EXTRA EXERCISE FROM JOHN BIRDS BOOK

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Section 3

Trigonometry

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Chapter 22

Introduction to
trigonometry

22.1 Trigonometry By Pythagoras’ theorem: e2 = d2 + f 2
Hence 132 = d2 + 52
Trigonometry is the branch of mathematics that deals 169 = d2 + 25
with the measurement of sides and angles of trian- Thus d2 = 169 − 25 = 144
gles, and their relationship with each other. There are i.e.
many applications in engineering where knowledge of √
trigonometry is needed. d = 144 = 12 cm
EF = 12 cm

22.2 The theorem of Pythagoras Problem 2. Two aircraft leave an airfield at the
same time. One travels due north at an average
With reference to Fig. 22.1, the side opposite the speed of 300 km/h and the other due west at an
right angle (i.e. side b) is called the hypotenuse. The average speed of 220 km/h. Calculate their distance
theorem of Pythagoras states: apart after 4 hours

‘In any right-angle triangle, the square on the N B
hypotenuse is equal to the sum of the squares on the WE
other two sides.’

Hence b2 = a2 + c2 S 1200 km

A CA
880 km

c b Figure 22.3
B aC
After 4 hours, the first aircraft has travelled
Figure 22.1 4 × 300 = 1,200 km, due north, and the second aircraft
has travelled 4 × 220 = 880 km due west, as shown in
Problem 1. In Fig. 22.2, find the length of EF. Fig. 22.3. Distance apart after 4 hour = BC.
From Pythagoras’ theorem:
D e ϭ 13 cm
f ϭ 5 cm F BC2 = 12002 + 8802

E d = 1440000 + 774400 and
√
Figure 22.2
BC = 2214400

Hence distance apart after 4 hours = 1488 km

188 Engineering Mathematics

Now try the following exercise

7. Figure 22.5 shows a cross-section of a com-

ponent that is to be made from a round bar.

Exercise 84 Further problems on the If the diameter of the bar is 74 mm, calculate
theorem of Pythagoras
the dimension x. [24 mm]

1. In a triangle CDE, D = 90◦ , CD = 14.83 mm x

and CE = 28.31 mm. Determine the length

of DE. [24.11 mm]

Section 3 2. Triangle PQR is isosceles, Q being a right f 74 mm 72 mm
Figure 22.5
angle. If the hypotenuse is 38.47 cm find (a)

the lengths of sides PQ and QR, and (b) the
value of ∠QPR. [(a) 27.20 cm each (b) 45◦]

3. A man cycles 24 km due south and then 20 km

due east. Another man, starting at the same

time as the first man, cycles 32 km due east

and then 7 km due south. Find the distance

between the two men. [20.81 km]

4. A ladder 3.5 m long is placed against a

perpendicular wall with its foot 1.0 m from

the wall. How far up the wall (to the nearest 22.3 Trigonometric ratios of
acute angles
centimetre) does the ladder reach? If the foot
(a) With reference to the right-angled triangle shown
of the ladder is now moved 30 cm further in Fig. 22.6:

away from the wall, how far does the top of (i) sin θ = opposite side
hypotenuse
the ladder fall? [3.35 m, 10 cm]
i.e. sin θ = b
5. Two ships leave a port at the same time. One c
travels due west at 18.4 km/h and the other adjacent side
due south at 27.6 km/h. Calculate how far
apart the two ships are after 4 hours. (ii) cosine θ =
[132.7 km] hypotenuse

6. Figure 22.4 shows a bolt rounded off at one i.e. cos θ = a
end. Determine the dimension h. [2.94 mm] c

h r ϭ 16 mm (iii) tangent θ = opposite side
R ϭ 45 mm adjacent side

Figure 22.4 i.e. tan θ = b
a

(iv) secant θ = hypotenuse
adjacent side

i.e. sec θ = c
a

(v) cosecant θ = hypotenuse
opposite side
c

i.e. cosec θ =
b

Introduction to trigonometry 189

(vi) cotangent θ = adjacent side Since cos X = 9 , then XY = 9 units and
opposite side 41

i.e. cot θ = a XZ = 41 units.
b Using Pyt√hagoras’ theorem: 412 = 92 +YZ2 from

which YZ = 412 − 92 = 40 unts.

c Thus, sin X = 40 tan X = 40 = 4 ,
and , 4
b 41 9 9

q a cosec X = 41 =1 1 , sec X = 41 = 5
4
Figure 22.6 40 40 99
(b) From above,
cot X = 9 Section 3
40

b Problem 4. If sin θ = 0.625 and cos θ = 0.500
determine the values of cosec θ, sec θ, tan θ and cot θ
(i) sin θ = c = b = tan θ,
cos θ a a

c cosec θ = 1 = 1 = 1.60
sin θ 0.625
i.e. tan θ = sin θ
cos θ sec θ = 1 = 1 = 2.00
a cos θ 0.500

(ii) cos θ = c = a = cot θ, tan θ = sin θ = 0.625 = 1.25
sin θ b b cos θ 0.500

c cot θ = cos θ 0.500
i.e. cot θ = cos θ = = 0.80
sin θ 0.625
sin θ
1 f (x) f (x)
8 8
(iii) sec θ = 7
cos θ 6 B
6
(iv) cosec θ = 1 4 B
sin θ 3A 4 C
2
(Note ‘s’ and ‘c’ go together) 2A u
1

(v) cot θ =
tan θ

Secants, cosecants and cotangents are called the recip- 0 24 6 8 0 24 6 8
rocal ratios. (a) (b)

Figure 22.8

9 Problem 5. Point A lies at co-ordinate (2, 3) and
Problem 3. If cos X = determine the value of point B at (8, 7). Determine (a) the distance AB,
(b) the gradient of the straight line AB, and (c) the
41 angle AB makes with the horizontal
the other five trigonometric ratios

Figure 22.7 shows a right-angled triangle XYZ. (a) Points A and B are shown in Fig. 22.8(a).
In Fig. 22.8(b), the horizontal and vertical lines
Z AC and BC are constructed. Since ABC is a
right-angled triangle, and AC = (8 − 2) = 6 and
41 BC = (7 − 3) = 4, then by Pythagoras’ theorem:

XY AB2 = AC2 + BC2 = 62 + 42
9 √

Figure 22.7 and AB = 62 + 42 = 52 = 7.211,
correct to 3 decimal places

190 Engineering Mathematics

(b) The gradient of AB is given by tan θ, 22.4 Fractional and surd forms of
trigonometric ratios
i.e. gradient = tan θ = BC = 4 = 2
AC 6 3 In Fig. 22.11, ABC is an equilateral triangle of side 2

(c) The angle AB makes with the horizontal is units. AD bisects angle A and bisects the side BC. Using
given by: tan−1 2 = 33.69◦
3 Pythagoras’ theorem on triangle ABD gives:

√
AD = 22 − 12 = 3.

Now try the following exercise √
sin 30◦ = BD = 1 , cos 30◦ = AD = 3
Section 3 Hence, AB 2 AB 2

Exercise 85 Further problems on and tan 30◦ = BD = √1
trigonometric ratios of acute AD 3
angles √
sin 60◦ = AD = 3 , cos 60◦ = BD = 1
1. In triangle ABC shown in Fig. 22.9, find sin A, AB 2 AB 2
cos A, tan A, sin B, cos B and tan B.
and tan 60◦ = AD = √
3
BD
B
5 3 A
A C 30° 30°

Figure 22.9
⎡ ⎤
3 4 3 ⎦⎥⎥ 2 2
⎢⎢⎣ sin A , cos A , tan 4 3
sin B = = A = 4
= 5 = 5 B = 60° 60°
4 3 BIDIC

, cos B , tan
5 53
Figure 22.11
2. For the right-angled triangle shown in Fig.
22.10, find:

(a) sin α (b) cos θ (c) tan θ P

a 17 45° 2 R
8 I

q 45°
QI
15

Figure 22.10 Figure 22.12

15 15 8 In Fig. 22.12, PQR is an isosceles triangle
(a) (b) (c) with P√Q = QR =√1 unit. By Pythagoras’ theorem,
17 17 15 PR = 12 + 12 = 2

3. If cos A = 12 find sin A and tan A, in fraction Hence,
13
sin 45◦ = 1 , cos 45◦ = 1 and tan 45◦ = 1
form. √ √
55 22

sin A = , tan A =
13 12

4. Point P lies at co-ordinate (−3, 1) and point A quantity that is not exactly expressib√le as a r√ational
Q at (5, −4). Determine (a) the distance PQ, number is called a surd. For example, 2 and 3 are
(b) the gradient of the straight line PQ, and (c) called surds because they cannot be expressed as a frac-
the angle PQ makes with the horizontal. tion and the decimal part may be continued indefinitely.
For example,
[(a) 9.434 (b) −0.625 (c) −32◦]
√√
2 = 1.4142135 . . . , and 3 = 1.7320508 . . .

Introduction to trigonometry 191

From above, 4. (tan 45◦)(4 cos 60◦ − 2 sin 60◦) √
tan 60◦ − tan 30◦ [2 − 3]
sin 30◦ = cos 60◦, sin 45◦ = cos 45◦ and
sin 60◦ = cos 30◦. 5. 1 + tan 30◦ tan 60◦ √1
3
In general,
22.5 Solution of right-angled Section 3
sin θ = cos(90◦ − θ) and cos θ = sin(90◦ − θ) triangles

For example, it may be checked by calculator that To ‘solve a right-angled triangle’ means ‘to find the
sin 25◦ = cos 65◦ , sin 42◦ = cos 48◦ and unknown sides and angles’. This is achieved by using
cos 84◦ 10 = sin 5◦ 50 , and so on. (i) the theorem of Pythagoras, and/or (ii) trigonometric
ratios. This is demonstrated in the following problems.
Problem 6. Using surd forms, evaluate:
3 tan 60◦ − 2 cos 30◦ Problem 7. In triangle PQR shown in Fig. 22.13,
tan 30◦ find the lengths of PQ and PR.

√ √ 3 P
3, and
From above, tan 60◦ = cos 30◦ = 2

tan 30◦ = √1 , hence
3

√ −2 √
33 3

3 tan 60◦ − 2 cos 30◦ = 2

tan 30◦ √1 Q 38° R
7.5 cm
3 Figure 22.13
√√ √
3 3− 3 2 3
= 1 =1
√√
33
√ tan 38◦ = PQ = PQ hence
√ 3 = 2(3) = 6 QR 7.5
=2 3
1 PQ = 7.5 tan 38◦ = 7.5(0.7813)

Now try the following exercise = 5.860 cm
cos 38◦ = QR = 7.5 hence
Exercise 86 Further problems on
fractional and surd form of PR PR
trigonometrical ratios 7.5 7.5

Evaluate the following, without using calculators, PR = cos 38◦ = 0.7880 = 9.518 cm
leaving where necessary in surd form: [Check: Using Pythagoras’ theorem
(7.5)2 + (5.860)2 = 90.59 = (9.518)2]

1. 3 sin 30◦ − 2 cos 60◦ 1 Problem 8. Solve the triangle ABC shown in
2 Fig. 22.14
2. 5 tan 60◦ − 3 sin 60◦ 7√
tan 60◦ 3 35 mm
2 AB
3. 3 tan 30◦
[1] 37 mm
C
Figure 22.14

192 Engineering Mathematics

To ‘solve triangle ABC’ means ‘to find the length AC Area of triangle XYZ
and angles B and C’.
1
sin C = 35 = 0.94595 hence = (base)(perpendicular height)
37
2
∠C = sin−1 0.94595 = 71.08◦ or 71◦5
= 1 = 1
∠B = 180◦ − 90◦ − 71.08◦ = 18.92◦ or 18◦55 (since (XY )(XZ) (18.37)(7.906)
angles in a triangle add up to 180◦) 22

sin B = AC hence = 72.62 mm2
37
Section 3 Now try the following exercise
AC = 37 sin 18.92◦ = 37(0.3242)
Exercise 87 Further problems on the
= 12.0 mm, solution of right-angled
triangles
or, using Pyth√agoras’ theorem, 372 = 352 + AC2, from
which, AC = 372 − 352 = 12.0 mm 1. Solve triangle ABC in Fig. 22.16 (i).

Problem 9. Solve triangle XYZ given ∠X = 90◦, B D 3 cm
∠Y = 23◦17 and YZ = 20.0 mm. Determine also its
area 4 cm E G H

It is always advisable to make a reasonably accurate A 35° C 41° 15.0 mm
sketch so as to visualize the expected magnitudes of 5.0 cm I
unknown sides and angles. Such as sketch is shown in F
Fig. 22.15 (iii)
(i) (ii)
∠Z = 180◦ − 90◦ − 23◦17 = 66◦43
sin 23◦17 = XZ hence XZ = 20.0 sin 23◦17 Figure 22.16
[BC = 3.50 cm, AB = 6.10 cm, ∠B = 55◦]
20.0
= 20.0(0.3953) 2. Solve triangle DEF in Fig. 22.16 (ii).
= 7.906 mm [FE = 5 cm, ∠E = 53.13◦, ∠F = 36.87◦]

cos 23◦17 = XY hence XY = 20.0 cos 23◦17 3. Solve triangle GHI in Fig. 22.16(iii).
20.0
= 20.0(0.9186) [GH = 9.841 mm, GI = 11.32 mm,
= 18.37 mm ∠H = 49◦]

4. Solve the triangle JKL in Fig. 22.17 (i) and
find its area.

J M 25°35Ј 3.69 m Q
6.7 mm P

K 32.0 mm N 8.75 m
51° L R
O (iii)
(i) (ii)

Z Figure 22.17
20.0 mm
KL = 5.43 cm, JL = 8.62 cm,
∠J = 39◦, area = 18.19 cm2

X 23°17Ј Y 5. Solve the triangle MNO in Fig. 22.17(ii) and
find its area.
Figure 22.15
MN = 28.86 mm, NO = 13.82 mm.
[Check: Using Pythagoras’ theorem ∠O = 64.42◦, area = 199.4 mm2
(18.37)2 + (7.906)2 = 400.0 = (20.0)2]

6. Solve the triangle PQR in Fig. 22.17(iii) and Introduction to trigonometry 193
find its area.
PR = 7.934 m, ∠Q = 65.06◦, (Note, ∠PRQ is also φ – alternate angles between
∠R = 24.94◦, area = 14.64 m2 parallel lines.)

