BASIC MATHEMATICS
1.1
FUNCTIONS BASICS
Washington, pp. 81 – 88; 94
1.


What is a function?
A function is a rule linking one set of numbers (the domain) to another set of numbers (the range).
o
A function can link a number from the domain to only one value in the range of the function.
Types of functions
o
Many-to-one (Many -> one)
o
One-to-one (One -> one)
More than one value from the domain linked to
Every value from the domain linked to only
the same value in the range
one value in the range
2
Example: y = x
Example: y = x 3
-2
1
-1
4
-2
-8
-1
-1
1
1
2
8
1
2


There is no such thing as ...
o
many-to many functions
o
one-to-many functions
Graphical tests
o
Is a relationship a function?
Do the vertical line test: Draw a vertical line through the curve. If it intercepts the curve once everywhere, the
relationship is a function
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o

2.

Domain, co-domain, range, dependent and independent variable
o
Get the definitions!
Composite functions
Definition: The composition of two functions f and g is a combination of f and g such that ( f  g ) ( x) = f ( g ( x) ) .
o
3.


What type of function is it?
Do the horizontal line test: Draw a horizontal line through the curve. If it intercepts the curve once everywhere, the
function is one-to-one function
In general, f ( g ( x) ) ≠ g ( f ( x) )
Inverse functions
Definition: Let f and g be two functions. If f ( g ( x) ) = x AND g ( f ( x) ) = x , then g is the inverse of f, and f is the inverse of g
Steps
5
y
o
Let y = f (x )
4
o
Swap x and y
3
o
Solve for y
o
Check for limitations
2
Function
o
Write down f −1 (x )
1
 NB! Only one  one functions have inverse functions!
 NB! Always check the domain of the inverse function
-3
-2
-1
1
2
3
4
5
x
 NB! A function and its inverse is symmetrical about the line y = x
-1
−1
 f −1 ( x)  f=
 NB! f =
[ f ( x)] x
Inverse
-2
-3
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1
f ( x)
 NB! Not all functions have inverse functions
o
Only one-to-one functions have unique inverse functions
o
Use the horizontal line test!
 NB! Invertible function: A function that can be inverted
4.
Sketches of inverse functions
 Swap the x and y values of the given function's critical points to get the critical points of the inverse function
 The graphs of a function and its inverse are reflections about the line y = x , that is, they are symmetrical about the line y = x
5.
Applications
•
Inverse trig functions
•
Inverse of exponential functions, that is, log functions
•
Inverse hyperbolic functions
SUPPLEMENTARY EXERCISE 1.1
1.
Apply the line tests to each of the following graphs determine whether the curve represents a function. If it is a function classify
it as either one-to-one or many-to-one. Motivate your classification.
1.1
1.2
1.3
1.4
 NB! f −1 ( x) ≠
y
y
y
y
x
x
x
x
2.
Given: f =
( x)
2 x − 4 and
2.1
g ( x=
)
+2.
Determine f  g and g  f .
x2
2
3.
2.2
Is g the inverse of f ? Why?
−1
Determine f , the inverse of the function f , in each of the following cases.
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3.1
4.
5.
5.1
6.
7.
8.
8.1
3
t
3.3
3.4
f ( y=
) 3y − 7
,t ≠ 4
f (s) =
2− 2 ,s ≠ 0
t −4
s
Use the function f, represented by f ( x=
) 3 x − 2 , to explain the following statement: "A function and its inverse is symmetrical
about the line y = x ."
Determine the inverse of each of the following functions.
x +1
5.2
5.3=
g ( x) = x5
u ( x=
)
3 − 2x
y
, x ≠ − 12
2x +1
Is the following function invertible? Motivate your answer.
f ( x) =x 3 + 4 x 2 + 3 x
1
Given that f ( x) =2 + x − 1 and g ( x) =
. Determine the domain and range of f ( x) and g ( x) .
x−2
x +1
, x ≠ 1 , and h( x) = x , x ≥ 0 . Determine the following.
Given that g ( x) =
x −1
8.2
g (π )
g (2t + 1)
f ( x=
)
3
2− x
3.2
=
f (t )
(
h ( x − 1) 2
)
8.3
h( x + 2)
9.
11.
In the following expression, express y explicitly as a function of x.
1− y
x=
.
1+ y
Determine the inverses of the functions defined below.
10.2 u ( x=
g ( x) = x5
)
3 − 2x
x +1
10.4
f ( x=
)
3 − e2 x
y=
2x +1
Given: f ( x) = x 2 , g ( x) = sin x and h( x) = cos x .
11.1
Determine the exact numerical values of: f ( g (0.3) ) + f ( h(0.3) ) .
11.2
Show that f ( h( x) ) − f ( g ( x) ) =
h(2 x) .
10.
10.1
10.3
