Mathematics for Electronic and Electrical Engineering
Self-study course MATH1055
MATH1055
MATHEMATICS FOR ELECTRONIC AND ELECTRICAL ENGINEERING
MODULE 0: COURSE DESCRIPTION
1. Introduction: Aims and Objectives
Mathematics is an essential tool for the engineer. In this course you are introduced to some
of the mathematical techniques which you will need in the rest of your engineering studies.
The aims and objectives of this course include the knowledge and understanding of the
material, but also other additional skills you are expected to develop through this self-paced
course. Having successfully completed this module, you will be able to:
• Demonstrate knowledge and understanding of basic di↵erential and integral calculus,
di↵erential equations, complex numbers, vectors and matrices, and be familiar with
partial di↵erentiation and some more advanced techniques of calculus;
• Show logical thinking in problem solving;
• Work more e↵ectively with self-study material;
• Demonstrate organisational and time-management skills;
• Critically analyse and solve some mathematical problems;
• Perform calculations in simple situations and work through some longer examples.
Outline
This is a self-study course, written around a textbook – there are no lectures. The
syllabus is divided into what we call modules, each of which is designed to cover a similar
amount of work. (Elsewhere in the University, a module refers to a course running over a
whole semester or year; here it is a topic, taking about one week.)
The following are the main points describing how this self-study course works. All yearspecific information (marking/testing room, timetable, people, etc.) can be found in the
Organisational Data document given to you separately.
• You have a list of modules (topics) to study and a recommended list of deadlines for the
topics.
• For each module (topic) you get a set of instructions detailing which section of the
textbook to read, which exercises to do, and a specimen test. All the modules are in
in this booklet, in the order they must be studied; all solutions are in a separate
solutions booklet, in the same order. Occasionally a module will be self-contained
(i.e. all the material is in the module itself with no reference to the textbook).
• After working through a module, and when you think you are ready, come to the testing
room and get a module test from the administrator. The test should take you about
20-30 minutes to complete.
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• After finishing the test, take it to the marking room, and sit down with any free marker.
They will mark your work and go over your answers with you, giving you feedback on
any errors. The mark you get counts towards the 20% coursework mark for this course.
This “marking” session is really about feedback and one-to-one teaching.
• After marking, take the test back to the administrator in the testing room to have
your mark recorded.
• That’s it. If you passed, go home and study the next module. If you didn’t pass, go home
and revise this module: you have one more opportunity at passing the test.
• In January there is a one-hour multiple-choice class test worth 5% of your final mark.
This test covers first half of the syllabus (the first 10 modules).
• In May/June there is a final exam, worth 75% of the mark. This covers the material of
the entire year.
2. Details
• TEXTBOOK: To be able to follow this self-study course, you need a copy of the
textbook. The main book is Modern Engineering Mathematics by Glyn James,
Pearson, 2015 (5th edition). However there are alternatives to this textbook: see
section 5. We recommend that you buy the textbook. Alternatively, the library holds
several copies, including a good number of e-book licences.
• STUDY: Make sure you go through all the work in each module, especially the specimen
test, which is similar in length and diculty to the module test you will take. Once
you complete the specimen test of the module, go through your work with the solutions
(that we provide), and self-mark your work. The module instructions are a minimal
amount of work necessary to learn a topic. You would of course benefit from doing
more than the minimum: there are many more exercises in the book and online, see
section 5 below.
• HELP: If you are having diculties, come and get help. For short queries, the markers
are glad to help. Come to any of the marking sessions and ask away. If there are many
people queuing to get their tests marked, they will get priority from the markers, but
if you come at the beginning of a session you will probably find a free marker. For
longer queries, come to the Engineering Maths Workshop (MATH1061). See
also section 8 below for more details about available help.
• TIMING: The module test is not timed, so time yourself: if you can do it in about 25
mins, that’s good. If you are taking longer than 35 minutes, this means you probably
haven’t prepared enough: keep that in mind for the next topic. The formula sheet
is available for the test (please bring a copy), as for the final exam (you will be given
a copy), so get familiar with what is in there. If you take a long time for a test and
don’t finish it or get it marked in one session, give it to the administrator who will give
it back to you when you come to the next session to finish the test or get it marked
(this should rarely be necessary). Note that no tests begin in the last half hour of
a session, so you cannot start a test after 12:15pm on Mondays or 11.15am
on Wednesdays. So please arrive early to avoid disappointment, as there may be
queues.
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• MARKS: Each test is worth about 1% of your final mark, so don’t worry too much
if you get a low mark every now and then. Go over the material again, think about
the feedback you got from the marker, make sure you know it for the class test in
January, and especially, for the final exam in May/June, that’s when it counts most.
The marks are whole numbers from 0 to 4. The typical pass mark is 3 (corresponding
to 75/100, or an A grade!). The maximum mark of 4 requires an answer that is
completely correct and clear, and markers are instructed to be strict. Marks 0, 1 or
2 means ‘fail’ and you have one more chance at taking a di↵erent test on the same
topic. If you fail the same module a second time, the best mark will be recorded and
you move on to the next topic. You can only retake a module if you fail, that is, you
cannot retake a test to improve a mark of 3 to a 4, as the test are not about scoring
a high mark but about checking your understanding, learning, and preparing yourself
for the final exam. See also section 7 below.
• ARRIVE EARLY: Make sure you turn up on time for the testing/marking sessions,
especially as you cannot take a test in the last half an hour, and queues may
build up at busy times.
• RECORDING YOUR MARK: Don’t walk o↵ without having your test marked, and
the mark recorded by the administrator. If you take an unmarked or unrecorded
test home, you will get a zero mark for that attempt. Please look at Grade Centre on
Blackboard regularly to check that your marks have been correctly recorded. It is your
responsibility to make sure that the correct mark has been recorded on Blackboard.
Please let the administrator know if there is a problem, but allow at least 1 week
since you did the test for the marks to be uploaded.
• MULTIPLE TESTS: You may do more than one test on the same session, time permitting: if you pass a test, you may take a test for the next module if you wish to.
However, if you fail a test (for the first time), you must go home, revise, and take
another test on the same module on a di↵erent day.
• ORDER OF MODULES: The modules (topics) that you will study are shown
below, in the order chosen by your Faculty. You must study the modules, and take the
tests, in this order (no skipping allowed). This is the material you are expected
to learn this year, to a greater or lesser degree, without ‘picking and choosing’ the
material you want to learn, and the material you want to discard: they are all equally
important. Moreover, some modules are prerequisites for later modules, or are needed
for other courses in your first year (see below), and all this have been reflected in the
order below.
Semester 1: 5, 3, 4, 14, 15, 7, 8, 9, 10, 11, online class test (January).
Semester 2: 16, 17, 6, 22, 12, 13, 18, 19, 24, 25, final exam (May/June).
Note: You may start Semester 2 modules in Semester 1, if you have already finished all
the previous modules. In fact, it is not impossible to finish all 20 modules by January!
• PREREQUISITES: Modules 1 and 2 on Algebra and Trigonometry are included in
the booklet for revision. It is assumed that you know this material. If you find that
you have forgotten some of the topics in these modules, you are strongly advised to
work through them! If you find yourself having to do a lot of catching up with these
prerequisites, get some help in the Engineering Maths Workshop (MATH1061)
(see above, or section 8 below).
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• PACE YOURSELF: The course is “self-paced” and you are not expected to follow a
strict timetable. However, the mathematics is used in the other engineering courses
you are taking in your department. You are expected to cover approximately one
module per week, in the order listed above. Your other lecturers will expect you
to have learned the relevant material by the weeks shown in the Organisational Data
Sheet. And, of course, you need to have advanced to Module 11 by the end of teaching
in December, for the multiple-choice online class test in January. Note that: the
University is closed on May Bank Holiday (a Monday); that you may miss test/marking
sessions due to field trips, illness, etc.; and that you must come another day if you fail
a module test for the first time. Please plan ahead and make sure you do not
run out of time for taking tests. Untaken tests get a mark of zero. Finally, note
that we send personal tutors a weekly progress report, and that if you do fall behind,
we may contact you and your tutor.
• REMEMBER: Taking and passing the module tests is not a goal but the means to a
goal: It is very important to plan your pace throughout the year so that you do
each module in a way that prepares you for the final exam. In particular, avoid
short-term memorisation of topics/exercises and make sure you are confident with the
material, including all types of exercises, before moving on.
3. People involved
(i) Co-ordinators: Dr Oscar Dias and Dr Nansen Petrosyan (both Mathematical Sciences)
are responsible for the course as a whole.
(ii) Academic Supervisors: Each of the self-paced sessions is overviewed by a Mathematics or Engineering sta↵ member (the coordinator of the self-paced session).
(iii) Administrators: responsible for the distribution and invigilation of the module tests,
and keeping an accurate record of your progress. Sometimes the queues get large:
please be always patient and polite! If you have feedback about the self-paced session
please provide it to the administrators: they always give it to the co-ordinators.
(iv) Markers: Mathematics or Engineering PhD students plus an academic supervisor in
each session. They will: (a) mark and discuss tests, and (b) answer all your short
mathematical questions. If you appreciate the explanations provided by a particular
marker you can choose to be marked by her/him (just give your way in the queue until
she/he is available). If you feel that your mark requires a revision please approach the
academic supervisor of the session.
(v) Mathematics helpers: Mathematics PhD students who run theMATH1061 Engineering Maths Workshop (Mondays & Wednesdays 3-6pm); they will answer all
your (short or long) questions and provide any mathematical help you may need.
4. List of modules (topics)
Here is the complete list of self-paced modules, in numerical order. You will study most of
them following an order that is not the natural numerical order displayed below (see the
Organisational Data sheet for the precise ordering or the item “ORDER OF MODULES” in section “2. Details” above), but feel free to dip into any of them at any time.
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In the Booklet and Blackboard you will find more than 20 modules. However, you will only
be tested and examined on the 20 modules that are in your syllabus (also identified precisely in the Organisational Data sheet). The leftover modules are kept in the Booklet
and Blackboard in case you find them useful in future years: of course, you are also welcome
to try them out for yourself right away! If you want help on any of the other modules, you
can ask a marker, or use the MATH1061 Engineering Maths Workshop, see section 8
below.
1. Algebra (revision of basic rules; equations; inequalities; partial fractions)
2. Trigonometry (revision of standard trigonometric functions and formulae)
3. Di↵erentiation I (basic rules; standard di↵erentials; Newton’s method for finding
roots; simple partial di↵erentiation)
4. Integration I (standard integrals; simple substitutions; integration by parts; numerical
integration)
5. Complex numbers I (Argand diagram; polar form; exponential form and Euler’s formula)
6. Di↵erential equations I (classification; separable first order ODEs; homogeneous second order ODEs with constant coecients)
7. Functions (functions and inverse functions; trigonometric and inverse trigonometric
functions; exponential and logarithmic functions; hyperbolic and inverse hyperbolic functions; di↵erentiation of inverse trigonometric and hyperbolic functions)
8. Di↵erentiation II (maxima, minima and points of inflection; curve sketching; parametric, implicit and logarithmic di↵erentiation; Maclaurin’s series; Taylor’s series)
9. Integration II (more advanced substitutions; applications including volumes of revolution, centroids, centres of gravity, mean values, arc length)
10. Integration III (integration of rational functions; improper integrals)
11. Integration IV (double integrals; polar integrals; triple integrals)
12. Di↵erential equations II (dx/dt = f (x/t); linear and exact first order ODEs)
13. Di↵erential equations III (inhomogeneous second order ODEs with constant coecients; free and forced oscillations)
14. Vectors I (basic properties; Cartesian components; scalar and vector products)
15. Vectors II (triple products; di↵erentiation and integration of vectors; vector equations
of lines and planes)
16. Matrices I (terminology; basic properties; determinants)
17. Matrices II (solving sets of linear equations; calculation of inverse using cofactor and
elimination methods)
18. Matrices III (rank of a matrix; eigenvalues and eigenvectors)
19. Further calculus I (chain rule for partial derivatives; higher partial derivatives; total
di↵erentials and errors)
20. Further calculus II (sequences and series; Rolle’s theorem; Taylor’s and Maclaurin’s
theorems; l’Hôpital’s rule)
21. Laplace transforms (definition; simple transforms and properties; solution of first and
second order linear ODEs with constant coecients)
22. Complex numbers II (complex trigonometric and hyperbolic functions; logarithm of
a complex number; de Moivre’s theorem; nth roots; simple loci)
23. Fourier series (periodic signals; whole-range Fourier series; even and odd functions)
24. Statistics I (probability; conditional probability; combinations and permutations; discrete and continuous random variables)
25. Statistics II (mean and standard error of sample data; normal distribution; sampling;
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confidence intervals; hypothesis testing)
26. Applications to electrical circuits (complex numbers and alternating currents; complex impedance; di↵erential equations for RLC circuits; forced oscillations and resonance;
complex solutions of di↵erential equations; phasors)
27. Further applications to electrical circuits (inverse of a matrix using elimination;
mesh analysis of circuits; node analysis of circuits)
5. What is in a Module?
A ‘Module’ (topic) consists of a batch of printed material which is to be used in conjunction
with either of the course textbooks, and from Module 3 onwards it is essential to have
easy access to a copy of a book (you will have to self-study the 20 modules listed in the
Organisational Data sheet and also in the item “ORDER OF MODULES” of section
“2. Details” above).
The main book, which will be referred to as J. throughout the modules, is Modern
Engineering Mathematics by Glyn James, Pearson 2015 (5th edition), ISBN 978-1292-08073-4 (print)
It is important that you get a copy of the 5th edition since the page and section numbers
displayed in the self-study guide of the Booklet referes strictly to this edition (not the 6th
edition when/if available, neither previous editions). Copies of the 3rd and 4th edition are
also available in the library.
Many of you arrive to the University coming from di↵erent schools, countries or programmes.
It is thus natural that some of you have a less strong mathematical background at the
beginning of the year. If this is your case, as a warm-up textbook to help you getting into
the pace before you move to the main book, we suggest that you first read/study the “easier”
textbook (with less mathematical background) referred to as S. throughout the modules:
Engineering Mathematics by K A Stroud, Palgrave 2007 (6th edition), ISBN 978403942463.
It is important that you should use this Stroud textbook only as a warm-up. It is then
fundamental that you proceed to the next stage and study the main textbook (James).
Indeed, although at the beginning of the year you might have a less strong mathematical
background, in the end of the year you must all have a similar strong mathematical
background! Moreover, Stroud does not cover all the material of the 20 modules! So do
not use Stroud as the single textbook! You will be excellent engineers and for that you need
to be strong at maths.
These two textbooks, J. and S. were chosen by a group of engineers and mathematicians to
be the most appropriate books for this course. However, no book (or set of lecture notes)
is perfect for everyone and the intention of the Modules is to guide you through the book,
sometimes supplementing its contents.
Each Module begins with Module Topics—a list of the main points covered in it. In
some Modules there may also be an introductory paragraph. This is followed by the Work
Scheme based on James (FIFTH edition), which is split into numbered sections. Most
of the sections will refer to particular parts of J., which you will be asked to read (usually for
background) or study (essential for the understanding of the Module). Some sections of the
work scheme, however, will contain material not in J. At various places in the work scheme
you will be asked to do Exercises, most of which will be taken from J. The handwritten
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Worked Solutions to these Exercises are included in the solutions booklet.
Some of you that begin the year with a less strong mathematical background may find
the book by Stroud, a programmed learning text, more appropriate to initiate the studies
(recall the ‘warm-up’ discussion above). This covers less of the syllabus, but for certain
Modules it can be very useful to first study this textbook and only then move to the main
textbook (James). Therefore, a Work Scheme based on Stroud (6th edition) is also
included in each Module.
After completing the work scheme you should attempt the Specimen Test, which is included
in the Module and is similar to the module test you will take in the testing session. Solutions
to the Specimen Test are included in the solutions booklet too. These tests show whether
you know the basic material in the Modules, although the examination Section B questions
(see Section 10) can be longer and more dicult.
Optional lecture notes which summarise the content of the Modules 1– 25 are on Blackboard and on the course website (see section 11). Using these is optional, but they may be
a good introduction to each module, although you still need to work through the modules
and sample problems in the self-study guide. You learn mathematics by doing!
A new edition of S. has been published in 2012, but the modules refer to the old 6th edition,
published in 2007. If you feel you need the gentler approach of Stroud, you can get it from
the Library or buy a second-hand copy of the sixth edition. Do not buy the new seventh
edition, as the references in the modules may not be accurate.
All the references to J. in Modules 1-27 are up to date for the 5th edition published in
2015. However, on Blackboard and on the course website (see section 11) you will also find
versions of the self-study guide for Modules 1-25 which refer to the previous editions of J.
and S., also available in the Library. These are equally well suitable for the course.
Pearson, the publisher of J., runs an online platform called MyMathLab. This can be
helpful for students who need extra practice, as they provide many additional exercises
with interactive step-by-step solutions. However, you will need to buy an access card from
Pearson, at a cost. You can find more information about MyMathLab on Blackboard.
We have video solutions to some of the problems in the Solutions booklet, available in
Blackboard. You may prefer watching someone solving the problem, particularly if you find
the handwritten solution not clear enough. The videos have been recorded by Mathematics
academic sta↵ volunteering their time.
6. How to study
You may work through the self-study guide in whatever manner you choose, but the following
notes and tips may be of some guidance. First of all, understanding is not an all or nothing
process; one understands at various levels. Thus, when studying a section of a book it is
suggested that you might adopt the following approach. First read the section quickly and
try to get a feeling for the scope of the material and its level. Then go back over it carefully,
more than once if necessary, each time trying to get a deeper understanding of what you are
reading. If you come across something on which you are stuck leave it for the time being
and carry on. Even leaving it for a day or two before going back to it can sometimes help.
If it does not, get help (see section 8) before going for a test.
You will get more out of your reading if you become actively involved in the sense of con7
stantly asking yourself questions such as ‘What is this all about?’, ‘Why is it done this way?’
and ‘Is this a significant or a trivial point?’. Simply underlining key words in the text or
jotting down thoughts and queries in the margin can be valuable ways of increasing your
concentration.
Another golden rule in studying is that a little and often is far better than a lot in one go.
So do not try to cram everything in just before taking a test. It is far better to spread your
learning over the week. The ‘cramming’ approach may just enable you to pass the test but
you will find that the knowledge gained will not stick and your will pay the consequences
later. Remember that if you pass the test without being fully prepared for it you are only
cheating yourself—the tests are primarily for your benefit and to tell you whether you have
properly understood material on which you will subsequently be examined and which is
essential for the proper understanding of your engineering work.
It is very important to go through the worked examples in J. - one often first understands a
piece of bookwork by seeing it in action in a particular instance. Try the Module exercises
- unless you can do the exercises you haven’t fully understood the bookwork! This is so
important that it is worth repeating: try the exercises before you look at the worked
solutions. Even if you do not get very far with some of them it will be of much greater
benefit to you if you try them on your own first. This way you can often isolate your diculty
and then, when you do see how something is done, it is much more likely to remember it.
It is far too easy to read through a solution thinking you understand what is going on but
then to find that you are completely unable to do anything like it yourself later on. Finally,
do the specimen test and check your answers! The specimen test is of the same length,
diculty and choice of material as the module test and will give you a very good idea of
how well you have understood the material.
7. Testing and Marking
When you feel competent in a module (topic), go to the testing room and ask the administrator for the appropriate test. Please show the administrator your student ID card. If
you do not have your student ID card you may not be allowed to take a test.
When you have collected the test, sit down, write your name and department/faculty on
the test sheet and answer the questions by writing on the test sheet. During the testing
sessions you MAY use a university approved calculator (please bring your own), and
consult the Formula Sheet (please bring a copy with you) but nothing else. The test
should take approximately 25 minutes to complete. Remember that the object of the test
is not to get the best possible score, but for you and the marker to check what you already
know, identify those points where you are struggling, and learning through the process.
Hence if you get stuck with something in the test do not spend too long over it, but rather
ask the marker later.
After you have finished the test bring it to the marking room. Choose any marker in the
marking room and ask him/her to mark your test. If you prefer you may choose the same
Marker every week. In order to allow adequate time for discussion you should take your test
sheet to the marking room at least 15 minutes before the end of the session. There may not
be sucient time to mark your test. In this case you should return the unmarked test to
the administrator and come back at the beginning of the next session. Please do not take
unmarked tests home.
The test will be marked and discussed with you and it will be given a mark of 0, 1, 2, 3
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or 4. The main purpose of the test is to discover whether you have sucient knowledge to
proceed to the next module. The discussion with the marker is the key feedback
in this module. It provides you with an opportunity to obtain help with any
diculties you may have. If you are given a 3 or 4, you have passed. Then take the
marked test back to the administrator who will record your mark. The marked test will be
returned to you; please keep it for the whole semester; it will be useful when you revise
for the exam, and in the unlikely case that we need to double-check your mark. If you
are given a mark of 0, 1 or 2 you have not passed, the administrator will record your mark
and you will be asked to return on a di↵erent day to take a di↵erent test on the same
topic. The marker may also suggest that you should attend the Engineering Mathematics
Workshop (MATH1061), see Section 8. A maximum of two attempts is allowed for each
module, but only if you fail (0, 1 or 2 marks); you don’t get a second attempt if you already
achieved a 3, since the tests are about learning and preparing you for the final exam, not
about achieving perfect scores. The marks awarded for each test (the higher of the two
marks if you retake a test) count towards your 20% coursework mark for the course. You
can check the correct entry of your marks in Blackboard Grade Center (please allow at least
2 weeks for the administrator to upload the marks). If you find any error, please show the
corresponding marked test to the administrator at your next visit to the testing room.
8. Available help
(i) Quick queries:
Except when they are engaged in marking tests the markers will be
available during the timetable periods in the marking room to help you with any points
that cause you diculty. The most appropriate time is usually in the first 30 minutes
of the session whilst most people are still taking their tests, or during the last half-hour
when most people have finished! Don’t be shy: come and ask for help, we are always
pleased when people keep us busy. Remember: there are no stupid questions!
(ii) All queries: If your problems require lengthy discussions then you should attend the
Engineering Mathematics Workshop (MATH1061). This runs Mondays and
Wednesdays 3-6pm throughout the teaching weeks of both semesters, and continues
in the May exam period until the final exam. There are two or three mathematics
helpers to answer your questions. There are also a few copies of the course texts
available for consultation. The Workshop is there principally to support this course
but it can also be used by any other student in the University with mathematical
queries. The Workshop has proved an extremely useful facility for first and second
year engineering students. If you experience diculties during the year, or your mathematical background is weak or rusty, then you are strongly advised to make use of the
Workshop. You can drop in any time it is open, for five minutes with a quick query,
or go along for the full three hours each session and work through the week’s module
with help readily available when you get stuck. It is there to help you: Use it!
9. Assessment
At the end the year, in May/June, you sit a two-hour written final examination. The final
exam covers the entire year’s material. It counts for 75% of your final mark, the selfpaced tests for 20%, and the one-hour multiple-choice online class test in January for 5%.
MATH1055 is core to your programme, so you must pass the course (a final mark of 40% or
more). If you fail, you will normally be required to take a referral/resit examination paper
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in August/September and your final mark will be 100% the referral exam mark (normally
capped at 40%). Please ask your personal tutor or the student oce if in doubt about
referral rules.
The format of the examination paper will be as follows: Part A with short multiple-choice
questions, similar to the multiple-choice questions in the online January Class Test, and
Part B containing longer questions. All questions are compulsory. You must be able to
do the longer questions in the Modules to be able to cope with the longer questions in the
examination. The rubric of the exam (the first page, containing the exam instructions) will
be published on Blackboard in advance.
10. Blackboard
All the material for this course is available on Blackboard.
In Blackboard you will find the Formula Sheet, this Course Description (= Module 0),
Organizational Data sheets, Modules 1-27, lecture notes for Modules 1-25, video solutions
for selected problems, and examination papers and solutions from recent years.
Whenever we find appropriate, we will also use Blackboard to send important announcements that might guide/help you in your studies. These Blackboard messages are directly
connected to your university email. Therefore it is important that you check your university email regularly.
11. Contact and Feedback
Please send any comments you have about the course (e.g. typographical errors in the
paperwork, topics which you feel could be better explained) to the respective Academic
Supervisors or to the Self-Study Course Coordinators:
Dr Oscar Dias, Mathematical Sciences, oce 54/6003, telephone 023 8059 5112, email
O.J.Campos-Dias@soton.ac.uk
Dr Nansen Petrosyan, Mathematical Sciences, oce 54/8001, telephone 023 8059 5111, email
N.Petrosyan@soton.ac.uk
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SCHOOL OF MATHEMATICS
MATHEMATICS FOR PART I ENGINEERING
Self-paced Course
MODULE 1
ALGEBRA
Module Topics
1. Simplifying expressions and algebraic functions
2. Rearranging formulae
3. Indices
4. Rationalising a denominator containing a square root
5. Linear and quadratic equations
6. Simultaneous linear equations
7. Inequalities
8. Partial fractions
Most modules are based on the book by James but the first two, on Algebra and Trigonometry, are selfcontained. Facility in carrying out algebraic manipulation is very important in your studies, and Module 1
covers most of the basic algebraic topics.
Many of you will know much of the material in the first two modules and so these will be mainly revision.
It is important, however, that you are able to carry out mathematical manipulations quickly and so, even
if the material is familiar, you are still strongly advised to work through the exercises to improve your
speed and accuracy. To some of you the discussed material will not all be revision, in which case you
should spend time working carefully through the modules. The book (not the course text) Introduction
to Engineering Mathematics by Croft, Davison and Hargreaves, published by Addison-Wesley,
contains a large number of chapters on the basic material covered in the first two modules and this book is
recommended if you need extra examples.
Work Scheme
Study the following sections, read carefully the worked Examples and do the stated Exercises. Solutions to
the Exercises are given towards the end of this module, starting on p.21.
1. Removing brackets
A basic rule in removing brackets from mathematical expressions is
a(b + c) = ab + ac.
The quantity a outside the bracket multiplies both the quantities b and c inside. Note that the meaning
of an expression depends crucially on including brackets, where appropriate. Omitting them in the above
expression, for instance, gives ab + c, which is not the same. You should also recall that
(a + b)c = ac + bc,
a(b  c) = ab  ac.
If more than one set of brackets is present the inner ones are removed first. Now go through the following
worked Example.
–1–
Example 1. Remove the brackets in the following and simplify the resulting expressions by combining like terms: (i) a  3  2(4b  2  3(3a  2b)),
(ii) (x  2)2 .
(i)
(ii)
= a  3  2(4b  2  9a + 6b) = a  3  8b + 4 + 18a  12b = 19a  20b + 1.
= (x  2)(x  2) = x(x  2)  2(x  2) = x2  2x  2x + 4 = x2  4x + 4.
In determining the above solution note that it was essential to use the following rules:
(positive) ⇥ (positive) = positive,
(positive) ⇥ (negative) = negative
(negative) ⇥ (positive) = negative,
(negative) ⇥ (negative) = positive.
The process of removing brackets is also commonly called expanding brackets.
***Do Exercise 1.
Remove brackets and simplify the following
(i)
x  1 + 2(3x  8),
(v)
(x  1)(x + 2)(x  3),
(ii)
2a + 3(c  a)  2(b  a),
(iii)
(y  3)3 .
(vi)
(x + 7)(x  5),
(iv)
(x + 4)2 ,
2. Factorisation This is the reverse of the process of removing brackets. You are given an algebraic
expression and you try to rewrite it as the product of factors. In many cases you may have an additional
implicit requirement that the factors contain only integers (positive or negative) - you will not always be
able to find factors satisfying this constraint.
With very simple expressions the factorisation is straightforward:
e.g.
t2 + t = t(t + 1).
5x  10 = 5(x  2),
Sometimes you must try to factorise a quadratic expression ax2 + bx + c, where a, b and c are numbers, into
a product of linear factors. For example, if you are asked to factorise the quadratic expression x2  4x  21
then you seek integers m and n such that
x2  4x  21 = (x  m)(x  n).
Since the right-hand side can be expanded to give
x(x  n)  m(x  n) = x2  nx  mx + mn = x2  (m + n)x + mn,
comparison with the original quadratic shows that you require
(i)
m + n = 4,
(ii)
mn = 21.
There are eight possible ways of satisfying condition (ii) with integers: m = 1, n = 21; m = 3, n = 7;
m = 7, n = 3; m = 21, n = 1; m = 1, n = 21; m = 3, n = 7; m = 7, n = 3; m = 21, n = 1. It is
easily seen that only the third pair of numbers satisfies condition (i). Hence, the required factorisation is
x2  4x  21 = (x  7)(x  (3)) = (x  7)(x + 3).
When the coecient of x2 is not unity then the above approach must be slightly modified. For the quadratic
2x2 + 5x  3 you write
2x2 + 5x  3 = (2x  m)(x  n)
and, after expanding the right-hand side and comparing coecients, require
2n + m = 5,
–2–
mn = 3.
It relatively easy to deduce that the above pair of equations has solution m = 1, n = 3, so the appropriate
factorisation is
2x2 + 5x  3 = (2x  1)(x + 3).
In many situations the quadratic does not factorise into linear factors with integer coecients. Show, for
instance, that x2 + x + 1 cannot be written in the form (x  m)(x  n) for any integers m and n.
With practice you will be able to spot the linear factors and write them down, although it is advisable to
remove the brackets in your answer to verify that you do indeed get back to the original expression. That is
to say, you should verify that your answer is correct.
***Do Exercise 2.
(i)
x2 + 7x + 12,
Factorise
(ii)
x2  2x  3,
(iii)
x3  25x,
(iv)
8 + 2x  x2 ,
(v)
2x2  3x  2.
3. Algebraic fractions The rules for adding, subtracting, multiplying and dividing arithmetic fractions
carry over into algebra. It is important you express fractions in their simplest form, by cancelling common
12x3
factors. The numerator and denominator of the quotient
both have factors of 3 and x, and cancelling
3x
these gives
12x3
= 4x2 .
3x
This answer applies only when x 6= 0, since division by zero is not allowable.
As a further example of simplifying fractions you can see that
3x3
3x2
=
.
2
3x + x
3x + 1
Note that the denominator 3x2 +x = x(3x+1) and so it is not divisible by 3 or x2 , just x or 3x+1. In general
a fraction (either arithmetic or algebraic) is expressed in its simplest form by factorising the numerator and
denominator separately and then cancelling any common factors.
x2  2x  15
x+3
On factorising the numerator the above quotient can be written
Example 2.
Simplify
(x  5)(x + 3)
= x  5,
x+3
provided x 6= 3. The latter restriction on x is necessary because in obtaining the answer x  5
the common factor x + 3 was cancelled by dividing both numerator and denominator by this factor.
Clearly this is only acceptable provided x + 3 6= 0, i.e. x 6= 3, and hence the quotient in Example 2
simplifies to x  5 only if x 6= 3. When x = 3 the original fraction has the value 0/0 which has no
meaning.
To add or subtract algebraic fractions you must first write each fraction in its simplest form by cancelling
any common factors. Next you have to determine the simplest algebraic expression that has the given
denominators as its factors - i.e. the lowest common denominator (LCD). Each fraction is then written with
this LCD as its denominator, before combining.
–3–
Example 3.
Express as single fractions
(i)
x y
+ ,
y
x
(ii)
2x + 1
3
+ 2
2
(x + 1)
x + 3x + 2
(i) LCD of denominators y and x is xy, so
x y
xx yy
x2 + y 2
+ =
+
=
.
y
x
xy
xy
xy
(ii) This time the LCD of denominators (x + 1)2 and x2 + 3x + 2 (= (x + 1)(x + 2)) is (x + 1)2 (x + 2),
hence
3
2x + 1
3(x + 2)
(2x + 1)(x + 1)
+
=
+
(x + 1)2
(x + 2)(x + 1)
(x + 1)2 (x + 2) (x + 2)(x + 1)2
3x + 6 + (2x2 + 2x + x + 1)
=
(x + 1)2 (x + 2)
2
2x + 6x + 7
=
.
(x + 1)2 (x + 2)
Note that the numerator in the above expression does not factorise into linear terms with integer coecients.
Example 4.
Express as a single fraction
x + 2 can be expressed
3
(x + 2)(2x  1)
3
2x2  x + 4x  2 + 3
=
+
=
2x  1
2x  1
2x  1
2x  1
2x2 + 3x + 1
=
,
2x  1
which can be written
(i)
1
1
+
,
x x1
3
2x  1
x+2
and the LCD of 1 and 2x  1 is 2x  1. Hence
1
x+2+
***Do Exercise 3.
x+2+
(2x + 1)(x + 1)
.
2x  1
Express as a single fraction
(ii)
2
1
+
,
2
(t + 2)
t+2
(iii)
x+
1
x2 + 1
,
(iv)
2x
x2 + x  2

