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Appendix (Complex Trigonometric Functions)
We define the trigonometric or circular functions
sin
z
,
cos
z
, etc., in terms of exponential
functions as follows:
Note that the usual formulas for exponentiation work with this definition.
The complex-valued basic trigonometric functions are also defined via exponential function as
follows:
sin
,
cos
sin
cos
tan
,
cot
cos
sin
sec
,
csc
cos
1
1
2
2
1
2
1
in
2
s
z
z
z
z
i
i
i
i
i
i
i
i
i
i
i
i
i
i
z
z
z
z
z
z
z
z
z
z
z
i
z
i
=
=
=
=
z
z
z
z
z
z
z
z
z
z
z
z
i
=
=
e
e
e
e
e
e
e
e
e
e
e
e
e
e
e
i
i
i
e
Many of the properties familiar in the case of real trigonometric functions also hold for the
complex trigonometric functions. For example, we have:
2
2
2
2
2
2
sin
cos
,
sec
tan
,
csc
cot
sin
sin ,
cos
cos ,
tan
tan
sin
sin
cos
cos
sin
,
cos
cos
sin
sin
cos
tan
t
1
1
1
1
an
tan
tan
tan
1
2
1
2
1
2
1
2
1
2
1
2
1
2
1
2
1
2
z
z
z
z
z
z
z
z
z
z
z
z
z
z
z
z
z
z
z
+
z
z
z
z
=
=
z
z
z
z
z
+
z
=
z
+
The following relations exit between the trigonometric or circular and the hyperbolic functions:
sin
sin ,
cos
cos ,
tan
tan
sin
sin ,
cos
cos ,
tan
tan
z
z
z
z
z
z
z
z
i
i
i
z
z
=
=
=
=
i
i
i
i
i
i
h
h
h
h
h
=
=
i
h
Example 19
(Finding of a Complex number)
:
In each part, express the value of the given trigonometric function in the form
a
b
i
:
a)
cos
i
b)
sin
2
i
c)
tan
2
i
IIUM, Faculty of Engineering,
Department Engineering in Science
Engineering Mathematics I
Semester 1, 2021/2022
Chapter I:
Complex Numbers
Lecturer
Associate Professor Dr. Abdurahim Okhunov
28
Solution:
a)
2
2
1
1
1
1
1
1
1
1
cos
5431
2
2
2
2
.
i i
i i
i
i
=
=
=
e
e
e
e
e
e
e
e
=
i
b)
In this case we have that
2
2
2
1
2
2
1 2
1 2
1 2
1
2
1
2
1
1
1
1
2
2
2
2
sin
cos
sin
cos
sin
.
.
.
1
1
2
2
1
2
2
2
2
2
1
0 9781 2 8062
1 4031 0 4891
2
.
.
2
2
i
i
i
i
i
i
i
i
i
i
i
i
i
i
e
e
e
e
e
e
i
i
i
i
i
i
i
e
e
e
e
e e
e
e
i
i
=
=
=
=
=
=
i
i
i
c)
In this case we have that
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
1
sin
2
2
2
2
1
2
tan
co
2
2
s
2
2
2
2
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
e
e
e
e
e
e
e
e
e
e
e
e
e
e
e
e
e
e
=
i
i
i
i
i
i
i
i
e
e
e
e
e
e
e
e
2
2
0 96
.
40
.
e
e
i
i
2
2
2
2
2
2
2
2
0
0
1
1
2
2
2
1
0
cos
sin
cos
sin
cos
sin
cos
sin
cos
sin
cos
sin
cos
sin
c
i
i
i
i
e
e
e
e
e
e
e
e
e
e
e
e
e
e
applying Eulor
i
i
i
i
i
i
formua
i
l
i
i
e
i
2
0
1
2
2
2
2
2
1
os
sin
1
1
1
1
1
i
i
e
e
e
e
e
i
i
i
2
2
1
e
e
2
2
2
2
2
2
.
e
e
e
i
i
e
e
e
Example 20
(Finding of a Complex number)
:
Express in the form
a
b
i
:
a)
6
cos
sin
2
2
i
and b)
10
1
3
IIUM, Faculty of Engineering,
Department Engineering in Science
Engineering Mathematics I
Semester 1, 2021/2022
Chapter I:
Complex Numbers
Lecturer
Associate Professor Dr. Abdurahim Okhunov
29
Solution:
6
a)
cos
sin
cos
sin
cos
si
6
6
2
2
2
2
3
3
1
0
1
n
.
i
i
i
i
10
10
10
10
10
1
3
2
2
3
3
3
3
1
3
1024
512
512
3
2
2
b)
cos
sin
cos
sin
.
i
i
i
i
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