CHAPTER 39 TRIGONOMETRIC WAVEFORMS
EXERCISE 165 Page 448
1.
Determine all of the angles between 0
°
and 360
°
whose sine is:
(a) 0.6792 (b) –0.1483
(a) Sine is positive in the 1st and 2nd quadrants
If sin
θ
= 0.6792, then
θ =
= 42.78
°
or 180
°
– 42.78
°
= 137.22
°
as shown in
diagram (i) below.
(i) (ii)
(b) Sine is negative in the 3rd and 4th quadrants
If sin
θ
= 0.1483, then
θ =
= 8.53
°
From diagram (ii) above, the two values where sin
θ
= –0.1483 are 180
°
+ 8.53
°
= 188.53
°
and
360
°
– 8.53
°
= 351.47
°
2.
Solve the following equations for values of x between 0
°
and 360
°
:
(a) x = cos
−
1 0.8739 (b) x = cos
−
1 (–0.5572)
(a) Cosine is positive in the 1st and 4th quadrants
x =
−
1 = 29.08
°
or 360
°
–29.08
°
= 330.92
°
as shown in diagram (i) below.
661 © 2014, John Bird

(i) (ii)
(b) Cosine is negative in the 2nd and 3rd quadrants
x =
−
1 = 56.14
°
hence, from diagram (ii) shown above, the two values of x for
which x = −
1
−
are: 180
°
– 56.14
°
= 123.86 and 180
°
+ 56.14
°
= 236.14
°
3.
Find the angles between 0
°
to 360
°
whose tangent is:
(a) 0.9728 (b) –2.3420
(a) Tangent is positive in the 1st and 3rd quadrants
θ =
= 44.21
°
or 180
°
+ 44.21
°
= 224.21
°
as shown in diagram (i) below.
(i) (ii)
(b) Tangent is negative in the 2nd and 4th quadrants
θ =
= 66.88
°
and from diagram (ii) shown above,
θ
= 180
°
– 66.88
°
= 113.12
°
and 360
°
– 66.88
°
= 293.12
°
4.
Solve, in the range 0
°
to 360
°
, giving the answers in degrees and minutes: cos −
1 (–0.5316) = t
Cosine is negative in the 2nd and 3rd quadrants.
= 57.886
°
or 57
°
53
′
as shown in the diagram below.
662 © 2014, John Bird

From the diagram, t = 180
°
– 57
°
53
′
= 122
°
7
′
and t = 180
°
+ 57
°
53
′
= 237
°
53
′
5.
Solve, in the range 0
°
to 360
°
, giving the answers in degrees and minutes: sin −
1
( −
0.6250
) = α
Sine is negative in the 3rd and 4th quadrants
If sin
θ
= 0.6250, then
α =
= 38.682
°
= 38
°
41
′
From the diagram above, the two values where sin
α
= –0.6250 are 180
°
+ 38
°
41
′
= 218
°
41
′
and
360
°
– 38
°
41
′
= 321
°
19
′
6.
Solve, in the range 0
°
to 360
°
, giving the answers in degrees and minutes: tan
−
1 0.8314 =
θ
θ
Tangent is positive in the 1st and 3rd quadrants
=
= 39.74
°
or 39
°
44
′
663 © 2014, John Bird

From the diagram, the two values of
θ
between 0
°
and 360
°
are:
39
°
44
′
and 180
°
+ 39
°
44
′
= 219
°
44
′
664 © 2014, John Bird

EXERCISE 166 Page 452
1. A sine wave is given by y = 5 sin 3 x . State its peak value.
Peak value = amplitude = maximum value = 5
2.
A sine wave is given by y = 4 sin 2 x . State its period in degrees.
Period =
360
°
= 180°
2
3. A periodic function is given by y = 30 cos 5 x . State its maximum value.
Maximum value = amplitude = peak value = 30
4.
A periodic function is given by y = 25 cos 3 x . State its period in degrees.
Period =
360
°
= 120°
3
5.
State the amplitude and period of y = cos 3 A and sketch the curve between 0
°
and 360
°
Amplitude = 1 and period =
360
°
3
A sketch of y = cos 3 A is shown below.
= 120°
6.
State the amplitude and period of y = 2 sin
5 x
2
and sketch the curve between 0
°
and 360
°
665 © 2014, John Bird

