Unit3 Trigonometric MATH2(D) Ikbn

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UNIT 3

TRIGONOMETRY

3.1 Introduction

Trigonometry (from Greek trigōnon "triangle" + metron "measure") is a branch

of mathematics that deals with triangles, particularly triangles in a plane where

one angle of the triangle is 90 degrees (right angled triangles). It specifically deals

with the relationships between the sides and the angles of triangles; the

trigonometric functions, and calculations based upon them. The insights of

trigonometry permeate other branches of geometry, such as the study of spheres

using spherical trigonometry.

Trigonometry has important applications in many branches of pure mathematics

as well as of applied mathematics and, consequently remains applicable in natural

sciences. Trigonometry is usually taught in secondary schools, often in a

precalculus course.

Objectives

At the end of the topic, you will be able to:

• Determine the values of trigonometric ratios for any acute angle.

• Verify trigonometric identities.

• Learn what a (trigonometric) identity is and how to solve trigonometric

equations.

•

Learn and memorize the basic identities involving sine and cosine that areresult of the definitions of tangent, cotangent, secant, and cosecant.

• Learn the sine and cosine of the negative of an angle measure.

69

http://en.wikipedia.org/wiki/Greek_language

http://en.wikipedia.org/wiki/Mathematics

http://en.wikipedia.org/wiki/Triangle_(geometry)

http://en.wikipedia.org/wiki/Plane_(mathematics)

http://en.wikipedia.org/wiki/Angle

http://en.wikipedia.org/wiki/Right_angled_triangle

http://en.wikipedia.org/wiki/Trigonometric_function

http://en.wikipedia.org/wiki/Sphere

http://en.wikipedia.org/wiki/Spherical_trigonometry

http://en.wikipedia.org/wiki/Branches

http://en.wikipedia.org/wiki/Pure_mathematics

http://en.wikipedia.org/wiki/Applied_mathematics

http://en.wikipedia.org/wiki/Science

http://en.wikipedia.org/wiki/Secondary_school

http://en.wikipedia.org/wiki/Precalculus

http://en.wikipedia.org/wiki/Greek_language

http://en.wikipedia.org/wiki/Mathematics

http://en.wikipedia.org/wiki/Triangle_(geometry)

http://en.wikipedia.org/wiki/Plane_(mathematics)

http://en.wikipedia.org/wiki/Angle

http://en.wikipedia.org/wiki/Right_angled_triangle

http://en.wikipedia.org/wiki/Trigonometric_function

http://en.wikipedia.org/wiki/Sphere

http://en.wikipedia.org/wiki/Spherical_trigonometry

http://en.wikipedia.org/wiki/Branches

http://en.wikipedia.org/wiki/Pure_mathematics

http://en.wikipedia.org/wiki/Applied_mathematics

http://en.wikipedia.org/wiki/Science

http://en.wikipedia.org/wiki/Secondary_school

http://en.wikipedia.org/wiki/Precalculus



3.2 Trigonometric ratio

Figure 3.1

In a right-angled triangle as shown in Figure 3.1,

(a) the hypotenuse is the side opposite to the right angle,

(b) the opposite side of the angle is the side opposite to the angle ,

(c) the adjacent side of the angle is the side beside the angle .

We define the trigonometric ratios as:

sine of angle asAB

AC

hypotenuse

opposite= - this ratio is denoted by θ sin .

cosine of angle asABBC

hypotenuseadjacent = - this ratio is denoted by cos .

tangent of angle asBC

AC

adjacent

opposite= - this ratio is denoted by θ tan .

Example 3.1

θ

opposite side

adjacent side

hypotenuse

B

A

C

70



From the figure below, find the values of θ θ sin,cos and θ tan .

Figure 3.2

Solution:

Given .17

8

hypotenuse

adjacentcos ===

PR

PQθ

To find the values of θ sin

andθ

tan , we use Pythagoras’ Theorem,

225

817 22

222

222

=−=

−=

+=

PQ PR RQ

PQ RQ PR

15

225

=

= RQ

Therefore,17

15

hypotenuse

oppositesin ===

PR

RQθ

Therefore,815

adjacentoppositetan ===

PQ RQθ .

Example 3.2

Given that25

24sin =θ , calculate the values of θ cos and θ tan .

71

θ

8

17

P

R

Q



Solution:

Figure 3.3

To find the values of θ cos and θ tan , we use Pythagoras’ Theorem,

49

2425 22

222

222

=−=

−=+= RQ PR PQ

PQ RQ PR

7

49

=

= PQ

Therefore,25

7

hypotenuse

adjacentcos ===

PR

PQθ

Therefore,7

24

adjacent

oppositetan ===

PQ

RQθ .

Example 3.3

Find the length of the side labeled z in each of the triangles below, given

3

1cos =θ .

72

θ

2425

P

R

Q

θ

z

9 cm



Figure 3.4

Solution:

.cm3

cm93

1

cm93

1

hypotenuse

adjacentcos

=

×=

===

z

z θ

Example 3.4Find the length of the side labeled y in each of the triangles below, given

8

5sin =θ .

Figure 3.5

Solution:

73

θ y

17.5 cm



.cm28

cm5.1758

8

5

cm5.17

cm5.17

8

5

hypotenuse

oppositesin

=

×=

=

===

y

yθ

Example 3.5

Find the length of the side labeled x in each of the triangles below and calculate

the value of θ tan . Given that5

3sin =θ

Figure 3.6

Solution:

To find the length of the side labeled x, we use Pythagoras’ Theorem,

16925

35 22

222

222

=−=−=

−=

+=

RQ PR PQ

PQ RQ PR

cm4

16

=∴

=

x

PQ

4

3

adjacent

oppositetan ===∴

PQ

RQθ .

Example 3.6

74

θ x

3 cm

5 cm

P

Q

R

x 17 cm8 cm

P RQ y

S



Figure 3.7

In the diagram, PQR is a straight line. Given that5

3cos = x and

17

15cos = y , find

the length of QR.