3x
(x  1)2
4. Formulae Physical quantities are often related to each other using formulae, e.g. the area of a circle,
A, is related to its radius, R, through the formula A = ⇡R2 .
To evaluate a formula you must substitute numbers in place of the symbols, remembering that you must be
very careful about the units you use.
In the formula A = ⇡R2 , A is called the subject of the formula since it appears by itself on one side of the
formula, and nowhere else. It is often necessary to rearrange a formula so that a di↵erent variable becomes the
subject. During this rearrangement, or transposition, you can carry out a number of di↵erent operations to
the formula provided you do the same thing to both sides. The possible operations are addition, subtraction,
multiplication, division (by a non-zero quantity) and taking ’functions’ of both sides (e.g. taking the square
or square root).
Example 5. An object with initial speed u and constant acceleration a travels a distance s in time
t given by s = ut + 12 at2 . Rearrange the formula so that the subject is a.
Subtracting ut from both sides gives
s  ut =
–4–
1 2
at ,
2
then multiplying both sides by 2 leads to
2(s  ut) = at2 .
Finally dividing both sides by t2 implies
2(s  ut)
.
t2
a=
***Do Exercise 4.
(i)
(iii)
(iv)
A = ⇡r2
Rearrange the following formulae to obtain expressions for the stated variable
for r,
(ii)
v 2 = u2 + 2as
1
1
1
+
for R,
=
R
R1
R2
s
l
T = 2⇡
for l.
g
for s,
(first express right-hand side as a single fraction),
5. Indices Products of a number can be written compactly using indices, or powers. For example,
81 = 3 ⇥ 3 ⇥ 3 ⇥ 3 and you write
81 = 34 .
The basic rules for manipulating indices are
am an = am+n ,
am
= amn ,
an
am =
1
am
1
(am =
or
am
),
(am )n = (an )m = amn ,
1
1
m
(am ) n = (a n )m = a n .
Note that
(am cn )3 = (am cn )(am cn )(am cn ) = am am am cn cn cn
= am+m+m cn+n+n = a3m c3n ,
and in general one obtains the result
(am cn )p = amp cnp .
Example 6.
3
Evaluate 16 2
Using the above rules
3
1
16 2 = (16 2 )3 = 43 = 4 ⇥ 4 ⇥ 4 = 64.
Check the answer using the xy button on your calculator.
Example 7.
 1
a 2 b3 c 2
1
acd 2
Simplify
3
=
 1
a 2 b3 c 2
1
acd 2
1
a2 b 2 c 2
1
acd 2
3
=a
21
3
2
b c
1
2 1
d
 12
3
2
= ab c
 12  12
–5–
d
=
ab 2
1
1
c2 d2
.
***Do Exercise 5.
Simplify
1
3
4
(i)
(81) ,
(v)
✓ 4 ◆ 13
a
,
a1
(ii)
1
2
2
a a ,
(vi)
(iii)
3
(a 3 )
,
a
(iv)
1
2
a 3 b 3 c(a2 c4 ),
x(xa )2 .
6. Rationalising a denominator involving a square root This section is concerned with rewriting
1
p so that no square roots appear in the denominator. The method for achieving
an expression such as
a b
p


