We note early that the notes below and onwards assume the basic algebra and arithmetic of a GCSE math course; everything past that point is built up here from the start.
The Number Line
Most of what follows rests on two ideas: the number line, which is where numbers live, and the set, which is how we talk about a collection of them at once.
We start with the number line. Here is an infinite line.
Figure 1.1. An infinite line.
The arrowheads are there to say that the line is meant to extend in both directions, to infinity. Now let us fix a point on this line (can be any point, but we do normally like it being in the middle), and label it 0.
Figure 1.2. A zero point fixed on the line.
This point represents the number zero, and it is the point against which every other number is measured. Numbers sitting to the left of 0 are called negative, numbers sitting to the right are called positive, and 0 itself is neither.
One thing is still missing; we can now say which side of 0 a point lies on, but not how far along it lies, so let us finally fix a unit of length.
Figure 1.3. A unit of length fixed on the line.
This unit of length is what we use, among other things, to compare how far the other numbers sit from zero.
The infinite line above, together with its fixed zero point and its fixed unit of length, is the (real) number line. The numbers represented by the points of the number line are called real numbers, and the collection of all of them is written R.
A few real numbers, and where they sit:
Figure 1.4. Seven real numbers on the number line.
The Natural Numbers
The natural numbers are the numbers we count with. Counting is done by incrementing a quantity, one object at a time, and on the number line an increment is a move to the right by the unit length. The distance we have moved from zero is then the number of times we moved, which is exactly the number of objects counted.
The natural numbers (sometimes called whole numbers) are the numbers represented by those points of the number line that can be reached by starting at 0 and moving right by a whole unit length some number of times.
Figure 1.5. The natural numbers. The point 0 is drawn hollow because we do not count it as one of them.
Remark (A Convention).
Whether 0 counts as a natural number is a matter of convention; but my personal preference is we do not count it, so N={1,2,3,…}, and when we do want zero alongside them we write N0={0,1,2,…}.
The Integers
The natural numbers count, but they cannot measure the difference between two counts; if an increment in quantity is a move to the right by the unit length, then a decrement is a move to the left by the unit length, and allowing both directions gives the integers.
The integers are the numbers represented by those points of the number line that can be reached by starting at 0 and moving in either direction by a whole unit length some number of times. The collection of all of them is written Z.
Figure 1.6. The integers.
Closure
One property that ordinary arithmetic assumes without comment has not been mentioned yet.
A collection of numbers is closed under an operation if carrying out that operation on any two of its members again produces a member of that same collection.
The real numbers are closed under addition, subtraction, multiplication and division: the sum, difference, product and ratio of any two real numbers is again a real number. Division is the one case needing a footnote, and we come back to it below.
Smaller collections need not behave so well: If a and b are natural numbers then so are a+b and ab, and if a and b are integers then so are a+b, a−b and ab; but by our definition the naturals are not closed under subtraction, since 1−2 is not a natural number, and the integers are not closed under division, since 1/2 is not an integer. Each failure is a reason to enlarge the collection we are working with.
Which of the four operations is the collection of negative integers closed under?
The Rational Numbers
Subtraction led us to the integers; division leads to the next enlargement. Where an integer is what we get by laying whole units end to end, a rational number is what we get by first cutting the unit into equal segments.
A rational number is a real number that can be written as a fraction whose numerator is an integer and whose denominator is a nonzero integer; that is, x is rational if and only if there are integers a and b with b=0 such that x=a/b. The collection of all rational numbers is written Q.
Placing them on the line takes a little more work than placing the integers did. Given a positive integer b, the number 1/b is the length obtained by breaking the unit length into b equal parts. For b=3:
Figure 1.7. The unit length broken into three equal parts.
The numbers of the form a/b with a an integer are then those reached by starting at 0 and moving left or right by 1/b units some number of times, with a deciding both the direction and the number of steps.
Figure 1.8. Thirds on the number line. Every integer is a whole number of thirds, three for each unit.
Nothing new happens when b is negative, because then −b is positive and a/b=−a/−b, so the picture for a negative denominator is the picture for a positive one.
Remark (Why the denominator cannot be zero).
In terms of the picture above, a/0 asks us to break the unit into zero equal parts, which does nothing at all: there is no piece left to step by, so no point on the line gets named. The same objection can be made in one line. Dividing by b is meant to undo multiplying by b, so if a/0 were a number q we would need a=q⋅0=0. If a=0 there is then no such q at all, and if a=0 then everyq qualifies. Either there is no answer or there is no way to choose between the answers, so we leave a/0 undefined.
The Rationals are Closed
The rationals are closed under addition, subtraction and multiplication as the integers are, and unlike the integers they are also closed under division. This is quick to see.
Let x and y be rational. By the definition we may write x=a/b and y=c/d, where a, b, c, d are integers and b=0, d=0. Then
x+y=ba+dc=bdad+bc.
The integers are closed under multiplication, so ad, bc and bd are integers; they are closed under addition, so ad+bc is an integer; and bd=0, since neither b nor d is zero. So x+y is an integer over a nonzero integer, which is to say it is rational.
The same follows for the rest. For the difference and the product,
x−y=bdad−bc,xy=bdac,
and both are again an integer over a nonzero integer. For the quotient, suppose also that y=0, so that c=0; then
Where exactly in the argument above was it used that b and d are nonzero? Would anything go wrong if a or c were zero?
Sets
The number line says what a single number is; to speak about many at once (the naturals, the integers, the solutions of an equation) we need the second idea.
A set is a collection of distinguishable objects, mathematical or otherwise.
The objects need not be numbers. The set of all animals in this building called Bob is a perfectly good set, and so is the set of letters appearing in the word “banana”, which is the collection consisting of b, a and n.
The objects collected in a set are its elements. If a is one of the objects collected in the set A, we say a is an element of A, or that a belongs to A, and write a∈A.
If a is not one of them we write a∈/A, read ”a is not in A”.
Let B be the set of animals in this building called Bob. If the cat is called Bob then the cat∈B, and if the dog is called Rex then the dog∈/B. Numbers behave the same way: −2∈Z, while −2∈/N.
Subsets
In the example above, every animal called Bob is also an animal in the building. That situation is common enough to deserve a name: given two sets A and B, it may happen that every element of A is also an element of B, which is to say that whenever x∈A we also have x∈B.
A set A is included in a set B, or is a subset of B, if every element of A is also an element of B. We write A⊂B.
If every element of A lies in B and every element of B lies in A, then the two collections have exactly the same members, and we call them equal: A=B precisely when A⊂B and B⊂A.
Remark.
Sometimes we want to say that A is contained in Band that at least one element of B is missing from A, so that the two are not equal. Then we call A a proper subset of B and write A⊊B.
The collections of numbers built up in the first half of this chapter now line up in a chain,
N⊊N0⊊Z⊊Q⊊R,
and every inclusion in it is proper: 0 lies in N0 but not in N, −1 lies in Z but not in N0, 1/2 lies in Q but not in Z, and 2 lies in R but not in Q; we do not prove the last of these here, but Exercise 1.6 does.
Remark (Decorating the symbols).
Three decorations on these symbols come up constantly. A star removes zero, so Z∗, Q∗ and R∗ are the integers, rationals and reals apart from 0. A subscript plus keeps the non-negative ones and a subscript minus the non-positive ones, giving Z+, Q+, R+ and Z−, Q−, R−. The two can be combined: R+∗ is the set of strictly positive real numbers.
It is worth pointing out how much the symbol ⊂ resembles the symbol <, and where the resemblance stops. Given two numbers a and b, exactly one of a<b, a=b, b<a holds; two numbers cannot be different and yet fail to be comparable. Sets are not like this. Two sets can differ without either one being contained in the other.
Let A={1,2} and B={2,3}. Then A=B, but A⊂B, since 1∈A and 1∈/B, and also B⊂A, since 3∈B and 3∈/A. Neither set comes first.
Remark.
As in the example, a slash through the symbol denies it: A⊂B says that A is not a subset of B, which is to say that at least one element of A escapes B. It is not used often.
Describing a set in a sentence, as we have been doing, is exact but slow. Two shorter notations do most of the work. The first is simply to list the elements, separated by commas and enclosed in curly brackets. This is called the roster method:
A={2,3,5,7}.
A list may be too long to write out, or infinite, or of a length depending on some variable. In those cases we write out enough of it to make the pattern plain and leave the rest implicit, using an ellipsis:
N={1,2,3,…},{2,4,6,…,100}.
Remark.
There is some genuine ambiguity here, since the reader is expected to work out what pattern the "…" is meant to suggest. Usually that is obvious. Sometimes it is not: {0,n,3,2,…} is meaningless.
In contrast to a list, the second notation describes the members implicitly by a rule. This is set-builder notation: we write
{x∣x has some property},
read “the set of all elements x such that x has that property”, the bar being read as “such that”. So the rationals, defined earlier in words, can be written
Q={baa∈Z and b∈Z with b=0}.
Each notation has its own strength. A list says exactly what is in the set and nothing about why those things and not others; a rule says exactly why and leaves you to work out what actually satisfies it. The set {2,3,5,7} names its four members at a glance and hides that they are the primes below 10.
Write each of these sets both ways, as a list and by a rule: the even integers between −6 and 6; the integers whose square is 9; the real numbers whose square is −1.
Two Special Sets
The Empty Set
Calling a set a collection of objects is slightly misleading, because it suggests there has to be something there. There does not.
Every set carries with it a test for membership: given an object, either it belongs or it does not. Nothing in that test requires any object to pass it. Consider the set of all real numbers greater than 5 but less than 3. Any such number would belong to the set, and there is no such number, so the collection is empty, but it is still well defined, because we can still answer the membership question for every object we are handed.
The set with no elements is called the empty set, written ∅.
Remark.
A set is well defined when, given any object, there is an objective rule deciding whether the object belongs to it or not: one of the two answers must happen, and not both. The sets we study are always of this kind. “The set of all large numbers” is not a set in this sense, since nothing decides whether 106 belongs to it.
Remark (The empty set is not zero).
The number 0 and the empty set are not the same thing. Take the set of all numbers that are neither positive nor negative. Exactly one number qualifies, so the set is {0}, which has one element and so is not empty. A box with a zero in it is not an empty box.
The Universe of Discourse
At certain times we wish to limit membership in a set not only by the rule but also by what is eligible for membership in the first place: when solving an equation about lengths, nobody intends “the colour blue” to be a candidate.
The universe of discourse, usually written U, is a set fixed in advance that contains every eligible object under discussion, so that for every eligible object b and every set A under discussion,
b∈U and A⊂U.
So, for instance, we may write {k∈{1,2,3,4}∣k⩾2}={2,3,4},
where the eligible objects are the four listed and the rule then keeps three of them.
Once a universe is fixed, every set we can speak of is caught between two bounds: no set can have fewer elements than the empty set, and none can contain an object the universe does not, so
∅⊂A⊂U.Figure 1.9. A set A drawn inside its universe U, which is the rectangle surrounding it. Here x∈A, while y∈U but y∈/A.
A picture of this kind, with sets drawn as regions and the universe as the rectangle around them, is called a Venn diagram. A Venn diagram helps us see an argument about sets, but it does not prove one.
The Arithmetic of Sets
Just as we combine two numbers to get a third, we can combine two sets to get a third. Doing so gives us an arithmetic of sets, and it has its own operations, its own rules, and its own closure.
Throughout this section every set is a subset of a fixed universe of discourse U, as above.
The union of sets A and B, written A∪B, is the set of all elements belonging to at least one of A and B:
A∪B={x∣x∈A or x∈B}.
Figure 1.10. The union A∪B is the whole of the shaded region.
The word “or” here is the inclusive one: it means at least one, not exactly one. If Tom goes to the shop, or Jerry does, or both of them do, then in every one of those three cases somebody from the union went.
The intersection of sets A and B, written A∩B, is the set of all elements belonging to both A and B at once:
A∩B={x∣x∈A and x∈B}.
Figure 1.11. The intersection A∩B is the overlap.
The intersection of two sets can be empty even when neither set is. Take U=Z, let A be the even integers and B the odd ones; no integer is both, so A∩B=∅. This is one of the reasons the empty set has to exist at all: without it, the intersection of two perfectly good sets would sometimes fail to be a set.
The complement of a set A, written A′, is the set of all elements of the universe U that are not in A:
A′={x∣x∈U but x∈/A}.
Figure 1.12. The complement A′ is everything inside the universe but outside A.
The complement depends entirely on the universe. We never speak of all the non-A’s in the world; only of all the elements of U which are not in A. Since everything under discussion lies in U anyway, we could equally have written A′=U∩A′, and it is sometimes useful to remember that U is there even when it is not written.
The relative complement of A in B, also called the difference and written B−A, is the set of all elements of B that are not in A:
B−A={x∣x∈B but x∈/A}.
Figure 1.13. The difference B−A: the part of B that does not overlap A.
The difference is not really a fourth operation, because it can be written with the three we already have:
B−A=B∩A′.
An element of B−A is one that is in B and is not in A, and “not in A” is exactly “in A′”.
Combining the Operations
Expressions can be built up out of these operations as freely as arithmetic expressions are built out of + and ⋅, with brackets settling the order. So A∩(B∪C) is read ”A intersect the union of B and C”, and moving the bracket to give (A∩B)∪C need not leave the set alone.
One combination splits a union into three pieces:
A∪B=(A∩B′)∪(A∩B)∪(B∩A′).
The three pieces on the right are disjoint (no element lies in two of them), and they name the three regions of the diagram: the part of A outside B, the overlap, and the part of B outside A.
Figure 1.14. A union broken into three disjoint parts.
What is True of All Sets
Not everything that looks natural is true. It is tempting to expect complementation to pass through a union the way a minus sign passes through a bracket, giving (A∪B)′=A′∪B′, and that is false in general. What is true is
(A∪B)′=A′∩B′,
which is one of De Morgan’s laws: to be outside both A and B is to be outside A and also outside B, and the union has turned into an intersection along the way.
What we want in this arithmetic are statements that hold for all sets, not ones that happen to come out right for a convenient choice of A and B.
Is A−B the same set as B−A? Compare with a−b and b−a for numbers.
Finally, both of the operations we started with are closed: given two sets, A∪B is a set and A∩B is a set, so the results can be combined again in their turn.
The Inclusion–Exclusion Principle
Write #(X) for the number of elements of a finite set X. The question of this section is what # does to a union, and the answer is not the one most people guess.
Counting a Union
A common mistake is to treat ∪ as though it were +. A union does combine two sets, but it is not numerically additive unless the sets are completely separate.
Suppose 20 students take Math 209, making up a set A, and 30 students take Bio 232, making up a set B. Intuition says #(A∪B)=20+30=50, and that is right if and only if no student takes both courses. Suppose 7 students take both. Those 7 are among the 20 and they are also among the 30, so adding 20+30 counts each of them twice, although each of them is one person. That leaves 13 taking only Math 209 and 23 taking only Bio 232.
Figure 1.15. The three regions hold 13, 7 and 23 students, so 43 students in all.
Counting the diagram region by region gives 13+7+23=43 distinct students, and the general rule follows.
For finite sets A and B,
#(A∪B)=#(A)+#(B)−#(A∩B),
and this is called the inclusion–exclusion principle.
Here 20+30−7=43, as counting the regions said it should be.
The intersection is what matters. Every element of A∩B contributes 1 to #(A) and another 1 to #(B) although it is only one element, so the sum #(A)+#(B) counts the intersection exactly twice, and subtracting it once puts things right. The wrong answer 50 minus the correct answer 43 is 7, which is the size of the intersection.
What the Two Sizes Alone Will Tell You
If all we are told is #(A)=20 and #(B)=30, we cannot recover #(A∪B); we can only bound it, since
#(A∪B)=50−#(A∩B).
At one extreme the sets are disjoint, #(A∩B)=0 and #(A∪B)=50. At the other, A sits entirely inside B, so #(A∩B)=20 and #(A∪B)=30. The union therefore has at least 30 and at most 50 elements, and by choosing the intersection we can arrange for any answer between the two.
Adding is Safe on Disjoint Regions
Behind all of this is one rule: numbers may be added when the regions they count are mutually exclusive. The three regions A∩B′, A∩B and A′∩B of Figure 1.14 do not overlap, so
#(A∪B)=#(A∩B′)+#(A∩B)+#(A′∩B),
and this addition needs no correction, because no element has been counted twice. The subtraction in the two-set formula is needed because #(A) and #(B) count overlapping regions.
