Download - Unit3 Trigonometric MATH2(D) Ikbn
7/29/2019 Unit3 Trigonometric MATH2(D) Ikbn
http://slidepdf.com/reader/full/unit3-trigonometric-math2d-ikbn 1/53
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
7/29/2019 Unit3 Trigonometric MATH2(D) Ikbn
http://slidepdf.com/reader/full/unit3-trigonometric-math2d-ikbn 2/53
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
7/29/2019 Unit3 Trigonometric MATH2(D) Ikbn
http://slidepdf.com/reader/full/unit3-trigonometric-math2d-ikbn 3/53
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
7/29/2019 Unit3 Trigonometric MATH2(D) Ikbn
http://slidepdf.com/reader/full/unit3-trigonometric-math2d-ikbn 4/53
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
7/29/2019 Unit3 Trigonometric MATH2(D) Ikbn
http://slidepdf.com/reader/full/unit3-trigonometric-math2d-ikbn 5/53
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
7/29/2019 Unit3 Trigonometric MATH2(D) Ikbn
http://slidepdf.com/reader/full/unit3-trigonometric-math2d-ikbn 6/53
.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
7/29/2019 Unit3 Trigonometric MATH2(D) Ikbn
http://slidepdf.com/reader/full/unit3-trigonometric-math2d-ikbn 7/53
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
7/29/2019 Unit3 Trigonometric MATH2(D) Ikbn
http://slidepdf.com/reader/full/unit3-trigonometric-math2d-ikbn 8/53
.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
7/29/2019 Unit3 Trigonometric MATH2(D) Ikbn
http://slidepdf.com/reader/full/unit3-trigonometric-math2d-ikbn 9/53
( ) ( )
( )
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
'
7/29/2019 Unit3 Trigonometric MATH2(D) Ikbn
http://slidepdf.com/reader/full/unit3-trigonometric-math2d-ikbn 10/53
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
7/29/2019 Unit3 Trigonometric MATH2(D) Ikbn
http://slidepdf.com/reader/full/unit3-trigonometric-math2d-ikbn 11/53
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
7/29/2019 Unit3 Trigonometric MATH2(D) Ikbn
http://slidepdf.com/reader/full/unit3-trigonometric-math2d-ikbn 12/53
(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
7/29/2019 Unit3 Trigonometric MATH2(D) Ikbn
http://slidepdf.com/reader/full/unit3-trigonometric-math2d-ikbn 13/53
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.
7/29/2019 Unit3 Trigonometric MATH2(D) Ikbn
http://slidepdf.com/reader/full/unit3-trigonometric-math2d-ikbn 14/53
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.
7/29/2019 Unit3 Trigonometric MATH2(D) Ikbn
http://slidepdf.com/reader/full/unit3-trigonometric-math2d-ikbn 15/53
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
7/29/2019 Unit3 Trigonometric MATH2(D) Ikbn
http://slidepdf.com/reader/full/unit3-trigonometric-math2d-ikbn 16/53
(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 +
7/29/2019 Unit3 Trigonometric MATH2(D) Ikbn
http://slidepdf.com/reader/full/unit3-trigonometric-math2d-ikbn 17/53
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
7/29/2019 Unit3 Trigonometric MATH2(D) Ikbn
http://slidepdf.com/reader/full/unit3-trigonometric-math2d-ikbn 18/53
We can read off from the graph (Figure 3.13):
)decimal2in(77.0310sin −≈°
)decimal2in(64.0310cos š
)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
7/29/2019 Unit3 Trigonometric MATH2(D) Ikbn
http://slidepdf.com/reader/full/unit3-trigonometric-math2d-ikbn 19/53
87
°0
°90
°270
°180
1st Quadrant2nd Quadrant
3rd Quadrant 4th Quadrant
θ −°180θ
°180
θ
°−180θ
θ °180
θ −°360
θ
°360
7/29/2019 Unit3 Trigonometric MATH2(D) Ikbn
http://slidepdf.com/reader/full/unit3-trigonometric-math2d-ikbn 20/53
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
7/29/2019 Unit3 Trigonometric MATH2(D) Ikbn
http://slidepdf.com/reader/full/unit3-trigonometric-math2d-ikbn 21/53
(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
7/29/2019 Unit3 Trigonometric MATH2(D) Ikbn
http://slidepdf.com/reader/full/unit3-trigonometric-math2d-ikbn 22/53
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
7/29/2019 Unit3 Trigonometric MATH2(D) Ikbn
http://slidepdf.com/reader/full/unit3-trigonometric-math2d-ikbn 23/53
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
7/29/2019 Unit3 Trigonometric MATH2(D) Ikbn
http://slidepdf.com/reader/full/unit3-trigonometric-math2d-ikbn 24/53
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
7/29/2019 Unit3 Trigonometric MATH2(D) Ikbn
http://slidepdf.com/reader/full/unit3-trigonometric-math2d-ikbn 25/53
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
7/29/2019 Unit3 Trigonometric MATH2(D) Ikbn
http://slidepdf.com/reader/full/unit3-trigonometric-math2d-ikbn 26/53
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
°150 is in the second quadrant.