2
2
2
2

y

xy
+
yx

y
b gives
=
x
.
Putting
x
=
a
and
y
=
this
uses
the
result
(x
+
y)
(x

y)
=
x
⇣
p ⌘
p
p ⌘⇣
2
a + b a  b = a  b. Hence, multiplying numerator and denominator by a + b you obtain
⇣
p ⌘
p
1 a+ b
1
a+ b
p = ⇣
p ⌘ = a2  b .
p ⌘⇣
a b
a b a+ b
1
p .
1+ 2
p
Multiply the numerator and denominator by 1  2 to give
Example 8.
Rationalise the denominator in
p 

p
p
p
1 1 2
1 2
1 2
p  =
p 

=
= 1 + 2.
12
1
1+ 2 1 2
Check your answer again using your calculator.
***Do Exercise 6. Rationalise the denominator in
1
6
p ,
p .
(ii) p
(i)
1 7
2+ 3
7. Linear equations Physical quantities are related by equations. You often need to solve an equation
in an unknown quantity, say x. That is to say you need to find the value, or values, of x which will make
both sides of the equation equal.
The simplest equations to solve are linear equations, in which x appears only to the first power, that is as
1
x, and not as x3 , x 4 , etc. The standard form is
ax + b = 0,
but linear equations often appear in non-standard form.
Example 9.
Solve the equation 3x + 10 = 4(1  x).
Multiplying out the right-hand side (RHS)
3x + 10 = 4  4x.
Adding 4x to each side gives
3x + 4x + 10 = 4,
–6–
and then subtracting 10 from both sides
7x = 4  10 = 6.
Dividing both sides by 7 then leads to the solution x = 6/7.
8. Quadratic equations
Quadratic equations have the standard form
ax2 + bx + c = 0, where a 6= 0.
If the quadratic expression on the left-hand side can be factorised then it is easy to write down the solution
of the quadratic equation.
Example 10.
Solve 6x2  13x  5 = 0.
The left-hand side factorises to give
(3x + 1)(2x  5) = 0,
and hence either 3x + 1 = 0 or 2x  5 = 0. Solving these linear equations leads to
x=
1
5
or x = .
3
2
There are two values of x, therefore, which satisfy the above quadratic equation.
Sometimes you may be given a quadratic which you find dicult, or impossible, to factorise. In these
situations you can always use the result:
the solutions of the equation ax2 + bx + c = 0 (with a 6= 0) are
p
b ± b2  4ac
x=
.
2a
Comparing the general equation with that considered in Example 10 it is seen that a = 6, b = 13, c = 5.
5
1
Substitute these values into the general formula and verify the solutions x =  , x = .
3
2
Note that if b2  4ac > 0 then its square root is also a non-zero real number, with the plus and minus signs
in the general formula leading to two distinct real roots.
When b2 = 4ac there is only one value of x which satisfies the equation. For instance, the equation x2 
6x + 9 = 0 can be written (x  3)2 = 0, with solution x = 3 (twice). x = 3 is known as a repeated root.
Finally, if b2  4ac < 0 then its square root is a complex number (see a later module). In this situation the
quadratic is said to possess complex roots.
A third method for solving quadratic equations involves completing the square. Since
(x + a)2 = (x + a)(x + a) = x2 + ax + ax + a2
= x2 + 2ax + a2 ,
by subtracting a2 from both sides it follows that
(x + a)2  a2 = x2 + 2ax.
Hence, by choosing a = 4, the expression x2 + 8x can be written
x2 + 8x = (x + 4)2  42 = (x + 4)2  16.
–7–
Confirm the above answer by removing the bracket. Completing the square is used in the example below.
Example 11.
Solve x2 + 6x + 6 = 0 by completing the square.
The method consists of rewriting the first two terms of this quadratic by completing the square. Using
the result proved above it follows that
x2 + 6x = (x + 3)2  32 = (x + 3)2  9.
Hence
x2 + 6x + 6 = (x + 3)2  9 + 6 = (x + 3)2  3,
and so the given quadratic equation can also be expressed
(x + 3)2  3 = 0.
Adding 3 to both sides gives
(x + 3)2 = 3
which, after taking square roots, becomes
p
x + 3 = ± 3.
p
p
p
The two solutions, therefore, are x = 3 ± 3 (i.e. x = 3 + 3 and x = 3  3).
You can verify the above solutions by using the general formula for the solution of a quadratic.
It is important to note that the technique of completing the square is also very useful in other areas. For
instance, you will certainly use the technique for some integration problems.
9. Solution of simultaneous linear equations A linear equation in two unknowns x and y has the
form
ax + by = c,
where a, b, c are constants and a and b are non-zero. If you have two such equations then you have two
simultaneous linear equations in two unknowns. A set of values for x and y which satisfies both equations is
called a solution of the pair of equations. To solve the system of equations it is necessary to eliminate one
of the unknowns by adding or subtracting appropriate multiples of the equations (see below).
Example 12.
Solve the set of equations
4x + 2y = 5
5x  3y = 2.
First you must decide which unknown to eliminate. Suppose the variable y is chosen, then it is
necessary to ensure the coecients of the y terms in both equations have equal magnitude. An
obvious way to achieve this is to multiply the first equation by 3 and the second equation by 2:
12x + 6y = 15
10x  6y = 4.
Adding the above equations leads to 12x + 10x = 15  4, which gives 22x = 11, or x = 12 . Substituting
back into the first given equation leads to
✓ ◆
1
4
+ 2y = 5.
2
This implies 2 + 2y = 5, i.e. 2y = 3, y = 32 .
Hence, the solution set is x = 12 , y = 32 .
You should always verify your answer by substituting back into both equations stated in the question.
–8–
10. Solving equations using graphs A variety of di↵erent types of equation can be solved using graphs.
By drawing a graph (or graphs) you can see whether the given equation has a solution and, if so, obtain
approximate values for each solution. Note that a graphical method may not give you precise values for these
solutions although the accuracy of the answer can usually be improved by drawing better graphs, using the
zoom facility on graphical calculators to home in on the points of intersection or by algebraic means.
Example 13.
Solve, by using graphs, (i) 3x + 10 = 4(1  x), (ii) 6x2  13x  5 = 0,
(iii) 4x + 2y = 5, 5x  3y = 2, (the equations solved algebraically in Examples 9, 10 and 12).
(i) In this case you can plot y = 3x + 10 and y = 4(1  x) and then look for the point of intersection.
Alternatively, you rearrange the terms as in Example 9 to obtain 7x + 6 = 0, plot y = 7x + 6 and look
for the value (or values) of x where it intersects the x-axis (where y = 0).
y
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y = 7x + 6
The lines y = 3x + 10 and
y = 4(1  x) intersect near x = 0.8.
20
A more accurate graph would provide
a better solution.
y = 3x + 10
10
In the alternative method the graph of
y = 7x + 6 is again a straight line, and
it is easily seen that y = 7x + 6 = 0
when x is approximately 0.8.
x
3
2
1
1
2
3
y = 4(1  x)
10
20
(ii)
y
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20
10
x
3
2
1
1
2
3
10
20
–9–
To find a solution of 6x2  13x  5 = 0
draw a graph of y = 6x2  13x  5
and look for the values of x, if any, for
which y = 0 (i.e. where the graph
intersects the x-axis).
The graph suggests that approximate solutions for x are 0.3 and
2.5. By zooming in and expanding the
graph near these points it is possible
to achieve more accurate solutions.
(You should observe from the graph that y never becomes 20, for instance, and hence the equation
6x2  13x  5 = 20, which can be written 6x2  13x + 15 = 0, has no real solutions (or roots).)
(iii) In this case you look for the point of intersection of the two graphs 4x + 2y = 5 and 5x  3y = 2.
First draw the graphs of both these equations.
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y
3
5x  3y = 2
2
Clearly the lines intersect at
approximately x = 0.5, y = 1.5.
1
These values can again be confirmed by drawing more accurate graphs
x
2
1
1
2
2
2
4x + 2y = 5
3
It is worth noting that if the two graphs are parallel then they do not intersect and the system has no
solution. This situation would arise if you were asked to solve the equations x + y = 1 and 2x + 2y = 5,
for instance.
Graphical methods can easily be extended to more complicated equations.
***Do Exercise 7.
(i)
4  2x = 3,
***Do Exercise 8.
(i)
(ii)
2x + y = 3,
(ii)
(i)
x3 = 2  x,
4(1  x) = 3(2x  1),
(iii)
(iv)
2
1
=
.
x
x+1
x2  9x + 14 = 0,
(iii)
x2  9x  8 = 0,
(iv)
8
= 6  x.
x+3
Solve the following sets of equations
(ii)
4x  y = 3,
***Do Exercise 10.
2
= 5,
x+1
Solve the equations
x2  x  12 = 0,
***Do Exercise 9.
(i)
Solve the equations
3x + 4y = 12,
9x + 2y = 9,
(iii)
4x  2y = 5
5x + 3y = 2
Solve the following equations (using graphs in both cases to illustrate your results):
(ii)
x3 = x.
11. Inequalities The statement that the number a is less than the number b is written a < b. This
means the same as the statement b is greater than a, which is written b > a. Every inequality, that is a
statement that one number is less than another, may be written using either the symbol < or the symbol
>. An inequality a < b has a geometric meaning; if a and b are represented by points on a number line with
– 10 –
positive numbers to the right of the origin, then the point with coordinate a lies to the left of the point with
coordinate b.
.
.
.
...................................................................................................................................................................................................................................................................................................
.
.
.
0
a
b
The symbol < in a < b is often called the sign (or direction or sense) of the inequality - reversing the sign
leads to a > b.
The statement a  b means that either a < b or a = b. Similarly a  b means that either a > b or a = b.
(You read the symbols  and  as “less than or equal to” and “greater than or equal to” respectively.) The
statement “a is positive and b is negative” is equivalent to a > 0 and b < 0; whereas “a non-negative” and
“b non-positive” may be written a  0 and b  0 respectively.
If a and b are two distinct numbers then one of them must be greater than the other; this is the first basic
rule for inequalities, rule I below. If a and b are both positive and given by non-terminating decimals you
pick the greater number by successively comparing digits. For example, if a = 12.47325 and b = 12.475111,
then a < b since the first digit in which they di↵er is the third digit after the decimal point and that digit
is greater for b.
A negative number is less than any positive number. If a and b are both negative it is useful to refer to the
number line if you are in any doubt about which is the greater. For example, 3.18 > 7.23.
.................................................................................................................................................................................................................................................................................................................................................................................................................................................
7.23
3.18
0
When working with inequalities there are a number of basic rules.
I. If a and b are two numbers, then one of the three statements a = b, a < b, a > b is true and
the other two are false.
II. If a < b and b < c then a < c.
III. If a < b then a + c < b + c
(i.e. adding or subtracting the same quantity to both sides of an inequality leaves the inequality
sign unchanged).
IV. If a < b and c > 0 then ac < bc
(multiplying or dividing both sides of an inequality by the same positive quantity does not
change the sign of the inequality).
V. If a < b and c < 0 then ac > bc
(multiplying or dividing both sides of an inequality by the same negative quantity reverses the
sign of the inequality).
The above rules can be illustrated using the number line. Rule III is equivalent to translating the figure
to the right or left, depending on the sign of c - translation preserves the order. Rule IV is equivalent to
magnifying (or shrinking) the number line - again the order is unchanged. However, you can easily show
that Rule V implies a magnification and reflection, with the order (and hence sign of inequality) reversed.
12. Solution of inequalities To solve an inequality you must find all numbers x for which the inequality is true. The actual steps taken in determining the solution are similar to those used in solving the
corresponding equation, but you must be careful in dealing with the sign of the inequality.
Example 14.
Solve 3  2x < 4x  5.
First you must move all terms involving x to the left-hand side and all numbers to the other side. To
achieve this you must remove all x terms from the right-hand side (by subtracting 4x) and all numbers
– 11 –
from the left-hand side (by subtracting 3). Hence
(3  2x)  4x  3 < (4x  5)  4x  3
i.e.
 6x < 8.
1
To obtain the inequality for x you must multiply by  (or divide by 6), remembering from Rule
6
V that you must also reverse the inequality
6x
8
>
6
6
4
i.e. x > .
3
The solution is a set of numbers (called the solution set) - it consists of all numbers greater than
4
.
3
If a and b are numbers such that a < b then the statement a < x < b is equivalent to the two statements
x > a and x < b. In writing the double inequality it is essential that a < b. Thus, the inequalities x > 1
and x < 2 can be expressed by the double inequality 1 < x < 2, but it is not possible to write x > 2 and
x < 1 as the double inequality 2 < x < 1 (since 2 is not less than 1).
You may find it helpful to observe that acceptable double inequalities, such as 1 < x < 2, represent a single
region on the number line, whereas the inequalities x > 2 and x < 1 give two distinct sections of the line.
The solution of a double inequality is equivalent to the solution of two simultaneous inequalities, each of
which is solved using the method discussed above.
Example 15.
Solve x  6 < 2x  5  x  3.
This means that you must find the values of x which satisfy the inequalities x  6 < 2x  5 and
2x  5  x  3.
Each of these is treated separately:
(x  6)  x < (2x  5)  x
(x  6)  x + 5 < (2x  5)  x + 5
i.e.
 1 < x.
(2x  5)  x  (x  3)  x
(2x  5)  x + 5  (x  3)  x + 5
x  2.
Hence the double inequality implies x > 1 and x  2. The solution set therefore consists of all
numbers x such that 1 < x  2.
The absolute value | a | of a number a is defined by
⇢
a if a  0
|a|=
a if a < 0.
Thus | 8 |= 8, | 8 |= (8) = 8. Note therefore that | a |> 0 if a 6= 0, and | a |= 0 if a = 0. It also follows
that
|ab| = |a| |b|,
1
1
 
,
 =
a
|a|
 a  |a|
 
.
 =
b
|b|
|a| is also often called the modulus of a or the magnitude of a.
– 12 –
The absolute value has a geometric interpretation and it is the distance from the origin to the corresponding
point on the number line. Hence |a| < 4 means 4 < a < 4, i.e.
 |a| !
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a
4
0
4
The quantity |a  b| is the distance between the points a and b on the number line, whatever the numerical
values of a and b. The corresponding graphs, of course, may vary:
e.g.
if a < 0, b > 0
if a > 0, b > 0
 |a  b| !
 |a  b| !
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a
Example 16.
0
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a
0
b
b
Solve |x  2| < 5.
From the comments above the inequality means 5 < x  2 < 5. This is a double inequality which is
equivalent to 5 < x  2 and x  2 < 5. Adding 2 to both sides of the first of these inequalities gives
3 < x whereas adding 2 to the second inequality leads to x < 7. Hence the solution set consists of
all numbers greater than 3 and less than 7; i.e. 3 < x < 7.
As discussed above the result has a geometrical interpretation. The quantity |x  2| is the distance of
the point x on the number line from the point 2, and this distance is less than 5.
 5 !  5 !
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.
.
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3
0
2
7
Hence the point x must satisfy 3 < x < 7.
Inequalities containing a quotient must be solved with care as shown below.
2x
< 4.
3+x
The denominator can be removed by multiplying both sides by 3 + x. However, the quantity 3 + x
can be positive or negative and the sign of the inequality depends on which of those values it takes.
Both situations must be treated separately.
Example 17.
Solve
Case (i) 3 + x > 0. In this case multiplying both sides by 3 + x gives
2  x < 4(3 + x)
2  x < 12 + 4x
2  12 < 4x + x
i.e.
 10 < 5x
x > 2.
– 13 –
It is important to note that when x > 2 it follows that x + 2 > 0, (or x + 3 > 1) and hence the
necessary condition 3 + x > 0 is always satisfied.
Case (ii) 3 + x < 0. Now you can again multiply both sides by 3 + x, but after doing so the inequality
must be reversed. This time, therefore, you obtain
2  x > 4(3 + x)
2  x > 12 + 4x
10 > 5x
i.e.
x < 2.
However, case (ii) requires 3 + x < 0, i.e. x < 3, and the latter inequality and x < 2 are both
satisfied only if x < 3. The final solution to the original inequality, therefore, is the set of values of
x satisfying x < 3 and x > 2.
13. Solving inequalities using graphs To solve an inequality of the form f (x) < g(x) you can draw
graphs of f (x) and g(x) and investigate where the first lies below the second. Generally this will be achieved
by finding the points of intersection, which involves solving the equation f (x) = g(x) (which has been
considered in previous sections).
Example 18.
Solve graphically the inequality x2  4x + 1 < 3.
The graphs of y = x2  4x + 1 and y = 3 are shown on the diagram and you can see that the solution
of the inequality consists of all values of x between the points P and Q.
y
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3
x
Q 5
1 P
3
P and Q are determined by solving x2  4x + 1 = 3, which reduces to x2  4x  2 = 0. Using the
formula gives
p
p
p
p
4 ± ((4)2  4(1)(2))
4 ± 24
4±2 6
x=
=
=
= 2 ± 6,
2(1)
2
2
p
p
and so the solution of the inequality is the set of all x satisfying 2  6 < x < 2 + 6.
Example 19.
Use graphical means to solve the inequality in Example 17.
2x
and y = 4. The first graph is not easy to obtain
3+x
but from plotting points, using a graphics calculator or using methods discussed in a later module on
graph plotting you can find the graph is in two parts as shown below:
Here you need to draw the graphs of y =
– 14 –
y
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10
4
y=4
x
6
3
P
3
10
The straight line y = 4 has also been added to the figure. You can easily observe that the solution
of the inequality consists of all values of x to the right of P and to the left of the asymptote x = 3.
2x
= 4, (i.e. 2  x = 4(3 + x) = 12 + 4x, 5x = 10, x = 2), you know that P is
From solving
3+x
situated at x = 2. Hence the solution set is all values of x satisfying x < 3 and x > 2 (which
agrees with the earlier result!).
***Do Exercise 11.
(i)
6x + 5  x  5,
(iv)
2x2  2 < x2  x,
14. Partial fractions
degree n has the form
Solve the inequalities
(ii)
5 < x  4 < 2  x,
(v)
1<
3x  1
< 2,
x3
(iii)
x + 10 < 2x  5  x  3,
(vi)
|2x  5| < 9,
(vii)
|4  3x|  3.
Polynomial functions arise in many engineering situations. A polynomial of
an xn + an1 xn1 + · · · + a2 x2 + a1 x + a0 ,
where an , · · · a0 are all constants. The degree is the highest power occurring in the polynomial. Hence,
x2 + 1 is a polynomial of degree 2 and x4 + x2 + 1 is a polynomial of degree 4.
In solving problems it is sometimes necessary to consider a quotient of polynomials (i.e. one polynomial
divided by another). Such quotients can be split into a sum of much simpler fractions, called partial
fractions, and this splitting is very important, for instance, in determining solutions to some integration
problems and in applying Laplace transform methods to di↵erential equations.
Given a quotient of polynomials in which the denominator has degree d and the numerator has degree n
then the quotient, or fraction, is proper if n < d and is said to be improper if n  d.
It can be shown that any polynomial with real cocients can be factorised into a product of linear and
quadratic factors (with all coecients real). In particular, therefore, the denominator can be expressed as a
product of such factors.
For the moment consider only proper fractions. Next, factorise the denominator, if it is not already in the
required form, so that it consists of a product of linear and quadratic factors. Each factor produces a partial
fraction of a particular form according to the following rules:
– 15 –
factor in denominator
partial fraction
ax + b
(linear)
A
ax + b
(ax + b)2
(repeated linear)
A
B
+
ax + b (ax + b)2
ax2 + bx + c
(quadratic)
Ax + B
ax2 + bx + c
(ax2 + bx + c)2
(repeated quadratic)
Ax + B
Cx + D
+
ax2 + bx + c (ax2 + bx + c)2
In all cases the unknown constants A, B, C... are determined by evaluating an identity at some chosen values
of x or by equating coecients, or from using some combination of the two methods (as shown in the
examples below).
Example 20.
Express as partial fractions
x5
x2 + 2x  3
The denominator can be factorised to give (x + 3)(x  1), and then there is a partial fraction corresponding to each linear factor. Hence,
x5
A
B
=
+
,
(x + 3)(x  1)
x+3 x1
where A and B are constants. Rewriting the right-hand side using a common denominator gives
x5
A(x  1) + B(x + 3)
=
,
(x + 3)(x  1)
(x + 3)(x  1)
and multiplying both sides by (x + 3)(x  1), or by equating numerators (since the denominators are
identical), leads to the equation
x  5 = A(x  1) + B(x + 3).
The latter equation must hold for all values of x, and in these situations is often called an identity.
Substituting particular values for x into the equation gives linear equations in A and B (most of which
will be related to the others). Obviously you can save a lot of e↵ort by choosing suitable values for x
and with linear factors it is usually best to choose values which ensure the linear factors are zero, in
turn. For the above, therefore, the convenient choices are x = 1 and x = 3. Consider these in turn:
x=1
1  5 = A(0) + B(4)
4 = 4B,
x = 3
i.e.
B = 1.
 3  5 = A(3  1) + B(0)
8 = 4A,
Thus the partial fractions are
i.e.
A = 2.
2
1
x5
=