Remark (Finite sets only).
From here on every set is finite. The formula becomes troublesome for infinite sets, because we cannot reliably subtract one infinity from another. Take A=N. If B is the set of even natural numbers then A−B is the odd ones, which is infinite; but if B is instead the set of natural numbers greater than 10, then A−B={1,2,…,10}, which has exactly 10 elements. In both cases A and B are infinite, so no arithmetic on their sizes alone can tell those two answers apart.
What is the chance of drawing a card that is either a heart or a face card from a standard deck of 52?
There are 13 hearts, making a set H, and 16 face cards (aces, kings, queens and jacks), making a set F. The trap is to answer 13+16=29. Four cards, the ace, king, queen and jack of hearts, lie in both sets, so #(H∩F)=4 and
#(H∪F)=13+16−4=25.
So the chance is 25/52, not the 29/52 the intuitive count gives.
The name says what is happening: include, then exclude, then include again. Follow an element that lies in all three sets. The first line counts it three times, once in each of #(A), #(B) and #(C). The second line subtracts it three times, once for each pair, which leaves it counted zero times. The third line adds it back once. It ends up counted once, which is correct. An element in exactly two of the sets is counted twice on the first line, subtracted once on the second and untouched by the third, so it too is counted once.
The diagram gets there another way, by filling the seven disjoint regions from the inside out. Put 20 in the centre. Each pair intersection already contains that 20, so the pair-only regions hold 35−20=15, 40−20=20 and 50−20=30. Each single set contains the three regions just filled, so the parts taken by one language alone hold
90−(15+20+20)=35,80−(15+20+30)=15,110−(20+20+30)=40.Figure 1.16. The seven disjoint regions of F∪G∪S, filled from the centre outwards.
Adding the seven regions, which is safe because they do not overlap,
35+15+40+15+20+30+20=175,
agreeing with the formula. The diagram carries more than the total: it also shows, for instance, that 15 students took German and neither of the other two.
More Than Three
The pattern continues, with the signs alternating. Add the sets taken one at a time; subtract the intersections taken two at a time; add those taken three at a time; subtract those taken four at a time; and carry on alternating until the intersection of all of them has been used.
In a group of 50 people, 30 read the news online and 25 read it on paper. What is the largest the overlap could be, and what is the smallest, given that everybody reads it one way or the other?
Write out the four-set formula in full. How many terms does it have, and how many of them carry a minus sign?
The Real Numbers
The first section of this chapter named the real numbers, as the points of the number line, and left it at that. The reason for leaving it there is that the real numbers are notoriously awkward to pin down, and doing it properly is the business of an analysis course rather than this one. For now the following will serve.
A real number is a number that can be written as a decimal expansion, possibly one that never terminates and never repeats. These are exactly the numbers represented by the points of the number line: every point is a real number, and every real number is a point.
There also exist irrational numbers, meaning real numbers not contained in Q. One example is the ratio between the circumference of a circle and its diameter, which is known as π; another is 2. The set of real numbers describes all physical quantities that can be represented on a line, and R is the set of all rational and irrational numbers together.
The Rules of Arithmetic
The operations + and ⋅ in R — and so also in N, Z and Q — satisfy the following properties. Every manipulation in the rest of these notes is built out of them.
Let a,b,c∈R.
Commutativity.a+b=b+a and a⋅b=b⋅a.
Associativity.(a+b)+c=a+(b+c) and (a⋅b)⋅c=a⋅(b⋅c).
Distributivity.(a+b)⋅c=a⋅c+b⋅c and a⋅(b+c)=a⋅b+a⋅c.
Neutral elements.a+0=0+a=a and a⋅1=1⋅a=a.
Inverses. There is a number −a with a+(−a)=0, and if a=0 there is a number 1/a with a⋅1/a=1.
We call 0 the neutral element with respect to + and 1 the neutral element with respect to ⋅, and we call −a the negative of a and 1/a the reciprocal of a. None of these properties is claimed for subtraction or division, and in general several of them fail there; that is the point of Exercise 1.9.
What Follows from the Rules
Much of ordinary algebra follows from these five rules. We record the pieces we shall need.
If a+b=0 then b=−a and a=−b. Adding −a to both sides of a+b=0 gives −a+a+b=−a, and the left-hand side is 0+b=b, so b=−a. Adding −b to both sides instead gives a=−b in the same way.
As a matter of convention, we shall write
a−binstead ofa+(−b).
With that convention, associativity and commutativity mean that a sum involving three terms may be written in many ways, all of them the same number:
(a−b)+c=(a+(−b))+c=a+(−b+c)=a+(c−b)=(a+c)−b,
so we may also write this sum as a−b+c=a+c−b.
−(−a)=a. By definition −a is the number that adds to a to give 0; but (−a)+a=0 says exactly that a is the number which adds to −a to give 0, and that number is −(−a).
−(a+b)=−a−b. We show this by adding −a−b to a+b: rearranging, (a+b)+(−a−b)=(a+(−a))+(b+(−b))=0+0=0, so −a−b is the negative of a+b.
0⋅a=0. Here distributivity does the work:
0⋅a+a=0⋅a+1⋅a=(0+1)⋅a=1⋅a=a.
Adding −a to both sides gives 0⋅a+a−a=a−a=0, and the left-hand side is simply 0⋅a+0=0⋅a. So 0⋅a=0.
(−1)⋅a=−a. We show this the same way:
(−1)⋅a+a=(−1)⋅a+1⋅a=(−1+1)⋅a=0⋅a=0,
and a number that adds to a to give 0 is the negative of a.
−(ab)=(−a)b. What has to be shown is that (−a)b is the negative of ab, which amounts to showing that the two add to zero; and by distributivity ab+(−a)b=(a+(−a))b=0⋅b=0.
−(ab)=a(−b). We show this by the same computation with the factors the other way round: ab+a(−b)=a(b+(−b))=a⋅0=0.
(−a)(−b)=ab. Applying the previous two facts in turn, (−a)(−b)=−(a(−b))=−(−(ab))=ab.
With the rules of arithmetic written down we can say precisely how fractions behave, since every rule about them comes out of those five.
Throughout this section m, n, r, s are integers, and any letter standing in a denominator is nonzero.
The rule for cross-multiplying says that
nm=srif and only ifms=rn.
This is the rule that turns a question about two fractions into a question about two integers, and we use it whenever two fractions have to be compared.
The cancellation rule says that for a nonzero integer a,
anam=nm.
To test the equality we apply the rule for cross-multiplying, so what has to hold is (am)n=m(an), and that is true by associativity and commutativity. For instance,
54=(−2)(5)(−2)(4)=−10−8.
In dealing with quotients of integers which may be negative, it is useful to observe that
n−m=−nm,
which cross-multiplying turns into (−m)(−n)=mn, something we already know. Both are equal to −m/n, and for this reason we shall write
−nm=n−m=−nm
without worrying about which of the three is meant.
A fraction r/s of positive integers is in lowest form if r and s have no common divisor other than 1.
Any positive rational number has an expression as a fraction in lowest form. Start from any way of writing it as a quotient of positive integers m/n. We know 1 is a common divisor of m and n, and any common divisor is at most equal to m or n, so among all common divisors there is a greatest one; call it d and write m=dr and n=ds with r and s positive integers. Cancelling d,
nm=dsdr=sr,
and r and s have no common divisor left, since one would have made d larger.
Cancellation also explains addition. Given m/n and r/s we have seen that
nm=snsm,sr=nsnr,
so both now have the common denominator sn, and adding is then adding numerators:
nm+sr=nsms+nr.
Multiplication needs no such preparation, since
nm⋅sr=nsmr.
These are the formulas the closure argument earlier in the chapter used, and the two of them together are why Q is closed under all four operations while Z is not.
Put 84/126 in lowest form, and check the answer by cross-multiplying against the original.
Rearranging a Relation
Suppose three numbers are related by
a+b=c.
Adding −b to both sides gives a=c−b: the b has crossed the equals sign and changed sign. That move, together with the corresponding one for multiplication, is most of elementary equation solving.
By inspection we see that these four expressions are all different in nature, and by investigating each of them in turn we shall identify the differences.
Equations
Substituting 1 for x in the left-hand side (LHS) and right-hand side (RHS) of [1] separately, we find
LHS=(1+2)2=32=9,RHS=2+7=9,
so LHS = RHS when x=1. Substituting 2 for x in the same way, however,
LHS=(2+2)2=16,RHS=4+7=11,
so LHS = RHS when x=2. Now rearranging the original expression gives
x2+4x+4=2x+7⟹x2+2x−3=0⟹(x+3)(x−1)=0,
from which LHS = RHS if and only if either x+3=0, that is x=−3, or x−1=0, that is x=1. The equality holds for no other value of x.
An expression of this type, in which the two sides are equal only for a number of distinct values of the unknown quantity, is called an equation. The process of finding those values is called solving the equation.
Sets give us a tidy way of stating the answer. The problem “solve (x+2)2=2x+7” becomes “find the solution set of (x+2)2=2x+7”, and we write
S={−3,1}.
Remark.
The same question can also be answered in set-builder notation, where
S={x∈R∣(x+2)2=2x+7},
which is a statement of a different kind: the list tells us what the set contains but not the property its members share, while the rule tells us the property but not the members. Solving the equation is exactly the work of turning the second description into the first.
Recalling that two sets are equal when each is contained in the other, this tells us what it means for two equations to be the same problem: two equations are equivalent when they have the same solution set. Each rearrangement above replaced an equation by an equivalent one, and that is what entitled us to read the answer off the final line.
Are x2=x and x=1 equivalent? Compare their solution sets.
Identities
Now take [2]. Substituting 1 for x in both sides,
LHS=(1+2)2=9,RHS=12+4⋅1+4=9,
and substituting −1 for x as before,
LHS=(−1+2)2=1,RHS=(−1)2+4(−1)+4=1.
Whatever other numerical value we substitute for x we find that LHS = RHS, so it appears that the two sides agree for all values of x. Expanding the bracket, as we did above, says the same thing in algebra; the following geometric illustration says it a third way, without any algebra at all. Consider a square of side x+2 units.
Figure 1.17. One square, counted two ways.
Since these two squares are identical, their areas are identical. The left one has area (x+2)2, and the right one has been cut into four pieces of areas x2, 2x, 2x and 4. Hence (x+2)2=x2+4x+4 for all values of x, and we say that (x+2)2 is identical tox2+4x+4.
A relationship whose two sides are equal for any value of the unknown quantity is called an identity. Using the symbol ≡ for “is identical to”, the relationship above is written
(x+2)2≡x2+4x+4.
The two sides of an identity are two forms of the same expression. We shall use the identity symbol whenever we are dealing with an identity relationship, and we strongly recommend that the reader does the same.
State which of the following are equations and which are identities.
x2−9=(x−3)(x+3)
p2+2p−3=3−2p−p2
y−1=y1
q2−12q=q−11+q+11
Inequalities
The third relationship, x−2>1, is obviously different from the first two. Reading from left to right, the symbol > means “is greater than”, and < means “is less than”.
By inspection we see that for x−2 to have a value greater than one, x must have a value greater than three:
A relationship between two expressions using one of <, >, ⩽, ⩾ is called an inequality.
Here the number line does the explaining. Consider a line as being made up of adjacent points; then all the real values the variable x can take are represented by positions of points on that line, and the position of a point to the left of a second point corresponds to a value of x less than the value at the second point:
−99<1,−10<−5,…
The values of x given by the statement x>3 can then be represented by a section of this line.
Figure 1.18. A hollow circle leaves the endpoint out; a solid one takes it in.
From this we see that not all values of x satisfy the inequality, but that there is an infinite set of values which do: the solution of an inequality is a range, or several ranges, of values of the variable involved. Note that x=3 is not included in the range, and this is what the hollow circle records. For x⩾3, which means x is greater than or equal to 3, the value x=3 is included in the range, and the circle is filled in.
Find the range of values of x for which the following inequalities are true, and illustrate the range on a number line.
x+1⩽−1
0⩾x−4
3<4−x
Expressions
The fourth of the four, (x+2)2, differs from the other three: it asserts nothing. It is a term, not a relationship, so there is nothing to solve and nothing to check. It has a value once x is given, and that is all.
Powers
We have been writing x2 and (x+2)2 without comment, on the strength of school algebra. We set the notation down properly, because we are about to extend it beyond whole-number exponents.
Let a∈R and n∈N. Then ”a raised to the power of n” is defined as
an=n occurrences of aa⋅a⋯a,
where a is called the base and n the exponent.
Power Rules
Let a,b∈R and m,n∈N. Then
an⋅am=an+m,an⋅bn=(a⋅b)n,(an)m=anm.
Each of the three is a matter of counting the factors. For the first, writing both powers out gives n copies of a followed by m more, which is n+m copies in all:
an⋅am=n copiesa⋯a⋅m copiesa⋯a=an+m.
For the second, n copies of a beside n copies of b can be paired off, one a to one b, by commutativity, giving n copies of a⋅b:
For a>0 and a rational exponent r=p/q with p∈Z and q∈N, ”a raised to the power of r” is defined as
ar=(ap)1/q.
For a=0, this definition is used only when p>0; expressions with a zero base and a non-positive rational exponent are undefined.
The power rules above hold with m and n rational whenever all the expressions involved are defined; in particular, they may be used freely with positive bases.
the last step by commutativity. The other two go the same way.
The first identity is also the picture we drew for (x+2)2, with the 2 replaced by b: the big square has area (a+b)2, and it decomposes into smaller rectangles of areas a2, ab, ab and b2.
Use the third identity to compute 101×99 in your head.
Surds
Expressions such as 4 and 25 have exact numerical values, namely 4=2 and 25=5. Expressions such as 2, 3 and 5 cannot be written as exact terminating or repeating decimals. The root form itself is exact. We might say that
2=1.4correct to 2 significant figures,
or that
2=1.4142136correct to 8 significant figures,
but we cannot write down an exact terminating or repeating decimal equal to 2. Such numbers are the irrational numbers met earlier, and it is often convenient to leave them in the exact forms 2, 3, and so on.
A root left in its root form, rather than replaced by a decimal approximation, is called a surd.
Remark.
Recall that 2 means the positive square root of 2. So although x2=4 is solved by both x=2 and x=−2, we still have 4=2.
Surds occur frequently in solutions, so it is useful to be able to simplify them. The tool is the power rule ab=ab: look for a square factor and take it outside.
When the solution to a problem comes out containing surds, it is accepted practice to leave the answer in surd form, simplified as far as possible, unless an approximation has been asked for. Simplifying a fractional answer can often be managed by removing the surds from the denominator, and this process is called rationalising the denominator.
The expansion of (5−27)(5+27) above shows how this works in general: a bracket of the form (a+b)(a−b) gives a2−b2, and squaring removes a square-root surd in a or in b. So a denominator a+b is rationalised by multiplying above and below by a−b.
An integer is even if it is twice an integer, and odd if it is one more than twice an integer. In set notation, the even integers and the odd integers are
E={a∈Z∣a=2k for some k∈Z},O={a∈Z∣a=2k+1 for some k∈Z}.
Every integer belongs to exactly one of E and O: dividing by 2 leaves a remainder of either 0 or 1, and no integer can be written in both forms.
The Parity of a Sum
Let a and b be positive integers.
If a is even and b is even, then a+b is even.
If a is even and b is odd, then a+b is odd.
If a is odd and b is even, then a+b is odd.
If a is odd and b is odd, then a+b is even.
We show the first. If a and b are even then a=2k and b=2l for some integers k and l, so by distributivity
a+b=2k+2l=2(k+l),
and k+l is an integer, so a+b is twice an integer. The other three are left as problems below; each is the same computation with a +1 carried along.
The Parity of a Square
Let a be a positive integer. If a is even then a2 is even, and if a is odd then a2 is odd.
If a=2k then a2=(2k)2=4k2=2(2k2), which is even. If a=2k+1 then
a2=(2k+1)2=4k2+4k+1=2(2k2+2k)+1,
which is one more than twice an integer, and so odd.
This can also be read backwards: if a2 is even then a is even. Every integer is either even or odd, and not both. If a were odd, what we have just shown would make a2 odd; but a2 is even, so the odd case is ruled out, and a must be even. Exercise 1.6 uses this step.
Taking inspiration from the previous exercise, show that p is not a rational number when p is prime. (Reminder: an integer is called prime when it has exactly two divisors.)
Let A be the set whose elements are a, b, c and d.