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8.4
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12.
13.
1
, x ≠ 0 . If g ( x=
) x 2 + 1 , for what values of x is f  g defined?
x
Consider f ( x=
) 3 x − 5 . Determine:
Let f ( x) =
13.1
13.2
f  f −1
f −1  f
13.3 Is it always true that f  g = g  f ? Is it ever true that f  g = g  f ?
ANSWERS 1.1
1.1 Many-to-one function
1.2
Not a function
1.3
2.1
1.4 One-to-one function
2.1 =
f ( g ( x) ) x=
; g ( f ( x) ) x
3.1
f −1 ( x)= 2 − x3
3.2
−1
f=
(t )
4t
,t ≠ 1
t −1
3.3
3.4
f −1 ( y ) =
y+7
3
4.
Discussion - graph
5.1
5.2
3 − x2
1− x
5.3
f −1 ( x)
x ≠ 12
=
2
2x −1
No. The function fails the horizontal-line test
6.
Many-to-one function
Yes.
=
f −1 ( s )
3
,s < 2
2−s
g −1 ( x) = 5 x
u −1 ( x) =
y
10
8
6
4
2
-4
-3
-2
-1
1
2
x
-2
-4
{ x / x ≥ 1, x ∈ R} ; Range of f ( x)= { y / y ≥ 2, y ∈ R}
Domain of g ( x) = { x / x ∈ R, x ≠ 2} ; Range of g ( x) = { y / y ∈ R, y ≠ 0}
7.
Domain of f ( x)=
8.1
g (π ) =
8.4
h ( x − 1)
(
π +1
π −1
2
x 1
) =−
8.2
9.
t +1
g (2t + 1) =
t
1− x
, x ≠ −1
y=
1+ x
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8.3
h( x + 2)=
x +2
10.1
g −1 ( x) = x 5
1
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10.2
11.1
12.
13.1
3 − x2
1− x
10.3 =
y −1 ( x)
, x ≠ 12 10.4
2
2x −1
11.2
f ( g (0.3) ) + f ( h(0.3) ) =
1
f  g is defined for all real values of x.
13.2
13.3
f −1  f = x
f  f −1 = x
u −1 ( x) =
1
ln 3 − x 2
2
Show that …
−1
f=
( x)
No; Yes.
1.2 ANGLE UNITS AND COORDINATE SYSTEMS
Washington, pp. 240 - 249
1.
•
•
The radian as angle unit
1
of a revolution
1° = 360
o Sexagesimal system (base 60): 1 degree = 60 minutes; 1 minute = 60 seconds
 12°13'15" ≈ 12.2208°
 30.689° ≈ 30°41' 20.4"
 Use your calculator!
 Application: GPS coordinates
1 grad = 1 decimal degree (not used in this course)
•
• 1 radian = angle at centre of circle subtended by an arc s equal in length to the radius r
o The radian (or rad) is NOT an SI unit; it is a derived SI unit, similar to the newton
180° =π rad
°
π
o 1° = 180
rad and 1 rad = 180
π
Why is it OK to leave out the unit and write, for example, π3 instead of π3 rad, but WRONG to leave out the unit in 60° ?
•
o θ = rs metres
: the units cancel; rad is a dimensionless quantity, and thus, when working with angles, π3 = π3 rad
metres
o 60 means "60 radians" which is completely different from 60° !
Remember to check the calculator mode when solving trig equations!
•
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o Always give angles in the unit specified in the question. For example, sin
=
x 0.5, x ∈ [0; π ] means the angles must be in
radians
1 revolution
360°
2π radians π
1 revolution per
minute
rads/sec
=
= = =
1 minute
60 seconds 60 seconds 30
•
Memorize the following table.
Degrees
0°
30°
π
Radians
0
6
•
•
•
•
•
45°
60°
90°
120°
135°
150°
π
4
π
3
π
2
2π
3
3π
4
5π
6
180°
π
270°
3π
2
360°
2π
2.
Polar coordinates
Coordinate system: A reference system used to specify the position of a point/object
Cartesian/rectangular coordinate system (x;y): Specify the position of a point in terms of its perpendicular distance from
two (or three) mutually perpendicular axes
• Polar coordinate system (r;θ): Specify the position of a point in terms of its distance from a fixed point, called the origin or
pole, and an angle, called the polar angle, from a fixed direction
o Polar axis/ray: axis in a fixed direction, usually coinciding with the x axis
o Radial coordinate r: Distance from the pole
 r ≥ 0 in this course
o Angular coordinate 2: Angle between r and the polar axis
 Anticlockwise positive; clockwise negative
 θ ∈ (−π ; π ] are called principal values
Graphs in polar coordinate systems
o There is special graph paper available for sketches in polar coordinate systems consisting of concentric circles and
radial lines as shown in the supplementary exercises
Conversion between coordinate systems
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o By hand and on your calculator
o From polar to rectangular:
=
x r=
cos θ ; y r sin θ
o From rectangular to polar: r =x 2 + y 2 ; θ =
tan −1
•
•
•
( )
y
x
Uniqueness of coordinates
o Every point (x;y) in a rectangular system is unique
o In a polar coordinate system a point may have more than one representation (r ;θ ) =
(r ;θ + 2π k ), k ∈ 