.
x2 + 2x  3
x+3 x1
– 16 –
Example 21.
Express as partial fractions
The denominator factorises again:
x
x2 + 6x + 9
x2 + 6x + 9 = (x + 3)2 , so
x
A
B
,
=
+
x2 + 6x + 9
x + 3 (x + 3)2
and multiplying both sides by (x + 3)2 leads to the identity
x = A(x + 3) + B.
With a linear factor it is convenient to choose x = 3, in which case
3 = A(0) + B,
i e.
B = 3.
The remaining coecient could be found by substituting a di↵erent value for x (for example, with
x = 0, 0 = A(3) + B, 3A = B = (3) = 3, A = 1). Alternatively you can compare
coecients of powers of x on both sides of the identity x = A(x + 3) + B.
For the x term, equating coecients gives 1 = A, as above. From the constant terms (i.e. x0 terms)
it follows that 0 = 3A + B, which is identically satisfied by A = 1, B = 3, as expected.
The final result, therefore, is
x
1
3
.
=

x2 + 6x + 9
x + 3 (x + 3)2
Let us now consider a more complicated example.
Example 22.
Express as partial fractions
3x2 + 5x + 2
(x2 + 4x + 5)(x  1)
The quadratic x2 + 4x + 5 does not split into a product of linear factors with real coecients, so the
denominator is already in its simplest form. The rules stated earlier tell you that for this example the
appropriate partial fractions are
3x2 + 5x + 2
(x2 + 4x + 5)(x  1)
=
A
Bx + C
+
.
x  1 x2 + 4x + 5
Writing the right-hand side in terms of the common denominator (x1)(x2 +4x+5) and then equating
numerators, or multiplying throughout by the product (x  1)(x2 + 4x + 5), gives
3x2 + 5x + 2 = A(x2 + 4x + 5) + (Bx + C)(x  1).
Choose x = 1 (to make the linear factor zero)
3 + 5 + 2 = A(1 + 4 + 5) + (Bx + C)(0),
10 = 10A,
A = 1.
Before equating coecients, multiplying out the right-hand side leads to
3x2 + 5x + 2 = A(x2 + 4x + 5) + Bx(x  1) + C(x  1)
= A(x2 + 4x + 5) + Bx2  Bx + Cx  C.
– 17 –
Equating coecients now gives
x2
3=A+B =1+B
(since A = 1)
i.e. B = 2.
x
5 = 4A  B + C = 4(1)  2 + C
C = 5  4 + 2 = 3.
All three constants have now been calculated but comparing constant terms on both sides of the
equation provides a check on your answers:
2 = 5A  C,
constant
and this equation is satisfied by A = 1, C = 3.
Examples 20-22 have involved partial fractions for proper fractions.
Attention is now fixed on improper fractions. Recall that the fraction (or quotient) is improper if the degree
of the polynomial in the numerator, n say, is greater than the degree of the polynomial in the denominator,
x4 + 1
d. For example, the fraction 2
is improper, since the numerator has degree 4 and the denominator
x +1
degree 2.
It can be shown that
x4 + 1
2
= x2  1 + 2
.
x2 + 1
x +1
The corresponding general result can be written
polynomial degree  (d  1)
polynomial degree n
= polynomial degree (n  d) +
.
polynomial degree d
orig. denominator degree d
Note that for the specific example and the general result the final term on the right-hand side is a proper
fraction. Considering the right-hand side of the general expression, both the first term and the numerator
in the final term can be determined by long division, but partial fractions would still usually be needed for
this final term. An alternative, and often simpler, approach is not to use long division but determine the
polynomial of degree (n  d) as part of the partial fraction process. This second method is illustrated below.
(2x + 1)(x + 1)
in terms of partial fractions.
2x  1
Here the numerator is quadratic, the denominator is linear and hence the general result above gives
Example 23.
Express
const
(2x + 1)(x + 1)
= linear +
.
2x  1
2x  1
Note that the denominator of the final term above is linear, and hence its numerator (which is of lower
order) must be a constant. To be more precise
C
(2x + 1)(x + 1)
= Ax + B +
,
2x  1
2x  1
and, after multiplying both sides by 2x  1, the identity obtained is
(2x + 1)(x + 1) = (Ax + B)(2x  1) + C.
Choosing x = 12 ,
✓ ✓ ◆
◆✓
◆
1
1
2
+1
+ 1 = (Ax + B)(0) + C
2
2
– 18 –
2
✓ ◆
3
= C,
2
C = 3.
Equating coecients in the identity:
x2
2 = 2A,
A = 1.
constant
1 = B + C,
B = C  1 = 2.
2 + 1 = A + 2B,
[check term in x :
Hence
satisfied.]
(2x + 1)(x + 1)
3
=x+2+
,
2x  1
2x  1
which is the reverse of Example 4.
***Do Exercise 12.
(i)
(iv)
1
,
(x  1)(x  2)
x2
,
(x  1)2
Express in partial fractions
(ii)
(v)
x
(x2 + 5x + 4)
,
(iii)
x3  1
.
(x + 1)(x + 3)
– 19 –
1
(x2 + 1)(x2 + 4)
,
Specimen Test 1
1.
Expand (x + 1)(3  2x)
2.
Simplify
3.
Express as a single fraction
4.
Solve
5.
Solve x2 + 3x + 1 = 0
6.
Solve the set of equations
1
x(x 3 )4
1
x2
2
3

x+1 x+2
3
=2
x+1
4x  y =
9
x + 3y = 1
7.
Solve 5  x  3(x  1)
8.
Solve |2 + x| < 3
9.
Express in partial fractions
x2 + 1
x(x  1)
– 20 –
Worked solutions to Exercises
1. (i)
(ii)
(iii)
(iv)
(v)
= x  1 + 6x  16 = 7x  17
= 2a + 3c  3a  2b + 2a = a  2b + 3c
= x(x  5) + 7(x  5) = x2 + 2x  35
= (x + 4)(x + 4) = x(x + 4) + 4(x + 4) = x2 + 8x + 16
= (x  1)(x(x  3) + 2(x  3))
= (x  1)(x2  x  6)
= x(x2  x  6)  1(x2  x  6)
= x3  x2  6x  x2 + x + 6
= x3  2x2  5x + 6
(vi)
= (y  3)(y  3)(y  3) = (y  3)(y(y  3)  3(y  3))
= y(y 2  6y + 9)  3(y 2  6y + 9)
= y 3  9y 2 + 27y  27
2. (i)
x2 + 7x + 12 = (x + 4)(x + 3)
[ (x  m)(x  n) = x2  (m + n)x + mn requires mn = 12, m + n = 7. Solution is m = 4, n = 3
or equivalently m = 3, n = 4.]
(ii)
x2  2x  3 = (x  3)(x + 1)
[as above : m = 3, n = 1].
(iii)
(iv)
(v)
x3  25x = x(x2  25) = x(x  5)(x + 5)
8 + 2x  x2 = (x2  2x  8) = (x  4)(x + 2) = (4  x)(x + 2)
2x2  3x  2 = (2x + 1)(x  2)
[(2x  m)(x  n) = 2x2  (2n + m)x + mn ) mn = 2, 2n + m = 3, leading to the solution n = 2,
m = 1.]
3. (i)
LCD of x and x  1 is x(x  1)
1(x)
x1+x
2x  1
1(x  1)
+
=
=
Solution =
x(x  1) (x  1)(x)
x(x  1)
x(x  1)
(ii)
LCD of (t + 2)2 and (t + 2) is (t + 2)2
1
2(t + 2)
2t + 5
1 + 2t + 4
Solution =
+
=
=
(t + 2)2
(t + 2)(t + 2)
(t + 2)2
(t + 2)2
(iii)
x+
(iv)
x2 + x  2 = (x + 2)(x  1)
1
x
1
= + 2
x2 + 1
1
x +1
LCD of 1 and x2 + 1 is x2 + 1
x(x2 + 1)
1
x3 + x + 1
Solution =
+ 2
=
2
1(x + 1)
x +1
x2 + 1
LCD of (x + 2)(x  1) and (x  1)2 is (x + 2)(x  1)2
2x(x  1)
3x(x + 2)
x(2(x  1)  3(x + 2))

=
(x + 2)(x  1)(x  1) (x  1)2 (x + 2)
(x + 2)(x  1)2
x(x  8)
x(x + 8)
=
=
(x + 2)(x  1)2
(x + 2)(x  1)2
Solution =
– 21 –
4. (i)
A
= r2 , r = ±
⇡
(ii)
v 2  u2 = 2as,
(iii)
(iv)
A
⇡
(positive if distance)
v 2  u2
2a
LCD of R1 and R2 is R1 R2
1
1(R2 )
1(R1 )
R2 + R1
=
+
=
R
R1 (R2 ) R2 (R1 )
R1 R2
R1 R2
R=
R1 + R2
s
l
T
l
T2
=
=
,
2⇡
g 4⇡ 2
g
l=
5. (i)
r
s=
gT 2
4⇡ 2
3
3
(81) 4 = (34 ) 4 = 33 = 27
1
5
(ii)
a 2 a2 = a 2
(iii)
(a 3 )3
1
a1
=
= a2 = 2
a
a
a
1
(iv)
(v)
1
2
7
2
= a 3 b 3 c(a2 c4 ) = a 3 b 3 c5
✓ 4 ◆ 13
1
a
1
= (a3 ) 3 = a1 =
1
a
a
(vi)
x(xa )2 = x(x2a ) = x2a+1
6. (i)
=
(ii)
7. (i)
p
p
p
p
(1 + 7)
6(1 + 7)
6
6(1 + 7)
p
p =
=
= (1 + 7)
17
6
(1  7) (1 + 7)
p
p
p
p
p
p
( 2  3)
2 3
1
p
p
p =
= 2+ 3
= p
2

3
( 2 + 3) ( 2  3)
2x = 4  3 = 1 ) x =
1
2
3
5
(ii)
2 = 5(x + 1) = 5x + 5, 5x = 3, x = 
(iii)
4  4x = 6x + 3, 2x = 1, x = 
(iv)
2=
8. (i)
x2  x  12 = (x  4)(x + 3) = 0 ) x = 4 or  3
(ii)
(iii)
(iv)
1
2
x
, 2(x + 1) = x, 2x + 2 = x, x = 2
x+1
x2  9x + 14 = (x  7)(x  2) = 0 ) x = 7 or 2
p
p
(9) ± ((9)2  4(1)(8))
9 ± 113
=
= 9.8151; or  0.8151.
x=
2
2
8 = (6  x)(x + 3) = 6x + 18  x2  3x = 18 + 3x  x2
) x2  3x  10 = (x  5)(x + 2) = 0 ) x = 5 or  2
– 22 –
9. (i) Adding the given equations leads to 6x = 6, x = 1. Substituting this value into the first given equation
then implies 2(1) + y = 3, so y = 1. Hence solution is x = 1, y = 1.
(ii) Eliminate y
3x + 4y = 12
(first equation)
18x + 4y = 18
(second equation ⇥ 2)
Subtracting second equation from first, 15x = 12  (18) = 30, which implies x = 2. Substituting
this solution back into the original first equation gives
3(2) + 4y = 12,
4y = 12  (6) = 18,
y=
18
9
= .
4
2
9
.
2
(iii) This time we choose to eliminate x
Hence solution is x = 2, y =
(first equation ⇥ 5)
(second equation ⇥ 4)
20x  10y = 25
(subtracting)
20x + 12y = 8
 22y = 25  (8) = 33
3
y=
2
3
Substituting back into first equation in question 4x  2( ) = 5,
2
3
1
Thus solution is x = , y =  .
2
2
4x = 5  3 = 2,
x=
1
.
2
10. (i) The graphs of y = 2  x and y = x3 are shown below. Clearly they intersect at x = 1 approximately. In
fact this is a solution since when x = 1 both 2  x and x3 are equal (= 1). No other solution is possible.
..... y
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Graph for Problem 10(i)
Graph for Problem 10(ii)
x
x
1
1
+1
+1
(ii) Graphs of y = x3 and y = x are displayed above. These graphs intersect at x = 0, x = 1, x = 1
approximately. It is obvious by substitution that all three of the latter values are the exact points of
intersection.
– 23 –
Algebraically, at the points of intersection x3 = x. Hence
x3  x = x(x2  1) = x(x  1)(x + 1) = 0
with solutions x = 0, +1, 1.
11. (i)
6x + 5  x  5
6x + 5  x  5  x  5  x  5
5x  10,
(ii) Equivalent to
5 < x  4
) x  2
x  4 < 2  x.
and
First inequality (add 4 to both sides)
Second inequality (add x + 4 to both sides)
which implies
 5 + 4 < x  4 + 4,
i.e.  1 < x.
(x  4) + x + 4 < (2  x) + x + 4
2x < 6,
i.e. x < 3.
For this example, therefore, the solution set is 1 < x < 3.
(iii) Equivalent to x + 10 < 2x  5 and 2x  5  x  3.
First inequality
Second inequality
(x + 10)  x + 5 < (2x  5)  x + 5, i.e. 15 < x.
(2x  5)  x + 5  (x  3)  x + 5, i.e. x  2.
A solution must satisfy both inequalities simultaneously and this is not possible. Therefore, there is NO
solution.
(iv)
(2x2  2)  x2 + x + 2 < (x2  x)  x2 + x + 2
x2 + x < 2
x2 + x  2 < 0
(x + 2)(x  1) < 0
For the product of brackets to be negative, the quantities in the two brackets must have opposite signs.
Hence x + 2 > 0 and x  1 < 0,
Thus x > 2 and x < 1,
or x + 2 < 0 and x  1 > 0.
or x < 2 and x > 1.
The second pair of inequalities is not possible, so the required solution is 2 < x < 1.
3x  1
3x  1
3x  1
and
< 2. The graphs of y =
, y = 1 and y = 2 are shown on
x3
x3
x3
the next page. The graphs show that the double inequality is never satisfied when x > 3. At points of
intersection
1(x  3) = 3x  1, 2x = 2, x = 1
(v) Equivalent to 1 <
3x  1 = 2(x  3) = 2x  6, x = 5
Therefore the solution set is 5 < x < 1.
– 24 –
y
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3
2
1
x
8
6
4
2
2
4
6
8
Graph for Problem 11(v)
(vi) The stated inequality implies
9 < 2x  5 < 9
which is equivalent to  9 < 2x  5 and 2x  5 < 9.
First inequality above gives  9 + 5 < 2x,
Second inequality implies 2x < 9 + 5,
x > 2.
x < 7.
2 < x < 7.
Thus the solution set is



5 9
5
[Alternatively you could divide the original inequality by 2 to give x   < . Distance of x from
2
2
2
9
must be less than leading to the same answer].
2
(vii) In this case the given inequality leads to the double inequality 3  4  3x  3.
The above is equivalent to
 3  4  3x
and
4  3x  3.
The first inequality gives 3x  4 + 3,
x
7
3
Second inequality implies 3x  4  3,
x
1
.
3
1
7
Hence the solution set is  x  .
3


 3



4
4
4
[Alternatively   x  1, or x    1, which means that the distance of x from must be less
3
3
3
than or equal to 1, giving the earlier answer).
– 25 –
12. (i)
1
A
B
A(x  2) + B(x  1)
=
+
=
(x  1)(x  2)
x1 x2
(x  1)(x  2)
comparing numerators gives the identity
choosing x = 1,
choosing x = 2,
Hence the answer is
1 = A(x  2) + B(x  1)
1 = A(1) ) A = 1
1 = B(1) ) B = 1
1
1

.
x2 x1
(ii) Denominator factorises to (x + 4)(x + 1), hence
x
A
B
A(x + 1) + B(x + 4)
=
+
=
(x + 4)(x + 1)
x+4 x+1
(x + 4)(x + 1)
comparing numerators leads to identity
Answer is
4
3
x+4