7. A ladder rests against the top of the perpendic- Problem 10. An electricity pylon stands on
horizontal ground. At a point 80 m from the base of
ular wall of a building and makes an angle of the pylon, the angle of elevation of the top of the
pylon is 23◦. Calculate the height of the pylon to
73˚ with the ground. If the foot of the ladder is the nearest metre

2 m from the wall, calculate the height of the Figure 22.21 shows the pylon AB and the angle of
elevation of A from point C is 23◦ and
building. [6.54 m] Section 3
tan 23◦ = AB = AB
8. Determine the length x in Fig. 22.18. BC 80
[9.40 mm]
A
10 mm
56°

x C 23° B
80 m
Figure 22.18

Figure 22.21

22.6 Angle of elevation and Hence, height of pylon AB = 80 tan 23◦
depression
= 80(0.4245) = 33.96 m
(a) If, in Fig. 22.19, BC represents horizontal ground = 34 m to the nearest metre.
and AB a vertical flagpole, then the angle of ele-
vation of the top of the flagpole, A, from the point Problem 11. A surveyor measures the angle of
C is the angle that the imaginary straight line AC
must be raised (or elevated) from the horizontal elevation of the top of a perpendicular building as
CB, i.e. angle θ. 19◦. He move 120 m nearer the building and finds
the angle of elevation is now 47◦. Determine the
A
height of the building

Cq B The building PQ and the angles of elevation are shown
in Fig. 22.22

Figure 22.19 In triangle PQS, tan 19◦ = x h
P + 120

f hence h = tan 19◦(x + 120)

QR i.e. h = 0.3443(x + 120) (1)

Figure 22.20 P

(b) If, in Fig. 22.20, PQ represents a vertical cliff and h 47° R 19° S
R a ship at sea, then the angle of depression of Q x 120
the ship from point P is the angle through which
the imaginary straight line PR must be lowered Figure 22.22
(or depressed) from the horizontal to the ship, i.e.
angle φ.

194 Engineering Mathematics

In triangle PQR, tan 47◦ = h Hence
x
(2) 129.9 + x = 75 = 75 = 206.0 m
hence h = tan 47◦(x), i.e. h = 1.0724x tan 20◦ 0.3640

Equating equations (1) and (2) gives: from which,

0.3443(x + 120) = 1.0724x x = 206.0 − 129.9 = 76.1 m

0.3443x + (0.3443)(120) = 1.0724x Thus the ship sails 76.1 m in 1 minute, i.e. 60 s, hence,

(0.3443)(120) = (1.0724 − 0.3443)x speed of ship = distance = 76.1
m/s
41.316 = 0.7281x time 60
Section 3 41.316
= 76.1 × 60 × 60
x = = 56.74 m 60 × 1,000 km/h
0.7281
= 4.57 km/h
From equation (2), height of building,
h = 1.0724x = 1.0724(56.74) = 60.85 m

Problem 12. The angle of depression of a ship Now try the following exercise

viewed at a particular instant from the top of a 75 m Exercise 88 Further problems on angles of
vertical cliff is 30◦. Find the distance of the ship elevation and depression

from the base of the cliff at this instant. The ship is 1. If the angle of elevation of the top of a vertical
30 m high aerial is 32◦, how far is it to the
sailing away from the cliff at constant speed and 1
aerial? [48 m]
minute later its angle of depression from the top of
the cliff is 20◦. Determine the speed of the ship in

km/h

2. From the top of a vertical cliff 80.0 m high the

30° angles of depression of two buoys lying due
A west of the cliff are 23◦ and 15◦, respectively.

20° How far are the buoys apart? [110.1 m]

75 m 3. From a point on horizontal ground a surveyor
B
30° 20° D measures the angle of elevation of the top of
Cx a flagpole as 18◦40 . He moves 50 m nearer

Figure 22.23 to the flagpole and measures the angle of ele-
vation as 26◦22 . Determine the height of the
Figure 22.23 shows the cliff AB, the initial position of the
ship at C and the final position at D. Since the angle of flagpole. [53.0 m]
depression is initially 30◦ then ∠ACB = 30◦ (alternate
angles between parallel lines). 4. A flagpole stands on the edge of the top of
a building. At a point 200 m from the build-
tan 30◦ = AB = 75 ing the angles of elevation of the top and
BC BC bottom of the pole are 32◦ and 30◦ respectively.
Calculate the height of the flagpole.
[9.50 m]

hence BC = 75 = 75 5. From a ship at sea, the angle of elevation of the
tan 30◦ 0.5774
top and bottom of a vertical lighthouse stand-
= 129.9 m ing on the edge of a vertical cliff are 31◦ and
26◦, respectively. If the lighthouse is 25.0 m
= initial position of ship from base of cliff
high, calculate the height of the cliff.

[107.8 m]

In triangle ABD, 6. From a window 4.2 m above horizontal ground
tan 20◦ = AB = 75 = 75 the angle of depression of the foot of a building
BD BC + CD 129.9 + x

Introduction to trigonometry 195

across the road is 24◦ and the angle of elevation secant 32◦ = 1 = 1.1792
of the top of the building is 34◦. Determine, cos 32◦
correct to the nearest centimetre, the width of
the road and the height of the building. cosecant 75◦ = 1 = 1.0353
sin 75◦
[9.43 m, 10.56 m]
cotangent 41◦ = 1 = 1.1504
7. The elevation of a tower from two points, one tan 41◦
due east of the tower and the other due west
of it are 20◦ and 24◦, respectively, and the two secant 215.12◦ = cos 1 = −1.2226
points of observation are 300 m apart. Find the 215.12◦
height of the tower to the nearest metre.
[60 m] cosecant 321.62◦ = sin 1 = −1.6106 Section 3
321.62◦

cotangent 263.59◦ = tan 1 = 0.1123
263.59◦

22.7 Evaluating trigonometric ratios Problem 13. Evaluate correct to 4 decimal places:
of any angles
(a) sine 168◦14 (b) cosine 271.41◦
(c) tangent 98◦4

The easiest method of evaluating trigonometric func- (a) sine 168◦14 = sine 14◦ = 0.2039
tions of any angle is by using a calculator. The 168
following values, correct to 4 decimal places, may be 60
checked:
(b) cosine 271.41◦ = 0.0246
sine 18◦ = 0.3090,
sine 172◦ = 0.1392 (c) tangent 98◦4 = 4◦ = −7.0558
sine 241.63◦ = −0.8799, tan 98
cosine 56◦ = 0.5592 60
cosine 115◦ = −0.4226,
cosine 331.78◦ = 0.8811 Problem 14. Evaluate, correct to 4 decimal
tangent 29◦ = 0.5543, places:
tangent 178◦ = −0.0349
tangent 296.42◦ = −2.0127 (a) secant 161◦ (b) secant 302◦29

(a) sec 161◦ = 1 = −1.0576
cos 161◦
To evaluate, say, sine 42◦23 using a calculating means
23◦ (b) sec 302◦29 = 1 = 1
cos 302◦29 29◦
finding sine 42 since there are 60 minutes in 1 cos 302
60 60

degree.

23 = 0.3833˙, thus 42◦23 = 42.3833˙◦ = 1.8620
60

Thus sine 42◦23 = sine 42.3833˙◦ = 0.6741, correct to Problem 15. Evaluate, correct to 4 significant
figures:
4 decimal places.
(a) cosecant 279.16◦ (b) cosecant 49◦7
Similarly, cosine 72◦38 38◦
= cosine 72 = 0.2985,
60
correct to 4 decimal places.

Most calculators contain only sine, cosine and tan- (a) cosec 279.16◦ = sin 1 = −1.013
gent functions. Thus to evaluate secants, cosecants and 279.16◦
cotangents, reciprocals need to be used.
(b) cosec 49◦7 = 1 = 1 = 1.323
The following values, correct to 4 decimal places, may sin 49◦7 7◦
be checked: sin 49
60

196 Engineering Mathematics

Problem 16. Evaluate, correct to 4 decimal (b) cosec−1 1.1784 = sin−1 1
places:
(a) cotangent 17.49◦ (b) cotangent 163◦52 1.1784

= sin−1 0.8486…

= 58.06◦ or 58◦4 or 1.013 radians

(a) cot 17.49◦ = 1 = 3.1735 (c) cot−1 2.1273 = tan−1 1
tan 17.49◦
2.1273

cot 163◦52 1 1 = tan−1 0.4700…
= tan 163◦52 = 52◦
(b) = 25.18◦ or 25◦11 or 0.439 radians
tan 163
60
Section 3
= −3.4570 Problem 20. Evaluate the following expression,
correct to 4 significant figures:
Problem 17. Evaluate, correct to 4 significant
figures: 4 sec 32◦10 − 2 cot 15◦19
(a) sin 1.481 (b) cos (3π/5) (c) tan 2.93 3 cosec 63◦8 tan 14◦57

(a) sin 1.481 means the sine of 1.481 radians. Hence By calculator:
a calculator needs to be on the radian function.
sec 32◦10 = 1.1813, cot 15◦19 = 3.6512
Hence sin 1.481 = 0.9960 cosec 63◦8 = 1.1210, tan 14◦57 = 0.2670

(b) cos(3π/5) = cos 1.884955… = −0.3090 Hence
(c) tan 2.93 = −0.2148 4 sec 32◦10 − 2 cot 15◦19
3 cosec 63◦8 tan 14◦57
Problem 18. Evaluate, correct to 4 decimal
places: = 4(1.1813) − 2(3.6512)
(a) secant 5.37 (b) cosecant π/4 3(1.1210)(0.2670)
(c) cotangent π/24
= 4.7252 − 7.3024 = −2.5772
0.8979 0.8979

(a) Again, with no degrees sign, it is assumed that 5.37 = −2.870, correct to 4 significant figures.
means 5.37 radians.

Hence sec 5.37 = 1 = 1.6361 Problem 21. Evaluate correct to 4 decimal places:
cos 5.37 (a) sec (−115◦) (b) cosec (−95◦17 )

(b) cosec(π/4) = 1 = 1

sin (π/4) sin 0.785398 . . .

= 1.4142 (a) Positive angles are considered by convention to be
anticlockwise and negative angles as clockwise.
(c) cot (5π/24) = 1 = 1

tan (5π/24) tan 0.654498 . . . Hence −115◦ is actually the same as 245◦ (i.e.
360◦ − 115◦).
= 1.3032

Problem 19. Determine the acute angles: Hence sec(−115◦) = sec 245◦ = cos 1
245◦
(a) sec−12.3164 (b) cosec−1 1.1784
(c) cot−1 2.1273 = −2.3662

(a) sec−1 2.3164 = cos−1 1 (b) cosec(−95◦47 ) = 1 = −1.0051
sin −95 47◦
2.3164

= cos−1 0.4317… 60

= 64.42◦ or 64◦25 or 1.124 radians

Introduction to trigonometry 197

Now try the following exercise

In Problems 15 to 18, evaluate correct to 4 signifi-
cant figures.

Exercise 89 Further problems on 15. 4 cos 56◦19 − 3 sin 21◦57 [1.097]
evaluation trigonometric
ratios 11.5 tan 49◦11 − sin 90◦ [5.805] Section 3
16. 3 cos 45◦ [−5.325]
In Problems 1 to 8, evaluate correct to 4 decimal [0.7199]
places: 5 sin 86◦3
17. 3 tan 14◦29 − 2 cos 31◦9
1. (a) sine 27◦ (b) sine 172.41◦ (c) sine 302◦52
[(a) 0.4540 (b) 0.1321 (c) −0.8399] 6.4 cosec 29◦5 − sec 81◦
18. 2 cot 12◦
2. (a) cosine 124◦ (b) cosine 21.46◦
(c) cosine 284◦10 19. Determine the acute angle, in degrees and
[(a) −0.5592 (b) 0.9307 (c) 0.2447]
minutes, correct to the nearest minute, given
3. (a) tangent 145◦ (b) tangent 310.59◦ by: sin−1 4.32 sin 42◦16
(c) tangent 49◦16 7.86 [21◦42 ]
[(a) −0.7002 (b) −1.1671 (c) 1.1612]
20. If tan x = 1.5276, determine sec x, cosec
4. (a) secant 73◦ (b) secant 286.45◦ x, and cot x. (Assume x is an acute angle)
(c) secant 155◦41 [1.8258, 1.1952, 0.6546]
[(a) 3.4203 (b) 3.5313 (c) −1.0974]
In Problems 21 to 23 evaluate correct to 4 signifi-
5. (a) cosecant 213◦ (b) cosecant 15.62◦ cant figures.
(c) cosecant 311◦50
[(a) −1.8361 (b) 3.7139 (c) −1.3421] (sin 34◦27 )( cos 69◦2 ) [0.07448]
21. (2 tan 53◦39 )
6. (a) cotangent 71◦ (b) cotangent 151.62◦
(c) cotangent 321◦23 22. 3 cot 14◦15 sec 23◦9 [12.85]
[(a) 0.3443 (b) −1.8510 (c) −1.2519]
cosec 27◦19 + sec 45◦29 [−1.710]
2π 23. 1 − cosec 27◦19 sec 45◦29
7. (a) sine (b) cos 1.681 (c) tan 3.672
24. Evaluate correct to 4 decimal places:
3
[(a) 0.8660 (b) −0.1010 (c) 0.5865] (a) sin(−125◦) (b) tan(−241◦)
(c) cos(−49◦15 )
π
8. (a) sine (b) cosec 2.961 (c) cot 2.612 [(a) −0.8192 (b) −1.8040 (c) 0.6528]

8 25. Evaluate correct to 5 significant figures:
[(a) 1.0824 (b) 5.5675 (c) −1.7083]
(a) cosec (−143◦) (b) cot(−252◦)
In Problems 9 to 14, determine the acute angle (c) sec(−67◦22 )
in degrees (correct to 2 decimal places), degrees
and minutes, and in radians (correct to 3 decimal [(a) −1.6616 (b) −0.32492 (c) 2.5985]
places).

9. sin−1 0.2341 [13.54◦, 13◦32 , 0.236 rad] 22.8 Trigonometric approximations
for small angles
10. cos−1 0.8271 [34.20◦, 34◦12 , 0.597 rad]
If angle x is a small angle (i.e. less than about 5◦) and is
11. tan−1 0.8106 [39.03◦, 39◦2 , 0.681 rad] expressed in radians, then the following trigonometric
approximations may be shown to be true:
12. sec−1 1.6214 [51.92◦, 51◦55 , 0.906 rad]
(i) sin x ≈ x
13. cosec−1 2.4891 (ii) tan x ≈ x
[23.69◦ , 23◦41 , 0.413 rad] (iii) cos x ≈ 1 − x2

14. cot−1 1.9614 [27.01◦, 27◦1 , 0.471 rad] 2

198 Engineering Mathematics By calculator, sin 5◦ = 0.08716, thus sin x ≈ x,
and tan 5◦ = 0.08749, thus tan x ≈ x,
For example, let x = 1◦ , i.e. 1 × π = 0.01745 cos 5◦ = 0.99619;
180
since x = 0.08727 radians,
radians, correct to 5 decimal places. By calculator,
sin 1◦ = 0.01745 and tan 1◦ = 0.01746, showing that: x2 0.087272
sin x = x ≈ tan x when x = 0.01745 radians. Also, 1− =1− = 0.99619 showing that:
cos 1◦ = 0.99985; when x = 1◦, i.e. 0.001745 radians, 22

Section 3 1 − x2 = 1 − 0.017452 = 0.99985, x2
22 cos x = 1 − when x = 0.0827 radians.

correct to 5 decimal places, showing that 2

x2 If sin x ≈ x for small angles, then sin x ≈ 1, and this
cos x = 1 − when x = 0.01745 radians. x

2 relationship can occur in engineering considerations.
Similarly, let x = 5◦ , i.e. 5 × π = 0.08727 radians,

180
correct to 5 decimal places.

