INTRODUCTORY COURSE...5.2 The cosine rules 26 5.3 Solving oblique triangles 27 5.4 Exercises 28 6....

INTRODUCTORY

COURSE

MATHEMATICS

FACULTY OF ENGINEERING TECHNOLOGY

GROUP T LEUVEN CAMPUS

Content Algebra

1. Real numbers 3 1.1 The power of a real number with an integer exponent 3 1.2 The nth root of a real number 4 1.3 The power of a real number with a rational exponent 5 1.4 The power of a real number with an integer exponent 6 1.5 Rationalizing the denominator 7 1.6 The logarithm of a real number 7 1.7 Exercise 9

2. Polynomials with real coefficients in 1 variable x 11 2.1 Definition 11 2.2 Zeroes of a polynomial function 11 2.3 Special products 11 2.4 Polynomial division 12 2.5 Exercises 16

3. Equations and inequalities 17 3.1 Equations in IR 17 3.2 Inequalities in IR 20 3.3 Exercises 26

4. Absolute value of a real number 28 4.1 Definition 28 4.2 Property 29 4.3 Operations 29 4.4 Application: graph of a real absolute value function 29 4.5 Exercises 31

5. Matrices and determinants 32 5.1 Matrices 32 5.2 Determinants 37 5.3 Exercises 41

6. Systems 43 6.1 Systems of n equations in n unknowns 43 6.2 Systems of linear inequalities in 1 unknown 45 6.3 Exercises 46

Analytic Geometry

1. Vector and lines 3 1.1 Vectors 3 1.2 Lines 5 1.3 Parallel lines 8 1.4 The Euclidian vectorspace 9 1.5 Angles 11 1.6 Orthogonality of 2 lines 12

1.7 Distance from a point to a line (O.N.B.) 13 1.8 Exercise 14

2. Conic sections 19 2.1 Introduction 19 2.2 The circle 20 2.3 The parabola 21 2.4 Exercises 23

Calculus

1. Relations functions, mappings, 1-1 mappings 3 2. Expanding IR 4 3. Continuity of a function in IR 5

3.1 Continuity at a point 5 3.2 Right-and left-continuity 6

4. Limits 7 4.1 Limits in real number 7 4.2 Right-hand and left-hand limits 9 4.3 Limits at infinity 9 4.4 Infinite limits 9 4.5 Infinite limits at infinity 10 4.6 Elementary rules to calculate limits 10 4.7 Indeterminate cases 11 4.8 Exercises 14

5. Differentiation 17 5.1 The derivative in a point 17 5.2 Geometrical interpretation of the derivative 17 5.3 Left-hand and right-hand derivative in a point 18 5.4 A vertical tangent 19 5.5 Differentiation rules 19 5.6 Exercises 21

6. Indefinite integrals 24 6.1 Antiderivative functions 24 6.2 The indefinite integral 24 6.3 Properties 24 6.4 Basic integrals 25 6.5 Integration by substitution 25 6.6 Exercise 26 6.7 Integration by parts 29 6.8 Exercises 29

7. Definite integrals 30 7.1 The fundamental theorem of calculus 30 7.2 Properties of the definite integral 30 7.3 The substitution method 31 7.4 Partial integration 31 7.5 Exercise 32

Trigonometry

1. Angles 4 1.1 The trigonometric circle 4 1.2 Oriented angles 4 1.3 Conversion between radians and degrees 6

2. The trigonometric numbers 7 2.1 Definitions 7 2.2 Some special angles and their trigonometric numbers 8 2.3 Sign variation for the trigonometric numbers by quadrant 9 2.4 Pythagorean identities 9 2.5 Special pairs of angles 10 2.6 Exercises 12

3. The trigonometric functions 14 3.1 Periodic functions 16 3.2 Even and odd functions 16 3.3 Sine function 17 3.4 Cosine function 17 3.5 Tangent function 18 3.6 Cotangent function 18 3.7 The secantfunction 19 3.8 The cosecant function 19 3.9 Exercises 20

4. Right triangles 21 4.1 Formulas 21 4.2 Exercises 23

5. Oblique triangles 25 5.1 The sine rules 25 5.2 The cosine rules 26 5.3 Solving oblique triangles 27 5.4 Exercises 28

6. Extra’s 29 6.1 Special lines in a triangle 29 6.2 Isosceles triangles 31 6.3 Equilateral triangles 32 6.4 Exterior angles 32

7. Trigonometric formulas 33 7.1 Sum and difference formulas 33 7.2 Double-angle formulas 34 7.3 Half-angle formulas 35 7.4 Trigonometric numbers in terms of tan α/2 35 7.5 Conversions sum/difference of angles into product of angles and vice versa 36 7.6 Exercise 37

Algebra

Dr. Caroline Danneels

1 Real numbers ........................................................................................................................3

1.1 The power of a real number with an integer exponent............................................................... 3

1.2 The nth rooth of a real number .................................................................................................... 4

1.3 The power of a real number with a rational exponent ............................................................... 5

1.4 Properties of extraction of roots ................................................................................................. 6

1.5 Rationalizing the denominator ................................................................................................... 7

1.6 The logarithm of a real number.................................................................................................. 7

1.7 Exercices .................................................................................................................................... 9

2 Polynomials with real coefficients in 1 variable x ...........................................................11

2.1 Definition ................................................................................................................................. 11

2.2 Zeroes of a polynomial function .............................................................................................. 11

2.3 Special products ....................................................................................................................... 11

2.4 Polynomial division ................................................................................................................. 12

2.5 Exercices .................................................................................................................................. 16

3 Equations and inequalities.................................................................................................17

3.1 Equations in IR ........................................................................................................................ 17

3.2 Inequalities in IR ...................................................................................................................... 20

3.3 Exercises .................................................................................................................................. 26

4 Absolute value of a real number .......................................................................................28

4.1 Definition ................................................................................................................................. 28

4.2 Property .................................................................................................................................... 29

4.3 Operations ................................................................................................................................ 29

4.4 Application: graph of a real absolute value function ............................................................... 29

4.5 Exercises .................................................................................................................................. 31

5 Matrices and determinants ................................................................................................32

5.1 Matrices ................................................................................................................................... 32

5.2 Determinants ............................................................................................................................ 37

5.3 Exercises .................................................................................................................................. 41

6 Systems ................................................................................................................................43

6.1 Systems of n equations in n unknowns .................................................................................... 43

6.2 Systems of linear inequalities in 1 unknown ........................................................................... 45

6.3 Exercises .................................................................................................................................. 46

_____________________________________________________________________________ Algebra 3

1 Real numbers

1.1 The power of a real number with an integer exponent

n0

00

a and n : a a.a....a (n factors)

a : a 1

∀ ∈ ∀ ∈ =

∀ ∈ =

ℝ ℕ

ℝ

( ) ( )n -1-n -1 n0 n

1 a and n : a = a = a =

a∀ ∈ ∀ ∈ℝ ℕ

Properties: a,b and m,n∈ ∈ℝ ℝ

m n m+na a a⋅ =

mm-n

n

aa

a=

( )n n nab a b= ⋅

n n

n

a a

b b

=

( )nm m na a ⋅=

Example

( ) ( )-3 -2-2 4 3 3

5-5 -2

2

a a ab aab

bb

a

⋅ ⋅ =

Rule of signs:

if n is even, then ( )n n-a a=

if n is odd, then ( )n n-a -a=

_____________________________________________________________________________ Algebra 4

1.2 The nth rooth of a real number

The nth ( )0n ∈ ℕ root of a real number a is each real number x of which the nth power is a

or n0 x, a , n : x is the nth root of a x a∀ ∈ ∀ ∈ ⇔ =ℝ ℕ

if n is odd and a∈ℝ then a has in ℝ one nth root, written as:

Examples

3 8 2= since 32 8=

3 8 2− = − since 3( 2) 8− = −

If n is even and :

a has 2 nth roots in ℝ which are each others opposite and are

denoted and .

We agree upon to represent a positive real number; so = 2 > 0.

Example: 4 has 2 square roots and

a = 0 then a has in ℝ one nth root namely 0

then a has in ℝ no nth root.

Convention:

n na a=

n na a− = −

We will only treat with a +∈ℝ , since the other forms can be rewritten to this form since

if < 0 and is even then does not exist

if < 0 and is odd then we write as - || with | |

n

n n

a n a

a n a a a +

∈ ℝ

an

a∈ IR0+

an − a

n

an

164

4 = 2 − 4 = −2

a∈ IR0−

an

_____________________________________________________________________________ Algebra 5

1.3 The power of a real number with a rational exponent

Definition: m

n mn0 0 a , m , n : a a+∀ ∈ ∀ ∈ ∀ ∈ =ℝ ℤ ℕ

Use:

1. 1

nna a=

2. m

-n

n m

1a

a=

3. mp m

np nmp mnp na a a a= = =

Calculation rules: +0 a, b en q, q' :∀ ∈ ∀ ∈ℝ ℚ

q q' q+q'a a a⋅ =

qq-q'

q'

aa

a=

( )q q qa b a b⋅ = ⋅

q q

q

a a

b b

=

( )q'q q q'a a ⋅=

_____________________________________________________________________________ Algebra 6

1.4 Properties of extraction of roots

1.4.1 Multiplication and division of radicals

+ n n n0 a, b , n : a b a b∀ ∈ ∀ ∈ ⋅ =ℝ ℕ

n+ + n

0 0 n

a a a , b ; n :

b b∀ ∈ ∀ ∈ ∀ ∈ =ℝ ℝ ℕ

Examples

44 48 2 16 2= =

4 123 9

12 53 12 4

a aa

a a= =

1.4.2 Exponentiation and extraction of roots of radicals

( )m n+ mn0 a , n , m : a a∀ ∈ ∀ ∈ ∀ ∈ =ℝ ℕ ℕ

+ m nn m n m0 a ; m,n : a a a⋅∀ ∈ ∀ ∈ = =ℝ ℕ

Examples

( )23 2 238 8 2 4= = =

3 38 8 2= =

1.4.3 Adding and subtraction of radicals

( )+ n n n n0 a , n , p, q, r : p a q a r a p q r a∀ ∈ ∀ ∈ ∀ ∈ + − = + −ℝ ℝ ℝ

Example:

243 + 5 33 − 813 = 2 33 + 5 33 −3 33 = 4 33

_____________________________________________________________________________ Algebra 7

1.5 Rationalizing the denominator

a a b

bb=

1 a b

a ba ± b=

−∓

3 32 23

3 3

1 a a b b

a ba b

⋅ +=±±

∓

1.6 The logarithm of a real number

Let a be the base of a logarithmic system 0(a > 0, a 1), x , y ≠ ∈ ∈ℝ ℝ

Then log yay x x a= ⇔ =

We call y the logarithm of x to base a.

The fact that loga x is the inverse of ax can be expressed with the following two identities:

Logarithmic identy 1: x

alog a x=

Logarithmic identy 2: alog xa x=

Names:

• If a = e (number of Euler), the logarithm is called a natural logarithm, notation ln x.

ln 1 = 0 ln e = 1

• If a = 10, the logarithm is called a Briggs logarithm, notation log x.

log 1 = 0 log 10 = 1

Properties:

1. If the base a is larger than 1 (a > 1), the function loga is increasing everywhere. If the base a is between 0 and 1 (0 < a < 1), then the function loga is decreasing everywhere.

2. Because a1 = a, logarithmic identity 1 above implies loga a = 1

3. Because a0 = 1, logarithmic identity 1 above implies

loga1= 0

_____________________________________________________________________________ Algebra 8

Operations:

0 x, y : log x y = log x + log ya a a+∀ ∈ ⋅ℝ

0

x x, y : log = log x log y

ya a a+∀ ∈ −ℝ

10 x : log (x ) = log x a a+ −∀ ∈ −ℝ

0 x ; z : log (x ) = z log x za a

+∀ ∈ ∀ ∈ ⋅ℝ ℤ

n0 0

1 x ; n : log x = log x

na a+∀ ∈ ∀ ∈ℝ ℕ

_____________________________________________________________________________ Algebra 9

1.7 Exercices

1.7.1 Simplify ( +0a,b,c ∈ ℝ )

1. ( )

( ) ( )

27 2 3 4

3 210 3 19

a a b c

abc b c

−

− 9 16

6

b c

a

2. ( ) ( )

3 2

2 2

2 3

2 2

− −

−+

− + −

1 18

3. ( ) ( )

( ) ( ) ( )

3 23 4

64 4 2

2a b 4a

a a 2ab

− −

− − − 3

2 a

4. n n n 1 n 22 a b+ + n 2 2ab ab

5. 4 5

46 8

32a b

81c d 4

2 2

2ab 2b 3cd c

6. 2 4( 25a) b a, b IR− − ∈

1.7.2 Calculate ( +0a,b ∈ ℝ )

1. 3 4 3 128a b 4 b 2a

2. a a a 8 7 a

3. 4 4

3

4a

2a 6 4 2a

4. 3

3 2

5 a 30

5a 60a× 6

3

2a

5. 1,54 8

6.

32 43a

−

12 a

−

_____________________________________________________________________________ Algebra 10

1.7.3 Rationalize the denominator ( +a ∈ ℝ )

1.

2.

3.

1.7.4 Calculate, even without knowledge of a

1. 3 log 4

log 2a

a

⋅ 6

2. 35 25 5 1

log log 3log log36 21 6 2a a a a+ − + 0

1

a − b

3+ 5

5 −1

1

3+ 7

_____________________________________________________________________________ Algebra 11

2 Polynomials with real coefficients in 1 variable x

2.1 Definition

n n-1n n-1 1 0a x + a x ... a x a+ + + with n n-1 1 0

n

a ,a ,....,a ,a

n ,a 0

∈∞ ∈ ≠ ℕ

is a polynomial of degree n in x,

notation V(x).

2.2 Zeroes of a polynomial function

a ∈ ℝ is a zero of the associated polynomial function f : : x V(x)→ →ℝ ℝ if the polynomial expression evaluated to a, equals 0.

In symbols: a is a zero of V(x) V(a) = 0.

Example

is a zero of V(x) = 3 26x 7x 7x 1− − −

since 3 2

1 1 1 1V 6 7 7 1 0

2 2 2 2 − = ⋅ − − ⋅ − − ⋅ − − =

2.3 Special products

If A, B and C are polynomials:

⇔

−1

2

A + B( ) A −B( ) = A2 − B2

A + B( )2 = A 2 + 2AB + B2 A − B( )2 = A 2 − 2AB + B2

A + B+ C( )2 = A 2 + B2 + C2 + 2AB+ 2BC+ 2AC

A + B( )3 = A3 + 3A2B+ 3AB2 + B3

A − B( )3 = A3 −3A2B+ 3AB2 − B3

A 3 − B3 = A − B( ) A2 + AB+ B2( )

A 3 + B3 = A + B( ) A2 − AB+ B2( )

_____________________________________________________________________________ Algebra 12

2.4 Polynomial division

For any polynomials A(x) and B(x) ( )B(x) 0, degree A(x) degree B(x)≠ ≥ , there exist unique

polynomials Q(x) and R(x) such that

A(x) = B(x)Q(x) + R(x) and degree R(x) < degree B(x).

2.4.1 Polynomial long division

A(x) (dividend) B(x) (divider)

Q(x)(quotient)

R(x)(remainder)

If R(x) = 0 we say that B(x) evenly divides A(x).

Example

x4 − 2 x3 + 9 x2 − 2 x − 2 x2 + 2

− 6 x4 − 12 x2 6 x2 − 2 x − 3

− 2 x3 − 3 x2 − 2 x − 2

2 x3 + 4 x

− 3 x2 + 2 x − 2

3 x2 + 6

2 x + 4

⇒⇒⇒⇒ 4 3 2

22 2

6 2 9 2 2 2 46 2 3

2 2

x x x x xx x

x x

− + − − += − − ++ +

_____________________________________________________________________________ Algebra 13

2.4.2 Division of a polynomial by x-d with d ∈∈∈∈ IR: IR: IR: IR: synthetic division (Horner algorithm)

Example:

Q(x) = 4x2 − 8x + 11

R(x) = − 16

⇒3

24 5 6 164 8 11

2 2

x xx x

x x

− + = − + −+ +

2.4.3 Divisibility by x-d with d ∈∈∈∈ IR

2.4.3.1 Remainder theorem

If a polynomial f(x) is divided by x – d, then the remainder is equal to V(d).

Consequence: V(x) is divisible by x−d

Examples

V(x) = 3x2 – 5x + 7 is not divisible by x – 2 since V(2) = 9 ≠ 0

V(x) = 2x3 + 3x2 – 5x + 12 is divisible by x + 3 since V(–3) = 0

2.4.3.2 Factoring polynomials

Factoring polynomials is a very important issue in algebra. To factor a polynomial put your polynomial in standard form, from highest to lowest power:

V(x) = anxn + … + a0

After you have factored out the greatest common factor and special products, you can apply the Rational Root Test.

The Rational Root (or Rational Zeroes) Test is a handy way of obtaining a list of useful first guesses when you are trying to find the zeroes (roots) of a polynomial. Given a polynomial with integer coefficients, the possible (or potential) zeroes are found by listing the factors of the constant (last) term over the factors of the leading coefficient, thus forming a list of fractions. This listing gives you a list of potential rational (fractional) roots to test - hence the name of the Test.

4x3 − 5x + 6( ): x + 2( )

4 0 − 5 6

− 2 − 8 16 − 22

4 − 8 11 − 16

⇔ R = V d( ) = 0

_____________________________________________________________________________ Algebra 14

Caution: Don’t make the Rational Root Test out to be more than it is. It doesn’t say those rational numbers are roots, just that no other rational numbers can be roots. And it doesn’t tell you anything about whether some irrational or even complex roots exist. The Rational Roots Test is only a starting point.