Solution:

To find the length of the PR, we use Pythagoras’ Theorem,

22564289

817 22

222

222

=−=−=

−= +=SP SR PR PRSP SR

cm15

225

=

= PR

.cm3

40

3

5cm8

5

3cm8

cm8

5

3

hypotenuse

adjacentcos

=×=

=

====

SQ

SQSQ

SP x

To find the length of the PQ, we use Pythagoras’ Theorem,

9

102464

9

1600

83

40 2

2

222

222

=−=

−=

−=+=

SP SQ PQ

PQSP SQ

75



.cm3

32

9

1024

=

= PQ

So that, to find the length of the QR, we use PQ PR−

cm.3

13

cm3

32cm15

=

−=

−= PQ PRQR

Practice 3.1

In the diagram, PQR and SRT are straight lines. Given that5

4sin = x and 1tan = y ,

the length of SRT , in cm is

Solution

( )

( )

.cm4

cm5

5

4

hypotenuse

oppositesin

=

×=

===

SR

SR x

To find the length of the QR, we use Pythagoras’ Theorem,

( ) ( )

( )91625

52

222

2

=−=−=

−=

+=

SRQS QR

QS

( )

cm3=

=QR

So that, to find the length of the PQR, we use QR PQ +

76



( ) ( )

( )

cm.5

cm3

=+=+= PQR

Then, find the length of the RT

( ) cm51

cm.51

adjacent

oppositetan

=×=∴

====

RT

RT

PR

RT y

Finally, to find the length of the SRT , we use RT SR +

( ) ( )

( )

cm.9

cm4

=+=+=SRT

3.2.1 Converting the Units of Measurement of Angles.

Angles are measured in the units of degrees ( )° and minutes ( )' . The

relationship between degrees and minutes is shown below

Example 3.7

Convert °4.21 into degrees and minutes.

Solution:

( )

.2421

2421

604.021

4.0214.21

'

'

'

°=

+°=

×°+°=

°+°=°

Example 3.8

Convert '632° into degrees.

77

°

=

=°

60

11

'601

'



Solution:

.1.32

1.032

60

1632

632632

'

''

°= °+°=

°

×+°=

+°=°

Practice 3.2

1. Convert the unit of the following angles into degrees and minutes.

(a) °9.19

Solution

( ) ( )

( )[ ]( ) ( )

'5419

'

9.019

9.19

°=

+°=

×°+°=

°+°=°

2. Convert the unit of the following angles into degrees.

(a) '4877°

Solution

( ) ( )

( ) ( )

°=

°+°=

°

×+°=

+°=°

8.77

60

14877

4877

'

''

2.2.5 Finding the Values of Tangent, Sine and Cosine of

°°° 60and45,30 without Using a Scientific Calculator.

Angles °°° 60and45,30 are special angles. To find the values of tangent, sine

and cosine of °°° 60and45,30 , use the following equilateral triangle to help

you.

78



Figure 3.8 Figure 3.9

From the triangles in Figure above, the table below is obtained

Table 3.1

Example 3.9

Without using a scientific calculator (by using Figure 3.8 and Figure 3.9), find the

values of

(a) °30tan . (b) °45sin . (c) °60cos .

(d) °45tan . (e) °30sin . (f) °60sin .

Solution:

(a) From the Figure 3.8,3

1

adjacent

opposite30tan ==°

(b) From the Figure 3.9,2

1

hypotenuse

opposite45sin ==°

(c) From the Figure 3.8,2

1

hypotenuse

adjacent60cos ==°

(d) From the Figure 3.9, 11

1

adjacent

opposite45tan ===°

(e) From the Figure 3.8,2

1

hypotenuse

opposite30sin ==°

°30 °45 °60

Tan31 1 3

Sin2

1

2

1

2

3

Cos2

3

2

1

2

1

79

°60

32

1

°30

°45

12

1

°45



(f) From the Figure 3.8,2

3

hypotenuse

opposite60sin ==°

Practice 3.3

By using Figure 3.8 and Figure 3.9, fill in the blank to find the values of

(a) °60tan . (b) °45cos . (c) °30cos .

Solution

(a) From the Figure 3.8,( )

( )3

adjacent60tan ==°

(b) From the Figure 3.9,( )

( )

==°hypotenuse

adjacent45cos

(c) From the Figure 3.8, ( )( )2

adjacent30cos ==°

2.3.5 Finding the Values of Tangent, Sine and Cosine Using a

Scientific Calculator.

Example 3.10

By using a scientific calculator, find the values of (a) °27tan (b) °2.52sin (c)

'4270cos °

Solution:

(a) Press 27

Screen display :

510.027tan =°∴ (in 3 decimal places)

(b) Press °2.52

Screen display :

80

tan 27

0.509525449

make sure that the modeused is the ‘Degree

Mode’. Press MODE

MODE MODE MODE1 to set the calculator to

the ‘Degree Mode’.

sin 52.20.790155012

,,,° ,,,° ,,,° ,,,°

0.330514392



790.02.52sin =°∴ (in 3 decimal places)

(c) Press70

42

Screen display :

331.0'4270cos =°∴ (in 3 decimal places)

Practice 3.4

By using a scientific calculator, find the values of

(a) °80cos . (b) °1.35tan . (c)'

3271cos ° .

Solution

(a) =°80cos .

(b) =°1.35tan .

(c) =°'3271cos .

3.4 Finding the Angles Using a Scientific Calculator.

Example 3.11

Find the value of in each of the following cases.

(a) 4

3

tan =θ

. (b)8.0sin =θ

. (c)2677.0cos

=θ

.

Solution:

(a)

Press 3 4

81

SHIFT a b/c

3 436.86989765

Press after the answer in

degrees to display the angle in

degrees and minutes.



