Circular Functions · 2008-12-10 · Circular Functions Objectives To use radians and degrees for...

P1: FXS/ABE P2: FXS

9780521740524c16.xml CUAU021-EVANS August 23, 2008 13:21

C H A P T E R

16Circular Functions

ObjectivesTo use radians and degrees for the measurement of angle.

To convert radians to degrees and vice versa.

To define the circular functions sine, cosine and tangent.

To explore the symmetry properties of circular functions.

To find standard exact values of circular functions.

To understand and sketch the graphs of circular functions.

16.1 Measuring angles in degrees and radiansThe diagram shows a unit circle, i.e. a circle of radius 1 unit.

–1

–1

1

1

0

y

xAC

D

B

The circumference of the unit circle = 2� × 1

= 2� units

∴ the distance in an anticlockwise direction around the

circle from

A to B = �

2units

A to C = � units

A to D = 3�

2units

Definition of a radianIn moving around the circle a distance of 1 unit from A to P,

the angle POA is defined. The measure of this angle is 1 radian.

–1

–1

11c

1

0

y

xA

P

1 unitOne radian (written 1c) is the angle subtended at the

centre of the unit circle by an arc of length 1 unit.

445Cambridge University Press • Uncorrected Sample Pages • 2008 © Evans, Lipson, Wallace TI-Nspire & Casio ClassPad material prepared in collaboration with Jan Honnens & David Hibbard

SAMPLE

P1: FXS/ABE P2: FXS


446 Essential Mathematical Methods 1 & 2 CAS

Note: Angles formed by moving anticlockwise around the circumference of the unit

circle are defined as positive. Those formed by moving in a clockwise direction are said to be

negative.

Degrees and radiansThe angle, in radians, swept out in one revolution of a circle is 2�c.

∴ 2�c = 360◦

∴ �c = 180◦

∴ 1c = 180◦

�or 1◦ = �c

180

Example 1

Convert 30◦ to radians.

Solution

1◦ = �c

180

∴ 30◦ = 30 × �

180= �c

6

Example 2

Convert�c

4to degrees.

Solution

1c = 180◦

�

∴ �c

4= � × 180

4 × �= 45◦

Note: Often the symbol for radian, c, is omitted. For example, angle 45◦ is written as�

4rather

than�c

4.

Cambridge University Press • Uncorrected Sample Pages • 2008 © Evans, Lipson, Wallace TI-Nspire & Casio ClassPad material prepared in collaboration with Jan Honnens & David Hibbard

SAMPLE

P1: FXS/ABE P2: FXS


Chapter 16 — Circular Functions 447

Using the TI-NspireTo change 32 degrees to radians, type

32◦�Rad as shown.

The degree symbol ◦ is found in the

catalog (k 4). The �Rad command

can be found in the catalog (k 1 ).

To change 2 radians to degrees, type

2r�DD as shown.

The radian symbol r is found in the

catalog (k 4). The �DD command can

be found in the catalog (k 1 ).

Note: If the calculator is already in radian mode you can change 32◦ to radians by

simply typing 32◦ followed by enter. Likewise, if the calculator is already in degree

mode, simply type 2r followed by enter to obtain the result in degrees.

Using the Casio ClassPadWhen using the CAS calculator to change 32o to

radians, ensure your calculator is in Radian mode

and enter 32o, then EXE. (The degree symbol is

found in the mth—TRIG menu on the keyboard.)

The answer can be displayed in exact notation, as

shown, or highlight the solution and click to

convert to decimal.

To change 0.5 radians to degrees click

Interactive—Transformation—to and enter 0.5.

Exercise 16A

1 Express the following angles in radian measure in terms of �:Example 1

a 60◦ b 144◦ c 240◦

d 330◦ e 420◦ f 480◦


SAMPLE

P1: FXS/ABE P2: FXS



2 Express, in degrees, the angles with the following radian measures:Example 2

a2�

3b

5�

6c

7�

6d 0.9�

e5�

9f

9�

5g

11�

9h 1.8�

3 Use a calculator to convert the following angles from radians to degrees:

a 0.6 b 1.89 c 2.9 d 4.31

e 3.72 f 5.18 g 4.73 h 6.00

4 Use a calculator to express the following in radian measure:

a 38◦ b 73◦ c 107◦ d 161◦

e 84.1◦ f 228◦ g 136.4◦ h 329◦

5 Express, in degrees, the angle with the following radian measure:

a −�

3b −4� c −3� d −�

e5�

3f

−11�

6g

23�

6h

−23�

6

6 Express each of the following in radian measure, in terms of �:

a −360◦ b −540◦ c −240◦

d −720◦ e −330◦ f −210◦

16.2 Defining circular functions: sine and cosineConsider the unit circle.

The position of point P on the circle can be described by

relating the cartesian coordinates x and y and the angle �.

The point P on the circumference corresponding to an

angle � is written P(�). θc

–1

–1

1

1

1

0

y

y

xx

P(θ)

Many different angles will give the same point P on

the circle, so the relation linking an angle to the

coordinates is a many-to-one function. There are,

in fact, two functions involved and they are called

sine and cosine, and they are defined as follows:

The x-coordinate of P, x = cosine �, � ∈ R

The y-coordinate of P, y = sine �, � ∈ R

Note: These functions are usually written in an abbreviated

form as follows:

x = cos �

y = sin �

–1

–1 1

1

0

y

x

P(θ) = (cosθ, sinθ)

sinθ

cosθ

θNote: cos (2� + �) = cos � and sin (2� + �) = sin �,

as adding 2� results in a return to the same point

on the unit circle.


SAMPLE

P1: FXS/ABE P2: FXS



Example 3

Evaluate sin � and cos �.

Solution

In moving through an angle of �, the position is P(�), which is (−1, 0)

∴ cos � = −1

sin � = 0

Example 4

Evaluate sin

(−3�

2

)and cos

(−�

2

).

Solution

sin

(−3�

2

)= 1

cos(−�

2

)= 0

Example 5

Evaluate sin

(5�

2

)and sin

(7�

2

).

Solution

sin

(5�

2

)= sin

(2

1

2�

)= sin

(2� + �

2

)= sin

(�

2

)= 1

sin

(7�

2

)= sin

(3

1

2�

)= sin

(2� + 3�

2

)= sin

(3�

2

)= −1

Example 6

Evaluate sin

(9�

2

)and cos (27�).

Solution

sin

(9�

2

)= sin

(4� + �

2

)= sin

(�

2

)= 1

cos (27�) = cos (26� + �) = cos � = −1


SAMPLE

P1: FXS/ABE P2: FXS



Exercise 16B

1 For each of the following angles, t, determine the values of sin t and cos t:Examples 5, 6

a t = 0 b t = 3�

2c t = −3�

2d t = 5�

2

e t = −3� f t = 9�

2g t = 7�

2h t = 4�

2 Evaluate using your calculator (Check that your calculator is in Rad mode):

a sin 1.9 b sin 2.3 c sin 4.1 d cos 0.3

e cos 2.1 f cos (−1.6) g sin (−2.1) h sin (−3.8)

3 For each of the following angles, �, determine the values of sin � and cos �:Example 6

a � = 27� b � = −5�

2c � = 27�

2d � = −9�

2

e � = 11�

2f � = 57� g � = 211� h � = −53�

16.3 Another circular function: tangentAgain consider the unit circle.

If a tangent to the unit circle at A is drawn then the

y-coordinate of C, the point of intersection of the

extension of OP and the tangent, is called

tangent � (abbreviated to tan �).

By considering the similar triangles OPD and OCA:

tan �

1= sin �

cos �

∴ tan � = sin �

cos �–1

–1

1

1

0

y

B

x

P(θ)

sin θ

tanθ

cos θ

AD

C (1, y)

θ

Now when cos � = 0, tan � is undefined.

Hence tan � is undefined when � = ±�

2, ±3�

2, ±5�

2, . . .

∴ Domain of tan = R \ {�: cos � = 0}

Example 7

Evaluate, using a calculator:

a tan 1.3 b tan 1.9 c tan (−2.8) d tan 59◦ e tan 138◦

Solution

a tan 1.3 = 3.6 (Don’t forget calculator must be in RAD mode)

b tan 1.9 = −2.93 (cos 1.9 is negative)

c tan (−2.8) = 0.36 (cos −2.8 and sin −2.8 both negative ∴ tan is positive)

d tan 59◦ = 1.66 (Calculate in DEG mode)

e tan 138◦ = −0.9


SAMPLE

P1: FXS/ABE P2: FXS



Exercise 16C

1 Evaluate:

a tan � b tan (−�) c tan

(7�

2

)

d tan (−2�) e tan

(5�

2

)f tan

(−�

2

)

2 Use a calculator to find correct to 2 decimal places:Example 7

a tan 1.6 b tan (−1.2) c tan 136◦ d tan (−54◦)

e tan 3.9 f tan (−2.5) g tan 239◦

3 For each of the following values of � find tan �:

a � = 180◦ b � = 360◦ c � = 0

d � = −180◦ e � = −540◦ f � = 720◦

16.4 Reviewing trigonometric ratiosFor right-angled triangles:

Hypotenuse, H

Adjacent side, A

Opposite side, O

θ

sin � = O

H

cos � = A

H

tan � = O

A

Applying these trigonometric ratios to the right-angled

triangle, OAB, in the unit circle:

–1

–1 1

1

1

0

y

y

xx

B

A

θ

sin � = O

H= y

1= y

cos � = A

H= x

1= x

tan � = O

A= y

x= sin �

cos �

For 0 < � <�

2, the functions sin, cos and tan are defined

by the trigonometric ratios and are the same as the respective

circular functions introduced earlier.