Example

Factor the polynomial: 3 2 2x 3x 11x 6− − +

1. The factors of the leading coefficient (2) are {1,2} 2. The factors of the constant term (6) are {1,2,3,6} 3. Therefore the possible rational zeroes are ±1,2,3 or 6 divided by 1 or 2: ±1/1, 2/1, 3/1,

6/1, 1/2, 2/2, 3/2, 6/2 reduced: ±1, 2, 3, 6, 1/2, 3/2

4. Check for which values V(x) equals 0: as soon as you find one value which satisfies this criterium, apply synthetic division to find the reduced polynomial that you’ll use to find the remaining roots: V(1) ≠ 0 V(−1) ≠ 0 V(−2) ≠ 0 V(2) ≠ 0 V(3) = 0

This means that x – 3 is a factor of the given polynomial. The reduced polynomial can be produced by applying the synthetic division:

2 −3 −11 6 d = 3 6 9 −6

2 3 −2 0

We find: 2x3 − 3x2

− 11x + 6 = (x – 3)(2x2 + 3x – 2)

5. Again, search for the remaining candidate roots in the list ±1, 1/2, 2: As we have already checked ±1 and ±2 (see above), we check V(1/2) = 0 Applying synthetic division to 2x2 + 3x – 2 gives: 2x2 + 3x – 2 = (x – 1/2)(2x + 4)

6. All together: 2x3

−3x2 − 11x + 6 = (x – 3)(x – 1/2)(2x + 4)

= (x – 3)(2x – 1)(x + 2)

_____________________________________________________________________________ Algebra 15

2.4.3.3 Coefficientrules

1. A polynomial of degree n is divisible by (x – 1) if the sum of the coefficients (included the constant term) equals zero.

Example

x5 – 2x3 + 2x2 – 1 is divisible by (x – 1)

2. If the sum of the coefficients of the terms with odd power of x equals the sum of the coefficients of the terms with even power of x (included the constant term), then the polynomial is divisible by (x + 1).

Example

x5 + x4 – x3 + x2 – x – 3 is divisible by (x + 1)

_____________________________________________________________________________ Algebra 16

2.5 Exercices

Simplify (be efficient):

1. ( )( ) ( )2x 2 x 2 x 4+ − + 4 x 16−

2. ( ) ( )23x 2 9x 6x 4+ − + 3 27x 8+

3. ( ) ( )4 3 29x x 1 : x x 1+ + − + + 2 Q 9x 10x 19,R 29x 20= − − − = +

4. ( ) ( )2 320x 7x 3x 2 : x 3 − + − − 2 Q 7x x,R 2= − − = −

(use synthetic division)

5. ( ) ( )3 28x 20x +2x 2 : 2x 1 − − − 2 Q 4x 8x 3,R 5= − − = −

(use synthetic division)

6. For which value of n ∈ ℝ the polynomial 3 2(3x 2nx 5nx 10)+ − + is divisible by (x + 1)? Determine for the obtained value of n the quotient.

2 n 1,Q 3x 5x 10= − = − +

7. Factor: 4 3 28x 20x 18x 7x 1− + − + ( ) ( )3x 1 2x 1− −

_____________________________________________________________________________ Algebra 17

3 Equations and inequalities

3.1 Equations in IR

3.1.1 Definition

An equation is an expression that represents the equality of two expressions involving constants and/or variables.

3.1.2 Solving an equation

To solve an equation means to find the value or values of the unknown quantity (or quantities) that satisfy the equation. The solution of an equation is the set of all values which, when substituted for unknowns, make an equation true. Let’s call this set S.

Example

2 2S(7x 2 5x) 1,

7 − = = −

Two equations are equivalent if their solution set is the same.

Example

( ) 94x 3 0 3x

4 + = ⇔ − =

since 9 3

S(4x 3 0) S 3x4 4

+ = = − = = −

Properties:

Following properties make it possible to simplify and solve equations:

1. ( ) ( )A B A C B C= ⇔ + = + . We will apply this property to bring over terms.

2. ( ) ( ) 0A B mA mB with m = ⇔ = ∈ℝ

3. ( ) ( )A B C 0 A 0 B 0 C 0= ⇔ = ∨ = ∨ =i i

4. ( ) ( )A C B C A B C 0= ⇔ = ∨ =i i

_____________________________________________________________________________ Algebra 18

Avoid following errors:

a) If you apply A B A C B C= ⇒ =i i , then it’s possible that you import solutions.

Example

2

2

2

x 41 2x

x 2 x 2x (x 2) 4 2x(x 2)

x 5x 6 0

(x 3)(x 2) 0

(x 3 x 2)

imported

+ = +− −

⇒ + − = + −

⇔ − + =⇔ − − =⇔ = ∨ =

↓

So you should solve this equation as follows:

{ }

2x (x 2) 4 2x(x 2) x 2

x 3

S 3

⇔ + − = + − ∧ ≠⇔ =⇒ =

Remember: before removing the denominator which contains the unknown, mention that the donominator is different from zero.

b) If you apply A C B C A B= ⇒ =i i then it is possible that you loose solutions.

Example

( ) ( ) ( )( )x 1 x 2 3x 2 x 2

x 1 3x 2

2x 3 0

3x :1 solution is lost

2

− + = + +⇔ − = +⇔ + =

⇔ = −

So you should solve this equation as follows:

( )( )( ) ( )x 2 2x 3 0

x 2 0 2x 3 0

3x 2 x

23

S , 22

⇔ + + =

⇔ + = ∨ + =

⇔ = − ∨ = −

⇒ = − −

_____________________________________________________________________________ Algebra 19

3.1.3 Discussion of a linear equation that contains parameters

Consider ax b 0; a, b+ = ∈ℝ

1. a ≠ 0 ax + b = 0 ⇔ x = −b/a (= the only solution) 2. a = 0 0x + b = 0 ⇔ 0x = −b

if b 0 then S = (false equation)

if b = 0 then S = (identical equation)

≠ ∅ ℝ

Example

px – m2 = 3x – 9 p and m are real parameters

⇔ (p – 3)x = m2 – 9

1. 2m - 9

p 3 1 solution: x = p - 3

≠ ⇒

2. 2p = 3 0 x = m 9⇒ −

if m = 3 m = 3 then 0x = 0 S = IR

if m 3 m 3 then S =

∨ − ⇒ ≠ ∧ ≠ − ∅

3.1.4 Solving a second degree equation in 1 unknown

Standard form: 2ax bx c 0+ + = met 0a ;b,c∈ ∈ℝ ℝ

Determine the discriminant 2b 4ac∆ = −

If

1 2

1 2

> 0 then ax² + bx + c = 0 has roots

b b x and x

2a 2a = 0 then ax² + bx + c = 0 has roots

- x = x =

∆

− − ∆ − + ∆= =

∆

two different

two equal

b

2a < 0 then ax² + bx + c = 0 has roots

∆ no real

If b is even, then one can use the simplified formulas:

If b = 2b' then 2 24b' 4ac 4(b ' ac) 4 '∆ = − = − = ∆

⇒ 1 2

b ' ' b ' 'x , x

a a

− − ∆ − + ∆= =

_____________________________________________________________________________ Algebra 20

3.2 Inequalities in IR

3.2.1 Definition

An inequality is a statement that a mathematical expression is less than or greater than some other expression A < B (or A ≤ B, A > B, A ≥ B). A and B are expressions of which at least one contains a variable.

3.2.2 Solving inequalities

To solve an inequality means to find the value or values of the unknown that satisfy the inequality. Let’s call this set of solutions S.

Example

] [ ] [S(2x(x 1) 0) ,0 1,− > = −∞ ∪ +∞

Two inequalities are equivalent if their solution set is the same.

Properties:

Following properties make it possible to simplify and solve inequalities:

1. (A B) (A C B C)< ⇔ + < +

2. ( ) 0

0

mA mB if m A B

mA mB if m

+

−

< ∈< ⇔ > ∈

ℝ

ℝ

In an inequality we never remove the variable from the denominator!

_____________________________________________________________________________ Algebra 21

3.2.3 Discussion of the linear inequality that contains parameters

Consider ax b 0;a, b+ > ∈ℝ

ax b 0 ax b+ > ⇔ > −

1. b ba 0 ax b x S x | x

a a > > − ⇔ > − = ∈ > −

ℝ

2. b ba 0 ax b x S x | x

a a < > − ⇔ < − = ∈ < −

ℝ

3. b 0 S

a 0 ax b 0x bb 0 S

> == > − ⇔ > − ≤ = ∅

ℝ

Example

px 2m 3 2x p (p,m )

(p 2)x 2m p 3

− + < − ∈⇔ − < − −

ℝ

Discussion:

1. 2m p 3

p 2 S x xp 2

− −> = ∈ < − ℝ

2. 2m p 3

p 2 S x xp 2

− −< = ∈ > − ℝ

3. if 2m 5 0 S

p 2 0x 2m 5if 2m 5 0 S

− > == < − − ≤ = ∅

ℝ

_____________________________________________________________________________ Algebra 22

3.2.4 Solving quadratic inequalities in 1 unknown

Consider 20ax bx c 0 with a ;b,c+ + ≥ ∈ ∈ℝ ℝ

We first examine the evolution of the sign of the left hand side. Therefore we look at the graph of the function . This function represents a parabola with axis of symmetry parallel to the y-axis.

The intersection points of the parabola with the y-axis are obtained by solving the set of

equations: 2y ax bx c

x 0

= + +

= . ⇒ { }S (0,c)=

The intersection points of the parabola with the x-axis are obtained by solving the set of

equations: 2y ax bx c

y 0

= + +

= :

b b

if > 0 ,0 and ,02a 2a

− − ∆ − + ∆∆ ⇒

are the intersection points with the x-axis

if = 0 ∆ the parabola touches the x-axis in ,02

b

a −

if < 0 ∆ doesn’t intersect with the x-axis

Furthermore we know that if a > 0 the concave side of the parabola is directed upwards and if a < 0 the concave side of the parabola is directed downwards.

y = ax2 + bx+ c

_____________________________________________________________________________ Algebra 23

Summary:

This summary leads us to the following sign chart of y = ax2 + bx + c :

x sign of y

sign of a 0 opposite sign

of a 0 sign of a

1. If a > 0 ⇒ 21 2S(ax bx c 0) ] , x ] [x , [+ + ≥ = − ∞ ∪ +∞

2. If a < 0 ⇒ 21 2S(ax bx c³ 0) [x , x ]+ + ≥ =

Example: 2x 2x 3 0− − ≤

⇒ roots of the left hand side of this inequality are –1 , 3

⇒ sign chart:

x 2x 2x 3− − + 0 − 0 +

⇒ S = [−1,3]

a > 0

a < 0

D > 0 D = 0 D < 0

x x

1 2

x = x 1 2

x = x 1 2

x x 1 2

x1 x2

x1 x2

_____________________________________________________________________________ Algebra 24

3.2.5 Solving rational inequalities

Solving rational inequalities is very similar to solving polynomial inequalities. But because rational expressions have denominators (and therefore may have points where they're not defined), you have to be a little more careful in finding your solutions.

To solve a rational inequality, you first find the zeroes (from the numerator) and the undefined points (from the denominator). You use these zeroes and undefined points to divide the number line into intervals. Then you find the sign of the rational on each interval.

Example 1: 3

2

x(1 x)(x 3)f (x) 0

(3 x) (2x 3)

− += >− +

⇒ Sign chart:

x 3− 3

2− 0 1 3

x − − − − − 0 + + + + +

1 x− + + + + + + + 0 − − −

3(x 3)+ − 0 + + + + + + + + +

2(3 x)− + + + + + + + + + 0 +

(2x 3)+ − − − 0 + + + + + + +

− 0 + − 0 + 0 − −

⇒ ( ) ] [3S f (x) 0 3, 0,1

2 > = − − ∪

Example 2: 2

2

(x 1)(2x x 1)f (x) 0

x x 6

− + += ≤+ −

First find the zeroes of the nominator and of the denominator:

zeroes nominator: 1; zeroes denominator: -3, 2

⇒ Sign chart:

x -3 1 2

− − − 0 + + +

+ + + + + + +

+ 0 − − − 0 +

− + 0 − +

⇒ ] [ ] [S(f (x) 0) , 3 1,2≤ = −∞ − ∪

f x( )

x −1

2x2 + x + 1

x2 + x − 6

f x( )

_____________________________________________________________________________ Algebra 25

Example 3: 4 3

3 2

x x x 1f (x) 0

x x x

− + − += <+ +

First, factor the nominator and the denominator: 2

2

(x 1)(x 1)( x x 1)f (x)

x(x x 1)

− + − + −=+ +

⇒ Sign chart:

x −1 0 1

− − − − − 0 +

− 0 + + + + +

x − − − 0 + + +

− − − − − − −

+ + + + + + +

+ 0 − + 0 −

⇒ ] [ ] [S(f (x) 0) 1,0 1,< = − ∪ + ∞

Example 4: x 2 x 3

x 1 x 1

+ +≤+ −

x 2 x 3 3x 50 0

x 1 x 1 (x 1)(x 1)

+ + − −⇔ − ≤ ⇔ ≤+ − + −

⇒ Sign chart:

x 1− 1

3x 5− − + 0 − − − − −

x 1+ − − − 0 + + +

x 1− − − − − − 0 +

f (x) + 0 − + −

] [5S(f (x) 0) , 1 1,

3

⇒ ≤ = − − ∪ − +∞

x −1

x +1

− x2 + x −1

x 2 + x + 1

f x( )

−5

3

_____________________________________________________________________________ Algebra 26

3.3 Exercises

3.3.1 Solve

1. ( ) ( ) ( )( )x 5 3x 7 x 5 5x 3− + = − + { }2,5

2. ( )2x 5 9 0− − = { }2,8

3. 2

6 1 2

3x 1 x 3x x− =

− −

∅

4. 1 1 2

x 2 x 2 3+ =

− + { }-1,4

5. ( )

12x 2x 3 120

3x 1 2 5 1 3x

−= −− −

173,

6

6. 2x 7x 3(11 a x) 8a 0+ + − ⋅ + =

For which value of a ∈ ℤ there exists 1 solution? Determine the solution for that value.

a 1, x 5= − = −

7. 3x 3

4x 32 5

− > − 24

x25

>

8. 4x 3 x 1 x 5

5 8 2

− + −− > 71

x7

> −

9. 3x 1 2x

3x 2 2x 3

+ <+ −

2 3 3

, ,3 11 2

− − ∪ +∞

10. 1 2 3

x 1 x 2 x 3+ ≤

+ − − ] [ [ [ ] [, 1 1,2 3,−∞ − ∪ ∪ +∞

11. ( ) ( ) ( )2 2x 2 2x x 3 x 1 2x 0− + − − − > ] [3, 1, 2

2 −∞ − ∪

12. x 1 x 1

x 1 x 1

− +<+ −

] [ ] [1,0 1,− ∪ +∞

_____________________________________________________________________________ Algebra 27

3.3.2 Determine the number of solutions for which value(s) of m and solve

1. 2m x 2 4x m− = −

2. x x

2 0p m p m

+ − =+ −

3. ( )2p p x 2x m p 2− = − ⋅ −

4. mx 1 x m− ≤ +

5. 2x 4 m x 8+ > ⋅ +

6. 2m (x 4) 4 x− < −

7. 2p x 5m 2m m x m p⋅ − + < ⋅ − ⋅

_____________________________________________________________________________ Algebra 28

4 Absolute value of a real number

4.1 Definition

We define the absolute value (or modulus) of a real number x as follows:

x x if x 0

x :x x if x 0

= ≥∀ ∈ = − ≤ℝ

Remark: 1) for each real number different from 0 one of the 2 parts of the functiondefinition is applicable. Only for 0 the 2 parts are applicable since they both give the same functionvalue.

2) x +∈ℝ

Graph:

To obtain the graph of y = you can reflect that part of the graph of y = x that is under the x-axis around the x-axis.

x

x

_____________________________________________________________________________ Algebra 29

4.2 Property

x , a : x a

a x a

+∀ ∈ ∀ ∈ ≤

− ≤ ≤

ℝ ℝ

⇕

4.3 Operations

x, y : x y x y∀ ∈ ⋅ = ⋅ℝ

110x : x x

−−∀ ∈ =ℝ

0

xxx , y :

y y∀ ∈ ∈ =ℝ ℝ

x, y : x y x y∀ ∈ + ≤ +ℝ

4.4 Application: graph of a real absolute value function

Given: f (x) x 3= −

Asked: make a graph of f

Solution:

Method 1: starting from the definition of f without absolute value signs:

( )( )x 3 if x 3 0

f (x)x 3 if x 3 0

= − − ≥= = − − − ≤

x 3 if x 3 (1)f (x)

3 x if x 3 (2)

= − ≥⇔ = = − ≤

(1) represents a halfline through (3, 0), (4, 1)

(2) represents a halfline through (3, 0), (2, 1)

f x( ) = x − 3

_____________________________________________________________________________ Algebra 30

Method 2: starting from the graph of y = x − 3

• y = x − 3 is represented by a line through (0, −3), (3, 0)

• to get the graph of y x 3= − , reflect that part of the graph of y = x − 3 which is under the

x-axis around the x-axis.

_____________________________________________________________________________ Algebra 31

4.5 Exercises

Draw the function graph of following functions in an orthogonal coordinate system:

1. y 2x 1= +

2. y 2 x 3 4= + +

3. 2y x 4= −

4. 2y 3x 4x 1= − + +

5. 2y x 9 9= − +

6. y x 2 x 2= − + +

7. 2y x 1 x 4= − − −

_____________________________________________________________________________ Algebra 32

5 Matrices and determinants

5.1 Matrices

5.1.1 Definition

An m×n-matrix A is a rectangular array of m×n real (or complex) numbers, containing m rows and n columns.

In general:

( )

11 12 13 1n

ij 21 22 23 2n

mxn

m1 m2 m3 mn

a a a ... a

a a a a ... a=

... ... ... ... ...Aa a a ... a

The horizontal and vertical lines in a matrix are called rows and columns, respectively. The numbers in the matrix are called its entries or its elements. Entries are often denoted by a variable with two subscripts, as shown in the general notation above. The first index denotes the row, while the second one denotes the column.

To specify a matrix's size, a matrix with m rows and n columns is called an m-by-n matrix or m × n matrix, while m and n are called its dimensions.

The set of all m×n matrices is denoted as m n×ℝ or as m n×ℂ depending on the elements aij are real or complex numbers.

5.1.2 Equal matrices

Two matrices A and B are called equal if and only if:

• They have the same dimensions,

• Elements on the same position are equal.

( ) ( )ij ij ij ij

i 1,2,3,...mA a and B b then A B a b with

j 1,2,3,...m

== = = ⇔ = =

There are a number of operations that can be applied to modify matrices called matrix addition, scalar multiplication and transposition. These form the basic techniques to deal with matrices.