Screen display :

°=∴ 870.36θ (in 3 decimal places) – in degrees

'5236°=∴θ – in degrees and minutes

(b) 8.0sin =θ

.753

130.53'

°=∴

°=∴

θ

θ

(c) 2677.0cos =θ

.2874

473.74

'°=∴

°=∴

θ

θ

Practice 3.5

Find the value of for 7065.0sin =θ by using a scientific calculator.

Solution

7065.0sin =θ

=∴

=∴−

θ

θ 7065.0sin1

=∴θ

2.4.5 Reciprocal ratios

In addition to the three trigonometrically ratios there are three reciprocal ratios,

namely:

82

.sincos

tan1cot

,cos

1sec

,sin

1cos

θ

θ

θ θ

θ θ

θ θ

==

=

=ec

(in 3 decimal places) – in degrees.

in degrees and minutes.



The values of these for a given angle can also be found using a calculator by

finding the appropriate trigonometric ratio and then pressing the reciprocal key (If

°63cot , press 1/ °63tan in your calculator).

Example 3.12

By using a scientific calculator, find the values of

(a) °12cot (b) °37sec (c) °71cosec

Solution:

(a) 705.412tan

112cot =

°=°

(b) 252.137cos

137sec =

°=°

(c) 058.171sin

171cos =

°=°ec (in 3 decimal places).

Example 3.13

Without using a scientific calculator (by using Figure 3.8 and Figure 3.9), find the

values of

(a) °45sec (b) °30cot (c) °60cosec

Solution:

(a)2

2

1

1

45cos

145sec =

=

°=°

83



(b)3

3

1

1

30tan

130cot =

=

°=°

(c) 3

2

23

1

60sin

160cos =

=

°

=°ec

Practice 3.6

1. Without using a scientific calculator (by using Figure 3.8 and Figure 3.9), fill in

the following brackets.

(a) °60cot . (b) °30sec . (c) °45cosec .

Solution

(a) ( ) ( ) 3

11160cot ===°

(b) ( ) ( ) 3

21130sec ===°

(c) ( ) ( )2

1145cos ===°ec

3.3 Trigonometric Equation

3.3.1 Positive and Negative Angles

Positive angle is an angle measured in the anticlockwise direction from the

positive x-axis as shown in the circular diagram.

Figure 3.10

Negative angle is an angle measured in the clockwise direction from the positive

x-axis as shown in the circular diagram.

84

x

y

0

θ +



Figure 3.11

The triangle moves all round the circle so we can measure the sine, cosine and

tangent for angles up to °360 .

Figure 3.12

Figure 3.13

x°360°0

°270

°180 1

-1

-1

The circle has a

radius of 1 unit.

85

x

y

0

θ −

y°901

x

°310 °0

°270

°180 1

-1

-1

y

1 °90

P

oppositehypotenuse

adjacent



We can read off from the graph (Figure 3.13):

)decimal2in(77.0310sin −≈°

)decimal2in(64.0310cos ≈°

)decimal2in(19.1310tan −≈° .

The following diagram shows the signs of the trigonometric values for each

quadrant between °0 and °360 .

Figure 3.14

86

°0

°90

°270

°180

1st Quadrant

All are positive

2nd Quadrant

Only sin is positive

3rd Quadrant

Only tan is positive

4th Quadrant

Only cos is positive



87

°0

°90

°270

°180

1st Quadrant2nd Quadrant

3rd Quadrant 4th Quadrant

θ −°180θ

°180

θ

°−180θ

θ °180

θ −°360

θ

°360



Figure 3.15

Example 3.14

Represent each of the following angles using a circular diagram and state the

quadrant where the angle is in.

(a) °500 (b) π 9

22radians

(c) °−505 (d) π 9

2− radians

Solution:

(a) The positive angle of °500 is represented by the following circular

diagram (Figure 3.16).

Figure 3.16

°+°=° 140360500

Based on the above circular diagram (Figure 3.16), the positive angle of

°500 is in the second quadrant.

88

x

y

°140



(b) The positive angle of π 9

22radians is represented by the following

circular diagram (Figure 3.17).

Figure 3.17

°+°=

+= 80360rad9

42rad

9

22π π π

Based on the above circular diagram (Figure 3.17), the positive angle of

π 9

22radians is in the first quadrant.

(c) The negative angle of °505 is represented by the following circular

diagram (Figure 3.18).

89

x

y

rad

rad

x°− 505

y

°−145



Figure 3.18

°−°−=°− 145360505

Based on the above circular diagram (Figure 3.18), the negative angle of

°505 is in the third quadrant.

(d) The negative angle of π 92 radians is represented by the following

circular diagram (Figure 3.19).

Figure 3.19

°== 04rad9

2rad9

2π π

Based on the above circular diagram (Figure 3.19), the negative angle of

π 92 radians is in the fourth quadrant.

Practice 3.7

90

x

y

rad



Represent each of the following angles using a circular diagram and state the quadrant

where the angle is in.

(a) °595 (b) °870

(c)π

4

15

radians (d)π

6

11

− radians

(a) (b)

(c) (d)

3.3.2 Radian Measure

An alternative unit of measure of an angle is the radian. If a straight line of length

r rotates about one end so that the other end describes an arc of length r , the line is

said to have rotated through 1 radian = 1 rad.

91

x

y y

x

x

y

x

y



Figure 3.20

Because the are described when the line rotates through a full angle is the

circumference of a circle which measures r π 2 , the number of radians in a full

angle is π 2 rad. Consequently, relating degrees to radians we see that:

To convert from degrees to radians, divide by °360 and multiply by 2 times pi (

π ). To convert from radians to degrees, divide by 2 times pi (π ) and multiply

by °360 .

Example 3.15

Write

(a) °30 (b) °225

in radians.