SAMPLE

P1: FXS/ABE P2: FXS



Exercise 16D

1 Find the value of the pronumeral for each of the following:

a

8

3θ

b

5

25°x

c

6

25°

x

d

10

50°x

e

6

5

θf

10

20°x

g

65°x

5

h

70°

x 7

i

40°x

5

2 a Use your calculator to find a and b correct to 4 decimal places.

b Hence find the values of c and d.

c i Use your calculator to find cos 140◦ and

sin 140◦.

ii Write cos 140◦ in terms of cos 40◦.

40°140°

x

y

1(c, d) (a, b)

16.5 Symmetry properties of circular functionsThe coordinate axes divide the unit circle into four

quadrants. The quadrants can be numbered,

anticlockwise from the positive direction of the

x-axis, as shown.x

y

Quadrant 2

Quadrant 3 Quadrant 4

Quadrant 1

0Relationships, based on symmetry, between

circular functions for angles in different quadrants

can be determined.


SAMPLE

P1: FXS/ABE P2: FXS




By symmetry:

sin(π – θ) = b = sinθcos(π – θ) = –a = – cosθtan(π – θ) =

–a = – tanθb


sin(2π – θ) = –b = –sinθcos(2π – θ) = a = cosθ

sin(π + θ) = –b = –sinθcos(π + θ) = –a = –cosθ

tan(π + θ) = –a = tanθ–btan(2π – θ) = – a = –tanθb

P(π – θ)

P(2π – θ)

P(θ) = (cosθ, sinθ) = (a, b)

P(π + θ)

(0, b)

(0, –b)

(–a, 0)

(a, 0)

θ

0

a

Note: These relationships are true for all values of �.

Signs of circular functionsThese symmetry properties can be summarised for the

signs of sin, cos and tan for the four quadrants as follows:

1st quadrant: All are positive (A)

2nd quadrant: Sin is positive (S)

3rd quadrant: Tan is positive (T)

4th quadrant: Cos is positive (C).

S A

T C

y

x

Negative of anglesBy symmetry:

cos (−�) = cos �

sin (−�) = −sin �

tan (−�) = −sin �

cos �= −tan �

–1

–1

10

1

P(–θ)

P(θ)

x

y

–θθ

Example 8

If sin x = 0.6, find the value of:

a sin (� − x) b sin (� + x) c sin (2� − x) d sin (−x)


SAMPLE

P1: FXS/ABE P2: FXS



Solution

a sin (� − x)

= sin x

= 0.6

b sin (� + x)

= −sin x

= −0.6

c sin (2� − x)

= −sin x

= −0.6

d sin (−x)

= −sin x

= −0.6

Example 9

If cos x◦ = 0.8, find the value of:

a cos (180 − x)◦ b cos (180 + x)◦ c cos (360 − x)◦ d cos (−x)◦

Solution

a cos (180 − x)◦

= −cos x◦

= −0.8

b cos (180 + x)◦

= −cos x◦

= −0.8

c cos (360 − x)◦

= cos (x)◦

= 0.8

d cos (−x)◦

= cos x◦

= 0.8

Exercise 16E

1 If sin � = 0.42, cos x = 0.7 and tan � = 0.38, write down the values of:Example 8

a sin (� + �) b cos (� − x) c sin (2�− �) d tan (�− �)

e sin (�− �) f tan (2�− �) g cos (� + x) h cos (2� − x).

2 a If sin x◦ = sin 60◦ and 90◦ < x◦ < 180◦, find the value of x.

b If sin x◦ = −sin 60◦ and 180◦ < x◦ < 270◦, find the value of x.

c If sin x◦ = −sin 60◦ and −90◦ < x◦ < 0◦, find the value of x.

d If cos x◦ = −cos 60◦ and 90◦ < x◦ < 180◦, find the value of x.

e If cos x◦ = −cos 60◦ and 180◦ < x◦ < 270◦, find the value of x.

f If cos x◦ = cos 60◦ and 270◦ < x◦ < 360◦, find the value of x.

3 a If cos x = −cos(�

6

)and

�

2< x < �, find the value of x.

b If cos x = −cos(�

6

)and � < x <

3�

2, find the value of x.

c If cos x = cos(�

6

)and

3�

2< x < 2�, find the value of x.

4 Write down the values of:

a a = cos (� − �)

b b = sin (� − �)

c c = cos (−�)

d d = sin (−�)

e tan (� − �)

f tan (−�)

y

x

(c, d )

(a, b)

10

π – θ

θθ–

√32

12

,


SAMPLE

P1: FXS/ABE P2: FXS



5 Write down the values of:

a d = sin (� + �)

b c = cos (� + �)

c tan (� + �)

d sin (2� − �)

e cos (2� − �)

y

x

(c, d )

–

π + θ

1

θ√32

12

,

6 If sin x◦ = 0.7, cos �◦ = 0.6 and tan �◦ = 0.4, write down the values of:Example 9

a sin (180 + x)◦ b cos (180 + �)◦ c tan (360 − �)◦ d cos (180 − �)◦

e sin (360 − x)◦ f sin (−x)◦ g tan (360 + �)◦ h cos (−�)◦

16.6 Exact values of circular functionsA calculator can be used to find the values of the circular functions for different values of �.

For many values of � the calculator gives an approximation. We consider some values of �

such that sin, cos and tan can be calculated exactly.

Exact values for 0(0◦) and�

2(90◦)

From the unit circle:

When � = 0, sin � = 0

cos � = 1

tan � = 0

When � = �

2, sin

(�

2

)= 1

cos(�

2

)= 0

tan(�

2

)is undefined.

–1

–1

0θ

1

1

cosθsinθ

x

y

Exact values for�

6(30◦) and

�

3(60◦)

Consider an equilateral triangle ABC of side length 2 units.

In �ACD, by Pythagoras’ Theorem:

DC =√

AC2 − AD2

=√

3

sin 30◦ = AD

AC= 1

2

cos 30◦ = CD

AC=

√3

2

tan 30◦ = AD

CD= 1√

3

sin 60◦ = CD

AC=

√3

2

cos 60◦ = AD

AC= 1

2

tan 60◦ = CD

AD=

√3

1=

√3

C

A D B

22

11

30° 30°

60° 60°


SAMPLE

P1: FXS/ABE P2: FXS



Exact values for�

4(45◦)

AC =√

12 + 12 =√

2

sin 45◦ = BC

AC= 1√

2

cos 45◦ = AB

AC= 1√

2

tan 45◦ = BC

AB= 1

C

1

145°

A B

As an aid to memory, the exact values for circular functions can be tabulated.

Summary

� (�◦) sin � cos � tan �

0 0 1 0

�

6(30◦)

1

2

√3

2

1√3

�

4(45◦)

1√2

1√2

1

�

3(60◦)

√3

2

1

2

√3

�

2(90◦) 1 0 undefined

Example 10

Evaluate:

a cos 150◦ b sin 690◦

Solution

a cos 150◦ = cos (180 − 30)◦

= −cos 30◦

= −√

3

2

b sin 690◦ = sin (2 × 360 − 30)◦

= sin (−30)◦

= −1

2


SAMPLE

P1: FXS/ABE P2: FXS



Example 11

Evaluate:

a cos

(5�

4

)b sin

(11�

6

)

Solution

a cos

(5�

4

)= cos

(� + �

4

)

= −cos(�

4

)(by symmetry)

= − 1√2

b sin

(11�

6

)= sin

(2� − �

6

)

= −sin(�

6

)(by symmetry)

= −1

2

Exercise 16F

1 Without using a calculator, evaluate the sin, cos and tan of each of the following:Example 10

a 120◦ b 135◦ c 210◦ d 240◦ e 315◦

f 390◦ g 420◦ h −135◦ i −300◦ j −60◦

2 Write down the exact values of:Example 11

a sin

(2�

3

)b cos

(3�

4

)c tan

(5�

6

)

d sin

(7�

6

)e cos

(5�

4

)f tan

(4�

3

)

g sin

(5�

3

)h cos

(7�

4

)i tan

(11�

6

)

3 Write down the exact values of:

a sin

(−2�

3

)b cos

(11�

4

)c tan

(13�

6

)d tan

(15�

6

)

e cos

(14�

4

)f cos

(−3�

4

)g sin

(11�

4

)h cos

(−21�

3

)


SAMPLE

P1: FXS/ABE P2: FXS



16.7 Graphs of sine and cosineGraphs of sine functionsA table of values for y = sin x is given below.

x −� −3�

4−�

2−�

40

�

4

�

2

3�

4�

5�

4

3�

2

7�

42�

9�

4

5�

2

11�

43�

y 0 − 1√2

−1 − 1√2

01√2

11√2

0 − 1√2

−1 − 1√2

01√2

11√2

0

A calculator can be used to plot the graph of y = sin x, (−� ≤ x ≤ 3�). Note that Radian

mode must be selected.