_____________________________________________________________________________ Algebra 33

5.1.3 Addition of matrices with the same dimension

The sum A+B of two m-by-n matrices A and B is calculated entrywise:

( ) ( ) ( )ij ij ij ija b a b where 1 i m and 1 j n+ = + ≤ ≤ ≤ ≤

Example

2 4 6 1 4 0 2 1 4 4 6 0 3 8 6

3 5 7 0 2 3 3 0 5 2 7 3 3 7 4

+ + + + = = − + + −

Properties:

1.Closure: mxn mxnA,B : A B∈ + ∈ℝ ℝ

2. Commutative property: A + B = B + A

3. Associative property: (A + B) + C = A + (B + C)

4. The zero Matrix (all aij = 0) is the identity element. A + 0 = A

5. Inverse element: for each matrix A=(aij) there exists –A=(−aij) such that A + (−A) = 0

⟹ m n, is a commutative group× +ℝ

5.1.4 Scalar multiplication of a matrix and a real number

ij ijr : r(a ) (r a )∀ ∈ =ℝ i

r is called a scalar. This operation is called scalar multiplication.

Example

1 0 3 0

3 2 3 6 9

0 1 0 3

= − −

_____________________________________________________________________________ Algebra 34

Properties:

1. Closure: m n m nr , A : r A× ×∀ ∈ ∀ ∈ ⋅ ∈ℝ ℝ ℝ

2. 1 is identity element: m nA :1 A A×∀ ∈ ⋅ =ℝ

3. Associative property: m nr,s , A : (r s)A r(s A)×∀ ∈ ∀ ∈ ⋅ = ⋅ℝ ℝ

4. Scalar multiplication is distributive over the addition in m n : r(A B) r A r B× + = ⋅ + ⋅ℝ

5. Scalar multiplication is distributive over the addition in : (r s)A r A s A+ = ⋅ + ⋅ℝ

Since m n,× +ℝ is also a commutative group we call m n, ,+×ℝ ℝ is a real vector space.

5.1.5 Transposed matrix

The transpose of an m-by-n matrix A is the n-by-m matrix AT formed by turning rows into columns and vice versa:

(AT)ij = Aji

11 12 1n 11 21 m1

21 22 2n 12 22 m2T

m1 m2 mn 1n 2n mn

a a ... a a a ... a

a a ... a a a ... aA= A =

... ... ... ... ... ... ... ...

a a ... a a a ... a

5.1.6 Matrix multiplication

Multiplication of two matrices is defined only if the number of columns of the left matrix is the same as the number of rows of the right matrix. If A=(aij) is an m-by-n matrix and B=(bij) is an n-by-p matrix, then their matrix product C=A∙B is the m-by-p matrix of which entries are given by dot-product of the corresponding row of A and the corresponding column of B:

( )ijC c= where p

ij i1 1j i2 2 j ip pj ik kjk 1

c a b a b ... a b a b , i 1...m, j 1...n=

= + + + = = =∑ .

_____________________________________________________________________________ Algebra 35

Example

11 12 1311 12

11 12 13 21 22 2321 22

21 22 23 31 32 3331 32

41 42 4341 42

c c ca a

b b b c c ca a=

b b b c c ca a

c c ca a

4 x 2 2 x 3 4 x 3

with

Properties:

1. Associative property: (A∙B)∙C = A∙(B∙C)

2. Right distributivity: A∙(B + C) = A∙B + A∙C

3. Left distributivity: (A + B)∙C = A∙C + B∙C

4. Not-commutative property: A∙B ≠ B∙A in general.

Example

[ ] [ ] [ ]1 1 1 2 3

1 2 3 1 9 while 1 1 2 3 1 2 3

2 2 2 4 6

= =

5.1.7 Square matrices

A square matrix is a matrix which has the same number of rows and columns. An n-by-n matrix is known as a square matrix of order n. The elements aii with i = 1, 2, 3, ... n form the main diagonal of the matrix.

Any two square matrices of the same order can be added and multiplied.

c11 = a11 b11 + a12 b21

c23 = a21 b13 + a22 b23

_____________________________________________________________________________ Algebra 36

Special square matrices:

Identitymatrix of order n:

The identity matrix or unit matrix of order n is the n-by-n square matrix with ones on the main diagonal and zeros elsewhere. It is denoted by In:

ii

ij

a 1 i, j 1,2,...,n :

a 0 if i j

=∀ = = ≠

The important property of matrix multiplication of identity matrix is that for m-by-n A:

m nI A A I A⋅ = ⋅ =

Zero matrix of order n:

The zero matrix of order n is the n-by-n square matrix with all its elements 0. It is denoted by On. The zero matrix acts as an absorbing element for matrix multiplication:

for m-by-n A:

m n m,nO A A O O⋅ = ⋅ =

Om,n is the zero m-by-n matrix with all its entries 0.

Zero Divisors:

There exist matrices A and B for which A∙B = 0 and A ≠ 0 en B ≠ 0. Such matrices are called zero divisors.

Example: 1 1 1 2 0 0

1 1 1 2 0 0

− = −

Invertible matrix:

A square matrix A is called invertible or non-singular if there exists a matrix B such that A∙B = In. This is equivalent to B∙A = In. Moreover, if B exists, it is unique and is called the inverse matrix of A, denoted A−1.

_____________________________________________________________________________ Algebra 37

Diagonal matrix:

If all entries outside the main diagonal are zero, A is called a diagonal matrix.

5.2 Determinants

5.2.1 Definitions

The determinant is a special number associated with any square matrix. The determinant of a matrix A is denoted det(A), or without parentheses det A or |A| or |a|. An alternative notation, used in the case where the matrix entries are written out in full, is to denote the determinant of a matrix by surrounding the matrix entries by vertical bars instead of the usual brackets or parentheses.

The determinant of a square 2-by-2 matrix 11 12

21 22

a a

a a

is given by 11 1211 22 12 21

21 22

a aa a a a

a a= −

To calculate the determinant of a square 3-by-3 matrix A

A = 11 12 13

21 22 23

31 32 33

a a a

a a a

a a a

continue as follows:

Minor of an element:

To find the minor of aij

1. Cross out the entries that lie in the corresponding row i and column j. 2. Rewrite the matrix without the marked entries. 3. Obtain the determinant �ij of this new 2-by-2 matrix.

M ij is called the minor for element aij.

Cofactor of an element:

�ij = (−1)i+j Mij

The cofactor �ij is independent of the elements of the ith row and the elements of the jth column.

_____________________________________________________________________________ Algebra 38

Example

12 13 12 132 121

32 33 32 33

a a a a( 1)

a a a aα += − = −

Definition:

The determinant of a 3-by-3 matrix is the real number obtained by taking the elements of a chosen row or column and multiplying each term by the corresponding cofactor and then adding these products.

So, for a chosen fixed row value i, the determinant can be calculated emanating from the ith row:

|A| = ai1 . αi1 + ai2 . αi2 + ai3 . αi3 + ... ain . αin

Example

3 2 1

A = 2 1 2

3 2 4

−

1 111

1 2( 1) 4 4 8

2 4α += − = + =

−

1 212

2 2( 1) (8 6) 2

3 4α += − = − − = −

1 313

2 1( 1) 4 3 7

3 2α += − = − − = −

−

⟹ det A = 3∙8 + 2∙(−2) + 1∙(−7) = 13 (emanating from the first row).

_____________________________________________________________________________ Algebra 39

5.2.2 Properties of determinants

1. A square matrix A and its transposed AT have the same determinant.

det A = det AT

2. If B results from the matrix A by interchanging two rows (or columns) of A, then det B = − det A.

Consequences:

• a determinant with 2 equal rows (columns) is zero.

• the sum of the products of the elements of a row (column) and the corresponding cofactors of another row (column) is zero. Example: check the previous example:

21 11 22 12 23 13a a a 2 8 1 ( 2) 2 ( 7) 0α α α+ + = ⋅ + ⋅ − + ⋅ − =

3. If the elements of a row (column) can be written as a sum of 2 elements, the determinant can be considerd in the same way as the sum of 2 determinants.

' '1 1 1 1 1 1 1 1 1 1

' '2 2 2 2 2 2 2 2 2 2

' '3 3 3 3 3 3 3 3 3 3

a a b c a b c a b c

a a b c a b c a b c

a a b c a b c a b c

++ = ++

Example

2 1 2 1 2 2 1 1 2 1

0 2 1 2 0 1 2 2 1 2 13

1 4 2 4 1 2 4 4 2 4

++ = + =

− + − − − −

4. If we multiply one row (column) with a constant, the determinant of the new matrix is the determinant of the old one multiplied by the constant.

1 1 1 1 1 1

2 2 2 2 2 2

3 3 3 3 3 3

a kb c a b c

a kb c k a b c

a kb c a b c

=

Consequences:

• a common factor of all the elements of a row (column) of A can be ‘taken outside the determinant’.

• a determinant in which one row (column) is a multiple of another row (column) is zero.

_____________________________________________________________________________ Algebra 40

5. To any row of A we can add any multiple of any other row without changing det(A).

1 1 1 1 1 1 1

2 2 2 2 2 2 2

3 3 3 3 3 3 3

a b c a kb kb c

a b c a kb kb c

a b c a kb kb c

+= +

+

6. det (A.B) = det A . det B

7. The previous properties allow us to calculate a determinant by reducing the order: by applying the previous properties try to get a row or column which contains only one value different from zero. Example

6 9 2

2 3 1

3 5 2

=r1 − r3

3 4 0

2 3 1

3 5 2

=r3 − 2r2

3 4 0

2 3 1

−1 −1 0

= −13 4

−1 −1= −1 −3+ 4( ) = −1

_____________________________________________________________________________ Algebra 41

5.3 Exercises

1. Calculate x, y, z if 2

2

4 x x y+5z =

y 0 2x+5 x+y+z

Solution: x = 2, y = −3, z = 1

2. Fill in the missing numbers:

2 . 5 1 . 6

. 3 2 3 . 11 5

4 3 . 3 4 .

+ =

3. If A is a p-by-q matrix and B is a r-by-s matrix, under what condition both matrixproducts exist?

Solution: q = r and s = p

4. Following matrices are given:

0 11 2 1 0 1

A = 1 0 B = C = 3 4 2 1 3

2 2

Calculate A∙B, (A∙B)∙C, B∙C, A∙(B∙C) and check the associative property.

5. Proof that (A∙B)T = BT∙AT if

g ha b c

A = , B = i jd e f

k l

6. Determine all matrices B for which A∙B = B∙A if 1 2

A = 3 4

Solution: 2

x z3

z x z

+

7. Which relationship exists between matrices B (of order 3) and A∙B if

1.

1 0 0

A = 0 k 0

0 0 1

2.

0 1 0

A = 1 0 0

0 0 1

_____________________________________________________________________________ Algebra 42

8. Calculate the following determinants (by reducing the order):

1.

3 2 0

4 2 1

1 3 4

−−

2.

2 1 1 3

0 0 3 1

2 3 4 1

1 1 0 1

3.

1 0 0 0 a

0 0 0 a 1

0 0 a 1 0

0 a 1 0 0

a 1 0 0 0

4.

9 5 1

2 3 1

1 6 2

−−

−

5.

8 1 9 1

6 3 4 5

5 4 7 2

2 3 4 1

6.

a b c

c a b

b c a

Solutions:

49 −5 1 + a5 10 20 3 3 3a b c 3a b c+ + − ⋅ ⋅

9. Proof that:

a b c k a b c

p q r k p q r

x y z x k y k z

⋅⋅ =

⋅ ⋅

_____________________________________________________________________________ Algebra 43

6 Systems

6.1 Systems of n equations in n unknowns

6.1.1 Matrixnotation

11 1 12 2 1n n 1

21 1 22 2 2n n 2

n1 1 n2 2 nn n n

a x + a x +...+ a x = b

a x + a x +...+ a x = b

...

a x +a x +...+ a x = b

ij ia ,b∈ ∈ℝ ℝ

Set

11 12 1n

21 22 2n n n

n1 n2 nn

a a ... a

a a ... aA =

... ... ... ...

a a ... a

×

∈

ℝ

1 1

2 2n 1 n 1

n n

x b

x bX = , B =

... ...

x b

× ×

∈ ∈

ℝ ℝ

⟹ Matrixnotation: A∙X = B

Example

-2x + 5y - z = 1 -2 5 -1 x 1

x + z = 0 A = 1 0 1 , X = y , B = 0

-2y + 3z = -2 0 -2 3 z -2

⇒

_____________________________________________________________________________ Algebra 44

6.1.2 Solving a system of equations by elimination

The goal is to reduce the system until you are left with an equivalent system in which the first equation contains n unknown, the second equation contains n - 1 unknown, the third n – 2, the last but one contains 2 unknown, the last contains 1 unknown. This is done by combining equations to cancel out unknown. When you have reduced the system, you solve the last equation, and back-substitute this value into either of the other equations and then solve the remaining system for the other unknown.

Example

6.1.3 Solving a system of equations by the method of Cramer

The system A.X = B has one unique solution ⇔ | A | ≠ 0 .

This solution is given by

ii

Ax

A=

With Ai is the n-by-n matrix defined as:

Ai =

11 12 1 i-1 1 1 i+1 1n

21 22 2 i-1 2 2 i+1 2n

n1 n2 n i-1 n n i+1 nn

a a ... a b a ... a

a a ... a b a ... a

... ... ... ... ... ... ... ...

a a ... a b a ... a

x + y − z = 9 1 −3

x − 2y − 3z = 1 −1

3x + 6y + z = 37 1

⇒

x + y − z = 9

3y + 2z = 8 −1

3y + 4z = 10 1

⇒

x + y − z = 9

3y + 2z = 8

2z = 2

⇒

x + y = 9+1

3y = 8 − 2

z =1

⇒

x = 10− 2

y = 2

z = 1

⇒

x = 8

y = 2

z = 1

_____________________________________________________________________________ Algebra 45

Example

Take the example of 6.1.2.

A =

1 1 1

1 2 3

3 6 1

− − −

⇒ |A| = − 6

A1 =

9 1 1

1 2 3

3 6 1

− − −

⇒ |A1| = − 48

A2 =

1 9 1

1 1 3

3 37 1

− −

⇒ |A2| = − 12

A3 =

1 1 9

1 2 1

3 6 37

−

⇒ |A3| = − 6

1 32A AA

x 8, y 2,zA A A

⇒ = = = = =

6.2 Systems of linear inequalities in 1 unknown

Solve each inequality separately and the solution of the system is formed by those values which satisfy all inequalities.

Example

x 3 5234 2x

x2 3 92x x 2

x 515 3

− − < − − < ⇔ − <− <

⟹ S = 23

x x 59

∈ − < <

ℝ

- 23/9 5

_____________________________________________________________________________ Algebra 46

6.3 Exercises

6.3.1 Solve the following systems of equations

1.

x 2y 3z 2

2x y 3z 1

4x 7y 3z 7

− + − = + + = + + =

9 3

k,1 k,k5 5

− +

2.

y 3z 1

2x 2y z 2

4x 5y z 5

− + = + + = + − =

∅

3.

3x y z 16

x 3y 4z 72

x y z 8

− + = + + = − + = −

{ }(12,20,0)

4.

x 2y 1

3y 5z 11

2z 3x 13

+ = − = − − = −

{ }(5,-2,1)

5.

1 1 3

x y 2

1 1 2

y z

1 1 1

z x

+ = + =

+ =

4 4

4, ,5 3

6.3.2 Solve the following systems of inequalities

1. 2x 4 0

(x 3)(x 5) 0

− ≥

− + < ] [ [ [-5,-2 2,3∪

2.

3 1

3x 2 x1

0(4 x 1)(x 3)

< − <

− +

1

0,4

3. 3x 17

1 25

−− < < ] [4,9

Analytic Geometry

Team Mathematics


1 Vectors and lines ................................................................................................................... 3

1.1 Vectors ........................................................................................................................................ 3

1.2 Lines ............................................................................................................................................ 5

1.3 Parallel lines ................................................................................................................................ 8

1.4 The euclidian vectorspace ........................................................................................................... 9

1.5 Angles ....................................................................................................................................... 11

1.6 Orthogonality of 2 lines ............................................................................................................ 12

1.7 Distance from a point to a line (O.N.B.) ................................................................................... 13

1.8 Exercises ................................................................................................................................... 14

2 Conic sections ...................................................................................................................... 19

2.1 Introduction ............................................................................................................................... 19

2.2 The circle .................................................................................................................................. 20

2.3 The parabola.............................................................................................................................. 21

2.4 Exercises ................................................................................................................................... 23

_____________________________________________________________________________ Ananlytic Geometry 3

1 Vectors and lines

1.1 Vectors

1.1.1 The concept of a vector

π is the set of all points of the plane. Take in the plane π a fixed origin O. A translation t in the plane is now completely determined by the image point P of O by this translation: t(O) = P. The translation is determined by point P or also by any couple of points (A,B) so that t(A)=B.

With a translation t, we associate a free vector. The vector AB��

is the set of couples of points

(A,B) such that t(A)=B. AB CD=��

if there is a translation t such that t(A)=B and t(C)=D.

A free vector can now be represented by means of an arrow with origin in O and with terminus

point t(O)=P. Now each point P in the plane can be seen as the terminus point of P OP=��

. P��

is called a bound vector.

The vector O��

is called the zero vector.

The plane in which we consider each point a terminus point of bound vectors is called πo.

1.1.2 Basis and coordinate system

Let’s choose in πo xE��

en yE��

which together with O��

form a triangle.

For each P��

in πo there is only one 2( , )x y ∈ℝ such that x yP xE yE= +��

.

{ },x yE E��

is called a basis of πo.

(x,y) are called the cartesian coordinates of P��

with respect to the basis { },x yE E��

. We emphasize

the notion “coordinate” is related to a basis.

OEx is the x-axis; OEy is the y-axis; together they represent the coordinate axes of the xy-coordinate system.

If (a1,a2) are the coordinates of A��

we also write this vector as A(a1,a2) or 1 2( , )A OA a a= =��

. This

means that we will use the word vector (with starting point the origin) and the word point in the xy-plane for the same concept.


1.1.3 Operations with vectors

In πo we define the following operations:

1. The addition of 2 vectors A��

and B��

is performed by adding the 2 associated vectors OA��

and OB��

using the parallelogram rule. The result C��

is determined by its terminus point.

From the figure one can easily see that, if ( )1 2,A a a=��

and ( )1 2,B b b=��

, then the sum

vector is: ( )1 2 1 2,C A B a a b b= + = + +��

.

A

B

C

x

y

1a

2a

1b

2b

O

2. The multiplication of a vector A��

of πo and a scalar k ∈R is a multiple of A��

: k A��

.

Furthermore, if k > 0, k A��

has the same sense as A��

, and if k < 0, k A��

has the opposite sense.