Write A in roster form.
How many subsets does A have? How many of them have exactly two elements? List those by the roster method.
If a fair coin is tossed four times, is there a fifty-fifty chance of getting two heads and two tails? How is that question related to the previous part?
Now let A have six members instead. How many subsets does it have, and how many of them have exactly two elements, exactly three, exactly four? Two of those three counts agree; say why.
The universe of discourse decides what an answer looks like.
Write the solution set of x4=1 in set-builder notation, and then in roster form when the universe is (i) the complex numbers, (ii) the real numbers, (iii) the positive real numbers, (iv) the even integers. (Reminder: the complex numbers are the numbers a+bi with a and b real and i2=−1.)
Write the solution set of (2x−3)(x+1)(x+7)=0 in set-builder notation, and then in roster form when the universe is (i) the rational numbers, (ii) the real numbers, (iii) the integers, (iv) the positive integers.
For real numbers a and b we say a is less than b, and write a<b, if and only if b−a is positive, that is, if b=a+h for some positive h. If a<b we also say b is greater than a and write b>a. Using only this definition and the rules of arithmetic, show the following.
If a<b then a+c<b+c, and also a−c<b−c.
If a<b and c<d then a+c<b+d. Is a−c<b−d also true in this case? Explain.
If a<b and c is positive then ac<bc. What happens if the restriction that c be positive is dropped?
If a<b and a and b are either both positive or both negative, which is to say ab>0, then 1/b<1/a.
Check Yourself
Fresh questions on the whole chapter — none of them is worked out above. Do each on paper first; the box only tells you whether you got there.
Answers are checked in your browser, as often as you like. Nothing is sent anywhere and
nothing is kept but your own work. A formula may be written with the symbols themselves or
with ~ & | -> <-> ^, and \and, \or, \to expand as you type.
Let a be a positive integer whose square a2 is odd. What follows about a?
answerone of these
Lesson 2
Functions, and Equations
Taught
Everything here rests on the arithmetic and algebra of the first lesson: the rules of arithmetic, powers and roots, surds, and the difference between an equation and an identity. What is new is the idea of a function, and the algebra that idea makes possible.
A function, defined for all numbers, is an association which to each number associates another number. If we denote the function by f, the association is written
f:x↦f(x),
and f(x) is called the value of the function at x, or the image of x under f.
Thinking of the value given to x as an input and the corresponding value of the function as an output, we may read f:r↦2r+1 as “the function which, when we input a value for r, gives as output the value 2r+1“.
Figure 2.1. A function as a rule turning an input into an output. What makes it a function is that each input has exactly one output.
Using f as the symbol for a function, we write f(x) for a function of x and f(r) for a function of r, so the two associations above may equally be written f(x)=(x+2)2 and f(r)=2r+1. To represent the value when x=1 we write f(1). So for f:x↦(x+2)2,
f(1)=(1+2)2=9,f(3)=(3+2)2=25,f(−2)=(−2+2)2=0.
Both x and f(x) vary, but x may be given any value while the value of f(x) depends on it. So x is called the independent variable and f(x) the dependent variable.
The association which to each number x associates the number 4 is the constant function with value 4. More generally, for a given number c, the association
x↦cfor all numbers x
is the constant function with value c.
Remark (A warning about language).
It is convenient, and slightly incorrect, to write a sentence like “let f(x) be such and such a function”. The trouble is that x is not quantified: what is meant is the function whose value at a number x is such and such. We shall try to avoid the loose phrasing at least while the idea is new.
The examples so far have been given by formulas. Functions may be defined quite arbitrarily, and no formula is required: to describe a function amounts to giving its values at all numbers for which it is defined.
Remark (What the values are).
We have adopted the convention that the values of a function are numbers. It is not a universal convention, and it is a useful one here.
The Algebra of Functions
Functions defined on the same set can be added and multiplied, and when they are they obey the rules of arithmetic from the first lesson.
Let f and g be functions defined on the same set S. Their sumf+g is the function whose value at an element x of S is
(f+g)(x)=f(x)+g(x).
Associativity of addition for numbers gives associativity for functions at once: for any f, g, h defined on S,
(f+g)+h=f+(g+h),
and commutativity for numbers gives f+g=g+f in the same way.
There is a zero function, whose value at every x in S is 0, and which we also denote by 0. For every f defined on S,
f+0=0+f=f.
If f is defined on S then minus f, written −f, is the function whose value at x is −f(x).
So functions satisfy the same basic rules for addition as numbers do, and the same holds for multiplication. The productfg of two functions defined on S is the function whose value at x is
(fg)(x)=f(x)g(x),
and this product is commutative and associative for the same reason as before. If 1 denotes the constant function with value 1 for all x in S, then
1f=fand0f=0.
Multiplication distributes over addition. Reading the definitions off one after another,
If every term of a function has the form axn, where a is a constant and n is a non-negative integer, the function is a polynomial. The highest power of x occurring in it is its degree.
The functions x2, 2x, 3x5−7x+6 and (x−4)2 are polynomials; x, x−2 and 1/x are not, since their exponents are not non-negative integers (Definition 1.29 gives the meaning of x1/2, and it is not a whole power).
The polynomial 5x6−7x3+6x has degree 6.
Fractional Functions
A fractional function is one of the form
q(x)p(x)
with p and q polynomials, where q is not the zero polynomial. Its domain consists of the real numbers for which q(x)=0. It is called proper if the degree of the numerator is less than the degree of the denominator, and improper if the degree of the numerator is greater than or equal to that of the denominator.
An improper numerical fraction such as 79 may be written as a whole number plus a proper fraction,
79=77+2=1+72,
and an improper algebraic fraction splits the same way. The fraction x2+1x2−1 is improper, its numerator and denominator having equal degree, and
It is often useful to run this backwards: to take a fractional function and write it as a sum of separate fractions with simpler denominators. This is called decomposing the function into partial fractions. If the original fraction is proper then the partial fractions are proper too.
The shape of the decomposition is fixed by the factors of the denominator, and the constants in it are found by substitution or by comparing coefficients.
Linear Factors
A linear denominator carries a single constant on top.
The denominators are identical, so the numerators must be as well:
x+3=A(x+4)+B(x−2),
and this holds for every value of x. Substituting x=2 kills the B term and gives 5=6A, so A=65. Substituting x=−4 kills the A term and gives −1=−6B, so B=61. Therefore
(x−2)(x+4)x+3=6(x−2)5+6(x+4)1.
A Quadratic Factor
A quadratic denominator that does not factorise carries a linear numerator, so that the partial fraction is still proper.
each numerator chosen so that its fraction is proper. Clearing denominators,
x−3=A(x2+1)+(Bx+C)(x−1).(∗)
Substituting x=1 gives −2=2A, so A=−1. There is no value of x making x2+1 zero, so A cannot be eliminated the same way, and the remaining constants come from comparing coefficients. On the left the coefficient of x2 is 0; on the right it is A+B. Hence B=−A=1. Substituting x=0 in (∗) gives −3=A−C=−1−C, so C=2. Therefore
(x−1)(x2+1)x−3=x−1−1+x2+1x+2.
Either route reaches the constants: substituting values chosen to eliminate terms, or comparing the coefficients of each power. In practice a mixture of the two is quickest.
so a repeated linear factor is better written as two fractions. We therefore look for
(x+1)(x−2)2x−1=x+1A+x−2B+(x−2)2D,
that is, x−1=A(x−2)2+B(x+1)(x−2)+D(x+1). Substituting x=2 gives 1=3D, so D=31. Substituting x=−1 gives −2=9A, so A=−92. Comparing the coefficients of x2 gives 0=A+B, so B=92. Therefore
(x+1)(x−2)2x−1=−9(x+1)2+9(x−2)2+3(x−2)21.
In general a repeated factor (ax+b)2 in a denominator gives rise to two partial fractions,
ax+bAand(ax+b)2B,
and a factor (ax+b)3 gives rise to three, with denominators ax+b, (ax+b)2 and (ax+b)3.
Improper Fractions
An improper fraction has to be divided out first, leaving a whole part and a proper remainder.
with a,b,c real numbers, is a quadratic equation. Its solutions are called its roots.
Solving by Factorising
If the left-hand side factorises, the equation is solved by setting each factor to zero, since a product of two numbers is zero only when one of them is.
The left-hand side factorises, so the equation becomes
(2x−1)(x−3)=0,
from which either 2x−1=0, giving x=21, or x−3=0, giving x=3.
Losing a Solution
Solutions can be lost if a step is taken carelessly. Consider 2x2−14x=0 and two ways of handling it.
Dividing through by 2x gives x−7=0, so x=7.
Factorising instead gives 2x(x−7)=0, so either 2x=0 or x−7=0, giving x=0 or x=7.
The first route lost the solution x=0, and it lost it because the equation was divided by the common factor x. Dividing by a constant factor is correct and desirable; dividing by a factor containing the unknown throws away every solution that makes the factor zero. This is worth remembering when solving any equation, quadratic or otherwise.
Half the coefficient of x is −45, so adding (45)2=1625 to both sides makes the left-hand side a perfect square:
(x−45)2=1625−21=1617.
Taking square roots,
x−45=±417,sox=45±17.
Since 17=4.123 to four significant figures, x=2.28 or x=0.22.
The Quadratic Formula
Completing the square on the general equation gives a formula for the roots once and for all.
Let a,b,c be real with a=0. The solutions of ax2+bx+c=0 are
x=2a−b±b2−4ac,
provided b2−4ac is positive or zero. If b2−4ac is negative the equation has no solution in the real numbers.
We obtain this by completing the square as in the example. Solving the equation amounts to solving ax2+bx=−c, and dividing by a makes this
x2+abx=−ac.
To complete the square on the left we want x2+abx=x2+2sx, so s=2ab. Adding s2=4a2b2 to both sides gives
(x+2ab)2=4a2b2−ac=4a2b2−4ac.
If b2−4ac is negative the right-hand side is negative and cannot be the square of a real number, so the equation has no real solution. If b2−4ac is positive or zero we may take the square root, and
x+2ab=±2ab2−4ac,
which rearranges into the formula.
Remark.
The formula is worth committing to memory. Read it aloud like a line of verse: ”x equals minus b, plus or minus the square root of b squared minus four ac, all over two a.”
The Discriminant
The quantity under the root decides how many roots there are.
For the equation ax2+bx+c=0 the number b2−4ac is called the discriminant.
If the discriminant is positive, b2−4ac can be evaluated and the two signs give two different values: the equation has two real distinct roots. If it is zero, both signs give the same value x=−2ab: the equation has one repeated root, also called equal roots. If it is negative, b2−4ac has no real value and the equation has no real roots. To summarise, ax2+bx+c=0
has two real distinct roots if b2−4ac>0;
has equal roots if b2−4ac=0;
has no real roots if b2−4ac<0.
Figure 2.2. The three cases. The roots are the points where the curve meets the horizontal axis, and the discriminant records how many such points there are.
For 4x2−7x+3=0 the discriminant is (−7)2−4(4)(3)=49−48=1>0, so there are two distinct real roots.
For x2+ax+a2=0 the discriminant is a2−4a2=−3a2, which is negative for every a=0, so there are no real roots; when a=0 the discriminant is 0 and the equation x2=0 has the repeated root x=0.
For x2−px−q2=0 the discriminant is (−p)2−4(1)(−q2)=p2+4q2, which is positive unless p and q are both zero, so there are two distinct real roots.
Show that the roots of ax2+(a+b)x+b=0 are real for all values of a and b, and find a relationship between p and q for which the roots of px2+qx+1=0 are equal.
The Sum and the Product of the Roots
The roots of a quadratic can be described without being found.
Let α and β be the roots of ax2+bx+c=0. Then (x−α)(x−β)=0 has the same solutions, and expanding it gives
x2−(α+β)x+αβ=0.
Dividing the original equation by a gives
x2+abx+ac=0.
Both are quadratics in which the coefficient of x2 is 1 and which are satisfied by exactly α and β, so the remaining coefficients agree:
The roots of 2x2−3x+6=0 have sum −(−23)=23 and product 26=3. Conversely, a quadratic whose roots have sum 7 and product 10 may be written x2−7x+10=0.
The roots of 2x2−7x+4=0 are α and β. Find α1+β1 and αβ1, and write down the equation whose roots are α1 and β1.
From the equation, α+β=27 and αβ=2. Expressing the first quantity in terms of these,
α1+β1=αβα+β=27/2=47,αβ1=21.
The required equation has roots summing to 47 and multiplying to 21, so it is
x2−47x+21=0,that is4x2−7x+2=0.
Remark.
This method works only when each new root depends in the same way on each old one. It applies to roots α2 and β2, or α1 and β1; it does not apply to roots α+β and α−β.
The roots of 3x2+2x−1=0 are α and β. Without solving the equation, write down and simplify the equation whose roots are
α1 and β1;
2α and 2β;
α2 and β2.
Logarithms
If we have to find x in ab=x, we can simply work out ab. To solve an equation like xa=b we need the a-th root of b. When the unknown is the index, as in ax=b, neither will do, and this is where logarithms come in.
A logarithm is another word for an index, read backwards. Since 23=8, we may say that 3 is the power to which the base 2 must be raised to obtain 8, or that 3 is the logarithm which, with base 2, gives 8. This is written
3=log28.
For a positive base, ax has its usual real-number meaning for every real x, agreeing with the rational powers defined in Lesson 1. For a>1 these powers increase as x increases; for 0<a<1 they decrease, and in either case they take every positive value exactly once.
Let a>0 with a=1, and let b>0. The logarithm of b to base a, written logab, is the number x for which
ax=b.
For rational x, these powers agree with those of Definition 1.29; for other real x, use the standard extension described above.
The conditions are what make the definition work. When a>0 and a=1, the powers ax take every positive value, each exactly once, so for b>0 there is exactly one real number x with ax=b. In particular, two equal powers of a have equal indices.
Since 32=9, the number 2 is the logarithm which with base 3 gives 9, so log39=2. Since (51)−2=25, we have log1/525=−2.
The base may be any admissible number. Common logarithms have base 10, and it is usual to omit the base and write lg for them, so that
lg5=0.6990,lg100=2,lg0.01=−2,
these saying respectively that 100.6990=5, that 102=100 and that 10−2=0.01. Any base other than 10 must be stated.
Since 104=10000, lg10000=4, and a common logarithm gives a rough count of the digits in a number: a whole number with d digits lies between 10d−1 and 10d, so its common logarithm lies between d−1 and d. Numerical values of logarithms are found with a calculator, unless the answer can be spotted, as with lg10000.
Two values follow straight from the definition. Since a0=1, loga1=0 for every base a. Since the power to which a must be raised to give ak is k, loga(ak)=k, and in particular logaa=1. Read the other way round, the definition says that
alogab=b.
The Laws of Logarithms
Three rules cover all the manipulation, and each comes from a rule for indices.
Let b,c>0, and let logab=x and logac=y, so that ax=b and ay=c.
For the first rule, bc=axay=ax+y, so x+y=logabc, that is
logab+logac=logabc.(1)
For the second, cb=ayax=ax−y, so x−y=logacb, that is
logab−logac=logacb.(2)
For the third, let n be any real number and put z=logabn, so that az=bn=(ax)n=anx, whence z=nx, that is
nlogab=logabn.(3)
The identities (1), (2) and (3) are the three laws of logarithms. Written with a single symbol for the base they read
logxy=logx+logy,logyx=logx−logy,logxy=ylogx.
The laws can also be read straight from the identity alogab=b. For the first,
alogab+logac=alogabalogac=bc=alogabc,
and since two equal powers of a have equal indices, logab+logac=logabc. The second follows in the same way, and for the third
anlogab=(alogab)n=bn=alogabn.
Remark (A warning).
The base is the same throughout, and there is no law for a sum: logab+logac is logabc, and it is not loga(b+c). For instance lg2+lg5=lg10=1, while lg(2+5)=lg7=0.8451. Very few functions f satisfy f(b+c)=f(b)+f(c), so the idea that they do is one to get rid of.
Express each of the following in terms of loga, logb and logc.
logabc;
logcab;
logbca3.
Changing the Base
Tables and calculators supply logarithms to base 10, so a logarithm to another base has to be converted before it can be evaluated. Take log72 and call it x. Then 7x=2, and taking common logarithms of both sides,
xlg7=lg2,sox=lg7lg2=0.84510.3010=0.3562.
