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TEACH YOURSELF MATHS
LEAVING CERTIFICATE HONOURS
77 LESSONS
WEB SOLUTIONS FOR PAPERS 2001 TO 2006EVERY FORMULA INCLUDED
EASY TO FOLLOW STEPS
FULL LIST OF PROOFS
STUDY PLAN PROVIDED
Tony KellyKieran Mills
EVERYTHING YOU NEED TO KNOW
THE EASY WAY TO LEARN MATHS
MATHS IS ABOUT DOING QUESTIONS AND GETTING THE
RIGHT ANSWER. YOU LEARN MATHS BY DOING. IT IS
LIKE PLAYING A SPORT OR A MUSICAL INSTRUMENT –THE MORE YOU PRACTISE THE BETTER YOU GET AND
THE MORE ENJOYMENT YOU GET .
© Tony Kelly & Kieran Mills
Printed by Print4less
Published by Student Xpress 2006www.studentxpress.ie
All rights reserved.No part of this publication may be reproduced, copiedor transmitted in any form or by any means withoutwritten permission of the publishers or else under theterms of any licence permitting limited copying issuedby the Irish Copyright Licensing Agency, The Writers’Centre, Parnell Square, Dublin 1.
HOW DO YOU USE THIS BOOK?
1. HAVE A LOOK AT THE LAY-OUT OF THE LEAVING CERT. PAPER
PAPER 1 (Do 6 out of 8)1. Algebra2. Algebra3. Complex Numbers & Matrices4. Sequences & Series5. Sequences & Series6. Differentiation & Applications7. Differentiation & Applications8. Integration
PAPER 2SECTION A (Do 5 out of 8)1. Circle2. Vectors3. Line4. Trigonometry5. Trigonometry6. Discrete Maths7. Discrete MathsSECTION B (Do 1 out of 4)8. Further Calculus
You can start studying any section you wish but it is advisable to start withAlgebra and Trigonometry as these areas contain the fundamentals that areused in most other sections.
2. STUDY PLAN
Once you have selected your question go to the study plan so you can keeptrack of your learning by filling in the boxes as you progress.
3. STARTING THE LESSON
Now start reading through the lesson. Have pen and paper at the ready writingdown all the formulae. Many of the formulae you need to use are contained inthe official Department of Education table book [Pages 6, 7, 9, 41 and 42]. Theformulae you need to know for the Leaving Cert. from the table book pagesare at the end of this book. There is a colour code for formulae:
...... 8
...... 8
...... 8
Formula in the table book
Similar formula in the table book
Formula you need to learn
COMPLEX NUMBERS & MATRICES (Question 3, Paper 1)
1. Complex Number Algebra2. Complex Number Equations
LESSONS START PROOFS/FORMULAE LC QUESTIONS FINISH
3. De Moivre’s Theorem4. Matrix Algebra5. Matrix Equations
10/12 (4.00) 10/12 (6.00)
11/12 (4.00) 11/12 (6.00)
12/12 (5.00) 10/12 (8.00)
Now see if you can do the Leaving Cert. questions. All the parts of the Leav-ing Cert. from 2001 to 2006 based on each lesson are listed at the end of thelesson. You can get them on the website along with their solutions.Do the questions on your own and then check the solutions. If you get a wronganswer or get stuck then start again. You have only succeeded when you cancomplete each question from start to finish on your own.Be honest with yourself. You know when you have succeeded. Feel the satis-faction that you are accomplishing your tasks. It is hard work and takes a longtime. But, at least you have a plan and you know how to go forward to suc-ceed. Tick the study plan when you are finished.
4. START DOING QUESTIONS
WEBSITE: www.studentxpress.ieThis website is a valuable resource containing lots ofmathematical material. All the Leaving Cert. papers from2001 to 2006 are on the website with all the solutions. Thebook can also be purchased online.