Chapter 23

Trigonometric waveforms

23.1 Graphs of trigonometric 1.0 y ϭ sin A
functions
y
0.5

By drawing up tables of values from 0◦ to 360◦, graphs 0 30 60 90 120 150 180 210 240 270 300 330 360 A°
of y = sin A, y = cos A and y = tan A may be plotted. 0.5
Values obtained with a calculator (correct to 3 deci-
Ϫ1.0
mal places — which is more than sufficient for plotting (a)
graphs), using 30◦ intervals, are shown below, with the

respective graphs shown in Fig. 23.1.

(a) y = sin A 1.0 y ϭ cos A
y

0.5

A 0 30◦ 60◦ 90◦ 120◦ 150◦ 180◦ 0 30 60 90 120 150 180 210 240 270 300 330 360 A°
sin A 0 0.500 0.866 1.000 0.866 0.500 0 Ϫ0.5

A 210◦ 240◦ 270◦ 300◦ 330◦ 360◦ Ϫ1.0
sin A −0.500 −0.866 −1.000 −0.866 −0.500 0 (b)

(b) y = cos A

A0 30◦ 60◦ 90◦ 120◦ 150◦ 180◦ 4 y ϭ tan A
y
cos A 1.000 0.866 0.500 0 −0.500 −0.866 −1.000 150 330
2 30 60 90 120 180 210 240 270 300 360
A 210◦ 240◦ 270◦ 300◦ 330◦ 360◦
0
cos A −0.866 −0.500 0 0.500 0.866 1.000 Ϫ2 A°

(c) y = tan A

A 0 30◦ 60◦ 90◦ 120◦ 150◦ 180◦ Ϫ4
(c)
tan A 0 0.577 1.732 ∞ −1.732 −0.577 0

A 210◦ 240◦ 270◦ 300◦ 330◦ 360◦ Figure 23.1

tan A 0.577 1.732 ∞ −1.732 −0.577 0

From Fig. 23.1 it is seen that: 23.2 Angles of any magnitude

(i) Sine and cosine graphs oscillate between peak Figure 23.2 shows rectangular axes XX and YY inter-
secting at origin 0. As with graphical work, measure-
values of ±1. ments made to the right and above 0 are positive, while
those to the left and downwards are negative. Let 0A be
(ii) The cosine curve is the same shape as the sine free to rotate about 0. By convention, when 0A moves
curve but displaced by 90◦. anticlockwise angular measurement is considered pos-
itive, and vice versa. Let 0A be rotated anticlockwise
(iii) The sine and cosine curves are continuous and
they repeat at intervals of 360◦; the tangent

curve appears to be discontinuous and repeats at
intervals of 180◦.

200 Engineering Mathematics

90° Let 0A be further rotated so that θ4 is any angle in the
Y fourth quadrant and let AE be constructed to form the

Quadrant 2 Quadrant 1 right-angled triangle 0AE. Then
ϩ ϩ
− +
sin θ4 = + = − cos θ4 = + = +

180° Ϫ ϩ 0° −
A X 360° tan θ4 = + = −
X' 0

Ϫ Ϫ The above results are summarized in Fig. 23.4. The
Quadrant 3 Quadrant 4 letters underlined spell the word CAST when starting
in the fourth quadrant and moving in an anticlockwise
Section 3 Y' direction.
270°

Figure 23.2 90°

Quadrant 2 90° Quadrant 1

Sine All positive

A ϩA 0° 180° 0°
ϩ ϩ 360° 360°
180°
ϩ q1 ϩ
D Ϫ q2 q4 E B Tangent Cosine

C q3 0 ϩϪ

Ϫϩ

AA

Quadrant 3 Quadrant 4 270°

270°

Figure 23.3 Figure 23.4

so that θ1 is any angle in the first quadrant and left per- In the first quadrant of Fig. 23.1 all of the curves have
pendicular AB be constructed to form the right-angled
triangle 0AB in Fig. 23.3. Since all three sides of the tri- positive values; in the second only sine is positive; in the
angle are positive, the trigonometric ratios sine, cosine
and tangent will all be positive in the first quadrant. third only tangent is positive; in the fourth only cosine
(Note: 0A is always positive since it is the radius of a
circle.) is positive — exactly as summarized in Fig. 23.4. A

Let 0A be further rotated so that θ2 is any angle in the knowledge of angles of any magnitude is needed when
second quadrant and let AC be constructed to form the finding, for example, all the angles between 0◦ and 360◦
right-angled triangle 0AC. Then
whose sine is, say, 0.3261. If 0.3261 is entered into a cal-
+ − culator and then the inverse sine key pressed (or sin−1
sin θ2 = + = + cosθ2 = + = − key) the answer 19.03◦ appears. However, there is a
second angle between 0◦ and 360◦ which the calculator
tan θ2 = + = −
− does not give. Sine is also positive in the second quad-

Let 0A be further rotated so that θ3 is any angle in the rant [either from CAST or from Fig. 23.1(a)]. The other
third quadrant and let AD be constructed to form the angle is shown in Fig. 23.5 as angle θ where θ = 180◦
−19.03◦ = 160.97◦ . Thus 19.03◦ and 160.97◦ are the
right-angled triangle 0AD. Then angles between 0◦ and 360◦ whose sine is 0.3261 (check
that sin 160.97◦ = 0.3261 on your calculator).
− −
+ + Be careful! Your calculator only gives you one of

these answers. The second answer needs to be deduced

from a knowledge of angles of any magnitude, as shown

in the following worked problems.

sin θ3 = = − cosθ3 = = −

tan θ3 = − = + Problem 1. Determine all the angles between
− 0◦ and 360◦ whose sine is −0.4638

Trigonometric waveforms 201

90° y ϭ tan x
y

S A 1.7629 180° 270° 360° x
180° 19.03° q 240.44
0 90°
T 19.03° 0° 60.44
360°

C

270° Figure 23.8 Section 3

Figure 23.5 90°

The angles whose sine is −0.4638 occurs in the third and S A
fourth quadrants since sine is negative in these quadrants q
— see Fig. 23.6. 0°
180° 360°

q

y y ϭ sin x TC
1.0 270°

207.63° 332.37° Figure 23.9

0 90° 180° 270° 360° x
Ϫ0.4638
whose tangent is 1.7629 are 60.44◦ and 180◦ + 60.44◦,
Ϫ1.0 i.e. 240.44◦

Figure 23.6 Problem 3. Solve the equation

From Fig. 23.7, θ = sin−1 0.4638 = 27.63◦. Mea- cos−1 (−0.2348) = α for angles of α between 0◦
sured from 0◦, the two angles between 0◦ and 360◦ and 360◦
whose sine is −0.4638 are 180◦ + 27.63◦, i.e. 207.63◦
and 360◦ − 27.63◦, i.e. 332.37◦ Cosine is positive in the first and fourth quadrants and
(Note that a calculator only gives one answer, i.e. thus negative in the second and third quadrants — from
−27.632588◦) Fig. 23.5 or from Fig. 23.1(b).
In Fig. 23.10, angle θ = cos−1 (0.2348) = 76.42◦
90°
90°
SA

180° 0° SA
360°
q q

TC 180° q 0°
270° q 360°

Figure 23.7 T C

Problem 2. Determine all the angles between 0◦ 270°
and 360◦ whose tangent is 1.7629
Figure 23.10
A tangent is positive in the first and third quadrants —
see Fig. 23.8. Measured from 0◦ , the two angles whose cosine
is −0.2348 are α = 180◦ − 76.42◦ i.e. 103.58◦ and
From Fig. 23.9, θ = tan−1 1.7629 = 60.44◦ Mea- α = 180◦ + 76.42◦, i.e. 256.42◦
sured from 0◦, the two angles between 0◦ and 360◦

202 Engineering Mathematics

Section 3 Now try the following exercise If all horizontal components such as OS are projected
on to a graph of y against angle x◦ , then a cosine wave
Exercise 90 Further problems on angles is produced. It is easier to visualize these projections by
of any magnitude redrawing the circle with the radius arm OR initially in
a vertical position as shown in Fig. 23.12.
1. Determine all of the angles between 0◦ and
360◦ whose sine is: From Figs. 23.11 and 23.12 it is seen that a cosine
curve is of the same form as the sine curve but is
(a) 0.6792 (b) −0.1483 displaced by 90◦ (or π/2 radians).
(a) 42.78◦ and 137.22◦
(b) 188.53◦ and 351.47◦ 23.4 Sine and cosine curves

2. Solve the following equations for values of x Graphs of sine and cosine waveforms
between 0◦ and 360◦:
(i) A graph of y = sin A is shown by the broken line
(a) x = cos−1 0.8739 in Fig. 23.13 and is obtained by drawing up a
(b) x = cos−1 (−0.5572) table of values as in Section 23.1. A similar table
may be produced for y = sin 2A.
(a) 29.08◦ and 330.92◦
(b) 123.86◦ and 236.14◦ A◦ 0 30 45 60 90 120
2A 0 60 90 120 180 240
3. Find the angles between 0◦ to 360◦ whose sin 2A 0 0.866 1.0 0.866 0 −0.866
tangent is:
A◦ 135 150 180 210 225 240
(a) 0.9728 (b) −2.3418 2A 270 300 360 420 450 480
(a) 44.21◦ and 224.21◦ sin 2A −1.0 −0.866 0 0.866 1.0 0.866
(b) 113.12◦ and 293.12◦

A◦ 270 300 315 330 360

23.3 The production of a sine and 2A 540 600 630 660 720
cosine wave
sin 2A 0 −0.866 −1.0 −0.866 0
In Fig. 23.11, let OR be a vector 1 unit long and free
to rotate anticlockwise about O. In one revolution a cir- A graph of y = sin 2A is shown in Fig. 23.13.
cle is produced and is shown with 15◦ sectors. Each
radius arm has a vertical and a horizontal component. (ii) A graph of y = sin 1 A is shown in Fig. 23.14 using
For example, at 30◦ , the vertical component is TS and 2
the horizontal component is OS. the following table of values.

From trigonometric ratios, A◦ 0 30 60 90 120 150 180

1 A 0 15 30 45 60 75 90
2
1
sin 2 A 0 0.259 0.500 0.707 0.866 0.966 1.00

TS TS A◦ 210 240 270 300 330 360
, 105 120 135 150 165 180
sin 30◦ = = i.e. TS = sin 30◦ 1 A 0.966 0.866 0.707 0.500 0.259 0
2
TO 1 1
sin 2 A

and cos 30◦ = OS = OS i.e. OS = cos 30◦
,
TO 1
(iii) A graph of y = cos A is shown by the broken line

in Fig. 23.15 and is obtained by drawing up a

The vertical component TS may be projected across table of values. A similar table may be produced
to T S , which is the corresponding value of 30◦ on the
graph of y against angle x◦. If all such vertical compo- for y = cos 2A with the result as shown.

nents as TS are projected on to the graph, then a sine (iv) A graph of y = cos 1 A is shown in Fig. 23.16
2
wave is produced as shown in Fig. 23.11. which may be produced by drawing up a table

of values, similar to above.

Trigonometric waveforms 203

90° y
60° 1.0
120° y ϭ sin x
150° T 1.5 T'

180° O R S' 120° 210° Angle x °
270° S 360° 30° 60° 270° 330°
210°
240° 330° Ϫ1.5 Section 3

300° Ϫ1.0

Figure 23.11

Figure 23.12 y ϭ sin A y y ϭ sin A y ϭsin 1 A
1.0 2
y y ϭ sin 2A
1.0

0 90° 180° 270° 360° A° 0 90° 180° 270° 360° A°

Ϫ1.0 Ϫ1.0

Figure 23.13 Figure 23.14

Periodic time and period y = cos 2A repeat themselves every 180◦ (or π
(i) Each of the graphs shown in Figs. 23.13 to 23.16 radians); thus 180◦ is the period of these wave-
will repeat themselves as angle A increases and forms.
are thus called periodic functions. (iii) In general, if y = sin pA or y = cos pA (where p
(ii) y = sin A and y = cos A repeat themselves every is a constant) then the period of the waveform is
360◦ (or 2π radians); thus 360◦ is called the 360◦/p (or 2π/p rad). Hence if y = sin 3A then the
period of these waveforms. y = sin 2A and period is 360/3, i.e. 120◦ , and if y = cos 4A then
the period is 360/4, i.e. 90◦

204 Engineering Mathematics

y y ϭ cos A y ϭ cos 2A y
1.0 y ϭ sin 3A

1.0

0 90° 180° 270° 360° A°
Ϫ1.0
0 90° 180° 270° 360° A°

Section 3 Ϫ1.0 Figure 23.17

Figure 23.15 y
y y ϭ 3 sin 2A
1.0
3

y ϭcos 1 A y ϭ cos A
2

0 90° 180° 270° 360° A° 0 90° 180° 270° 360° A°
Ϫ3

Ϫ1.0 Figure 23.18

Figure 23.16 Problem 6. Sketch y = 4 cos 2x form x = 0◦ to
x = 360◦
Amplitude
Amplitude = 4 and period = 360◦/2 = 180◦ .
Amplitude is the name given to the maximum or peak A sketch of y = 4 cos 2x is shown in Fig. 23.19.
value of a sine wave. Each of the graphs shown in
Figs. 23.13 to 23.16 has an amplitude of +1 (i.e. they y
oscillate between +1 and −1). However, if y = 4 sin A, 4 y ϭ 4 cos 2x
each of the values in the table is multiplied by 4 and
the maximum value, and thus amplitude, is 4. Simi- 0 90° 180° 270° 360° x °
larly, if y = 5 cos 2A, the amplitude is 5 and the period is
360◦/2, i.e. 180◦ Ϫ4

Problem 4. Sketch y = sin 3A between A = 0◦
and A = 360◦

Amplitude = 1 and period = 360◦/3 = 120◦ Figure 23.19
A sketch of y = sin 3A is shown in Fig. 23.17.
Problem 7. Sketch y =2 sin 3 A over one cycle
Problem 5. Sketch y = 3 sin 2A from A = 0 to 5
A = 2π radians
Amplitude = 2; period = 360◦ = 360◦ × 5 = 600◦
Amplitude = 3 and period = 2π/2 = π rads (or 180◦) 3
A sketch of y = 3 sin 2A is shown in Fig. 23.18. 3
5