The same argument in general changes base a to base b. If logac=x then ax=c, and taking logarithms to base b gives xlogba=logbc, that is
logac=logbalogbc.
The identity alogab=b gives this in one line as well:
blogbc=c=alogac=(blogba)logac=blogba⋅logac,
and equating the indices of b gives logbc=logba⋅logac, which rearranges to the same identity.
Taking c=b in the change-of-base identity, and using logbb=1, gives the special case
logab=logba1.
Exponential Equations
An exponential equation is one in which the unknown appears as an index. The third law brings the index down where it can be reached.
Taking common logarithms of both sides and using the third law,
xlg5=lg10,sox=lg5lg10=0.69901=1.43
to three significant figures.
By the definition of a logarithm, the exact solution of ax=b is x=logab, so the solution above is x=log510. Taking logarithms is how such a number is evaluated.
Taking logarithms is no help when the unknown appears in a sum of powers, since log(22x+3⋅2x) cannot be simplified. What works instead is a substitution which turns the equation into a quadratic.
Putting everything on the same base, 9x=(32)x=(3x)2, so the equation is a quadratic in 3x:
(3x)2−12(3x)+27=0,that is(3x−3)(3x−9)=0.
So 3x=3 or 3x=9, which give x=1 or x=2. The last step, going back from the values of 3x to the values of x, is easily forgotten, and an answer that stops at 3x=3 or 9 loses marks in an examination.
The polynomials of the first chapter deserve a closer look. A function f defined for all numbers is a polynomial if there are numbers a0,a1,…,an such that for all x
The function f with f(x)=3x5−2x+1 is a polynomial, and f(1)=3−2+1=2. The function g with g(x)=21x4+3x2−x+5 is a polynomial, and
g(2)=21⋅24+3⋅22−2+5=8+12−2+5=23.
Coefficients and Degree
When a polynomial can be written as in (1) we say it is of degree at most n, and if an=0 we should like to say it has degree n. Some care is needed before we may. Could the same function also be written
f(x)=bmxm+⋯+b0
with a different top power? Could it happen, say, that
7x5−5x4+2x+1=x6−17x3+x+1
for every number x? Looking at the two sides settles nothing, and with large coefficients there is no easy test. If the answer were yes then the degree could not be defined at all, since in the example we would not know whether to call it 5 or 6. The answer is no, and we come to it below by way of the roots.
Let f(x)=x2−3x+2. Then f(1)=0, so 1 is a root, and f(2)=0, so 2 is a root as well.
Let f(x)=ax2+bx+c with a=0. If b2−4ac=0 the polynomial has the single root −2ab, and if b2−4ac>0 it has the two distinct roots
2a−b+b2−4acand2a−b−b2−4ac.
These are the quadratic equations of the previous chapter, said in the language of roots.
Taking Out a Root
Let f be a polynomial of degree at most n and let c be a root of it. Then there is a polynomial g of degree at most n−1 with
f(x)=(x−c)g(x)for all x.
We show this by rewriting f in powers of x−c rather than powers of x. Write f(x)=a0+a1x+a2x2+⋯+anxn and substitute
x=(x−c)+c
for x throughout. Each k-th power ((x−c)+c)k expands into a sum of powers of (x−c) multiplied by numbers, so there are numbers b0,b1,…,bn with
f(x)=b0+b1(x−c)+b2(x−c)2+⋯+bn(x−c)n
for all x. Setting x=c makes every term after the first vanish, and f(c)=0, so b0=0. We may therefore take the factor (x−c) out of what is left:
f(x)=(x−c)(b1+b2(x−c)+⋯+bn(x−c)n−1),
and g(x)=b1+b2(x−c)+⋯+bn(x−c)n−1 is the polynomial wanted.
Remark (The leading coefficient survives).
Look more closely at that g. Expanding ((x−c)+c)k produces one term (x−c)k and others involving lower powers of x−c, so the highest power of x−c to appear anywhere is the n-th, and it comes only from ((x−c)+c)n, carrying the coefficient an. Hence bn=an, and g has the form
g(x)=anxn−1+lower terms.
How Many Roots There Can Be
Let f be a polynomial and let a0,…,an be numbers with an=0 such that f(x)=anxn+an−1xn−1+⋯+a0 for all x. Then f has at most n roots.
We show this by taking the roots out one at a time. Let c1,c2,…,cr be distinct roots of f and suppose r⩾n. Write f(x)=(x−c1)g1(x) with g1 of degree at most n−1. Then
0=f(c2)=(c2−c1)g1(c2),
and c2=c1, so g1(c2)=0: that is, c2 is a root of g1. The same argument applies to g1, giving g1(x)=(x−c2)g2(x) with g2(c3)=0 and g2 of degree at most n−2. Continuing in this way until what is left is constant,
f(x)=(x−c1)(x−c2)⋯(x−cn)c
for all x, and by the remark above c=an=0. So if x is not one of c1,…,cn then no factor vanishes and f(x)=0. There is therefore no room for a further root.
takes the value 0 at every number. Put di=ai−bi and suppose some di is non-zero; let m be the largest index for which this happens, so that
0=dmxm+⋯+d0
for all x, with dm=0. A polynomial written this way has at most m roots, and this one has every number for a root. That is impossible, so every di is zero.
So there is only one way of writing a polynomial in the form (1): the numbers an,…,a0 are determined by the function. They are called the coefficients of f, the number an is the leading coefficient when an=0, and a0 is the constant term. The worry raised above is settled, and a polynomial with an=0 may safely be said to have degree n.
where we have written down the coefficient of every power up to the fifth, those of x4 and x2 being zero. The leading coefficient is 4 and the constant term is −20, so f has degree 5.
Remark (Vanishing at a point, and vanishing everywhere).
It often happens that f(x)=0 for some x, that is, that f has a root. This does not make f the zero polynomial. We call f the zero polynomial, or say it is identically zero, only when f(x)=0 for all numbers x, which happens exactly when all of its coefficients are 0. A polynomial with even one non-zero coefficient is not the zero polynomial, however many roots it may have.
Factorising
The previous chapter determines all the roots of a polynomial of degree 2. For higher degrees it is much harder. Formulas using radicals exist for degrees 3 and 4, and it is a classical result that no such formula can exist in general from degree 5 upwards.
For a polynomial with integer coefficients one may look for the rational roots, and a great deal of time is often spent factoring to find them. It is unusual for a polynomial to factorise so obligingly, and in the quadratic case the systematic answer is the formula rather than a search for factors. Two cases of factorising are worth having.
which tells us that 327=19⋅17+4, and 4 is the remainder. The first digit 1 was chosen as the largest integer whose product with 17 is at most 32; multiplying, subtracting and bringing down the 7 left 157; then 9 was chosen as the largest integer whose product with 17 is at most 157, and subtracting 153 left 4. Since 4<17 we stop.
In general, for positive integers n and d there are an integer q⩾0 and an integer r with 0⩽r<d such that n=qd+r. Note that although the procedure is called long division, it uses only multiplication and subtraction.
The same procedure works for polynomials, and is called the Euclidean algorithm: for non-zero polynomials f and g there are polynomials q and r, the degree of r being smaller than the degree of g, such that
Each step goes as follows. We take 4x first because 4x⋅x2 is the term of highest degree in f; multiplying 4x by x2+1 gives 4x3+4x, which we write underneath with matching powers aligned, and subtracting leaves −3x2−3x+2. We then take −3, because (−3)⋅x2 is the term of highest degree in what is left; multiplying gives −3x2−3, and subtracting leaves −3x+5. This has degree 1, smaller than the degree of g, so the computation is finished.
The reason the procedure works is visible in the steps. Writing f3=f, the multiplier 4x was chosen so that f3(x)−4x(x2+1) has degree 2, the term 4x3 cancelling; call the result f2. The multiplier −3 was then chosen so that f2(x)−(−3)(x2+1) has degree 1, the term −3x2 cancelling. Putting the two together,
Long division gives a second route to the factor taken out earlier. Dividing f by x−c,
f(x)=q(x)(x−c)+r(x)
where r has degree smaller than 1 and so is a constant, say a. Evaluating both sides at x=c gives 0=0+a, so a=0: the remainder vanishes and x−c is a factor.
The Remainder on Division
That last observation gives a remainder without any division, so we state it on its own.
If f is a polynomial which, on division by x−a, gives quotient Q(x) and remainder R, then
f(x)=(x−a)Q(x)+R,
and substituting a for x makes the first term vanish and gives
R=f(a).
So when f(x) is divided by x−a the remainder is f(a).
Put f(x)=x4−3x3+4x2−8. The constant term is −8, whose divisors are ±1,±2,±4,±8, so those are the only whole numbers worth trying. Now
f(1)=1−3+4−8=−6=0,
so x−1 is not a factor, while
f(−1)=1+3+4−8=0,
so x+1 is a factor. Taking it out, by inspection or by long division,
x4−3x3+4x2−8=(x+1)(x3−4x2+8x−8).
Now put g(x)=x3−4x2+8x−8. Since g(−1)=−1−4−8−8=0, the factor x+1 does not repeat; since g(2)=8−16+16−8=0, the factor x−2 appears, and
x3−4x2+8x−8=(x−2)(x2−2x+4).
The discriminant of x2−2x+4 is 4−16=−12<0, so it has no real roots and no linear factors. Therefore
x4−3x3+4x2−8=(x+1)(x−2)(x2−2x+4).
Note that after the first factor has been taken out it must be tried again on what remains, since repeated factors are common.
The Factors of a3±b3
Read as a polynomial in a, the expression a3−b3 vanishes when a=b, so a−b is a factor of it; and a3+b3 vanishes when a=−b, so a+b is a factor of that. Carrying out the division gives
Each expansion is the one above it multiplied by (1+x), which is to say the one above it added to the one above it shifted up a power. So the coefficient of any power of x is the sum of the coefficients of the same power and the preceding power in the line before. Writing the coefficients alone as a triangular array, and continuing the rule as far as we please, gives Pascal’s triangle.
Figure 2.3. Pascal’s triangle. Each entry is the sum of the two above it, as the arrows show for 10+10=20.
Row five of the triangle therefore gives
(1+x)5=1+5x+10x2+10x3+5x4+x5.
Expanding a Power
Once the coefficients are known, any binomial power follows by substitution.
A function defined for all numbers is even if f(x)=f(−x) for every x, and odd if f(x)=−f(−x) for every x. Determine which of the following are even, which are odd, and which are neither.
Use Pascal’s triangle to write down the expansion of (1+2)4, and simplify it to the form p+q2 with p and q whole numbers. Do the same for (1−2)4, and add your two answers.
Check Yourself
Fresh questions on the whole chapter — none of them is worked out above. Do each on paper first; the box only tells you whether you got there.
Answers are checked in your browser, as often as you like. Nothing is sent anywhere and
nothing is kept but your own work. A formula may be written with the symbols themselves or
with ~ & | -> <-> ^, and \and, \or, \to expand as you type.
In the expansion of (a+b)5, what is the coefficient of a2b3?
answerone of these
Lesson 3
Graphs, and Coordinate Geometry
Taught
Functions and Mappings
Mappings
Consider the function f defined by f:x↦2x+1. If we input the value 2 for x, f gives 5 as output, and we say that this function maps2 to 5, written 2↦5. Similarly f maps −2 to −3, −1 to −1 and 0 to 1. Here one value of the independent variable x maps to just one value of the dependent variable f(x), and no value of f(x) is reached from two different values of x.
Now consider f:x↦x2−2x+6. This function maps −1 to 9, 0 to 6, 1 to 5, 2 to 6 and 3 to 9. One value of x again gives just one value of f(x), but a value of f(x) is not necessarily obtained from only one value of x: both 0 and 2 map to 6.
Figure 3.1. Mapping diagrams for the two functions. On the left every output is reached from exactly one input; on the right 6 and 9 are each reached twice.
A mapping is one-one if each output arises from exactly one input, and many-one if at least one output arises from more than one input. A mapping which sends some input to more than one output is one-many.
So x↦2x+1 is a one-one mapping and x↦x2−2x+6 a many-one mapping.
We may regard a function as a rule for mapping a number a to a number b. The strict meaning of the word, however, is restricted to those relationships in which one input value gives rise to just one output value, and a function in the sense of Definition 2.1 has this built in, since it associates to each number a single other number. Both mappings above are functions. The relationship x↦±x is not: input 4 and it gives two outputs, 2 and −2. It is one-many, a perfectly good mapping for positive values of x, and not a function.
Graphs of Functions
Take again f(x)=x2−2x+6. For any chosen value of x the corresponding value of f(x) can be calculated.
x
−1
0
1
2
3
4
f(x)
9
6
5
6
9
14
Arranged in order of increasing x, the values of f(x) show a pattern, and the pattern is easier to see when the results are displayed graphically: we plot the values of f(x) on a vertical number line against the corresponding values of x on a horizontal one.
Figure 3.2. The curve y=x2−2x+6, with its values at the integers from −2 to 4 marked.
Figure 3.2 was drawn through values of x one unit apart. Values taken from a larger range, or closer together, all lie on the same smooth curve, so the curve represents every pair of related values of x and f(x). From it we can read off properties of the function.
The lowest point on the curve is where x=1 and f(x)=5. Although x, as the independent variable, may be given any real value, the corresponding value of f(x) is always greater than 5, or equal to 5 when x=1. We say that the function has a least value of 5.
The curve is symmetrical about the line x=1: two values of x equidistant from 1, of the form 1+a and 1−a, give the same value of f(x).
Conversely, every value of f(x) above the least corresponds to two values of x, of the form 1+a and 1−a.
All three can be confirmed algebraically. Completing the square on the right-hand side gives
f(x)=x2−2x+6=(x−1)2+5.
For the first, (x−1)2 is a squared quantity and so is never negative. Its least value is 0, which occurs when x=1, so the least value that f(x) can have is 5, at x=1.
For the second, take two values of x placed symmetrically on either side of x=1:
f(1+a)=a2+5,f(1−a)=(−a)2+5=a2+5,
so values of x symmetrical about x=1 give the same value of f(x).
For the third, the values of x giving any particular value c of f(x) solve x2−2x+6−c=0, and by the quadratic formula the roots of this equation are
x=1+c−5andx=1−c−5.
For these to be real we need c⩾5, which is the least value once more, and the two values of x corresponding to one value of f(x) are symmetrical about x=1.
Domain and Range
For f(x)=x2−2x+6 we have assumed that any real value of x may be input. A full definition of a function must state the set of permissible input values, so the function illustrated above is fully defined as
The set of input values for which a function is defined is its domain. Once the domain is fixed there is a corresponding set of output values, called the range of the function, or its image-set.
The domain is not always the whole of R, as the examples below show. The analysis above also showed that, for the domain x∈R, the output values were limited: the range of f:x↦x2−2x+6, x∈R, is f(x)⩾5.
Sketch the function f:x↦2x+1, x∈R, and state its range.
The graph is the straight line through (0,1) rising two units for each unit to the right. As x can be any real number the line is infinite in both directions, so the range of f is also the whole of R.
For the function of the previous example, let A be the point on the line for which x=2 and B the point for which x=−1. Define the function represented by the segment AB, and state its range.
The mapping represented by AB is x↦2x+1, but only for values of x from −1 to 2. So the function represented by AB is
f:x↦2x+1,x∈R,−1⩽x⩽2,
and its range is the set of values of f(x) corresponding to −1⩽x⩽2, that is −1⩽f(x)⩽5.
State the image-set of the function f(x)=2x+1, x∈{0,1,2,3,4}.
For this function there are just five input values, and the corresponding output values form the image-set, which is {1,3,5,7,9}. The graphical representation of this function is not a line but the set of five points.
Figure 3.3. One rule on two different domains: the segment AB of the second example, and the five points of the third.
Remark.
When the domain of a function is not stated, it is taken to be x∈R.
Sketch the graph of f:x↦x2, x∈R, and state its range. Then redefine the domain as x⩾0 and sketch the graph that now represents f.
Each of the mappings a↦a2+5, b↦±b and c↦−c has for its domain the non-negative real numbers. State which of them are functions.
The Quadratic Function
Functions of similar form usually have properties in common, so their graphs are usually similar in shape. Knowing the common characteristics of a family of functions lets us sketch any one of them without calculating a table of values.
A function of the form f(x)=ax2+bx+c, where a, b and c are constants and a=0, is a quadratic function.
The function x2−2x+6 analysed above is one. Completing the square on the right-hand side of the general form gives
f(x)=a(x+2ab)2+4a4ac−b2.