Use this book anytime you are doing Maths. It contains everything you need toknow to do the Leaving Cert. honours papers. So get working and enjoy.
PROOFThe proofs needed for the Leaving Cert. are inboxes as shown. Learn any proofs in the lessonand learn formulae with a red circle. Tick thestudy plan when you know them.
If you need more practice and detail you can get extra help from the two textbooks on which this revision book is based. These books are available inEason bookshops and online (www.studentxpress.ie).The Magical Bag of Mathematical Tricks, Paper 1The Magical Bag of Mathematical Tricks, Paper 2
AL
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FOR
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quat
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tegr
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est
LESS
ON
SST
AR
TL
C Q
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3. M
acla
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DIS
CR
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ATH
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1. D
iffer
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atio
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tistic
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LESS
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SST
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FIN
ISH
3. C
ount
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4. S
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5. H
ard
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4. M
axim
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FOR
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1
ALGEBRA
ALGEBRA (Q 1 & 2, PAPER 1)
LESSON No. 1
IN THIS LESSON YOU WILL REVISE THE BASICS OF ALGEBRA.
SOME BASICS
1. Difference of 2 squares[B] FACTORS: You need to remember these factors.
[A] MULTIPLYING OUT BRACKETS
....... 1a b a b a b2 2− = + −( )( )
Ex. x y x y x y x y2 2 2 281 9 9 9− = − = + −( ) ( ) ( )( )
3. Difference of 2 cubes
2. Sum of 2 cubes
a b a b a ab b3 3 2 2+ = + − +( )( ) ....... 2
Ex. x y x y x y x xy y3 3 3 3 2 28 2 2 2 4+ = + = + − +( ) ( ) ( )( )
a b a b a ab b3 3 2 2− = − + +( )( ) ....... 3
Ex. 8 27 2 3 2 3 4 6 93 3 3 3 2 2x y x y x y x xy y− = − = − + +( ) ( ) ( )( )
[C] MULTIPLICATION OF FRACTIONS
Ex. ( )( )
( )( ) ( )( )
a ba b
b ab a
a ba b
b ab a b a
++
×−−
=++
×−
+ −=
2
2 2
2
1
Multiply tops and multiply bottoms and/or cancel. Remember to factoriseeverything.
It is very useful to be able to expand squares and cubes of brackets.
( ) ( )( )x y x y x y x xy y+ = + + = + +2 2 22 [Coefficients: 1, 2, 1]
( )x y x x y xy y+ = + + +3 3 2 2 33 3 [Coefficients: 1, 3, 3, 1]
2
ALGEBRA
[E] POWERS
1. a a am n m n× = +
2. aa
am
nm n= −
3. a0 1=
POWER RULES
7. ab
ab
m
n
p mp
np
⎛
⎝⎜
⎞
⎠⎟ =
8. ab
ba
m
n
p n
m
p⎛
⎝⎜
⎞
⎠⎟ =
⎛
⎝⎜
⎞
⎠⎟
−
[F] SURDS
Ex. 23 1
23 1
3 13 1
2 3 12
3 1−
=−
×++
=+
= +( )
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LEAVING CERT. QUESTIONS
LESSON 1: SOME BASICS
2006 1 (a)2005 1 (b)2004 1 (a), 1 (b) (ii), 1 (c)2003 1 (a)2001 1 (a), 2 (b) (ii)
Ex. 41
72 3 5
41 1
72 5 12 2x x x x x x x−
−− −
=+ −
−− +( )( ) ( )( )
[D] ADDING FRACTIONS
You must use the lowest common denominator. Remember to factorise thedenominators first.
=− − −
− + −=
−− + −
4 2 5 7 12 5 1 1
132 5 1 1
( ) ( )( )( )( ) ( )( )( )
x xx x x
xx x x
4. aa
nn
− =1
5. ( )a am n mn=
6. ( )a b a bm n p mp np=
Ex. Express 3 3 315
15
15+ + in the form 3
pq where p q, .∈Z
SOLUTION
3 3 3 3 3 315
15
15
15
65+ + = × =
You are often asked to rationalise the denominator of a surd. This meansgetting rid of surds in the denominator. Multiply above and below by theconjugate of the denominator.