1
3
x+1
choosing
x = 1,
choosing
x = 4,
x = A(x + 1) + B(x + 4)
1
 1 = B(3), B = 
3
4
 4 = A(3), A =
3
.
(iii)
1
Ax + B
Cx + D
(Ax + B)(x2 + 4) + (Cx + D)(x2 + 1)
=
+
=
(x2 + 1)(x2 + 4)
x2 + 1
x2 + 4
(x2 + 1)(x2 + 4)
Equating numerators here leads to the identity
1 = (Ax + B)(x2 + 4) + (Cx + D)(x2 + 1)
Equating co-ecients of powers of x :x3 ,
A+C =0
2
x ,
B+D =0
x,
4A + C = 0
const.
1 = 4B + D
The only solution of the first and third equations is A = C = 0. From the second equation it follows
1
that D = B, so substituting into the fourth equation leads to 3B = 1, B = .
3
Hence the solution is
1
3
x2 + 1

1
3
x2 + 4
.
(iv) Numerator has degree 2, denominator has degree 2; hence division gives constant plus remainder. That
is to say
Bx + C
D
E
x2
=A+
=A+
+
(x  1)2
(x  1)2
x  1 (x  1)2
[Note that you do not need to calculate the intermediate constants B and C.]
Thus
x2
A(x  1)2 + D(x  1) + E
=
2
(x  1)
(x  1)2
which gives the identity x2 = A(x  1)2 + D(x  1) + E
– 26 –
SCHOOL OF MATHEMATICS
MATHEMATICS FOR PART I ENGINEERING
Self-paced Course
MODULE 2
TRIGONOMETRY
Module Topics
1. The measurement of angles and conversion between degrees and radians
2. The elementary trigonometric ratios and relations between them
3. The solution of general triangles
4. Trigonometric ratios for various angles and graphs of the elementary trigonometric functions
5. Other trigonometric ratios and relationships between them
6. Various useful formulae involving multiple angles
7. Trigonometric equations
This module is again self-contained. It is another long module but it should be mainly revision for you.
Work Scheme
Study the following sections, read carefully the worked Examples and do the stated Exercises. Solutions to
the Exercises are given towards the end of this Module, starting on p.21.
1. The measurement of angles Angles are usually measured either in degrees or in radians. There
are, of course, 360 degrees (360 ) in a complete revolution, 180 in a ‘straight line’ angle and 90 in a
right-angle.
By definition, if the angle ✓ in figure 1 is measured
in radians,
..........................
..........
.........
.......
.......
...... ............
......
...... ....
.....
.. .
.....
.
.
..... .... ...
.
..
.... ...... .
.
.
.
.
.
...
..
.
.
...
.
.
.
...
....
.
..
.
...
.
.
... ...
.......
...
..... .....
... ...
.
.
....
.
...
..
.
.. ...
.
.
.
...
...
............................................................. .........
...
...
...
..
.
...
...
...
...
...
...
...
...
...
.
.
.....
.
.....
.....
......
.....
.......
......
......
.........
.........................................
r
L
✓
Figure 1
Length of arc
L
= ,
radius
r
a ratio which is independent of the units in which both L and r are measured.
then
✓=
Since the length of the complete circumference of a circle of radius r is 2⇡r, in a complete revolution there
are 2⇡r/r = 2⇡ radians.
Conversion from degrees to radians and vice-versa is straightforward since 360 is equivalent to 2⇡ radians:
⇡
To convert from degrees to radians, multiply by
.
180
180
.
To convert from radians to degrees, multiply by
⇡
–1–
In trigonometry it is usually most convenient to work in radians and, therefore, you may assume that all
angles are measured in radians unless stated otherwise. If angles are measured in degrees then the usual
notation will be used (e.g. 45 ).
When using a pocket calculator, you must be very careful that it is operating in the correct—i.e. radian—
mode unless you specifically intend to use degrees. Remember, though, that an accuracy of four decimal
places in describing an angle in radians is equivalent to an accuracy of only two or three places in degrees.
The answers to many problems in this Module are given in both radians and degrees. These answers are
(hopefully!) accurate to four places. However, if you convert one answer directly to the other by using the
appropriate scale factor, because of rounding errors in the calculation you may well find an apparent error
in the last one or two decimal places.
Example 1. Convert 60 to radians.
⇡
⇡
60 is 60 ⇥
= radians.
180
3
⇡
Example 2. Convert radians to degrees.
6

⇡
⇡
180
= 30 .
radians is ⇥
6
6
⇡
One or two frequently used conversions are worth committing to memory:
30 =
⇡
,
6
45 =
⇡
,
4
60 =
⇡
,
3
90 =
⇡
,
2
180 = ⇡ ,
360 = 2⇡ .
(ii) 3⇡/2,
(iii) 7⇡/3.
For other cases, use your pocket calculator (or tables) if necessary.
***Do Exercise 1.
Express the following in degrees:
(i) ⇡/4,
(N.B. In all the Exercises in this module give numerical answers correct to 4 decimal places (i.e. 4 d.pl.),
using degrees or radians as appropriate.)
***Do Exercise 2.
Express the following in radians: (i) 18 ,
2. The three elementary trigonometric ratios
gent of an acute angle are defined to be:
Sine of angle =
side opposite
,
hypotenuse
Tangent of angle =
(ii) 35 ,
(iii) 350 ,
(iv) 420 .
In a right-angled triangle, the sine, cosine and tan-
Cosine of angle =
adjacent side
hypotenuse
side opposite
.
adjacent side
The notations used for the three ratios are respectively sin, cos and tan. Thus, in the right-angled triangle
shown in figure 2 (the side labelled c, opposite the right-angle is, of course, the hypotenuse):
b
sin ✓ = ,
c
cos ✓ =
a
b
, tan ✓ = .
c ...............
a
..
..... ....
.....
...
.....
.....
...
.....
.
.
.
...
.
...
.
.
.
.
...
....
.
.
.
.
...
....
.
.
.
...
.
...
.
.
.
.
...
....
.
.
.
.
...
...
.
.
.
.
...
....
.
.
.
.
...
....
.
.
.
.
...
...
.
.
.
.
....
........
.
.
.
.
.............
.... ...
.
.
.
.
.
.
..
...
.
...
.
.
.
.
.
.
.
...............................................................................................................
c
✓
a
Figure 2
–2–
b
Once again, assuming of course that all lengths are measured in the same units, these ratios are independent
of which units are used. This means that, given a right-angled triangle, if the length of any side and the
size of a second angle are known or if the lengths of any two sides are known then the triangle can be
‘solved’—that is to say all other sides and angles of the triangle can be found.
Example 3.
Solve the triangle shown in figure 2 if a = 2 cm and ✓ = ⇡/6.
Since tan ✓ = b/a, cos ✓ = a/c, it follows that b = a tan ✓ and c = a/ cos ✓ and hence, putting in the
given values for a and ✓, we see that
b = 2 tan
⇣⇡⌘
6
= 1.1547 cm
and
c=
2
= 2.3094 cm .
cos(⇡/6)
[Note that we do not need trigonometry to calculate the third angle—the sum of the angles of a
triangle is ⇡ and so the value of the third angle is ⇡/3. Also, having found b we could have used
Pythagoras’ theorem to give us c.]
Example 4.
Solve the triangle shown in figure 2 if a = 2 cm and b = 1 cm.
Firstly we use Pythagoras’ theorem to find c:
c=
p
a2 + b2 =
p
22 + 12 =
p
5 = 2.2361 cm .
Next, since
b
= .5 ,
a
the inverse tangent button on your pocket calculator gives ✓ = .4636 or 26.5651 . The remaining
angle is then ⇡  ( 12 ⇡ + .4636) = 1.1072 or 180  (90 + 26.5651 ) = 63.4349 .
tan ✓ =
3. Important relations between sin, cos and tan Referring again to figure 2 you can see that, by
Pythagoras’ theorem,
c2 = a2 + b2 .
However, a = c cos ✓ and b = c sin ✓ and hence
c2 = c2 cos2 ✓ + c2 sin2 ✓ ,
from which it follows that
cos2 ✓ + sin2 ✓ = 1 .
Notice the notation carefully. We write, for example, cos2 ✓ to mean (cos ✓)2 . [The quantity cos ✓2 has no
meaning since it could represent cos(✓2 ) or (cos ✓)2 ].
Example 5.
Verify the formula cos2 ✓ + sin2 ✓ = 1 for the angle ✓ in Example 4.
Since ✓ = .4636, your pocket calculators tell you that cos2 ✓ = .8, whilst sin2 ✓ = .2. Hence cos2 ✓ +
sin2 ✓ = 1.
Another relationship between cosine and sine is obvious from figure 2. The angles between a and c and
between b and c add to ⇡/2 since the third angle of the triangle is ⇡/2 and we know that the angles of a
triangle add to ⇡. It follows immediately that
⌘
 ✓ = sin ✓ ,
2
⌘
⇣⇡
 ✓ = cos ✓ .
sin
2
cos
⇣⇡
–3–
In addition it is easily shown that
tan ✓ =
since
tan ✓ =
sin ✓
,
cos ✓
b
b/c
sin ✓
=
=
.
a
a/c
cos ✓
You can check for yourself that the above three formulae involving ✓ are true for the numbers given in
Example 4.
The four major results in this section are identities—that is to say, they hold for all values of ✓. They are
important and you should be sure to memorise them. They are repeated as numbers 4, 5 and 7 of the ‘useful
results’ at the end of this module.
4. The solution of general triangles
lowing three pieces of information:
If you are given any triangle together with any one of the fol-
(a) the lengths of all three sides
(b) the lengths of two sides and the size of any angle
(c) the length of one side and the size of any two angles
you can still ‘solve’ the triangle completely although in the second case, unless the given angle lies between
the two sides, the solution may not be unique. To do this you can use two rules called respectively the
‘sine’ rule and the ‘cosine’ rule. In the notation used in figure 3 (which is the conventional notation with a
opposite A etc.), the ‘sine’ rule is written
a
b
c
=
=
sin A
sin B
sin C
and the ‘cosine’ rule gives the three formulae
a2 = b2 + c2  2bc cos A,
b2 = c2 + a2  2ca cos B,
c2 = a2 + b2  2ab cos C.
A
.....
.... ................
.........
.....
.........
.....
.........
....
.
.
.........
.
...
.........
.
.
.
.........
...
.
.
.........
.
...
.........
.
.
.
.........
...
.
.
.........
.
...
.........
.
.
.
.........
...
.
.
.........
.
...
.........
.
.
.
.........
...
.
.
.........
.
..
.
.
.
.
...........................................................................................................................................................................................................................
c
B
b
a
C
Figure 3
The table below shows you which formula you need for which situation.
Facts known
Rule to use
3 sides
2 sides and the angle between
cosine
2 sides and either of the other angles sine
1 side and any 2 angles
The sine and cosine rules appear on the Formula Sheet (and in the Data Book) but it is useful to remember
them. You do not need to know the proofs of the two rules but they are included below since they are very
straightforward.
–4–
A
.........
.... .. .....
.... . .........
.....
.... ..
.....
.... ..
.
.
.
.....
.
..
.....
....
.....
.
....
.
.
.
.
.....
.
...
.....
.
.
.
.
.....
..
...
.
.
.....
.
.
...
.....
.
.
.
.
.
.....
...
.
.
.....
.
.
.....
...
..
.
.
.
.....
.
...
.
.
.....
.
.
.
.....
...
.
.
.
.
.....
...
..
.
.....
.
.
.....
.
...
.
.
.
.
.....
.
...
.....
.
.
.
.
.....
.
...
.
.
.....
.
..
..........................................................................................................................................................................................
c
B
h
b
a
F
Figure 4
C
In figure 4, you can see that the height, h, of the triangle is given equivalently by h = c sin B and h = b sin C.
Hence c sin B = b sin C and
c
b
=
.
sin B
sin C
The remaining part of the sine rule is proved similarly or by symmetry.
To prove the cosine rule we see, again looking at figure 4,
c2 = h2 + (a  CF )2 = b2 sin2 C + (a  b cos C)2
= b2 sin2 C + a2 + b2 cos2 C  2ab cos C .
However, cos2 C + sin2 C = 1, and hence c2 = a2 + b2  2ab cos C.
Once again, the other two results follow similarly or by symmetry.
Example 7.
Solve the triangle ABC if a = 12, b = 8, c = 10.
Since all three sides but no angles are known, we must use a cosine formula.
c2 = a2 + b2  2ab cos C tells us that
a 2 + b2  c 2
2ab
144 + 64  100
=
2 ⇥ 12 ⇥ 8
108
=
= 0.5625 .
192
cos C =
Hence, using a calculator, C = .9734, (= 55.7711 ).
We could find the other angles in a similar way, but we now know one angle and so a simpler alternative
is to use the sine formula. Since
c
a
=
,
sin A
sin C
a sin C
sin A =
c
12 ⇥ .8268
=
= .9922 .
10
Hence A = 1.4455, (= 82.8192 ).
To find the remaining angle we could use the cosine or the sine formula, but obviously the easiest way is
to remember that the angles of a triangle add to ⇡, so that B = ⇡ AC = ⇡ 1.4455.9734 = .7227,
(= 41.4097 ). Notice that these results have been obtained retaining 10 digits throughout on a pocket
calculator. However, at each stage they have been stated to only four places. If you use the four place
answers (instead of the full 10 place ones) to perform your subsequent calculations, you will obtain
an answer which is certainly not accurate to four places. Try it and see!
–5–
Example 8.
Solve the triangle ABC if b = 2 cm, c = 1 cm, C = 20 .
Given two sides and non-included angle, we must use the sine formula. In some situations, and this
is one of them, the given information does not produce a unique triangle (that is one solution only)
and so it is always advisable to draw a reasonably accurate diagram before proceeding further. It is
easily seen that in this example two solutions are possible.
A
A ..................
.......
..... ...................
..........
.....
..........
.....
.
.
.
.
..........
.....
..........
..........
.....
.
.
.
.
..........
...
.
.
..........
.
.
...
.
.
.
... .................
.
...
.
..........
.
.
.
...
..........
...
.
.
.
..
...
.
.....................................................................................................................................................................................................
..... ..........
..... ..........
.....
..........
.....
..........
.....
..........
.....
..........
.....
..........
.....
..........
.....
... ..................
.....
..........
..
.....
....
..
.....
.......................................................................................................
20
B
Figure 5 (a)
20
C
B
C
Figure 5 (b)
2
b
sin C = sin 20 = .6840
c
1
Hence B = 43.1602 (= .7533), see figure 5(b), or B = 180  43.1602 = 136.8398 (= 2.3883), as
shown on figure 5(a). It follows that
sin B =
A = 180  B  C = 180  43.1602  20 = 116.8398 , (= 2.0392)
or 180  136.8398  20 = 23.1602 , (= .4042).
Finally we may find a from the cosine formula.
p
p
a = b2 + c2  2bc cos A = 22 + 12  2 ⇥ 2 ⇥ 1 ⇥ cos 116.8398 = 2.60883 cm
p
or 22 + 12  2 ⇥ 2 ⇥ 1 ⇥ cos 23.1602 = 1.1499 cm.
5. The area of a triangle The area of a triangle is half that of a rectangle on the same base—that is
to say, it is one half of the length of its base times its height.
Hence, using figure 4 and remembering that h = b sin C,
Area of 4ABC =
1
1
1
⇥ base ⇥ height = ⇥ a ⇥ h = ab sin C .
2
2
2
In a similar way it can also be shown that
Area of 4ABC =
1
1
bc sin A = ca sin B .
2
2
Although these formulae could be remembered and used it is usually easier to proceed from first principles
by calculating the height directly (see example below).
Example 9.
Find the area of the triangle ABC if b = 1 cm, c = 2 cm, C = 20 .
This is the same triangle as in Example 8 but with the lengths of the given sides interchanged. But
this time figure 6 confirms the answer is unique.
A
....................................
...................
....
...................
... ..................
..................
.
.
.
.
.
.
..........
.
.
.
.
.
.
.
.
.
.
.
.
...
..........
............
..
...
..................
.................................................................................................................................................................................................................................
20
B
C
Figure 6
sin B =
b
1
sin C = sin 20 = .1710
c
2
–6–
Hence B = 9.8466 . It follows that
A = 180  B  C = 180  9.8466  20 = 150.1534 .
This enables us to find a from the cosine formula:
p
p
a = b2 + c2  2bc cos A = 12 + 22  2 ⇥ 1 ⇥ 2 ⇥ cos 150.1534 .
Using a calculator we deduce a = 2.9102 cm.
The height h is easily found from
h
= sin 20 , and hence the area of the triangle is given by
1
1
⇥a⇥h
2
1
= ⇥ 2.9102 ⇥ 1 ⇥ sin 20
2
= .4977 cm2 .
Area of 4ABC =
***Do Exercise 3.
Solve the right-angled triangle shown below in each of the following cases:
A
.......
..... .
..... ....
.....
.
.
.
...
.
....
...
.....
.....
...
.....
.
.
.
...
.
....
.
.
.
...
.
....
.
.
.
.
...
...
.
.
.
.
...
....
.
.
.
.
...
....
.
.
.
.
...
...
.
.
.
.
...
....
.
.
.
.
....
....
.
.
.
.
.
.
.
.
.
.
.
.
.
.
. ...
...
.
.
.
.
.
.
.
......................................................................................................
c
a
B
(i) a = 1 m, B = ⇡/3;
***Do Exercise 4.