–2–3

–1

1

–1 1

0

2 4 5 6 7 8 9x

y

3π–π π π– – –

4 2 4 π4

π π2

3π4

5π4

3π2

7π4

9π 5π4

3π2π2

11π4

3

–1

√2

1

√2

y = sin x

Observations from the graph of y = sin xThe graph repeats itself after an interval of 2� units. A function which repeats itself

regularly is called a periodic function and the interval between the repetitions is called the

period of the function (also called the wavelength).

Thus sin x has a period of 2� units.

The maximum and minimum values of sin x are 1 and −1 respectively.

The distance between the mean position and the maximum position is called the

amplitude. The graph of y = sin x has an amplitude of 1.

Graphs of cosine functionsA table of exact values for cos x, −� ≤ x ≤ 3�, is as shown:

x −� −3�

4−�

2−�

40

�

4

�

2

3�

4�

5�

4

3�

2

7�

42�

9�

4

5�

2

11�

43�

y −1−1√

20

1√2

11√2

0−1√

2−1

−1√2

01√2

11√2

0−1√

2−1


SAMPLE

P1: FXS/ABE P2: FXS



Using the TI-NspireA graph of y = cos x for −� ≤ x ≤ 3�

can be plotted in a Graphs & Geometry

application by entering

f1(x) = cos(x) | −� ≤ x and ≤ 3�.

Using the Casio ClassPadA graph of y = cos x for −� ≤ x ≤ 3� can be

plotted on the CAS calculator.

Enter y1 as shown, tick to select and click

to produce the graph. (The window shown

is produced by clicking in the graph window

and selecting Zoom—Auto.)

Observations from the graph of y = cos xPeriod = 2�

Amplitude = 1

The graph of y = cos x is the graph of y = sin x translated�

2units to the left and parallel

to the x-axis.


SAMPLE

P1: FXS/ABE P2: FXS



Sketch graphs of y = a sin (nt), y = a cos (nt)

Example 12

On separate axes draw graphs of the functions:

a y = 3 sin (2t), 0 ≤ t ≤ � b y = 2 cos (3t), 0 ≤ t ≤ 2�

3

Solution

a

t 0�

4

�

2

3�

4�

y = 3 sin (2t) 0 3 0 −3 0

b

t 0�

6

�

3

�

2

2�

3

y = 2 cos (3t) 2 0 −2 0 2

–3

–2

–1

0

1

3

2

y

π4

π2

3π4

π t

y = 3 sin(2t ), 0 ≤ t ≤ π

–2

–1

0

1

2

y

π6

π3

π2

2π3

t

2π3

y = 2 cos(3t), 0 ≤ t ≤

Observations

Function Amplitude Period

y = 3 sin (2t) 3 �

y = 2 cos (3t) 22�

3


SAMPLE

P1: FXS/ABE P2: FXS



Comparing these results with those for y = sin t and y = cos t, the following general rules can

be stated for a and n positive:

Function Amplitude Period

y = a sin (nt) a2�

n

y = a cos (nt) a2�

n

From the above it can be seen the transformation which takes the graph of y = sin t to

the graph of y = 3 sin (2t) has the following result on some important points of the graph

of y = sin t:

(0, 0) → (0, 0);(�

2, 1

)→

(�

4, 3

); (�, 0) →

(�

2, 0

);

(3�

2, 1

)→

(3�

4, 3

); (2�, 0) → (�, 0)

Also, it can be seen that the transformation which takes the graph of y = cos t to y = 2 cos (3t)

has the following result on some important points of the graph of y = cos t:

(0, 1) → (0, 2);(�

2, 0

)→

(�

6, 0

); (�, −1) →

(�

3, −2

);

(3�

2, 0

)→

(�

2, 0

); (2�, 1) →

(2�

3, 2

)

Note: The graph of y = 3 sin (2t) can be obtained from the graph of y = sin t by applying two

dilations. If f (t) = sin t, the graph of y = f (t) is transformed to the graph of y = 3f (2t). From

this it can be recognised that the sequence of transformations is:

dilation of factor 12 from the y-axis

dilation of factor 3 from the t-axis

The point with coordinates (t, y) is mapped to the point with coordinates

(t

2, 3y

)

In general, for a and n positive numbers, the following are important properties of the

functions f (t) = a sin (nt) and g(t) = a cos (nt):

The maximal domain of each of the functions is R.

The amplitude of each of the functions is a.

The period of each of the functions is2�

n.

The graph of y = a sin (nt) ( y = a cos (nt)) is obtained from the graph of

y = sin t ( y = cos t) by a dilation of factor a from the t-axis and a factor of1

nfrom the y-axis. The point with coordinates (t, y) is mapped to the point with

coordinates

(t

n, ay

).

The range of each function is [−a, a].


SAMPLE

P1: FXS/ABE P2: FXS

9780521740524c16a.xml CUAU021-EVANS August 23, 2008 13:58


Example 13

For each of the following functions with domain R state the amplitude and period:

a f (t) = 2 sin (3t) b f (t) = −1

2sin

(t

2

)c f (t) = 4 cos (3�t)

Solution

a Amplitude is 2

Period = 2�

3

b Amplitude is1

2

Period = 2� ÷ 1

2= 4�

c Amplitude is 4

Period = 2�

3�= 2

3

Example 14

Sketch the graphs of:

a y = 2 cos (2�) b y = 1

2sin

( x

2

)Show one complete cycle.

Solution

a The graph of y = 2 cos (2�) is obtained

from the graph of y = cos � by a

dilation of factor 2 from the �-axis and

by a dilation of factor1

2from the y-axis.

The period = 2�

2= � and the amplitude

is 2.

b The graph of y = 1

2sin

x

2is obtained

from the graph of y = cos x by a

dilation of factor1

2from the x-axis

and by a dilation of factor 2 from the

y-axis. The period = 2� ÷ 1

2= 4�

and the amplitude is1

2.

–2

0

2

y

π4

π2

π3π4

θ 0π

1

2

1

2–

2π 3π 4π

y

x

Example 15

Sketch the following graphs for x ∈ [0, 4�]:

a f (x) = −2 sin( x

2

)b y = −cos (2x)


SAMPLE

P1: FXS/ABE P2: FXS



Solution

a The graph of f (x) = −2 sin( x

2

)

is obtained from the graph of

y = 2 sin( x

2

)by a reflection

in the x-axis.

The period is 4� and the

amplitude is 2.

b The graph of y = −cos (2x) is

obtained from the graph of

y = cos (2x) by a reflection in

the x-axis.

The period is � and the

amplitude is 1.

0

–2

2

2π 4π

y

x0

–1

π 4π

y

x

1

3π2π


functions f (t) = −a sin (nt) and g(t) = −a cos (nt):

The amplitude of each of the functions is a.

The period of each of the functions is2�

n.

The graph of y = −a sin (nt) ( y = −a cos (nt)) is obtained from the graph of

y = a sin (nt) (a cos (nt)) by a reflection in the t-axis.

The range of each function is [−a, a].

Remember that sin (−x) = −sin x and cos (−x) = cos x. Hence when reflected in the y-axis the

graph of y = cos x transforms onto itself and the graph of y = sin x transforms onto the graph

of y = −sin x.

Example 16

Sketch the graph of f : [0, 2] →R, f (t) = 3 sin (�t)

Solution

–3

1

232

t21

3

Exercise 16G

1 Write down i the period and ii the amplitude of each of the following:Example 13

a 2 sin � b 3 sin (2�) c1

2cos (3�)


SAMPLE

P1: FXS/ABE P2: FXS



d 3 sin

(1

2�

)e 4 cos (3�) f −1

2sin (4�)

g −2 cos

(1

2�

)h 2 cos(�t) i −3 sin

(�t

2

)

2 Sketch the graph of each of the following, showing one complete cycle. State the amplitudeExample 14

and period.

a y = 3 sin (2x) b y = 2 cos (3�) c y = 4 sin

(�

2

)

d y = 1

2cos (3x) e y = 4 sin (3x) f y = 5 cos (2x)

g y = −3 cos

(�

2

)h y = 2 cos (4�) i y = −2 sin

(�

3

)

3 Sketch the graph of:Example 15

a f (x) = sin (2x) for x ∈ [−2�, 2�] b f (x) = 2 sin( x

3

)for x ∈ [−6�, 6�]

c f (x) = 2 cos (3x) for x ∈ [0, 2�] d f (x) = −2 sin (3x) for x ∈ [0, 2�]

4 Sketch the graph of f : [0, 2�] →R, f (x) = 5

2cos

(2x

3

). Hint: For endpoints find f (0) and

f (2�).