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
7/29/2019 Unit3 Trigonometric MATH2(D) Ikbn
http://slidepdf.com/reader/full/unit3-trigonometric-math2d-ikbn 27/53
155.1−=
(b)
Trigonometric Functions Reason
°°=°=° 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
°150 is in the second quadrant.
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)
Trigonometric Functions Reason
°°=°
150cos
150sin150tan
θ
θ θ
cos
sintan =
°−°=
30cos
30sin Based on Figure 3.14, the positive angle of
°150 is in the second quadrant.
From Figure 3.15, 2nd Quadrant
°<<° 18090θ
Only sin is positive
( )
( )
( )
( )θ θ −°−=°=°
°−°=°∴−°=
180coscos
30sin150sin
150180sin150sin
θ180sinθsin
( )
( )°−=°°−°−=°∴
30cos150cos
150180cos150cos
95
7/29/2019 Unit3 Trigonometric MATH2(D) Ikbn
http://slidepdf.com/reader/full/unit3-trigonometric-math2d-ikbn 28/53
( )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
°150 is in the second quadrant.
From Figure 3.15, 2nd Quadrant
°<<° 18090 θ
Only sin is positive
( )θ θ −°−= 180tantan
( )
( )°−=°
°−°−=°∴
30tan150tan
150180tan150tan
3
1−=
To find the values of tangent of °30 , use
Figure 3.8
(b)
Trigonometric Functions Reason
°=°
240sin
1240cos ec
θ θ
sin
1cos =ec
°−=
60sin
1 Based on Figure 3.14, the positive angle of
°240 is in the third quadrant.
From Figure 3.15,
3rd Quadrant °<<° 270180 θ
Only tan is positive
( )( )( )°−=°
°−°−=°∴°−−=
60sin240sin
180240sin240sin
180sinsin θ θ
( )2/31
−=
To find the values of sine of °60 , use
Figure 3.8
3
2−=
(c)
Trigonometric Functions Reason
96
7/29/2019 Unit3 Trigonometric MATH2(D) Ikbn
http://slidepdf.com/reader/full/unit3-trigonometric-math2d-ikbn 29/53
°=°
315tan
1315cot
θ θ
tan
1cot =
°−=
45tan
1 Based on Figure 3.14, the positive angle of
°315 is in the fourth quadrant.
From Figure 3.15, 4th Quadrant
°<<° 360270 θ
Only cos is positive
( )θ θ −°−= 360tantan
( )
( )°−=°
°°−°−=°∴
45tan315tan
315360tan315tan
1
1
1
1−=
−= To find the values of tangent of °45 , use
Figure 3.9
Or
Trigonometric Functions Reason
°°=
°=°
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
Only cos is positive
( )
( )
( )
( )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
7/29/2019 Unit3 Trigonometric MATH2(D) Ikbn
http://slidepdf.com/reader/full/unit3-trigonometric-math2d-ikbn 30/53
(a) °155sec (b)
− π 6
13cosec
Solution:
(a)
Trigonometric Functions Reason
°=°
155cos
1155sec
θ θ
cos
1sec =
°−=
25cos
1 Based on Figure 3.14, the positive angle of
°155 is in the second quadrant.
From Figure 3.15,
2nd Quadrant °<<° 18090 θ
Only sin is positive
( )θ θ −°−= 180coscos
( )
( )°−=°°−°−=°∴
25cos155cos
155180cos155cos
906.0
1
−= To find the values of cosine of °25 , use
calculator.104.1−=
(b)
Trigonometric Functions Reason
( )°−=
− 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
7/29/2019 Unit3 Trigonometric MATH2(D) Ikbn
http://slidepdf.com/reader/full/unit3-trigonometric-math2d-ikbn 31/53
1. 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) °240sec (b) °240tan
Solution
(a)
Trigonometric Functions Reason=°240sec
θ θ
cos
1sec =
= Based on Figure 3.14, the positive angle of
°240 is in the third quadrant.
From Figure 3.15,
3rd
Quadrant °<<° 270180 θ
Only tan is positive
( )
( )
( )°−=°°−°−=°∴
°−−=
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
°240 is in the third quadrant.
From Figure 3.15,
3rd Quadrant °<<° 270180 θ
Only tan is positive
( )
( )( )°=° °−°=°∴
°−=
60tan240tan180240tan240tan
180θtanθtan
= To find the values of tangent of °60 , use
Figure 3.83=
99
7/29/2019 Unit3 Trigonometric MATH2(D) Ikbn
http://slidepdf.com/reader/full/unit3-trigonometric-math2d-ikbn 32/53
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
°260 is in the third quadrant.