A sketch of y =2 sin 3 A is shown in Fig. 23.20.
5

y Trigonometric waveforms 205

2 y ϭ sin 3 A 45°
5 y ϭ cos A
y ϭ cos(Aϩ45°)
0 180° 360° 540° 600° A°
y

0 90° 180° 270° 360° A°

Ϫ2 Ϫ1.0 45° Section 3
Figure 23.20 Figure 23.22

Lagging and leading angles Problem 8. Sketch y = 5 sin(A + 30◦) from
A = 0◦ to A = 360◦
(i) A sine or cosine curve may not always start at 0◦.
Amplitude = 5 and period = 360◦/1 = 360◦.
To show this a periodic function is represented 5 sin(A + 30◦) leads 5 sin A by 30◦ (i.e. starts 30◦
earlier).
by y = sin(A ± α) or y = cos(A ± α) where α is a A sketch of y = 5 sin(A + 30◦) is shown in Fig. 23.23.
phase displacement compared with y = sin A or
y = cos A. 30°
y
(ii) By drawing up a table of values, a graph of
y = sin(A − 60◦) may be plotted as shown in 5
Fig. 23.21. If y = sin A is assumed to start at 0◦
then y = sin(A − 60◦) starts 60◦ later (i.e. has a
zero value 60◦ later). Thus y = sin(A − 60◦) is
said to lag y = sin A by 60◦

y 60° y ϭ 5 sin A
y ϭ sin A y ϭ 5 sin(Aϩ30°)
y ϭ sin(AϪ60°)
1.0

0 90° 180° 270° 360° A°
30°
0 90° 180° 270° 360° A°
Ϫ5
Ϫ1.0
Figure 23.23
60°
Problem 9. Sketch y = 7 sin(2A − π/3) in the
Figure 23.21 range 0 ≤ A ≤ 360◦

(iii) By drawing up a table of values, a graph of Amplitude = 7 and period = 2π/2 = π radians.
y = cos(A + 45◦) may be plotted as shown in In general, y = sin(pt − α) lags y = sin pt by α/p, hence
Fig. 23.22. If y = cos A is assumed to start 7 sin(2A − π/3) lags 7 sin 2A by (π/3)/2, i.e. π/6 rad
at 0◦ then y = cos(A + 45◦) starts 45◦ earlier or 30◦
(i.e. has a maximum value 45◦ earlier). Thus A sketch of y = 7 sin(2A − π/3) is shown in Fig. 23.24.
y = cos(A + 45◦) is said to lead y = cos A by 45◦
Problem 10. Sketch y = 2 cos(ωt − 3π/10) over
(iv) Generally, a graph of y = sin(A − α) lags y = sin one cycle
A by angle α, and a graph of y = sin(A + α) leads
y = sin A by angle α.

(v) A cosine curve is the same shape as a sine curve
but starts 90◦ earlier, i.e. leads by 90◦. Hence

cos A = sin (A + 90◦)

206 Engineering Mathematics

y π/6 y ϭ 7sin 2A 5. y = 7 sin 3x 7 960◦
7 y ϭ 7sin(2AϪπ /3) 28 ,

6. y = 6 sin(t − 45◦) 2
7. y = cos(2θ + 30◦)
[6, 360◦]

0 90° 180° 270° 360° A° [4, 180◦]
π/2 π 3π/2 2π
Ϫ7
π /6

Section 3 Figure 23.24 23.5 Sinusoidal form A sin (ωt ± α)

Amplitude = 2 and period = 2π/ω rad. In Fig. 23.26, let OR represent a vector that is free to
rotate anticlockwise about O at a velocity of ω rad/s.
2 cos(ωt − 3π/10) lags 2 cos ωt by 3π/10ω seconds. A rotating vector is called a phasor. After a time t sec-
ond OR will have turned through an angle ωt radians
A sketch of y = 2 cos(ωt − 3π/10) is shown in (shown as angle TOR in Fig 23.26). If ST is constructed
Fig. 23.25. perpendicular to OR, then sin ωt = ST /OT, i.e. ST = OT
sin ωt.

3π/10w rads w rads/s y y ϭ sin w t
y 1.0

2 y ϭ 2 cos wt
y ϭ 2 cos (wt Ϫ3π/10)
wt 90° 180° 270° 360°
0 SR
0 w t π/2 π 3π/2 2π w t

0 π/2w π/w 3π/2w 2π/w t Ϫ1.0

Figure 23.26

Ϫ2 If all such vertical components are projected on to a
Figure 23.25 graph of y against ωt, a sine wave results of amplitude
OR (as shown in Section 23.3).
Now try the following exercise If phasor OR makes one revolution (i.e. 2π radians) in
T seconds, then the angular velocity, ω = 2π/T rad/s,
Exercise 91 Further problems on sine and
cosine curves from which, T = 2π/ω seconds

In Problems 1 to 7 state the amplitude and period T is known as the periodic time.
of the waveform and sketch the curve between 0◦ The number of complete cycles occurring per second is
and 360◦ . called the frequency, f

1. y = cos 3A [1, 120◦] Frequency = number of cycles
2. y = 2 sin 5x [2, 144◦] second

2 [3, 90◦] 1ω
3. y = 3 sin 4t [3, 720◦ ] = = Hz

θ T 2π
4. y = 3 cos ω
i.e. f = Hz
2 2π

Hence angular velocity, ω = 2πf rad/s

Trigonometric waveforms 207

Amplitude is the name given to the maximum or peak Frequency, f = 1 = 1 = 50 Hz
T 0.02
value of a sine wave, as explained in Section 23.4. The 180 ◦

amplitude of the sine wave shown in Fig. 23.26 has an Phase angle, α = 0.27 rad = 0.27 ×
π
amplitude of 1.
A sine or cosine wave may not always start at 0◦. = 15.47◦ or 15◦28 leading

To show this a periodic function is represented by i = 30 sin(100πt)
y = sin(ωt ± α) or y = cos(ωt ±α), where α is a phase
displacement compared with y = sin A or y = cos A. Problem 12. An oscillating mechanism has a Section 3
A graph of y = sin(ωt − α) lags y = sin ωt by angle maximum displacement of 2.5 m and a frequency of
α, and a graph of y = sin(ωt + α) leads y = sin ωt by 60 Hz. At time t = 0 the displacement is 90 cm.
angle α. Express the displacement in the general form
A sin (ωt ± α)
The angle ωt is measured in radians

i.e. rad (t s) = ωt radians
ω
s

hence angle α should also be in radians.
The relationship between degrees and radians is:

360◦ = 2π radians or 180◦ = π radians Amplitude = maximum displacement = 2.5 m

Hence 1 rad = 180 = 57.30◦ and, for example, Angular velocity, ω = 2π f = 2π(60) = 120π rad/s
π
Hence displacement = 2.5 sin(120πt + α) m
71◦ = 71 × π = 1.239 rad
180 When t = 0, displacement = 90 cm = 0.90 m

Summarising, given a general sinusoidal function Hence, 0.90 = 2.5sin(0 + α)
y = Asin(ωt ± α), then:
i.e. sin α = 0.90 = 0.36
(i) A = amplitude Hence 2.5
(ii) ω = angular velocity = 2π f rad/s
(iii) 2π = periodic time T seconds α = sin−10.36 = 21.10◦

ω = 21◦6 = 0.368 rad
(iv) ω = frequency, f hertz
Thus, displacement = 2.5 sin (120πt + 0.368) m
2π
(v) α = angle of lead or lag (compared with

y = A sin ωt)

Problem 11. An alternating current is given by Problem 13. The instantaneous value of voltage
i = 30 sin (100πt + 0.27) amperes. Find the in an a.c. circuit at any time t seconds is given by
amplitude, periodic time, frequency and phase v = 340 sin (50πt − 0.541) volts. Determine the:
angle (in degrees and minutes)
(a) amplitude, periodic time, frequency and phase
i = 30 sin(100 πt + 0.27)A, hence amplitude = 30 A. angle (in degrees)
Angular velocity ω = 100 π, hence
(b) value of the voltage when t = 0
2π 2π 1 (c) value of the voltage when t = 10 ms
periodic time, T = = = (d) time when the voltage first reaches 200 V, and
ω 100π 50 (e) time when the voltage is a maximum

= 0.02 s or 20 ms Sketch one cycle of the waveform

208 Engineering Mathematics

(a) Amplitude = 340 V A sketch of v = 340 sin(50πt − 0.541) volts is
Angular velocity, ω = 50π shown in Fig. 23.27.
Hence periodic time, T = 2π = 2π = 1
ω 50π 25 Voltage v v ϭ 340 sin(50πt Ϫ0.541)
= 0.04 s or 40 ms v ϭ 340 sin 50πt
340
Frequency f = 1 = 1 = 25 Hz 291.4
T 0.04
200
Phase angle = 0.541 rad = 0.541 × 180
Section 3 π 0 10 20 30 40 t (ms)

= 31◦ lagging v = 340 sin(50πt) Ϫ175.1 7.447 13.44

(b) When t = 0, Ϫ340

v = 340 sin(0 − 0.541) Figure 23.27
= 340 sin(−31◦) = −175.1 V
Now try the following exercise
(c) When t = 10 ms,
10 Exercise 92 Further problems on the
sinusoidal form A sin(ωt ± α)
then v = 340 sin 50π 103 − 0.541
= 340 sin(1.0298) In Problems 1 to 3 find the amplitude, periodic
= 340 sin 59◦ = 291.4 volts time, frequency and phase angle (stating whether
it is leading or lagging sin ωt) of the alternating
(d) When v = 200 volts, quantities given.

then 200 = 340 sin(50πt − 0.541) 1. i = 40 sin(50πt + 0.29) mA
200
= sin(50πt − 0.541) 40, 0.04 s, 25 Hz, 0.29 rad
340 (or 16◦37 ) leading 40 sin 5πt

Hence (50πt − 0.541) = sin−1 200 2. y = 75 sin(40t − 0.54) cm
340
75 cm, 0.157 s, 6.37 Hz, 0.54 rad
= 36.03◦ or 0.6288 rad (or 30◦56 ) lagging 75 sin 40t
50πt = 0.6288 + 0.541
3. v = 300 sin(200πt − 0.412) V
= 1.1698
300 V, 0.01 s, 100 Hz, 0.412 rad
Hence when v = 200 V, (or 23◦36 ) lagging 300 sin 200πt

time, t = 1.1698 = 7.447 ms 4. A sinusoidal voltage has a maximum value of
50π 120 V and a frequency of 50 Hz. At time t = 0,
the voltage is (a) zero, and (b) 50 V.
(e) When the voltage is a maximum, v = 340 V Express the instantaneous voltage v in the form
v = A sin(ωt ± α).
Hence 340 = 340 sin(50πt − 0.541)
(a) v = 120 sin 100πt volts
1 = sin(50πt − 0.541) (b) v = 120 sin(100πt + 0.43)volts

50πt − 0.541 = sin−1 1 = 90◦ or 1.5708 rad

50πt = 1.5708 + 0.541 = 2.1118

Hence time, 2.1118
t = = 13.44 ms

50π

Trigonometric waveforms 209

5. An alternating current has a periodic time of multiples of f . Thus, if the supply frequency is 50 Hz, Section 3
25 ms and a maximum value of 20 A. When then the thid harmonic frequency is 150 Hz, the fifth
time t = 0, current i = −10 amperes. Express 250 Hz, and so on.
the current i in the form i = A sin(ωt ± α).
i = 20 sin 80πt − π amperes A complex waveform comprising the sum of the
6 fundamental and a third harmonic of about half the
amplitude of the fundamental is shown in Fig. 23.28(a),
6. An oscillating mechanism has a maximum dis- both waveforms being initially in phase with each other.
placement of 3.2 m and a frequency of 50 Hz. If further odd harmonic waveforms of the appropriate
At time t = 0 the displacement is 150 cm. amplitudes are added, a good approximation to a square
Express the displacement in the general form wave results. In Fig. 23.28(b), the third harmonic is
A sin(ωt ± α). shown having an initial phase displacement from the
fundamental. The positive and negative half cycles of
[3.2 sin(100πt + 0.488)m] each of the complex waveforms shown in Figs. 23.28(a)
and (b) are identical in shape, and this is a feature of
7. The current in an a.c. circuit at any time t waveforms containing the fundamental and only odd
seconds is given by: harmonics.

i = 5 sin(100πt − 0.432) amperes

Determine (a) the amplitude, periodic time, Complex Complex
frequency and phase angle (in degrees) (b) the waveform waveform
value of current at t = 0, (c) the value of cur-
rent at t = 8 ms, (d) the time when the current is v Fundamental v Fundamental
first a maximum, (e) the time when the current Third
first reaches 3A.
harmonic

0 t0 t

Third

harmonic

Sketch one cycle of the waveform showing (a) (b)

relevant points. Complex Complex

⎡⎤ waveform waveform
(a) 5 A, 20 ms, 50 Hz, Fundamental
Fundamental
⎢⎢⎣⎢⎢⎢⎢ 24◦45 lagging ⎥⎦⎥⎥⎥⎥⎥ v Second v Second
−2.093 A
4.363 A harmonic
6.375 ms
(b) harmonic t
(c) 0 t0
(d)
A

(e) 3.423 ms (c) (d)

23.6 Waveform harmonics Complex Complex
waveform waveform
Let an instantaneous voltage v be represented by Fundamental
v = Vm sin 2π f t volts. This is a waveform which varies Fundamental
sinusoidally with time t, has a frequency f , and a v Second v t
maximum value Vm. Alternating voltages are usually
assumed to have wave-shapes which are sinusoidal harmonic
where only one frequency is present. If the waveform is
not sinusoidal it is called a complex wave, and, what- 0 t0
ever its shape, it may be split up mathematically into
components called the fundamental and a number of B Second Third
harmonics. This process is called harmonic analysis. Third harmonic harmonic
The fundamental (or first harmonic) is sinusoidal and harmonic
has the supply frequency, f ; the other harmonics are (f)
also sine waves having frequencies which are integer (e)

Figure 23.28

A complex waveform comprising the sum of the
fundamental and a second harmonic of about half the
amplitude of the fundamental is shown in Fig. 23.28(c),
each waveform being initially in phase with each other.
If further even harmonics of appropriate amplitudes are
added a good approximation to a triangular wave results.
In Fig. 23.28(c), the negative cycle, if reversed, appears
as a mirror image of the positive cycle about point A.