Whatever value x takes, 4a4ac−b2 is constant, equal to K say, and (x+2ab)2⩾0 as it is a squared quantity. So the function has the general form
f(x)=K+a×(zero or a positive quantity).
If a is positive, f(x) is at least equal to K: it has a least value of 4a4ac−b2, occurring when x=−2ab.
If a is negative, f(x) can never be greater than K: it has a greatest value of 4a4ac−b2, occurring when x=−2ab.
So x=−2ab is the value of x corresponding to the greatest or least value of f(x). Taking two values of x symmetrical about it, x=−2ab±k,
f(−2ab+k)=ak2+K=f(−2ab−k),
that is, input values of x symmetrical about x=−2ab give as output the same value of f(x). From this analysis we deduce that the curve representing f(x)=ax2+bx+c is symmetrical about the line x=−2ab, called the axis of the curve, and that it turns at a least value when a>0 and at a greatest value when a<0.
Figure 3.4. The two alternative graphs of a quadratic function, each symmetrical about its dashed axis.
The curve representing any particular quadratic function can now be sketched from this information.
Here a=2 and b=−7, so the axis of the curve is x=−2ab=47, and as a>0 the function has a least value,
4a4ac−b2=84(2)(−4)−(−7)2=8−32−49=−881.
So f(x) has a least value of −881 when x=47. To locate the curve accurately on the axes we need one more pair of corresponding values, and f(0) is the easiest to find: f(0)=−4.
Figure 3.5. The curve y=2x2−7x−4.
A quicker sketch is available when the function factorises. Here
f(x)=2x2−7x−4=(2x+1)(x−4).
The coefficient of x2 is positive, so f(x) has a least value. When f(x)=0 the corresponding values of x are the roots of (2x+1)(x−4)=0, namely x=−21 and x=4, and the average of these, 47, gives the value of x about which the curve is symmetrical.
Remark.
This quicker method is suitable only when the function factorises.
Find the greatest or least value of each function, and the value of x at which it occurs. Then sketch the graph of the third, showing its axis of symmetry clearly.
x2−3x+5;
3−2x−x2;
(1+x)(2−x).
Inequalities
Consider the real numbers 5 and 2, for which 5>2. The introduction of an extra term on both sides leaves the inequality sign unchanged:
5+4>2+4,that is9>6,and5−4>2−4,that is1>−2.
Multiplication of both sides by a positive number also leaves the sign unchanged: 5×2>2×2, that is 10>4. However, if we multiply both sides by a negative number the inequality is no longer true. Multiplying by −2, the left-hand side becomes −10 and the right-hand side becomes −4, and −10<−4. This illustrates the general fact that multiplication or division of both sides by a negative number reverses the inequality sign.
To summarise, if a and b are real numbers such that a>b, then
a+k>b+k for all real values of k;
ak>bk for positive values of k;
ak<bk for negative values of k.
These are the rules for manipulating an inequality.
An inequality that involves a quadratic function is a quadratic inequality.
An example is (x−2)(2x+1)>0. The range of values of x satisfying it can be found graphically. Let f(x)=(x−2)(2x+1); the inequality asks for the values of x at which the curve y=f(x) lies above the x-axis.
Figure 3.6. The curve y=(x−2)(2x+1), with the parts above the x-axis drawn boldly.
From the sketch we see that f(x)>0, where the curve is above the x-axis, for values of x greater than 2 and for values less than −21. Therefore the ranges of values of x that satisfy (x−2)(2x+1)>0 are
Find the range of values of k for which the equation x2−kx+(k+3)=0 has real roots.
Real roots, equal or distinct, require a discriminant which is not negative:
(−k)2−4(k+3)⩾0,that isk2−4k−12⩾0.
Let f(k)=k2−4k−12=(k−6)(k+2). Its graph has a least value and meets the axis at k=−2 and k=6, and it lies on or above the axis outside these two values. So the equation has real roots for
Find the set of values of p for which f(x)=x2+3px+p is greater than zero for all real values of x.
f(x) is a quadratic function of x whose coefficient of x2 is positive, so f(x) has a least value. So if f(x)>0 for all x, the least value of f(x) has to be greater than zero. Completing the square on the right-hand side gives
f(x)=(x+23p)2+p−49p2,
so the least value of f(x) is p−49p2, and for f(x)>0 for all x we need
p−49p2>0,that is4p−9p2>0,that isp(4−9p)>0.
Let g(p)=p(4−9p). Its graph has a greatest value and meets the axis at p=0 and p=94, and it is positive between them. Therefore f(x)>0 for all real x for the set of values of p given by 0<p<94.
A function in which the variable appears as an exponent, that is as an index, is an exponential function. The functions 2x, 3x and 10x are all exponential functions of x.
Consider the function f(x)=2x, for which the following table shows corresponding values of x and f(x).
x
−10
−5
−4
−3
−2
−1
0
1
2
3
4
5
10
2x
10241
321
161
81
41
21
1
2
4
8
16
32
1024
From this table we see that
f(x)>0 for all real values of x;
as x increases, f(x) increases at a rapidly accelerating rate;
f(x)=1 when x=0;
as x decreases, through x=−10,−100,…, f(x) quickly becomes numerically smaller, and we say that as x approaches minus infinity, f(x) approaches the value zero. This is written x→−∞, f(x)→0.
From these observations the sketch of f(x)=2x is drawn.
Figure 3.8. The curve y=2x and its asymptote, the x-axis.
The curve approaches the x-axis but never actually touches it or crosses it.
A line which a curve approaches more and more closely, without ever reaching it, is an asymptote to the curve.
So the x-axis is an asymptote to the curve y=2x. Another way of expressing the behaviour of f(x) for negative values of x is that f(x) approaches a limiting value, or limit, of zero as x approaches minus infinity. This property is written
x→−∞lim2x=0.
Any function of the form ax, where a>1, is represented by a curve similar to that deduced for 2x.
A function whose numerator and denominator are both polynomials is a rational function. These are the fractional functions of the last lesson, under their other name.
For example x1, x−2x and x+1x2−7 are rational functions of x. Consider the function f(x)=x1 and the following table of corresponding values.
x
−3
−2
−1
0
1
2
3
4
10
x1
−31
−21
−1
undefined
1
21
31
41
101
From this table we see that
for x>0, f(x)>0, and as x→∞, f(x)→0;
for x<0, f(x)<0, and as x→−∞, f(x)→0, so that x and f(x) always have the same sign;
for x=0, f(0)=01, which has no finite value and is said to be undefined.
In such circumstances we investigate the behaviour of f(x) as x approaches zero. Now x can approach zero in two ways: it can decrease from positive values towards zero, which is to approach zero from above, or increase from negative values towards zero, which is to approach zero from below.
x
1
0.1
0.01
0.001
x1
1
10
100
1000
x
−1
−0.1
−0.01
−0.001
x1
−1
−10
−100
−1000
From these tables we see that as x decreases to zero, f(x)→∞, and as x increases to zero, f(x)→−∞. From these observations we can draw the sketch representing f(x)=x1.
Figure 3.9. The curve y=x1. Both axes are asymptotes.
This curve has two asymptotes, the horizontal and the vertical axes. All the curves looked at so far have been unbroken, or continuous, but this curve has a break, or discontinuity, at the point where x=0 and f(0) is undefined. In general, if f(x) is undefined for a finite value of x, x=a say, then the curve representing f(x) has a discontinuity where x=a.
The function x1 does not approach a unique value as x approaches zero: it goes to ∞ or to −∞ depending on whether x approaches zero from above or from below. In this case we say that x→0limx1does not exist. In general, for x→alimf(x) to exist with a value k, say, f(x) must approach k both as x approaches a from above and as x approaches a from below.
Logarithmic Functions
For values of a>0 with a=1, any function of the form logax, loga(2x+1), and so on, is a logarithmic function, the logarithm being that of Definition 2.21. Consider the function f(x)=log2x and the following table of corresponding values.
x
−1
0
41
21
1
2
4
8
log2x
does not exist
undefined
−2
−1
0
1
2
3
From this table we see that
when x=−1, f(−1)=log2(−1)=b say; but there is no real value of b for which 2b=−1, since every power of 2 is positive. So f(−1) does not exist, and all negative values of x lead to the same conclusion: log2x does not exist for negative values of x;
for x>1, f(x)>0, and as x→∞, f(x)→∞;
for x=0, f(0) is undefined, so we investigate the behaviour of f(x) as x approaches zero from above. As x decreases to zero, f(x)→−∞, and for 0<x<1, f(x)<0.
From these observations we can sketch the graphical representation of f(x)=log2x.
Figure 3.10. The curve y=log2x.
Any function of the form logax with a>1 has a graph of similar shape. Note that, while logax does not exist for negative values of x, the value of logax can itself be negative.
Simple Variations of Functions
The curve representing f(x)=2x is now familiar, and from it we can obtain the graphs of some simple variations.
g(x)=2x+1. Comparing g(x)=2x+1 with f(x)=2x, for any one value of x, g(x) is one unit greater than f(x). So the curve representing g(x) is the same shape as that representing f(x) but raised vertically by one unit. In general, the curve representing f(x)+c is the curve representing f(x) raised vertically by c units.
g(x)=2−x. For g(x) an input x=a gives the output 2−a, while for f(x) the input x=−a gives the same output. So g(a)=f(−a), that is g(x)=f(−x) for x∈R, and the curve representing g(x) is the same as that representing f(x) with the negative and positive values of x transposed. In general, the curve representing f(−x) is the reflection in the vertical axis of the curve representing f(x).
g(x)=−2x. Here g(x)=−f(x), so the curve representing g(x) is the same shape as that for f(x) with the positive and negative values of f(x) transposed. In general, the curve representing −f(x) is the reflection in the horizontal axis of the curve representing f(x).
Figure 3.11. The three variations of y=2x, each drawn against the faint original.
Write down the values of f(x)=(21)x corresponding to x=2,4,6 and to x=−2,−4,−6. From these values deduce the behaviour of f(x) as x→∞ and as x→−∞, and sketch the graph of f(x), marking any asymptote. Which variation of 2x is this curve?
Inverse Functions
Consider the mapping f:x↦2x, x∈{2,3,4}. Under this function the domain {2,3,4} maps to the image-set {4,6,8}.
It is possible to reverse this mapping: we can map each member of the image-set {4,6,8} back to the corresponding member of the domain by halving it, 4↦2, 6↦3, 8↦4. Expressed as an algebraic relationship, if x∈{4,6,8} then x↦21x maps 4↦2, 6↦3, 8↦4.
This reverse mapping is a one-one mapping, so it is a function in its own right, and it is called the inverse function of f. Denoting this inverse function by f−1, we see that f−1:x↦21x, x∈{4,6,8}, reverses the mapping f:x↦2x, x∈{2,3,4}. In fact f:x↦2x can be reversed for all real values of x, so if f is the function defined by f:x↦2x, x∈R, then f−1 is the function which reverses this mapping, defined by
f−1:x↦21x,x∈R.
Now consider the function f:x↦x2, x∈R. This is a many-one mapping: there are two values of x which map to one value of f(x), both 2 and −2 mapping to 4, for example. We can reverse this mapping by taking the positive and the negative square root of each member of the image-set, which in algebraic form is the mapping x↦±x. However, this is a one-many mapping, so it is not a function, and we say that f:x↦x2, x∈R, does not have an inverse function.
If, however, we restrict the domain of f to x⩾0, redefining the function as f:x↦x2, x⩾0, then it becomes a one-one mapping. The reverse mapping x↦x is also one-one, so the function f:x↦x2, x⩾0, does have an inverse, namely
A function f maps the domain of f to the image-set of f. If the reverse mapping, of the image-set of f to the domain of f, is a function, it is called the inverse function of f and is denoted by f−1.
Remark.
If f defines a one-one mapping then f−1 exists, but if f defines a many-one mapping then f−1 does not exist.
The Graphs of Functions and Their Inverses
The graphs of f(x)=2x and f−1(x)=21x for x∈R, and of f(x)=x2 and f−1(x)=x for x⩾0, are sketched below. Observing these, we see that in each case the graph of f−1(x) is the reflection of the graph of f(x) in the line y=x.
Figure 3.12. Two functions and their inverses, each pair reflected in the dashed line y=x.
This is true for the graph of any function f and its inverse f−1. The line y=x is the graph of f:x↦x, and this function is its own inverse. So if the graph representing a function is known, the graph representing its inverse can be sketched. Even when a function f does not possess an inverse, the curve representing f can still be reflected in the line y=x, but in that case the reflected graph does not represent a function.
Given the function f(x)=2x, x∈R, find f−1 as a function of x and sketch the graph of f−1.
The function f maps x to 2x. To find f−1 we have to reverse this process, that is map values of 2x back to values of x. If 2x=w, say, then taking logarithms to base 2 of each side gives x=log2w. Hence the relationship x↦w can be expressed as w↦log2w, and this is a one-one mapping for w>0, so it is a function, and it reverses the mapping x↦2x. Replacing the variable w by the variable x, the inverse of f(x)=2x is
f−1(x)=log2x,x>0.
Figure 3.13. The curve y=2x and its inverse y=log2x, reflected in the dashed line y=x.
In the same way, for any base a>0 with a=1, the function logax from the positive numbers to the real numbers is the inverse of the function ax from the real numbers to the positive numbers.
Each of the following functions has the domain x∈R. Determine which of them have an inverse, and where f−1 exists express it as a function of x.
3x;
2x+1;
x2−4.
Coordinate Geometry
Locating a Point
Graphical methods lend themselves particularly well to the investigation of the geometric properties of many kinds of curves. We restrict ourselves to plane figures, those that can be described fully using only two dimensions, and to represent any figure on a graph we need, as a start, a simple and unambiguous way of describing the position of a point.
Consider the problem of describing the location of a town, Birmingham say. There are many ways in which this can be done, but all require reference to at least one known place and known directions, called a system, or frame, of reference. Within this frame of reference two measurements, or coordinates, are needed to locate the town precisely.
Figure 3.14. The position of B described in two alternative ways.
In the first description the system of reference is the fixed point O and the direction due north from O, and the coordinates of B are 50 km from O on a bearing of 36∘52′. In the second the system of reference is the pair of directions due east and due north from the fixed point O, and the coordinates of B are 30 km east of O and 40 km north of O. The two systems most often used for mathematical analysis are basically similar to these two practical systems.
Polar Coordinates
The system of reference is a fixed point O, called the pole, and a fixed direction from O, the line Ox, called the initial line. The coordinates of a point P are the distance of P from O and the angle OP makes with Ox, measured in an anticlockwise sense from Ox. These coordinates are written as an ordered pair (r,θ), that is (distance, angle).
Cartesian Coordinates
The system of reference is a fixed point O, the origin, and a pair of perpendicular lines through O. It is usual to draw these lines horizontally and vertically; the horizontal line is called the x-axis and the vertical line the y-axis. The coordinates of a point P are the directed distances of P from O parallel to the axes: a positive coordinate is a distance measured in the positive direction of the axis, and a negative coordinate is a distance in the opposite direction. The coordinates are given as an ordered pair (a,b), with the x-coordinate, or abscissa, first and the y-coordinate, or ordinate, second.
Figure 3.15. The point (2,30∘) in polar coordinates, and the points (1,4) and (−2,−1) in Cartesian coordinates.
Coordinate geometry is the name given to the analysis, using graphical methods, of geometric properties. The properties of straight lines and of many curves are most simply found using Cartesian coordinates, so this system of reference is used more frequently than any other. For this analysis we need to refer to three types of points:
fixed points whose coordinates are known, such as the point (4,5);
fixed points whose coordinates are not known numerically, referred to as the points (x1,y1), (x2,y2), and so on, or (a,b);
points which are not fixed, called general points, a general point being referred to as the point (x,y).
It is conventional to use the letters P, Q, R for general points and A, B, C for fixed points. To avoid distorting the shape of a curve when drawing it on a Cartesian plane, the two axes are graduated using identical scales.
Represent on one diagram the points whose polar coordinates are (1,45∘), (3,90∘), (1,150∘) and (2,200∘), and on another the points whose Cartesian coordinates are (4,2), (−1,5), (0,3) and (−2,−5).
The Length of the Line Joining Two Points
Figure 3.16. The points A(1,2) and B(3,4), with N(3,2) completing a right-angled triangle.
From the diagram we see that the length of the line joining A(1,2) and B(3,4) can be found by Pythagoras, using the point N(3,2):
AB2=AN2+BN2=(3−1)2+(4−2)2=8,soAB=8=22.