3
ALGEBRA
Ex. If x bx c2 + + is a factor of x p3 − show that (i) b p3 = , (ii) bc = p.SOLUTION
The long division process is useful in the type of question outlined in the example.
The remainder has to be zero ⇒ = +R x0 0
⇒ = =b c bc p2 and . Now prove what they are asking you.
If ( )x k− is a factor of f (x) then k is a root of f (x) = 0,i.e. f (k) = 0 and vice versa.
The factor theorem states that:
x b−x x x p3 20 0+ + −
∓ ∓ ∓x bx cx3 2
− − −bx cx p2
± ± ±bx b x bc2 2
( ) ( )b c x bc p2 − + −
LESSON No. 2
IN THIS LESSON YOU WILL LEARN THE DIVISION PROCESS WHICH IS USEFUL IN OBTAINING
CERTAIN RESULTS.
DIVISION
LEAVING CERT. QUESTIONS
LESSON 2: DIVISION
2006 1 (c) 2005 1 (c)2003 2 (b) (ii) 2001 1 (c)
LESSON No. 3
IN THIS LESSON YOU WILL LEARN ABOUT THE FACTOR THEOREM. THIS THEOREM PROVIDES AMETHOD FOR SOLVING ALL KINDS OF EQUATIONS.
FACTOR THEOREM
x bx c2 + +
Ex. ( )x −1 is a factor of f x x x x( ) .= + − −2 2 13 2
∴ =x 1 is a root or a solution of f x( ) = 0
∴ = + − − = + − − =f ( ) ( ) ( ) ( )1 2 1 1 2 1 1 2 1 2 1 03 2
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This type of problem can also be done by lining up coefficients. Both methodsare outlined on the website.
4
ALGEBRA
PROOF OF FACTOR THEOREM
f x ax bx cx d( ) = + + +3 2
f k ak bk ck d( ) = + + +3 2
∴ − = − + − + −f x f k a x k b x k c x k( ) ( ) ( ) ( ) ( )3 3 2 2
= − + + + + +( ){ }x k ax akx ak bx bk c2 2 = −( ) ( )x k g x
∴ = + −f x f k x k g x( ) ( ) ( ) ( )
(i) f k f x x k g x( ) ( ) ( ) ( )= ⇒ = −0 ∴ −x k is a factor.
(ii) x k− is a factor ⇒ =f k( ) .0
Can you prove the factor theorem for a quadratic equation f x ax bx c( ) ?= + +2
[B] PROPERTIES OF ROOTS α β,
These are equations of the form y f x ax bx c= = + + =( ) .2 0[A] METHODS OF SOLUTION
x b b aca
=− ± −2 4
2
You need to be able to prove the factor theorem.
LEAVING CERT. QUESTIONS
LESSON 3: FACTOR THEOREM
2006 1 (b) 2004 1 (b) (i)2003 1 (b) 2001 1 (b)
LESSON No. 4
IN THIS LESSON YOU WILL LEARN ABOUT QUADRATIC EQUATIONS, HOW TO MAKE NEW
EQUATIONS USING THE ROOTS AND HOW TO INTERPRET THE MEANING OF THE ROOTS.
QUADRATIC EQUATIONS
Sum S: α β+ = − =−b
a21
nd.
st.
Product P: αβ = =ca
31
rd.
st.