(ii) a = 2 m, B = 10 ;
b
C
(iii) a = 3 m, b = 1 m.
Solve the triangle shown below in the following cases:

(i) c = 5.1 m, B = 47 , A = 70 ;
(ii) a = 12.1 m, b = 17.2 m, c = 18.4 m;
(iii) b = 7 m, c = 3 m, C = 10 .
A
.....................
.........
....
.........
.....
.........
.....
.........
....
.
.
.........
.
...
.........
.
.
.
.........
...
.
.
.........
.
...
.........
.
.
.
.........
...
.
.
.........
.
...
.........
.
.
.
.........
...
.
.
.........
.
...
....
.
.
.
........................................................................................................................................................................................................
c
B
***Do Exercise 5.
b
a
C
Find the area of each of the triangles in Exercise 4.
6. Trigonometric ratios for special angles up to ⇡/2 The angles 0 , 30 , 45 , 60 and 90 (i.e. 0,
⇡/6, ⇡/4, ⇡/3 and ⇡/2) occur frequently in trigonometric work and it is helpful to memorise the corresponding
values of sine, cosine and tangent for each of these angles. Alternatively, you should be able to sketch the
appropriate triangle and deduce the value of the ratios (see below), or use your calculator!
(a) Angle 0
A
.......
................................ ...
...............................
...
................................
................................
...
................................
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
...
.
.
.
...
....................
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
..
.
.
.
.
......
...................................................................................................................................................................................................................................................................................................................
c
✓
B
a
Figure 7
–7–
b
C
In this case the triangle looks like figure 7 as ✓ approaches 0. Clearly b approaches 0 and as this happens a
a
b
b
and c become equal. However sin ✓ = , cos ✓ = , tan ✓ = , and so we conclude that
c
c
a
sin 0 = 0,
cos 0 = 1,
tan 0 = 0.
(b) Angle ⇡/6 or ⇡/3 When ✓ = ⇡/3 the 4ABC forms half of an equilateralp triangle—see
p figure 8.
Hence, if AB = 2 units, BC = 1 unit and Pythagoras’ theorem tells us that AC = 22  12 = 3 units.
A
.....
.........
... .. ..
... .... ...
.
.
... .... ...
.
..
...
..
...
...
...
.
...
.
...
.
...
.
.
...
.
.
.
.
.
.
.
.
...
... ....................... ...
.
....
...
...
..
.
...
.
.
...
.
.
.
.
.
.
.
...
..
.
.
.
.
...
...
....
..
.
.
.
.
...
..
...
.
.
...
.
.
.
.
.
.
...
...
....
..
.
.
...
.
.
.
..
.
.
.
.
...
..
...
.
.
.
.
...
.
.
.
.
........
...
....
.. ........
.
.
.
.
...
...
.
.
...
.
.
.
...
...
.
.
.
.
.
.
.
...
..
...
...
.
...
............................................................................................... .... .... .... .... .... .... .... .... .... .... ...
⇡/6
p
2
3
2
⇡/3
1
B
It follows that
1
⇡
cos = = .5,
3
2
1
C
Figure 8
p
3
⇡
sin =
= .8660,
3
2
tan
⇡ p
= 3 = 1.7321.
3
From the same triangle we can also read o↵ the results for an angle of ⇡/6:
p
⇡
3
cos =
= .8660,
6
2
(c) Angle ⇡/4
sin
⇡
1
= = .5,
6
2
tan
⇡
1
= p = .5774.
6
3
This time the picture is as in figure 9, with a = b = 1 and c =
p
2.
A
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p
2
1
1
Figure 9
C
⇡/4
B
It follows that:
cos
⇡
1
= p = .7071,
4
2
sin
⇡
1
= p = .7071,
4
2
tan
⇡
= 1.
4
(d) Angle ⇡/2 This case is much the same as (a), but with the triangle the other way round, as shown in
figure 10. This time, however, it is the angle at A which approaches 0 so that a approaches zero length and
b and c eventually become equal. Thus
cos
⇡
= 0,
2
–8–
sin
⇡
= 1.
2
As ✓ becomes nearer and nearer to ⇡/2 you can also see that tan(⇡/2) gets larger and larger. We write
tan ✓ ! 1 as ✓ !
⇡
.
2
If you calculate tan(⇡/2), your calculator will show an error.
A
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c
b
✓
B a C
Figure 10
7. Angles between ⇡/2 and ⇡ cos ✓, sin ✓ and tan ✓ have so far only been defined when ✓ is an acute
angle, although in using your calculator to evaluate some of your results you have used larger angles.
If ✓ lies between ⇡/2 and ⇡, figure 11 shows what we must do.
A.....
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... ............. ..........
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....... .... .... .... .... .... .... ...... .... .... ..................................................................................................
.....
F
↵ ✓
B
Figure 11
C
x
Choose BC to be along the positive x-axis of a coordinate system with B at the origin.
We define the three trigonometrical ratios cos ✓, sin ✓ and tan ✓ to have the same numerical values respectively
as those of cos ↵, sin ↵ and tan ↵ but, since BF is in the negative direction of the x-axis (shown in the figure
along the direction of BC), we put a minus sign in front of each of the ratios which involve this length.
Notice that other distances are still treated as positive. Thus:
cos ✓ = 
BF
=  cos ↵ ,
BA
sin ✓ =
FA
= + sin ↵ ,
BA
tan ✓ = 
FA
=  tan ↵ .
BF
Since ↵ + ✓ = ⇡, it follows that ↵ = ⇡  ✓ and the above formulae can be written
cos ✓ =  cos(⇡  ✓),
sin ✓ = sin(⇡  ✓),
–9–
tan ✓ =  tan(⇡  ✓).
8. Angles between ⇡ and 2⇡ and > 2⇡ or < 0
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.....
✓
F
C
↵ B
x
✓
B
C
F
↵
x
A
A
Figure 13
Figure 12
To find the trigonometric ratios for these angles, we simply extend what we did in section 7. Let us again
assume that BC is along the positive x-axis of a coordinate system with B at the origin. Then the magnitudes
of the trigonometric ratios for the angles AB̂C shown in figures 12 and 13 are the same as for the acute
angles that AB makes with this x-axis, but an appropriate sign must also be attached. As in section 7, this
sign is calculated by looking at the signs attached to the lines BF and AF (note AB is always assumed to
have a positive sign). If BF is to the right, it is positive; if it is to the left, it is negative. If AF is up, it is
positive; if it is down, it is negative.
Thus in figure 12 BF is to the left and AF is down: hence cosines will be negative, sines negative and
tangents (negative over negative) positive whilst, in figure 13, BF is to the right and AF down: hence
cosines will be positive, sines negative and tangents negative.
Finally, if the angle is in excess of 2⇡ we have gone full circle (at least once!) and we simply knock o↵ as
many whole number multiples of 2⇡ (complete revolutions) as we need in order to obtain an angle between
0 and 2⇡ and find the appropriate trigonometric ratio for this. Thus, for example,
sin
⇣
⇡⌘
⇡
9⇡
= sin 2 ⇥ 2⇡ +
= sin = 1 .
2
2
2
Likewise, if the angle is less than 0, we have to add as many whole number multiples of 2⇡ as is necessary
to put the angle into the range 0 to 2⇡. Thus
cos(✓ + 2n⇡) = cos ✓,
sin(✓ + 2n⇡) = sin ✓,
tan(✓ + 2n⇡) = tan ✓,
if n is an integer. In fact tan(✓ + n⇡) = tan ✓, but that you will see later!
Most people find it easiest to remember which signs are positive by drawing the following diagram (figure 14)
in which A stands for ‘All’, S stands for ‘Sine’, T stands for ‘Tangent’ and C stands for ‘Cosine’. All other
signs are negative.
⇡/2
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S
⇡
T
A
C
3⇡/2
Figure 14
– 10 –
0, 2⇡,...
Notice and remember the very useful results that:
cos(✓) = cos ✓,
Example 10.
sin(✓) =  sin ✓,
tan(✓) =  tan ✓.
Evaluate sin(2⇡/3).
2⇡/3 lies in the range from ⇡/2 to ⇡. Figure 14 therefore shows that sin(2⇡/3) is positive and so
sin(2⇡/3) = + sin(⇡  2⇡/3)
= sin(⇡/3). Using the results obtained with figure 8 it follows that
p
sin(2⇡/3) = sin(⇡/3) = 3/2 = .8660. Check that your calculator gives the latter value for both
sin(2⇡/3) and sin(⇡/3).
Example 11.
Evaluate cos 2.
2 is larger than ⇡/2 (= 1.5708) and smaller than ⇡ (= 3.1416). Hence it follows from figure 14 that
cos 2 is the negative of the cosine of ⇡  2 (= 1.1416). Using your calculator verify that the values for
cos 2 and  cos(⇡  2) are both equal to .4161
Example 12.
Evaluate cos(190 ).
190 (= 3.3161) is between ⇡ and 3⇡/2. Hence, looking at figure 14, its cosine is the negative of the
cosine of 3.3161⇡. Check the values given by your calculator for both cos(3.3161) and  cos(3.3161
⇡). You’ll find that they are both the same, namely .9848.
Example 13.
Evaluate tan 5.
5 is bigger than 3⇡/2 (= 4.7123) and smaller than 2⇡ (= 6.2831). Hence, from figure 14, tan 5 =
 tan(2⇡  5). Again, your calculator should give the same answer for both tan 5 and  tan(2⇡  5),
namely 3.3805.
It may be that you are more used to using degrees than radians—in that case you may prefer to check
the range in which 5 lies by converting it to degrees (286.4789 ) and noting that this lies between
270 and 360 .
Example 14.
Evaluate sin(33⇡/4).
The angle is outside the range 0 to 2⇡ and so we have to subtract whole number
multiples of 2⇡ to
p
bring it into range. 33⇡/4 = 4 ⇥ 2⇡ + 14 ⇡. Hence sin(33⇡/4) = sin(⇡/4) = 1/ 2.
Check that the answer (.7071) given by your calculator agrees whichever way you do the calculation—
that is to say, if your calculator allows you do this type of calculation directly!
Example 15.
Evaluate tan(⇡/4).
tan(⇡/4) =  tan(⇡/4) = 1.
Example 16.
Evaluate sin(4⇡).
4⇡ is equivalent to 0 for our purposes—two negative revolutions takes us back to where we started—
and so sin(4⇡) = sin 0 = 0.
– 11 –
9. Graphs of trigonometric functions With a graph-plotting calculator you can very easily see what
the graphs of cos ✓, sin ✓ and tan ✓ look like for ✓ in the range (2⇡, 2⇡). If you don’t have such a calculator,
the previous sections give sucient values of each function for you to be able to make a good plot. The three
graphs are:
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+1
✓
2⇡
3⇡/2
⇡
⇡/2
⇡/2
⇡
3⇡/2
2⇡
1
Figure 15: Graph of cos ✓
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.
+1
✓
2⇡
3⇡/2 ⇡
⇡/2
⇡/2
⇡
3⇡/2
2⇡
1
Figure 16: Graph of sin ✓
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2
1
✓
2⇡
3⇡/2 ⇡
⇡/2
⇡/2
⇡
3⇡/2
2⇡
1
2
Figure 17: Graph of tan ✓
As we saw in previous sections, outside the range (0, 2⇡) each curve repeats itself every 2⇡, the extensions
joining on smoothly to the curves shown in each case. We say that these curves ‘have period 2⇡’. In fact,
the tangent curve repeats itself every ⇡—has period ⇡—a fact which means that we have to take particular
– 12 –
care if we are asked to find the angle in the range (0, 2⇡) which has a particular value for its tangent: there
are two possible answers! We shall discuss this in more detail later.
Notice that, if the cosine curve is displaced to the right by ⇡/2, the sine curve is obtained whereas if the sine
curve is displaced to the right by ⇡/2 then the negative of the cosine curve is derived. This means that
sin
⇣⇡
2
⌘
+ ✓ = cos ✓ ,
cos
⇣⇡
2
⌘
+ ✓ =  sin ✓ ,
which are two useful results.
Notice also that the tangent curve becomes infinite (either in the positive or negative direction) at the points
✓ = 3⇡/2, ⇡/2 , ⇡/2 , 3⇡/2.
***Do Exercise 6. (i) Use your calculator to find the sine, cosine and tangent of each of the following:
(a) 140 , (b) 7, (c) 700 .
(ii) Use your calculator to find (a) cos(20 ), (b) sin(250 ), (c) tan(450 ), (d) sin(10).
(iii) Show by direct calculation that (a) cos2 120 + sin2 120 = 1, (b) cos2 410  sin2 410 = cos 820 , (c)
2 sin 1.5 cos 1.5 = sin(3).
(iv) Given that ⇡/2  ↵  ⇡ and sin ↵ = 24/25, find cos ↵ and tan ↵ without using your calculator (check
your answers with the calculator).
10. Other trigonometric ratios There are three other ratios that you will meet. These are the reciprocals of (i.e. one over) the cosine, sine and tangent and are called respectively the secant, cosecant and
cotangent. In the standard notation:
sec ✓ ⌘
1
,
cos ✓
cosec ✓ ⌘
1
,
sin ✓
cot ✓ ⌘
1
.
tan ✓
Be very careful of your notation. cos1 ✓ is used to mean the angle whose cosine is ✓ not 1/ cos ✓, and
similarly for sin1 ✓ and tan1 ✓. Thus, for example, sec ✓ = (cos ✓)1 , not cos1 ✓. cosn ✓ means (cos ✓)n
whenever n 6= 1 but not when n = 1. Awkward, isn’t it?
It is worth pointing out that the above diculty with notation is the main reason why the angle whose cosine
is ✓ is often denoted by arc cos ✓, instead of cos1 ✓.
Sketch for yourselves the graphs of the functions sec ✓, cosec ✓ and cot ✓, noticing in particular that, whilst
1  cos ✓  1 and 1  sin ✓  1 for all values of ✓, the functions sec ✓ and cosec ✓ are always either  1
or  1. Also notice that, unlike tan ✓, cot ✓ is infinite at 0 and ⇡ but finite at ⇡/2 and 3⇡/2.
Example 17.
Evaluate sec 3.14, cosec 3.14 and tan 3.14 to 6 d.pl.
sec 3.14 = 1/ cos 3.14 = 1/  .999999 = 1.000001.
cosec 3.14 = 1/ sin 3.14 = 1/.001593 = 627.883190.
cot 3.14 = 1/ tan 3.14 = 1/  .001593 = 627.882394.
Notice how nearly numerically equal the values of sin 3.14 and tan 3.14 (and therefore of cosec 3.14
and cot 3.14) are for these values of ✓ so close to ⇡.
11. Two further identities You will remember that we proved in section 3 that cos2 ✓ + sin2 ✓ = 1 and
that tan ✓ = sin ✓/ cos ✓ for all values of ✓. If we divide the first equation by cos2 ✓ we find
cos2 ✓
sin2 ✓
1
+
=
,
2
2
cos ✓ cos ✓
cos2 ✓
– 13 –
and making use of the second equation then gives
1 + tan2 ✓ = sec2 ✓ .
In a similar way (dividing the first equation by sin2 ✓ instead of cos2 ✓), we find
cot2 ✓ + 1 = cosec2 ✓ .
These two formulae are useful and, if possible, should be remembered, although they do appear on the
Formula Sheet (and in the Data Book). The important point is to know that formulae exist connecting
tan2 ✓ to sec2 ✓ (and cot2 ✓ to cosec2 ✓).
12. Trigonometric ratios for sums and di↵erences of angles It is often necessary to be able to express cos 2✓ in terms of cos ✓, a formula which is a special case of that which gives cos(A + B) in terms of
the cosines and sines of A and B alone.
From figure 18 we can easily show that
cos(A + B) = cos A cos B  sin A sin B .
In the figure, the perpendicular, AF , from A onto BC has F as its foot and the extension of BC is of length
e. In addition, the angle AĈF is of size A + B since it is an exterior angle of the triangle ABC.
A
......
..........
............ ..
....... ..... ..
....... .........
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.
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..
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....................................................................................................................................................................................................................................
A
y
b
h
x
B
B
A+B
a
e
C
.
.......................................................................................................
F
.
...................................................................................................
.
l
Figure 18
In triangle ACF ,
cos(A + B) =
la
e
=
.
b
b
However, from triangle ABF ,
l = (x + y) cos B ,
where x = a cos B and y = b cos A. It follows that
(x + y) cos B  a
(a cos B + b cos A) cos B  a
=
b
b
a(cos2 B  1)
a
=
+ cos A cos B =  sin2 B + cos A cos B
b
b
cos(A + B) =
and, since the sine rule tells us that
we can replace
sin B
sin A
=
,
b
a
a
sin B by sin A and conclude that
b
cos(A + B) = cos A cos B  sin A sin B
– 14 –
as required.
If in this formula we replace A by (⇡/2)  A, we obtain a corresponding formula for sin(A  B) and if in the
two formulae we now have we replace B by B, we obtain two more formulae for cos(A  B) and sin(A + B).
From your point of view, the proofs are relatively unimportant. What matters is that you know the following
formulae are on the Formula Sheet (and in the Data Book) and you are able to use them.
cos(A + B) = cos A cos B  sin A sin B
sin(A + B) = sin A cos B + cos A sin B
cos(A  B) = cos A cos B + sin A sin B
sin(A  B) = sin A cos B  cos A sin B .
Notice the somewhat unexpected arrangement of the signs in the above equations.
The corresponding formulae for tan(A ± B) are less important since they can be obtained, if needed, from
those for cos(A ± B) and sin(A ± B) by division. For example
tan(A + B) =
sin(A + B)
,
cos(A + B)
and the expressions for sin(A + B) and cos(A + B) can then be used.
These formulae for sums and di↵erences of angles are very useful in determining a number of the formulae
quoted earlier. For example, using the formula for cos(A  B), we obtain
cos(⇡  ✓) = cos ⇡ cos ✓ + sin ⇡ sin ✓
= 1 ⇥ cos ✓ + 0 ⇥ sin ✓ =  cos ✓ .
13. Double angle formulae These are the formulae mentioned at the beginning of the previous note.
If we put both A and B equal to ✓ in the formulae for cos(A + B) and sin(A + B), we obtain
cos 2✓ = cos2 ✓  sin2 ✓ ,
sin 2✓ = 2 sin ✓ cos ✓ .
These results are highly important and must be committed to memory. The former may be written (using
sin2 ✓ + cos2 ✓ = 1) in the alternative forms:
cos 2✓ = cos2 ✓  sin2 ✓ = 2 cos2 ✓  1 = 1  2 sin2 ✓ .
These alternative identities are also very important and you must either learn them or be able to derive
them quickly.
Example 18.
Prove that sin(C  D) + sin(C + D) = 2 sin C cos D.
sin(C  D) = sin C cos D  cos C sin D and
sin(C + D) = sin C cos D + cos C sin D .
By adding these two equations we obtain the required result.
– 15 –
14. Sum and di↵erence formulae If in Example 18 we put C = 12 (A + B) and D = 12 (A  B), we
obtain
1
1
sin A + sin B = 2 sin (A + B) cos (A  B) .
2
2
This is one of a set of four formulae which are useful in some integration problems in calculus, but not worth
remembering—just know where to find them (on the Formula Sheet and in the Data Book) or how to deduce
them if you need them! The other three are:
1
1
(A + B) sin (A  B) ,
2
2
1
1
cos A + cos B = 2 cos (A + B) cos (A  B) ,
2
2
1
1
cos A  cos B = 2 sin (A + B) sin (A  B) ,
2
2
sin A  sin B = 2 cos
Notice the negative sign in the last of these.
Example 19.
(a)
(b)
Express the following as products: (a) sin 6✓ + sin 4✓,
(b) cos 18✓  cos 2✓.
1
1
(6✓ + 4✓) cos (6✓  4✓) = 2 sin 5✓ cos ✓ .
2
2
1
1
cos 18✓  cos 2✓ = 2 sin (18✓ + 2✓) sin (18✓  2✓) = 2 sin 10✓ sin 8✓ .
2
2
sin 6✓ + sin 4✓ = 2 sin
Example 20.
Express cos 7✓ cos 3✓ as a sum or di↵erence of two cosines.
Here we use the formula
cos A + cos B = 2 cos
1
1
(A + B) cos (A  B) .
2
2
We put 7✓ = 12 (A + B) and 3✓ = 12 (A  B) so that, after multiplying both equations by 2, we obtain
A + B = 14✓ and A  B = 6✓. Adding these two equations and dividing by two we see that A = 10✓.
It follows that B = 4✓. Thus
cos 7✓ cos 3✓ =
***Do Exercise 7.
(ii) tan(A + B) =
1
(cos 10✓ + cos 4✓) .
2
Prove the following identities:
tan A + tan B
,
1  tan A tan B
(iii) tan 2✓ =
(i) (sin ✓ + cos ✓)2 = 1 + sin 2✓,
2 tan ✓
,
1  tan2 ✓
(iv)
tan ✓ + cot ✓
= cosec ✓.
sec ✓
***Do Exercise 8.
↵).
If sin ↵ = 5/13, where ↵ is acute, find tan ↵, cot ↵ and cosec ↵ (without calculating
***Do Exercise 9.
and b.
If tan ✓ = a/b, where ✓ is acute, find expressions for sin ✓ and sec2 ✓ in terms of a
***Do Exercise 10. By writing 3✓ = 2✓ + ✓, and then using the identities for sums of angles and double
angles, show that (i) sin 3✓ = 3 sin ✓  4 sin3 ✓, (ii) cos 3✓ = 4 cos3 ✓  3 cos ✓.
***Do Exercise 11. Using the formulae for sums and di↵erences of angles show that✓
◆
⇣⇡
⌘
⇣⇡
⌘
3⇡
(i) sin
+ ✓ = cos ✓, (ii) cos
+ ✓ =  sin ✓, (iii) sin (⇡  ✓) = sin ✓, (iv) cos
+ ✓ = sin ✓.
2
2
2
– 16 –
***Do Exercise 12.
(i) sin 4✓ + sin ✓,
Express as products of sines and/or cosines:
(ii) sin 8✓  sin 6✓,
***Do Exercise 13.
(i) 2 sin 6✓ cos 2✓,
(iii) cos 12✓ + cos 10✓,
(iv) cos ✓  cos 2✓,
(v) sin (⇡/3) + sin (⇡/4).
Express as sums or di↵erences of trigonometric functions:
(ii) 2 cos 5✓ cos 3✓,
(iii) 2 cos 4✓ sin ✓,
(iv) 2 sin 7✓ sin 5✓.
15. Simple trigonometric equations It is important to realise that, since sin ✓ and cos ✓ repeat themselves every 2⇡, even a simple equation like
1
cos ✓ =
2
is satisfied by an infinite number of values of ✓. By referring back to figure 15, or using your graphics
calculator, you will see that not only is 13 ⇡ a solution, but so are 2⇡ + 13 ⇡, 4⇡ + 13 ⇡, 2⇡ + 13 ⇡ and so on.
In fact it’s worse than this—even between ✓ = 0 and ✓ = 2⇡ there are two di↵erent solutions, as you can
again see by looking at the graph of cos ✓ in figure 15 (or on your graphics calculator) or by using the ‘SATC’
diagram in figure 14. Either method shows you that 2⇡  13 ⇡ = 53 ⇡ is also a solution. We can therefore write
the general solution of this equation as
✓=
1
⇡ + 2n⇡ ,
3
or
✓=
5
⇡ + 2n⇡ ,
3
where n is any whole number (positive, negative or zero).
It makes things simpler, to define a unique inverse to each of the trigonometric functions and then calculate
all other solutions to equations like the one above from it. Pocket calculators have the same problem. They
cannot come up with several di↵erent inverses at the same time and, instead, have to choose one and leave
you to calculate the rest.
Calculators use the following convention, and you should do the same: cos1 x is defined to be that solution
of cos ✓ = x which lies in the range 0  ✓  ⇡, sin1 x is that solution of sin ✓ = x which lies in the range
 12 ⇡  ✓  12 ⇡ and tan1 x is that solution of tan ✓ = x which lies in the range  12 ⇡  ✓  12 ⇡. Notice the
ranges are not the same for each function: this is necessary in order to ensure that each possible value is
covered once and once only as can be seen from figures 15-17.
Example 21.
Find the general solution of cos ✓ = .3.
Your calculator will tell you that cos1 (.3) is 1.8755. This is the solution of the equation which lies
in the range [0, ⇡]—in fact between 12 ⇡ and ⇡. The graph then shows you that 2⇡  1.8755 = 4.4077
is also a solution, giving you a second answer between 0 and 2⇡. The general solution is then
✓ = 1.8755 + 2n⇡ ,
or
✓ = 4.4077 + 2n⇡
for any integer n (positive, negative or zero).
p
Example 22. Find the general solution of sin ✓ =  12 2.
p
sin1 ( 12 2) =  14 ⇡ = .7854 since this is the appropriate value lying in the range [ 12 ⇡, 12 ⇡]. Your
calculator will give you the same answer. As you can see from the graph, the two solutions lying
between 0 and 2⇡ are ✓ = ⇡ + 14 ⇡ and 2⇡  14 ⇡, i.e. ✓ = 54 ⇡ and ✓ = 74 ⇡. The general solution is then
✓=
5⇡
+ 2n⇡ ,
4
or
for integer n.
– 17 –
✓=
7⇡
+ 2n⇡
4
Example 23.
Find the solution of tan ✓ = 1.6 which lies between ⇡ and 2⇡.
Your calculator gives you tan1 1.6 = 1.0122. This is the value which lies in [ 12 ⇡, 12 ⇡]. The tangent
function (see the graph) repeats itself every ⇡. The general solution of the equation is tan1 ✓ =
1.0122 + n⇡ and the solution we require, obtained by taking n = 1, is thus 4.1538.
Some trigonometric equations can readily be solved using the double angle formulae (section 13) and simple
algebra.
Example 24.
0  ✓  2⇡.
Find the solutions to the equation 6 cos 2✓ + 5 cos ✓ + 4 = 0 which lie in the range
The double angle formulae tell us that cos 2✓ = 2 cos2 ✓  1 and hence the given equation can be
written
6(2 cos2 ✓  1) + 5 cos ✓ + 4 = 0
or
12 cos2 ✓ + 5 cos ✓  2 = 0 ,
which is a quadratic equation in cos ✓. This equation factorises to give
(3 cos ✓ + 2)(4 cos ✓  1) = 0
and hence either 3 cos ✓ + 2 = 0 or 4 cos ✓  1 = 0. It follows that cos ✓ = 2/3 or cos ✓ = 1/4 (results
which could also have been found from the quadratic equation using the formula).
According to the calculator, these solutions give
✓ = cos
1
✓
2