For example, (1; π4 ), (1; 94π ) , (1; − 74π ) all represent the same point

Work therefore with principal values θ ∈ (−π ; π ]
Uses
o In complex numbers to determine roots and logs of negative numbers
o In vectors in two dimensions
o By air traffic controllers to describe the position of a plane
o In calculus
o In modelling such as the groundwater flow equation
Other coordinate systems
o The polar coordinate system is a two-dimensional system
o The system can be extended to a three-dimensional system (beyond the scope of this course)
 The cylindrical coordinate system
 The spherical coordinate system
 The following website has more information as well as links to other sites, should you be interested
http://en.wikipedia.org/wiki/Coordinate_system
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SUPPLEMENTARY EXERCISE 1.2
1.
Sketch the following pairs of graphs on one system of axes on the interval indicated.
1.1 =
1.2
p ( x) = cos x; q ( x) = cos( x + π4 ); x ∈ [0; 2π ]
f ( x) 2sin x=
; g ( x) sin 2 x; x ∈ [−π ; π ]
2.
Solve for x if x ∈ [0; 2π ] and
2.1
2.2
2.3
sin x = −0.356
cos x = −0.513
2 tan x = −0.5
3.
Convert the following angles in degrees to radians without using a calculator.
3.1
3.2
3.3
3.4
25°
178°
320°
35°30 '
4.
Convert the following angles in radians to degrees.
π
7π
π
4.1
4.2
4.3
4.4
3π
5
13
7
5.
Convert 3 500 revs/min to rads/sec.
6.
Convert, by hand and with a calculator, the point (2; π3 ) from polar to Cartesian coordinates.
7.
Represent the point with rectangular coordinates (1;-2) in polar coordinates with angle in degrees.
8.
Determine the polar coordinates of the points with the following rectangular coordinates. Give angles in radians.
8.1
8.2
8.3
8.4
(2; 2)
(−2;3)
(2 2; −2)
(−1; − 3)
ANSWERS 1.2
2.1
3.1
4.1
5.
3.506;5.919
5π
36
25.714°
366.519 = 259 π
2.2
3.2
4.2
6.
Basic Mathematics: Functions basics
2.109; 4.174
71π
360
252°
(1;1.732)
2.3
3.3
4.3
7.
2.897; 6.038
89π
90
540°
(2.237;-63.435 ° )
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16π
9
3.4
4.4
13.846°
03/06/20
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8.1
(2.828; π4 )
(3.464; −0.615)
8.2
8.3
(2; −2.094)
8.4
(3.606; 2.159)
1.3 THE MODULUS FUNCTION
Washington, pp 3, 477 – 480
1.
Absolute values
Definition: The absolute value or modulus of a real number a, denoted by a , is the numerical value of the number without its sign.
Examples:
2 =2
π
−π =
−7.6 =
7.6
2 = 2
2.
a)
Basic equations and inequalities
x = 3 ⇒ x = 3 or − x =
3 , that is, x = 3 or x = −3
b)
x < 2 ⇒ x < 2 or − x < 2 ⇒ x < 2; x > −2 , that is −2 < x < 2
3.
Applications
•
In complex numbers: z =x + jy ⇒ mod( z ) =z = x 2 + y 2
•
Magnitude of a vector: a = x i + yj ⇒ a =
•
The limitations on the binomial formula when expanding (a + b) n if n is NOT a non-negative integer: b < a
•
Indicate the distance between two numbers
o
x =3 ⇔ x =3 or x =−3 ⇔ x is 3 units form 0
x2 + y 2
Distance = 3
-3
Example:
x − 2 =5 ⇒ x is a distance 5 from 2. Why? Solving for x yields x − 2 = 5 ⇒ x = 7
y
Definition:
 x if
f ( x=
) x= 
− x if
1
2
6
7
Distance = 5
-3
The modulus function
0
Distance = 5
5 ⇒ x =−3 as represented in Figure 1.
or −( x − 2) =
4.
-1
-2
Distance = 3
-2
-1
0
1
2
3
4
5
Figure 1 Absolute value as distance
5
4
x≥0
x<0
3
2
1
-5
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-4
-3
-2
-1
1
2
3
4
5
x
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3
An example of a piecewise defined function because the function is defined in two "pieces": y = x on the interval x ∈ [0; ∞)
and y = − x on the interval x ∈ (−∞;0)
• Sometimes called the absolute value function
• Domain: x ∈  ;
Range: y ∈ [0; ∞)
5.
More graphs
•
• Take note of the translations.
• MUST indicate intercepts with axes and coordinates of maximum/minimum point
• Maximum/minimum point is called the salient point
SUPPLEMENTARY EXERCISE 1.3 (KA Motsepe)
1.
Determine the value of the following:
1.2
1.3
1.4
1.1
3−4
3− 4
4−3
4−3
2.
4.
3.1
4.
4.1
Is a − b = a − b for all values of a and b? Motivate your answer using suitable mathematical examples.
Sketch the graphs of the following functions:
3.2
3.3
3.4
k ( x)= x − 1 3.5
f ( x)= x + 1
g ( x=
) x +1
h( x ) = 1 − x
Calculate x if
4.2
4.3
x = 13
x = −5
2x < 6
5.
Explain, in your own words, the meaning of x − 1 =
3.
6.
Consider the function g ( x) = x − 2 − 3 .
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m( x)= 1 − x
03/06/20
Page 11
6.1
Calculate the intercepts of the graph of g(x) with the axes.
6.2
Calculate g (5) and g (−5) .
6.3
Sketch the graph of g(x) for x ∈ [−5;5] .
ANSWERS 1.3
1.1
1
1.2
1
1.3
-1
1.4
2.
No. For example, 2 − 3 ≠ 2 − 3
3.1
3.2
4.1
5.
6.1
6.3
No solution. Absolute values is always positive
x is a distance of three units from 1, that is, x = 4 or x = −2 .
y-intercept: (0;-1); x-intercepts: (-1;0), (5;0)
y
1
3.3
3.4 & 3.5
4.2
x = ± 13
6.2
g (5)= 0; g (−5)= 4
4.3
−3 < x < 3
4
3
2
1
-5
-4
-3
-2
-1
1
2
3
4
5
x
-1
-2
-3
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1.4 TRIG BASICS