Solution:

(a)66

1

360

230 π π

π ==

°×°

(b)45

45

3602225 π

π π ==

°×°

You should be able to memorize some standard conversions like below:

°0 °30 °45 °60 °90 °120 °150 °180 °225 °270 °330 °360

92

r r

r

1 radian

rad0.01751thatSo

rad....2831.6

rad2360

=°

=

=° π



0

6

π

4

π

3

π

2

π

3

2π

6

5π π

4

5π

2

3π

6

11π π 2

Table 3.2

Example 3.16

Convert6

11π radians to degrees.

Solution

°=

°×

330

2

3606

11

π

π

.

Practice 3.8

1. Convert4

5π radians to degrees.

Solution

[ ]

[ ]

°=

×

2254

5π

.

2. Match the degrees below with the suitable radians.

93



3.3.3 Trigonometric Functions of Any Angle

Example 3.17

Given that 5.0150sin =° and 866.0150cos −=° , find the values of

(a) °150sec (b) °150cot

Solution:

(a)

Trigonometric Functions Reason

°=°

150cos

1150sec

θ θ

cos

1sec =

866.0

1

−= Based on Figure 3.14, the positive angle of


From Figure 3.15, 2nd Quadrant

°<<° 18090 θ

Only sin is positive0.86601cos −=°5 got from the given

value of cosine of °015 .

Degrees

°0

°30

°45

°60

°90

°120

°150

°180

Radians

•6

π

•6

5π

•3

π

• 0

• π

•2

π

•4

π

•3

2π

94



155.1−=

(b)


°°=°=° 150sin150cos

150tan1150cot

θ θ

θ

θ θ θ

sincos

cos

sin1

tan1cot =

==

5.0

866.0−=

732.1−=

Based on Figure 3.14, the positive angle of


From Figure 3.15,

2nd Quadrant °<<° 18090 θ

Only sin is positive0.86601cos −=°5 and 0.501sin =°5

got from the given values of cosine and

sine of °015 .

Example 3.18

Using the values of the trigonometric ratios of the angles °° 45,30 and °60 ,

find the value of each of the following trigonometric functions. State your answers

in terms of surds (square roots).

(a) °150tan (b) °240cosec (c) °315cot

Solution:(a)


°°=°

150cos

150sin150tan

θ

θ θ

cos

sintan =

°−°=

30cos

30sin Based on Figure 3.14, the positive angle of



°<<° 18090θ


( )

( )

( )

( )θ θ −°−=°=°

°−°=°∴−°=

180coscos

30sin150sin

150180sin150sin

θ180sinθsin

( )

( )°−=°°−°−=°∴

30cos150cos

150180cos150cos

95



( )2/3

2/1

−= To find the values of sine and cosine of

°30 , use Figure 3.8

3

1−=

Or

Trigonometric Functions Reason°−=° 30tan150tan Based on Figure 3.14, the positive angle of



°<<° 18090 θ


( )θ θ −°−= 180tantan

( )

( )°−=°

°−°−=°∴

30tan150tan

150180tan150tan

3

1−=

To find the values of tangent of °30 , use

Figure 3.8

(b)


°=°

240sin

1240cos ec

θ θ

sin

1cos =ec

°−=

60sin

1 Based on Figure 3.14, the positive angle of


From Figure 3.15,

3rd Quadrant °<<° 270180 θ


( )( )( )°−=°

°−°−=°∴°−−=

60sin240sin

180240sin240sin

180sinsin θ θ

( )2/31

−=

To find the values of sine of °60 , use

Figure 3.8

3

2−=

(c)


96



°=°

315tan

1315cot

θ θ

tan

1cot =

°−=

45tan


°315 is in the fourth quadrant.

From Figure 3.15, 4th Quadrant

°<<° 360270 θ


( )θ θ −°−= 360tantan

( )

( )°−=°

°°−°−=°∴

45tan315tan

315360tan315tan

1

1

1

1−=

−= To find the values of tangent of °45 , use

Figure 3.9

Or


°°=

°=°

315sin

315cos

315tan

1315cot

θ

θ

θ

θ θ θ

sin

cos

cos

sin

1

tan

1cot =

==

°−°=

45sin

45cos Based on Figure 3.14, the positive angle of

°315 is in the fourth quadrant.

From Figure 3.15, 4th Quadrant

°<<° 360270 θ


( )

( )

( )

( )θ360cosθcos −°=°−=°

°−°−=°∴−°−=

45sin315sin

315360sin315sin

360sinsin θ θ

( )

( )°=°°−°=°∴

45cos315cos

315360cos315cos

( ) 12/1

2/1

−=−=To find the values of sine and cosine of

°45 , use Figure 3.9

Example 3.19

Express each of the following trigonometric functions in terms of the

trigonometric ratios of acute angles. Hence, find each value using a calculator.

97



(a) °155sec (b)

− π 6

13cosec

Solution:

(a)


°=°

155cos

1155sec

θ θ

cos

1sec =

°−=

25cos



From Figure 3.15,

2nd Quadrant °<<° 18090 θ


( )θ θ −°−= 180coscos

( )

( )°−=°°−°−=°∴

25cos155cos

155180cos155cos

906.0

1

−= To find the values of cosine of °25 , use

calculator.104.1−=

(b)


( )°−=

− 390cos6

13

cos ecec π

°×−=

− 1806

13

cos6

13

cos ecec π

( )( )°−

=°−390sin

1390cosec

θ θ sin

1cos =ec

( ) ( )°−=

°− 390sin

1

390sin

1 The formula ( ) ( )θ θ sinsin −=− is used.

( )°−=

30sin

1

5.0

1

−=

To find the values of sine of °30 , use

calculator.2−=

Practice 3.9

98



1. Using the values of the trigonometric ratios of the angles °° 45,30 and °60 ,



(a) °240sec (b) °240tan

Solution

(a)

Trigonometric Functions Reason=°240sec

θ θ

cos

1sec =

= Based on Figure 3.14, the positive angle of


From Figure 3.15,

3rd

Quadrant °<<° 270180 θ


( )

( )

( )°−=°°−°−=°∴

°−−=

60cos240cos

180240cos240cos

180coscos θ θ

= To find the values of cosine of °60 , use

Figure 3.82−=

(b)

Trigonometric Functions Reason=°240tan Based on Figure 3.14, the positive angle of


From Figure 3.15,

3rd Quadrant °<<° 270180 θ


( )

( )( )°=° °−°=°∴

°−=

60tan240tan180240tan240tan

180θtanθtan

= To find the values of tangent of °60 , use

Figure 3.83=

99



2. Express the following trigonometric functions below in terms of the trigonometric

ratios of acute angles. Hence, find each value using a calculator.