3
◆
= 2.3005
or
✓ = cos
1
✓ ◆
1
= 1.3181 ,
4
and from figure 14, or figure 15, it is clear that each of the above solutions leads to a corresponding
second solution between 0 and 2⇡. Hence there are four solutions between 0 and 2⇡:
✓ = 2.3005 ,
✓ = 2⇡  2.3005 = 3.9827 ,
✓ = 1.3181 ,
✓ = 2⇡  1.3181 = 4.9651 .
Equations of the form sin A ± sin B = 0 or
16. Slightly more dicult trigonometric equations
cos A ± cos B = 0 can be solved by using the appropriate sum or di↵erence formula.
Example 25.
Find the roots of the equation sin 7✓ = sin ✓ for 0  ✓  12 ⇡.
Here we use the formula for the di↵erence of two sines:
1
1
sin 7✓  sin ✓ = 2 cos (7✓ + ✓) sin (7✓  ✓) = 2 cos 4✓ sin 3✓ .
2
2
Hence, equating this to zero, we see that either cos 4✓ = 0 or sin 3✓ = 0.
It follows from the first of these equations (look at the graphs!) that 4✓ = 12 ⇡ + n⇡ and from the
second that 3✓ = n⇡. Hence ✓ = 18 ⇡ + 14 n⇡ or 13 n⇡ and, picking out the solutions which are in the
required range, we see (n = 0, 1 in the first equation) that ✓ = 18 ⇡ or 38 ⇡ or (n = 0, 1 in the second
equation) ✓ = 0 or 13 ⇡. Thus there are four solutions in the required range.
Finally, equations of the form a cos ✓ ± b sin ✓ = 0 can be solved by expressing the left-hand side in one of
the forms A sin(✓ ± ↵) or A cos(✓ ± ↵) for suitably chosen A and ↵. This is an extremely important form for
the Engineer since it is expressed in terms of ‘phase’ (↵) and ‘amplitude’ (A).
– 18 –
Solve the equation 3 sin ✓ + 4 cos ✓ = 3 for 0  ✓  2⇡.
Example 26.
First notice that
A sin(✓ + ↵) = A cos ↵ sin ✓ + A sin ↵ cos ✓ .
Hence, if this is to be identically equal to the left-hand side of the given equation,
A cos ↵ = 3
and
A sin ↵ = 4 .
Squaring and adding, we see that
A2 (sin2 ↵ + cos2 ↵) = 32 + 42
so that A = 5. It follows that cos ↵ = 35 and sin ↵ = 45 , so that, by calculator, ↵ = .9273. Notice
carefully that the signs of cos ↵ and sin ↵ force the solution for ↵ to lie in a single quadrant, and
hence there is only one solution to this pair of equations in [0, 2⇡]. If we tried to use just one of
the equations instead of both, we would get two solutions, one wrong since it would not satisfy the
second equation!
Our equation now simplifies to
5 sin(✓ + .9273) = 3
or
sin(✓ + .9273) = 3/5 .
Once more using our calculator and looking at the sine graph, we find that ✓ + .9273 = .6435 or
⇡  (✓ + .9273) = .6435. That is to say ✓ = .2838 or ✓ = +1.5708, with general solutions ✓ =
.2838 + 2n⇡ or ✓ = 1.5708 + 2n⇡.
Finally, choice of n = 1 in the first equation or n = 0 in the second gives the required solutions:
✓ = 5.9994 or 1.5708—i.e. ⇡/2.
Find the values of ✓ in the range 0  ✓  2⇡ which satisfy the following equations:
***Do Exercise 14.
(i) sin ✓ = .87,
(ii) tan ✓ = 1.39,
***Do Exercise 15.
(i) sin 2✓ = .81,
Find all values of ✓ which satisfy the following equations:
(ii) cos 3✓ = .49,
***Do Exercise 16.
(i) sin 2✓ = sin 60 ,
(iii) tan 12 ✓ = .63
Find all values of ✓ in the range 0  ✓  180 which satisfy the following equations:
(ii) cos 3✓ = cos 120 ,
***Do Exercise 17.
***Do Exercise 18.
(iii) tan2 ✓ = 13 .
Find all solutions of the following equations which lie in the range 0  ✓  ⇡:
(i) 2 sin2 ✓ + sin ✓  1 = 0,
(i) cos ✓ = cos 4✓,
(iii) cos ✓ = .22
(ii) 16 tan2 ✓  24 tan ✓ + 9 = 0,
(iii) sin 2✓ = sin ✓.
Find all solutions of the following equations which lie in the range 0  ✓  2⇡:
(ii) sin 3✓ + sin ✓ = 0,
(iii) 7 sin ✓  8 cos ✓ = 9,
– 19 –
(iv) 5 cos ✓ + 12 sin ✓ = 4.
Specimen Test 2
⇡
radians in degrees
6
1.
Express
2.
In the triangle ABC find AB (correct to 4 decimal places)
........
...... ..
...... ....
......
..
......
.
.
.
.
...
.
.
......
...
......
.
.
.
.
...
.
....
.
.
.
.
...
.
....
.
.
.
.
...
.
....
.
.
.
.
...
.
....
.
.
.
.
...
.
....
.
.
.
.
...
.
....
.
.
.
.
...
.
....
.
.
.
.
...
.
....
.
.
.
.
...
.
....
.
.
.
.
...
.

....
.
.
.
.
...
.
....
.
.
.
.
..
.
.
.....................................................................................................................................................
C
8
43
A
B
3.
If sin ✓ = 0.3 find
(i) cosec ✓,
(ii) cos ✓
4.
In the triangle ABC find AC (correct to 4 decimal places)
C
........
..... .........
.....
......
.....
......
.....
.....
.
.
.
.
.....
......
.....
.....
.....
.
.
.
.
.
.....
.....
.
.
.....
.
.
.....
....
.
.
.
.
.....
....
.
.
.....
.
.
....
.....
.
.
.
.
.....
...
.
.
.
.
....
.
.....
.
.
.
.

 .............
....
.
.
.
.
.....
....
.
.
.
.
.
........................................................................................................................................................................................
8
35
40
A
5.
B
In the triangle P QR find 
Q
............
... .............
..........
...
..........
...
..........
...
..........
..........
...
..........
...
..........
...
..........
...
..........
..........
...
..........
...
..........
...
..........
...
..........
..........
...
..........
...
..........
...
..........
...
...
....................................................................................................................................................................................................
12
6

10
P
6.
Sketch the graph of sin x for 2⇡ < x < 2⇡
7.
Express sin(⇡ + ✓) in terms of sin ✓
8.
Find all values of ↵ which satisfy cos ↵ = 0.5
9.
Express cos 4✓ + cos 2✓ as a product of trigonometric functions
– 20 –
R
Worked solutions to Exercises
Note carefully that, whilst alternative answers have been given in degrees and radians, these have been
calculated independently. Owing to rounding errors, the last decimal place(s) may not agree if you try
to convert directly from an answer given in degrees to the corresponding answer in radians or vice versa.
◆
180 
180 ⇡
=
= 45 ,
⇡ 4
4
✓
◆
7⇡
180 7⇡
(iii)
= 420 .
=
3
⇡ 3
⇡
=
1. (i)
4
✓
⇡
⇡
18 =
= .3142,
180
10
⇡
350 = 6.1087,
(iii) 350 =
180
2. (i) 18 =
(ii)
(ii)
(iv)
3⇡
=
2
✓
180 3⇡
⇡ 2
◆
= 270 ,
⇡
35 = .6109,
180
⇡
420 =
420 = 7.3304.
180
35 =
1
1
a
=
=
= 2 m,
cos B
cos(⇡/3)
.5
b = a tan B = tan(⇡/3) = 1.7321 m,
✓
◆
1
1
1
1
A=⇡
⇡ + B = ⇡  ⇡  ⇡ = ⇡.
2
2
3
6
3. (i) c =
a
2
2
=
=
= 2.0309 m,

cos B
cos(10 )
.9848
b = a tan B = 2 tan(10 ) = .3527 m,
(ii) c =
A = 180  (90 + B) = 180  90  10 = 80 .
p
p
(iii) c = a2 + b2 = 10 m = 3.1623 m,
b
1
tan B = = . Hence, since B < 12 ⇡, B = tan1 .3333 = .3218 (= 18.4349 ).
a✓ 3
◆
1
1
⇡ + B = ⇡  .3218 = 1.2490, (= 71.5651 ).
A=⇡
2
2
4. (i) Firstly, A + B + C = 180 . Hence, C = 180  47  70 = 63 . Now we can use the sine rule to give
b
c
a
=
=
implies that
us a and b:
sin A
sin B
sin C
5.1 ⇥ .9397
c sin B
5.1 ⇥ .7314
c sin A
=
= 5.3787 m, b =
=
= 4.1862 m.
a=
sin C
.8910
sin C
.8910
(ii) Here we must use the cosine formula: a2 = b2 + c2  2bc cos A tells us that
b2 + c 2  a 2
17.22 + 18.42  12.12
cos A =
=
= .7710
2bc
2 ⇥ 17.2 ⇥ 18.4
and so A = 39.5594 , (= .6904).
In the same way, the angle cos B is given by
c2 + a 2  b2
18.42 + 12.12  17.22
cos B =
=
= .4247
2ca
2 ⇥ 18.4 ⇥ 12.1
and so B = 64.8655 , (= 1.1321).
Finally, C = 180  A  B = 180  39.5594  64.8655 = 75.5751 , (= 1.3190).
(iii) Given two sides and non-included angle, we must use the sine formula—and the answer is not unique
(draw a diagram to confirm this)!
b
7
sin B = sin C = sin 10 = .4052
c
3
– 21 –
Hence B = 23.9023 , (= .4172) or B = 180  23.9023 = 156.0977 , (= 2.7244). It follows that
A = 180 B C = 180 23.9023 10 = 146.0977 , (= 2.5499) or 180 156.0977 10 = 13.9023 ,
(= .2426).
Finally we may find a from the cosine formula.
p
p
p
a = b2 + c2  2bc cos A = 72 + 32  2 ⇥ 7 ⇥ 3 ⇥ cos 146.0977 = 92.8596 = 9.6364 m
p
p
or 72 + 32  2 ⇥ 7 ⇥ 3 ⇥ cos 13.9023 = 17.2303 = 4.1509 m.
The area in each case is given by, for example 12 ⇥ base ⇥ height = 12 ⇥ a ⇥ b sin C = 12 ab sin C. Thus the
answers are
5.
(i) 12 ⇥ 5.3787 ⇥ 4.1862 ⇥ sin 63 = 10.0311 m2 ,
(ii) 12 ⇥ 12.1 ⇥ 17.2 ⇥ sin 75.5751 = 100.7795 m2 ,
(iii) 12 ⇥ 9.6364 ⇥ 7 ⇥ sin 10 = 5.8567 m2 or 12 ⇥ 4.1509 ⇥ 7 ⇥ sin 10 = 2.5228 m2 .
6. (i) (a) .6428, .7660, .8391, (b) .6570, .7539, .8714, (c) .3420, .9397, .3640.
(ii) (a) .9397, (b) .9397, (c) 1, (d) .5440.
(iii) (a) cos2 120 + sin2 120 = .25 + .75 = 1,
(b) cos2 410  sin2 410 = .4132  .5868 = .1736 = cos 820
(c) 2 sin 1.5 cos 1.5 = 2 ⇥ .9975 ⇥ .0707 = .1411 = sin(3).
p
p
(iv) cos2 ↵ + sin2 ↵ = 1. Hence cos ↵ = ± 1  sin2 ↵ = ± 1  (24/25)2 = ±7/25. However, ↵ is obtuse
and so lies between 90 and 180 . It follows that cos ↵ is negative and so cos ↵ = 7/25.
sin ↵
tan ↵ =
= 24/7. The sign is correct for an obtuse angle.
cos ↵
7. (i) (sin ✓ + cos ✓)2 = (sin2 ✓ + cos2 ✓) + 2 sin ✓ cos ✓ = 1 + 2 sin ✓ cos ✓ = 1 + sin 2✓ .
(ii)
tan(A + B) =
sin(A + B)
sin A cos B + cos A sin B
=
cos(A + B)
cos A cos B  sin A sin B
On dividing top and bottom of the right hand side by cos A cos B we obtain
tan(A + B) =
tan A + tan B
.
1  tan A tan B
(iii) Putting A = B = ✓ in part (ii), the result follows immediately.
(iv)
tan ✓ + cot ✓
(sin ✓/ cos ✓) + (cos ✓/ sin ✓)
=
sec ✓
(1/ cos ✓)