In general, if A(x1,y1) and B(x2,y2) are any two points, the point N(x2,y1) gives AN=x2−x1 and NB=y2−y1, and by Pythagoras AB2=AN2+NB2. Therefore the length of the line joining A(x1,y1) to B(x2,y2) is
AB=(x2−x1)2+(y2−y1)2.
This formula still holds when some, or all, of the coordinates are negative. For A(−2,2) and B(3,−1), AN=3−(−2)=5 and NB=−1−2=−3, and AB=25+9=34.
The Midpoint of the Line Joining Two Points
Figure 3.17. The midpoint M(2,3) of the line joining A(1,1) and B(3,5).
Let M be the midpoint of the line joining A(1,1) and B(3,5). The foot of the perpendicular from M to the x-axis lies halfway between the feet of the perpendiculars from A and B, so the x-coordinate of M is
1+21(3−1)=21(3+1)=2.
Similarly the y-coordinate of M is 1+21(5−1)=21(5+1)=3. Therefore M is the point (2,3).
In general, if M is the midpoint of the line joining A(x1,y1) and B(x2,y2), the x-coordinate of M is x1+21(x2−x1)=21(x1+x2), the arithmetic mean of the x-coordinates of A and B, and likewise its y-coordinate is the arithmetic mean of the y-coordinates. So the coordinates of M are
(2x1+x2,2y1+y2).
This formula also holds when some, or all, of the coordinates are negative: the midpoint of the line joining A(−3,−2) and B(1,3) is (21(−3+1),21(−2+3))=(−1,21).
The gradient of a straight line is a measure of its slope with respect to the x-axis. It is the increase in the y-coordinate divided by the increase in the x-coordinate between one point on the line and another point on the line.
Consider the line passing through the points A(2,3) and B(6,1). From A to B the y-coordinate decreases by 2, that is increases by −2, and the x-coordinate increases by 4, so the gradient of AB is 4−2=−21. From B to A the gradient is
increase in xincrease in y=−42=−21,
so it does not matter in which order the two points are considered, provided they are considered in the same order when calculating the increases in both x and y.
Consider now the straight line through A(1,2) and B(4,3). From A to B the increase in the y-coordinate is 1 and the increase in the x-coordinate is 3, so the gradient of AB is 31. If C and D are any two other points on the same line, the right-angled triangle they make with lines parallel to the axes is similar to the one made by A and B, so it gives the same ratio: the gradient of a line may be found from any two points on the line.
From the two examples we see that the gradient of a line may be positive or negative. A positive gradient indicates an uphill slope with respect to the positive direction of the x-axis, that is a line which makes an acute angle with the positive sense of the x-axis. A negative gradient indicates a downhill slope, a line which makes an obtuse angle with it.
Figure 3.18. A positive gradient rises to the right; a negative gradient falls.
In general, the gradient of the line passing through A(x1,y1) and B(x2,y2) is
the increase in the x-coordinatethe increase in the y-coordinate=x2−x1y2−y1.
As the gradient of a straight line is the increase in y divided by the increase in x from one point on the line to another, the gradient measures the increase in y per unit increase in x, that is the rate of increase of y with respect to x.
Parallel Lines
If l1 and l2 are parallel lines, they are equally inclined to the positive direction of the x-axis, so the gradient triangles of the two lines are similar and give the same ratio: parallel lines have equal gradients.
Perpendicular Lines
Consider two perpendicular lines whose gradients are m1 and m2. Draw through the origin a line OS parallel to the first and a line OR parallel to the second, drop S to T on the x-axis and R to Q on the y-axis. Because OS and OR are perpendicular, the triangle OQR is a copy of the triangle OTS turned through a right angle, so the triangles are similar and
OTST=OQQR.Figure 3.19. Perpendicular lines OS and OR, and the similar triangles OTS and OQR.
But the gradient of OS is OTST=m1, and as OR rises to the left the gradient of OR is −QROQ=m2. Since OTST and QROQ are reciprocals,
m1m2=OTST×(−QROQ)=−1.
So the product of the gradients of perpendicular lines is −1, or, if one line has a gradient m, the gradient of any line perpendicular to it is −m1.
Show that the point (−76,0) is on the median through A of triangle ABC, where A, B, C are the points (2,4), (−2,3), (1,−2). If also the point (a,b) is on this median, find a relationship between a and b.
If AD is the median through A, then D is the midpoint of BC, that is the point (−21,21). If E(−76,0) is on AD, the gradients of AD and AE should be equal. Now
gradient of AD=2+214−21=57,gradient of AE=2+764−0=57,
so E is on the median AD. A condition that P(a,b) should be on AD is that the gradient of AP equals the gradient of AD:
Find the gradients of the lines passing through (−1,−3) and (−2,1), and through (3,−2) and (−1,4).
Determine, by comparing gradients, whether the points (0,−1), (1,1) and (2,3) are collinear, that is whether they lie on the same straight line.
Determine whether AB is parallel or perpendicular to CD, where A, B, C, D are (0,−1), (1,1), (1,5), (−1,1).
Equations and Regions
The Cartesian system of reference provides a means of defining the position of any point in a plane, and this plane is called the xy plane. In general x and y are independent variables, each able to take any value independently of the value of the other, unless some restriction is placed on them.
Consider the set of points for which x=2. As the value of y is not restricted, these points all lie on the line parallel to the y-axis passing through (2,0). So the equation x=2 defines this line in the xy plane; x=2 is called the equation of the line, which is briefly referred to as “the line x=2”.
Now consider the set of points for which x>2. All points to the right of the line x=2 have an x-coordinate greater than 2, so the inequality x>2 defines the region of the xy plane to the right of the line. Similarly x<2 defines the region to its left.
Figure 3.20. The region x>2, with its boundary line drawn broken.
Remark.
The region defined by x>2 does not include the line x=2. When a region does not include the points on its boundary lines, these are drawn as broken lines; when it does include them, they are drawn as solid lines.
Now consider the function f(x)=(x−3)(x+1), and the curve representing it drawn on the xy plane. If P, Q and R are points on a line x=x1, with Q on the curve, the y-coordinate of Q is f(x1). The y-coordinate of P, above Q, is greater than that of Q, and the y-coordinate of R, below Q, is less. This argument applies for all values of x1. Therefore the inequality y>(x−3)(x+1) defines the set of points in the region above the curve, and the inequality y<(x−3)(x+1) defines the region below it, whereas only for points on the curve is y=(x−3)(x+1). This last is called the equation of the curve, and the curve is often referred to simply as the curve y=(x−3)(x+1).
Figure 3.21. The region y>(x−3)(x+1), with the points P, Q, R on one vertical line.
An equation such as x2−7x+3=0 contains only one variable, and its solution comprises a finite set of values of x. An equation containing two variables, such as y=(x−3)(x+1), has as its solution an infinite set of ordered pairs (x,y). If A is the solution set of the equation y=f(x), the elements of A are the coordinates (x,y) of all points on the curve y=f(x), and conversely the coordinates of points not on the curve are not elements of A. So for a point P on the curve and a point Q off it, P∈A but Q∈/A.
In general, if f(x) is any function of x, then in the xy plane
y=f(x) defines the curve representing f(x), and is called the equation of that curve;
y>f(x) and y<f(x) define the regions of the plane above and below that curve.
Determine whether the points (5,11) and (−2,−20) are on the curve y=(x−4)(x+6).
Substituting 11 for y in the left-hand side of the equation gives 11, and substituting 5 for x in the right-hand side gives (5−4)(5+6)=11. The two sides are equal, so (5,11) is a member of the solution set of y=(x−4)(x+6) and is on the curve.
Substituting −20 for y in the left-hand side gives −20, and substituting −2 for x in the right-hand side gives (−2−4)(−2+6)=−24. So the left-hand side is greater than the right-hand side, and (−2,−20) is not a point on the curve: it lies above it.
Draw a sketch to show the region of the xy plane defined by the inequalities 0⩽x⩽2, y⩾0, y⩽x2.
The relationship 0⩽x⩽2 contains two inequalities, x⩾0 and x⩽2, which must be considered separately. Taking each inequality in turn, we shade out the region that it excludes.
y⩾0 is the line y=0 and the region above the x-axis, so we shade out the region below the x-axis.
y⩽x2 is the curve y=x2 and the region below it, so we shade out the region above the curve.
x⩾0 is the y-axis and the region to its right, so we shade out the region to the left of the y-axis.
x⩽2 is the line x=2 and the region to its left, so we shade out the region to the right of this line.
Combining these four, the unshaded region, including the boundary lines, is the set of points that satisfies all the given inequalities.
Figure 3.22. The region 0⩽x⩽2, y⩾0, y⩽x2. Here the wanted region is shaded, rather than the regions excluded.
The Equation of a Particular Curve
So far in our work on functions and graphs we have begun with a function and deduced from its properties the curve that represents it. The reverse process, in which we begin with a curve given geometrically and deduce its equation, is as follows.
Consider the circle whose centre is the point C(4,2) and whose radius is 2. Any point P on the circumference of this circle is such that PC=2; any point Q inside the circle satisfies CQ<2; and any point R outside the circle satisfies CR>2. The distance of any point (x,y) from C is (x−4)2+(y−2)2, so the coordinates (x,y) of P must satisfy the equation
(x−4)2+(y−2)2=4.Figure 3.23. The circle with centre C(4,2) and radius 2, with P on it, Q inside and R outside.
This equation defines the set of points on the circumference of the circle, and so is the equation of the circle. Similarly the coordinates of Q satisfy the inequality (x−4)2+(y−2)2<4, so this inequality defines the region inside the circle, and the inequality (x−4)2+(y−2)2>4 defines the region outside it.
The Straight Line
Straight lines play an important part in any geometric analysis. A straight line may be defined in many ways, for example as
the line which passes through the origin and has a gradient of 21, or
the line which passes through the points (2,1) and (−4,−2).
For the first, if P(x,y) is a point on the line other than the origin, then the gradient of OP is 21. The gradient of OP is x−0y−0=xy, so the coordinates of P satisfy
xy=21,or2y=x,
and 2y=x is the equation of the line.
Remark.
For any point Q(x,y) above P, y>21x, that is 2y>x. Therefore the inequality 2y>x defines the region above the line, and similarly 2y<x defines the region below it.
For the second, a point P(x,y) is on the line through A(2,1) and B(−4,−2) exactly when the gradient of PA equals the gradient of AB. The gradient of PA is x−2y−1 and the gradient of AB is 2−(−4)1−(−2)=21, so the coordinates of P satisfy
x−2y−1=21,or2y=x.
These apparently different definitions give the same line. It is conventional to use integers for coefficients whenever possible.
Consider the more general case of the line whose gradient is m and which passes through the origin. For a point P(x,y) on this line other than the origin, the gradient of OP is m, so the coordinates of P satisfy xy=m, or y=mx; the final equation also includes the origin.
Generalising even further to cover any straight line, consider the line whose gradient is m and which cuts the y-axis at a directed distance c from the origin. The number c is called the intercept on the y-axis. With A the point (0,c), a point P(x,y) is on the line exactly when the gradient of AP is m, so the coordinates of P satisfy
x−0y−c=m,ory=mx+c.Figure 3.24. The line y=mx+c, with intercept c on the y-axis and gradient m.
This is called the standard form of the equation of a straight line. It follows that
an equation of the form y=mx+c represents a straight line with gradient m and intercept c on the y-axis;
any equation involving a linear relationship between x and y, that is ax+by+c=0 where a and b are constants not both zero, is the equation of a straight line.
Write down the gradient of the line 3x−4y+2=0, and find the equation of the line through the origin which is perpendicular to the given line.
Writing 3x−4y+2=0 in standard form gives y=43x+21, so the gradient of the given line is 43. So the gradient of the perpendicular line is −34. The required line passes through the origin, that is it has zero intercept on the y-axis, so its equation is
This line can be located accurately in the xy plane once two points on the line are known. The intercepts on the axes can be found by inspection: x=0 gives y=23, and y=0 gives x=−3. So the line passes through (0,23) and (−3,0).
The Line with Gradient m Through the Point (x1,y1)
If P(x,y) is any point on the line with gradient m passing through A(x1,y1), then the gradient of AP is m. Therefore the coordinates of P satisfy
Find the equation of the line with gradient −31 passing through (2,−1).
Substituting −31 for m, 2 for x1 and −1 for y1 gives the equation of the line as
y−(−1)=−31(x−2),that isx+3y+1=0.
Alternatively, as any straight line has an equation y=mx+c, the equation of this line can be written y=−31x+c. As the point (2,−1) lies on the line, its coordinates satisfy the equation, so −1=−32+c and c=−31. Therefore the equation is y=−31x−31, or x+3y+1=0.
Remark.
The worked examples necessarily contain a lot of explanation, but this should not mislead the reader into thinking that solutions need be equally long. The temptation to overwork a problem should be avoided, particularly in coordinate geometry, where problems are basically simple. With a little practice, either of the methods above gives the equation of a line directly.
The Line Through (x1,y1) and (x2,y2)
When x1=x2, the gradient of the line through (x1,y1) and (x2,y2) is x2−x1y2−y1, so by the previous section its equation is
y−y1=x2−x1y2−y1(x−x1).
For example, the line through (1,−2) and (3,5) has equation
y+2=3−15−(−2)(x−1),that is7x−2y−11=0.
Intersection
If two curves cut at a point A, then A is called a point of intersection of the curves. The coordinates of A satisfy both equations, so A can be found by solving the equations simultaneously.
so x=23 or x=1. Substituting these in y=x−1 gives y=21 and y=0. Therefore A and B are the points (23,21) and (1,0).
In general, the coordinates of the points of intersection of two curves y=f(x) and y=g(x) can be found from the simultaneous solution of the equations y=f(x) and y=g(x).
Find the equation of the line through (1,2) which is perpendicular to the line 3x−7y+2=0.
Writing 3x−7y+2=0 in standard form gives y=73x+72, showing that the given line has a gradient of 73. So the required line has gradient −37 and passes through (1,2). Using y−y1=m(x−x1) gives its equation as
y−2=−37(x−1),that is7x+3y−13=0.
Note that the line perpendicular to 3x−7y+2=0 has an equation 7x+3y−13=0: the coefficients of x and y have been transposed, and the sign between the x and y terms has changed. In fact, given the line ax+by+c=0, any line perpendicular to it has an equation
bx−ay+k=0,
and this property of perpendicular lines can be used to shorten the working of problems.
A, B and C are the points (0,4), (2,3) and (−2,−1). Find the circumcentre of triangle ABC.
The circumcentre of a triangle is the point of intersection of the perpendicular bisectors of its sides.
AC has gradient 0−(−2)4−(−1)=25, and its midpoint is (−1,23). Therefore the perpendicular bisector of AC has gradient −52 and passes through (−1,23), so its equation is
y−23=−52(x+1),that is4x+10y−11=0.(1)
Similarly the gradient of AB is 2−03−4=−21, and its midpoint is (1,27). Therefore the perpendicular bisector of AB has gradient 2 and passes through (1,27), so its equation is
y−27=2(x−1),that is4x−2y+3=0.(2)
Subtracting (2) from (1) gives 12y−14=0, so y=67, and then 4x=2y−3=−32 gives x=−61. Therefore the circumcentre of triangle ABC is the point (−61,67).
Find the equation of the line passing through (−1,3) and (−4,−3).
Find the equation of the line through (1,−2) perpendicular to 2x−3y+6=0.
Draw a sketch showing the region of the xy plane defined by y<2 and y>(x−2)(x+2).
Remark (Summary).
If A and B are the points (x1,y1) and (x2,y2), then
the length of AB is (x2−x1)2+(y2−y1)2;
the midpoint of AB is the point (21(x1+x2),21(y1+y2));
when x1=x2, the gradient of AB is x2−x1y2−y1. A vertical line has no finite gradient and has an equation of the form x=constant.
The equation y=mx+c defines the straight line with gradient m and intercept c on the y-axis. The inequality y>mx+c defines the region of the xy plane above that line, and y<mx+c the region below it.
If lines l1 and l2 have equations y=m1x+c1 and y=m2x+c2, then l1 and l2 are parallel if m1=m2, and, when both gradients are finite and non-zero, perpendicular if m1m2=−1. A horizontal line is perpendicular to a vertical line. The equation of any line perpendicular to ax+by+c=0 is of the form bx−ay+k=0.