5 For each of the following give a sequence of transformations which takes the graph of

y = sin x to the graph of y = g(x), and state the amplitude and period of g(x):

a g(x) = 3 sin x b g(x) = sin (5x) c g(x) = sin( x

3

)

d g(x) = 2 sin (5x) e g(x) = −sin (5x) f g(x) = sin (−x)

g g(x) = 2 sin( x

3

)h g(x) = −4 sin

( x

2

)i g(x) = 2 sin

(−x

3

)

6 Sketch the graph of:Example 16

a f : [0, 2] → R, f (t) = 2 cos (�t) b f : [0, 2] → R, f (t) = 3 sin (2�t)

7 a On the one set of axes sketch the graph of f : [0, 2�] → R, f (x) = sin x and

g: [0, 2�] → R, g(x) = cos x.

b By inspection from the graph state the values of x for which sin x = cos x.

16.8 Sketch graphs of y = a sinn(t ± ε) andy = a cosn(t ± ε)In this section translations of graphs of functions of the form f (t) = a sin (nt) and

g(t) = a cos (nt) in the direction of the t-axis are considered. When a translation of�

4units in

the positive direction of the t-axis is applied to the graph of y = f (t), the resulting image has

equation y = f(

t − �

4

). That is, the graph of f (t) = a sin (nt) is mapped to the graph with

equation y = a sin n(

t − �

4

).


SAMPLE

P1: FXS/ABE P2: FXS



Example 17

On separate axes draw the graphs of the following functions. Use a calculator to help establish

the shape. Set the window appropriately by noting the range and period.

a y = 3 sin 2(

t − �

4

),

�

4≤ t ≤ 5�

4b y = 2 cos 3

(t + �

3

), −�

3≤ t ≤ �

3

Solution

a y = 3 sin 2(

t − �

4

),

�

4≤ t ≤ 5�

4Note that the range is [−3, 3] and

the period is �.

b y = 2 cos 3(

t + �

3

), −�

3≤ t ≤ �

3Note that the range is [−2, 2] and

the period is2�

3.

y

t

3210

–1–2–3

π4

π2 4

3π 5π4

π

–2

–10

1

2

y

π3

–π3

–π6

tπ6

Observations1 For y = 3 sin 2

(t − �

4

), amplitude = 3, period = �. The graph is the same shape as

y = 3 sin (2t) but is translated�

4units in the positive direction of the t-axis.

2 For y = 2 cos 3(

t + �

3

), amplitude = 2, period = 2�

3. The graph is the same shape

as y = 2 cos (3t) but is translated�

3units in the negative direction of the t-axis.

In each case ε has the effect of translating the graph parallel to the t-axis (ε is called the phase).

Note: To determine the sequence of transformations needed, the techniques of Section 6.10 can

also be used.

The graph of y = sin t is transformed to the graph of y = 3 sin 2(

t − �

4

).

Write the second equation asy′

3= sin 2

(t ′ − �

4

). From this it can be seen that y = y′

3and t = 2

(t ′ − �

4

).

Hence y′ = 3y and t ′ = t

2+ �

4. Hence the sequence of transformations is:

dilation of factor 3 from the t-axis

dilation of factor1

2from the y-axis and

translation of�

4units in the positive direction of the t-axis.

The observation that the graph of y = f (t) is transformed to the graph of y = 3 f(

2(

t − �

4

)),

where f (t) = sin t, also yields this information.


SAMPLE

P1: FXS/ABE P2: FXS



Exercise 16H

1 Sketch the graph of each of the following, showing one complete cycle. State the periodExample 17

and amplitude, and the greatest and least values of y.

a y = 3 sin(

� − �

2

)b y = sin 2(� + �) c y = 2 sin 3

(� + �

4

)

d y = √3 sin 2

(� − �

2

)e y = 3 sin (2x) f y = 2 cos 3

(� + �

4

)

g y = √2 sin 2

(� − �

3

)h y = −3 sin (2x) i y = −3 cos 2

(� + �

2

)

2 For the function f : [0, 2�] →R, f (x) = cos(

x − �

3

):

a find f (0), f (2�) b sketch the graph of f

3 For the function f : [0, 2�] →R, f (x) = sin 2(

x − �

3

):

a find f (0), f (2�) b sketch the graph of f

4 For the function f : [−�, �] →R, f (x) = sin 3(

x + �

4

):

a find f (−�), f (�) b sketch the graph of f

5 Find the equation of the image of y = sin x for each of the following transformations:

a Dilation of factor 2 from the y-axis followed by dilation of factor 3 from the x-axis

b Dilation of factor1

2from the y-axis followed by dilation of factor 3 from the x-axis

c Dilation of factor 3 from the y-axis followed by dilation of factor 2 from the x-axis

d Dilation of factor1

2from the y-axis followed by a translation of

�

3units in the positive

direction of the x-axis

e Dilation of factor 2 from the y-axis followed by a translation of�

3units in the negative

direction of the x-axis

16.9 Solution of trigonometric equations

Example 18

Find all solutions to the equation sin � = 1

2for � ∈ [0, 4�].

Solution

–1

0

1

y

π6

12

θπ

y =y = sinθ

2π 3π 4π

It is clear from the graph that

there are four solutions in the

interval [0, 4�].

The solution for x ∈[0,

�

2

]is

x = �

6.


SAMPLE

P1: FXS/ABE P2: FXS



This solution can be obtained

from a knowledge of exact values or

by using sin−1 on your calculator.

The second solution is obtained by

symmetry. The function is positive in the

second quadrant and sin (� − �) = sin �.

Therefore x = 5�

6is the second solution.

–1

0

1

y

π6

5π6

13π6

17π6

12

θ

y =y = sinθ

It can be seen that further solutions can

be achieved by adding 2�, as

sin � = sin (� + 2�).

Thus � = 13�

6and

17�

6are also solutions.

Example 19

Find two values of x:

a sin x = −0.3 in the range 0 ≤ x ≤ 2� b cos x◦ = −0.7 in the range 0 ≤ x ≤ 360

Solution

a First we must solve the equation sin x = 0.3.

Use your calculator to find the solution for x ∈[0,

�

2

]; x = 0.30469 . . .

Now the value of sin is negative for P(x) in the 3rd and 4th quadrants. From the

symmetry relationships (or from the graph of y = sin x):

3rd quadrant: x = � + 0.30469 . . .

= 3.446 (correct to

3 decimal places)

4th quadrant: x = 2� − 0.30469 . . .

= 5.978 (correct to

3 decimal places)

∴ if sin x = −0.3, x = 1.875, or x = 5.978

( , 0.3)

( , –0.3)( , –0.3)

b First we solve the equation cos x◦ = 0.7.

Use your calculator to find the solution for x ∈ [0◦, 90◦].

Now the value of cos is negative for P(x) in the 2nd and 3rd quadrants.

2nd quadrant: x = 180 − 45.57

= 134.43

3rd quadrant: x = 180 + 45.57

= 225.57

∴ if cos x◦ = −0.7, x = 134.43, 225.57

(0.7, )

(–0.7, )

(–0.7, )


SAMPLE

P1: FXS/ABE P2: FXS



Example 20

Find all the values of � between 0 and 360 for which:

a cos �◦ =√

3

2b sin �◦ = −1

2c cos �◦ − 1√

2= 0

Solution

a cos �◦ is positive, ∴ P(�◦) lies in the 1st or 4th quadrants.

cos �◦ =√

3

2� = 30 or 360 − 30

� = 30 or 330

b sin �◦ is negative, ∴ P(�◦) is in the 3rd or 4th quadrants.

sin �◦ = −1

2� = 180 + 30 or 360 − 30

� = 210 or 330

c cos �◦− 1√2

= 0

∴ cos �◦ = 1√2

and since cos �◦ is positive, P(�◦) lies in the 1st or 4th quadrants.

∴ cos �◦ = 1√2

� = 45 or � = 360 − 45

� = 45 or 315

Using the TI-NspireFor Example 20a, make sure the

calculator is in degree mode.

Complete as shown.


SAMPLE

P1: FXS/ABE P2: FXS



Using the Casio ClassPadEnsure the mode is set to Degrees (Deg in the status

bar at the bottom of the Main screen.)

Enter the equation

cos(x) =√

(3)/2 | 0 ≤ x ≤ 360.

Use your stylus to highlight only cos(x) = √(3)/2,

then tap Interactive—Equation/inequality—solve

and ensure the variable is set to x.

Example 21

Solve the equation sin (2�) = −√

3

2for � ∈ [−�, �].

Solution

0

y

–π2

π2

y = sin(2θ)

θπ–π

√32y = –

It is clear that there are four solutions.

To solve the equation, let x = 2�.

Note:

If � ∈ [−�, �]

then 2� = x ∈ [−2�, 2�]

Consider the equation sin x = −√

3

2for x ∈ [−2�, 2�].