You can solve quadratic equations by1. Factorisation: Works sometimes, OR
2. Use formula 4: Always works. ....... 4
....... 5
....... 6
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5
ALGEBRA
The following results are useful whengenerating new quadratics from oldquadratics:
Forming a quadratic equation given its roots:
x x2 0− + =S P
Ex. Form a quadratic equation that has roots of 3 and −1.SOLUTION
Roots: α β= = −3 1,
Sum S: α β+ = − =3 1 2 and Product P: αβ = − = −( )( )3 1 3
Equation: x x x x2 20 2 3 0− + = ⇒ − − =S P
α β α β αβ2 2 2 2+ = + −( )
α β α β α αβ β3 3 2 2+ = + − +( )( )
α β α β α β4 4 2 2 2 2 22+ = + −( )
[C] NATURE OF THE ROOTS AND GRAPHS OF QUADRATICS
1. If b ac2 4 0− > ⇒ ≠α β[Two different real roots.]
2. b ac2 4 0− = ⇒ =α β[Two real roots that are the same.]
3. b ac2 4 0− < ⇒α β, not real.[There are no real roots (complex roots).]
X
Y
X
Y
X
Y
α β=
βα
....... 7
REMEMBER: If b ac2 4 0− ≥ ⇒ Real roots.
If b ac2 4 0− < ⇒ Unreal or complex roots.
The nature of the roots is determined by the expression under the square rootsign in formula 4. It is b ac2 4− and is called the DISCRIMINANT.
6
ALGEBRA
LEAVING CERT. QUESTIONS
LESSON 5: CUBIC EQUATIONS
2005 2 (b) 2002 1 (b)
1. Guess at a root α (it must divide exactly into the constant term).2. Form a factor from the root ( ).x −α3. Factorise fully using division process and solve.
STEPS
LEAVING CERT. QUESTIONS
LESSON 4: QUADRATIC EQUATIONS
2006 2 (b)2004 2 (b) (ii), 2 (c) (ii)2003 1 (c), 2 (c) (ii)2002 1 (a), 1 (c), 2 (c)2001 2 (c)
These are equations of the form y f x ax bx cx d= = + + + =( ) .3 2 0Method of solution: Guess at a root using the factor theorem.The basic method for solving a cubic equation is:
LESSON No. 5
IN THIS LESSON YOU WILL LEARN HOW TO SOLVE CUBIC EQUATIONS. THIS INVOLVES GUESS-ING A ROOT AND THEN USING THE FACTOR THEOREM.
CUBIC EQUATIONS
Ex. Solve 2 2 1 03 2x x x+ − − = given it has an integer root.SOLUTION
1. f ( ) ( ) ( ) ( )1 2 1 1 2 1 1 2 1 2 1 0 13 2= + − − = + − − = ⇒ is a root.
2. ( )x −1 is a factor.
3. 2 2 1 1 2 3 13 2 2x x x x x x+ − − ÷ − = + +( ) by division.Solve the quadratic by factorisation or using the formula.
∴ + − − = − + + =2 2 1 1 1 2 1 03 2x x x x x x( )( )( )
∴ = − −x 1 112, ,
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7
ALGEBRA
LEAVING CERT. QUESTIONS
LESSON 6: FUNCTIONS
2006 2 (c) 2005 2 (c)2004 2 (c) (i) 2002 2 (b) (ii)
A function simply takes in a number (object), performs an operation on it andsends out a new number (object). The operation is determined by the rule of thefunction.
x Operation f x y( ) =
Function, f
LESSON No. 6
IN THIS LESSON YOU WILL LEARN ABOUT FUNCTIONS. ON THE LEAVING CERT. PAPER
FUNCTIONS ARE USED EXTENSIVELY AND APPLIED IN ALL KINDS OF SITUATIONS.