Exercises
Questions marked with an examining board are taken from past A-level papers: JMB is the Joint Matriculation Board, U of L the University of London, C Cambridge, and AEB the Associated Examining Board.
In Utopia, assume income is non-negative. Income tax on earnings is calculated as follows: the first £10,000 is tax free, the next £10,000 is taxed at 5%, and the remaining income is taxed at 10%.
Taking income as input and tax payable as output, state whether these rules for calculating tax constitute a function. If they do, state the implied domain and range.
If £I is income and £T is the tax payable, express the mapping I↦T as formulae of the form T=f(I), stating the values of I for which each is valid.
Determine, for each of the expressions f(x)=x2+4x−6 and g(x)=−x2−8x+2, the range, or ranges, of values of x for which it is positive. Give your answers correct to two places of decimals, and explain briefly the reasons for your answers. (C)
State the range of values of x for which 2x2+5x−12 is negative.
The value of the constant a is such that the quadratic function f(x)=x2+4x+a+3 is never negative. Determine the nature of the roots of the equation af(x)=(x+2)(a−1), and deduce the value of a for which this equation has equal roots. (AEB, 1973)
If a>0, show that the quadratic expression ax2+bx+c is positive for all real values of x when b2<4ac. Hence find the range of values of p for which the quadratic function
f(x)=4x2+4px−(3p2+4p−3)
is positive for all real values of x. Illustrate your result by making sketch graphs of f(x) for each of the cases p=0 and p=1. (U of L)
If a is a positive constant, find the set of values of x for which a(x2+2x−8) is negative. Find the value of a if this function has a least value of −27.
Find two quadratic functions of x which are zero at x=1, which take the value 10 when x=0, and which have a greatest value of 18. Sketch the graphs of these two functions. (U of L)
Find the set of values of k for which f(x)=3x2−5x+k is greater than unity for all real values of x. Show that, for all k, the least value of f(x) occurs when x=65, and find k if this least value is zero. (U of L)
ABCD is a quadrilateral, where A, B, C and D are the points (3,−1), (6,0), (7,3) and (4,2). Show that the diagonals bisect each other at right angles, and hence find the area of ABCD.
A circle of radius two units, with its centre at the origin, cuts the x-axis at A and B and cuts the positive y-axis at C. Show that AB subtends a right angle at C. If D(a,b) is a point on the circumference of the circle, find a relationship between a and b.
A line is drawn through the point A(1,2) to cut the line 2y=3x−5 at P and the line x+y=12 at Q, with P between A and Q. If AQ=2AP, find the coordinates of P and Q. (U of L)
Check Yourself
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At which points do the curves y=x2 and y=x(2−x) intersect?
answerone of these
Lesson 4
Limits, and Differentiation
Taught
Limits
The expression
x−2(x+1)(x−2)
is not defined when x=2, where it would read 00. For every other value of x the factor x−2 cancels, and the expression is simply x+1. Its graph is the line y=x+1 with one point missing: a hole at x=2.
Figure 4.1. The graph of y=x−2(x+1)(x−2), which is the line y=x+1 with a hole at (2,3).
The expression is well defined for all values of x other than 2, so we can legitimately ask what happens to it as x approaches 2. Since it equals x+1 at all those values, it approaches 3. We say that its limit as x approaches 2 is 3, and write
x→2limx−2(x+1)(x−2)=3.
The value at x=2 itself plays no part. The limit is about the values of the expression near 2, and here there is no value at 2 at all.
Let f(x) be defined for all values of x near a, except possibly at a itself. We say that f(x)tends to the limitk as x tends to a, and write
x→alimf(x)=k,
if f(x) gets and stays as close to k as we want, provided x is sufficiently close to a.
It is not enough for the values to keep getting closer to k. The numbers 2.1,2.01,2.001,… get closer to 2, but each is also closer to 1 than the one before, and we want to say that they approach 2 and not 1. The difference is that they get as close to 2 as we want, however small we set the allowed distance, whereas every one of them stays more than 1 away from 1.
One-Sided Limits
Define f(x) to be x−1 when x is negative and x+1 when x is positive. Its graph is two half-lines with a jump between them at x=0.
Figure 4.2. The graph of f(x), equal to x−1 for negative x and to x+1 for positive x.
The limit of f(x) as x→0 does not exist. We can make x as close to 0 as we want, but f(x) does not get as close as we want to any one value: it is near −1 when x is negative and near 1 when x is positive. We met the same difficulty with x1 in the last lesson, where the behaviour depended on whether x approached zero from above or from below; here both sides settle down, but to different values. We write
x→0+limf(x)=1andx→0−limf(x)=−1,
the first being the limit as x tends to 0 from above, or from the right, and the second the limit from below, or from the left. The limit x→alimf(x) exists only when these two one-sided limits exist and are equal.
Limits at Infinity
We can also take limits as x tends to infinity. If g(x)=x1, then g(x) approaches 0 as x grows, and we write
x→∞limg(x)=0.
As x tends to minus infinity g(x) also approaches 0, so x→−∞limg(x)=0 as well. For a limit at infinity, ”x is sufficiently close to a” in the definition of a limit is replaced by ”x is large enough”, and for minus infinity by ”x is negative and large enough in size”.
The line touching the curve at A is the tangent to the curve at A.
The line through A perpendicular to the tangent at A is the normal to the curve at A.
Figure 4.3. A chord AB, and the tangent and normal at A.
A straight line has the same gradient all along it. The gradient of a curve, which measures its slope, changes continually as we move along it. Suppose that as we move along the curve towards A, the gradient stops changing at A and stays constant from then on. We would then be moving along a straight line, and that line is the tangent at A. So the gradient of the curve at A is the same as the gradient of the tangent at A.
The gradient of a curve at a point is the gradient of the tangent to the curve at that point. It measures the rate of increase of y with respect to x at that point.
If B is another point on the curve, not too far from A, the gradient of the chord AB is an approximate value for the gradient of the tangent at A, and the closer B is to A the better the approximation.
An approximate value can also be found by plotting the curve, drawing the tangent by eye and measuring its gradient. This method has to be used when we know the coordinates of a finite number of points on a curve but not its equation, as with the data from an experiment. When the equation of the curve is known we want an accurate method, so that we can take the analysis of curves and functions further.
Approaching the Tangent
Take the parabola y=x2 and the point (1.5,2.25) on it. The tangent there, drawn on the left of Figure 4.4, seems to rise about 3 units for each unit it moves to the right, so the gradient looks to be about 3. Is it exactly 3, and how would we find out?
Figure 4.4. The tangent to y=x2 at (1.5,2.25), and a close-up in which the dashed chord to (1.6,2.56) runs almost along the tangent.
To approximate it, take a second point on the parabola close to the first, such as (1.6,2.56), where 2.56=1.62 because the point lies on y=x2. The line through (1.5,2.25) and (1.6,2.56) is found as in the last lesson. Its gradient is
1.6−1.52.56−2.25=0.10.31=3.1,
and its equation is y=3.1x−2.4. So our approximation to the gradient is 3.1, and the close-up in Figure 4.4 shows how near the chord runs to the tangent.
The approximation cannot be made perfect by taking both points to be (1.5,2.25), since infinitely many lines pass through a single point and that tells us nothing. It can be made better. The chord through (1.5,2.25) and (1.501,2.253001) has gradient
1.501−1.52.253001−2.25=0.0010.003001=3.001,
so the gradients really do seem to be approaching 3. As B approaches A, the gradient of the chord AB approaches the gradient of the tangent at A, that is
B→Alim(gradient of chord AB)=gradient of tangent at A.
Find the gradient of the curve y=x(2x−1) at the point A where x=1.
When x=1, y=1, so A is the point (1,1). We calculate the gradient of a chord AB and observe what happens to it as B approaches A. Take a succession of points B1,B2,B3,… where x=1.5,1.25,1.125,…, each time halving the remaining difference between the x-coordinates of A and B.
B
B1
B2
B3
B4
B5
x
1.5
1.25
1.125
1.0625
1.03125
gradient of AB
4
3.5
3.25
3.125
3.0625
As B approaches A the gradient of the chord AB approaches 3, and we deduce that the gradient of the curve at A is 3.
Figure 4.5. The chords AB1, AB2 and AB3 on y=x(2x−1), turning towards the tangent at A(1,1).
The Delta Prefix
This numerical method is unsatisfactory, not least because of the amount of calculation involved, and in the end it only suggests the value 3. So instead of placing B at particular positions we introduce a variable quantity for the difference between the x-coordinates of A and B.
A variable quantity prefixed by δ means a small increase in that quantity. So δx is a small increase in x, and δy is a small increase in y.
Remark.
The letter δ is only a prefix. It cannot be treated as a factor: δx is one quantity, not δ multiplied by x.
Return to y=x(2x−1) and the point A(1,1). If δx is the increase in the x-coordinate in moving from A to B, then the x-coordinate of B is 1+δx. Every point on the curve satisfies y=x(2x−1), so the y-coordinate of B is
(1+δx)[2(1+δx)−1]=(1+δx)(2δx+1).
The gradient of AB is the increase in y divided by the increase in x:
δx(1+δx)(2δx+1)−1=δx3δx+2(δx)2=3+2δx.
As B approaches A, the difference between their x-coordinates approaches zero, that is δx→0. Therefore
gradient at A=δx→0lim(3+2δx)=3.
Dividing by δx is allowed because a limit only involves values of δx near zero, never zero itself.
The same calculation settles the question for the parabola. Writing h for the small increase in x from 1.5, the gradient of y=x2 at (1.5,2.25) is
Use the method of the last example to find the gradient of each curve at the point indicated.
y=(x+1)(x−1), where x=2;
y=2x(x−4), where x=0;
y=(x+2)(x−1), where x=−3.
The Gradient Function
We found that the gradient of y=x(2x−1) is 3 at the point where x=1. Instead of a fixed point, take A to be any point (x,y) on the curve. Its y-coordinate can be written x(2x−1), since both coordinates satisfy the equation of the curve. Let B be another point on the curve such that the increase in the x-coordinate in moving from A to B is δx. Then B is the point [x+δx,(x+δx)(2x+2δx−1)], and
gradient of chord AB=δx(x+δx)(2x+2δx−1)−x(2x−1)=δx2x2+4xδx+2(δx)2−x−δx−2x2+x=δx4xδx−δx+2(δx)2=4x−1+2δx.
Then the gradient at any point A on the curve is
δx→0lim(4x−1+2δx)=4x−1.
So the function 4x−1 gives the gradient at every point on the curve y=x(2x−1). The gradient at a particular point, that is the rate of increase of y with respect to x there, is found by substituting the x-coordinate of the point into 4x−1. At x=1 this gives 3, as before.
The function 4x−1 is called the gradient function of y=x(2x−1), and the process of deriving it is called differentiation with respect to x. Since 4x−1 was derived from x(2x−1), it is also called the derivative, or derived function, of x(2x−1). Using dxd as a symbol for “the derivative with respect to x of”, we write
dxd[x(2x−1)]=4x−1,
or, when y=x(2x−1),
dxdy=4x−1.
Here dxdy means the derivative of y with respect to x, and is sometimes called the differential coefficient of y. As an alternative notation we can use the symbol D for “the derivative of”, so that D[x(2x−1)]=4x−1, and D is referred to as the differential operator.
Since 4x−1 represents the rate of increase of y with respect to x, dxdy also represents the rate of increase of y with respect to x. Similarly dtdv means the derivative of v with respect to t, or the rate of increase of v with respect to t.
Use the method of the last example to differentiate each of the following with respect to x.
y=x2;
y=x4;
y=5x2;
y=x1.
Differentiating from First Principles
Consider any curve y=f(x). Let A[x,f(x)] be any point on it, and let δx be the increase in the x-coordinate in moving from A to another point B on the same curve, so that B is the point [x+δx,f(x+δx)]. Let δy be the corresponding increase in the y-coordinate, that is
δy=f(x+δx)−f(x).Figure 4.6. The increases δx and δy in moving from A to B along y=f(x).
Let f be a function and c a number. The derivative of f at c is the gradient of the graph of f at the point where x=c, defined by
h→0limhf(c+h)−f(c).
A function for which this limit exists is differentiable at c. The derivative of f(x) is written f′(x) or dxdf(x), and when y is a function of x that we would graph, it is written dxdy.
The limit exists when the graph looks as though it has a slope at the point. If the graph had a spike there, it would not have a well-defined slope and the limit would not exist.
Using this definition to differentiate a function is called differentiating from first principles, and it can be used for any function, including functions we have not yet met. Fortunately it is not always necessary to go back to first principles, because whole families of functions can be differentiated by rules.
Remark.
We cannot tell how fast something is going from a single photograph of it. But we can photograph it at two nearby moments and divide the distance it moved by the time between the photographs, which gives a good approximation to its speed, and the closer the two moments, the better the approximation. This is how instruments that measure speed work, and it is the gradient of a chord standing in for the gradient of a tangent.
Rules of Differentiation
Differentiating a Constant
The equation y=c, where c is a constant, represents a straight line parallel to the x-axis, so it has zero gradient:
dxd(c)=0.
Differentiating ax
The equation y=ax, where a is a constant, represents a straight line with gradient a, so
dxd(ax)=a.
Differentiating xn
The table collects results from the examples and problems above.
f(x)
x
x2
x3
x4
x−1
dxdf(x)
1
2x
3x2
4x3
−x−2
From this table it appears that to differentiate a power of x we multiply by that power and then subtract one from the power, that is
dxd(xn)=nxn−1.(1)
This is the power rule.
The Power Rule for Whole Numbers
If we replace 1.5 by x and do the same calculation as for the parabola, the algebra gives
(x+h)2−x2=x2+2xh+h2−x2=2xh+h2,
and after dividing by h the remaining h vanishes in the limit, so the parabola y=x2 has gradient 2x at each value of x. Expanding further powers shows a pattern:
In each line the term with h to the first power is nxn−1h, and every other term contains h2 or a higher power of h. When n is a positive whole number we show that this always happens by using the binomial expansion. The coefficients of (x+h)n are those of (1+x)n, the numbers in row n of Pascal’s triangle, so
(x+h)n=xn+(second number in row n)xn−1h+(terms containing h2,h3,…).
The second number in a row is the sum of the two numbers above it, which are 1 and the second number of the row before. So it goes up by 1 from each row to the next, and as it is 1 in row one, it is n in row n. Therefore
h(x+h)n−xn=nxn−1+(terms containing h,h2,…),
and as h→0 this tends to nxn−1.
The power rule, deduced here for whole numbers, is valid for all powers of x, including fractional and negative powers, although this cannot be shown at this stage, and for the time being we take it on trust. Under the power conventions in Lesson 1, if n is not a whole number then we use xn only where it is defined, and for the fractional powers considered here this means x>0. The rule needs the power to be a fixed number: it does not find the gradient of xx, where the power changes with x.
Gradients do not multiply in the way they add. So in order to differentiate at this stage, any product must be expanded and any quotient must be divided out, to give terms which are added or subtracted, and no rule should be assumed that has not been established.
Differentiate (x−3)(2x+5) and x(x−1) with respect to x.
Find the gradient of the curve y=(2x−3)(x+1) at the point where x=0.
Find the coordinates of the point on the curve y=x at which the gradient is 2.
Tangents and Normals
The tangent to a curve at a point passes through that point and has the gradient of the curve there, so its equation comes from the line with a given gradient through a given point. When the tangent has non-zero gradient m, the normal passes through the same point with the perpendicular gradient−m1. A horizontal tangent has a vertical normal.
Take y=x2 at (1.5,2.25) once more. The tangent is the line of gradient 3 through (1.5,2.25), which is
y−2.25=3(x−1.5).
Rearranging, 4y−12x+9=0 is the equation of the tangent.
The normal to y=x2 at (1.5,2.25) is the line through that point perpendicular to the tangent, so its gradient is −31 and its equation is
y−2.25=−31(x−1.5),that isy+31x−411=0.
Multiplying through by 12 to clear the fractions gives 12y+4x−33=0.
Figure 4.7. The parabola y=x2 with its tangent 4y−12x+9=0 and its normal 12y+4x−33=0 at (1.5,2.25).
Find the equation of the tangent to the curve y=x2−3x+2 at the point where it cuts the y-axis.
The curve cuts the y-axis where x=0, at (0,2). The gradient of the tangent there is the value of dxdy when x=0. As y=x2−3x+2, dxdy=2x−3, so when x=0 the gradient of the curve is −3.
