Mathematics for IT

This article covers topics such as trigonometric functions, coordinate geometry, complex numbers, and matrices. It includes solved problems and examples for each topic.

MATHEMATICS FOR IT
[Author Name(s), First M. Last, Omit Titles and Degrees]
[Institutional Affiliation(s)]

Question One: Trigonometric Functions; Angles
(a)
(i) 7/sin 35=10/sin A
Sin A= (sin 35*10)/7
A=55⁰
(ii) s2 =72+102
s2=49+100
s2=149
s= √ 149=12.21 cm
(b) For a small Ɵ, a triangle as shown below is noticed
i. With a small value of Ɵ, the opposite angle i.e. a is very small with the adjacent angle i.e. b as
well as the hypotenuse (h) being relatively large. Hence,
Sin Ɵ=a/h ≈ 0
tan Ɵ=a/b≈ 0

besides, it should be noticed that h and b are almost the same height with h being only slightly
longer and hence
cos Ɵ=b/h≈ 1
b≈h and hence a/h≈a/b and thus sin Ɵ≈tan Ɵ
ii. When the C=9⁰;
10
sin 81 = AB
sin 9
AB= 10 sin 9
sin 81 =4.173 units
10
sin 81 = AC
sin 90
AC= 10 sin 9 0
sin 81 =10.12 units
Question 2: Applications of Trigonometric Functions
(a) Scalene triangle. Neither of the sides nor the angles is equal
(b) a2 =b2+c2-2bc cos A
=202+302-2*30*20 cos 25
a=√212.4=14.57 units
(c) b
sin B = a
sin A

30
sin B = 14.57
sin 25
sin B= 30 sin 25
14.57 =0.871
B=60.57⁰
(d) c
sin C = a
sin A
30
sin C = 14.57
sin 25
sin C= 20sin 25
14.57 =0.581
C=35.52⁰
(e) Area of the triangle
A= √s (s−a)(s−b)( s−c)
S=p/2= (20+30+14.57)/2=32.3
A= √ 32.3 (32.3−30)(32.3−20)(32.3−14.57)
√ 32.3× 2.3 ×12.3 ×17.73=127.28 sq. units
Question 3: Trigonometric Functions of Real Numbers
(a)

(i)15⁰
π Radians=180⁰
15⁰= 15× π
180 =0.8333 π
(ii) 240⁰
π Radians =180⁰
240⁰= 240× π
180 =1.333 π
-425⁰
π Radians=180⁰
(iii)-425⁰=−425 × π
180 =-2.361π
(b) Conversion to degrees
(i) π
3
π=1800; π
3 = π
3 ×180 × 1
π =60⁰
(ii) 5 π
8
π=1800; 5 π
8 = 5 π
8 ×180 × 1
π =112.5⁰

(iii) −5 π
4
π=1800; −5 π
4 =−5 π
4 ×180 × 1
π =-225⁰
Question 4: Coordinate Geometry in two dimensions
(a) y=x2 +4 x−8
y=3 x−2
y=y; x2+4 x−8=3 x−2
x2−x−6=0
x= 1± √ 5
2 =1 ± 2.236
2
x=1.618∨−0.618
When x=1.618, y=3(1.618)-2=2.854 while when x=-0.618, y=3(-0.618)-2=-3.854. The
coordinates of intersection are thus (1.618, 2.854) and (-0.618, -3.854)
(ii) y=x2 +4 x−8; y=2 x−9
x2+ 4 x−8=2 x−9
x2+ 2 x +1=0
x= 1± √4−4
2 = 1± 0
2
x=0.5

When x=0.5, y=2(0.5)-2=-1. The coordinates of intersection are thus (0.5, -1)
(b)
(i) y=x2−4 x +4−1; x2+ y2=9
y=y; x2−4 x+3= √ −x2+ 9
(x2−4 x+3)2=±(x¿ ¿2+ 9)¿
x2−x=0 ; x=1
But x2+ y2=9 ;1+ y2=9 ; y=2.828
The coordinates are (1, 2.828)
(ii) Coordinates are (1, 2.8) and (0,0)
m= y2− y1
x−x1
= 2.8−0=2.8
1−0
2.8= y −2.8
x−1 ; y=2.8x
Question 5: Coordinate Geometry in three dimensions
(i) Sphere
(ii)¿
¿
¿