210 Engineering Mathematics

In Fig. 23.28(d) the second harmonic is shown with an sum of the fundamental, a second harmonic and a third
initial phase displacement from the fundamental and the harmonic are shown with initial phase displacement.
positive and negative half cycles are dissimilar. The positive and negative half cycles are seen to be
dissimilar.
A complex waveform comprising the sum of the fun-
damental, a second harmonic and a third harmonic is The features mentioned relative to Figs. 23.28(a) to
shown in Fig. 23.28(e), each waveform being initially (f) make it possible to recognise the harmonics present
‘in-phase’. The negative half cycle, if reversed, appears in a complex waveform.
as a mirror image of the positive cycle about point B.
In Fig. 23.28(f), a complex waveform comprising the

Section 3

Chapter 24

Cartesian and polar
co-ordinates

24.1 Introduction r = x2 + y2 and θ = tan−1 y are the two formulae we
x
There are two ways in which the position of a point in
a plane can be represented. These are need to change from Cartesian to polar co-ordinates.
The angle θ, which may be expressed in degrees or
(a) by Cartesian co-ordinates, i.e. (x, y), and radians, must always be measured from the positive
x-axis, i.e. measured from the line OQ in Fig. 24.1. It is
(b) by polar co-ordinates, i.e. (r, θ), where r is a suggested that when changing from Cartesian to polar
‘radius’ from a fixed point and θ is an angle from co-ordinates a diagram should always be sketched.
a fixed point.
Problem 1. Change the Cartesian co-ordinates
(3, 4) into polar co-ordinates

24.2 Changing from Cartesian into A diagram representing the point (3, 4) is shown in
polar co-ordinates Fig. 24.2.

In Fig. 24.1, if lengths x and y are known, then the yP
length of r can be obtained from Pythagoras’ theorem
(see Chapter 22) since OPQ is a right-angled triangle.

Hence r2 = (x2 + y2) r
4

from, which r = x2 + y2 q
O
y x
P 3

r Figure 24.2
y
√
q From Pythagoras’ theorem, r = 32 + 42 = 5 (note that
O
Qx −5 has no meaning in this context). By trigonometric
x
ratios, θ = tan−1 4 = 53.13◦ or 0.927 rad
[note that 53.13◦ 3
= 53.13 × (π/180) rad = 0.927 rad.]

Figure 24.1 Hence (3, 4) in Cartesian co-ordinates corresponds
to (5, 53.13◦) or (5, 0.927 rad) in polar co-ordinates.
From trigonometric ratios (see Chapter 22),

from which tan θ = y Problem 2. Express in polar co-ordinates the
x position (−4, 3)

θ = tan−1 y
x

212 Engineering Mathematics

A diagram representing the point using the Cartesian Problem 4. Express (2, −5) in polar co-ordinates.
co-ordinates (−4, 3) is shown in√Fig. 24.3.
From Pythagoras’ theorem r = 42 + 32 = 5 A sketch showing the position (2, −5) is shown in Fig.

By trigonometric ratios, α= tan−1 3 = 36.87◦ or 24.5.
4
0.644 rad. √
r = 22 + 52 = 29 = 5.385
Hence θ = 180◦ − 36.87◦ = 143.13◦
correct to 3 decimal places

or θ = π − 0.644 = 2.498 rad. α = tan−1 5 = 68.20◦ or 1.190 rad
2
Py
Section 3
y

r
3

a q q2 x
4 Ox Oa

Figure 24.3 r 5

Hence the position of point P in polar co-ordinate P
form is (5, 143.13◦) or (5, 2.498 rad).

Problem 3. Express (−5, −12) in polar Figure 24.5
co-ordinates
Hence θ = 360◦ − 68.20◦ = 291.80◦
A sketch showing the position (−5, −12) is shown in or θ = 2π − 1.190 = 5.093 rad
Fig. 24.4.
Thus (2, −5) in Cartesian co-ordinates corresponds
r = 52 + 122 = 13 to (5.385, 291.80◦) or (5.385, 5.093 rad) in polar

α = tan−1 12 = 67.38◦ or 1.176 rad. co-ordinates.
5
and Now try the following exercise
θ = 180◦ + 67.38◦ = 247.38◦
Hence
or θ = π + 1.176 = 4.318 rad.

y x Exercise 93 Further problems on changing
form Cartesian into polar
5q co-ordinates
aO
In Problems 1 to 8, express the given Cartesian
co-ordinates as polar co-ordinates, correct to 2
decimal places, in both degrees and in radians.

12 r 1. (3, 5)

[(5.83, 59.04◦) or (5.83, 1.03 rad)]

P 2. (6.18, 2.35)
[(6.61, 20.82◦) or (6.61, 0.36 rad)]
Figure 24.4
3. (−2, 4)
Thus (−5, −12) in Cartesian co-ordinates corre- [(4.47, 116.57◦) or (4.47, 2.03 rad)]
sponds to (13, 247.38◦) or (13, 4.318 rad) in polar
co-ordinates. 4. (−5.4, 3.7)
[(6.55, 145.58◦) or (6.55, 2.54 rad)]

Cartesian and polar co-ordinates 213

y

5. (−7, −3) O rϭ4 y
[(7.62, 203.20◦) or (7.62, 3.55 rad)] Figure 24.7 q ϭ 32° x

6. (−2.4, −3.6) x
[(4.33, 236.31◦) or (4.33, 4.12 rad)]
Section 3
7. (5, −3)
[(5.83, 329.04◦) or (5.83, 5.74 rad)]

8. (9.6, −12.4)
[(15.68, 307.75◦) or (15.68, 5.37 rad)]

24.3 Changing from polar into Problem 6. Express (6, 137◦) in Cartesian
Cartesian co-ordinates co-ordinates

From the right-angled triangle OPQ in Fig. 24.6. A sketch showing the position (6, 137◦) is shown in
cos θ = x and sin θ = y Fig. 24.8.
rr
from trigonometric ratios x = r cos θ = 6 cos 137◦ = −4.388

Hence x = r cos θ and y = r sin θ which corresponds to length OA in Fig. 24.8.
y = r sin θ = 6 sin 137◦ = 4.092
y
P which corresponds to length AB in Fig. 24.8.

r y By
Qx
q rϭ6
O q ϭ 137°

x A Ox

Figure 24.6 Figure 24.8
Thus (6, 137◦) in polar co-ordinates corresponds to
If length r and angle θ are known then x = r cos θ and (−4.388, 4.092) in Cartesian co-ordinates.
y = r sin θ are the two formulae we need to change from (Note that when changing from polar to Cartesian
polar to Cartesian co-ordinates. co-ordinates it is not quite so essential to draw a sketch.
Use of x = r cos θ and y = r sin θ automatically produces
Problem 5. Change (4, 32◦) into Cartesian the correct signs.)
co-ordinates.
Problem 7. Express (4.5, 5.16 rad) in Cartesian
A sketch showing the position (4, 32◦) is shown in co-ordinates.
Fig. 24.7.
A sketch showing the position (4.5, 5.16 rad) is shown
Now x = r cos θ = 4 cos 32◦ = 3.39 in Fig. 24.9.
and y = r sin θ = 4 sin 32◦ = 2.12
x = r cos θ = 4.5 cos 5.16 = 1.948
Hence (4, 32◦) in polar co-ordinates corresponds to
(3.39, 2.12) in Cartesian co-ordinates. which corresponds to length OA in Fig. 24.9.
y = r sin θ = 4.5 sin 5.16 = −4.057

214 Engineering Mathematics x 3. (7, 140◦) [(−5,362, 4.500)]
4. (3.6, 2.5 rad) [(−2.884, 2.154)]
y 5. (10.8, 210◦) [(−9.353, −5.400)]
6. (4, 4 rad) [(−2.615, −3.207)]
q ϭ 5.16 rad 7. (1.5, 300◦) [(0.750, −1.299)]
A

O
r ϭ 4.5

B

Figure 24.9

Section 3 8. (6, 5.5 rad) [(4.252, −4.233)]

which corresponds to length AB in Fig. 24.9. 9. Figure 24.10 shows 5 equally spaced holes
on an 80 mm pitch circle diameter. Calculate
Thus (1.948, −4.057) in Cartesian co-ordinates cor- their co-ordinates relative to axes Ox and Oy in
responds to (4.5, 5.16 rad) in polar co-ordinates. (a) polar form, (b) Cartesian form. Calculate
also the shortest distance between the centres
24.4 Use of R → P and P → R of two adjacent holes.
functions on calculators
y
Another name for Cartesian co-ordinates is rectangu-
lar co-ordinates. Many scientific notation calculators Ox
possess R → P and P → R functions. The R is the first
letter of the word rectangular and the P is the first let-
ter of the word polar. Check the operation manual for
your particular calculator to determine how to use these
two functions. They make changing from Cartesian to
polar co-ordinates, and vice-versa, so much quicker and
easier.

Now try the following exercise

Exercise 94 Further problems on changing Figure 24.10
polar into Cartesian
co-ordinates [(a) (40, 18◦), (40, 90◦), (40, 162◦), (40, 234◦)
and (40, 306◦)
In Problems 1 to 8, express the given polar (b) (38.04, 12.36), (0, 40), (−38.04, 12.36),
co-ordinates as Cartesian co-ordinates, correct to (−23.51, −32.36), (23.51, −32.36)
3 decimal places. (c) 47.02 mm]

1. (5, 75◦) [(1.294, 4.830)]

2. (4.4, 1.12 rad) [(1.917, 3.960)]

Revision Test 6

This Revision test covers the material contained in Chapter 22 to 24. The marks for each question are shown in
brackets at the end of each question.

1. Fig. R6.1 shows a plan view of a kite design. 6. Sketch the following curves labelling relevant

Calculate the lengths of the dimensions shown points: (a) y = 4 cos(θ + 45◦)

as a and b. (4) (b) y = 5 sin (2t − 60◦) (6)

2. In Fig. R6.1, evaluate (a) angle θ (b) angle α (5) 7. Solve the following equations in the

3. Determine the area of the plan view of a kite shown range 0◦ to 360◦ (a) sin −1 (−0.4161) = x Section 3

in Fig. R6.1. (4) (b) cot −1(2.4198) = θ (6)

20.0 cm a 8. The current in an alternating current cir-
42.0 cm a cuit at any time t seconds is given by:
i = 120 sin(100πt + 0.274) amperes. Determine

b 60.0 cm (a) the amplitude, periodic time, frequency and
q phase angle (with reference to 120 sin 100πt)

(b) the value of current when t = 0
(c) the value of current when t = 6 ms
(d) the time when the current first reaches 80 A

Sketch one cycle of the oscillation. (17)

Figure R6.1 9. Change the following Cartesian coordinates into

4. If the angle of elevation of the top of a 25 m per- polar co-ordinates, correct to 2 decimal places, in

pendicular building from point A is measured as both degrees and in radians: (a) (−2.3, 5.4)
27◦, determine the distance to the building. Cal-
(b) (7.6, −9.2) (6)

culate also the angle of elevation at a point B, 20 m 10. Change the following polar co-ordinates into

closer to the building than point A. (5) Cartesian co-ordinates, correct to 3 decimal

5. Evaluate, each correct to 4 significant figures: places: (a) (6.5, 132◦) (b) (3, 3 rad) (4)
(a) sin 231.78◦ (b) cos 151◦16 (c) tan 3π (3)
8

Chapter 25

Triangles and some
practical applications

25.1 Sine and cosine rules Cosine rule

To ‘solve a triangle’ means ‘to find the values of With reference to triangle ABC of Fig. 25.1, the cosine
unknown sides and angles’. If a triangle is right-angled,
trigonometric ratios and the theorem of Pythagoras rule states:
may be used for its solution, as shown in Chapter 22.
However, for a non-right-angled triangle, trigono- a2= b2 + c2 − 2bc cos A
metric ratios and Pythagoras’ theorem cannot be used.
Instead, two rules, called the sine rule and cosine rule, or b2= a2 + c2 − 2ac cos B
are used.
or c2= a2 + b2 − 2ab cos C
Sine rule
The rule may be used only when:
With reference to triangle ABC of Fig. 25.1, the sine
rule states: (i) 2 sides and the included angle are initially given, or
(ii) 3 sides are initially given.
abc
== 25.2 Area of any triangle

sin A sin B sin C The area of any triangle such as ABC of Fig. 25.1 is
given by:

(i) 1 × base × perpendicular height, or
2

A (ii) 1 ab sin C or 1 ac sin B or 1 bc sin A, or
cb 2 2 2
√
(iii) s(s − a)(s − b)(s − c)

where s = a + b + c
2

BaC

Figure 25.1 25.3 Worked problems on the
solution of triangles and their
The rule may be used only when: areas

(i) 1 side and any 2 angles are initially given, or Problem 1. In a triangle XYZ, ∠X = 51◦,
∠Y = 67◦ and YZ = 15.2 cm. Solve the triangle and
(ii) 2 sides and an angle (not the included angle) are
initially given. find its area

Triangles and some practical applications 217

The triangle XYZ is shown in Fig. 25.2. Since Applying the sine rule:
the angles in a triangle add up to 180◦, then
z = 180◦ − 51◦ − 67◦ = 62◦. 22.31 17.92
sin 78◦51 = sin C
X
17.92 sin 78◦51
from which, sin C = = 0.7881
22.31
51°
z y Hence C = sin−1 0.7881 = 52◦0 or 128◦0 (see Chap-

ters 22 and 23).

67° Since B = 78◦51 , C cannot be 128◦0 , since
128◦0 + 78◦51 is greater than 180◦.
c ϭ 17.92 mmY x ϭ 15.2 cm Z

Figure 25.2 Section 3Thus only C = 52◦0 is valid.

Applying the sine rule: Angle A = 180◦ − 78◦51 − 52◦0 = 49◦9

15.2 = y = z Applying the sine rule:
sin 51◦ sin 67◦ sin 62◦
a 22.31
Using 15.2 y sin 49◦9 = sin 78◦51
sin 51◦ = sin 67◦ and transposing gives:
22.31 sin 49◦9
y = 15.2 sin 67◦ = 18.00 cm = XZ from which, a = sin 78◦51 = 17.20 mm
sin 51◦
Hence A = 49◦9 , C = 52◦0 and BC = 17.20 mm.
15.2 z
Using sin 51◦ = sin 62◦ and transposing gives: Area of triangle ABC = 1 ac sin B
2
15.2 sin 62◦
z = sin 51◦ = 17.27 cm = XY = 1 (17.20)(17.92) sin 78◦ 51 = 151.2 mm2
2

Area of triangle XYZ = 1 xy sin Z Problem 3. Solve the triangle PQR and find its
2
area given that QR = 36.5 mm, PR = 26.6 mm and
= 1 (15.2)(18.00) sin 62◦ = 120.8 cm2 ∠Q = 36◦
2

(or area = 1 xz sin Y = 1 (15.2)(17.27) sin 67◦
2 2

= 120.8 cm2) Triangle PQR is shown in Fig. 25.4.

It is always worth checking with triangle problems P
that the longest side is opposite the largest angle, and
vice-versa. In this problem, Y is the largest angle and r q ϭ 29.6 mm
XZ is the longest of the three sides.