0x

y

y = sinx

√32y = –

–2π 2π

The first quadrant solution to the equation

sin x =√

3

2is x = �

3.

Symmetry gives the solutions to

sin x = −√

3

2for x ∈ [0, 2�] as

x = � + �

3and x = 2� − �

3

i.e. x = 4�

3or x = 5�

3


SAMPLE

P1: FXS/ABE P2: FXS



The other two solutions are obtained by subtracting 2�:4�

3− 2� and

5�

3− 2�

∴ the required solutions for x are −2�

3or −�

3or

4�

3or

5�

3

∴ the required solutions for � are −�

3or −�

6or

2�

3or

5�

6

Using the TI-NspireComplete as shown.

Using the Casio ClassPadEnsure the mode is set to Radians (Rad in the status bar

at bottom of the Main screen).

Enter the equation

sin(2x) =√

(3)/2 | −� ≤ x ≤ �.

With your stylus highlight only sin(2x) = −√(3)/2

then tap Interactive—Equation/inequality—solve

and ensure the variable is set to x.

Exercise 16I

1 Without using a calculator, find all the values of x between 0 and 2� for each of theExample 18

following:

a√

2 sin (x) + 1 = 0 b√

2 cos (x) − 1 = 0


SAMPLE

P1: FXS/ABE P2: FXS



2 Find, correct to 2 decimal places, all the values of x between 0 and 2� for which:Example 19a

a sin x = 0.8 b cos x = −0.4 c sin x = −0.35

d sin x = 0.4 e cos x = −0.7 f cos x = −0.2

3 Without using a calculator, find all the values of � between 0◦ and 360◦ for each of theExample 20

following:

a cos �◦ = −√

3

2b sin �◦ = 1

2c cos �◦ = −1

2d 2 cos (�◦) + 1 = 0 e 2 sin �◦ = √

3 f√

2 sin (�◦) − 1 = 0

4 Find all the values of x between 0 and 4� for which:

a sin x = 0.6 b sin x = − 1√2

c sin x =√

3

2

5 Find all the values of x between −� and � for which:

a cos x = − 1√2

b sin x =√

3

2c cos x = −1

2

6 a Sketch the graph of f : [−2�, 2�] → R, f (x) = cos x.

b On the graph mark the points which have y-coordinate1

2and give the associated

x-values.

c On the graph mark the points which have y-coordinate −1

2and give the associated

x-values.

7 Solve the following equation for � ∈ [0, 2�]:Example 21

a sin (2�) = −1

2b cos (2�) =

√3

2c sin (2�) = 1

2

d sin (3�) = − 1√2

e cos (2�) = −√

3

2f sin (2�) = − 1√

2

8 Solve the following equations for � ∈ [0, 2�]:

a sin (2�) = −0.8 b sin (2�) = −0.6

c cos (2�) = 0.4 d cos (3�) = 0.6

16.10 Sketch graphs of y = a sin n(t ± ε) ± band y = a cos n(t ± ε) ± bTranslations parallel to the y-axis are now considered.

Example 22

Sketch each of the following graphs. Use a calculator to help establish the shape.

a y = 3 sin 2(

t − �

4

)+ 2 for

�

4≤ t ≤ 5�

4

b y = 2 cos 3(

t + �

3

)− 1 for −�

3≤ t ≤ �

3


SAMPLE

P1: FXS/ABE P2: FXS



Solution

543210

–1

–π3

–π6

π6

π3

t

y

π3

– 1y = 2cos3 t +

t

y

1

0–1

–2

–3

π4

π2

3π4

5π4

π

π4y = 3sin2 t – + 2

Observations1 The graph of y = 3 sin 2

(t − �

4

)+ 2 is the same shape as the graph of

y = 3 sin 2(

t − �

4

)but it is translated 2 units in the positive direction of the y-axis.

2 Similarly, the graph of y = 2 cos 3(

t + �

3

)− 1 is the same shape as the graph of

y = 2 cos 3(

t + �

3

)but it is translated 1 unit in the negative direction of the y-axis.

In general, the effect of ±b is to translate the graph ±b units parallel to the y-axis.

Finding axis intercepts

Example 23

Sketch the graphs of each of the following for x ∈ [0, 2�]. Clearly indicate axis intercepts.

a y = √2 sin (x) + 1 b y = 2 cos (2x) − 1 c y = 2 sin 2

(x − �

3

)− √

3

Solution

a To determine the axis intercepts, the equation√

2 sin (x) + 1 = 0 must be solved.√2 sin (x) + 1 = 0

∴ sin x = − 1√2

∴ x = � + �

4or 2� − �

4

∴ x = 5�

4or

7�

4

∴ Intercepts:

(5�

4, 0

),

(7�

4, 0

)y

π2

3π2

7π4

5π4

x2ππ

√2 + 1

√2 sin (x) + 1

–√2 + 1

(0, 1)

0

y =


SAMPLE

P1: FXS/ABE P2: FXS



b 2 cos (2x) − 1 = 0

∴ cos (2x) = 1

2

∴ 2x = �

3or

5�

3or

7�

3or

11�

3

∴ x = �

6or

5�

6or

7�

6or

11�

6

∴ Intercepts:(�

6, 0

),

(5�

6, 0

),

(7�

6, 0

),

(11�

6, 0

)

1

0

–1

–2

–3

x

y

π6

5π

2π67π6

11π6

y = 2 cos (2x) – 1

c sin 2(

x − �

3

)=

√3

2

∴ 2(

x − �

3

)= �

3or

2�

3or

7�

3or

8�

3

∴ x − �

3= �

6or

�

3or

7�

6or

4�

3

∴ x = �

2or

2�

3or

3�

2or

5�

3

∴ Intercepts:(�

2, 0

),

(2�

3, 0

),

(3�

2, 0

),

(5�

3, 0

)

0

π2

2π3

3π2

5π3

π 2πx

y

y = 2 – √3

y = –2 – √3

(0, –2√3)

–√3

y = 2 sin 2π3

x – – √3


SAMPLE

P1: FXS/ABE P2: FXS



Exercise 16J

1 Sketch the graphs of each of the following for x ∈ [0, 2�]. List the x-axis intercepts of eachExample 23

graph for this interval:

a y = 2 sin (x) + 1 b y = 2 sin (2x) − √3 c y = √

2 cos (x) + 1

d y = 2 sin (2x) − 2 e y = √2 sin

(x − �

4

)+ 1

2 Sketch the graphs of each of the following for x ∈ [−2�, 2�]:

a y = 2 sin (3x) − 2 b y = 2 cos 3(

x − �

4

)c y = 2 sin (2x) − 3 d y = 2 cos (2x) + 1

e y = 2 cos 2(

x − �

3

)− 1 f y = 2 sin 2

(x + �

6

)+ 1

3 Sketch the graphs of each of the following for x ∈ [−�, �]:

a y = 2 sin 2(

x + �

3

)+ 1 b y = −2 sin 2

(x + �

6

)+ 1

c y = 2 cos 2(

x + �

4

)+ √

3

16.11 Further symmetry properties and thePythagorean identityComplementary relationships

sin(�

2− �

)= a

and since a = cos �

sin(�

2− �

)= cos �

y

x

P(θ)θ

θb

a

b

a

0

Pπ2

– θ

π2

– θSimilarly

cos(�

2− �

)= b

and since b = sin �

cos(�

2− �

)= sin �

sin(�

2+ �

)= a = cos �

cos(�

2+ �

)= −b = −sin �

0

y

xa

a

b

b

θ

θ

P(θ)

P π 2

+ θ

+ θπ2


SAMPLE

P1: FXS/ABE P2: FXS



Example 24

If sin � = 0.3 and cos � = 0.8 find the values of:

a sin(�

2− �

)b cos

(�

2+ �

)c sin (−�)

Solution

a sin(�

2− �

)= cos �

= 0.8

b cos(�

2+ �

)= −sin �

= −0.3

c sin (−�) = −sin �

= −0.3

The Pythagorean identityConsider a point, P(�) on the unit circle.

By Pythagoras’ Theorem

0x

y

cosθ

P(θ)

M

1

1

1

–1

–1

θsinθ

OP 2 = OM 2 + MP 2

∴ 1 = (cos �)2 + (sin �)2

Now (cos �)2 and (sin �)2 may be written

as cos2 � and sin2 �.

∴ 1 = cos2 � + sin2 �

As this is true for all values of � it is called

an identity. In particular this is called the

Pythagorean identity.

cos2 � + sin2 � = 1

Example 25

Given that sin x = 3

5and

�

2< x < �, find cos x and tan x.

Solution

1 = cos2 x + sin2 x.