From Figure 3.15, 3rd Quadrant
°<<° 270180 θ
Only tan is positive
( )( )( )°−=°
°−°−=°∴ °−−=
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
7/29/2019 Unit3 Trigonometric MATH2(D) Ikbn
http://slidepdf.com/reader/full/unit3-trigonometric-math2d-ikbn 33/53
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.
7/29/2019 Unit3 Trigonometric MATH2(D) Ikbn
http://slidepdf.com/reader/full/unit3-trigonometric-math2d-ikbn 34/53
(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.
7/29/2019 Unit3 Trigonometric MATH2(D) Ikbn
http://slidepdf.com/reader/full/unit3-trigonometric-math2d-ikbn 35/53
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
°=360 x is repeated for each complete cycle. Hence, the
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
=°
=°
−=°
=°
=°
7/29/2019 Unit3 Trigonometric MATH2(D) Ikbn
http://slidepdf.com/reader/full/unit3-trigonometric-math2d-ikbn 36/53
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
°=180 x is repeated for each complete cycle. Hence, the
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=+•
=+•
=+•
7/29/2019 Unit3 Trigonometric MATH2(D) Ikbn
http://slidepdf.com/reader/full/unit3-trigonometric-math2d-ikbn 37/53
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
7/29/2019 Unit3 Trigonometric MATH2(D) Ikbn
http://slidepdf.com/reader/full/unit3-trigonometric-math2d-ikbn 38/53
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)
Trigonometric Identities Reason
LHS:
( ) 1tanseccos 222 −− y y yec
LHS represents left-hand side
( ) 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 −
LHS represents left-hand side
z
z z
z
z z
2
22
cos
sintan
cos
sintan
=∴
=
106
7/29/2019 Unit3 Trigonometric MATH2(D) Ikbn
http://slidepdf.com/reader/full/unit3-trigonometric-math2d-ikbn 39/53
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)
Trigonometric Identities Reason
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
7/29/2019 Unit3 Trigonometric MATH2(D) Ikbn
http://slidepdf.com/reader/full/unit3-trigonometric-math2d-ikbn 40/53
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
Trigonometric Identities Reason
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
7/29/2019 Unit3 Trigonometric MATH2(D) Ikbn
http://slidepdf.com/reader/full/unit3-trigonometric-math2d-ikbn 41/53
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
7/29/2019 Unit3 Trigonometric MATH2(D) Ikbn
http://slidepdf.com/reader/full/unit3-trigonometric-math2d-ikbn 42/53
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
7/29/2019 Unit3 Trigonometric MATH2(D) Ikbn
http://slidepdf.com/reader/full/unit3-trigonometric-math2d-ikbn 43/53
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
7/29/2019 Unit3 Trigonometric MATH2(D) Ikbn
http://slidepdf.com/reader/full/unit3-trigonometric-math2d-ikbn 44/53
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
7/29/2019 Unit3 Trigonometric MATH2(D) Ikbn
http://slidepdf.com/reader/full/unit3-trigonometric-math2d-ikbn 45/53
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
7/29/2019 Unit3 Trigonometric MATH2(D) Ikbn
http://slidepdf.com/reader/full/unit3-trigonometric-math2d-ikbn 46/53
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
7/29/2019 Unit3 Trigonometric MATH2(D) Ikbn
http://slidepdf.com/reader/full/unit3-trigonometric-math2d-ikbn 47/53
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
7/29/2019 Unit3 Trigonometric MATH2(D) Ikbn
http://slidepdf.com/reader/full/unit3-trigonometric-math2d-ikbn 48/53
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 ,
find the value of each of the following trigonometric functions. State your answers
in terms of surds (square roots).
(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
trigonometric equations.
(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
trigonometric equations.
(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
7/29/2019 Unit3 Trigonometric MATH2(D) Ikbn
http://slidepdf.com/reader/full/unit3-trigonometric-math2d-ikbn 49/53
(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
7/29/2019 Unit3 Trigonometric MATH2(D) Ikbn
http://slidepdf.com/reader/full/unit3-trigonometric-math2d-ikbn 50/53
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
7/29/2019 Unit3 Trigonometric MATH2(D) Ikbn
http://slidepdf.com/reader/full/unit3-trigonometric-math2d-ikbn 51/53
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
7/29/2019 Unit3 Trigonometric MATH2(D) Ikbn
http://slidepdf.com/reader/full/unit3-trigonometric-math2d-ikbn 52/53
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
7/29/2019 Unit3 Trigonometric MATH2(D) Ikbn
http://slidepdf.com/reader/full/unit3-trigonometric-math2d-ikbn 53/53