°260cosec

Solution

Trigonometric Functions Reason=°260cos ec

θ θ

sin

1cos =ec

= Based on Figure 3.14, the positive angle of


From Figure 3.15, 3rd Quadrant

°<<° 270180 θ


( )( )( )°−=°

°−°−=°∴ °−−=

80sin260sin

180260sin260sin180sinsin

θ θ

= To find the values of sine of °80 , use

calculator.015.1−=

3.3.4 Simple Trigonometric Equations

The steps to solve simple trigonometric equations are as follows:

1. Determine the quadrants the angle should be in based on the given

trigonometric equation.

2. Find the basic angle using a scientific calculator.3. Determine the range of values of the required angles, for examples the

range of values of θ 2 or θ 3 .

4. Determine the values of angles in those quadrants.

Example 3.20

100



Find all the angles between °0 and °360 that satisfy each of the following

trigonometric equations.

(a) 5427.0sin =θ (b) 6725.1tan −=θ

(c) 7123.02cos = x (d) ( ) 6582.0102sin −=°− x

Solution:

(a) 5427.0sin =θ

Basic °=∠ 87.32

°°=∴ 13.147,87.32θ

Reason:

5427.0sin =θ is positive in the first and second quadrants.

(b) 6725.1tan −=θ

Basic °=∠ 12.59 (ignore the negative sign of 1.6725, when finding the

basic angle using a calculator.)

°°=∴ 88.300,88.120θ

(c) 7123.02cos = x

Basic °=∠ 58.44

°°°°=∴ 42.675,58.404,42.315,58.442 x

°°°°=∴ 71.337,29.202,71.157,29.22 x

101

Press

0.5427

°−° 87.32180

is negative in the second and fourth quadrants. foquadrants.

°−° 12.59180 °−° 12.59360

°−° 58.44360 ( )°−°+° 58.44360360°+° 58.44360

is positive in the first and fourth quadrants. foquadrants.



(d) ( ) 6582.0102sin −=°− x

Basic °=∠ 16.41

°°°°=°−∴ 84.678,16.581,84.318,16.221102 x

°+°°+°°+°°+°=∴ 1084.678,1016.581,1084.318,1016.2212 x

°°°°=∴ 84.688,16.591,84.328,16.2312 x

°°°°=∴ 42.344,58.295,42.164,58.115 x

Practice 3.10

Find all the angles between °0 and °360 that satisfy 9015.0cos =θ .

Solution

9015.0cos =θ

Basic ( )=∠

( ) ( )°°=∴ ,θ

3.4 Trigonometric Identities

3.4.1 Graphs of the Functions of Sine, Cosine and Tangent

The sketch of the graph of y = sin x is as shown below.

102

period

°+° 16.41180 °+° 16.221360 °−° 16.41360°+° 84.318360

is negative in the third and fourth quadrants. foquadrants.



Graph 3.1

Based on the above Graph 3.1 of x y sin= ,

The shape of the graph of x y sin= from °=0 x to

°=360 x is repeated for each complete cycle. Hence, the

function x y sin= is periodic with a period of °360 .

The maximum and minimum values of the function

x y sin= are 1 and -1 respectively.

The sketch of the graph of y = cos x is as shown below.

Graph 3.2

Based on the above Graph 3.2 of x y cos=

,The shape of the graph of x y cos= from °=0 x to


function x y cos= is periodic with a period of °360 .

The maximum and minimum values of the function

x y cos= are 1 and -1 respectively.

The sketch of the graph of y = tan x is as shown below.

103

0360sin

1270sin0180sin

190sin

00sin

=°

−=°

=°

=°

=°

period

1360cos

0270cos

1180cos

090cos

10cos

=°

=°

−=°

=°

=°



Graph 3.3

Based on the above Graph 3.3 of x y tan= ,

The shape of the graph of x y tan= from °=0 x to


function x y tan= is periodic with a period of .180°

The function x y tan= does not have any maximum or

minimum values. As x approaches

°°°° 630,450,270,90 and so on, the function

x y tan

= approaches ∞ (positive or negative).

3.4.2 Basic Trigonometric Identities

Three basic trigonometric identities are:

Basic identities are also known as Pythagorean identities.

Proof for the identities 1cossin 22 =+ θ θ

104

period period

Asymptote

0360tan

270tan

0180tan

90tan

00tan

=°

±∞=°

=°

±∞=°

=°

θ θ

θ θ

θ θ

22

22

22

coscot1

sec1tan

1cossin

ec=+•

=+•

=+•



Based on the right-angled triangle in the above diagram, using the Pythagoras’

Theorem:

222 cba =+

2

2

2

2

2

2

c

c

c

b

c

a=+ (the whole equation is divided by .2c )

Therefore, 1cossin 22 =+ θ θ

Proof for the identities θ θ 22 sec1tan =+

From 1cossin 22 =+ θ θ

θ θ

θ

θ

θ

22

2

2

2

cos

1

cos

cos

cos

sin=+ (the whole equation is divided by .cos2θ )

Therefore, θ θ 22sec1tan =+

Proof for the identities θ θ 22 coscot1 ec=+

From 1cossin 22 =+ θ θ

θ θ

θ

θ

θ

22

2

2

2

sin

1

sin

cos

sin

sin=+ (the whole equation is divided by θ

2sin )

Therefore, θ θ 22

coscot1 ec=+

Example 3.21

Prove each of the following trigonometric identities.