If ( )1 2,A a a=��

, then ( )1 2,k A ka ka=��

.

1.1.4 Midpoint of a line segment

The midpoint of a line segment [AB] with ( )1 2,A a a=��

and ( )1 2,B b b=��

is determined by

1 1 2 2,2 2 2

a b a bA BM

+ ++ = =

��


1.2 Lines

1.2.1 Vector equation

Take a line e through the origin. Every other point S on that line is the terminus point of a

vector S��

that determines the direction of e in a unique way. Such a vector is called a direction

vector of e. If S��

is a direction vector of e, then every kS��

with 0k ∈R also is.

O

S

e

Take a line 1e parallel to e and two different points P��

and 0P��

on 1e .

0P�

P�

S�

0P P−� �

1e

e

If S�

is a direction vector of e , then 0P P kS− =��

or 0P P kS= +��

for a certain k ∈R .

Otherwise, for every k , the terminus point of the vector 0P kS+��

is on the line 1e .

The vector equation 0P P kS= +��

is therefore a necessary and sufficient condition for a point to

lie on the line 1e (given by a point 0P and the direction S�

).

0P P kS= +��

with k ∈R

is called the vector equation of the line.


1.2.2 The parameter equations

Replacing the vectors by their coordinates in 2R with ( , )P x y=

�, 0 0 0( , )P x y=�

and ( , )S a b=�

gives

0 0( , ) ( , ) ( , )x y x y k a b= + with k ∈R

Furthermore

0 0( , ) ( , )x y x ka y kb= + + with k ∈R

which is equivalent with

0

0

x x ka

y y kb

= + = +

with k ∈R

These equations are called parameter equations of a line through a given point 0 0( , )x y with

direction numbers ( , )a b , the coordinates of the direction S�

. It is important to see that the above

equations are not unique since both 0P�

and S�

may be chosen.

1.2.3 Cartesian equation(s)

So we have two ways to give the equation of a line: the vector equation and the parameter equations. A third possibility, although less general, is the Cartesian equation(s). They may be obtained starting from the parameter equations. From the parameter equations, k may be eliminated:

0 0x x y y

a b

− −= if 0ab ≠

or

0 0( )b

y y x xa

− = − if 0≠a

Both equations are called the Cartesian equation of the line. The ratio b

a (in front of x) is called

the slope m.

If 0a = , the parameter equations are

0

0

x x

y y kb

= = +

with k ∈R

The second of these equations simply says that y may be any real number. Since there is no limitation on y, this equation is superfluous. The other equation then becomes the Cartesian equation:

0x x=

This line is parallel to the y-axis. The simplest direction numbers are (0,1). The slope m is not defined, since 0a = .

In the same way, if 0b = , the cartesian equation becomes 0y y= . This line is parallel to the

x-axis. The simplest direction numbers are (1,0). Its slope is m = 0.


Notice that ( , ) (0,0)a b = is impossible. The zero vector is excluded as direction vector of a line.

If the line is given by two points 0 0 0( , )P x y en 1 1 1( , )P x y then a direction vector of the line is

given by 1 0−� �P P , so we obtain:

0 1 0( )P P k P P= + −� � � �

with k ∈R

as vector equations,

0 1 0

0 1 0

( )

( )

x x k x x

y y k y y

= + − = + −

met k ∈R

as parameter equations,

0 0

1 0 1 0

x x y y

x x y y

− −=− −

as cartesian equation if 01 xx ≠ and .

If 01 xx = the equation is similar to a = 0; if 01 yy = the equation is similar to b= 0.

So, in general, the Cartesian equation of a line in 2ℝ takes the form

0Ax By C+ + =

with A and B not both zero at the same time.

Otherwise it can be proven that each equation of this form in 2ℝ respresents a line.

What is the slope of this line?

Find a set of direction numbers.

01 yy ≠


1.3 Parallel lines

If is given :

line e1 with direction vector 1S , direction numbers (a1,b1), slope m1, equation A1x+B1y+C1=0,

line e2 with direction vector 2S , direction numbers (a2,b2), slope m2, equation A2x+B2y+C2=0

e e 1

2 O S S

1 2

then

e1 // e2

12 SkS =⇔ with 0∈k R

==

⇔12

12

kbb

kaa with 0∈k R

12 mm =⇔

2 1

2 1

A kA

B kB

=⇔ =

with 0∈k R

Use:

Given line e with equation Ax+By+C=0 and point P0(x0,y0)

Determine the equation of line f // e and through P0

⇒ f has equation: A(x- x0)+B(y- y0)=0


1.4 The euclidian vectorspace

1.4.1 Definitions

If 1 2( , )A a a��

and 1 2( , )B b b��

in are given, then the scalar product of these two vectors is

1 1 2 2A B a b a b⋅ = + ∈��

ℝ

Then these vectors are perpendicular if

A B A . B 0⊥ ⇔ =��

Then the norm of the vector is

2 21 2A A A a a 0= ⋅ = + >

��

Then A��

is a normalized vector if

1=A

Then the distance between 1 2( , )A a a and ( )1 2,B b b is

( ) ( )222

211),( babaBABAd −+−=−=

A

B

A - B

O

π 0


1.4.2 Normalizing a vector

If OA ≠��

then

a

AE

A=��

��

has length 1 and thus is a unit vector in the same sense as A��

.

This process is called normalizing a vector.

1.4.3 Orthonormal basis (O.N.B.)

{ },x yE E��

is an orthonormal basis

==

⊥⇔

1yx

yx

EE

EE


1.5 Angles

First we want to mention that angles are positive angles if the terminal ray rotates counterclockwise around the vertex from the initial ray.

1.5.1 The angle of a line with the x-axis

(analytical expressions hold in an orthonormal basis)

In the following figure, we see in the right triangle that

mm ==1

tanα

In other words, the slope of the line is equal to the tangent of the angle between the x-axis and the line.

α

E y

E x

y

x

e

0

1

m

1.5.2 The angle between two lines

If line e1 has direction vector 1S��

, direction numbers (a1,b1) and slope m1

and line e2 has direction vector 2S��

, direction numbers (a2,b2) and slope m2

then the angle α between these lines is equal to the acute angle determined by the direction vectors of the lines:

1 2 1 2 1 2

2 2 2 21 2 1 2 1 2

cosS S a a b b

S S a a b bα

⋅ += =

+ ⋅ +

��

��


1.6 Orthogonality of 2 lines

If line e1 has direction vector 1S , direction numbers (a1,b1), slope m1, equation

1 1 1 0A x B y C+ + =

and line e2 has direction vector2S , direction numbers (a2,b2), slope m2 , equation 2 2 2 0A x B y C+ + =

then (the analytical expression with respect to an orthonormal basis is):

e

e

1

2 O

S

S

1

2

e1 ⊥ e2

2 1S S⇔ ⊥��

02121 =+⇔ bbaa

12

1

mm −=⇔

02121 =+⇔ BBAA

Use:

Find the equation of line f ⊥ e with equation 0Ax By C+ + = and which passes through point

( )0 0 0,P x y .

⇒ f has equation: ( ) ( )0 0 0B x x A y y− − − =


1.7 Distance from a point to a line (O.N.B.)

Consider in a point 0 0 0( , )P x y and a line : 0e Ax By C+ + = .

To find the (orthogonal) distance from 0P��

to the line e , one may proceed in the following way:

• Construct the line r through 0P��

perpendicular to e .

• Find the intersection S��

of r and e .

• The distance between 0P��

and e is then 0( , )d P S��

.

So:

r may be given by means of:

0

0

x x kAk

y y kB

= +∈ = +ℝ

To find the intersection S��

of r and e, the parameter equations of r is substituted in the equation of : 0e Ax By C+ + = . Then this equation is solved for k:

( ) ( )0 0 0A x kA B y kB C+ + + + =

gives the value of the parameter kS which corresponds to the intersection S��

:

0 02 2S

Ax By Ck

A B

+ += −+

So the distance is:

( ) ( )( ) ( )( )2 20 0 0 0 0 0, S Sd P S P S x x k A y y k B= − = − + + − +��

( ) 2 20, Sd P S k A B= +��

and finally

0 00 0 2 2

( , ) ( , )Ax By C

d P e d P SA B

+ += =

+

��

π 0


1.8 Exercises

1. Given are A(5,-1), B(-1,5), C(-7,2). a. Determine the coordinates of the midpoints of the sides of the triangle ABC.

( ) 7 12,2 , 4, , 1,

2 2 − −

b. Determine the coordinates of the centroid (or geometrical center or center of area) of the triangle ABC.

( )1,2−

2. Determine the direction numbers and the slope of the lines determined as follows: a. through (-2, 7) and (1, -8) (3, -15) and m = -5 b. line x (k,0) and m = 0 c. line y (0,k) and m does not exist! d. the line with equation: 2 4 0x y− + = (1, 2) and m = 2

e. the line with equation: 3

4y x= (4, 3) and

3

4m =

f. the line with equation: 2 3y x= + (1, 2) and m = 2

3. Make a drawing of the lines with the following equations: a. 4y x= b. 2 3 0x y+ = c. 4 2 5 0x y+ + =

4. Determine the angle which the following lines make with the x-axis in an orthonormal coordinate system. a. 5 0y x− + = α = 45°;α = 135°

b. 3 5 0y x− − = α = 60°;α = 120°

5. Proof that the figure formed by A(-2,1), B(-1,4), C(5,6), D(4,3) is a parallelogram.

6. Given line e with equation 2 4x y+ = a. does A(4,1) belong to e ? No b. does B(4,0) belong to e ? Yes c. determine the abscissa of the point on e with ordinate -5 x = 14

d. determine the ordinate of the point C with abscissa 1 3

2y =

e. draw line e f. find the intersection points with the x-axis and the y-axis (4, 0) and (0, 2)


7. Given line e with equation: 3 2 0ax y+ + = with a∈ℝ . Determine, if possible, a such that:

a. e passes through (2, 0) a = -1 b. e passes through (0,0) impossible c. e is parallel to line f with equation 3 5 0x y− − = a = -9 d. e is parallel to the x-axis a = 0 e. e is parallel to the y-axis impossible

f. the x-intercept of e is +3 2

3a = −

g. the y-intercept of e is +5 impossible

h. the y-intercept of e is a∈ℝ

8. Find the equation of the line : a. whose slope is m = -2 and passes through the point (3, 4) 2 10 0y x+ − = b. which passes through the points (2, 3) and (5, 1) 3 2 13 0y x+ − = c. which passes through the origin and the point (2, 6) 3 0y x− = d. which passes through the points (3, 5) and (7, 5) 5y = e. which passes through the points (-2, 4) and (-2, 1) 2x = − f. which passes through the points (1, 3) and (-2, -6) 3 0y x− = g. with slope m = 2 and whose y-intercept is +4 2 4 0y x− − = h. whose y-intercept is +3 and x-intercept is +2 3 2 6 0x y+ − = i. whose y-intercept is -6 and which passes through the point (2, 4) 5 6 0y x− + = j. which passes through the point (-2, 6 ) and is parallel to the line e: 3 2 5 0x y+ − =

2 3 6 0y x+ − =

k. which passes through the point (0, 3) and is parallel to the line through the points (2,0) and (5, 2) 3 2 9 0y x− − =

l. which passes through the point (2, 3) and is parallel to the y-axis x = 2 m. which passes through the point (-2, 4) and is parallel to the x-axis y = 4

9. Given line e with equation : ( 2) ( )y a x a b= − + + with ,a b ∈ℝ . Determine a and b such that: a. e passes through the points (1, 3) and (3, 5) a = 3 and b = -1 b. e passes through the point (2, -3) and is parallel to line f: 5 0x y+ + = a = 1 and b = -2 c. the x-intercept of e is -2 and e has (2, -10) as direction numbers a = -3 and b = -7 d. e passes through (0, 0) en (-2, -4) a = 4 and b = -4

10. Determine the equation of line e (in an O.N.B.): a. which passes through the point (4, -3) and forms an angle of 45° with the x-axis

7 0; 1 0y x y x− + = + − = b. whose y-intercept is +3 and forms an angle of 30° with the positive x-axis

3 3 9 0;3 3 9 0y x y x− − = + − =

−2

3


11. Determine the equation of line e which passes through the point (2, 4): a. whose positive x-intercept is double the positive y-intercept 2 10 0x y+ − = b. whose negative x-intercept and positive y-intercept are equal 2 0x y− + =

12. Determine a such that e // f:

: ( 1) ( 2) 1 0

: ( 1) (2 4 ) 0

e a x a y

f a x a y a

− − + + =+ + − + =

a = 0 or a = 3

13. Examine if the following points are collinear:

a. (8, 3), (-6, -3), (15, 6) Yes b. (1, 1), (4, -1), (-5, 5) Yes

14. Determine a such that the points (2, 3), (a, 2) and (a + 2, a - 3) are collinear. Determine the equation of the line passing through these points.

3 and 5 0

4and 2 8 0

a y x

a y x

= + − == + − =

15. Determine the points of intersection of the following lines:

a. : 2 4

:3 7

e x y

f x y

+ = + =

(2, 1)

b. : 5 3 1 0

: 2 8 0

e x y

f x

+ − = + =

(-4, 7)

c. : 6 3 4

3: 2 3 0

2

e y x

f x y

− = − + =

coinciding lines

16. Give the general equation of all lines passing through (2, 1). 1 ( 2)y m x− = −

17. Give the general equation of all lines with slope 2. 2y x q= +

18. Proof that the 3 medians of a triangle are concurrent. 19. For A(2,0), B(1,1), C(1,2) given in an O.N.B., determine A B⋅��

2

A C⋅��

2

C B⋅��

3 A A⋅��

4 B B⋅��

2 C C⋅��

5

20. Proof that the points A(4,6), B(2,-4), C(-2,2) form a right triangle.


21. Determine the equation of the line perpendicular to

a. : 5 6 0e x y+ − = and passing through (0, 0) 15 3

5 0 and ,13 13

x y − =

b. : 2 5 0f x y− + = and passing through (2, -6) 2 10 0 and ( 4, 3)x y+ + = − − Determine also the intersection points.

22. Determine the equation of the line perpendicular to the line with equation 2 5 4 0x y+ + = in the intersection point with the x-axis. 5 2 10 0x y− + =

23. A line has slope 2. Determine the equation of the line perpendicular to that line and passing through the point (4,1). 2 6 0y x+ − =

24. Determine the equation of the perpendicular bisector of the line segment [AB] with A(3,-1)and B(1,3). 2 0y x− =

25. The equations of the sides of a triangle are: 3 4 14 0;4 13 0; 5 8 0x y x y x y− + = + − = + − = . Determine the equations of the altitudes and the coordinates of the orthocenter.

5 5 0

3 4 15 0

4 10 0

30 55,

19 19

y x

y x

y x

− + =+ − =− − =

26. Determine the length of the sides of the triangle formed by A(5, -3), B(-1, 5), C(-7, 2). 10;13;3 5

27. Determine the distance from the origin to A(-2, 4) and to the line : 3 5e x y− =

102 5;

2

28. Determine the distances from A(-3, 4), B(2, 1), C(-3, 2) to the line : 5 0e x y− + =

2;3 2;0

29. Determine the distances from A(5, 1) to B(1, -2), C(-2, 2) and to the line BC.

5;5 2;5


30. Determine the distance from A(-4, 4) to the line : 2 3 0e x y+ − = using a. the formula b. the construction method

7 5

5

31. Determine the distance from A(3, -5) to the line perpendicular to 022: =+− yxe and

through B(1, 1). 2 5

32. Determine the distance from the origin to the perpendicular bisector of the line segment [AB] with A(-9, 7), B(15, -3).

2

33. Determine the distance between the parallel lines with equations

2 3 0 and 2 2 0x y x y+ + = + − = . 5 34. If : 2 1 0; : 2 3 0e x y f x y+ + = + − = , determine p such that the distances from A(1,p) to the

lines e and f are equal.

1 and 5

3p p= − =

35. Determine the equation of the line through A(2, 3) and at a distance 3 from the origin.

3 and 5 12 39 0y y x= + − =

36. Determine the equation of the line through A(-5, 1) such that the distance from B(3, -1) and C(-3, 2) to that line is equal.

2 3 0 and 10 5 0y x y x+ + = + − =

37. Determine the equation of the line parallel to line : 5 12 11 0e x y+ − = and at a distance 1 from A(-2, 1).

5 12 15 0

5 12 11 0

x y

x y

+ − =+ + =


2 Conic sections

2.1 Introduction

Parabolas, ellipses and hyperbolas are conic sections. A conic section is a curve which is obtained by the intersection of a cone with a plane. Circles are also conic sections, since they are special cases of ellipses. The type of a conic section depends on the angle between the intersecting plane and the cone.

Figure 1 snijding van een kegel door een vlak

Left: to obtain an ellipse, the angle between the plane and the axis is larger than the angle between the axis of the cone and a line on the cone.

Middle: to obtain a parabola, the angle between the plane and the axis of the cone is equal to the angle between the axis and a line on the cone.

Right: to obtain an hyperbola the angle between the plane and the axis of the cone is smaller than the angle between the plane and a line on the cone.

In the following paragraphs we give the Cartesian equations in standard form of a circle and a parabola in an orthonormal coordinate system. The equations represent the condition which points P(x,y) must satisfy to be on the graph of the conic section.


2.2 The circle

Although the circle is only a special case of an ellipse, yet we will first pay attention to its definition and its equation.

Definition:

A circle C is the set of points P(x,y) which are on a constant distance R from a fixed point M(x0,y0). This fixed distance R is called the radius, the fixed point M(x0,y0) the center of the circle.

2 2 20 0

( , )

( ) ( )

P d P M R

x x y y R

∈ ⇔ =⇔ − + − =

C

If the center M is in the origin, the equation of the cirlce is 2 2 2x y R+ =

The general equation of the circle is:

022 =++++ CByAxyx

Be aware that not all equations of this form represent a circle (see exercise 3).


2.3 The parabola

Definition:

A parabola P is the set of points P(x,y) for which the distance to a fixed line d is equal to the distance to a fixed point F which is not on d. The point F is called the focal point (focus) of the parabola, the line d the directrix.

( , ) ( , )Q d Q d d Q F∈ ⇔ =P

The equation of the parabola P with focal point F(p,0) and directrix :d x p= − is:

2 4y p x=

This is easily found from the definition:

Q

F

d y

x T

Figure 2 parabool

2 2

2 2 2

2

( , ) ( , )

( ) ( 0)

( ) ( 0) ( )

4

Q d Q d d Q F

x p y x p

x p y x p

y p x

∈ ⇔ =

⇔ − + − = +

⇔ − + − = +⇔ =

P

So the equation of the parabola is 2 4y p x= .