For sin x = 3

5, 1 = cos2 x + 9

25

Then cos2x = 1 − 9

25

= 16

25

Therefore cos x = ±4

5


SAMPLE

P1: FXS/ABE P2: FXS



But x is in the second quadrant, hence cos x = −4

5.

tan x = sin x

cos x

= 3

5÷ −4

5

= 3

5× −5

4

= −3

4

Exercise 16K

1 If sin x = 0.3, cos � = 0.6 and tan � = 0.7, find the values of:Example 24

a cos (−�) b sin(�

2+ �

)c tan (−�) d cos

(�

2− x

)

e sin (−x) f tan(�

2− �

)g cos

(�

2+ x

)h sin

(�

2− �

)

i sin

(3�

2+ �

)j cos

(3�

2− x

)

2 a Given that 0 < � <�

2and cos � = sin

�

6, find the value of �.

b Given that 0 < � <�

2and sin � = cos

�


c Given that 0 < � <�

2and cos � = sin

�


d Given that 0 < � <�

2and sin � = cos

3�


3 Given that cos x = 3

5and

3�

2< x < 2�, find sin x and tan x.Example 25

4 Given that sin x = 5

13and

�

2< x < �, find cos x and tan x.

5 Given that cos x = 1

5and

3�

2< x < 2�, find sin x and tan x.

16.12 The tangent functionA table of values for y = tan x is given below. Use a calculator to check these values, and plot

the graph of y = tan (x).

x –� −3�

4−�

2−�

40

�

4

�

2

3�

4�

5�

4

3�

2

7�

42�

9�

4

5�

2

11�

43�

y 0 1 ud −1 0 1 ud −1 0 1 ud −1 0 1 ud −1 0


SAMPLE

P1: FXS/ABE P2: FXS



6 9

π 2π 3π

8754321

1

x = x = x = x =

0–1–2–3

–1

–π

–π

2

π2

3π2

5π2

x

y

Note: x = −�

2,

�

2,

3�

2and

5�

2are asymptotes.

The x-axes intercepts where sin x = 0, are x = 0 or � or 2� etc.

In general, x = k�, where k is an integer.

Observations from the graphThe graph repeats itself every � units, i.e. the period of tan is �.

Range of tan is R.

The equation of the asymptotes are of the form y = (2k + 1)�

2, where k is an integer.

The x-axis intercepts occur for x = k�, where k is an integer.

Transformations of y = tan xConsider a dilation of factor

1

2from the y-axis and a dilation of factor 3 from the x-axis being

applied to the graph of y = tan x.

(x, y) →(

1

2x, 3y

). If the image of (x, y) under the transformation is (x ′, y′) then x ′ = 1

2x

and y′ = 3y.

Hence x = 2x ′ and y = y′

3. Thus the graph of y = tan x is transformed to the graph of

y′

3= tan 2x ′; that is, it is transformed to the graph of y = 3 tan 2x.

The period of the graph will be�

2.


function f (t) = a tan (nt):

The period of the function is�

n.

The graph of y = a tan (nt) is obtained from the graph of y = tan t by a dilation of

factor a from the t-axis and a factor of1

nfrom the y-axis.

The range of the function is R.

The asymptotes have equations x = (2k + 1)�

2n, where k is an integer.

The t-axis intercepts are x = k�

n, where k is an integer.


SAMPLE

P1: FXS/ABE P2: FXS



Example 26

Sketch the graph of each of the following for x ∈ [−�, �]:

a y = 3 tan (2x) b y = −2 tan (3x)

Solution

a y = 3 tan (2x)

Period = �

n= �

2

Asymptotes: x = (2k + 1)�

4, k ∈ Z

Axes intercepts: x = k�

2, k ∈ Z

b y = −2 tan (3x)

Period = �

n= �

3

Asymptotes: x = (2k + 1)�

6, k ∈ Z

Axes intercepts: x = k�

3, k ∈ Z

y

x

x =x =x =–3π

4x =

–π

–π–π

4

π4

π π22

3π4

0

y

x

x =

–2π–π –π3

0 π π33

2π3

5π6

x =–5π6

x =π2

x =π6

x =–π2

x =–π6

Solution of equations

Example 27

Solve each of the following equations for x ∈ [−�, �]:

a tan x = −1 b tan (2x) = √3 c 2 tan (3x) = 0

Solution

a tan x = −1

x = 3�

4or

−�

4

b tan (2x) = √3

2x = �

3or

4�

3or

−2�

3or

−5�

3

x = �

6or

4�

6or

−2�

6or

−5�

6

= �

6or

2�

3or

−�

3or

−5�

6

c 2 tan (3x) = 0

3x = −3� or −2� or −� or 0 or � or 2� or 3�

Hence x = −� or−2�

3or

−�

3or 0 or

�

3or

2�

3or �


SAMPLE

P1: FXS/ABE P2: FXS



Example 28

Sketch the graph of y = tan (2x) + 1 for x ∈ [−�, �].

Solution

The graph of y = tan (2x) + 1 is formed from the graph of y = tan (2x) by a translation

of 1 unit in the positive direction of the y-axis.

The y-axis intercept occurs when x = 0. When x = 0 then y = 0.

For the x-axis intercepts consider tan (2x) + 1 = 0

This implies tan (2x) = −1

Hence 2x = 3�

4or

−�

4or

7�

4or

−5�

4

and x = 3�

8or

−�

8or

7�

8or

−5�

8

The asymptotes are the same as those for y = tan (2x). That is, x = (2k + 1)�

4, k ∈ Z.

x =x = x =x =–3π4

3π4

–π4

π4

0 1x

y

–5π8

(π, 1)(–π, 1)

3π8

7π8

–π8

Exercise 16L

1 For each of the following state the period:

a y = tan (4x) b y = tan

(2x

3

)c y = −3 tan (2x)

2 Sketch the graph of each of the following for x ∈ [−�, �]:Example 26

a y = tan (2x) b y = 2 tan (3x) c y = −2 tan (3x)


SAMPLE

P1: FXS/ABE P2: FXS



3 Solve each of the following equations for x ∈ [−�, �]:Example 27

a 2 tan (2x) = 2 b 3 tan (3x) = √3

c 2 tan (2x) = 2√

3 d 3 tan (3x) = −√3

4 Sketch the graph of each of the following for x ∈ [−�, �]:Example 28

a y = 3 tan (x) + √3 b y = 2 tan (x) + 2 c y = 3 tan (x) − 3

16.13 Numerical methods with a CAS calculator

Using the TI-Nspire

Example 29

Solve the quationx

2= sin x , giving your answer correct ot 2 decimal places.

Use Solve( ) (b 31) as shown.

Press / · to obtain the answer as

a decimal number.

Fitting dataConsider the points (1, 2.08), (2, 2.3),

(3, 0.49), (4, −1.77) and (6, −0.96).

Enter the data in a Lists &

Spreadsheet application as shown.


SAMPLE

P1: FXS/ABE P2: FXS



Choose Sinusoidal Regression (b

41 C ) from the list of available

regressions and complete as shown.

This now gives the values of a, b, c

and d, and the equation has been entered

in f1(x).

The curve can be shown in a Graphs

& Geometry application together with

the Scatter Plot (b 34) using an

appropriate Window (b 4).


SAMPLE

P1: FXS/ABE P2: FXS



Using the Casio ClassPad

Example 29

Solving the equation x/2 = sin(x) is done

numerically by drawing the graph of the each

side of the equation, then finding the

intersection on the calculator.

Note: The window has been set for

0 ≤ x ≤ 2� and −2 ≤ y ≤ 2 in order to see

where all the solutions lie.

Use Zoom—Box to zoom on the non-zero

solution, then with the graph window selected

(bold box), select G-solve—Intersect to

obtain the numerical solution.

Fitting dataConsider the points (1, 2.08), (2, 2.3),

(3, 0.49), (4, −1.77) and (6, −0.96).

From the Menu select and enter the

data in lists 1 and 2 as shown.

Select Calc—Sinusoidal Reg and check

the entries are correct.

Note: Set Copy Formula to y1 as this will

put the formula for the graph drawn in the

section automatically for later use if

required.


SAMPLE

P1: FXS/ABE P2: FXS



Note the formula, then click OK again to

produce the graph.

Exercise 16M

1 Solve each of the following equations for x correct to 2 decimal places:Example 29

a cos x = x b sin x = 1 − x

c cos x = x2 d sin x = x2

2 For each of the following sets of data find a suitable trigonometric rule (model).

a� 0

�

4

�

2

3�

4�

y 1 2.4 −1 2.4 1

b� 0 0.2 0.4 0.6 0.8

y 0 1.77 2.85 2.85 1.77

c� 0 0.2 0.4 0.6 0.8

y 5 2.18 0.34 0.13 1.62

16.14 General solution of circular function equationsSolution of equations of circular functions has been discussed in Section 16.9 for functions

over a restricted domain. In this section, we consider the general solutions of such equations

over the maximal domain for each function.

If the equation of a circular function has one or more solutions in one ‘cycle’, then it will

have corresponding solutions in each cycle of its domain, i.e. there will be an infinite number

of solutions.

For example, if cos x = a, then the solution in the interval [0, �] is given by:

x = cos−1(a)

By the symmetry properties of the cosine function, other solutions are given by:

−cos−1(a), ±2� + cos−1(a), ±2� − cos−1(a), ±4� + cos−1(a), ±4� − cos−1(a), . . .