If y = 2 sin
5 x
2
, amplitude = 2 and period =
360
°
5
= 144
°
2
A sketch y = 2 sin
5 x
is shown below
2
7.
State the amplitude and period of y = 3 sin 4 t and sketch the curve between 0
°
and 360
°
Amplitude = 3 and period =
360
°
4
A sketch of y = 3 sin 4 t is shown below
= 90°
8.
State the amplitude and period of y = 5 cos
θ
2
and sketch the curve between 0
°
and 360
°
Amplitude = 5 and period =
360
°
1
= 720°
A sketch of y = 5 cos
θ
2
is shown below
2
666 © 2014, John Bird

9.
State the amplitude and period of y =
7
2 sin
3 x
and sketch the curve between 0
°
and 360
°
8
Amplitude =
7
2
= 3.5 and period =
360
°
3
= 960°
8
A sketch of y =
7
2 sin
3 x
is shown below
8
10.
State the amplitude and period of y = 6 sin( t – 45
°
) and sketch the curve between 0
°
and 360
°
If y = 6 sin(t – 45
°
), amplitude = 6 and period =
360
°
1
= 360°
A sketch y = 6 sin( t – 45
°
) is shown below.
667 © 2014, John Bird

11.
State the amplitude and period of y = 4 cos(2
θ
+ 30
°
) and sketch the curve between 0
°
and 360
°
If y = 4 cos(2
θ
+ 30
°
), amplitude = 4 and period =
360
°
= 180
°
2
A sketch y = 4 cos(2
θ
+ 30
°
) is shown below
(Note that y = 4 cos(2
θ
+ 30
°
) leads y = 4 cos 2
θ
by
30
°
2
= 15
°
)
12. The frequency of a sine wave is 200 Hz. Calculate the periodic time.
Periodic time T =
1
=
1 f 200
s = 5 ms
13 . Calculate the frequency of a sine wave that has a periodic time of 25 ms.
1
Frequency f =
T
=
1
−
3
= 40 Hz
14. Calculate the periodic time for a sine wave having a frequency of 10 kHz.
Periodic time T =
1 f
=
1
×
3
s
= 100 μs or 0.1 ms
15. An alternating current completes 15 cycles in 24 ms. Determine its frequency?
668 © 2014, John Bird

24
One cycle is completed in
15 ms = 1.6 ms Hence, periodic time T = 1.6 ms
1
Frequency f =
T
=
1
−
3
= 625 Hz
16. Graphs of y
1
= 2 sin x and y
2
= 3 sin( x + 50
°
) are drawn on the same axes. Is y
2
lagging or
leading y
1
? y
2
is leading y
1
by 50°
17. Graphs of y
1
= 6 sin x and y
2
= 5 sin( x – 70
°
) are drawn on the same axes. Is y
1
lagging or
leading y
2
? y
2
is lagging y
1
by 70° hence, y
1
is leading y
2
by 70°
669 © 2014, John Bird

EXERCISE 167 Page 454
1.
Find the (a) amplitude, (b) frequency, (c) periodic time and (d) phase angle (stating whether it is
leading or lagging sin
ω t ) for: i = 40 sin(50
π t + 0.29) mA.
If i = 40 sin(50
π t + 0.29) mA, then amplitude = 40 mA,
ω
= 50
π
rad/s = 2
π f from which, frequency, f =
50
π
2
π
= 25 Hz
periodic time, T =
1
=
1
= 0.040
s or 40 ms f 25 phase angle = 0.29 rad leading or 0.29
×
180
°
π
= 16.62
°
leading or 16
°
37
′
leading
2.
Find the (a) amplitude, (b) frequency, (c) periodic time and (d) phase angle (stating whether it is
leading or lagging sin
ω t ) for: y = 75 sin(40 t – 0.54) cm.
If y = 75 sin(40 t – 0.54) cm, then amplitude = 75 cm
ω
= 40 rad/s = 2
π f from which, frequency, f =
40
2
π
= 6.37 Hz
periodic time, T =
1
=
1 f 6.37
= 0.157 s phase angle = 0.54 rad lagging or 0.54
×
180
°
π
= 30.94
°
lagging or 30
°
56
′
lagging
3.
Find the (a) amplitude, (b) frequency, (c) periodic time and (d) phase angle (stating whether it is
leading or lagging sin
ω t ) for: v = 300 sin(200
π t – 0.412) V.
If v = 300 sin(200
π t – 0.412) V, then amplitude = 300 V
ω
= 200
π
rad/s = 2
π f from which, frequency, f =
200
π
2
π
= 100 Hz
periodic time, T =
1
=
1
= 0.010 s or 10 ms f 100 phase angle = 0.412 rad lagging or 0.412
×
180
°
π
= 23.61
°
lagging or 23
°
36
′
lagging
670 © 2014, John Bird