Washington, pp. 544 – 554
1.
Function
Name
Definition
The basics
sin A =
y
r
(−∞; ∞)
Range
[−1;1]
Period
360°
Even/odd
Odd
Derivative
d
(sin x)
dx
= cos x
Domain
∫ sin xdx
Integral
cot x
Cotangent
tan x
Tangent
cos x
Cosine
sin x
Sine
x
r
y
x
(−∞; ∞)
[−1;1]
360°
Even
d
(cos x)
dx
= − sin x
x ≠ ±90°; ±270°;
(−∞; ∞)
180°
Odd
d
(tan x)
dx
= sec 2 x
1
tan A
x
=
y
x ≠ ±0°; ±180°; ±360°;
(−∞; ∞)
180°
Odd
d
(cot x)
dx
= − csc 2 x
∫ cos xdx
∫ sec xdx
∫ csc xdx
cos A =
tan A =
2
=
− cos x + c = sin x + c = tan x + c
⊕
⊕
coth A =
2
=
− cot x + c
csc x / cosec x
Cosecant
1
1
csc A =
sec A =
sin A
cos A
r
r
=
=
y
x
x ≠ ±90°; ±270°; x ≠ ±0°; ±180°; ±360°;
(−∞; −1] ∪ [1; ∞)
(−∞; −1] ∪ [1; ∞)
360°
360°
Even
Odd
d
d
(sec x)
(csc x)
dx
dx
= sec x tan x
= − csc x cot x
∫ sec x tan xdx
∫ csc x cot xdx
= sec x + c
=
− csc x + c
sec x
Secant
Note the reciprocal trig functions sec x , csc x = cosec x and cot x .
Reciprocal ≠ inverse in this context!!!
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EXAMPLES – memorize!
θ
sin θ
0°
0
30°
45°
60°
1
2
1
2
csc θ
3
2
Undefined
2
2
2
3
1
cos θ
1
1
2
1
2
0
sec θ
3
2
1
2
3
2
Undefined
tan θ
0
1
3
2
1
3
Undefined
cot θ
Undefined
3
1
1
3
0
90°
1
=
sin 23° 0.3907  ⇒=
csc 23=
° sin123° 2.5593 ≈ 2.559
1
cos 97° = −0.1218 ⇒ sec 97° = cos97
° = −8.2055 ≈ −8.206
1
tan 263
=
° 8.1443 ⇒ cot 263
=
° tan=
0.1227  ≈ 0.123
263°
⊕
⊕
Use the x −1 key on your calculator!!!!
Note that rounding takes place in the last step only.
1
DON'T round off the value of=
=
2.5575 ≈ 2.558 - this answer is
sin 23° 0.3907  ≈ 0.391 and then determine the inverse
0.391
INCORRECT!!!!!!
Remember, sin −1 x ≠
1
1
since
= csc x
sin x
sin x
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2.
Graphs of trig functions and their reciprocals
⊕
Note the asymptotes (vertical lines approached but never intercepted)
Memorize the properties of all six graphs (the intercepts with axes, positions and values of maxima and minima; asymptotes)
More identities/properties
1 + cot 2 x =
csc 2 x
cos 2 x + sin 2 x =
1
cos 2 x − sin 2 x =
cos 2 x
1 + tan 2 x =
sec 2 x
sin x
x
cot x = cos
tan x = cos
sin(− x) =
− sin x
cos(− x) =
cos x
sin x
x
Compound angles
tan x ± tan y
tan( x ± y ) =
sin( x=
cos( x ± y ) =
cos x cos y  sin x sin y
± y ) sin x cos y ± cos x sin y
1  tan x tan y
3.
4.
Double angles
sin 2 x = 2sin x cos x
cos 2 x = 1 − 2sin 2 x
=
cos 2 x 2 cos 2 x − 1
1
⇒ sin x cos x =
2 sin 2 x
1
⇒ sin 2 x =
2 (1 − cos 2 x )
1
⇒ cos 2 x =
2 (1 + cos 2 x )
Inverse trig functions
• Note the asymptotes!
• Take note of the domain and range of the inverse functions – we are interested in functions!
Function
Name
sin −1 x
Inverse sine
cos −1 x
Inverse cosine
Basic Mathematics: Functions basics
tan −1 x
Inverse tangent
cot −1 x
Inverse
cotangent
sec −1 x
Inverse secant
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csc −1 x
Inverse cosecant
03/06/20
Page 15
Domain
y = sin −1 x
⇔x=
sin y
−1 ≤ x ≤ 1
y = cos −1 x
⇔x=
cos y
−1 ≤ x ≤ 1
y = tan −1 x
⇔x=
tan y
All real numbers
Range
− π2 ≤ y ≤ π2
0≤ y ≤π
− π2 < y < π2
Definition
y = cot −1 x
⇔x=
cot y
All real
numbers
0< y <π
y = sec −1 x
⇔x=
sec y
x ≤ −1 or x ≥ 1
y = csc −1 x
⇔x=
csc y
x ≤ −1 or x ≥ 1
0 ≤ y < π2 or
− π2 ≤ y < 0 or
< y ≤π
d
sec −1 f ( x) 
dx 
f '( x)
=
2
f ( x) [ f ( x)] − 1
0 < y ≤ π2
d
csc −1 f ( x) 
dx 
− f '( x)
=
2
f ( x) [ f ( x)] − 1
π
2
Derivative
d
sin −1 f ( x) 
dx 
f '( x)
=
2
1 − [ f ( x)]
Integral
∫ a −u
d
cos −1 f ( x) 
dx 
− f '( x)
=
2
1 − [ f ( x)]
-
du
2
2
= sin −1 ( ua ) + c
d
 tan −1 f ( x) 
dx
f '( x)
=
2
1 + [ f ( x)]
-
du
∫u +a
2
=
1
a
d
cot −1 f ( x) 
dx
− f '( x)
=
2
1 + [ f ( x)]
2
tan
2
−1
( ua ) + c
-
du
∫ u a −u
=
1
a
2
sec −1 ( ua ) + c
⊕
The shaded information is enrichment only – there is no need to memorize it now (Mathematics II work!).
⊕
cot −1 x ≠
1
since cot −1 x = tan −1 ( 1x ) . Can you prove this?
tan −1 x
Basic Mathematics: Functions basics
© Tshwane University of Technology: EL Voges
03/06/20
Page 16
5.
Graphs of inverse trig functions
Basic Mathematics: Functions basics
© Tshwane University of Technology: EL Voges
03/06/20
Page 17
TRIGONOMETRIC WAVES (SINUSOIDS)
6.
•
•
•
•
•
•
Formulas for waves
General equation:
=
y R sin(ω t ± α )
o
R: amplitude – watch out for the unit!
o
ω: angular velocity – rad/s
o
α: Phase angle/phase
Period: time taken to complete one cycle
2π
o
seconds
T=
ω
Frequency: number of cycles in one second
ω
Hz ⇒ ω = 2πf
o
f =
2π
Compare y = sin t and y1 = sin(t − π4 ) :
o
y1 = sin(t − π4 ) is y = sin t shifted π 4 radians to the right
o
y1 = sin(t − π4 ) lags y = sin t by π 4 since y1 peaks after y
Compare y = sin t and y1 = sin(t + π4 ) :
o
y1 = sin(t + π4 ) is y = sin t shifted π 4 radians to the left
o =
y1 sin(t + π4 ) leads y = sin t by π 4 since y1 peaks before y
In general:
o
R sin(ωt − α ) lags R sin(ωt ) by α ω
Basic Mathematics: Functions basics
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03/06/20
Page 18
R sin(ωt + α ) leads R sin(ωt ) by α ω
Time displacement: this lead/lag time