Problem 2. Solve the triangle ABC given 36°
B = 78◦51 , AC = 22.31 mm and AB = 17.92 mm. Q p ϭ 36.5 mm R

Find also its area Figure 25.4
Applying the sine rule:

Triangle ABC is shown in Fig. 25.3. 29.6 36.5
sin 36◦ = sin P
A
36.5 sin 36◦
from which, sin P = = 0.7248
29.6

b ϭ 22.31 mm P = sin−1 0.7248 = 46◦27 or 133◦33
P = 46◦27 and Q = 36◦
78°51′ C Hence R = 180◦ − 46◦27 − 36◦ = 97◦33
Ba When P = 133◦33 and Q = 36◦
then R = 180◦ − 133◦33 − 36◦ = 10◦27
Figure 25.3 When
then

218 Engineering Mathematics

Thus, in this problem, there are two separate sets of In Problems 3 and 4, use the sine rule to solve the
results and both are feasible solutions. Such a situation triangles DEF and find their areas.
is called the ambiguous case.
3. d = 17 cm, f = 22 cm, F = 26◦
Case 1. P = 46◦27 , Q = 36◦, R = 97◦33 , p = 36.5 mm D = 19◦48 , E = 134◦12 ,
and q = 29.6 mm e = 36.0 cm, area = 134 cm2

From the sine rule: 4. d = 32.6 mm, e = 25.4 mm, D = 104◦22
E = 49◦0 , F = 26◦38 ,
r = 29.6 f = 15.09 mm, area = 185.6 mm2
sin 97◦33 sin 36◦

Section 3 from which, 29.6 sin 97◦33
r = sin 36◦ = 49.92 mm

Area = 1 pq sin R = 1 (36.5)(29.6) sin 97◦ 33 In Problems 5 and 6, use the sine rule to solve the
2 2 triangles JKL and find their areas.

= 535.5 mm2

Case 2. P = 133◦33 , Q = 36◦, R = 10◦27 , p = 36.5 mm 5. j =⎡3J.8=5 c4m4,◦k29=,3L.2=3 cm, K = 36◦ ⎤
and q = 29.6 mm 99◦31 , l = 5.420
cm,

From the sine rule: ⎢⎢⎣⎢OarReaJ==61.13353◦ cm2 = 8◦29 , ⎥⎦⎥⎥
31 , L
r 29.6
sin 10◦27 = sin 36◦ l = 0.811 cm, area = 0.917 cm2

from which, 29.6 sin 10◦2 6. k = 46 mm, l = 36 mm, L = 35◦ ⎤
r = sin 36◦ = 9.134 mm ⎡ K = 47◦8 , J = 97◦52 ,

Area = 1 pq sin R = 1 (36.5)(29.6) sin 10◦27 ⎢⎣⎢ j = 62.2 mm, area = 820.2 mm2 ⎥⎦⎥
2 2 OR K = 132◦52 , J = 12◦8 ,

= 97.98 mm2 j = 13.19 mm, area = 174.0 mm2

Triangle PQR for case 2 is shown in Fig. 25.5.

133°33′ 25.4 Further worked problems on
the solution of triangles
P 29.6 mm and their areas
9.134 mm

Q 36.5 mm R

36° 10°27′

Figure 25.5 Problem 4. Solve triangle DEF and find its area

Now try the following exercise given that EF = 35.0 mm, DE = 25.0 mm and
∠E = 64◦

Exercise 95 Further problems on the Triangle DEF is shown in Fig. 25.6.
solution of triangles and
their areas D

In Problems 1 and 2, use the sine rule to solve the f ϭ 25.0 mm e
triangles ABC and find their areas.
64°
1. A = 29◦, B = 68◦, b = 27 mm. E d ϭ 35.0 mm F
C = 83◦, a = 14.4 mm, c = 28.9 mm,
area = 189 mm2 Figure 25.6

2. B = 71◦26 , C = 56◦32 , b = 8.60 cm. Applying the cosine rule:
A = 52◦2 , c = 7.568 cm,
a = 7.152 cm, area = 25.65 cm2 e2 = d2 + f 2 − 2df cos E
i.e. e2 = (35.0)2 + (25.0)2 − [2(35.0)(25.0) cos 64◦]

= 1225 + 625 − 767.1 = 1083

Triangles and some practical applications 219

√ negative, angle A would be obtuse, i.e. lie between 90◦
from which, e = 1083 = 32.91 mm and 180◦).
Applying the sine rule:

32.91 25.0 Applying the sine rule:
sin 64◦ = sin F
9.0 7.5
25.0 sin 64◦ sin 79.66◦ = sin B

from which, sin F = 32.91 = 0.6828 sin B = 7.5 sin 79.66◦ = 0.8198
9.0
from which,

Thus ∠F = sin−1 0.6828 Hence B = sin−1 0.8198 = 55.06◦
= 43◦4 or 136◦56
and C = 180◦ − 79.66◦ − 55.06◦ = 45.28◦ Section 3
F = 136◦56 is not possible in this case since √
136◦56 + 64◦ is greater than 180◦. Thus only F = 43◦4
is valid. Area = s(s − a)(s − b)(s − c), where

∠D = 180◦ − 64◦ − 43◦4 = 72◦56 s = a + b + c = 9.0 + 7.5 + 6.5 = 11.5 cm
22

Hence

Area of triangle DEF = 1 d f sin E area = 11.5(11.5 − 9.0)(11.5 − 7.5)(11.5 − 6.5)
2

= 1 (35.0)(25.0) sin 64◦ = 393.2 mm2 = 11.5(2.5)(4.0)(5.0) = 23.98 cm2
2

Problem 5. A triangle ABC has sides a = 9.0 cm, Alternatively,
b = 7.5 cm and c = 6.5 cm. Determine its three
angles and its area area = 1 ab sin C = 1 (9.0)(7.5) sin 45.28◦
2 2

= 23.98 cm2

Triangle ABC is shown in Fig. 25.7. It is usual first Problem 6. Solve triangle XYZ, shown in Fig.
to calculate the largest angle to determine whether the 25.8, and find its area given that Y = 128◦,
triangle is acute or obtuse. In this case the largest angle XY = 7.2 cm and YZ = 4.5 cm
is A (i.e. opposite the longest side).
X
A
y
c ϭ 6.5 cm b ϭ 7.5 cm z ϭ 7.2 cm

B a ϭ 9.0 cm C 128°
Y x ϭ 4.5 cmZ
Figure 25.7
Figure 25.8
Applying the cosine rule:
a2 = b2 + c2 − 2bc cos A Applying the cosine rule:

from which, y2 = x2 + z2 − 2xz cos Y
2bc cos A = b2 + c2 − a2 = 4.52 + 7.22 − [2(4.5)(7.2) cos 128◦]
= 20.25 + 51.84 − [−39.89]
and cos A = b2 + c2 − a2 = 20.25 + 51.84 + 39.89 = 112.0
2bc √

= 7.52 + 6.52 − 9.02 = 0.1795 y = 112.0 = 10.58 cm
2(7.5)(6.5)
Applying the sine rule:
Hence A = cos−1 0.1795 = 79.66◦ (or 280.33◦, which
is obviously impossible). The triangle is thus acute 10.58 = 7.2
angled since cos A is positive. (If cos A had been sin 128◦ sin Z

from which, sin Z = 7.2 sin 128◦ = 0.5363
10.58

220 Engineering Mathematics

Hence Z = sin−10.5363 = 32.43◦ (or 147.57◦ which, A section of the roof is shown in Fig. 25.9.
here, is impossible).
B
X = 180◦ − 128◦ − 32.43◦ = 19.57◦

Area = 1 xz sin Y = 1 (4.5)(7.2) sin 128◦ 33° 40°
2 2 A 8.0 m C

= 12.77 cm2 Figure 25.9

Angle at ridge, B = 180◦ − 33◦ − 40◦ = 107◦

Section 3 Now try the following exercise From the sine rule:

Exercise 96 Further problems on the 8.0 = a
solution of triangles and sin 107◦ sin 33◦
their areas
from which, a = 8.0 sin 33◦ = 4.556 m
In Problems 1 and 2, use the cosine and sine rules sin 107◦
to solve the triangles PQR and find their areas.
Also from the sine rule:
1. q = 12 cm, r = 16 cm, P = 54◦
p = 13.2 cm, Q = 47.35◦, 8.0 = c
sin 107◦ sin 40◦
R = 78.65◦, area = 77.7 cm2
2. q = 3.25 m, r = 4.42 m, P = 105◦ from which, 8.0 sin 40◦
c = sin 107◦ = 5.377 m
p = 6.127 m, Q = 30.83◦,
R = 44.17◦, area = 6.938 m2 Hence the roof slopes are 4.56 m and 5.38 m, correct
to the nearest centimetre.
In Problems 3 and 4, use the cosine and sine rules
to solve the triangles XYZ and find their areas. Problem 8. A man leaves a point walking at
6.5 km/h in a direction E 20◦ N (i.e. a bearing of
3. x = 10.0 cm, y = 8.0 cm, z = 7.0 cm. 70◦). A cyclist leaves the same point at the same
X = 83.33◦, Y = 52.62◦, Z = 44.05◦, time in a direction E 40◦ S (i.e. a bearing of 130◦)
area = 27.8 cm2
travelling at a constant speed. Find the average
4. x = 21 mm, y = 34 mm, z = 42 mm.
Z = 29.77◦, Y = 53.50◦, Z = 96.73◦, speed of the cyclist if the walker and cyclist are
area = 355 mm2
80 km apart after 5 hours

25.5 Practical situations involving After 5 hours the walker has travelled 5 × 6.5 = 32.5 km
trigonometry (shown as AB in Fig. 25.10). If AC is the distance the
cyclist travels in 5 hours then BC = 80 km.
There are a number of practical situations where the
use of trigonometry is needed to find unknown sides 20°
and angles of triangles. This is demonstrated in the N
following worked problems.
32.5 km B
Problem 7. A room 8.0 m wide has a span roof WE
which slopes at 33◦ on one side and 40◦ on the
other. Find the length of the roof slopes, correct to A 40°
the nearest centimetre
S 80 km
b

C

Figure 25.10

Applying the sine rule:

80 = 32.5
sin 60◦ sin C
sin C = 32.5 sin 60◦ = 0.3518
from which, 80

Triangles and some practical applications 221

Hence C = sin−1 0.3518 = 20.60◦ (or 159.40◦, which Problem 10. In Fig. 25.12, PR represents the
is impossible in this case). inclined jib of a crane and is 10.0 m long. PQ is
4.0 m long. Determine the inclination of the jib to
B = 180◦ − 60◦ − 20.60◦ = 99.40◦. the vertical and the length of tie QR

Applying the sine rule again: R

from which, 80 b Q
sin 60◦ = sin 99.40◦ 120°

80 sin 99.40◦ 4.0 m
b = sin 60◦ = 91.14 km

Since the cyclist travels 91.14 km in 5 hours then 10.0 m Section 3

average speed = distance = 91.14 = 18.23 km/h
time 5

Problem 9. Two voltage phasors are shown in P
Fig. 25.11. If V1 = 40 V and V2 = 100 V determine Figure 25.12
the value of their resultant (i.e. length OA) and the
Applying the sine rule:
angle the resultant makes with V1

A PR PQ
V2 ϭ 100 V sin 120◦ = sin R

from which, sin R = PQ sin 120◦
PR
45°
OB = (4.0) sin 120◦ = 0.3464
10.0
V1 ϭ 40 V
Hence ∠R = sin−1 0.3464 = 20.27◦ (or 159.73◦, which
Figure 25.11 is impossible in this case).

Angle OBA = 180◦ − 45◦ = 135◦ ∠P = 180◦ − 120◦ − 20.27◦ = 39.73◦, which is the
Applying the cosine rule: inclination of the jib to the vertical.

OA2 = V12 + V22 − 2V1V2 cos OBA Applying the sine rule:
= 402 + 1002 − {2(40)(100) cos 135◦}
= 1600 + 10 000 − {−5657} 10.0 QR
= 1600 + 10 000 + 5657 = 17 257 sin 120◦ = sin 39.73◦

The resultant from which, length of tie, QR = 10.0 sin 39.73◦
√ sin 120◦

OA = 17 257 = 131.4 V = 7.38 m

Applying the sine rule:

131.4 = 100 Now try the following exercise
sin 135◦ sin AOB
Exercise 97 Further problems on practical
from which, sin AOB = 100 sin 135◦ = 0.5381 situations involving
trigonometry
131.4
1. A ship P sails at a steady speed of 45 km/h in
Hence angle AOB = sin−1 0.5381 = 32.55◦ (or 147.45◦, a direction of W 32◦ N (i.e. a bearing of 302◦)
which is impossible in this case). from a port. At the same time another ship Q

Hence the resultant voltage is 131.4 volts at
32.55◦ to V1

222 Engineering Mathematics

leaves the port at a steady speed of 35 km/h in 6. A laboratory 9.0 m wide has a span roof that
a direction N 15◦ E (i.e. a bearing of 015◦). slopes at 36◦ on one side and 44◦ on the other.
Determine their distance apart after 4 hours.
Determine the lengths of the roof slopes.
[193 km]
[6.35 m, 5.37 m]
2. Two sides of a triangular plot of land are 52.0 m
and 34.0 m, respectively. If the area of the plot 7. PQ and QR are the phasors representing the
is 620 m2 find (a) the length of fencing required
to enclose the plot and (b) the angles of the alternating currents in two branches of a cir-
triangular plot.
[(a) 122.6 m (b) 94.80◦, 40.66◦, 44.54◦] cuit. Phasor PQ is 20.0 A and is horizontal.

3. A jib crane is shown in Fig. 25.13. If the tie Phasor QR (which is joined to the end of PQ
rod PR is 8.0 long and PQ is 4.5 m long deter-
Section 3 mine (a) the length of jib RQ and (b) the angle to form triangle PQR) is 14.0 A and is at an
between the jib and the tie rod. angle of 35◦ to the horizontal. Determine the
[(a) 11.4 m (b) 17.55◦]
resultant phasor PR and the angle it makes with
R
phasor PQ [32.48 A, 14.31◦]

25.6 Further practical situations
involving trigonometry

130° P Problem 11. A vertical aerial stands on

horizontal ground. A surveyor positioned due east

of the aerial measures the elevation of the top as
48◦. He moves due south 30.0 m and measures the
elevation as 44◦. Determine the height of the aerial

Q

In Fig. 25.16, DC represents the aerial, A is the initial

Figure 25.13 position of the surveyor and B his final position.