4.
A sinusoidal voltage has a maximum value of 120 V and a frequency of 50 Hz. At time t = 0, the
voltage is (a) zero, and (b) 50 V. Express the instantaneous voltage v in the form v = A sin(
ω t ±
α
).
Let v = A sin(
ω t
± α
) = 120 sin(2
π ft +
φ
) = 120 sin(100
π t +
φ
) volts, since f = 50 Hz
(a) When t = 0, v = 0 hence, 0 = 120 sin(0 +
φ
), i.e. 0 = 120 sin
φ
from which, sin
φ
= 0 and
φ
= 0
Hence, if v = 0 when t = 0, then v = 120 sin 100
π t volts
(b) When t = 0, v = 50 V hence, 50 = 120 sin(0 +
φ
)
50
from which,
120
= sin
φ
and
φ = sin
−
1

50


120

=
24.624
°=
24.624
×
π
180
=
0.43 rad
Hence, if v = 50 when t = 0, then v = 120 sin(100
π t + 0.43) volts
5.
An alternating current has a periodic time of 25 ms and a maximum value of 20 A. When
time = 0, current i = –10 amperes. Express the current i in the form i = A sin(
ω t ±
α
).
If periodic time T = 25 ms, then frequency, f
Angular velocity,
ω
= 2
π t =2
π
(40) = 80
π
rad/s
=
1
=
T
Hence, current i = 20 sin(80
π t +
φ
)
When t = 0, i = –10, hence –10 = 20 sin
φ
1
−
3
= 40 Hz from which, sin
φ
=
−
10
= −
20
0.5
and
φ = sin ( 0.5) 30
Thus, i
=
20 sin 80
π t
−
π
6
A or i
= ( π t
−
0.524
)
A or
−
π
6 rad
6.
An oscillating mechanism has a maximum displacement of 3.2
m and a frequency of 50 Hz. At
time t = 0 the displacement is 150 cm. Express the displacement in the general form A sin(
ω t ±
α
).
Displacement, s = A sin(
ω t ±
α
) where A = 3.2
m and
ω =
2
π f
=
2
π ×
50 100
π
671 © 2014, John Bird

i.e. s = 3.2 sin(100
πt + α
)
At time t = 0, displacement = 150 cm = 1.5 m
Hence, 1.5 = 3.2 sin(100
πt + α
) = 3.2 sin
α i.e.
1.5
3.2
= sin
α and
α
= sin
−
1

1.5

= 27.953° = 0.488 rad

3.2

Hence, displacement = 3.2 sin(100πt + 0.488) m
7.
The current in an a.c. circuit at any time t seconds is given by:
i = 5 sin(100
π t – 0.432) amperes
Determine the (a) amplitude, frequency, periodic time and phase angle (in degrees)
(b) value of current at t = 0
(c) value of current at t = 8 ms
(d) time when the current is first a maximum
(e) time when the current first reaches 3 A.
Sketch one cycle of the waveform showing relevant points.
(a) If i = 5 sin(100
π t – 0.432) mA, then amplitude = 5 A,
ω
= 100
π
rad/s = 2
π f from which, frequency, f =
100
π
2
π
= 50 Hz
periodic time, T =
1
=
1
= 0.020
s or 20 ms f 50
and phase angle = 0.432 rad lagging or 0.432
×
180
°
π
= 24.75
°
lagging or 24
°
45
′
lagging.
(b) When t = 0, i = 5 sin(–0.432) = –2.093 A (note that –0.432 is radians)
(c) When t = 8 ms, i = 5 
π ( × −
3
) −
0.432
 = 5 sin (2.081274) = 4.363 A
(d) When the current is first a maximum, 5 = 5 sin(100
π t – 0.432)
i.e. 1 = sin(100
π t – 0.432)
672 © 2014, John Bird

and 100
π t – 0.432 = −
1
=
(again, be sure your calculator is on
radians)
from which, time t =
100
π
= 0.006375
(e) When i = 3 A, 3 = 5 sin(100
π t – 0.432) s or 6.375 ms
i.e.
3
5
= sin(100
π t – 0.432)
and 100
π t – 0.432 = sin −
1
3
5
=
0.6435
from which, time t =
100
π
A sketch of one cycle of the waveform is shown below.
= 0.003423
s or 3.423 ms
Note that since phase angle
φ
= 24.75
°
, in terms of time
φ t
then
24.75
≡
φ t
360 20
from which,
φ t
= 1.375 ms
Alternatively,
φ t
=
φ
ω
=
0.432
100
π
= 1.375 ms, as shown in the sketch.
673 © 2014, John Bird