The x-axis is the symmetry axis of the parabola. The point F is the focal point, the line d is the directrix and the point halfway between the focal point and the directrix is the top of the parabola.

It is important to see that the above equation is only valid for this specific choice of the coordinate system and of the position of the focal point and the directrix. If the parameter p is negative, the focal point is at the left of the top and the directrix at the right side. The parabola will open to the negative x-axis. The equation is the same.

If we select a horizontal directrix, the parabola will be oriented vertically. Again, depending on the sign of p, the parabola is open to the positive y-axis if p > 0 and is open to the negative y-axis if p < 0.


So there are four possibilities for a parabola with its top in the origin and one of the coordinate axes as symmetry axis.

2 4y p x= with 0p > 2 4y p x= with 0p <

2 4x p y= with 0p > 2 4x p y= with 0p <

If the parabola is shifted to an arbitrary position of its top T(x0,y0) one finds two possible equations:

Horizontal symmetry axis: 2

0 0( ) 4 ( )y y p x x− = − with 0 0( , )F x p y+ and 0:d x x p= −

Vertical symmetry axis: 2

0 0( ) 4 ( )x x p y y− = − with 0 0( , )F x y p+ and 0:d y y p= −

The general equation of a parabola with symmetry axis parallel to a coordinate axis is:

Horizontal symmetry axis:

CByAyx ++= 2

Vertical symmetry axis:

CBxAxy ++= 2


2.4 Exercises

(we work in an orthonormal coordinate system)

1. Determine the equation of the circle with center M(a, b) and radius R. a. 0, 0, 5a b R= = = 2 2 5x y+ =

b. 4, 1, 2a b R= = = 2 2( 4) ( 1) 4x y− + − =

2. Determine the center and radius of the following circles: a. 2 2 8 6 0x y x y+ − − = ( )4,3 5M and R =

b. 2 23 3 2 3 1 0x y x y+ − + + = 1 1 1

,3 2 6

M and R − =

c. 2 216 16 8 15 0x y x+ − − = 1

,0 14

M and R =

d. ( )2 236 48 36 227 0x y x y+ − + − = 2 1

, 73 2

M and R − =

3. Examine if following equations represent circles. If so, determine the center and the radius. a. 2 2 6 14 59 0x y x y+ − + + = no circle

b. 2 216 16 8 64 335 0x y x y+ + − − = 1

,2 ,54

C −

c. 2 24 4 12 40 109 0x y x y+ − + + = 3

, 5 ,02

C −

4. Find the equation of the circle which passes through (3, 3) and (5, 7) and has its center on the line : 5a x y− =

2 2 16 6 48 0x y x y+ − − + =

5. Find the equation of the circle which has the line segment joining the points A(5,6) and B(-1,0) as diameter.

2 2 4 6 5 0x y x y+ − − − =

6. Find the equation of the circle through P(-3,4) and concentric with the circle2 2: 3 4 1 0c x y x y+ + − − =

2 2 3 4 0x y x y+ + − =

7. Find the equation of the circle circumscribing the triangle ABC with A(2, 2), B(6, -2), C(-3, -5).

2 22 2 5 11 28 0x y x y+ − + − =


8. A circle has its center in M(3, 0) and passes through P(1, 1). Determine: a. the equation of the circle. 2 2( 3) 5x y− + =

b. the equation of the circle with the same center and double area. ( )2 23 10x y− + =

9. Determine the focal point and the directrix of the following parabolas and make a drawing. a. 2 8 0y x− = (2,0); 2x = −

b. 2 6 0y x+ = 3 3

,0 ;2 2

x − =

c. 22 0x y+ = 1 1

0, ;8 8

y − =

d. 22 0x y− = 1 1

0, ;8 8

y = −

10. Find the equation of the parabola with the x-axis as the symmetry axis and with the top in the origin and through the point A(-1, 3).

2 9y x= −

11. Find the equation of the parabola with focal point 1

0,2

−

and directrix 1

:2

d y = .

2 2 0x y+ =

12. Find the equation of the parabola with focal point (7, -2) and directrix : 3d x = . 2 4 8 44 0y y x+ − + =

13. Find the equation of the parabola with focal point (7, -2) and as directrix d the bisector of the second quadrant.

2 2 2 28 8 106 0x y xy x y+ − − + + =

14. Make a drawing of:

a. 23 24 2 50 0x x y− − + = ( ) ( )2 24 1

3x y− = −

b. 24 40 3 100 0y y x+ − + = ( )2 35

4y x+ =

Calculus


1 Relations, functions, mappings, 1-1 mappings ..................................................................3

2 Expanding IR ........................................................................................................................4

3 Continuity of a function in IR .............................................................................................5

3.1 Continuity at a point ................................................................................................................... 5

3.2 Right- and left-continuity ........................................................................................................... 6

4 Limits .....................................................................................................................................7

4.1 Limits in a real number .............................................................................................................. 7

4.2 Right-hand and left-hand limits ................................................................................................. 9

4.3 Limits at infinity ........................................................................................................................ 9

4.4 Infinite limits .............................................................................................................................. 9

4.5 Infinite limits at infinity ........................................................................................................... 10

4.6 Elementary rules to calculate limits ......................................................................................... 10

4.7 Indeterminate cases .................................................................................................................. 11

4.8 Exercises .................................................................................................................................. 14

5 Differentiation .....................................................................................................................17

5.1 The derivative in a point .......................................................................................................... 17

5.2 Geometrical interpretation of the derivative ............................................................................ 17

5.3 Left-hand and right-hand derivative in a point ........................................................................ 18

5.4 A vertical tangent ..................................................................................................................... 19

5.5 Differentiation rules ................................................................................................................. 19

5.6 Exercises .................................................................................................................................. 21

6 Indefinite integrals .............................................................................................................24

6.1 Antiderivative functions ........................................................................................................... 24

6.2 The indefinite integral .............................................................................................................. 24

6.3 Properties ................................................................................................................................. 24

6.4 Basic integrals .......................................................................................................................... 25

6.5 Integration by substitution ....................................................................................................... 25

6.6 Exercises .................................................................................................................................. 26

6.7 Integration by parts .................................................................................................................. 29

6.8 Exercises .................................................................................................................................. 29

7 Definite integrals ................................................................................................................30

7.1 The fundamental theorem of calculus ...................................................................................... 30

7.2 Properties of the definite integral ............................................................................................. 30

7.3 The substitution method ........................................................................................................... 31

7.4 Partial integration ..................................................................................................................... 31

7.5 Exercices .................................................................................................................................. 32

Calculus 3

1 Relations, functions, mappings, 1-1 mappings

In mathematics, a relation from A to B is a collection of ordered pairs (x,y) where x A∈ and y B∈ , in other words it is a subset of A x B.

The domain of the relation R is the set of all the first numbers of the ordered pairs. In other words: dom R { }| ( , )x x y R= ∈ .

The range of the relation R is the set of the second numbers in each pair. In other words: range R { }| ( , )y x y R= ∈ .

If one switches the order of the pairs of a relation, then one gets the inverse relation from B to A.

A function is a special type of relation where each x-value has one and only one y-value, so no

x-value can be repeated. All functions are relations but not all relations are functions.

The x-number is called the independent variable, and the y-number is called the dependent

variable because its value depends on the x-value chosen.

A function where for every element x of A there is exactly one f(x) in B is called a mapping

from A to B. Then dom f = A.

A mapping from A to B is called a 1-1 mapping when the inverse is also a mapping from B to A.

Calculus 4

2 Expanding IR

Given the set of real numbers IR, we add two elements −∞ en +∞ , and call the new set the

expanded set of real numbers.

Notatie: { },IR IR= ∪ −∞ +∞

It is clear that :x IR x∀ ∈ −∞ < < +∞ .

Properties:

0

0

1. : ( ) ( )

( ) ( )

: ( ) ( )

( ) ( )

: ( ) ( )

( ) ( )

x IR x x

x x

x IR x x

x x

x IR x x

x x

+

−

∀ ∈ + −∞ = −∞ = −∞ ++ +∞ = +∞ = +∞ +

∀ ∈ ⋅ +∞ = +∞ = +∞ ⋅⋅ −∞ = −∞ = −∞ ⋅

∀ ∈ ⋅ +∞ = −∞ = +∞ ⋅⋅ −∞ = +∞ = −∞ ⋅

2. ( ) ( )

( ) ( )

( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

if n is odd

1 10 0

n

n

+∞ + +∞ = +∞−∞ + −∞ = −∞−∞ ⋅ +∞ = −∞ = +∞ ⋅ −∞+∞ ⋅ +∞ = +∞ = −∞ ⋅ −∞

+∞ = +∞

−∞ = −∞

= =+∞ −∞

So, the following expressions have no meaning, they are indefinite:

+∞( ) + −∞( ) ; −∞( ) + +∞( ) ; 0.+∞ ; + ∞. 0 ; 0.−∞ ; −∞.0 ;+∞+∞

;+∞−∞

;−∞+ ∞

;−∞−∞

Calculus 5

3 Continuity of a function in IR

3.1 Continuity at a point

Definition:

a function f: IR → IR is continuous in a point a if the graph of the function does not show a

jump in a.

If the function is not continuous in a, it is defined as discontinuous in a.

Examples

f is continuous in a f is discontinuous in a

f is not left-continuous in a

f is not right-continuous in a

f is discontinuous in a f is discontinuous in a

f is not left-continuous in a f is right-continuous in a


f is discontinuous in a

f is left-continuous in a

f(a)

a

b

f(a)

a

Y

X

Y

X

X

Y

X

Y

a a

c

f(a)

b

f(a)

b

X

Y

a

bf(a)

Calculus 6

It is clear that the point a must belong to the domain of the function, otherwise continuity is not

possible.

The following mathematical definition can be understood in terms of the previous intuitive

definition of a continuous function.

is continuous in 0, 0 : ( ) ( )

a domff a

x a f x f aε δ δ ε∈⇔ ∀ > ∃ > − < ⇒ − <

3.2 Right- and left-continuity

is right-continuous in 0, 0 : ( ) ( )

is left-continuous in 0, 0 : ( ) ( )

a domff a

a x a f x f a

a domff a

a x a f x f a

ε δ δ ε

ε δ δ ε

∈⇔ ∀ > ∃ > ≤ < + ⇒ − <

∈⇔ ∀ > ∃ > − < ≤ ⇒ − <

Calculus 7

4 Limits

4.1 Limits in a real number

The concept ‘limit’ allows us to examine the behavior of a function in the close neighbourhood

of a point. The function value in this point is without importance.

Assume that a is a real number which is an adherent point of dom f. This means that in each interval around a, 0] , [,a a IRε ε ε +− + ∈ , there are numbers of dom f which are different from a.

In other words, f is not necessarily defined in a, but it is defined in other points arbitrary close

to a.

We say that a function has a limit b at an input a if the values f(x) come arbitrarily close to b for

x-values which come sufficiently close to a.

Notation: lim ( )

x af x b

→=

Since lim ( )

x af x b

→= describes the behavior of the function f in the environment of the point a,

the concept lim ( )x a

f x→

is obviously meaningless when a is an isolated point of dom f.

Definition:

If f(x) is defined in an open interval around a, except possibly at a itself, then

lim ( ) 0, 0 :0 ( )x a

f x b IR x a f x bε δ δ ε→

= ∈ ⇔ ∀ > ∃ > < − < ⇒ − <

Note that the limit equals the function value in a if the function is continuous in a:

lim ( ) ( ) is continuous in x a

f x f a f a→

= ⇔

Calculus 8

Examples

f is continuous in a f is discontinuous in a

f is not left-continuous in a

f is not right-continuous in a lim ( ) ( )

x af x f a

→= lim ( )

x af x b

→=

f is discontinuous in a f is discontinuous in a

f is not left-continuous in a f is right-continuous in a


lim ( ) lim ( )

lim ( ) does not exist

x a x a

x a

f x c f x b

f x> <

→ →

→

= ≠ =

⇒

lim ( ) ( ) lim ( )


x a x a

x a

f x f a f x b

f x> <

→ →

→

= ≠ =

⇒

f is discontinuous in a

f is left-continuous in a

lim ( ) lim ( ) ( )


x a x a

x a

f x b f x f a

f x> <

→ →

→

= ≠ =

⇒

f(a)

a

b

f(a)

a

Y

X

Y

X

X

Y

X

Y

a a

c

f(a)

b

f(a)

b

X

Y

a

bf(a)

Calculus 9

4.2 Right-hand and left-hand limits

We say the left-hand (or right-hand) limit of f(x) as x approaches a is b (or the limit of f(x) as x

approaches a from the left (or the right) is b) if the values f(x) come arbitrarily close to b for x-

values which come sufficiently close to a from the left-hand side (right-hand side).

] [

] [

lim ( ) 0, 0 : , ( )

lim ( ) 0, 0 : , ( )

x a

x a

f x b x a a f x b

f x b x a a f x b

ε δ δ ε

ε δ δ ε

>

<

→

→

= ⇔ ∀ > ∃ > ∈ + ⇒ − <

= ⇔ ∀ > ∃ > ∈ − ⇒ − <

The following notations are also commonly used:

lim ( ) , lim ( ) , ( )

x aaf x b f x b f a

− ↑= = −

4.3 Limits at infinity

Limits at infinity represent the behavior of function values f(x) if |x| ever increases. To define

limits at +∞, we demand that dom f contains at least a half line of the form [a,+∞[ . To define

limits at -∞, we demand that dom f contains at least a half line of the form ]-∞,a]

We say the limit of f(x) as x approaches (negative) infinity is b if the values f(x) come arbitrarily

close to b for arbitrarily large x-values.

lim ( ) 0, 0 : ( )

lim ( ) 0, 0 : ( )

x

x

f x b m x m f x b

f x b m x m f x b

ε ε

ε ε

→+∞

→−∞

= ⇔ ∀ > ∃ > > ⇒ − <

= ⇔ ∀ > ∃ > < − ⇒ − <

4.4 Infinite limits

We say the limit of f(x) as x approaches a is (negative) infinity, if the values f(x) become

arbitrarily large for x-values which come sufficiently close to a .

lim ( ) 0, 0 : 0 ( )

lim ( ) 0, 0 : 0 ( )

x a

x a

f x n x a f x n

f x n x a f x n

δ δ

δ δ

→

→

= +∞ ⇔ ∀ > ∃ > < − < ⇒ >

= −∞ ⇔ ∀ > ∃ > < − < ⇒ < −

Calculus 10

4.5 Infinite limits at infinity

We say the limit of f(x) as x approaches (negative) infinity is (negative) infinity, if the values

f(x) become arbitrarily large for arbitrarily large x-values.

lim ( ) 0, 0 : ( )

lim ( ) 0, 0 : ( )

lim ( ) 0, 0 : ( )

lim ( ) 0, 0 : ( )

x

x

x

x

f x n m x m f x n

f x n m x m f x n

f x n m x m f x n

f x n m x m f x n

→+∞

→+∞

→−∞

→−∞

= +∞ ⇔ ∀ > ∃ > > ⇒ >

= −∞ ⇔ ∀ > ∃ > > ⇒ < −

= +∞ ⇔ ∀ > ∃ > < − ⇒ >

= −∞ ⇔ ∀ > ∃ > < − ⇒ < −

4.6 Elementary rules to calculate limits

Given two real functions f and g for which the limit in a IR∈ exists, e.g.

1 2 1 2lim ( ) and lim ( ) with ,x a x a

f x L g x L L L IR→ →

= = ∈ then

1. [ ] 1 2lim ( ) ( ) lim ( ) lim ( )

x a x a x af x g x f x g x L L

→ → →+ = + = +

2. 1lim ( ) lim ( )x a x a

k f x k f x k L→ →

⋅ = ⋅ = ⋅

3. [ ] 1 2lim ( ) ( ) lim ( ) lim ( )x a x a x a

f x g x f x g x L L→ → →

⋅ = ⋅ = ⋅

4. 12

2

lim ( )( )lim if 0

( ) lim ( )x a

x ax a

f x Lf xL

g x g x L→

→→

= = ≠

5. 1lim ( ) lim ( )x a x a

f x f x L→ →

= =

6. ( ) ( ) ( )1lim ( ) lim ( )nn n

x a x af x f x L

→ →= =

7. 1lim ( ) lim ( )n nnx a x a

f x f x L→ →

= =

8. limx a

k k→

=

Calculus 11

4.7 Indeterminate cases

4.7.1 The indeterminate case 00

Rational function f (x)

g(x):

( ) 0lim

( ) 0x a

f x

g x→

=

means that lim ( ) ( ) 0x a

f x f a→

= = and lim ( ) ( ) 0x a

g x g a→

= = . So this means that

both f(x) and g(x) are divisible by (x – a), so it is possible to write 1( ) ( ) ( )f x x a f x= − and

1( ) ( ) ( )g x x a g x= − .

therefore 1 1

1 1

( ) ( ) ( )( ) 0lim lim lim

( ) 0 ( ) ( ) ( )x a x a x a

x a f x f xf x

g x x a g x g x→ → →

− = = = −

Example 2

22 2 2

4 ( 2)( 2) 2lim lim lim 4

( 2)( 3) 35 6x x x

x x x x

x x xx x→ → →

− − + += = = −− − −− +

Irrational form: make nominator and/or denominator rational.

Example

( )3 3

3

3

1 2 ( 1 2)( 1 2)lim lim

3 ( 3)( 1 2)

1 4lim

( 3)( 1 2)

1lim

( 1 2)

1

4

x x

x

x

x x x

x x x

x

x x

x

→ →

→

→

+ − + − + +=− − + +

+ −=

− + +

=+ +

=

Calculus 12

4.7.2 The indeterminate case ∞∞∞∞∞∞∞∞

Rational function:

1

1 1 01

1 1 0

...lim lim

...

n n nn n n

P p px xP p p

a x a x a x a a x

b x b x b x b b x

−−

−→∞ →∞−

+ + + +=

+ + + +

n

p

aif

b

if (sign of has to be determined)

= 0 if

n=p

n > p

n < p

=

∞=∞

Irrational function:

Determine the highest power of x in the nominator and denominator and simplify.