FUNCTIONS
Ex. Let f x x kmx
( ) ,=+2 2
where k and m are constants and m ≠ 0. Show
that f km f km
( ) .= ⎛⎝⎜
⎞⎠⎟
SOLUTION
f km km km km
k mkm
k mm
( ) ( )( )
( ) ( )=
+=
+=
+2 2 2 2
2
2
2
1 1
f km
km
kk
mm
k mm
km
km
m⎛⎝⎜
⎞⎠⎟ =
+=
+× =
+( )( )
( ) ( )2 2 2 1 2
2
2
2
2 1 1
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8
ALGEBRA
[A] QUADRATICS: ax bx c2 0+ + ≤
STEPS
1. Get all terms on one side and zero on the other side.2. Solve the corresponding equation to get the roots α β, .3. Carry out the region test. Use the roots in ascending order to form
regions: ← ↔ →α β Choose a nice number in each region to testthe inequality using the test box.
4. Based on the region test write down the solutions.
LESSON No. 7
IN THIS LESSON YOU WILL LEARN ABOUT DEALING WITH THREE TYPES OF INEQUALITIES:QUADRATIC, RATIONAL AND MODULUS INEQUALITIES.
INEQUALITIES
Ex. Solve x x2 5 6> − .SOLUTION
1. x x2 5 6 0− + >
2. Solve x x x x x2 5 6 0 3 2 0 3 2− + = ⇒ − − = ⇒ =( )( ) ,
Roots: α β= =2 3,3. Region test:
0( )( )− − >3 2 0
Correct
2.5( )( )− ⋅ ⋅ >0 5 0 5 0
Wrong
4( )( )1 2 0>Correct
2 3
4. ∴ < >x x2 3,
Region Test on ( )( )x x− − >3 2 0 .....Test Box
[B] RATIONALS: P xQ x
( )( )
> 0
STEPS
1. Multiply both sides by the denominator squared unless you arecertain that it is positive.
2. Get all terms on one side and take out the highest common factor.3. Solve the corresponding equation.4. Do region test on the roots in ascending order on Test Box.5. Based on the region test write down the solutions.
9
ALGEBRA
Ex. Solve xx
x−+
< − ∈12
1, R.
SOLUTION
1. xx−+
< −12
1 ⇒ − + < − +( )( ) ( )x x x1 2 1 2 2 [Multipy by ( )x + 2 2 ]
2. ⇒ + − + + <( )[( ) ( )]x x x2 1 2 0
−3( )( )− − <1 5 0
Wrong
−1( )( )1 1 0− <
Correct
0( )( )2 1 0<
Wrong
3. Solve ( )( ) ,x x x+ + = ⇒ = − −2 2 1 0 2 12
Roots: α β= − = −2 12,
4. Region test:Region Test on ( )( )x x+ + <2 2 1 0 .....Test Box
−2 − 12
5. ∴− < < −2 12x
[C] MODULUS ax b c+ >
STEPS
1. Solve the corresponding modulus equality.2. Do region test on roots in ascending order on Test Box.3. Based on the region test write down the solutions.
Ex. Solve 2 7 4 1x x− ≤ − .SOLUTION
1. Solve 2 7 4 1 3 43x x x− = − ⇒ = − ,
2. Region test:
Region Test on 2 7 4 1x x− ≤ − .....Test Box
−4− ≤ −15 17
Correct
0
− ≤ −7 1Wrong
2
− ≤3 7Correct
−3 43
3. ∴ ≤ − ≥x x3 43,
10
ALGEBRA
STEPS
1. Eliminate one letter.2. Solve for the other.3. Substitute into either of the original equations to get
second unknown.
[A] 2 LINEARS (x, y)
LEAVING CERT. QUESTIONS
LESSON 7: INEQUALITIES
2002 2 (b) (i) [Quadratic]2004 2 (b) (i) [Rational]2005 2 (a) [Modulus]2003 2 (b) (i) [Modulus]2001 2 (b) (i) [Modulus]
LESSON No. 8
IN THIS LESSON YOU WILL LEARN HOW TO SOLVE DIFFERENT TYPES OF SIMULTANEOUS
EQUATIONS.