EXERCISE 168 Page 459
1.
A complex current waveform i comprises a fundamental current of 50 A r.m.s. and frequency 100
Hz, together with a 24% third harmonic, both being in phase with each other at zero time. (a) Write down an expression to represent current i . (b) Sketch the complex waveform of current using harmonic synthesis over one cycle of the fundamental.
(a) Fundamental current: r.m.s. =
1
2
× maximum value
from which, maximum value = 2 r.m.s.
= ×
= 70.71 A
Hence, fundamental current is: i
1
= 70.71 sin 2
π
(100) t = 70.71 sin 628.3
t A
Third harmonic: amplitude = 24% of 70.71 = 16.97 A
Hence, third harmonic current is: i
3
= 16.97 sin 3(628.3) t = 16.97 sin 1885 t A
Thus, current i = i
1
+ i
3
= (70.71 sin 628.3
t + 16.97 sin 1885 t ) amperes
(b) The complex waveform for current i is shown sketched below:
2.
A complex voltage waveform v comprises a 212.1
V r.m.s. fundamental voltage at a frequency of
50 Hz, a 30% second harmonic component lagging the fundamental voltage at zero time by
π
/2 rad, and a 10% fourth harmonic component leading the fundamental at zero time by
π
/3 rad. (a) Write down an expression to represent voltage v . (b) Sketch the complex voltage waveform using harmonic synthesis over one cycle of the fundamental waveform.
674 © 2014, John Bird

(a) Voltage,
212.1
v =
0.707
π t
+
(0.30)
212.1
0.707


π t
−
π
2


+
(0.1)
212.1
0.707
volts
i.e. v = 300 sin 314.2
t + 90 sin 628.3
t
−
π
2

+
30 sin 1256.6
t
+
π
3
(b) The complex waveform representing v is shown sketched below


volts
π t
+
π
3


3.
A voltage waveform is represented by: v = 20 + 50 sin
ω t + 20 sin(2
ω t –
π
/2) volts.
Draw the complex waveform over one cycle of the fundamental by using harmonic synthesis.
One waveform of v = 20 + 50 sin
ω t + 20 sin(2
ω t –
π
/2) volts is shown sketched below using harmonic synthesis.
675 © 2014, John Bird

4.
Write down an expression representing a current i having a fundamental component of amplitude
16 A and frequency 1 kHz, together with its third and fifth harmonics being respectively one-fifth
and one-tenth the amplitude of the fundamental, all components being in phase at zero time. Sketch
the complex current waveform for one cycle of the fundamental using harmonic synthesis.
Fundamental current i
1
= 16 sin 2
π
(1000) t = 16 sin 2000
πt = 16 sin 2π
10 3 t
Third harmonic i
3
=
1
5
(16) sin 3(2000
πt) = 3.2 sin 6000πt = 3.2 sin 6π
10 3 t
1
Fifth harmonic i
5
=
10
(16) sin 5(2000
πt) = 1.6 sin 10 000πt = 1.6 sin π
10 4 t
Hence, current, i
= 16 sin 2π
10 3 t + 3.2 sin 6
π
10 3 t + 1.6 sin
π
10 4 t
676 © 2014, John Bird

A sketch of current is shown below
5.
A voltage waveform is described by:
v = 200 sin 377 t + 80 sin(1131 t +
π
4
) + 20 sin(1885 t –
π
3
) volts
Determine (a) the fundamental and harmonic frequencies of the waveform, (b) the percentage third harmonic and (c) the percentage fifth harmonic. Sketch the voltage waveform using harmonic synthesis over one cycle of the fundamental.
(a) From the fundamental voltage, 377 =
ω
1
=
2
π f
1
i.e. fundamental frequency, f
1
=
377
2
π
= 60 Hz
From the 3rd harmonic voltage, 1131 =
ω
3
=
2
π f
3
i.e. 3rd harmonic frequency, f
3
=
1131
2
π
= 180 Hz
677 © 2014, John Bird

From the 5th harmonic voltage, 1885 =
ω
5
=
2
π f
5
i.e. 5th harmonic frequency, f
5
=
1885
2
π
= 300 Hz
(b) Percentage 3rd harmonic =
80
200
×
100% = 40%
(c) Percentage 5th harmonic =
20
200
×
100% = 10%
The complex waveform representing v is shown sketched below:
678 © 2014, John Bird