(a) x x

xsin1

sin1

cos2

+=−

.

(b) ( ) y y y yec2222cot1tanseccos =−− .

105

a

θ

c

b



Solution:

(a)

Trigonometric Identities Reason

LHS:

x

x

sin1

cos2

−

LHS represents left-hand side

x

x

sin1

sin12

−− Since 1cossin 22 =+ θ θ ,

Thus θ θ 22 sin1cos −=

( )( )

x

x x

sin1

sin1sin1

−−+ Factorize the numerator

( ) xsin1+ = RHS RHS represents right-hand side

(b)


LHS:

( ) 1tanseccos 222 −− y y yec


( ) 11cos 2 − yec Since θ θ 22 sec1tan =+ ,

Thus 1tansec 22 =− θ θ

1cos 2− yec Since θ θ

22 coscot1 ec=+ ,

Thus 1coscot 22 −= θ θ ec

y2cot = RHS RHS represents right-hand side

Practice 3.11

Prove the following trigonometric identities by fill in the blank.

z z z z 2222sintansintan =−

Solution

Trigonometric Identities ReasonLHS:

z z 22 sintan −


z

z z

z

z z

2

22

cos

sintan

cos

sintan

=∴

=

106



z

z z z 2

222

cos

cossinsin −

( ) z

z z 2

22

cos

cos1sin −

Since1cossin 22 =+ θ θ

,Thus θ θ

22 cos1sin −=

( ) z z

z 2

2

2

sincos

sin

z z 22 sintan = RHS RHS represents right-hand side

Example 3.22

Find all the angles that satisfy each of the following equations for .3600 °≤≤° x

(a) 1costan2 = x x .

(b) 0tansin2 =− x x .

Solution:

(a)


1costan2 = x x -

1coscos

sin2 =

x

x

x x

x x

cos

sintan =

1coscos

sin2 =

x

x

x The xcos cancels

1sin2 = x

2

1sin = x

Divide by 2

.150,302

1sin

1 °°=

=∴ −

x

(b)

Trigonometric Identities Reason0tansin2 =− x x -

0cos

sinsin2 =

−

x

x x

x

x x

cos

sintan =

107



0cos

sin

cos

cossin2=

−

x

x

x

x x Multiply through by xcos

0sincossin2 =− x x x The xcos cancels( ) 01cos2sin =− x x Factorize

Either 0sin = x or

5.0cos01cos2 =⇒=− x x

Either one or the other factor equals zero.

( ) .360,180,00sin1 °°°==∴ −

x

or

( ) .300,605.0cos 1 °°==∴ − x

Practice 3.12

Fill in the blank and find all the angles that satisfy the following equation for

.3600 °≤≤° x

4cos5sin2 2 =+ x x

Solution


4cos5sin2 2 =+ x x -

( ) 4cos52 =+ x Since 1cossin 22 =+ x x

x x 22 cos1sin −=( ) 4cos5 =+ x Multiply out

02cos5cos22 =+− x x Set LHS = 0

( ) ( ) 0= Factorize

Either 2cos = x (impossible) or

( )=⇒=− x x cos01cos2

Either one or the other factor equals zero.

( ) ( ) ( ) °°==∴ −,5.0cos

1 x

108



3.5 Application of trigonometry

Marine sextants like this are used to measure the angle of the sun or stars with

respect to the horizon. Using trigonometry and a marine chronometer , the position

of the ship can then be determined from several such measurements.

There are an enormous number of applications of trigonometry and trigonometric

functions. For instance, the technique of triangulation is used in astronomy to

measure the distance to nearby stars, in geography to measure distances between

landmarks, and in satellite navigation systems. The sine and cosine functions are

fundamental to the theory of periodic functions such as those that describe sound

and light waves.

Fields which make use of trigonometry or trigonometric functions include

astronomy (especially, for locating the apparent positions of celestial objects, in

which spherical trigonometry is essential) and hence navigation (on the oceans, in

aircraft, and in space), music theory, acoustics, optics, analysis of financial

markets, electronics, probability theory, statistics, biology, medical imaging (CAT

scans and ultrasound), pharmacy, chemistry, number theory (and hence

cryptology), seismology, meteorology, oceanography, many physical sciences,

land surveying and geodesy, architecture, phonetics, economics, electrical

engineering, mechanical engineering, civil engineering, computer graphics,

cartography, crystallography and game development.