4. A building site is in the form of a quadrilateral From triangle ACD, tan 48◦ = DC , from which
as shown in Fig. 25.14, and its area is 1510 m2. AC
Determine the length of the perimeter of the DC
site. [163.4 m] AC = tan 48◦

D

28.5 m 34.6 m
72°

52.4 m 75°

48°
CA

Figure 25.14 44° 30.0 m

5. Determine the length of members BF and EB B
in the roof truss shown in Fig. 25.15.
[BF = 3.9 m, EB = 4.0 m] Figure 25.16
DC
E
Similarly, from triangle BCD, BC = tan 44◦
4m 4m
For triangle ABC, using Pythagoras’ theorem:
F D
2.5 m 50° 50° 2.5 m BC2 = AB2 + AC2

A 5m B 5m C

DC 2 DC 2
tan 44◦ tan 48◦
Figure 25.15 = (30.0)2 +

Triangles and some practical applications 223

DC2 11 = 30.02 (b) Figure 25.18 shows the initial and final positions of
tan2 44◦ − tan2 48◦ the crank mechanism. In triangle OA B , applying
the sine rule:
DC2(1.072323 − 0.810727) = 30.02

DC2 = 30.02 = 3440.4 30.0 = 10.0 O
0.261596 sin 120◦ sin A B
√
from which, 10.0 sin 120◦
Hence, height of aerial, DC = 3340.4 sin A B O =
= 58.65 m.
30.0
= 0.2887

Problem 12. A crank mechanism of a petrol 30.0 cm A A′ Section 3
engine is shown in Fig. 25.17. Arm OA is 10.0 cm 120°
long and rotates clockwise about 0. The connecting
rod AB is 30.0 cm long and end B is constrained to 50°
move horizontally 10.0 cm

B B′ O

30.0 cm A Figure 25.19

10.0 cm Hence A B O = sin−1 0.2887 = 16.78◦ (or 163.22◦
50° which is impossible in this case).
Angle OA B = 180◦ − 120◦ − 16.78◦ = 43.22◦
BO

Figure 25.17 Applying the sine rule:

(a) For the position shown in Fig. 25.17 determine 30.0 OB
the angle between the connecting rod AB and sin 120◦ = sin 43.22◦
the horizontal and the length of OB.
from which, OB = 30.0 sin 43.22◦
(b) How far does B move when angle AOB sin 120◦
changes from 50◦ to 120◦?

= 23.72 cm

(a) Applying the sine rule: Since OB = 35.43 cm and OB = 23.72 cm then

AB AO BB = 35.43 − 23.72 = 11.71 cm
sin 50◦ = sin B
from which, Hence B moves 11.71 cm when angle AOB
AO sin 50◦ changes from 50◦ to 120◦
sin B =

AB
10.0 sin 50◦
= = 0.2553

30.0

Hence B = sin−1 0.2553 = 14.78◦ (or 165.22◦, Problem 13. The area of a field is in the form of a
quadrilateral ABCD as shown in Fig. 25.19.
which is impossible in this case). Determine its area

Hence the connecting rod AB makes an angle B 42.5 m
of 14.78◦ with the horizontal. 39.8 m C

Angle OAB = 180◦ − 50◦ − 14.78◦ = 115.22◦ 56°

Applying the sine rule: 114° 62.3 m
30.0 OB A
sin 50◦ = sin 115.22◦ 21.4 m

from which, OB = 30.0 sin 115.22◦ D
Figure 25.19
sin 50◦

= 35.43 cm

224 Engineering Mathematics

A diagonal drawn from B to D divides the quadrilateral 5. A surveyor, standing W 25◦ S of a tower
into two triangles. measures the angle of elevation of the top of
the tower as 46◦30 . From a position E 23◦S
Area of quadrilateral ABCD from the tower the elevation of the top is
37◦15 . Determine the height of the tower if the
= area of triangle ABD + area of triangle BCD distance between the two observations is 75 m.
[36.2 m]
= 1 (39.8)(21.4) sin 114◦ + 1 (42.5)(62.3) sin 56◦
2 2 6. Calculate, correct to 3 significant figures, the
co-ordinates x and y to locate the hole centre
= 389.04 + 1097.5 = 1487 m2 at P shown in Fig. 25.21.
[x = 69.3 mm, y = 142 mm]
Section 3 Now try the following exercise
P
Exercise 98 Further problems on practical
situations involving
trigonometry

1. Three forces acting on a fixed point are repre-

sented by the sides of a triangle of dimensions y

7.2 cm, 9.6 cm and 11.0 cm. Determine the

angles between the lines of action and the three

forces. [80.42◦, 59.38◦, 40.20◦] 116° 140°

x 100 mm

2. A vertical aerial AB, 9.60 m high, stands on Figure 25.21
ground which is inclined 12◦ to the horizontal.

A stay connects the top of the aerial A to a point

C on the ground 10.0 m downhill from B, the 7. An idler gear, 30 mm in diameter, has to be

foot of the aerial. Determine (a) the length of fitted between a 70 mm diameter driving gear

the stay, and (b) the angle the stay makes with and a 90 mm diameter driven gear as shown

the ground. [(a) 15.23 m (b) 38.07◦] in Fig. 25.22. Determine the value of angle θ

between the centre lines. [130◦]

3. A reciprocating engine mechanism is shown

in Fig. 25.20. The crank AB is 12.0 cm long 90 mm
dia
and the connecting rod BC is 32.0 cm long.

For the position shown determine the length

of AC and the angle between the crank and the

connecting rod. [40.25 cm, 126.05◦] 30 mm
dia
99.78 mm θ
70 mm
4. From Fig. 25.20, determine how far C moves, dia

correct to the nearest millimetre when angle
CAB changes from 40◦ to 160◦, B moving in

an anticlockwise direction. [19.8 cm]

B Figure 25.22
40°
A 8. 16 holes are equally spaced on a pitch circle

C

of 70 mm diameter. Determine the length of

the chord joining the centres of two adjacent

Figure 25.20 holes. [13.66 mm]

Chapter 26

Trigonometric identities and
equations

26.1 Trigonometric identities i.e. 1+ b 2 c2
Hence
=
aa

A trigonometric identity is a relationship that is true 1 + tan2 θ = sec2 θ (3)
for all values of the unknown variable.
Dividing each term of equation (1) by b2 gives:
tan θ = sin θ cot θ = cos θ sec θ = 1
cos θ sin θ cos θ a2 b2 c2 (4)
b2 + b2 = b2 (5)
cosec θ= 1 and cot θ = 1
sin θ tan θ a2 c2
i.e. + 1 =

bb

are examples of trigonometric identities from Chapter 22. Hence cot2 θ + 1 = cosec2 θ
Applying Pythagoras’ theorem to the right-angled trian-
gle shown in Fig. 26.1 gives: Equations (2), (3) and (4) are three further examples of
trigonometric identities.
a2 + b2 = c2 (1)

c 26.2 Worked problems on
b trigonometric identities

q

a

Figure 26.1 Problem 1. Prove the identity
Dividing each term of equation (1) by c2 gives: sin2 θ cot θ sec θ = sin θ

a2 + b2 = c2 With trigonometric identities it is necessary to start with
c2 c2 c2 the left-hand side (LHS) and attempt to make it equal
to the right-hand side (RHS) or vice-versa. It is often
a2 b 2 useful to change all of the trigonometric ratios into sines
i.e. + = 1 and cosines where possible. Thus

cc

(cos θ)2 + (sin θ)2 = 1

Hence cos2 θ + sin2 θ = 1 (2) LHS = sin2 θ cot θ sec θ

Dividing each term of equation (1) by a2 gives: = sin2 θ cos θ 1

a2 b2 c2 sin θ cos θ
a2 + a2 = a2
= sin θ (by cancelling) = RHS

226 Engineering Mathematics

Problem 2. Prove that: Problem 5. Prove that:
tan x + sec x
=1 1 − sin x = sec x − tan x
tan x 1 + sin x
sec x 1 +

sec x

LHS = tan x + sec x LHS = 1 − sin x
sec x 1 + tan x 1 + sin x

sec x = (1 − sin x)(1 − sin x)
(1 + sin x)(1 − sin x)
Section 3 sin x + 1
cos x⎛ cos x
= sin x ⎞ (1 − sin x)2
= (1 − sin2 x)
1 ⎜⎝1 + cos x ⎠⎟
cos x 1

cos x Since cos2 x + sin2 x = 1 then 1 − sin2 x = cos2 x

sin x + 1

=1 cos x cos x LHS = (1 − sin x)2 = (1 − sin x)2
cos x 1 + sin x 1 (1 − sin2 x) cos2 x

cos x = 1 − sin x = 1 − sin x
cos x cos x cos x
sin x + 1

= cos x = sec x − tan x = RHS
1 [1 + sin x]
cos x

= sin x + 1 cos x Now try the following exercise

cos x 1 + sin x

= 1 (by cancelling) = RHS Exercise 99 Further problems on
trigonometric identities
Problem 3. Prove that: 1 + cot θ = cot θ
1 + tan θ Prove the following trigonometric identities:

cos θ sin θ + cos θ 1. sin x cot x = cos x
1+
LHS = 1 + cot θ = = sin θ 2. √ 1 = cosec θ
1 + tan θ sin θ cos θ + sin θ 1 − cos2 θ
1 + sin θ
cos θ cos θ
3. 2 cos2 A − 1 = cos2 A − sin2 A
= sin θ + cos θ cos θ
sin θ cos θ + sin θ 4. cos x − cos3 x = sin x cos x
sin x
= cos θ = cot θ = RHS
sin θ 5. (1 + cot θ)2 + (1 − cot θ)2 = 2 cosec2 θ

Problem 4. Show that: 6. sin2 x( sec x + cosec x) = 1 + tan x
cos2 θ − sin2 θ = 1 − 2 sin2 θ cos x tan x

From equation (2), cos2 θ + sin2 θ = 1, from which, 26.3 Trigonometric equations
cos2 θ = 1 − sin2 θ
Hence, LHS = cos2 θ − sin2 θ Equations which contain trigonometric ratios are called
trigonometric equations. There are usually an infinite
= (1 − sin2 θ) − sin2 θ number of solutions to such equations; however, solu-
tions are often restricted to those between 0◦ and 360◦.
= 1 − sin2 θ − sin2 θ A knowledge of angles of any magnitude is essential in

= 1 − 2 sin2 θ = RHS

Trigonometric identities and equations 227

the solution of trigonometric equations and calculators 26.4 Worked problems (i) on
cannot be relied upon to give all the solutions (as shown trigonometric equations
in Chapter 23). Figure 26.2 shows a summary for angles
of any magnitude. Problem 6. Solve the trigonometric equation:
5 sin θ + 3 = 0 for values of θ from 0◦ to 360◦
90°

Sine All positive 5 sin θ + 3 = 0, from which
(and cosecant) sin θ = −3/5 = −0.6000
positive 0°
360° y y ϭ sinq Section 3
180° Cosine 1.0 216.87° 323.13°
(and secant)
Tangent positive 0 90° 180° 270° 360° q
(and cotangent) Ϫ0.6
positive Ϫ1.0 (a)
90°
270° SA

Figure 26.2

Equations of the type a sin2A + b sin A + c = 0

(i) When a = 0, b sin A + c = 0, hence sin A = − c
b
and A = sin−1 − c
b
There are two values of A between 0◦ and 180° a 0°
a 360°
360◦ that satisfy such an equation, provided
c
−1 ≤ ≤ 1 (see Problems 6 to 8).
b
(ii) When b = 0, a sin2 A + c = 0, hence sin2 A = − c , TC
270°
a (b)

sin A = − c and A = sin−1 − c Figure 26.3
aa

If either a or c is a negative number, then the value

within the square root sign is positive. Since when

a square root is taken there is a positive and nega- Hence θ = sin−1(−0.6000). Sine is negative in the third
tive answer there are four values of A between 0◦
and 360◦ which satisfy such an equation, provided and fourth quadrants (see Fig. 26.3). The acute angle
−1 ≤ c ≤ 1 (see Problems 9 and 10). sin−1 (0.6000) = 36.87◦ (shown as α in Fig. 26.3(b)).
Hence θ = 180◦ + 36.87◦, i.e. 216.87◦ or θ = 360◦ −
b 36.87◦, i.e. 323.13◦

(iii) When a, b and c are all non-zero: Problem 7. Solve: 1.5 tan x − 1.8 = 0 for
a sin2 A + b sin A + c = 0 is a quadratic equation 0◦ ≤ x ≤ 360◦

in which the unknown is sin A. The solution of

a quadratic equation is obtained either by fac-

torising (if possible) or by using the quadratic

formula: √ 1.5 tan x − 1.8 = 0, from which
sin A = −b ± b2 − 4ac tan x = 1.8 = 1.2000
2a
1.5
(see Problems 11 and 12). Hence x = tan−1 1.2000

(iv) Often the trigonometric identities Tangent is positive in the first and third quadrants (see
cos2 A + sin2 A = 1, 1 + tan2 A = sec2 A and Fig. 26.4).
cot2 A + 1 = cosec2A need to be used to reduce
The acute angle tan−1 1.2000 = 50.19◦.
equations to one of the above forms (see Problems
Hence, x = 50.19◦ or 180◦ + 50.19◦ = 230.19◦
13 to 15).

228 Engineering Mathematics

y 2. 3 cosec A + 5.5 = 0
y ϭ tan x [A = 213.06◦ or 326.94◦]

1.2 180° 270° 360° 3. 4(2.32 − 5.4 cot t) = 0
0 90° 230.19° x [t = 66.75◦ or 246.75◦]
50.19°

(a) 26.5 Worked problems (ii) on
trigonometric equations
90°
Section 3 S A Problem 9. Solve: 2 − 4 cos2 A = 0 for values of
180° A in the range 0◦ < A < 360◦
50.19° 0°
50.19° 360°

T C 2 − 4 cos2 A = 0, from which cos2A = 2 = 0.5000
270° 4

(b) √
Hence cos A = 0.5000 = ±0.7071 and
Figure 26.4 A = cos−1 (±0.7071)

Problem 8. Solve: 4 sec t = 5 for values of t Cosine is positive in quadrant one and four and negative
between 0◦ and 360◦ in quadrants two and three. Thus in this case there are
four solutions, one in each quadrant (see Fig. 26.6).The
4 sec t = 5, from which sec t = 5 = 1.2500 acute angle cos−1 0.7071 = 45◦.
4
Hence A = 45◦, 135◦, 225◦ or 315◦
Hence t = sec−1 1.2500

90° y y ϭ cos A
1.0
SA
0.7071

180° 36.87° 0° 135° 225°
36.87° 360°
0 45° 180° 315° 360° A°

TC Ϫ0.7071
Ϫ1.0
270°
(a)
Figure 26.5
90°
Secant (=1/cosine) is positive in the first and
fourth quadrants (see Fig. 26.5). The acute angle SA
sec−1 1.2500 = 36.87◦. Hence
180° 45° 45° 0°
t = 36.87◦ or 360◦ − 36.87◦ = 323.13◦ 45° 45° 360°

TC

Now try the following exercise 270°
(b)

Exercise 100 Further problems on Figure 26.6
trigonometric equations

Solve the following equations for angles between Problem 10. Solve: 1 cot2 y = 1.3 for
0◦ and 360◦ 0◦ < y < 360◦ 2

1. 4 − 7 sin θ = 0 [θ = 34.85◦ or 145.15◦] 1 cot2 cot2 y
2
y = 1.3, from which, = 2(1.3) = 2.6