Remark that

if

if

2x = x x > 0

= -x x < 0

Example

222

¨

51 2

5 2lim lim

3 3

lim

x x

x

x xxx x

x x

x

→∞ →∞

→ +∞

+ + + + =

2

51 2 x

x

+ +

3 x

¨

1

limx

x

→ −∞

=

2

51 2 x

x

− + +

3 x

1

3

=

4.7.3 The indeterminate case ∞ − ∞∞ − ∞∞ − ∞∞ − ∞

Polynomial:

( )11 1 0

¨ ¨lim ... limn n n

n n nx x

a x a x a x a a x−−→ ∞ → ∞

+ + + + =

Irrational function: Multiply and divide by the conjugate term.

Calculus 13

Example

( )

( )

( ) ( )( )

2

2

2 2

2

2

2

lim 4 3 1 2

/ lim 4 3 1 2

4 3 1 2 4 3 1 2/ lim 4 3 1 2 lim

4 3 1 2

3 1lim lim

4 3 1 2

x

x

x x

x x

x x x

a x x x

x x x x x xb x x x

x x x

xx

x x x

→∞

→+∞

→−∞ →−∞

→−∞ →−∞

+ − +

+ − + = +∞

+ − + + − −+ − + =

+ − −

−= =+ − −

13

x

x

−

2

3

43 14 2

x x

= −

− + − −

Calculus 14

4.8 Exercises

1. 2

4

16lim

3 12x

x

x→

−−

8

3

2. 2

23

7 12lim

4 3x

x x

x x→

− +− +

1

2−

3. 2

31

1lim

1x

x

x→

−−

2

3

4. 4 4

3 3limx a

x a

a x→

−−

4

a3

−

5. 3 2

21

2 2lim

2x

x x x

x x→

+ − −+ −

2

6. 3

3 21

3 2lim

2 2 2x

x x

x x x→

− +− + −

0

7. 4 3 2

3 23

2 2 16 24lim

6x

x x x x

x x x→−

− − ++ −

10−

8. 4 3 2

3 20

2 2 16 24lim

6x

x x x x

x x x→

− − ++ −

4−

9. limm m

x a

x a

x a→

−−

m 1m a −⋅

10. limm m

n nx a

x a

x a→

−−

m nma

n−

11. 2

2 2lim

2x

x

x→

+ −−

1

4

12. 3

6 3lim

3x

x x

x>→

+ −−

0

13. 0

lim1 1x

x

x x→ + − − 1

14. 2 2

5

4 5 25lim

5x

x x x

x>→

− − − −−

−∞

15. 32

6 3 4 1lim

2x

x x

x→

− − +−

0

Calculus 15

16. 1

2 3 3lim

3 2x

x x

x→

− + ++ −

2

17. 3 23

22

1 1 2lim

2x

x x x

x x→

− + + −−

1

2−

18. 4

3

2 1lim

2 1x

x

x→

− −− −

1

2

19. 3 33 2 32

2 2lim

3 4 4x

x x

x x x x x><

→

− −− − + −

3

3±

20. Evaluate a so that

2 2 2

2 25

8 8lim

2 3 12x

x ax x a a

x ax a→

+ − + =+ −

a 2=

21. 2 21

2 3lim

3 2 1x x x x><

→

− − + − ∞∓

22. 2 22

8 6lim

6 2x

x

x x x x→

+ − + − − −

7

15

23. 23

2

12

3lim1

4 3

x

xx

xx

x x

→−

+++

++ +

4

9

24. 22 3 1

lim2x

x x

x→+∞

+ − +−

1−

25. 3 3 2

3

8 2 4lim

1x

x x x

x x→∞

+ − −+ +

1;3

26. 3 31 8

lim1x

x

x→∞

−+

2−

27. ( )2lim 3 4x

x x x→∞

− − +

±∞

28. ( )3 3 2lim 3 1x

x x x→∞

+ + −

1

29. ( )2lim 1 4 1x

x x x→∞

+ − + +

−∞

Calculus 16

30. ( )lim 1x

x x x→∞

+ −

1

2

31. ( )2 2lim 2 4 9 1 4x

x x x x→−∞

+ + + − −

+∞

32. 3 2

22

2 8 4lim 4 5 1

4x

x x xx x

x→∞

− − + − + − −

9;

4−∞ −

33. ( )2 2lim 5 8 5 4x

x x x x x→∞

+ + − + −

6±

Calculus 17

5 Differentiation

5.1 The derivative in a point

Consider:

1 1

0 0

( ) ( )lim limx x

f f x x f x

x x∆ → ∆ →

∆ + ∆ −=∆ ∆

If this limit exists, in other words, if 0

limx

fIR

x∆ →

∆ ∈∆

, then we call the limit the derivative of the

function f in the point x1. We say that the function f is differentiable or has a derivative in x1.

.

Notation: 10

'( ) limx

ff x

x∆ →

∆=∆

or 1( )Df x or 1( )df x

dx

5.2 Geometrical interpretation of the derivative

Consider the curve with equation y = f(x), and 2 neighboring points 1 1( , ( ))A x f x and

1 1( , ( ))B x x f x x+ ∆ + ∆ on that curve. Then the difference ratio 1 1( ) ( )f f x x f x

x x

∆ + ∆ −=∆ ∆

is the

slope of the line AB. When 0x∆ → then the line AB rotates around A and reaches the tangent

at the function in the point A.

Conclusion: 1 11

0

( ) ( )'( ) lim

x

f x x f xf x

x∆ →

+ ∆ −=∆

is the slope of the tangent at f in the point

1 1( , ( ))A x f x .

This means that the equation of the tangent line at 1 1( , ( ))x f x of the function y = f(x) is given

by:

1 1 1( ) '( )( )y f x f x x x− = −

A

B

1x 1x x+ ∆ x

y

1( )f x x+ ∆

1( )f x f∆

x∆

( )y f x=

Calculus 18

Example

2

0

2 2

0

0

3 4

(1 ) (1)'(1) lim

(1 ) 3(1 ) 4 (1 3 1 4)lim

lim( 1) 1

x

x

x

y x x

f x ff

x

x x

xx

∆ →

∆ →

∆ →

= − +

+ ∆ −=∆

+ ∆ − + ∆ + − − ⋅ +=∆

= ∆ − = −

5.3 Left-hand and right-hand derivative in a point

One can consider in0

limx

f

x∆ →

∆∆

separately the right-hand limit and the left-hand limit. If they

exist, we call:

1 11

0 0

( ) ( )lim lim ( )x x

f x x f xff x

x x> >

+∆ → ∆ →

+ ∆ −∆ ′= =∆ ∆

the right-hand derivative in1x ,

1 11

0 0

( ) ( )lim lim ( )x x

f x x f xff x

x x< <

−∆ → ∆ →

+ ∆ −∆ ′= =∆ ∆

the left-hand derivative in1x .

A function is differentiable in x1 if and only if the left-hand derivative and the right-hand

derivative in x1 are equal.

If both one-sided derivatives in x1 exist but they are different, the function is not differentiable.

This means that there exist two tangent lines in x1. A point like that is called a corner point.

Example

211

4y x= − is continuous for each value of x, buth the right-hand derivative in x = 2, is

different from the left-hand derivative in that point, namely

2

2

lim '( ) 1

lim '( ) 1

x

x

f x

f x>

<

→

→

=

= −

Conclusion:

If f is continuous in x = x1 ⇒ f is differentiable in x = x1.

But one can prove that, if f is differentiable in x = x1 ⇒ f is continuous in x = x1.

Calculus 19

5.4 A vertical tangent

Example

The function 3y x= is continuous for each value of x.

We calculate the derivative of f in x = 0:

3

3 20 0 0

(0 ) (0) 1lim lim limx x x

f x f x

x x x∆ → ∆ → ∆ →

+ ∆ − ∆= = = +∞∆ ∆ ∆

So the function is not differentiable in x=0. We call this an undefined derivative. Geometrically

this means that the function has a vertical tangent in 0.

5.5 Differentiation rules

To calculate derivatives it is advisable to use the following differentiation rules rather than

using the limit-definition.

Let f and g be two functions that are differentiable in x.

1. the derivative of a constant function c

' 0c =

2. the derivative of the argument

' 1x =

3. the derivative of a som of functions ( ) '( ) '( ) '( )f g x f x g x+ = +

4. the derivative of a product of functions ( ) '( ) '( ) ( ) ( ) '( )f g x f x g x f x g x⋅ = ⋅ + ⋅

5. the derivative of a quotient of functions

'

2

'( ) ( ) ( ) '( )( ) if ( ) 0

( )

f f x g x f x g xx g x

g g x

⋅ − ⋅= ≠

6. the derivative of a power of a function

( )' 1( ) ( ) '( )n nf x n f x f x−= ⋅ ⋅

Calculus 20

7. the derivative of a composite function: the chain rule ( ) '( ) '( ( )) '( )g f x g f x f x= ⋅�

9. the derivative of trigonometric functions

2

2

(sin ) ' cos

(cos ) ' sin

1(tan ) '

cos1

(cot ) 'sin

x x

x x

xx

xx

== −

=

= −

Calculus 21

5.6 Exercises

Calculate the derivative of the following functions:

1. 3 2 1y x x x= − − + 2' 3 2 1y x x= − −

2. 3 21 34

8 4y x x= + − 23 3

'8 2

y x x= +

3. ( )22 4 3y x x= − + ( )( )2' 2 4 3 2 4y x x x= − + −

4. ( ) ( )23 6 3y x x x= + + ( )( )2 2' 3 6 2 13 18y x x x x= + + +

5. 2

2,51

5 32

yx x

=− + −

22

2,5 7,5'

15 3

2

xy

x x

−= − + −

6. 2

2

2 3

2 2

xy

x x

−=− +

( )

2

22

4 14 6'

2 2

x xy

x x

− + −=− +

7. 2 5 7

2

x xy

x

− +=−

( )2

2

4 3'

2

x xy

x

− +=−

8. 2

( 2)1

x xy

x

+=−

( )

2

22

2( 1)'

1

x xy

x

− + +=−

9. 31 23

y x xx

= + + 22

2' 1y x

x= + −

10. 3

2 1

xy

x=

−

( )( )2 2

22

3'

1

x xy

x

−=

−

11. 2 2y a x= − 2 2

'x

ya x

−=−

12. 218y x x= + −

2

2

18'

18

x xy

x

− −=−

13. 3 23y x x= +

2

3 2

3 6'

2 3

x xy

x x

+=+

14. 32 5 5y x x= + −

2

3

6 5'

2 2 5 5

xy

x x

+=+ −

Calculus 22

15. 24 1 2y x x x= − − − − 2

2 1' 4

2 2

xy

x x

−= −− −

16. ( ) 22 1 4y x x= − − 2

2

(2 1)' 2 4

4

x xy x

x

−= − −−

17. ( )2 21 4 3y x x x= + − + ( )( )2

2

1 3 9 4'

4 3

x x xy

x x

+ − +=

− +

18. 3 25 4 1y x x= − + ( )

( )223

2 5 2'

3 5 4 1

xy

x x

−=

− +

19. ( ) ( )32 245 6 4 2y x x x= − + +

3 2

24

35 30 52 24'

2 2

x x xy

x

− + −=+

20. 5x

yx

+= 2

10'

2 5

xy

x x

+= −+

21. 2

2

2 3 1

1

x xy

x x

− +=− + ( )

3 2

2 2

4 6 9 5'

2 1 1

x x xy

x x x x

− + −=− + − +

22. ( )5

3

1 2

xy

x=

+

( )( )7

3 1 3'

1 2

xy

x

−=

+

23. ( )( )( ) ( )

1 2

3 4

x xy

x x

− −=

− −

( )( )( ) ( )

2

3 3

2 10 11'

1 2 3 4

x xy

x x x x

− + −=− − − −

24. ( )2 2

2

2 3 9

4

x xy

x

− −=

4 2

3 2

3 2 36'

4 9

x xy

x x

− − +=−

25. ( )

( )

2 3 3

3

2 1 2

2

x xy

x

− −=

+

( )( )( ) ( )

3 2

24 33

2 1 14 2 4 22'

2 2

x x x xy

x x

− − + −=

+ −

26. 24siny x= ' 4sin 2y x=

27. cos 2 5cos 2y x x= − − ' 2sin 2 5siny x x= − +

28. 2cos2

xy = ' sin cos

2 2

x xy = − ⋅

29. sin cos

sin cos

x xy

x x

+=−

( )2

2'

sin cosy

x x

−=−

Calculus 23

30. siny x x= − ' 1 cosy x= −

31. sin 2 2siny x x= − ' 2cos2 2cosy x x= −

32. cosy x=

sin'

2 cos

xy

x

−=

33. 3 2cos 4y x= 3

8sin 4'

3 cos4

xy

x

−=

34. 4cos3 3sin 4y x x= + ' 12sin3 12cos4y x x= − +

35. cos sec 2y x x= + ' sin 2tan 2 sec2y x x x= − +

36. cos 2

sin

xy

x=

2

2sin sin 2 cos cos2'

sin

x x x xy

x

− ⋅ − ⋅=

37. 2 2

sin cos

3cos 2sin

x xy

x x

⋅=−

( )

4 4

2 2 2 2

3cos 2sin'

3cos 2sin 3cos 2sin

x xy

x x x x

+=− −

Calculus 24

6 Indefinite integrals

6.1 Antiderivative functions

Definition

Let us consider a function f: IR → IR : x → f(x).

An antidervative function F of the function f is any function with the property F’(x) = f(x).

Property 1

If F is an antiderivative of the function f, then F+k with k an arbitrary constant є IR, also is.

Property 2

If F1 and F2 are both antiderivatives of f, then they differ only in the constant.

Conclusion

If F is an antiderivative of f, then all antiderivatives of f can be found by adding to F an

arbitrary real constant.

6.2 The indefinite integral

The set of antidervatives of a function f is called the indefinite integral of f.

Notation:

{ }( ) ( ) | '( ) ( ) and k IRf x dx F x k F x f x= + = ∈∫

or

( ) ( )f x dx F x k= +∫

6.3 Properties

1. ( ) ( )k f x dx k f x dx=∫ ∫

2. ( )( ) ( ) ( ) ( )f x g x dx f x dx g x dx+ = +∫ ∫ ∫

Calculus 25

6.4 Basic integrals

1

, if n -11

nn x

x dx kn

+

= + ≠+∫

sin cosxdx x k= − +∫

cos sinxdx x k= +∫

2 tancos

dxx k

x= +∫

2 cotsin

dxx k

x= − +∫

6.5 Integration by substitution

Let f be a continuous function of x and x = g(t) is a differentiable function of t then:

( ) ( ( )) '( )f x dx f g t g t dt= ⋅∫ ∫

Calculus 26

6.6 Exercises

Calculate the indefinite integral of the following functions:

1. 3 x 33

4x x k+

2. 34 x 344

7x x k+

3. x x 22

5x x k+

4. 3 5 2x x 4 5 25

22x x k+

5. 3

1

x 3 23

2x k+

6. 9

1

x

8

1

8k

x− +

7. 34

3 2

x

x 1212

13x x k+

8. 2 7x+ 2 7x x k+ +

9. sin cosx x+ cos sinx x k− + +

10. 2

7

2cos x

7tan

2x k+

11. 1

1 cos2x−

1cot

2x k− +

12. 2

5

5 8 3x x

x

− +

2 3 4

5 8 3

2 3 4k

x x x− + − +

13. 3

2

1 cos

cos

x

x

− tan sinx x k− +

14. 1 cos2x+ 2 sinx k± +

15. ( )53x+

( )63

6

xk

++

16. cos2x 1

sin 22

x k+

Calculus 27

17. 2cos x 1 1

sin 22 4

x x k+ +

18. 2sin x 1 1

sin 22 4

x x k− +

19. ( )45x−

( )55

5

xk

−+

20. 6x+ ( )26 6

3x x k+ + +

21. 4

5x− 8 5x k− +

22. sin3x 1

cos33

x k− +

23. ( )52 7x −

( )62 7

12

xk

−+

24. 3 4x + ( )23 4 3 4

9x x k+ + +

25. 2

1

cos 4x

1tan 4

4x k+

26. 3

1

3 1x − ( )231

3 12

x k− +

27. ( )2 2x x− − ( )222 2

5x x k− − +

28. ( )3 2 6x x+ + ( )223 2 6

5x x k+ + +

29. 3sin x 3cos

cos3

xx k− + +

30. 3cos x 3sin

sin3

xx k− +

31. sin 2 cosx x⋅ 32cos

3x k− +

32. 4cos x 3 1 1

sin 2 sin 48 4 32

x x x k+ + +

Calculus 28

33. 4sin x 3 1 1

sin 2 sin 48 4 32

x x x k− + +

34. 1

1 cosx+ tan

2x

k+

35. 3x x+ ( )226 3

5x x x k+ − + +

36. 2 1x x− ( )( )221 15 12 8 1

105x x x x k− + + − +

37. ( ) 33 2x x+ + ( )( ) 332 4 15 2

28x x x k+ + + +

38. 2

3 1

x

x

+−

( )23 20 3 1

27x x k+ − +

Calculus 29

6.7 Integration by parts

Let u = f(x) and v = g(x) be differentiable functions.

Since ( )d u v u dv v du⋅ = +

it follows that ( )u dv d u v v du= ⋅ −

and therefore

u dv uv v du= −∫ ∫

6.8 Exercises

Evaluate the indefinite integral of the following functions:

1. 2 cosx x 2 sin 2 cos 2sinx x x x x k+ − +

2. cosx x sin cosx x k+ +

3. 3 sinx x 3 2cos 3 sin 6 cos 6sinx x x x x x x k− + + − +

4. 2cos x 1 1

sin cos2 2

x x x k+ +

5. 3cos x 21 2cos sin sin

3 3x x x k+ +

6. sin cosx x x 1 1

cos2 sin 24 8

x x x k− + +

7. 2cosx x 21 1 1sin 2 cos2

4 4 8x x x x k+ + +

Calculus 30

7 Definite integrals

7.1 The fundamental theorem of calculus

Let the function f be continuous on [a,b] and let F be an antiderivative of the function f on

[a,b] then

[ ]( ) ( ) ( ) ( ) ( ) |b

b baa

a

f x dx F b F a F x F x= − = =∫

Remark: The definite integral is a real number, the indefinite integral is a set of antiderivatives.

Example

2

0

sin 2I xdx

π

= ∫

1 1sin 2 sin 2 (2 ) cos2

2 2xdx xd x x k= = − +∫ ∫

Therefore [ ] ( )20

1 1cos2 1 1 1

2 2I x

π= − = − − − =

7.2 Properties of the definite integral

Let f and g be functions that are continuous on a finite interval [a,b] .