109

http://en.wikipedia.org/wiki/Sextant

http://en.wikipedia.org/wiki/Marine_chronometer

http://en.wikipedia.org/wiki/Uses_of_trigonometry

http://en.wikipedia.org/wiki/Triangulation

http://en.wikipedia.org/wiki/Astronomy

http://en.wikipedia.org/wiki/Geography

http://en.wikipedia.org/wiki/Satellite_navigation_system

http://en.wikipedia.org/wiki/Periodic_function

http://en.wikipedia.org/wiki/Light


http://en.wikipedia.org/wiki/Navigation

http://en.wikipedia.org/wiki/Music_theory

http://en.wikipedia.org/wiki/Acoustics

http://en.wikipedia.org/wiki/Optics

http://en.wikipedia.org/wiki/Electronics

http://en.wikipedia.org/wiki/Probability_theory

http://en.wikipedia.org/wiki/Statistics

http://en.wikipedia.org/wiki/Biology

http://en.wikipedia.org/wiki/Medical_imaging

http://en.wikipedia.org/wiki/CAT_scan


http://en.wikipedia.org/wiki/Ultrasound

http://en.wikipedia.org/wiki/Pharmacy

http://en.wikipedia.org/wiki/Chemistry

http://en.wikipedia.org/wiki/Number_theory

http://en.wikipedia.org/wiki/Cryptology

http://en.wikipedia.org/wiki/Seismology

http://en.wikipedia.org/wiki/Meteorology

http://en.wikipedia.org/wiki/Oceanography

http://en.wikipedia.org/wiki/Physical_science

http://en.wikipedia.org/wiki/Surveying

http://en.wikipedia.org/wiki/Geodesy

http://en.wikipedia.org/wiki/Architecture

http://en.wikipedia.org/wiki/Phonetics

http://en.wikipedia.org/wiki/Economics

http://en.wikipedia.org/wiki/Electrical_engineering


http://en.wikipedia.org/wiki/Mechanical_engineering

http://en.wikipedia.org/wiki/Civil_engineering

http://en.wikipedia.org/wiki/Computer_graphics

http://en.wikipedia.org/wiki/Cartography

http://en.wikipedia.org/wiki/Crystallography

http://en.wikipedia.org/wiki/Game_development

http://en.wikipedia.org/wiki/Image:Frieberger_drum_marine_sextant.jpg

http://en.wikipedia.org/wiki/Sextant

http://en.wikipedia.org/wiki/Marine_chronometer

http://en.wikipedia.org/wiki/Uses_of_trigonometry

http://en.wikipedia.org/wiki/Triangulation


http://en.wikipedia.org/wiki/Geography

http://en.wikipedia.org/wiki/Satellite_navigation_system

http://en.wikipedia.org/wiki/Periodic_function

http://en.wikipedia.org/wiki/Light


http://en.wikipedia.org/wiki/Navigation

http://en.wikipedia.org/wiki/Music_theory

http://en.wikipedia.org/wiki/Acoustics

http://en.wikipedia.org/wiki/Optics

http://en.wikipedia.org/wiki/Electronics

http://en.wikipedia.org/wiki/Probability_theory

http://en.wikipedia.org/wiki/Statistics

http://en.wikipedia.org/wiki/Biology

http://en.wikipedia.org/wiki/Medical_imaging



http://en.wikipedia.org/wiki/Ultrasound

http://en.wikipedia.org/wiki/Pharmacy

http://en.wikipedia.org/wiki/Chemistry

http://en.wikipedia.org/wiki/Number_theory

http://en.wikipedia.org/wiki/Cryptology

http://en.wikipedia.org/wiki/Seismology

http://en.wikipedia.org/wiki/Meteorology

http://en.wikipedia.org/wiki/Oceanography

http://en.wikipedia.org/wiki/Physical_science

http://en.wikipedia.org/wiki/Surveying

http://en.wikipedia.org/wiki/Geodesy

http://en.wikipedia.org/wiki/Architecture

http://en.wikipedia.org/wiki/Phonetics

http://en.wikipedia.org/wiki/Economics



http://en.wikipedia.org/wiki/Mechanical_engineering

http://en.wikipedia.org/wiki/Civil_engineering

http://en.wikipedia.org/wiki/Computer_graphics

http://en.wikipedia.org/wiki/Cartography

http://en.wikipedia.org/wiki/Crystallography

http://en.wikipedia.org/wiki/Game_development



Example 3.23

If the distance of a person from a tower is 100 m and the angle subtended by the

top of the tower with the ground is 30o, what is the height of the tower in meters?

Steps:

• Draw a simple diagram to represent the problem. Label it carefully and clearly

mark out the quantities that are given and those which have to be calculated.

Denote the unknown dimension by say h if you are calculating height or by x if

you are calculating distance.

• Identify which trigonometric function represents a ratio of the side about which

information is given and the side whose dimensions we have to find out. Set up

a trigonometric equation.

•

Substitute the value of the trigonometric function and solve the equation for theunknown variable.

Solution:

• AB = distance of the man from the tower = 100 m

• BC = height of the tower = h (to be calculated)

• The trigonometric function that uses AB and BC is tan A, where A = 30o.

So tan 30o

= BC / AB = h / 100

Therefore height of the tower h = 100 tan 30o = (100)

3

1= 57.74 m.

Example 3.24

110



Akmarul slid down a slope inclined at an angle of °15 to the ground. If the top of

the slope is at a height of 12 m from the ground, find the distance, correct to 1

decimal place, traveled by Akmarul.

Solution:

Let the distance traveled by Akmarul be x m.

2588.0

12

15sin

12

1215sin

=

°=

=°

x

x

= 46.4 m

Hence, the distance traveled by Akmarul was 46.4 m.

Example 3.25

12 m

°15

111

12 m

°15

m



A tree casts a horizontal shadow 38 m long. If a line were to be drawn from the

end of the shadow to the top of the tree it would be inclined to the horizontal at

°60 . The height of the tree is obtained as follows:

360tan shadowof length

treeof height=°=

So that

2438383shadowof length3treeof height=×=×=×=

m.

Example 3.26

You are stationed at a radar base and you observe an unidentified plane at an

altitude h = 5000 m flying towards your radar base at an angle of elevation = 30o.

After exactly one minute, your radar sweep reveals that the plane is now at an

angle of elevation = 60o maintaining the same altitude. What is the speed of the

plane?

Solution:

°60

m

112



In the figure, the radar base is at point A. The plane is at point D in the first sweep

and at point E in the second sweep. The distance it covers in the one minute

interval is DE.

From the figure,

tan DAC = tan 30o = DC / AC = h / AC.

Similarly,

tan EAB = tan 60o = EB / AB = h / AB.

Distance covered by the plane in one minute = DE = AC - AB

= (h / tan 30o) - (h / tan 60o)

= (5000 3 ) - (5000 / 3 ) = 5773.50 m.

The velocity of the plane is given by V

= distance covered / time taken

= DE / 60 = 96.23 m/s.

