 315° C 540° D 270° E 300°

A B C − D E −

A 3 B −2 C −3 D 2 E 1

A 6π B 4π C D 2π E 3

A [0, 2] B [0, 4] C [−2, 0] D [−1, 1] E [−2, 2]

A y = 3 cos B y = 3 sin 2x

C y = 2 cos D y = 3 cos 2x

E y = −3 cos 2x

5Cmmultiple choiceultiple choice

9π4

------

5Cmmultiple choiceultiple choice4π3

------3

2-------

12--- 3

2-------

1

2------- 1

2---

5C

5Csin

π4--- cos

π6--- tan

π3---

5D


5E


5E2π3

------

5E

5Emmultiple choiceultiple choice

y

x

–3

3

0 2 3 4ππ π π

x2---

x3---



20 Sketch the graphs of the following functions.

a y = −4 sin for the domain −2π ≤ x ≤ 4π

b y = 1.5 cos 2x for the domain −π ≤ x ≤ π

21 The number of rabbits in a national park is observed for one year. At any time, t months, after observation begins, the number of rabbits is modelled by the function

P = 2 − 0.8 sin where P is in

thousands.a Find:

i the minimum number of rabbits

ii the maximum number of rabbits

iii the equilibrium number of rabbits.

b Find:i the period of the functionii the amplitude of the functioniii the frequency of the function.

c Sketch the graph of the function.d Find the population after 5 months.e For how long is the population below 1600?f For how long is the population above 2100?

5E x2---

5F

πt6-----

testtest

CHAPTERyyourselfourself

testyyourselfourself

5


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Graphing periodic functions - mathsbooks.netmathsbooks.net/Maths Quest 11B for Queensland/Ch 05... · Chapter 5 Graphing periodic functions 195 The length of time that it takes for