1. ( ) ( )a b

b a

f x dx f x dx= −∫ ∫

2. ( ) ( )b b

a a

k f x dx k f x dx=∫ ∫ with k a real constant.

3. ( )( ) ( ) ( ) ( )b b b

a a a

f x g x dx f x dx g x dx+ = +∫ ∫ ∫

Calculus 31

7.3 The substitution method

Given the integral ( )b

a

f x dx∫ and f is continuous on [a,b].

In this method we apply the substitution rule to the integrand and convert the limits as well.

Example

2

0

sin 2I xdx

π

= ∫

let 2x = t then 2dx =dt

0 0

2

x t

x tπ π

= ⇒ = = ⇒ =

Therefore ( )00

sin 1 1cos cos cos0 1

2 2 2

t dtI t

ππ

π = = − = − − = ∫

7.4 Partial integration

If u = u(x) and v = v(x) are differentiable functions of x and let udv uv vdu= −∫ ∫ then

[ ]bb b

b

aa aa

u dv uv vdu u v v du = − = − ∫ ∫ ∫

Example

2

1

1I x x dx= −∫

Set ( )32

1 1 13

u x du dx

dv x dx v x dx x

= ⇒ = = − ⇒ = − = − ∫

⇒ ( ) ( ) ( ) ( ) ( )22 2

3 3 5

1 1 1

2 2 2 2 2 4 4 161 1 2 0 1 1 0

3 3 3 3 5 3 15 15I x x x dx x

= − − − = − − ⋅ − = − − = ∫

Calculus 32

7.5 Exercices

Evaluate the following definite integrals:

1. 3

1

xdx∫ 4

2. ( )2

2

2

3 2 4x x dx−

− +∫ 32

3. 0

sinxdxπ

∫ 2

4. 2

21

1dx

x∫

12

5. 2

2cos xdxπ

π∫

2π

6. 1

2

0

1 x dx−∫ 4π

7. 1

5

5x x dx−

−

+∫ 20815

−

8. 2

22 5

xdx

x− +∫ 0

9. 2 3

0

cos sinx xdxπ

∫ 4

15

10. 2

3

0

cos sinx xdx

π

∫ 14

Trigonometry


1 Angles .................................................................................................................................... 4

1.1 The trigonometric circle .............................................................................................................. 4

1.2 Oriented angles ........................................................................................................................... 4

1.3 Conversion between radians and degrees ................................................................................... 6

2 The trigonometric numbers ................................................................................................. 7

2.1 Definitons .................................................................................................................................... 7

2.2 Some special angles and their trigonometric numbers ................................................................ 8

2.3 Sign variation for the trigonometric numbers by quadrant ......................................................... 9

2.4 Pythagorean identities ................................................................................................................. 9

2.5 Examples ................................................................................................................................... 10

2.6 Special pairs of angles .............................................................................................................. 12

2.7 Exercises ................................................................................................................................... 14

3 The trigonometric functions .............................................................................................. 16

3.1 Periodic functions ..................................................................................................................... 16

3.2 Even and odd functions ............................................................................................................. 16

3.3 Sine function ............................................................................................................................. 17

3.4 Cosine function ......................................................................................................................... 17

3.5 Tangent function ....................................................................................................................... 18

3.6 Cotangent function .................................................................................................................... 18

3.7 The secantfunction .................................................................................................................... 19

3.8 The cosecant function ............................................................................................................... 19

3.9 Exercises ................................................................................................................................... 20

4 Right triangles .................................................................................................................... 21

4.1 Formulas ................................................................................................................................... 21

4.2 Exercises ................................................................................................................................... 23

5 Oblique triangles ................................................................................................................ 25

5.1 The sine rules ............................................................................................................................ 25

5.2 The cosine rules ........................................................................................................................ 26

5.3 Solving oblique triangles .......................................................................................................... 27

5.4 Exercises ................................................................................................................................... 28

6 Extra’s ................................................................................................................................. 29

6.1 Special lines in a triangle .......................................................................................................... 29

6.2 Isosceles triangles ..................................................................................................................... 31

6.3 Equilateral triangles .................................................................................................................. 32

6.4 Exterior angles .......................................................................................................................... 32

7 Trigonometric formulas ..................................................................................................... 33

7.1 Sum and difference formulas .................................................................................................... 33

7.2 Double-angle formulas .............................................................................................................. 34

7.3 Half-angle formulas .................................................................................................................. 35

7.4 Trigonometric numbers in terms of tan α/2 .............................................................................. 35

7.5 Conversions sum/difference of angles into product of angles and vice versa .......................... 36

7.6 Exercises ................................................................................................................................... 37

Trigonometry 4

1 Angles

1.1 The trigonometric circle

Take an x-axis and an y-axis (orthonormal) and let O be the origin. A circle centered in O and with radius = 1, is called a trigonometric circle or unit circle. Turning counterclockwise is the positive orientation in trigonometry (fig. 1).

1.2 Oriented angles

An angle is the figure formed by two rays that have the same beginning point. That point is called the vertex and the two rays are called the sides of the angle (also legs). If we call [OA the initial side of the angle and [OB the terminal side, then we have an oriented angle. This angle is referred to as ∠AOB and the orientation is indicated by an arrow from the initial side to the terminal side. We can draw the arrow also in the opposite direction, still starting from the initial side of the angle [OA. Both angles represent the same oriented angle. The angle ∠BOA is a different oriented angle which we call the opposite angle of ∠AOB (fig. 2).

Remark: an oriented angle is in fact the set of all angles which can be transformed to each other by a rotation and/or a translation.

+

x

y

Z

α

∠AOB

A

B

∠BOA

A

B

OO

fig. 1 : the trigonometric circle fig. 2 : angles ∠AOB and ∠BOA

The introduction of the trigonometric circle makes it possible to attach a value to each oriented angle ∠AOB, which we will call α from now on. Represent the oriented angle in the trigonometric circle and let the initial side of this angle coincide with the x-axis (see fig. 1). Then the terminal side intersects the trigonometric circle in point Z. Then Z is the representation of the oriented angle α on the trigonometric circle.

Trigonometry 5

fig. 3 : the four quadrants

If Z ∈ I: angle α belongs to the first quadrant.

If Z ∈ II: angle α belongs to the second quadrant.

If Z ∈ III: angle α belongs to the third quadrant.

If Z ∈ IV: angle α belongs to the fourth quadrant.

There are two commonly used units of measurement for angles. The more familiar unit of measurement is that of degrees. A circle is divided into 360 equal degrees, so that a right angle is 90°. Each degree is subdivided into 60 minutes and each minute into 60 seconds. The symbols °, ' and " are used for degrees, arcminutes and arcseconds.

In most mathematical work beyond practical geometry, angles are typically measured in radians rather than degrees.

An angle of 1 radian determines on the circle an arc with length the radius of the circle. Because the length of a full circle is 2πR, a circle contains 2π radians. Contrariwise, if one draws in the centre of a circle with radius R an angle of θ radians, then this angle determines an arc on the circle with length θ·R. Subdivisions of radians are written in decimal form.

Next to Z you can put an infinite number of values which differ from each other by an integer multiple of 360° or 2π, because you can make more turns in one or the other direction starting at the initial side of the angle and arriving at the terminal side of the angle (these angles are called coterminal). The set of all these values is called the measure of the oriented angle α. The principal value of α is that value which belongs to ]- 180°, 180°], resp. ]- π,π].

Trigonometry 6

1.3 Conversion between radians and degrees

Because 2π = 360° ,following conversion formulas can be applied:

°

360 rr rad

2

2g rad

360

g

π

π

° →

→

i

i

Remark: when an angle is represented in radians, one does only mention the value, not the term ‘rad’.

Trigonometry 7

2 The trigonometric numbers

2.1 Definitons

Consider the construction of the oriented angle α as described in the previous paragraph. The terminal side of the angle α intersects the unit circle in the point Z. The sinα can be defined as the y-coordinate of this point. The cosα can be defined as the x-coordinate of this point. In this way, we can find the sine or cosine of any real value of α (α ∈ ΙR). Conversely, the choice of a cosine-value and a sine-value define in the interval [0,2π[ only one angle. Overall there are an infinite number of solutions, which one can find by adding on multiples of 2π.

x

y

Z

α1

cosα

sinα

fig. 4 : Sine and cosine in the trigonometric circle

Beside sine and cosine other trigonometric numbers are defined as follows :

tangent : sin

tancos

ααα

= cotangent : cos

cotsin

ααα

=

secant : 1

seccos

αα

= cosecant : 1

cscsin

αα

=

Fig. 5 gives a graphical representation of the above trigonometric numbers in terms of distances associated with the unit circle.

Trigonometry 8

1

1

cosα

α

sinα

tanαcotα

secα

cscα

S1

S2

fig. 5 : the graphical representation of the trigonometric numbers in terms of distances associated with the unit circle

Consequently, the trigonometric numbers have values which are in the following areas:

sin α [ -1,1 ] cos α [ -1,1 ]

tan α ] -∞ , +∞ [ cot α ] -∞ , +∞ [

sec α ] -∞ , −1] ∪ [ 1, +∞ [ csc α ] -∞ ,−1] ∪ [ 1, +∞ [

2.2 Some special angles and their trigonometric numbers

α 0 30° = π/6 45° = π/4 60° = π/3 90° = π/2

sin α 0 1/2 2 /2 3 /2 1

cos α 1 3 /2 2 /2 1/2 0

tan α 0 1/ 3 1 3 ∞

cot α ∞ 3 1 1/ 3 0

sec α 1 2/ 3 2 2 ∞

csc α ∞ 2 2 2 /3 1

Trigonometric numbers of angles in the other quadrants we shell find through the use of the reference angle (see paragraph 2.6.2.)

∈ ∈

∈ ∈

∈ ∈

Trigonometry 9

2.3 Sign variation for the trigonometric numbers by quadrant

Inside a quadrant the trigonometric numbers keep the same sign (fig. 6).

+

++

--

+

+

+

-

-

+

+

+-

-

sine cosecant

cosine secant

tangent cotangent

fig.6 : sign variation for the trigonometric numbers by quadrant

2.4 Pythagorean identities

The basic relationship between the sine and the cosine is the Pythagorean or fundamental

trigonometric identity: 2 2cos sin 1α α+ =

This can be viewed as a version of the Pythagorean theorem, and follows from the equation x2 + y2 = 1 for the unit circle (see fig. 7):

2 2 2OP PZ OZ+ = with cos ; sin ; 1OP PZ OZα α= = =

Dividing the Pythagorean identity by either cos2 θ or sin2 θ yields the following identities: 2 2

2 2

1 tan sec

1 cot csc

α αα α

+ =

+ =

P

Q Z

α

Ο x

y

fig. 7 : the triangle OPZ

Trigonometry 10

2.5 Examples

2.5.1 Calculation of the trigonometric numbers

Given: sin 5 13α =

Asked: all other trigonometric numbers

Because the sine of this angle is positive, the angle is situated in the first or second quadrant. We determine the other trigonometric numbers as follows:

• from the Pythagorean trigonometric identity:

2 2cos 1 sin 1 25 169 144 169α α= − = − =

we get: cos 144 169 12 13α = ± = ±

• tan sin cos 5 12α α α= = ±

• cot 1 tan 12 5α α= = ±

• sec 1 cos 13 12α α= = ±

• csc 1 sin 13 5α α= =

The two possible solutions for some of the trigonometric numbers correspond with the values of these numbers according to the quadrant in which the angle is situated.

Summary :

quadrant sin cos tan cot sec csc

1st 5/13 12/13 5/12 12/5 13/12 13/5

2nd 5/13 -12/13 -5/12 -12/5 -13/12 13/5

Trigonometry 11

2.5.2 Proof the following identity

2 2 2 2sec csc sec cscα α α α+ =

Proof : 2 2

2 2

2 2

2 2

2 2

2 2

1 1sec csc

cos sin

sin cos =

cos sin

1 =

cos sin

= sec csc

α αα α

α αα α

α αα α

+ = +

+

Trigonometry 12

2.6 Special pairs of angles

The sines, cosines and tangents, cotangents of some angles are equal to the sines, cosines and tangents, cotangents of other angles.

2.6.1 Formulas

a. Supplementary angles ( = sum is π )

sin(π − α) = sin α cos(π − α) = - cos α

tan(π − α) = - tan α cot(π − α) = - cot α

b. Anti-supplementary angles ( = difference is π )

sin(π + α) = - sin α cos(π + α) = - cos α

tan(π + α) = tan α cot(π + α) = cot α

c. Opposite angles ( = sum is 2π )

sin(2π − α) = - sin α cos(2π − α) = cos α

tan(2π − α) = - tan α cot(2π − α) = - cot α

d. Complementary angles ( = sum is π / 2 )

sin(π/2 − α) = cos α cos(π/2 − α) = sin α

tan(π/2 − α) = cot α cot(π/2 − α) = tan α

α

y

x

-απ+α

π-α

fig. 8 : special pairs of angles

Trigonometry 13

2.6.2 Reference angles

The use of reference angles is a way to simplify the calculation of the trigonometric numbers at various angles.

Associated with every angle drawn in standard position (which means that its vertex is located at the origin and the initial side is on the positive x-axis) (except angles of which the terminal side lies “on” the axes, called quadrantal angles) there is an angle called the reference angle. The reference angle is the acute angle formed by the terminal side of the given angle and the x-axis. Angles in quadrant I are their own reference angles. For angles in other quadrants, reference angles are calculated this way:

Quadrant β (reference angle)

I β = α

II β = π – α

III β = α – π

IV β = - α

The reference angle and the given angle form a pair of angles to which you can apply the properties in the previous paragraph. Due to these properties, the value of a trigonometric number at a given angle is always the same as the value of that angle’s reference angle, except when there is a variation in sign. Because we know the signs of the numbers in different quadrants, we can simplify the calculation of a trigonometric number at any angle to the value of the number at the reference angle for that angle, to be found in the table in paragraph 2.2.

2.6.3 How to find all angles

To find the angle if given a certain trigonometric number, usually there are 2 solutions. Calculators give the most obvious solution, but in practical situations, there can be a second solution, or the second solution can be the only correct solution. In this case the user must adjust the solution given by the calculator.

The following table gives for positive and negative trigonometric numbers the quadrant in which the solution given by the calculator, is situated, and in the last column the quadrant of the second solution:

Input Calculator Second solution ---------------------------------------------------------------------------------------------------------

Positive sine or cosecant 1 2 Negative sine or cosecant 4 3 Positive cosine or secant 1 4 Negative cosine or secant 2 3 Positive tangent or cotangent 1 3 Negative tangent or cotangent 4 2

Trigonometry 14

2.7 Exercises

2.7.1 Determine for the given trigonometric numbers the other trigonometric numbers; do not determine the angle before.

1. sin 6 6α = − 2. csc 4 3α =

3. cot 13 6α = − 4. sec 25 24α =

2.7.2 Proof the following identities

1. 2 2 2 2csc cot csc cotα β β α+ = +

2. ( ) ( )( ) ( )

2 21 sin 1 sincos cot

sec 1 sec 1

α αα α

α α− +

=+ −

3. ( )2sec tan 1 sinsec tan

sec tan 1 sin

α α αα αα α α

+ += + =− −

4. ( )( ) ( )32 1 tan

1 cot sec 2 tantan

αα α α

α+

+ + =

2.7.3 Simplify the following expressions by applying the formulas of pairs of angles.

1. ( )

( )( ) ( )cos cos

sin cos23

sin sin 2 sin cos2 2 2

x xx x

x x x x

π π π ππ π ππ

+ − − + + − − + +

2. csc(2 )sec( ) sec(2 )csc( )

3 3csc sec sec csc

2 2 2 2

x x x x

x x x x

π π π ππ π π π

+ − − −− − + + −

Trigonometry 15

2.7.4 Determine the following trigonometric numbers. First find the reference angle, then apply the properties of special pairs of angles.

1. sin 120° 2. cos ( -135° )

3. tan 225° 4. 3cot

4

π −

5. 11tan

3

π

2.7.5 Solve in IR. Express the solution(s) in radians.

1. cos 5x = 3

2−

2. sin 5x = 3

2−

3. sin 2x = 3 sin x

4. sin x = 1

5 and x ∈ ,

2

π π

;

asked: sin 2x

5. 22 sin x = 3 cos x

6. tan ( 3x + 2 ) = 3

Trigonometry 16

3 The trigonometric functions

3.1 Periodic functions

Definiton : a function f : IR → IR is periodic

⇔

∃ p IRo : ∀ x dom f : x + p dom f ∧ f (x + p ) = f(x)

If p satisfies this definition, then all positive and negative numbers which are an integer multiple of p also satisfy this definition. Therefore we call the smallest positive number which satisfies this definition the period P of the function. Graphically this periodicity means that the form of the graph of f(x) repeats itself over subsequent intervals with length P.

3.2 Even and odd functions

A function f is called EVEN if:

∀ x dom f : - x dom f ∧ f (- x) = f(x)

Consequently two points with opposite x-values must have the same y-value. So the graph must be symmetric about the y-axis.

A function f is called ODD if:

∀ x dom f : - x dom f ∧ f (- x) = - f(x)

Consequently two points with opposite x-values must have opposite y-values. So the graph is symmetric about the origin.

Remark: we consider the argument of trigonometric functions always in terms of radians.

∈ ∈ ∈

∈ ∈

∈ ∈

Trigonometry 17

3.3 Sine function

sin : IR → [ -1,1 ] : x → sin x

The period of this function is 2π. This function is odd, as opposite angles have opposite sines.

fig. 9 : the sinusoïde

3.4 Cosine function

cos : IR → [ -1,1 ] : x → cos x

The period of this function is 2π. This function is even, as opposite angles have the same cosine.

fig. 10 : the cosinusoïde

-π π

-π π

Trigonometry 18

3.5 Tangent function

tan : IR \ π

kπ , k 2

+ ∈

ℤ → IR : x → tan x

The period of this function is π. This function is odd, as opposite angles have opposite tangents.

fig. 11 : the tangent function

3.6 Cotangent function

cot : IR \ { }kπ , k ∈ ℤ → IR : x → cot x

The period of this function is π. This function is odd, as opposite angles have also opposite cotangents.

fig. 12 : the cotangent function

π− 2

π− 2

π π

π− 2

π π

Trigonometry 19

3.7 The secant function

sec : IR \ π

kπ , k 2

+ ∈

ℤ → ] −∞, −1] ∪ [1,+∞[ : x → sec x

The period of this function is 2π. This function is even, as opposite angles have the same cosines and so the same secants.

fig. 13 : de secansfunctie

3.8 The cosecant function

csc : IR \ { }kπ , k ∈ ℤ → ] −∞,−1] ∪ [1,+∞[ : x → csc x

The period of this function is 2π. This function is odd, as opposite angles have opposite sines and so opposite cosecants.