time displacement = α ω

lag
Sketching curves: label the axes, indicate maxima and minima, starting and end point of curve
o
o
•
The facts, such as the sine and cosine rules, are not included here since they are assumed to be general knowledge!
8.
Applications
•
The sinusoid
=
f (t ) A cos(ωt + α ) , a function in the time domain is associated with the phasor F = Ae jα , a function in the
frequency domain. (Mathematics IV)
•
The simple harmonic motion of, for example, a swinging pendulum is written in the form of a sinusoid.
•
The Fourier series of Mathematics III is expressed as a combination of sinusoids.
SUPPLEMENTARY EXERCISE 1.4 (KA Motsepe)
1.
Simplify the following expressions:
1.2
1.1
sin θ .cot θ .sec θ
sec θ .cos θ + tan 2 θ
cos θ
1.4
1.5 cot 2 A ( sec 2 A − 1)
−1
sec θ
2.
Prove the following identities:
2.1
2.2
tan A.cosec 2 A cos 2 A = cot A
sec 2 θ .cot 2 θ − 1 =
cot 2 θ
1 + 2sin α .cos α 2.5
( sin α + cos α ) =
sin 3 A − cos3 A =(sin A − cos A) (1 + sin A.cos A )
2.4
2
sin 4 x − cos 4 x = sin 2 x − cos 2 x
1.3
1.6
sin x.sec x
1
1
+
csc A − 1 csc A + 1 7
2.3
1
(1 − sin θ )(1 + tan θ ) =
2
2.6
3.
3.1
4.
Use a calculator to evaluate the following. Round answers off to three decimal places where necessary.
3.2
2sec3 1120 − 1
2sec104.50 co sec104.50
Use a calculator to determine, accurately to one decimal place, the values of x ∈ 00 ; 3600  if:
4.1
cot x + 1.664 =
0
Basic Mathematics: Functions basics
4.2
3sec x − 5.103 =
0
2
4.3
© Tshwane University of Technology: EL Voges
2
sec x + 1.544 =
0
3
03/06/20
Page 19
π 
Suppose the equation y = −6 cos  t  models a buoy bobbing up and down in water. The equilibrium point is y = 0. Describe
3 
the location of the buoy when t = 0.
π

6.
The temperature, in 0C, of a room varies according to the sinusoidal function
=
T (t ) 5.8sin  (t − 11)  .
12

6.1
Determine: (using the S.I. units)
6.1.1 the amplitude
6.1.2 the period
6.1.3 the frequency and
6.1.4 the time displacement of the wave, stating whether it is a lead or lag time.
6.2
Sketch one cycle of the graph of T(t).
ANSWERS 1.4
1.1
1
1.2
1.3
1.4
− sin 2 θ
sec 2 θ
tan x
5.
1.5
4.1
5.
1
1.6
2 tan A sec A
4.2
54.00 or 306.00
149.00 or 329.00
6 units below the equilibrium point.
6.1.1 Amplitude = 5.8 0C
6.2
6.1.2
Tperiod = 24 s
3.1
4.3
-39.0
115.60 or 244.40
6.1.3
f =
1
Hz
24
3.2
-8.3
6.1.4 time displacement = 11 s (lag).
T
5.8
35
0
11
t
23
-5.8
Basic Mathematics: Functions basics
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03/06/20
Page 20
1.5 EXPONENTS AND LOGS
Washington pp. 313 – 330; 363 - 383
1.
•
EXPONENTIAL FUNCTION AND NATURAL LOGARITHM
x
=
y e=
exp( x)
o =
e lim (1 + 1x )
x
x →∞
o
e = 2.71828182845904523536 is an irrational number
• =
y log
=
ln x
e x
o
y = ln x : natural log
o =
y log
=
x log10 x : common log
•
Graphs
Figure 3 The influence of the base on the curve
•
If f ( x) = e x and g ( x) = ln x , then f −1 ( x) = g ( x) and g −1 ( x) = f ( x) , that is,
the exponential function is the inverse of the log function, and vice verse
Basic Mathematics: Functions basics
Figure 2 Exponential growth and decay
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03/06/20
Page 21
Figure 4 Exponential and log functions
•
Laws/properties
Exponential functions
m, n ∈ 
1. e m × e n =
em+ n
em
2. n = e m − n
e
3. ( e m ) = e mn
n
y
Logarithmic functions
x, y, a, b all positive
1. ln( xy
=
) ln x + ln y
4
x
2. ln  =
 ln x − ln y
 y
3
( ) = n ln x
3. ln x
n
1
4. n = e − n
e
1
5. e n = n e
4. ln e x = x
6. e n = n e m
7. e0 = 1
8. e1 = e
6. e x ln a = a x
2
1
5. eln x = x
m
5
7. ln e = 1
8. ln1 = 0
-2
log a x
9. log y x =
log a y
-1
1
2
3
4
5
x
-1
-2
3.
Applications
•
•
Compound interest
Radioactive decay, e.g. carbon dating, N (t ) = N 0 e kt
•
Exponential growth, e.g. population growth, N (t ) = N 0 e kt
•
Richter scale, measuring the magnitude of an earthquake,=
M log A − log A0
Basic Mathematics: Functions basics
© Tshwane University of Technology: EL Voges
03/06/20
Page 22
SUPPLEMENTARY EXERCISE 1.5 (KA Motsepe)
1.
Simplify the following into a single algebraic fraction:
3
(x − y )
1.1
y x (4 xy ) 2
+
x2
y2
2.
Evaluate the following. Round off to 4 decimal places where necessary
2.1
3(
3.
Express the following expressions as a single logarithm.
log3 80 − log3 4 )
1.2
without using a calculator.
−1
2.2
−1 −1
log 3 2 + π
5.4
1
ln(1 + x 2 ) + ln x − ln sin x
2
Change the subject in the following equations to x:
 e2x − 1 
ex −1