SIMULTANEOUS EQUATIONS
2 3 5 2x y+ = − ×( )
3 2 12 3x y− = ×( )13 26 2x x= ⇒ =
9 6 36x y− =4 6 10x y+ = − 2 2 3 5( ) + = −y
⇒ = −y 3→ →
STEPS
1. Eliminate a letter from the linear.2. Substitute into quadratic and solve for the other letter.3. Substitute these values into linear to get all solutions.
[B] LINEAR AND QUADRATIC
→ →y x= −2 5
x xy2 2+ =⇒ = −x 1
3 2,⇒ − − =3 5 2 02x x
x x x2 2 5 2+ − =( ) y = − − = −2 513
173( )
y = − = −2 2 5 1( )
SOLUTIONS: ( , ), ( , )− − −13
173 2 1
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11
ALGEBRA
LEAVING CERT. QUESTIONS
LESSON 8: SIMULTANEOUS EQUATIONS
2005 1 (a) [2 linears]2004 2 (a) [3 linears]2002 2 (a) [3 linears]2006 2 (a) [Linear and Quadratic]2003 2 (a) [Linear and Quadratic]2001 2 (a) [Linear and Quadratic]
[C] 3 LINEARS (x, y, z)
STEPS
1. Eliminate one letter using two equations.2. Eliminate the same letter using two others.3. Solve the resulting equations as for two equations in two unknowns.4. Work backwards to find all letters.
→
x y zx y zx y z
− + = −+ − =− − =
3 4 1 12 3 8 23 2 5
.......( ).........( )..........( )3
x y zx y z− + = −+ − = ×3 4 1 1
6 3 9 24 2 3.......( ).....( )
7 5 23 4x z− = .....( )
4 2 6 16 2 23 2 5 3
x y zx y z+ − = ×− − =
......( )..........( )
7 7 21 5x z− = .....( )
7 7 21 5x z− = .....( )
7 5 23 4x z− = .....( )SOLUTIONS: x y z= = =4 3 1, ,
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12
ALGEBRA
QUICK OVERVIEW OF ALGEBRA
....... 1a b a b a b2 2− = + −( )( )
a b a b a ab b3 3 2 2+ = + − +( )( ) ....... 2
a b a b a ab b3 3 2 2− = − + +( )( ) ....... 3
If ( )x k− is a factor of f (x) then k is a root of f (x) = 0,i.e. f (k) = 0 and vice versa.
x b b aca
=− ± −2 4
2
Sum S: α β+ = − =−b
a21
nd.
st.
Product P: αβ = =ca
31
rd.
st.
α β α β αβ2 2 2 2+ = + −( )
α β α β α αβ β3 3 2 2+ = + − +( )( )
α β α β α β4 4 2 2 2 2 22+ = + −( )
....... 4
....... 5
....... 6
....... 7
REMEMBER: If b ac2 4 0− ≥ ⇒ Real roots.
If b ac2 4 0− < ⇒ Unreal or complex roots.
x x2 0− + =S P
Difference of two squares
Sum of two cubes
Difference of two cubes
Factor Theorem (Proof required)
Formula for solving quadratic equations
Roots of Quadratic Equations
Forming a Quadratic Equation
130
THE TABLES
THE TABLES
You are allowed to use the official Department of Education tablebook in the exam hall. There is a lot of information in this book thatyou do not need. The information on the following pages has beenextracted from the official table book and is exactly what you needfor the Leaving Cert. Honours Maths papers.