Coordinates are (5, 5, -5); r= √25=5
(iii) No. The coordinates do not pass through the origin
(iv) )¿
Question 6: Complex Numbers
(a)
(i) ( 5−17 i ) +(17−24 i)
¿ ( 5−17i+17−24 i ) =5+17−17 i−24 i
¿ 22−41i
(ii) ( 5
4 + 1
3 i )−(2− 3
5 i)
¿ 5
4 + 1
3 i−2+ 3
5 i= 5
4 −2+ 1
3 i+ 3
5 i
¿− 3
4 + 14
15 i
(iii) ( 2−7 i ) ( 5+6 i ) =10+12 i−35 i−42 i2
¿ 10−23i+ 42=52−23 i
(iv) 2−7 i
5+6 i . 5−6 i
5−6 i =10−12i−35i+ 42i2
25−30 i+30 i−36 i2 =−32−47i
61
(b) As noticed from the definition of what complex number is:
z=(x1, y1)

w=(x2, y2) in which
z+w=(x1, y1) + (x2, y2)
=(x1+y1+x2+y2)
=(x2, y2) + (x1, y1)
=w+z
Question 7: Polar, trigonometric and exponential forms of complex numbers
(a)Converting into polar form
2+4i
Suppose z=a+ bi
The z may be written in polar form as
z=r (cos Ɵ+ i sin Ɵ)
Where r= √a2 +b2∧Ɵ=arctan ( b
a )
For this question: z= 2+4i
Hence: a=2, b=4
Thus: r=√ 22+ 42=√20

And Ɵ=arctan ( 4
2 )≅0.46
Hence the polar form of 2+4i is approximately 4.47(cos (0.46) + i sin (0.46))
9+9i
Suppose z=a+ bi
The z may be written in polar form as
z=r (cos Ɵ+ i sin Ɵ)
Where r= √a2 +b2∧Ɵ=arctan ( b
a )
For this question: z= 9+9i
Hence: a=9, b=9
Thus: r=√ 92 +92=√162
And Ɵ=arctan ( 9
9 )≅ 0.79
Hence the polar form of 9+9i is approximately 12.73(cos (0.79) + i sin (0.79))
-7+2i
Suppose z=a+ bi
The z may be written in polar form as
z=r (cos Ɵ+ i sin Ɵ)

Where r= √a2 +b2∧Ɵ=arctan ( b
a )
For this question: z= -7+2i
Hence: a=-7, b=2
Thus: r=√ (−7)2 +22=√ 53
And Ɵ=π −arctan ( 7
2 )≅ 3.142−1.29
Hence the polar form of -7+2i is approximately 7.28(cos (1.852) + i sin (1.852))
(b) Conversion to trigonometric form
(2 – 4𝑖)
z=a+bi=|z| (cos Ɵ)+i sin Ɵ); |z|= √ a2 +b2 where z=a+bi
|z|=√ 42+ 22=2√5
Ɵ=arctan ( 4
2 )≅0.46
2 √ 5 ( cos 0.46 ) +¿
9+9i
z=a+bi=|z| (cos Ɵ)+i sin Ɵ); |z|= √ a2 +b2 where z=a+bi
|z|=√ 92 +92=9√2

Ɵ=arctan ( 9
9 ) ≅0.79
9 √2 ( cos 0.79 ) +¿
-7+2i
z=a+bi=|z| (cos Ɵ)+i sin Ɵ); |z|= √ a2 +b2 where z=a+bi
|z|=√ (−7)2 +22=2√13.25
Ɵ=π −arctan ( 7
2 ) ≅ 1.852
2 √13.25 ( cos 1.852 ) +¿
(c) Exponential form
2+4i
∝= 2
4 =0.5 radians=28.6 degrees
Ɵ= 0.5
r = √22 +42=√20=4.472
=4.472 e0.5 j
9+9i
∝= 9
9 =1 radians=57.3 degrees