y = ln
4.2
y= x

e +1
 2 
Solve for x (to 3 decimal places where necessary) if:
5.2
3 x+2 − 3 x = 6 x
e2 x +3 − 7 =
0
1− x
x +3
x
5.5
5e − 4 =
40
2 + 2 = 6 x +1
5.7
2 ln( x − 2) − ln x = 0
5.8
5.10
5 1.12 x+3 = 200
5.13
3log x − 2 log
3.1
4.
4.1
5.
5.1
ln x − b ln z + a ln y
(
2.3
3.2
)
x
=
2
2
5.6
3 2 x − 3 x +1 − 31 = 30
1 + ln x = ln(1 + x)
log 2 x + log x = 2
5.9
(1.25) x = 2
5.11
2 2 x +1 − 3 × 2 x = 2
5.12
e x + e−x = 3
5.14
ln(5 − 2 x) =
−3
5.3
Log worksheet
=
=
log10 x
e x;log x
 ln x log
1.
Use the log laws to write the following expressions as a single log in its simplest form.
1.1
1.2
1.3
2 log b x + 12 log b y
log 2 x + log 2 y − log 2 z
1.4
1
2
log x + log x − log x
4
1
4
4
5
2 + ln 7
1
8
8
Basic Mathematics: Functions basics
1.5
log 4 x − log 4 y − 1
1.6
© Tshwane University of Technology: EL Voges
5log 3 (2 x − 1) − 5log 3 ( x − 2)
log 2 5 + log8 5
03/06/20
Page 23
2.
2.1
Write the following expressions as the sum and difference of logs without powers.
3x 2
9
log 2
2.2
2.3
log 3 2
y
x
(
)
log
100
x +1
(
)
2.6
 1 − x 2 cos x 

y = ln 
e3 x


3.3
log x 128 = 7
3.6
x = log 5.2 1
2.4
ln 2 x e3 x
3.
3.1
Solve for x. Round off to two decimal places where necessary.
3.2
log 2 x = 5
log x 10000 = 4
3.4
log x 81 = 4
3.7
4.
4.1
3.8
3 =7
9.1 = 3.2
Solve for x.
 Always try to simplify first!
 Always check your answer. Remember, in the real number system we can't determine the log of a negative number.
4.3
4.2
ln(2 x) + ln(3 x − e) − 2 =
ln 8
log(2 x +=
3) log(2 x − 1)
log(2 x + 8) =1 + log( x − 4)
4.4
ln x + ln( x − 1) =
5
2.5
3.5
ln (sin x) tan x 
x = log 5 125
x
x
ANSWERS 1.5
3
y 2 + 8 x3
1.1
3
1
x 2y 2
4.5
1.2
xy
y−x
3.1
 xy a 
ln  b 
 z 
5.2
5.7
5.12
ln( x + 2) + ln x 2 =
ln x
4.6
−1 + ln( x − 2) =ln x
0.2370
2.1
20
2.2
3.2
 x 2 (1 + x 2 ) 
ln 

 sin x 
4.1
 1+ y 
x = ln 

 1− y 
4.2=
x
−0.527
5.3
1.262
5.4
−1.175
5.5
4
5.8
5.13
2.902
25
5.9
5.14
3.106
2.475
5.10
±0.962
1
Basic Mathematics: Functions basics
2.3
1.2671
5.1
3
0.369
5.6
0.582
17.852
5.11
1
© Tshwane University of Technology: EL Voges
1
2
ln(2e y + 1)
03/06/20
Page 24
ANSWERS Worksheet
1.1
log 2 xyz
( )
log ( )
4
1.6
2.4
3.2
3.7
4.4
(
)
1.2
log b x 2 y
2.1
log 2 3 + log 2 x − log 2 y
2.5
3.3
3.8
4.5
tan x ln(sin x)
2
0.53
0.41
1.3
log 3 ( 2xx−−21 )
5
1.4
log x 2
1.5
2.2
2 − log 3 x
2.3
3
6
3.1
32
3.6
0
No solution
x
4y
4
3
log 2 5
2 − log( x + 1)
x ln 2 + 3x
10
1.77
12.69
2.6
3.4
4.1
4.6
ln(1 − x 2 ) + ln(cos x) − 3 x
3
3.5
3.62
4.2
No solution
4.3
More Logarithms
1.
Express the following expressions as a single logarithm.
1.1
ln x − b ln z + a ln y
2.
Evaluate the following: Round off to 4 decimal places where necessary
2.1
3(
2.2
log 3 2 + π
3.
3.3
Solve for x. Round answer off to 4 decimal places where necessary.
x
3.2
3log x − 2 log =
2
e2 x +3 − 7 =
0
2
3.4
5 (1.12 x+3 ) = 200
5e1− x − 4 =
40
3.5
ln(5 − 2 x) =
−3
3.1
log3 80 − log3 4 )
1.2
1
ln(1 + x 2 ) + ln x − ln sin x
2
without using a calculator.
Basic Mathematics: Functions basics
2.3
3.6
5
2 + ln 7
ln x + ln( x − 1) =
1
© Tshwane University of Technology: EL Voges
03/06/20
Page 25
Logarithms Answers
 xy a 
1.1
ln  b 
 z 
2.2
0.2370
3.2
−0.5270
3.5
2.4751
 x 2 (1 + x 2 ) 
ln 