131
THE TABLES
PAGE 6 & 7 OF THE TABLES
r
C
b
a
ch
C
b
a
h
h
r
r
h
r
l
CIRCLE
Length = 2π rArea = π r 2
SECTOR
Length = rθ θ( in radians)
Area = 12
2r θ θ( in radians)
TRIANGLE
Area = ahArea = ab Csin
PARALLELOGRAM
CYLINDER
Area of curved surface = 2π rhVolume = π r h2
CONE
Curved surface area = π rl
Volume = 13
2π r h
SPHERE
Area of surface = 4 2π r
Volume = 43
3π r
Area = 12 ah
Area = 12 ab Csin
r θ
132
THE TABLES
cos( ) cos cos sin sinA B A B A B+ = −
sin( ) sin cos cos sinA B A B A B+ = +
tan( ) tan tantan tan
A B A BA B
+ =+
−1
Compound Angle formulae
The formulae for cos( ), sin( ),A B A B− −
tan( )A B− can be obtained by changing thesigns in these formulae. You need to be able toprove cos( )A B± and sin( ).A B±
cos sin2 2 1A A+ =
tan sincos
A AA
=
sec tancos
2 221 1A A
A= + =
cottan
AA
=1
seccos
AA
=1
cosecsin
AA
=1
There are 6 trig functions. They can all be written in terms of sine and cosine.
A 0 π2π π
3π4
π6
cos A
sin A
tan A 0
0
1 −1
0
0
0
1
∞
12
32
3
12
12
1
32
12
13
A 0o 180o 90o 60o 45o 30o
cos( ) cos− =A A sin( ) sin− = −A A tan( ) tan− = −A A
This is how you deal with negative angles.
Use the Sine and Cosine rules to solve triangles.
Sine formula: a
Ab
Bc
Csin sin sin= =
Cosine formula: a b c bc A2 2 2 2= + − cos
[Radians]
[Degrees]
(cos , sin )A A
cos A
sin A
1
1
-1
-1
A
[Prove]
A
CB
bc
a
PAGE 9 OF THE TABLES
133
THE TABLES
cos cos sin2 2 2A A A= −
sin sin cos2 2A A A=
tan tantan
2 21 2A A
A=
−
These formulae are obtainedby replacing B by A in thecompound angle formulae.
cos tantan
2 11
2
2A AA
=−+
sin tantan
2 21 2A A
A=
+
cos ( cos )2 12 1 2A A= + sin ( cos )2 1
2 1 2A A= −
Use these when integrating trig squares.
Used to change products to sums, useful when integrating products of trig functions.
Used to change sums into products, useful when solving trig equations.
De Moivre’s Therorem
(cos sin ) cos sinθ θ θ θ+ = +i n i nn
2cos cos cos( ) cos( )A B A B A B= + + −
2sin cos sin( ) sin( )A B A B A B= + + −
2sin sin cos( ) cos( )A B A B A B= − − +
2cos sin sin( ) sin( )A B A B A B= + − −
sin sin sin( ) cos( )A B A B A B+ = + −2 2 2
sin sin cos( )sin( )A B A B A B− = + −2 2 2
cos cos cos( )cos( )A B A B A B+ = + −2 2 2
cos cos sin( )sin( )A B A B A B− = − + −2 2 2
134
THE TABLES
DIFFERENTIATION INTEGRATION
f x( ) ′ ≡f x ddx
f x( ) [ ( )]
xn nxn−1
ln x1x
cos x −sin xsin x cos xtan x sec2 xsec x sec tanx xcosec x −cosec cotx xcot x −cosec2 x
ex ex
eax aeax
sin−1 xa
12 2a x−
tan−1 xa
aa x2 2+
Products and Quotients:
y uv dydx
u dvdx
v dudx
= = +;
y uv
dydx
v dudx
u dvdx
v= =
−; 2
We take a > 0 and omit constants ofintegration.
f x( ) f x dx( )∫
x nn ( )≠ −1xn
n+
+
1
1
1x ln x
cos x sin xsin x −cos x
tan x ln sec x
ex ex
eax1a
eax
12 2a x−
sin−1 xa
12 2a x+
1 1
axa
tan−
cos2 x 12
12 2[ sin ]x x+
sin2 x 12
12 2[ sin ]x x−
Integration by parts:
u dv uv v du∫ ∫= −
PAGE 41 & 42 OF THE TABLES