cos ✓ sin ✓ (sin ✓/ cos ✓) + (cos ✓/ sin ✓)
=
cos ✓ sin ✓(1/ cos ✓)
sin2 ✓ + cos2 ✓
sin ✓
1
=
= cosec ✓ .
sin ✓
=
8.
Given that sin ↵ = 5/13, where ↵ is acute, the lengths of the side opposite ↵ and p
the hypotenuse
p are 5 and
13 respectively. Using Pyhthagoras’ theorem the length of the remaining side is 132  52 = 144 = 12.
Hence tan ↵ = 5/12, cot ↵ = 1/ tan ↵ = 12/5 and cosec ↵ = a/ sin ↵ = 13/5.
– 22 –
9.
Since tan ✓ = a/b and ✓ is acute, a and b are respectively
the side opposite to and the side
p by ✓ in a
p
right-angled triangle whose hypotenuse is of length a2 + b2 . It follows that sin ✓ = a/ a2 + b2 . In
addition 1 + tan2 ✓ = sec2 ✓ and hence sec2 ✓ = (a/b)2 + 1.
10. (i)
sin 3✓ = sin(2✓ + ✓)
= sin 2✓ cos ✓ + cos 2✓ sin ✓,
using the formula for sin(A + B)
2
2
= 2 sin ✓ cos ✓ + (1  2 sin ✓) sin ✓,
2
using the formulae for sin 2✓ and cos 2✓
2
= 2 sin ✓(1  sin ✓) + (1  2 sin ✓) sin ✓
= 3 sin ✓  4 sin3 ✓ .
(ii)
cos 3✓ = cos(2✓ + ✓)
= cos 2✓ cos ✓  sin 2✓ sin ✓,
2
using the formula for cos(A + B)
2
= (2 cos ✓  1) cos ✓  2 sin ✓ cos ✓,
2
using the formulae for sin 2✓ and cos 2✓
2
= (2 cos ✓  1) cos ✓  2(1  cos ✓) cos ✓
= 4 cos3 ✓  3 cos ✓ .
⌘
⇡
⇡
+ ✓ = sin cos ✓ + cos sin ✓ = 1 ⇥ cos ✓ + 0 ⇥ sin ✓ = cos ✓ .
2
2
2
⇣⇡
⌘
⇡
⇡
(ii) cos
+ ✓ = cos cos ✓  sin sin ✓ = 0 ⇥ cos ✓  1 ⇥ sin ✓ =  sin ✓ .
2
2
2
(iii) sin (⇡  ✓) = sin ⇡ cos ✓  cos ⇡ sin ✓ = 0 ⇥ cos ✓  (1) ⇥ sin ✓ = sin ✓ .
✓
◆
3⇡
3⇡
3⇡
(iv) cos
+ ✓ = cos
cos ✓  sin
sin ✓ = 0 ⇥ cos ✓  (1) ⇥ sin ✓ = sin ✓ .
2
2
2
11. (i) sin
⇣⇡
12. (i) sin 4✓ + sin ✓ = 2 sin 52 ✓ cos 32 ✓.
(ii) sin 8✓  sin 6✓ = 2 cos 7✓ sin ✓.
(iii) cos 12✓ + cos 10✓ = 2 cos 11✓ cos ✓.
(iv) cos ✓  cos 2✓ = 2 sin 32 ✓ sin( 12 ✓) = 2 sin 32 ✓ sin 12 ✓.
7
1
⇡ cos 24
⇡.
(v) sin 13 ⇡ + sin 14 ⇡ = 2 sin 24
13. (i) 2 sin 6✓ cos 2✓ = sin 8✓ + sin 4✓.
(ii) 2 cos 5✓ cos 3✓ = cos 8✓ + cos 2✓.
(iii) 2 cos 4✓ sin ✓ = sin 5✓  sin 3✓.
(iv) 2 sin 7✓ sin 5✓ =  cos 12✓ + cos 2✓.
14. (i) ✓ = 1.0552 , 2.0864 (= ⇡  1.0552), (= 60.4586 , 119.5414 ).
(ii) ✓ = .9472 , 4.0887 (= ⇡ + .9472), (= 54.2678 , 234.2678 ).
(iii) ✓ = 1.3490 , 4.9342 (= 2⇡  1.3490, (= 77.2910 , 282.7090 ).
15. (i) 2✓ = .9442 + 2n⇡ , 4.0858 + 2n⇡.
Hence ✓ = .4721 + n⇡ , 2.0429 + n⇡, (= 27.0480 + 180n , 117.0480 + 180n ).
– 23 –
(ii) 3✓ = 2.0829 + 2n⇡ , 4.2003 + 2n⇡.
Hence ✓ = .6943 + 23 n⇡ , 1.4001 + 23 n⇡, (= 39.7802 + 120n , 80.2198 + 120n ).
(iii) 12 ✓ = .5622 + n⇡.
Hence ✓ = 1.1244 + 2n⇡, (= 64.4219 + 360n ).
16. (i) sin 2✓  sin 60 = 2 cos(✓ + 30 ) sin(✓  30 ) = 0.
It follows that ✓ +30 = 90 +180n so that ✓ = 60 +180n , or ✓ 30 = 180n so that ✓ = 30 +180n .
Hence the values of ✓ in the required range are
✓ = 30 , 60 , (= .5236 , 1.0472).
(ii) cos 3✓  cos 120 = 2 sin( 32 ✓ + 60 ) sin( 32 ✓  60 ).
It follows that 32 ✓ +60 = 180n so that ✓ = 40 +120n , or 32 ✓ 60 = 180n , so that ✓ = 40 +120n .
Hence the values of ✓ in the required range are
✓ = 40 , 160 , 80 , (= .6981 , 2.7925 , 1.3963 ).
(iii) tan ✓ = ± p13 .
Hence the values of ✓ in the required range are
✓ = 30 , 150 , (= 16 ⇡ , 56 ⇡).
17. (i) The equation factorises to give (2 sin ✓  1)(sin ✓ + 1) = 0, so that either sin ✓ = 12 or sin ✓ = 1.
Hence the only solutions in the required range are
✓ = 16 ⇡ , 56 ⇡, (= 30 , 150 ).
(ii) The equation factorises to give (4 tan ✓  3)2 = 0, so that tan ✓ = 34 .
Hence there is just one solution in the range, namely ✓ = .6435, (= 36.8699 ).
(iii) Here, sin 2✓  sin ✓ = sin ✓(2 cos ✓  1) = 0, and so either sin ✓ = 0 or cos ✓ = 12 .
Hence the solutions in the required range are ✓ = 0 , ⇡ , 13 ⇡, (= 0 , 180 , 60 ).
18. (i) cos ✓  cos 4✓ = 2 sin 52 ✓ sin 32 ✓ = 0.
Thus 52 ✓ = n⇡ so that ✓ = 25 n⇡, or 32 ✓ = n⇡ so that ✓ = 23 n⇡.
It follows that the solutions in the required range are
✓ = 0 , 25 ⇡ , 45 ⇡ , 65 ⇡ , 85 ⇡ , 23 ⇡ , 43 ⇡ , 2⇡, (= 0 , 72 , 144 , 216 , 288 , 120 , 240 , 360 ).
(ii) sin 3✓ + sin ✓ = 2 sin 2✓ cos ✓ = 0.
It follows that 2✓ = n⇡ so that ✓ = 12 n⇡, or ✓ = 12 (2n + 1)⇡.
Hence the solutions in the required range are ✓ = 0 , 12 ⇡ , ⇡ , 32 ⇡ , 2⇡, (= 0 , 90 , 180 , 270 , 360 ).
p
p
p
p
(iii) Dividing by 72 + 82 = 113 and writing cos ↵ = 7/ 113, sin ↵ = 8/ 113, so that ↵ = .8520, we find
sin(✓  ↵) = p
9
.
113
Hence ✓  ↵ = 1.0097 + 2n⇡ or ✓  ↵ = 2.1319 + 2n⇡.
It follows that ✓ = 1.8617 + 2n⇡ or ✓ = 2.9839 + 2n⇡. The values in the required range are thus
✓ = 1.8617 , 2.9839, (= 106.6648 , 170.9668 ).
p
(iv) Dividing by 52 + 122 = 13, and writing sin ↵ = 5/13, cos ↵ = 12/13, so that ↵ = .3948, the equation
becomes
sin(✓ + ↵) = 4/13 .
Hence ✓ + ↵ = .3128 + 2n⇡ or ✓ + ↵ = 2.8288 + 2n⇡ and ✓ = .0820 + 2n⇡ or ✓ = 2.4340 + 2n⇡. It
follows that the values in the required range are ✓ = 6.2012 , 2.4340, (= 355.3003 , 139.4599 ).
– 24 –
Useful results
(a) Commit the following to memory. (Some of the results can easily be worked out from
graphs, the Formula Sheet, calculators etc. but it is often necessary to have the results at
your fingertips.)
⇡
180
180
To convert from radians to degrees, multiply by
.
⇡
2. Some useful values:
1. To convert from degrees to radians, multiply by
30 =
⇡
,
6
45 =
⇡
,
4
⇡
,
3
60 =
⇡
,
2
90 =
180 = ⇡ ,
360 = 2⇡ .
3. In the right-angled triangle shown sine, cosine and tangent are defined as follows:
sin ✓ =
b
side opposite
⌘ ,
hypotenuse
c
cos ✓ =
adjacent side
a
⌘ ,
hypotenuse
c
..
.......
..... ..
..... ....
......
..
......
.
.
.
.
...
......
...
.....
.
.
.
.
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...
.....
.
.
.
...
.
....
.
.
.
.
...
...
.
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.
.
...
.....
.
.
.
...
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....
.
.
.
.
...
...
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.
.
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...
.........
.
.
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..............
.... ...
.
.
.
.
.
.
...
. ...
...
.
.
.
.
.
.
.
........................................................................................................
c
✓
tan ✓ =
side opposite
b
⌘ .
adjacent side
a
b
a
4.
cos2 ✓ + sin2 ✓ = 1 .
5.
cos
⇣⇡
2
⌘
 ✓ = sin ✓ ,
sin
⇣⇡
2
⌘
 ✓ = cos ✓ .
6.
cos(✓) = cos ✓ ,
sin(✓) =  sin ✓ ,
7.
tan ✓ =
tan(✓) =  tan ✓ .
sin ✓
.
cos ✓
8. The area of a general triangle ABC is given by
Area of 4ABC =
1
1
⇥ base ⇥ height = ab sin C
2
2
or, of course, by either of the other two similar expressions.
9. The values of sine, cosine and tangents of certain important angles between 0 and ⇡/2 are:
✓
0
⇡/6
sin ✓
0
cos ✓
1
tan ✓
0
1/2
p
3/2
p
1/ 3
– 25 –
⇡/4
p
1/ 2
p
1/ 2
1
⇡/3
p
3/2
⇡/2
1/2
p
3
0
1
1
10. The sine and cosine functions have period 2⇡, the tangent function has period ⇡:
cos(✓ + 2n⇡) = cos(✓) ,
sin(✓ + 2n⇡) = sin ✓ ,
tan(✓ + n⇡) = tan ✓ ,
if n is an integer.
11. The other three trigonometric ratios are defined as follows:
sec ✓ ⌘
1
,
cos ✓
cosec ✓ ⌘
1
,
sin ✓
cot ✓ ⌘
1
.
tan ✓
12.
cos 2✓ = 2 cos2 ✓  1 ,
sin 2✓ = 2 sin ✓ cos ✓ .
13. The diagram to help you remember the signs of the trigonometric functions of ✓ for all real values
of ✓ is:
⇡/2
...................................
........
......
....
......
.....
.....
.....
...
....
...
.
...
.
...
.
...
.
.
...
....
...
....
..
...
...
.........................................................................................
...
..
...
...
....
...
...
.
...
..
...
.
.
.
.
.
...
...
....
....
...
.....
...
....
......
..
......
.......
.........................................
S
⇡
A
T
0, 2⇡,...
C
3⇡/2
(b) The following appear on the Formula Sheet but it is very useful to know them!
1. The ‘sine’ and ‘cosine’ formulae for a general triangle ABC are
a
b
c
=
=
,
sin A
sin B
sin C
a2 = b2 + c2  2bc cos A ,
with similar results giving b2 in terms of c, a, cos B and c2 in terms of a, b, cos C.
2. The graphs of the three elementary trigonometric functions between 2⇡ and 2⇡ are:
...
........
.......
....
..........
..........
.......
.....
....... .. ......
......
.....
..... ... .........
.
.
.....
.
.
..
...
...
..
....
...
.
.
.
.
.
.
.
.
...
.
...
..
...
...
....
...
...
...
...
...
...
...
...
...
...
...
...
...
..
..
...
.
.
.
.
.
.
...
..
..
..
.
...
........................................................................................................................................................................................................................................................................................................................................................................
.
...
...
...
...
...
....
...
..
.
.
...
...
...
.
.
.
.
...
...
...
....
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
.
.
.....
.
.
.
.
.
.
.
.
.
.
.
.
.....
..
.....
.....
......
....
...... ...........
...................
...........
......
...
...
+1
✓
2⇡ 3⇡/2 ⇡
⇡/2
⇡/2
1
Graph of cos ✓
– 26 –
⇡
3⇡/2
2⇡
....
.......
.......
...
.......
...................
.......................
...
.....
......
.....
.....
.....
.....
.
....
.....
.
...
.
...
...
..
...
.
.
.
.
.
.
...
...
.
..
..
.
.
.
.
.
.
.
...
...
. ....
.
...
...
.
.
.
...
. ....
.
...
.
.
.
.
.
...
...
. ...
.
...
.
.
.
...
...
....
.
.
.
.
.
.
.
.
.
.
.
........................................................................................................................................................................................................................................................................................................................................................
.
.
.
.
...
...
. ...
.
...
...
...
.
..
...
...
.
... ...
.
.
...
...
..
... ...
...
...
...
... ...
...
...
...
...
...
...
...
....
...
.....
.....
...
.
.
.
.
.
.
.
.
.
.
.
.....
.....
.....
.....
......
......
...
.................
.................
........
...
...
+1
✓
2⇡ 3⇡/2 ⇡
⇡/2
⇡/2
⇡
3⇡/2
2⇡
1
Graph of sin ✓
. .
. .
. .
. .
....
... ..
... ..
... ..
... ..
.......
........
... .
... .
... .
... .
.... ..
.... ..
.... ..
.... ..
....
.. ..
.. ..
.. ..
.. ..
........
... .
... .
... .
... .
..
... ..
... ..
... ..
... ..
...
..
..
..
..
....
....
....
....
...
..
..
..
..
..
..
..
..
...
.
.
.
.
...
...
...
...
...
..
..
..
..
...
....
....
....
....
..
..
..
..
...
...
...
...
...
.
.
.
.
.
.
.
.
...
.
.
.
.
.
.
.
.
.
.
..
..
..
..
..
.
.
.
..
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.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
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.
.
.
..
.
..
..
..
.
..
...
...
...
...
.
.
.
.
.
.
.
.
.. ...
..
..
..
.
.
.
..
.
.
.
.
. ...
.
.
.
.
.
.
.
.
..
.
.
.
.
.
.
..
..
.
..
.. ....
...
...
...
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.
.
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....
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..
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.
..............................................................................................................................................................................................................................................................................................................................................................
.
.
...
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.
...
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.
...
...
...
... ...
..
.
.
..
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.
.
..
.
. ...
.
.
..
.
.
.
.
.
...
.
.
.
..
..
..
..
..
..
.. ...
.
.
.
.
.
.
.
.
....
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
.
...
...
...
...
........
..
..
..
..
...
.
...
.
.
.
....
....
....
...
...
..
..
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
....
....
....
....
...
..
..
..
..
...
... ....
... ....
... ....
... ....
.
... ...
... ...
... ...
... ...
........
.
.
.
.
.
.
.
.
.
.
.
.
....
... ...
... ...
... ...
... ...
...
... ...
... ...
... ...
... ...
...
...
...
...
...
... ..
... ..
... ..
... ..
..
2
1
✓
2⇡ 3⇡/2 ⇡
⇡/2
⇡/2
⇡
3⇡/2
2⇡
1
2
Graph of tan ✓
3.
1 + tan2 ✓ = sec2 ✓ .
4.
1 + cot2 ✓ = cosec2 ✓ .
5.
cos(A + B) = cos A cos B  sin A sin B ,
sin(A + B) = sin A cos B + cos A sin B .
cos(A  B) = cos A cos B + sin A sin B ,
sin(A  B) = sin A cos B  cos A sin B .
(c) Also on the Formula Sheet are:
1.
1
1
sin A + sin B = 2 sin (A + B) cos (A  B) ,
2
2
1
1
sin A  sin B = 2 cos (A + B) sin (A  B) ,
2
2
1
1
cos A + cos B = 2 cos (A + B) cos (A  B) ,
2
2
1
1
cos A  cos B = 2 sin (A + B) sin (A  B) ,
2
2
Solutions to the specimen test are on the website.
– 27 –