 sin x 
1.2671
17.8520
2.2229
1
1.2
2.3
3.3
3.6
2.1
20
3.1
3.4
25
−1.1748
1.6 HYPERBOLIC FUNCTIONS
1.
The basics
Function
cosh x
sinh x
tanh x
coth x
Name
Hyperbolic cosine
Hyperbolic sine
Hyperbolic
tangent
sinh x
tanh x =
cosh x
e x − e− x
= x −x
e +e
Hyperbolic
cotangent
cosh x
coth x =
sinh x
e x + e− x
= x −x
e −e
Definition
cosh x =
e x + e− x
2
y
Graph
sinh x =
3
e x − e− x
2
6
y
y
1.0
y
sech x
csch x
Hyperbolic secant Hyperbolic
cosecant
1
1
sech x =
csch x =
cosh x
sinh x
2
2
= x −x
= x −x
e +e
e −e
3
y
2
0.8
2
0.6
y
0.9
0.8
4
0.7
1
0.4
2
0.6
1
2
0.2
-5
-4
-3
-2
-1
-0.2
-3
-2
1
-1
1
-2
2
2
3
4
5
x
0.4
-5
-4
-3
-2
-1
1
2
3
x
-0.4
3
4
-5
-4
-3
-2
1
-1
2
3
4
5
x
0.3
-1
-1
0.2
0.1
-0.6
-0.8
-4
0.5
1
-2
-5
-4
-3
-2
-1
0
1
2
3
4
5
x
-2
-1.0
-3
-3
Domain
Range
Derivative
-2
-1
(−∞; ∞)
[1; ∞)
d
(cosh x)
dx
= sinh x
0
1
2
3
x
(−∞; ∞)
(−∞; ∞)
d
(sinh x)
dx
= cosh x
-6
Basic Mathematics: Functions basics
(−∞; ∞)
(−1;1)
d
(tanh x)
dx
= sech 2 x
(−∞;0) ∪ (0; ∞)
(−∞; −1) ∪ (1; ∞)
d
(coth x)
dx
= −csch 2 x
(−∞; ∞)
(−∞;0) ∪ (0; ∞)
(0;1]
(−∞;0) ∪ (0; ∞)
d
d
(sech x)
(csch x)
dx
dx
= −sech x tanh x
= −csch x coth x
© Tshwane University of Technology: EL Voges
03/06/20
Page 26
5
x
∫ cosh xdx
Integral
∫ csch xdx
∫ sech xdx
∫ sinh xdx
2
∫ csch x coth xdx
∫ sech x tanh xdx
2
= sinh x + c
= coth x + c
=
−csch x + c
= tanh x + c
=
−sech x + c
= cosh x + c
⊕
The unit of the x in cos x is either radians or degrees; the x in cosh x may have any unit depending on the application
2.
Inverse hyperbolic functions
−1
Function
sinh x
cosh −1 x
tanh −1 x
coth −1 x
sech −1 x
csch −1 x
Inverse
Inverse hyperbolic Inverse hyperbolic Inverse hyperbolic Inverse hyperbolic Inverse
Name
hyperbolic sine
cosine
tangent
cotangent
secant
hyperbolic
cosecant
−1
−1
1
−
1
−
1
−
Definition cosh x
sinh x
tanh x
coth x
sech x
csch −1 x
(
= ln x + x 2 + 1
y
)
(
= ln x + x 2 − 1
Graph
2
1
)
1  x +1 
= ln 