Trigonometric identities and equations 229

√ Hence 3 cos θ − 2 = 0, from which,
Hence cot y = 2.6 = ±1.6125, and
y = cot−1 (±1.6125). There are four solutions, one in cos θ = 2 = 0.6667, or 2 cos θ + 3 = 0, from which,
each quadrant. The acute angle cot−1 1.6125 = 31.81◦. 3
3
Hence y = 31.81◦, 148.19◦, 211.81◦ or 328.19◦ cos = − 2 = −1.5000

The minimum value of a cosine is −1, hence the latter

expression has no solution and is thus neglected. Hence

Now try the following exercise θ = cos−1 0.6667 = 48.18◦ or 311.82◦

Exercise 101 Further problems on since cosine is positive in the first and fourth quadrants.
trigonometric equations
Section 3
Solve the following equations for angles between Now try the following exercise
0◦ and 360◦
Exercise 102 Further problems on
1. 5 sin2 y = 3 y = 50.77◦, 129.23◦, trigonometric equations
230.77◦ or 309.23◦
Solve the following equations for angles between
2. 5 + 3 cosec2D = 8 [D = 90◦ or 270◦] 0◦ and 360◦
3. 2 cot2 θ = 5
[θ = 32.32◦, 147.68◦, 1. 15 sin2 A + sin A − 2 = 0
212.32◦ or 327.68◦] A = 19.47◦, 160.53◦,
203.58◦ or 336.42◦
26.6 Worked problems (iii) on
trigonometric equations 2. 8 tan2 θ + 2 tan θ = 15
θ = 51.34◦, 123.69◦,
Problem 11. Solve the equation: 231.34◦ or 303.69◦

8 sin2 θ + 2 sin θ − 1 = 0, for all values of θ 3. 2 cosec2t − 5 cosec t = 12
between 0◦ and 360◦ [t = 14.48◦, 165.52◦, 221.81◦ or 318.19◦]

Factorising 8 sin2 θ + 2 sin θ − 1 = 0 gives 26.7 Worked problems (iv) on
(4 sin θ − 1) (2 sin θ + 1) = 0 trigonometric equations

Hence 4 sin θ − 1 = 0, from which,

sin θ = 1 = 0.2500, or 2 sin θ + 1 = 0, from which, Problem 13. Solve: 5 cos2 t + 3 sin t − 3 = 0 for
4 values of t from 0◦ to 360◦
1
sin θ = − 2 = −0.5000

(Instead of factorising, the quadratic formula can, of Since cos2 t + sin2 t = 1, cos2 t = 1 − sin2 t.
Substituting for cos2 t in 5 cos2 t + 3 sin t − 3 = 0 gives
course, be used). θ = sin−1 0.2500 = 14.48◦ or 165.52◦,
5(1 − sin2 t) + 3 sin t − 3 = 0
since sine is positive in the first and second quadrants, 5 − 5 sin2 t + 3 sin t − 3 = 0
or θ = sin−1 (−0.5000) = 210◦ or 330◦, since sine is
−5 sin2 t + 3 sin t + 2 = 0
negative in the third and fourth quadrants. Hence 5 sin2 t − 3 sin t − 2 = 0

θ = 14.48◦, 165.52◦, 210◦ or 330◦

Problem 12. Solve: Factorising gives (5 sin t + 2)(sin t − 1) = 0. Hence
6 cos2 θ + 5 cos θ − 6 = 0 for values of θ from 0◦ to
360◦ 5 sin t + 2 = 0, from which, sin t = − 2 = −0.4000, or
5
Factorising 6 cos2 θ + 5 cos θ − 6 = 0 gives sin t − 1 = 0, from which, sin t = 1.
(3 cos θ − 2)(2 cos θ + 3) = 0.
t = sin−1 (−0.4000) = 203.58◦ or 336.42◦, since sine

is negative in the third and fourth quadrants, or

230 Engineering Mathematics

t = sin−1 1 = 90◦. Hence Since the left-hand side does not factorise the quadratic
t = 90◦, 203.58◦ or 336.42◦ formula is used.

as shown in Fig. 26.7. Thus, −(−4) ± (−4)2 − 4(3)(−2)
cot θ =
1.0 y ϭ sin t
y 203.58° 336.42° 2(3)
√√
0 90° 270° 360° t = 4 ± 16 + 24 = 4 ± 40
Ϫ0.4
66
Ϫ1.0
Section 3 = 10.3246 or − 2.3246
66

Figure 26.7 Hence cot θ = 1.7208 or −0.3874
θ = cot−1 1.7208 = 30.17◦ or 210.17◦, since cotangent

is positive in the first and third quadrants, or
θ = cot−1 (−0.3874) = 111.18◦ or 291.18◦, since

cotangent is negative in the second and fourth quadrants.

Problem 14. Solve: 18 sec2 A − 3 tan A = 21 for Hence, θ = 30.17◦, 111.18◦, 210.17◦ or 291.18◦
values of A between 0◦ and 360◦
Now try the following exercise
1 + tan2 A = sec2 A. Substituting for sec2 A in
18 sec2 A − 3 tan A = 21 gives Exercise 103 Further problems on
trigonometric equations
18(1 + tan2 A) − 3 tan A = 21
Solve the following equations for angles between
i.e. 18 + 18 tan2 A − 3 tan A − 21 = 0 0◦ and 360◦
18 tan2 A − 3 tan A − 3 = 0
1. 12 sin2 θ − 6 = cos θ
Factorising gives (6 tan A − 3) (3 tan A + 1) = 0 θ = 48.19◦, 138.59◦,
221.41◦ or 311.81◦

Hence 6 tan A − 3 = 0, from which, tan A = 3 = 0.5000 2. 16 sec x − 2 = 14 tan2 x
6 [x = 52.93◦ or 307.07◦]
1
or 3 tan A + 1 = 0, from which, tan A = − 3 = −0.3333.

Thus A = tan−1(0.5000) = 26.57◦ or 206.57◦, since

tangent is positive in the first and third quadrants, or 3. 4 cot2 A − 6 cosec A + 6 = 0 [A = 90◦]
A = tan−1 (−0.3333) = 161.57◦ or 341.57◦, since tan-

gent is negative in the second and fourth quadrants. 4. 5 sec t + 2 tan2 t = 3
[t = 107.83◦ or 252.17◦]
Hence

A = 26.57◦, 161.57◦, 206.57◦ or 341.57◦ 5. 2.9 cos2 a − 7 sin a + 1 = 0
[a = 27.83◦ or 152.17◦]

Problem 15. Solve: 3 cosec2θ − 5 = 4 cot θ in the 6. 3 cosec2β = 8 − 7 cot β
range 0 < θ < 360◦ β = 60.17◦, 161.02◦,
240.17◦ or 341.02◦
cot2 θ + 1 = cosec2 θ, Substituting for cosec2 θ in
3 cosec2 θ − 5 = 4 cot θ gives:

3( cot2 θ + 1) − 5 = 4 cot θ
3 cot2 θ + 3 − 5 = 4 cot θ
3 cot2 θ − 4 cot θ − 2 = 0

Chapter 27

Compound angles

27.1 Compound angle formulae (a) sin (π + α) = sin π cos α + cos π sin α (from
the formula for sin (A + B))
An electric current i may be expressed as
i = 5 sin(ωt − 0.33) amperes. Similarly, the displace- = (0)( cos α) + (−1) sin α
ment x of a body from a fixed point can be expressed = −sin α
as x = 10 sin(2t + 0.67) metres. The angles (ωt − 0.33)
and (2t + 0.67) are called compound angles because (b) − cos (90◦ + β)
they are the sum or difference of two angles. = −[cos 90◦ cos β − sin 90◦ sin β]
= −[(0)( cos β) − (1) sin β] = sin β
The compound angle formulae for sines and cosines
of the sum and difference of two angles A and B are: (c) sin (A − B) − sin (A + B)
= [sin A cos B − cos A sin B]
sin(A + B)= sin A cos B + cos A sin B − [sin A cos B + cos A sin B]
= −2 cos A sin B
sin(A − B)= sin A cos B − cos A sin B
Problem 2. Prove that:
cos(A + B)= cos A cos B − sin A sin B cos (y − π) + sin y + π = 0
2
cos(A − B)= cos A cos B + sin A sin B

(Note, sin(A + B) is not equal to (sin A + sin B), and cos ( y − π) = cos y cos π + sin y sin π
so on.)

The formulae stated above may be used to derive two
further compound angle formulae:

tan(A + B)= tan A + tan B = (cos y)(−1) + ( sin y)(0)
1 − tan A tan B
= − cos y
tan(A − B)= tan A − tan B sin y + π = sin y cos π + cos y sin π
1 + tan A tan B 2
22
The compound-angle formulae are true for all values = (sin y)(0) + ( cos y)(1) = cos y
of A and B, and by substituting values of A and B into
the formulae they may be shown to be true. Hence cos ( y − π) + sin y + π
2

= (−cos y) + (cos y) = 0

Problem 1. Expand and simplify the following

expressions: Problem 3. Show that
ππ
(a) sin(π + α) (b) −cos(90◦ + β)
tan x + tan x − = −1
(c) sin(A − B) − sin(A + B) 44

232 Engineering Mathematics

tan x + tan π (c) tan (P + Q)

tan x + π = 4 = tan P + tan Q = (1.4025) + (2.0225)
4 π 1 − tan P tan Q 1 − (1.4025)(2.0225)
1 − tan x tan
4

(from the formula for tan(A + B)) = 3.4250 = −1.865
−1.8366
= tan x + 1 = 1 + tan x
1 − ( tan x)(1) 1 − tan x

Section 3 tan x − π tan x − tan π since tan π = 1 Problem 5. Solve the equation:
4 4 4 4 sin(x − 20◦) = 5 cos x for values of x between
= 1 + tan x tan π = 0◦ and 90◦
tan x − 1
1 + tan x 4 sin (x − 20◦) = 4[sin x cos 20◦ − cos x sin 20◦]
from the formula for sin (A − B)
4
= 4[sin x(0.9397) − cos x(0.3420)]
Hence, tan x + π tan x − π = 3.7588 sin x − 1.3680 cos x
44
Since 4 sin(x − 20◦) = 5 cos x then
= 1 + tan x tan x − 1 3.7588 sin x − 1.3680 cos x = 5 cos x
1 − tan x 1 + tan x Rearranging gives:

= tan x − 1 = −(1 − tan x) = −1 3.7588 sin x = 5 cos x + 1.3680 cos x
1 − tan x 1 − tan x
= 6.3680 cos x
Problem 4. If sin P = 0.8142 and and sin x = 6.3680 = 1.6942
cos Q = 0.4432 evaluate, correct to 3 decimal
places: (a) sin(P − Q), (b) cos(P + Q) and (c) cos x 3.7588
tan(P + Q), using the compound angle formulae
i.e. tan x = 1.6942, and
Since sin P = 0.8142 then x = tan−1 1.6942 = 59.449◦ or 59◦27
P = sin−1 0.8142 = 54.51◦
Thus cos P = cos 54.51◦ = 0.5806 and [Check: LHS = 4 sin(59.449◦ − 20◦)
tan P = tan 54.51◦ = 1.4025 = 4 sin 39.449◦ = 2.542
Since cos Q = 0.4432, Q = cos−1 0.4432 = 63.69◦
Thus sin Q = sin 63.69◦ = 0.8964 and RHS = 5 cos x = 5 cos 59.449◦ = 2.542]
tan Q = tan 63.69◦ = 2.0225
Now try the following exercise
(a) sin (P − Q)
= sin P cos Q − cos P sin Q Exercise 104 Further problems on
= (0.8142)(0.4432) − (0.5806)(0.8964) compound angle formulae
= 0.3609 − 0.5204 = −0.160
1. Reduce the following to the sine of one angle:
(b) cos (P + Q) (a) sin 37◦ cos 21◦ + cos 37◦ sin 21◦
= cos P cos Q − sin P sin Q (b) sin 7t cos 3t − cos 7t sin 3t
= (0.5806)(0.4432) − (0.8142)(0.8964) [(a) sin 58◦ (b) sin 4t]
= 0.2573 − 0.7298 = −0.473
2. Reduce the following to the cosine of one
angle:
(a) cos 71◦ cos 33◦ − sin 71◦ sin 33◦

Compound angles 233

ππ ππ (iii) If a = R cos α and b = R sin α, where a and b are
(b) cos cos + sin sin constants, then
3 4⎡ 3 4 R sin(ωt + α) = a sin ωt + b cos ωt, i.e. a sine
(a) cos 104◦ ≡ −cos76◦ ⎤ and cosine function of the same frequency when
added produce a sine wave of the same frequency
⎣π ⎦ (which is further demonstrated in Chapter 34).
(b) cos
12 (iv) Since a = R cos α, then cos α = a and since
R
3. Show that: Section 3
(a) sin x + π + sin x + 2π = b = R sin α, then sin α = b
33 R
√
3 cos x If the values of a and b are known then the values of
R and α may be calculated. The relationship between
(b) −sin 3π − φ = cos φ constants a, b, R and α are shown in Fig. 27.1.
2

4. Prove that:

π 3π
(a) sin θ + − sin θ − =
44
√
2(sin θ + cos θ) R b

(b) cos (270◦ + θ) = tan θ a
cos (360◦ − θ) a

5. Given cos A = 0.42 and sin B = 0.73 evaluate
(a) sin(A − B), (b) cos(A − B), (c) tan(A + B),
correct to 4 decimal places.
[(a) 0.3136 (b) 0.9495 (c) −2.4687]

In Problems 6 and 7, solve the equations for values Figure 27.1
of θ between 0◦ and 360◦
From Fig. 27.1, by Pythagoras’ theorem:
6. 3 sin(θ + 30◦) = 7 cos θ
[64.72◦ or 244.72◦] R = a2 + b2
and from trigonometric ratios:
7. 4 sin(θ − 40◦) = 2 sin θ
[67.52◦ or 247.52◦] α = tan−1 b
a
27.2 Conversion of a sin ωt + b cos ωt
into R sin(ωt + α) Problem 6. Find an expression for
3 sin ωt + 4 cos ωt in the form R sin(ωt + α) and
(i) R sin(ωt + α) represents a sine wave of maximum sketch graphs of 3 sin ωt, 4 cos ωt and
value R, periodic time 2π/ω, frequency ω/2π and R sin(ωt + α) on the same axes
leading R sin ωt by angle α. (See Chapter 23).
Let 3 sin ωt + 4 cos ωt = R sin(ωt + α)
(ii) R sin(ωt + α) may be expanded using the then 3 sinωt + 4 cos ωt
compound-angle formula for sin (A + B), where
A = ωt and B = α. Hence = R[sin ωt cos α + cos ωt sin α]
= (R cos α) sin ωt + (R sin α) cos ωt
R sin(ωt + α) Equating coefficients of sin ωt gives:
3 = R cos α, from which, cos α = 3
= R[ sin ωt cos α + cos ωt sin α]
R
= R sin ωt cos α + R cos ωt sin α Equating coefficients of cos ωt gives:

= (R cos α) sin ωt + (R sin α) cos ωt 4
4 = R sin α, from which, sin α =

R