π− 2

π− 2

π π

Trigonometry 20

fig. 14 : the cosecant function

3.9 Exercises

3.9.1 Determine the period of the following functions and draw their graph

1. f(x) = sin 2x

2. f(x) cos3

x =

3. f(x) cos2

xπ = +

π− 2

π− 2

π π

Trigonometry 21

4 Right triangles

4.1 Formulas

a b

c

a b

c

α β

γ

α β

γ

B

C

A

C

B A

fig. 15 : orthogonal triangles used to set up the formulas in this paragraph

In a right triangle with α as the right angle, the following formulas apply:

α = 2

π β + γ =

2

π a2 = b2 + c2

If we draw in the triangle above a circle segment with centre in B and radius a (see the first triangle in fig. 15), then we recognize a segment of a circle with radius a. The adjacent side of the right angle c and the opposite side b have resp. the following lengths:

c = a cos β and b = a sin β

In a similar way, by considering a circle segment with centre in C and radius a (see second triangle in fig. 15), we find:

b = a cos γ and c = a sin γ

In words :

The cosine of an acute angle is the ratio of the length of the adjacent rectangle side and the length of the hypotenuse.

The sine of an acute angle is the ratio of the length of the opposite rectangle side and the length of the hypothenuse.

Trigonometry 22

By division of the first two formulas we get:

b = c tan β or c = b cot β

If we do the same with the last two formulas, we get:

c = b tan γ or b = c cot γ

In words :

The tangent of an acute angle is the ratio of the length of the opposite rectangle side and the length of the adjacent rectangle side.

The cotangent of an acute angle is the ratio of the length of the adjacent rectangle side and the length of the opposite rectangle side.

Example

Given : α = 90° ; β = 13° ; b = 10

Asked : all missing angles and sides.

Solution:

γ = 90° - β = 90° - 13° = 77°

a = 10

sin sin13

b

β=

°= 44.5

c = 2 2a b− = 43.5

Trigonometry 23

4.2 Exercises

1. Given : ∆ABC with a = 45, α = 90° , β = 40°10'35"

Asked : the remaining sides and angles.

2. An extension ladder stands slantwise to a vertical wall on a horizontal floor. If the ladder is completely extended, it makes an angle of 53°18' with the floor; completely retracted the angle is 29°10', while the top at that moment leans at a heigth of 5 meter against the wall. If we assume that the foot of the ladder does not change, calculate

• the maximal height one can reach • the maximal length of the ladder

3. An incident ray is a ray of light that strikes a surface. The angle between this ray and the perpendicular or normal to the surface is the angle of incidence (α). The refracted ray or transmitted ray corresponding to a given incident ray represents the light that is transmitted through the surface. The angle between this ray and the normal is known as the angle of refraction (β). The relationship between the angles of incidence and refraction is given by Snell's Law: sin

sin

αβ

= n

Example (see figure below): A ray of light strikes an air-water interface at an angle of 30 degrees from the normal (α = 30°). The relative refractive index for the interface is 4/3. At which distance from P the ray of light hits the bottom if the water is 1 m deep. Remark: solve this exercise without calculating the angle β.

fig. 16 : illustration for exercise 3

Trigonometry 24

In mechanics you will deal with exercises in which forces must be calculated. In the following exercises such situations will be sketched. We will confine to the calculation of angles between bars..

4. Calculate: • the angle between FE and the horizontal plane • the angle between FC and the vertical plane


5. Calculate the angle between CD and DF


6. Calculate the angle between BC and CD


Trigonometry 25

5 Oblique triangles

First, remember that also for oblique triangles the sum of angles is 180°.

An oblique triangle is any triangle that is not a right triangle. It could be an acute triangle (all three angles of the triangle are less than right angles) or it could be an obtuse triangle (one of the three angles is greater than a right angle).

Using the formulas for right triangles, we can set up formulas for oblique triangles. Let us consider an oblique triangle ∆ABC with sides a, b and c and angles α, β and γ.

5.1 The sine rules

The altitude from A to the opposite side a intersects this side in point S. In this way the triangle is divided in two right angles with one common side AS, with length d. Use now the formulas of a right triangle in triangles ∆ABS and ∆ACS to calculate d.

a

B C

d

a1 a2

α

β γ

A

C B

c b

γ β S

fig. 20 : oblique triangle

d = c sin β and d = b sin γ

So we get: sin sin

b c

β γ=

Apply the same reasoning with the altitude from B to the opposite side b to divide the triangle in two right triangles and derive similar formulas in which occur a and the opposite angle α. Then we get:

SINE RULES : sin sin sin

a b c

α β γ= =

Trigonometry 26

5.2 The cosine rules

This identity can be derived in different ways. In fig. 20 S divides the side a in two parts with length a1 and a2. Then a1 and d can be written respectively as

a1 = b cos γ

d = b sin γ

In the triangle ∆ABS apply Pythagoras theorem:

2 2 2 2 22 1

2 2 2 21 1

2 2 2 2 2

2 2

c = d + a = d + (a-a )

= b sin γ + a + a - 2 a a

= b sin γ + a + b cosγ - 2 a b cos γ

= b + a - 2 a b cos γ

The same expression can be derived if S lies outside side a.

Then, similar expressions can be derived for the other angles.

Summarized, in this way we get:

COSINE RULES : a2 = b2 + c2 - 2 b c cos α

b2 = a2 + c2 - 2 a c cos β

c2 = a2 + b2 - 2 a b cos γ

These statements relate the lengths of the sides of a triangle to the cosine of one of its angles. For example, the first statement states the relationship between the sides of lengths a , b and c, where α denotes the angle contained between sides of lengths b and c and opposite to the side with length a.

These rules look like the Pythagorean theorem except for the last term, and if you deal with a right triangle, that last term disappears, so these rules are actually a generalization of the Pythagorean theorem.

Trigonometry 27

5.3 Solving oblique triangles

One of the most common applications of the trigonometry is solving triangles – finding missing sides and/or angles, given some information about a triangle. The process of solving triangles can be broken down into a number of cases. In these situations we will use 3 sorts of formulas, applicable in all triangles: • the sum of all angles is 180° • the sine rule : relates two sides to their opposite angles • the cosine rule : relates the three sides of the triangle to one of the angles. Naturally, the given information must be such that the given elements allow a triangle: • the sum of the given angles can not be larger than 180°, • and the sides must meet the triangle inequality which states that for any triangle, the sum of

the lengths of any two sides must be greater than the length of the remaining side. a. If you know one angle and the two adjacent sides.

Then, there is 1 solution: you can determine the opposite side by using the cosine rule, another angle by using the sine rule and the remaining angle as 180° minus the two already determined angles.

Attention: the sine rule gives two solutions for the second angle (supplementary angles). Test the solutions by verifying the properties of a triangle (see exercises).

b. If you know one side and the two adjacent angles.

Then there is 1 solution: the third angle is immediately known as 180° minus the two given angles; the two remaining sides can be determined by using the sine rule.

c. If you know all three sides of a triangle.

Then there is 1 solution: determine one angle by using a cosine rule, the second angle can be determined by using another cosine rule or by using the sine rule. The last angle can be determined by the property of triangles that the sum of all angles must be 180°.

d. If you know sides a and b and β (one of the adjacent angles). In this case, there can be 0, 1 or 2 solutions.

Determine the angle α by using the sine rule. You will get 0 (if sin α > 1) or 2 solutions (supplementary angles have the same sine). For each solution determine the missing angle γ, and then the length of side c by using the sine rule. Finally you test if each solution which you find is acceptable: you can not have negative angles or sides (see exercises).

Trigonometry 28

5.4 Exercises

1. A tower is seen from the ground under an angle of 21°. If one approaches the tower by 24 meter, then this angle becomes 35°. Determine the height of the tower.

2. Two planes depart from the same point, each in a different direction. The directions form an angle of 32°. The velocity of the first plane is 600 km/hour, the velocity of the second is 900 km/hour. Determine their mutual distance after one hour and a half.

3. The pole of a flag reaches up from a facade with an angle of 45° (see fig. 21). Five meter above the base point of the pole in the wall, there is a cable fixed to the wall with a length of 3,60 meter. At which distance of the base point, measured along the pole, the other end of the cable can be fixed.


4. Solve the previous exercise for a cable with a length of 2 m, respectively with a cable with a length of 8 m.

5. Three observers are at mutual distances of 2, 3 and 4 meters. Determine for each observer the angle under which he sees the other 2 observers.

6. A boat sails north and sees a lighthouse 40° eastwards. After having sailed 20 km, this angle has increased to 80°. Determine at both positions the distance from the boat to the lighthouse.

7. The following figure demonstrates a situation in mechanics. Determine the angle between the ropes AC and AD (d = 1 m).

fig. 22 : illustration of exercise 7

Trigonometry 29

6 Extra’s

6.1 Special lines in a triangle

6.1.1 Altitude

An altitude of a triangle is a straight line through a vertex and perpendicular to (i.e. forming a right angle with) the opposite side.

This opposite side is called the base of the altitude, and the point where the altitude intersects the base (or its extension) is called the foot of the altitude. The three altitudes intersect in a single point, called the orthocenter of the triangle. The orthocenter lies inside the triangle if and only if the triangle is acute.

H

fig. 23 : altitudes

6.1.2 Median

A median of a triangle is a straight line through a vertex and the midpoint of the opposite side, and divides the triangle into two equal areas.

The three medians intersect in a single point, the triangle's centroid. The centroid cuts every median in the ratio 2:1, i.e. the distance between a vertex and the centroid is twice the distance between the centroid and the midpoint of the opposite side.

Z

fig. 24 : medians

Trigonometry 30

6.1.3 Other lines

An angle bisector of a triangle is a straight line through a vertex which cuts the corresponding angle in half. The three angle bisectors intersect in a single point, which always lies inside the triangle.

A perpendicular bisector of a side of a triangle is a straight line passing through the midpoint of the side and being perpendicular to it, i.e. forming a right angle with it. The three perpendicular bisectors meet in a single point, the triangle's circumcenter; this point may also lie outside the triangle.

Trigonometry 31

6.2 Isosceles triangles

β

α

β

h bb

a

H

fig. 25 : isosceles triangle

In an isosceles triangle, two sides are equal in length. The unequal side is called its base and the angle opposite the base is called the "vertex angle". The equal sides are called the legs of the triangle. The base angles of an isosceles triangle are always equal.

Property : the altitude and the median from the vertical angle coincide.

Let’s call the altitude h, the legs b, the base a, the vertical angle α and the base-angle β:

then : h = b sin β and 2

a= b cos β

6.2.1 Exercises

1. Set up analog formulas which use the vertex angle.

2. Determine the value of the angles of an isosceles triangle with base 8 and legs 14.

3. Determine the length of the legs of an isosceles triangle with vertex angle 42° and base 12.

4. Determine the length of each side of an isosceles triangle with vertex angle 36° and vertex-altitude 28.

5. In an isosceles triangle with vertex angle 24° the orthocenter lies at a distance of 26 cm to the top. Determine all angles and sides.

6.3 Equilateral triangles

In an equilateral triangle all sides have the same length. Therefore all three angles are equal to each other, and thus 60°.

Property : the altitude from a certain angle coincides with the median from that angle. orthocenter and the centroid coincide.

6.3.1 Exercises

1. Determine the distance from the orthocenter/centroid to one of the vertices in terms of the length of the side.

2. Determine the length of an altitude in an equilateral triangle with side 28 cm.

3. The altitude of an equilateral triangle has length 8 cm.

6.4 Exterior angles

The exterior angle of an angle in a triangle is formed by extended from the other side adjacent to that angle. adjacent interior angle ACB are

The sum of the interior angle and the external angle on the same vertex is 180°. Therefore sum of all exterior angles is 360° or 2

Trigonometry

riangles

aa

a

H=Z

60°

60° 60°

fig. 26 : equilateral triangle

all sides have the same length. Therefore all three angles are equal to

the altitude from a certain angle coincides with the median from that angle. orthocenter and the centroid coincide.

Determine the distance from the orthocenter/centroid to one of the vertices in terms of the

he length of an altitude in an equilateral triangle with side 28 cm.

The altitude of an equilateral triangle has length 8 cm. Determine the length of the sides.

exterior angle of an angle in a triangle is formed by one side adjacent to that angle and a line extended from the other side adjacent to that angle. Clearly, the exterior angle

are supplementary. That is:

angle and the external angle on the same vertex is 180°. Therefore sum of all exterior angles is 360° or 2π.

32

all sides have the same length. Therefore all three angles are equal to

the altitude from a certain angle coincides with the median from that angle. The

Determine the distance from the orthocenter/centroid to one of the vertices in terms of the

he length of an altitude in an equilateral triangle with side 28 cm.

Determine the length of the sides.

one side adjacent to that angle and a line Clearly, the exterior angle ACD and the

angle and the external angle on the same vertex is 180°. Therefore the

Trigonometry 33

7 Tr igonometric formulas

In this paragraph, we discuss formulas involving the trigonometric numbers of a sum or difference of two angles, of a double or half angle, conversions between sums and products of sines and cosines….

As we don’t want you to learn these formulas by heart, it is important to understand their mutual connection, the way how one formula can be derived from another formula.

We also want to emphasize that the knowledge of these formulas facilitates solving integrals of trigonometric functions.

7.1 Sum and difference formulas

Let’s start with the addition formula for the sine. Then the other formulas can be derived in an easy way.

sin(α + β) = sin α cos β + cos α sin β (1) Replace β by -β , with sin(-β) = - sin β, cos(-β) = cos β, then we get:

sin(α - β) = sin α cos β - cos α sin β (2)

For the similar cosine formulas:

cos(α + β) = sin[2

π- (α + β)]

= sin [(2

π - α ) - β]

= sin(2

π - α) cos β - cos(

2

π - α) sin β

or: cos(α+ β) = cos α cos β - sin α sin β (3)

Again, replace β by -β, then we get:

cos(α - β) = cos α cos β + sin α sin β (4)

You see that the sine formulas keep the plus- or minus sign, but mix the trigonometric functions. The cosine formulas change the sign but hold the trigonometric functions together.

Trigonometry 34

Let’s divide (1) side by side by (3), and then divide the nominator and the denominator in the right hand side by (cos α cos β) :

tan tantan( )

1 tan tan

α βα βα β++ =

− (5)

And replace β by -β, then we get:

tan tantan( )

1 tan tan

α βα βα β−− =

+ (6)

7.2 Double-angle formulas

Substitute α = β in the previous sum formulas, then we find the double-angle formulas: :

cos 2α = cos2 α - sin2 α (7) sin 2α = 2 sin α cos α (8)

2

2 tantan 2

1 tan

ααα

=−

(9)

Two useful forms of (7) are derived by replacing cos2 α by 1 - sin2 α, resp. sin2 α by 1 - cos2 α :

cos 2α = 1 - 2 sin2 α (10) cos 2α = 2 cos2 α - 1 (11) And so:

sin2 α = 12 ( 1 - cos 2α ) (12)

cos2 α = 12 ( 1 + cos 2α ) (13)

Trigonometry 35

7.3 Half-angle formulas

Replace 2α by α in (10) and (11):

cos α = 1 - 2 sin2 α2 (14)

cos α = 2 cos2 α2 - 1 (15)

7.4 Tr igonometric numbers in terms of tan αααα/2

In (8) we divide and multiply the right hand side by sec 2 α . By replacing in the denominator

sec 2 α = 1 + tan 2 α (see $ 2.4.) and by simplifying the nominator (apply the definition of sec α), we get:

2

2 tansin 2

1 tan

ααα

=+

(16)

Replace α by 2

α:

2

2 tan2sin

1 tan2

α

α α=+

(17)

By performing the same operation on (7) we find :

2

2

1 tan2cos

1 tan2

α

α α

−=

+ (18)

and by replacing α by2

α in (9) , we get:

2

2 tan2tan

1 tan2

α

α α=−

(19)

Trigonometry 36

7.5 Conversions sum/difference of angles into product of angles and vice versa

In the right hand side of the sum formulas (1) and (2) we notice the same term (sin α cos β). By adding (1) and (2) side by side and bringing factor 2 to the other side, we get :

sin α cos β = 12 [ sin(α + β) + sin(α - β) ] (20)

In a similar way, by subtracting (2) from (1) side by side, we get :

cos α sin β = 12 [ sin(α + β) - sin(α - β) ] (21)

By doing the same with formulas (3) and (4) (adding, resp. subtracting side by side), we get:

cos α cos β = 12 [ cos(α + β) + cos(α - β) ] (22)

sin α sin β = 12 [ cos(α - β) - cos(α + β) ] (23)

These four formulas convert the product of two cosines and/or sines with a different argument into a sum. The reverse formulas we get by bringing factor ½ to the other side and by substitution:

α = p + q

2

β = p - q

2

This leads us to the formulas of Simpson:

sin p + sin q = 2 sin p + q

2 cos p - q

2 (24)

sin p - sin q = 2 cos p + q

2 sin p - q

2 (25)

cos p + cos q = 2 cos p + q

2 cos p - q

2 (26)

cos p - cos q = - 2 sin p + q

2 sin p - q

2 (27)

Trigonometry 37

7.6 Exercises

1. Proof the following identity in a triangle:

2 2 2sin sin sin 2sin sin cosα β γ α β γ+ − =

2. Calculate and/or simplify:

a. tan cot4 4

π πα α − + +

0

b. sin cos

sin cos

α αα α

−+

tan4

πα −

3. Write in terms of powers of sin α and/or cos α:

a. sin3α 33sin 4sinα α−

b. cos 4α 2 41 8cos 8cosα α− +

c. tan2

α

sin

cos 1

αα +

d. sin cos

2 2

cos sin2 2

α α

α α

+

−

1 sin

cos

αα

+

4. Factorize:

a. sin 3 sinα α− 2cos2 sinα α

b. cos4 cos5 cos6α α α+ + cos5 (2cos 1)α α +

c. tan sinα α− 22tan sin2

αα

d. 2 2cos cosβ α− sin( )sin( )α β α β+ −

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