2  x −1 
1  1+ x 
= ln 

2  1− x 
y
y
2.0
5
y
 1 + 1 − x2 
1
1 + x2 
= ln 
=
ln  +



x
x
x 



5
4
4
3
3
2
2
y
6
y
5
1.5
3
2
4
1
1
1
3
-5
-4
-3
-2
-1
1
-1
2
3
4
-2.0
1.0
-1.5
-1.0
-0.5
0.5
-1
1.5
2.0
x
-5
-4
-3
-2
1
-1
-1
-2
-2
-3
-3
2
3
4
5
x
-3
-2
1
-1
2
-1
1
0.5
-4
-2
1.0
0.0
0.0
0.5
1.0
1.5
2.0
2.5
3.0
3.5
4.0
x
-5
-2
-4
-5
0
0.0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
1.1
1.2
x
-3
(−∞; ∞)
[1; ∞)
(−∞; −1) ∪ (1; ∞)
(0;1]
(−∞;0) ∪ (0; ∞)
(−1;1)
Domain
(−∞; ∞)
[0; ∞)
(−∞; ∞)
(−∞;0) ∪ (0; ∞)
[0; ∞)
(−∞;0) ∪ (0; ∞)
Range
⊕
Because hyperbolic functions are defined in terms of exponential functions, the inverse hyperbolic functions are written in
terms of log functions.
⊕
The shaded information is enrichment only – there is no need to memorize it!
Basic Mathematics: Functions basics
© Tshwane University of Technology: EL Voges
03/06/20
Page 27
2
3.
Comparative identities
Osborn's rule
To obtain the formula for hyperbolic functions from the analogous identity for the trig function, replace each trig
function by the corresponding hyperbolic function AND change the sign of every product or implied product of
two sines.
Trigonometric identity
Hyperbolic identity
2
2
cos x + sin x =
1
cosh 2 x − sinh 2 x =
1
2
2
2
2
cos x − sin x =
cos 2 x
cosh x + sinh x =
cosh 2 x
2
cos 2 x = 1 − 2sin x
cosh 2 x = 1 + 2sinh 2 x
=
cosh 2 x 2 cosh 2 x − 1
=
cos 2 x 2 cos 2 x − 1
1 + tan 2 x =
sec 2 x
1 − tanh 2 x =
sech 2 x
1 + cot 2 x =
csc 2 x
coth 2 x − 1 =csch 2 x
sin 2 x = 2sin x cos x
sin( x=
± y ) sin x cos y ± cos x sin y
cos( x ± y ) =
cos x cos y  sin x sin y
tan x ± tan y
tan( x ± y ) =
1  tan x tan y
sin 2=
x
1
2
(1 − cos 2 x)
cos =
x
(1 + cos 2 x)
− sin x
sin(− x) =
cos(− x) =
cos x
2
4.
1
2
More relationships (enrichment)
Relationship between trig and hyperbolic functions:
sinh jθ = j sinh θ ⇔ sinh jθ = j sin θ
Basic Mathematics: Functions basics
sinh 2 x = 2sinh x cosh x
sinh(
=
x ± y ) sinh x cosh y ± cosh x sinh y
cosh(
=
x ± y ) cosh x cosh y ± sinh x sinh y
tanh x ± tanh y
tanh( x ± y ) =
1 ± tanh x tanh y
=
sinh 2 x
1
2
(cosh 2 x − 1)
=
cosh x
(cosh 2 x + 1)
sinh(− x) =
− sinh x
cosh(− x) =
cosh x
2
1
2
cosh θ =cos jθ ⇔ cosh jθ =cos θ
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03/06/20
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Series expansions:
cosh x =+
1
x2 x4
+ +
2! 4!
sinh x =x +
x3 x5
+ +
3! 5!
5.
On the calculator
1
1
1
sech x =
csch x =
coth x =
cosh x
sinh x
tanh x
6.
Applications
Important note: The two photographs here were obtained from http://en.wikipedia.org/wiki/Catenary and its use was approved
by the legal department of the Tshwane University of Technology.
 Catenaries (free hanging chains/cables supported at the end points only)
Typical equation: For some constants A and B,
Hanging chain from
y= A + cosh Bx
http://en.wikipedia.org/wiki/Catenary
 Arches in buildings such as the Gateway Arch in St. Louis, Missouri
Gateway Arch from
Typical equation: For some constants A and K depending on a specific
=
y A ( cosh Kx − 1)
building,
http://en.wikipedia.org/wiki/Catenary
 Potential difference E between a telegraph line and earth (James, 2001: 140) with A and B constants, x the distance from the
transmitting end, r the resistance per kilometre of the conductor and R the insulation resistance per kilometer


r 
r 
=
E A cosh  x
 + B sinh  x

 R
 R
Basic Mathematics: Functions basics
© Tshwane University of Technology: EL Voges
03/06/20
Page 29
7.
Motivation for the name hyperbolic function
Trig (or circular) functions
The points (cos t ;sin t ) lie on a circle with a radius of 1 unit
Hyperbolic functions
The points (cosh t ;sinh t ) lie on a right half of an equilateral
hyperbola
References
• James, Glyn. 2001. Modern Engineering Mathematics. 3rd ed. Pearson Education: London.
• http://en.wikipedia.org/wiki/Catenary
SUPPLEMENTARY EXERCISE 1.6
1.
Use a calculator to evaluate the following expressions correct to three decimal places.
cosh(2.5)
1.1
1.2
1.3
1.4
sinh 3.5
tanh(−2.1)
sech ( 23 )
coth(ln 3)
1.5
sech (−1.7)
1.10
1.6
cosh(−1.2)
2.
2.1
3.
4.
5.
Use the definitions of the hyperbolic functions to prove the following.
2.2
2.3
tanh(− x) =
− tanh x
cosh x − sinh x =
e− x
tanh 2 x + sech 2 x =
1
Are the hyperbolic functions periodic? Motivate your answer.
Classify the six hyperbolic functions as even or odd.
Rewrite each of the following expressions in terms of exponential functions. Simplify your answer as much as possible.
1.7
csch 2.5
1.8
coth 2.1
1.9
cosh 0.5
Basic Mathematics: Functions basics
© Tshwane University of Technology: EL Voges
03/06/20
Page 30
5.2
5.3
5.4
cosh 3x − sinh 3x
cosh 2 x + sinh 2 x
3cosh(ln x)
sinh(3ln x)
Solve the following equations, accurately to three decimal places.
6.2
sinh x = 3
4 tanh x − 1 =0
Use the definitions of sinh x and cosh x to solve for x in the equation cosh 2 x − sinh 2 x =
0.5 . Give your answer correctly to
THREE decimal places.
⊕
Practice on derivatives and integrals of hyperbolic functions will be included in the supplementary exercises on differentiation
and integration.
ANSWERS 1.6
1.1
16.543
1.2
-0.970
1.3
6.132
1.4
1.250
1.5
0.813
1.6
1.811
1.7
0.165
1.8
1.030
1.9
0.354
1.10 1.128
2.1
Proof
2.2
Proof
2.3
Proof
3.
No. Does not repeat at regular
intervals
4.
Even: cosh, sech;
Odd: tanh, coth, sinh, csch
5.
Rewrite each of the following expressions in terms of exponential functions. Simplify your answer as much as possible.
3x 3
1
x3
5.2
5.3
5.4
5.1
e −3x
e2 x
+
− 3
2 2x
2 2x
6.1
1.818
6.2
0.255
7.
0.347
5.1
6.
6.1
7.
Basic Mathematics: Functions basics
© Tshwane University of Technology: EL Voges
03/06/20
Page 31