SAMPLE

P1: FXS/ABE P2: FXS



In general, if cos x = a:

x = 2n� ± cos−1(a), where n ∈ Z and a ∈ [−1, 1].

Similarly, if tan x = a:

x = n� + tan−1(a), where n ∈ Z and a ∈ R.

If sin x = a:

x = 2n� + sin−1(a) or x = (2n + 1)� − sin−1(a), where n ∈ Z and a ∈ [−1, 1].

Note: An alternative and more concise way to express the general solution of sin x = a is:

x = n� + (−1)n sin−1(a), where n ∈ Z and a ∈ [−1, 1].

Example 30

Find the general solution to each of the following equations.

a cos x = 0.5 b√

3 tan (3x) = 1 c 2 sin x = √2

Solution

a x = 2n� ± cos−1(0.5)

= 2n� ± �

3

= (6n ± 1)�

3, n ∈ Z

b tan (3x) = 1√3

3x = n� + tan−1

(1√3

)

= n� + �

6

= (6n + 1)�

6

x = (6n + 1)�

18, n ∈ Z

c sin x = 1√2

x = 2n� + sin−1

(1√2

)or x = (2n + 1)� − sin−1

(1√2

)

= 2n� + �

4= (2n + 1)� − �

4

= (8n + 1)�

4, n ∈ Z = (8n + 3)�

4, n ∈ Z


SAMPLE

P1: FXS/ABE P2: FXS



Using the TI-NspireCheck that the calculator is in Radian mode.

a Use Solve( ) from the Algebra menu

(b 31) and complete as shown.

Note the use of1

2rather than 0.5 to

ensure that the answer is exact.

b Complete as shown.

c Complete as shown.

Using the Casio ClassPadCheck the calculator is in Radian mode.

a In enter and highlight the equation

cos(x) = 0.5. Select Interactive—

Equation/inequality—solve, then click

to solve the equation.

The solutions are

x = 2m� − �/3, 2n� + �/3, m, n ∈ Z .


SAMPLE

P1: FXS/ABE P2: FXS



b Enter and highlight the equation√3 tan(3x) = 1 and follow the steps as

in a.

The solutions are

x = (6m + 1)�

18, m ∈ Z

c Enter and highlight the equation√2 sin(x) = 1 and follow the steps as in a.

The solutions are

x = 2m� − �

4, 2n� + 3�

4, m, n ∈ Z .

Example 31

Find the first three positive solutions to each of the following equations.

a cos x = 0.5 b√

3 tan (3x) = 1 c 2 sin x = √2

Solution

a The general solution (from Example 30) is given by x = (6n ± 1)�

3, n ∈ Z .

When n = 0, x = ±�

3and, when n = 1, x = 5�

3or x = 7�

3.

The first three positive solutions of cos x = 0.5 are x = �

3,

5�

3,

7�

3.

b The general solution (from Example 30) is given by x = (6n + 1)�

18, n ∈ Z .

When n = 0, x = �

18and, when n = 1, x = 7�

18, and when n = 2, x = 13�

18.

The first three positive solutions of√

3 tan (3x) = 1 are x = �

18,

7�

18,

13�

18.

c The general solution (from Example 30) is given by x = (8n + 1)�

4or

x = (8n + 3)�

4, n ∈ Z .

When n = 0, x = �

4or

3�

4and, when n = 1, x = 9�

4or x = 11�

4.

The first three positive solutions of 2 sin x = √2 are x = �

4,

3�

4,

9�

4.

Exercise 16N

1 Find the general solution to each of the following equations.Example 30

a sin x = 0.5 b 2 cos (3x) = √3 c

√3 tan x = −3

2 Find the first two positive solutions to each of the following equations.Example 31

a sin x = 0.5 b 2 cos (3x) = √3 c

√3 tan x = −3

3 Find the general solution to 2 cos(

2x + �

4

)= √

2, and hence find all the solutions for x in

the interval (−2�, 2�).


SAMPLE

P1: FXS/ABE P2: FXS



4 Find the general solution to√

3 tan(�

6− 3x

)− 1 = 0, and hence find all the solutions for

x in the interval [−�, 0].

5 Find the general solution to 2 sin (4�x) + √3 = 0, and hence find all the solutions for x in

the interval [−1, 1].

16.15 Applications of trigonometric functions

Example 32

It is suggested that the height h(t) metres of the tide above mean sea level on 1 January at

Warnung is given approximately by the rule h(t) = 4 sin(�

6t)

, where t is the number of hours

after midnight.

a Draw the graph of y = h(t) for 0 ≤ t ≤ 24. b When was high tide?

c What was the height of the high tide? d What was the height of the tide at 8 am?

e A boat can only cross the harbour bar when the tide is at least 1 metre above mean sea

level. When could the boat cross the harbour bar on 1 January?

Solution

a

0 6 12 18 24x

y

–4

4 y = h(t)

We note: period = 2� ÷ �

6= 12

b High tide occurs when h(t) = 4

4 sin(�

6t)

= 4

implies sin(�

6t)

= 1

∴ �

6t = �

2,

5�

2∴ t = 3, 15

i.e. high tide occurs at 03.00 and

15.00 (3 pm).

c The high tide has height 4 metres above the mean height.

d h(8) = 4 sin

(8�

6

)= 4 sin

(4�

3

)= 4 × −√

3

2= −2

√3.

The water is 2√

3 metres below the mean height at 8 am.

e We first consider 4 sin�

6t = 1.

Thus sin�

6t = 1

4

∴ �

6t = 0.2526, 2.889, 6.5358, 9.172

∴ t = 0.4824, 5.5176, 12.4824, 17.5173

i.e. the water is at height 1 metre at 00:29, 05:31, 12:29, 17:31.

Thus the boat can pass across the harbour bar between 00:29 and 05:31 and

between 12:29 and 17:31.


SAMPLE

P1: FXS/ABE P2: FXS



Exercise 16O

1 The depth, D(t) metres, of water at the entrance to a harbour at t hours after midnight on a

particular day is given by D(t) = 10 + 3 sin

(�t

6

), 0 ≤ t ≤ 24.

a Sketch the graph of D(t) for 0 ≤ t ≤ 24. b Find the value of t for which D(t) ≥ 8.5.

c Boats which need a depth of w metres are permitted to enter the harbour only if the

depth of the water at the entrance is at least w metres for a continuous period of 1 hour.

Find, correct to 1 decimal place, the largest value of w which satisfies this condition.

2 The depth of water at the entrance to a harbour t hours after high tide is D metres, where

D = p + q cos (rt)◦ for suitable constants p, q, r. At high tide the depth is 7 m; at low tide,

6 hours later, the depth is 3 m.

a Show that r = 30 and find the values of p and q.

b Sketch the graph of D against t for 0 ≤ t ≤ 12.

c Find how soon after low tide a ship that requires a depth of at least 4 m of

water will be able to enter the harbour.

3 A particle moves on a straight line, OX, and its distance x metres from O at

time t (s) is given by x = 3 + 2 sin (3t).

a Find its greatest distance from O. b Find its least distance from O.

c Find the times at which it is 5 metres from O for 0 ≤ t ≤ 5.

d Find the times at which it is 3 metres from O for 0 ≤ t ≤ 3.

e Describe the motion of the particle.


SAMPLE

P1: FXS/ABE P2: FXS


Review


Chapter summary

Definition of a radian y

x

1

1

1c

1 unit

10–1

–1

One radian (written lc) is the angle formed at the centre

of the unit circle by an arc of length 1 unit.

1c = 180◦

�1◦ = �c

180

Sine and cosine y

xx

y

P(θ)

θc

10

1

–1

–1

1

x-coordinate of P(�) in unit circle,

x = cosine �, � ∈ R

y-coordinate of P(�) in unit circle,

y = sine �, � ∈ R

Abbreviated to x = cos �

y = sin �

Tangent y

B

A

tanθ

x

1

–1

–1

1

0

sinθθc

cosθ

If the tangent to the unit circle at A is drawn then the

y-coordinate of B is called tangent � (abbreviated to tan �).

Also by using similar triangles:

tan � = sin �

cos �

Circular functions and trigonometric ratiosy

y

x

x0

1

θ

Adjacentside, A

Hypotenuse,H Opposite

side, O

θ

sin � = O

H= y

1= y

cos � = A

H= x

1= x

tan � = O

A= y

x= sin �

cos �


SAMPLE

P1: FXS/ABE P2: FXS


Rev

iew


Symmetry properties of circular functions

Quadrant 2 (sin is positive)sin(π – θ) = b = sinθ

sin(π + θ) = –b = –sinθQuadrant 3 (tan is positive)

sin(2π – θ) = –b = –sinθQuadrant 4 (cos is positive)

y

x

–b

b

–1

–1

1

1θ

Quadrant 1 (all functions are positive)e.g. sinθ = b

Solutions of trigonometric equations of the type sin x◦ = a and cos x◦ = a

e.g. If cos x◦ = −0.7, find the two values of x in the range 0 ≤ x ≤ 360.