Ɵ=1
r = √92+92=√ 162=12.73
=12.73 e1 j
-7+2i
∝= 2
7 =0.286 radians=16.4 degrees
Ɵ=π−1.29=1.852
r = √72+22=√56=7.28
=7.28 e1.852 j
Question 8: Matrices
(a) A+C= (−2 4
5 7 ) + ( 2 2
3 4 ) =( 0 6
8 11 )
(b) B-D= (11 4 15
5 6 8
3 5 2 )− (1 6 5
7 2 4
8 9 3 )= ( 10 −2 10
−2 4 4
−5 −4 −1 )
(c) A.C= (−2 4
5 7 )∗(2 2
3 4 )= ( 8 12
31 38 )
(d) B.D= (11 4 15
5 6 8
3 5 2 )× (1 6 5
7 2 4
8 9 3 )=
(159 209 116
111 114 73
54 46 41 )

(e) (B+D). (B.D)= (12 10 20
12 12 12
11 14 5 )× (159 209 11 6
111 114 73
54 46 41 )=
( 4098 4568 2942
3888 4428 2760
3573 4125 2503 )
Question 9
(b) (50 30 20
70 15 15 )( A
B )=( 6
12 )
(ii) (70 30 0
30 40 30
0 30 70 )(C
E
G )=
( H
I
J )
Question 10

( 2 −5 9
−1 3 −7
6 −8 4 )( x
y
z )=
( 37
−31
8 )
AX=B
X=A-1B
The first step is to find the inverse of the matrix
Calculating the inverse of matrix A
[ 2 −5 9
−1 3 −7
6 −8 4 ]−1
=
[−11
3
−13
3
2
3
−19
6
−23
6
5
12
−5
6
−7
6
1
12 ] [ 37
−31
8 ]= [4
5
6 ]
x
y
z
= [4
5
6 ]
Question 11
(i) A
→
+B
→
=
−6
2 + 4
8= 2
10
(ii) C
→
−D
→
=
5
7+−8
−3= (−3
4 )
(iii) B
→
× C
→
=( 4
8 ) × ( 5
7 ) =( 20
40 )
(iv) A
→
× D
→
=(−6
2 ) × (−8
3 ) =( 48
−16 )

(v) ( A
→
−B
→
)∗( C
→
−D
→
)= (−6
2 −4
8 )∗( 5
7−−8
−3 )= (−10
−6 )∗( 13
10 )=( −100
−60 )
(b) Step 1: Joining the vectors head to tail
Step 2: Drawing the resultant vector from the tail of one of the vectors to the head of the other
Step 3: Should the vectors be at right angles, the magnitude of the resultant vector can be found
using Pythagoras Theorem
x= √ 1902 +202
=√36500=191.05 m
s
Question 12
Question 13

Converting complex number to exponential
-3.7997-9.25i
∝= 3.7997
9.25 =−1.96 radians=28.6 degrees
Ɵ=−1.96
r = √ (−3.997)2 +9.2=√100.05=10.002
=10.002 e−1.96 j
Question 14
Denominator: (1+2i)(3-4i)(5+6i)
=(1+2i)(3-4i)=3+2i-8i2=3+2i+8
=11+2i
(11+2i) (5+6i)=55+76i+12i2=55+76i-12
=43-76i
Numerator: (7-8i)(9+10i)=63-2i-80i2=143-2i
(143-2i)(11-12i)=1573-1738i+24i2=1549-1738i
N/D=(43-76i)/(1549-1738i)
¿ (43−76 i)
(1549−1738i) . (1549+1738i)
(1549+1738i)=198695−194738i
5420045

Question 15
(−2 −5 5
−2 3 −7
6 −2 1 )(x
y
z )=
( 31
−49
21 )
AX=B
X=A-1B
The first step is to find the inverse of the matrix
Calculating the inverse of matrix A
[−2 −5 5
−2 3 −7
6 −2 1 ]1
=
[ −11
152
−5
152
−5
36
−5
19
−4
19
−3
19
−7
76
−17
76
−2
19 ] [ 31
−49
21 ]= ( 3.5
1.1
19.8 )
( x
y
z )= ( 3.5
1.1
19.8 )

References
AbuEloun, N. N., & Naser, S. S. A. (2017). Mathematics intelligent tutoring system
Dymond, J. H., Marsh, K. N., Wilhoit, R. C., & Frenkel, M. D. (2016). Virial Coefficients of
Mixtures (No. Virial Coefficients of Mixtures)