Example 3.27

Two men on opposite sides of a TV tower of height 30 m notice the angle of

elevation of the top of this tower to be 45o and 60o respectively. Find the distance

between the two men.

Solution:

113



The situation is depicted in the figure with CD representing the tower and AB

being the distance between the two men.

For triangle ACD,

tan A = tan 60o = CD / AD.

Similarly for triangle BCD,

tan B = tan 45o = CD / DB.

The distance between the two men isAB = AD + DB

= (CD / tan 60o) + (CD / tan 45o)

= (30 / 3 ) + (30 / 1) = 47.32 m.

EXERCISE

114



Trigonometric Ratios

1.

Based on the given figure, find h and .

2.

Based on the given figure, find the length of RQ and RS correct to 1 decimal

place.

3. By using a scientific calculator, find the value of each of the following.

(a) °12tan (b) °1.35tan (c) '563sin °

4. By using a scientific calculator, find the value of in degrees and minutes for

each of the following cases.

(a)2

1tan =θ (b) 3.0sin =θ (c) 5298.0cos =θ

Trigonometric Equation

115

θ

°55

12 cm

20 cm

h

R

S

P Q°10

°30

5 m



5. Represent each of the following angles using a circular diagram and state the

quadrant where the angle is in.

(a) °440 (b) π 3

8radians (c) π

4

3− radians

6. Given that 3420.020sin =° and 9397.020cos =° , find the values of

(a) °20tan (b) °20cot

7. Given that 8660.03

2sin =π and 5.0

3

2cos −=π , find the values of

(a) π 3

2sec (b) π

3

2cosec

8. Express each of the following trigonometric functions in terms of the

trigonometric ratios of acute angles. Hence, find each value using a calculator.

(a) °235sec (b)

π 6

11cot (c) °488sin

(d) )880(cos °− (e) )672(tan °−

9. Using the values of the trigonometric ratios of the angles °° 45,30 and °60 ,



(a) °150cos (b) °225sin (c) °240tan

(d) π 4

7cos (e) − π

3

5cosec

10. Find all the values of for °<<° 3600 θ that satisfy each of the following


(a) 6137.0sin =θ (b) 7825.2tan =θ

(c) 7283.0cos −=θ (d) 3569.2tan −=θ

11. Find all the angles between °0 and °360 that satisfy each of the following


(a) 5293.02sin =θ (b) 4673.02cos −=θ

(c) ( ) 7402.030tan =°+ x (d) ( ) 8803.0352sin −=°− x

12. Without using tables or a calculator, find all the angles between °0 and °360

that satisfy each of the following trigonometric equations.

(a) °= 67cossin x (b) °= 42sincos x

116



(c) °= 87cottan x (d) °= 53cossec ec x

(e) °−= 42cossec ec x (f) °−= 63cos2sin x

Trigonometric Identities

13. Prove each of the following trigonometric identities.

(a) x x

x 2

2

2

sintan1

tan=

+

(b) x xec x x seccoscottan =+

(c) z z z ec

z ec 2secsincos

cos=

−

(d) θ θ θ

2cos2cos1

1

cos1

1ec=

−+

+

(e) z z z z 2222cotcoscoscot =−

(f) y y y

y 22

2

2

sincostan1

tan1−=

+−

(g) x x xec

xcos1

cotcos

sin+=

−

14. Find all the angles that satisfy each of the following equations for .3600 °≤≤° x

(a) x x cos2cot −=

(b) x x cottan16 =

(c) 01sin2sin3 2 =−− x x

(d) 1cossin2 += xec x

(e) x x cos1sec2 +=

(f) ( ) x x tan15sec32 +=

Answers to exercise:

1. h = 8.4 cm '5124°=θ

2. RQ = 6.5 m

RS = 3.3 m3. (a) 0.213 3. (b) 0.703

3. (c) 0.069

4. (a) '3426° 4. (b) '2717°

4. (c) °58

117



6. (a) 0.364 6. (b) 2.748

7. (a) -2 7. (b) 1.155

8. (a) -1.743 8. (b) -1.7328. (c) 0.788 8. (d) -0.940

8. (e) 1.111

9. (a)23− 9. (b)

21−

9. (c) 39. (d)

2

1

9. (e)3

2

10 (a) °° 14.142,86.37 10 (b) °° 23.250,23.70

10 (c) °° 26.223,74.136 10 (d) °° 99.292,99.112

11 (a)°°°° 02.254,98.195,02.74,98.15

11 (b)°°°° 07.301,93.238,07.121,93.58

11 (c) °° 51.186,51.6 11 (d)

°°°° 66.346,34.318,66.166,34.13812 (a) °° 157,23 12 (b) °° 312,48

12 (c) °° 183,3 12 (d) °° 323,37

12 (e) °° 228,132 12 (f)°°°° 5.346,5.283,5.166,5.103

14 (a) °°°° 330,270,210,90 14 (b)°°°° 96.345,04.194,96.165,04.14

14 (c) °°° 53.340,47.199,90 14 (d) °°° 330,210,90

14 (e) °° 360,0 14 (f)°°°° 57.341,43.243,57.161,43.63

Activity

1.

118



A ladder is leaned against a wall at an angle of °65 from the ground. The foot

of the ladder is 3.8 m from the foot of wall. Find

(a) the length of the ladder,

(b) the height of the ladder up the wall, correct to 1 decimal place.

2.

A crew on two yachts observe the light of a lighthouse each at an angle of

elevation of °

28 and °42 as shown in the above figure. If the two yachts andthe lighthouse lie on a straight line, find the distance between the two yachts to

the nearest metres.

°65

3.8 m

119

32 m

°42 °28



3.

An aeroplane takes off at an angle of °33 from the ground. How far will it

travel in the air to reach a height of 500 m? Given your answer to the nearest

metres.

Answer

1. (a) 9.0 m (b) 8.1 m

2. 25 m.

3. 918 m.

°33

500 m

120



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