If cos x◦ = 0.7 then x = 45.6

cos is negative in the 2nd and 3rd quadrants

∴ x = 180 − 45.6 = 134.4

and x = 180 + 45.6 = 225.6

Further symmetry properties

Negative angles: y

x–1 1

–1

0

1

θ–θ

cos (−�) = cos �

sin (−�) = −sin �

tan (−�) = − sin �

cos �= −tan �

Complementary angles:

sin(�

2− �

)= cos �, sin

(�

2+ �

)= cos �

cos(�

2− �

)= sin �, cos

(�

2+ �

)= −sin �


SAMPLE

P1: FXS/ABE P2: FXS


Review


Exact values of circular functions

� sin � cos � tan �

0 0 1 0

�

6

1

2

√3

2

1√3

�

4

1√2

1√2

1

�

3

√3

2

1

2

√3

�

21 0 undefined

Graphs of circular functions

y

1

0

–1

y

1

0

–1

y

1

–1 0

π2π

θ

Amplitude = 1Period = 2π

Amplitude is undefined Period = π

Amplitude = 1Period = 2π

y = tanθ

y = sinθ y = cosθ

θ

θ

2π

2ππ–π2

π4

π2

π2

3π2

3π2

5π2

π


SAMPLE

P1: FXS/ABE P2: FXS


Rev

iew


Graphs of circular functions of the type y = a sin n(t ± ε) ± b and y = a cos n(t ± ε) ± b

1

–1

–2

–3

–π6

–π3

π6

π3

y

t0

e.g. y = 2 cos 3(

t + �

3

)− 1

Amplitude, a = 2

Period = 2�

n= 2�

3

The graph is the same shape as y = 2 cos (3t)

but is translated:

i�

3units in the negative direction of the t-axis, and

ii 1 unit in the negative direction of the y-axis.

Pythagorean identity

cos2 � + sin2 � = 1

Multiple-choice questions

1 In a right-angled triangle, the two shorter side lengths are 3 cm and 4 cm. To the nearest

degree, the value of the smallest angle is

A 1◦ B 23◦ C 37◦ D 53◦ E 92◦

2 The minimum value of 3 − 10 cos (2x) is

A −13 B −17 C −23 D −7 E −10

3 The range of the function f : [0, 2�] → R, f (x) = 4 sin(

2x − �

2

)is

A R B [0, 4] C [−4, 0] D [0, 8] E [−4, 4]

4 The period of the graph of y = 3 sin

(1

2x − �

)+ 4 is

A � B 3 C 4� D � + 4 E 2�

5 The graph of y = sin x is dilated by factor1

2from the y-axis and translated

�

4units in the

positive direction of the x-axis. The equation of the image is

A y = sin

(1

2x + �

4

)B y = sin

(1

2x − �

4

)C y = 2 sin

(x − �

4

)

D y = sin(

2x − �

4

)E y = sin

(2

(x − �

4

))6 The period of the function f : R → R, where f (x) = a sin (bx) + c and a, b and c are positive

constants, is

A a B b C2�

aD

2�

bE

b

2�7 One cycle of the graph of function with equation y = tan ax has vertical asymptotes at

x = −�

6and x = �

6. A possible value of a is

A 6 B � C�

6D

1

3E 3


SAMPLE

P1: FXS/ABE P2: FXS


Review


8 The equation 3 sin (x) + 1 = b, where b is a positive real number, has one solution in the

interval [0, 2�]. The value of b is

A 1 B 1.5 C 2 D 3 E 4

9 The number of solutions of the equation b = a sin x, where x ∈ [−2�, 2�] and a and b are

positive real numbers with a > b, is

A 2 B 3 C 4 D 5 E 6

10 The depth of water, in metres, in a harbour at a certain point at time t hours is given by

D(t) = 8 + 2 sin

(�t

6

), 0 ≤ t ≤ 24. The depth of the water is first 9 m at

A t = 0 B t = 1 C t = 2 D t = 3 E t = 4

Short-answer questions (technology-free)

1 Change each of the following to radian measure in terms of �:

a 330◦ b 810◦ c 1080◦ d 1035◦ e 135◦

f 405◦ g 390◦ h 420◦ i 80◦

2 Change each of the following to degree measure:

a5�c

6b

7�c

4c

11�c

4d

3�c

12e

15�c

2

f −3�c

4g −�c

4h −11�c

4i −23�c

43 Give exact values of each of the following:

a sin

(11�

4

)b cos

(−7�

4

)c sin

(11�

6

)d cos

(−7�

6

)

e cos

(13�

6

)f sin

(23�

6

)g cos

(−23

3�

)h sin

(−17

4�

)

4 State the amplitude and period of each of the following:

a 2 sin

(�

2

)b −3 sin (4�) c

1

2sin (3�)

d −3 cos (2x) e −4 sin( x

3

)f

2

3sin

(2x

3

)

5 Sketch the graphs of each of the following (showing one cycle):

a y = 2 sin (2x) b y = −3 cos( x

3

)c y = −2 sin (3x)

d y = 2 sin( x

3

)e y = sin

(x − �

4

)f y = sin

(x + 2�

3

)

g y = 2 cos

(x − 5�

6

)h y = −3 cos

(x + �

6

)


SAMPLE

P1: FXS/ABE P2: FXS


Rev

iew


6 Solve each of the following equations for R:

a sin � = −√

3

2, � ∈ [−�, �] b sin(2�) = −

√3

2, � ∈ [−�, �]

c sin(

� − �

3

)= −1

2, � ∈ [0, 2�] d sin

(� + �

3

)= −1, � ∈ [0, 2�]

e sin(�

3− �

)= −1

2, � ∈ [0, 2�]

Extended-response questions

1 The number of hours of daylight at a point on the Antarctic Circle is given approximately by

d = 12 + 12 cos1

6�

(t + 1

3

), where t is the number of months which have elapsed since

1 January.

a Find d:

i on 21 June (t ≈ 5.7) ii on 21 March (t ≈ 2.7)

b When will there be 5 hours of daylight?

2 The temperature A◦C inside a house at t hours after 4 am is given by A = 21 − 3 cos

(�t

12

)

for 0 ≤ t ≤ 24, and the temperature B◦C outside the house at the same time is given by

B = 22 − 5 cos

(�t

12

)for 0 ≤ t ≤ 24.

a Find the temperature inside the house at 8 am.

b Write down an expression for D = A − B, the difference between the inside and outside

temperatures.

c Sketch the graph of D for 0 ≤ t ≤ 24.

d Determine when the inside temperature is less than the outside temperature.

3 At a certain time of the year the depth of water d m in the harbour at Bunk Island is given by

the rule d = 3 + 1.8 cos(�

6t)

, where t is the time in hours after 3 am.

a Sketch the graph of the function d = 3 + 1.8 cos(�

6t)

over a 24-hour period from

3 am to 3 am.

b At what time(s) does high tide occur for t ∈ [0, 24]?

c At what time(s) does low tide occur for t ∈ [0, 24]?

A passenger ferry operates between Main Beach and Bunk Island. It takes 50 minutes to

go from Main Beach to Bunk Island. The ferry only runs between the hours of 8 am

and 8 pm and is only able to enter the harbour at Bunk Island if the depth of water is at

least 2 metres.

d What is the earliest time the ferry should leave Main Beach so that it arrives at Bunk

Island and can immediately enter the harbour?


SAMPLE

P1: FXS/ABE P2: FXS


Review


e The time to go from Bunk Island to Main Beach is also 50 minutes. The minimum time

the ferry takes at Bunk Island harbour is 5 minutes. The minimum time at Main Beach

is also 5 minutes.

i What is the latest time the ferry can leave Main Beach to complete a round trip in

105 minutes?

ii How many complete round trips can the ferry make in a day?

4 The depth of water D at the end of Brighton pier t hours after low tide is given by the rule

D = p − 2 cos (rt), where p and r are suitable constants.

At low tide (t = 0) the depth is 2 metres; at high tide, which occurs 8 hours later, the depth

is 6 metres.

a Show that r = �

8and p = 4.

b Sketch the graph of D = 4 − 2 cos(�

8t)

for 0 ≤ t ≤ 16.

c If the first low tide occurs at 4 am, when will the next low tide occur?

d At what times will the depth be equal to 4 metres?

The poles that support the Brighton pier stand 7.5 metres above the sea bed.

e How much of a particular pole is exposed at:

i high tide? ii 2 pm?

Over the years mussels have attached themselves to the pole. A particular mussel is

attached 4 metres from the top of the pole so that some of the time it is exposed and

some of the time it is covered by water.

f For how long will the mussel be covered by water during the time from one low tide to

the next?


SAMPLE

Circular Functions · 2008-12-10 · Circular Functions Objectives To use radians and degrees for...

Documents

Transcript of Circular Functions · 2008-12-10 · Circular Functions Objectives To use radians and degrees for...