Therefore the tangent has gradient −3 and passes through (0,2). Its equation is y=−3x+2, that is
Find the equation of the normal to the curve y=x1 at the point where x=2, and the coordinates of the point where this normal cuts the curve again.
As y=x−1, dxdy=−x21. So when x=2, y=21 and dxdy=−41. Therefore the tangent at (2,21) has gradient −41, and the normal, the line perpendicular to the tangent, has gradient 4. The equation of the normal is
y−21=4(x−2),that is8x−2y−15=0.
The points of intersection of the normal and the curve are given by solving 8x−2y−15=0 and y=x1 simultaneously. Substituting for y,
8x−x2−15=0,that is8x2−15x−2=0.
We already know that the normal and the curve meet where x=2, so x−2 is a factor, and
(x−2)(8x+1)=0.
Therefore they meet again where x=−81 and y=−8, at the point (−81,−8).
Figure 4.8. The normal to y=x1 at (2,21) meets the other branch of the curve at (−81,−8).
A stationary value of a function f(x) is a value of f(x) at which its rate of change with respect to x is zero, that is, where
dxdf(x)=0.
A point on the curve y=f(x) at which dxdy=0 is a stationary point.
At a stationary value of f(x) the gradient of the curve y=f(x) is zero, so the tangent to the curve is parallel to the x-axis. The stationary values of f(x) are the y-coordinates of the points on y=f(x) at which the tangent is parallel to the x-axis.
The curve y=x3−2x2 in Figure 4.9 goes flat and turns round in two places, and the derivative tells us exactly where. We found that dxd(x3−2x2)=3x2−4x, and the graph is flat where
3x2−4x=0,that isx(3x−4)=0.
The solutions are x=0 and x=34, so it is exactly at these two values of x that the graph is flat and turns round, at (0,0) and (34,−2732). A zero derivative does not always mean that the graph turns round, as we shall see with y=x3.
Figure 4.9. The curve y=x3−2x2, with horizontal tangents at (0,0) and (34,−2732).
If f(x)=x3−3x2+2, then dxdf(x)=3x2−6x. At stationary values of f(x) this is zero, so
3x2−6x=0,3x(x−2)=0.
So the stationary values of x3−3x2+2 occur when x=0 and when x=2. They are
f(0)=2andf(2)=23−3(22)+2=−2.
For the curve y=x3−3x2+2, the gradient is zero at the points (0,2) and (2,−2).
The derivative also tells us which way a function is going. Where dxdy is positive the curve is rising, and we say that y is increasing; where dxdy is negative the curve is falling, and y is decreasing. For y=x3−2x2 the derivative x(3x−4) is positive for x<0, negative for 0<x<34 and positive again for x>34, which is what Figure 4.9 shows.
Find the values of x at which x3−12x+1 has stationary values.
Find the stationary value of 3x2−4x+2.
Find the coordinates of the point on y=(x−2)(x+3) at which the gradient is zero.
Turning Points
The gradient of a curve can be zero at several points. Close to one of these points, the shape of the curve belongs to one of the three kinds shown at A, B and C in Figure 4.10.
Figure 4.10. A maximum turning point A, a minimum turning point B, and a point C where the gradient is zero but the curve does not turn.
Moving along the curve in the positive direction of the x-axis:
near A the gradient changes from positive, through zero at A, to negative;
near B the gradient changes from negative, through zero at B, to positive;
at C the gradient is zero, but it does not change sign as we move through C, so the curve does not turn at C. What does change at C is the sense in which the curve is turning, from clockwise to anticlockwise.
A point on a curve at which the gradient changes from positive, through zero, to negative is a maximum turning point. Its y-coordinate is a maximum value of y, or of f(x) where y=f(x).
A point at which the gradient changes from negative, through zero, to positive is a minimum turning point. Its y-coordinate is a minimum value of y, or of f(x).
A point on a curve at which the sense of turning changes is a point of inflexion.
The gradient at a maximum or minimum turning point must be zero. The terms maximum value and minimum value do not mean the same as greatest value and least value: maxima and minima describe the behaviour of a function only in the immediate neighbourhood of its stationary values. In Figure 4.10, to the left of A, the curve falls below the minimum value at B.
Apart from C there are two other points of inflexion in Figure 4.10, one between A and B and another between B and C, so the gradient at a point of inflexion is not necessarily zero. A stationary point can therefore be a maximum, a minimum, or neither. The simplest example of the last kind is y=x3 at the origin: its gradient 3x2 is zero there, but the curve rises on both sides.
Figure 4.11. The curve y=x3, stationary at the origin without turning.
The Nature of a Stationary Value
We know how to find the points on y=f(x) at which f(x) has stationary values, but to tell them apart we need to look further, and there are several ways of doing this. In Figure 4.10, let A1 and A2 be points on the curve close to A and to its left and right respectively, and let B1, B2 and C1, C2 be placed in the same way about B and C.
Values of y
At A, a maximum, y at A1 and y at A2 are both less than y at A. At B, a minimum, y at B1 and y at B2 are both greater than y at B. At C, a point of inflexion, y at C1 is less than y at C and y at C2 is greater.
Maximum
Minimum
Inflexion
Values of y either side of the stationary value
both smaller
both larger
one smaller and one larger
The Sign of the Gradient
At A1 the gradient dxdy is positive, at A it is zero and at A2 it is negative. At B1 it is negative, at B zero and at B2 positive. At C1 it is positive, at C zero and at C2 positive again.
Maximum
Minimum
Inflexion
Sign of dxdy moving through the stationary value
+0−
−0+
+0+ or −0−
The Second Derivative
When passing through A, dxdy changes from positive to negative, so dxdy decreases as x increases, that is, the rate of increase of dxdy with respect to x is negative. The rate of increase of dxdy with respect to x would be written dxd(dxdy), and this clumsy notation is condensed to
dx2d2y,
called the second derivative of y with respect to x. Similarly, when passing through B, dxdy changes from negative to positive, so dxdy increases as x increases and dx2d2y is positive.
Maximum
Minimum
Sign of dx2d2y
negative (or zero)
positive (or zero)
Points of inflexion are not so easily dealt with by this method. In the smooth examples considered here, dx2d2y=0 at such points, but dx2d2y can also be zero at maxima and minima.
The three tables summarise three alternative methods for determining the nature of a stationary value. The third method fails if dx2d2y is found to be zero at the stationary value, and in that case one of the first two has to be used.
Find the points on y=x4+4x3−6 at which the gradient is zero, and determine the nature of these points.
y=x4+4x3−6,dxdy=4x3+12x2,dx2d2y=12x2+24x.
The gradient is zero when 4x3+12x2=0, that is 4x2(x+3)=0, so when x=−3 and when x=0.
When x=−3, dx2d2y=12(9)+24(−3)=36, which is positive, so y has a minimum value here. It is y=(−3)4+4(−3)3−6=−33, so (−3,−33) is a minimum turning point.
When x=0, y=−6 and dx2d2y=0, which is inconclusive. So we look at the sign of dxdy=4x2(x+3) on either side of the point where x=0.
x
−21
0
21
dxdy
+
0
+
The gradient does not change sign, so (0,−6) is a point of inflexion.
Finding the maximum and minimum turning points gives a general idea of the shape and position of the curve.
y=2x3+x2−4x+1,dxdy=6x2+2x−4,dx2d2y=12x+2.
At turning points 6x2+2x−4=0, that is 2(3x−2)(x+1)=0, so x=32 or x=−1.
When x=32, dx2d2y=12(32)+2>0 and y=2(32)3+(32)2−4(32)+1=−2717, so (32,−2717) is a minimum turning point.
When x=−1, dx2d2y=12(−1)+2<0 and y=2(−1)3+(−1)2−4(−1)+1=4, so (−1,4) is a maximum turning point.
As y=2x3+x2−4x+1, the curve cuts the y-axis at (0,1).
From these three results we can sketch the curve.
Figure 4.12. A sketch of y=2x3+x2−4x+1 from its turning points and its intercept on the y-axis.
In some cases the intercepts on the x-axis can be found as well, although the equation which gives them is not always easy to solve. Here it is 2x3+x2−4x+1=0.
A farmer has an adjustable electric fence that is 100 m long. It is used to enclose a rectangular grazing area on three sides, the fourth side being a fixed hedge. Find the maximum area that can be enclosed.
The length of the enclosure can be varied, up to 100 m, and the width and area then depend on the length chosen. Let the length SR be x m. Then as PS+SR+RQ=100,
PS=RQ=21(100−x),
and the area A of the enclosure, in square metres, is
A=x[21(100−x)]=50x−21x2.
Now A is a quadratic function of x with a negative coefficient of x2, so from what we know of quadratic functions it is greatest at x=50, halfway between its zeros at x=0 and x=100. So the maximum area that can be enclosed is 50×25=1250 square metres.
The maximum value of 50x−21x2 can also be found by differentiation. We have
dxdA=50−x,
and A has a stationary value when dxdA=0, that is when x=50. Now dx2d2A=−1, which is negative, so this is a maximum, and the maximum value is 25×50=1250 square metres.
Figure 4.13. The enclosure against the hedge, and its area A=50x−21x2 as the length x varies.
The second method, using differentiation, is necessary when finding maximum or minimum values of functions that are not quadratic. For quadratic functions the first method is preferable, since their properties let us find the maximum or minimum value by inspection. In this case the greatest value and the maximum value are the same.
A rectangle has a perimeter of 20 units. Find its greatest possible area.
The perimeter, the total length of the edges, is twice the short side plus twice the long side, so the two sides add up to 10. So if one side has length x, the other has length 10−x, and the area of the rectangle is
x(10−x)=10x−x2.
This is differentiable everywhere, with derivative 10−2x. At its greatest value, either its gradient is 0 or x is at one of the ends of its range, x=0 or x=10. At the ends the area is 0, so we solve 10−2x=0, which gives x=5. The area there is 10(5)−52=25, so the greatest area is 25 square units.
The point where the gradient is zero is a maximum and not a minimum, for three separate reasons.
x=5 is the only place where 10x−x2 can turn round, and at values of x on either side of it, such as 0 and 10, the area is less than 25. If the area were more than 25 at x=6, say, the curve would have to turn round again to reach (10,0), and it does not.
The gradient 10−2x is positive to the left of x=5 and negative to the right.
The graph of 10x−x2 is a parabola opening downwards, so its only flat point is a maximum.
Remark.
If f(x)=ax2+bx+c, then f′(x)=2ax+b, so f(x) has a stationary value where x=−2ab, on the axis of the curve. If α and β are the roots of ax2+bx+c=0, where the curve crosses the x-axis, then by the sum of the roots21(α+β)=−2ab, so the turning point of the curve has x-coordinate 21(α+β), halfway between the roots. A ball thrown through the air follows such a parabola, if we ignore air resistance.
Find the stationary values of each of the following functions and determine their nature.
2x3−3x2−12x;
(x−3)(2x+1);
x3+3.
The Number e
Exponential Growth
The curve y=2x of the last lesson increases as x increases. Its values at x=−3,−2,−1,0,1,2,3 are 81,41,21,1,2,4,8, and the graph grows quite fast. Differentiating from first principles shows how fast:
h2x+h−2x=h2x⋅2h−2x=2x⋅h2h−1,
and the factor h2h−1 does not involve x at all. Its values, and those of the corresponding factor for 3x, as h approaches zero are
h
0.1
0.01
0.001
0.0001
h2h−1
0.7177
0.6956
0.6934
0.6932
h3h−1
1.1612
1.1047
1.0992
1.0987
so the gradient of 2x is about 0.693×2x, and the gradient of 3x is about 1.099×3x.
The rate of change of an exponential function is proportional to its current value. The larger the function gets, the faster it grows, which makes it larger still; this runaway growth is called exponential growth. It is what happens with the spread of a disease, or with interest in a bank.
A base for which the factor is exactly 1 would give a function whose derivative is equal to the function itself. The factor is less than 1 for base 2 and greater than 1 for base 3, and there is a number between 2 and 3 for which it is exactly 1.
The number e, which is about 2.71828182845904523536…, is the base for which
dxd(ex)=ex.
With e in place of 2 the factor takes the values 1.0517, 1.0050, 1.0005 and 1.0001 for the four values of h in the table. That such a number exists, and that ex has exactly this derivative, we accept on trust at this stage, as we did the power rule for fractional powers.
Figure 4.14. The curve y=ex. At every point the gradient equals the height: the tangent at (0,1) has gradient 1, and the tangent at (1,e) has gradient e.
The natural logarithm of a positive number x is its logarithm to base e, written
lnx=logex.
So lnx is the number y for which ey=x. Since loga1=0 and loga(ak)=k for every base, straight from the definition of a logarithm, we have ln1=0, lne=1 and ln(ek)=k.
As a function from the positive numbers to the real numbers, lnx is the inverse of the function ex from the real numbers to the positive numbers, in the same way that log2x is the inverse of 2x. Its graph is the reflection of y=ex in the line y=x, and has the same shape as the graph of y=log2x.
Remark.
People often write logx with no base. If you are a physicist or an engineer this usually means log10x, but mathematicians usually mean lnx. In these notes base 10 is written lg, base e is written ln, and log without a base appears only in rules that hold for every base.
Find the coordinates of the stationary point of y=ex−3x, and determine its nature.
dxdy=ex−3,dx2d2y=ex.
The gradient is zero when ex=3, that is when x=ln3, and there y=eln3−3ln3=3−3ln3. Since ex is positive for every x, dx2d2y>0, so (ln3,3−3ln3), which is about (1.099,−0.296), is a minimum turning point.
Find the equations of the tangents to the curve y=(2x−1)(x+1) at the points where the curve cuts the x-axis, and find the point of intersection of these tangents.
Find the coordinates of the point on y=x2−7x+3 at which the gradient is 2. Hence find the equation of the normal to y=x2−7x+3 which is parallel to x+2y−1=0.
Show that (t,t1) lies on the curve y=x1 for all non-zero values of t. Find the equation of the tangent to y=x1 at (t,t1), and the area of the triangle enclosed by this tangent and the coordinate axes.
An open rectangular box is made from a square sheet of cardboard by removing a square from each corner and joining the cut edges. If the cardboard has edge 0.5 m, find the maximum volume of the box.
A cylinder is cut from a solid sphere of radius 5 cm, so that the curved edges of the cylinder reach the surface of the sphere. If the height of the cylinder is 2h, show that its volume is 2πh(25−h2), and find the maximum volume of such a cylinder.
The table shows the results of an experiment in which some hot liquid was left to cool, its temperature θ °C being measured at intervals of one minute.
Time t (minutes)
0
1
2
3
4
5
Temperature θ (°C)
100
93
87
83
80
78
What does dtdθ represent? By drawing a graph of these results, estimate the rate of decrease of the temperature after three minutes.
Show that the tangent to y=ex at the point where x=1 passes through the origin. Hence find the value of k for which the line y=kx is a tangent to y=ex.
Check Yourself
Fresh questions on the whole chapter — none of them is worked out above. Do each on paper first; the box only tells you whether you got there.
Answers are checked in your browser, as often as you like. Nothing is sent anywhere and
nothing is kept but your own work. A formula may be written with the symbols themselves or
with ~ & | -> <-> ^, and \and, \or, \to expand as you type.
When a line OP is pivoted at O and rotates from its initial position OP0 to a new position OP1, the angle P0OP1 is a measure of the rotation of OP.
Figure 5.1. The angle θ measures the rotation of OP from OP0 to OP1.
The angle θ is usually measured in one of two units.
The Degree
The ancient Babylonian mathematicians, who thought that the solar year was 360 days long, divided one complete revolution into 360 equal parts, each part now being known as one degree, 1∘. Using the degree as the unit of rotation, half a revolution corresponds to 180∘ and a quarter of a revolution, that is a right angle, corresponds to 90∘.
Angles smaller than a degree are usually given as decimal parts, so that half a degree is 0.5∘. In some fields, such as navigation, a degree is divided into 60minutes, written 60′, and each minute into 60seconds, written 60′′.
Converting in this way is not convenient for angles which are not simple fractions of a revolution. It would not be easy, for instance, to express 47∘34′ as a multiple of π. We would first have to express 34′ as a decimal part of a degree, 34′=0.567∘, and then use
47.567∘=47.567×180π=0.830 radians.
Calculations like this are tedious, and a calculator which offers the conversion avoids them.
To visualise the size of an angle of one radian, it helps to remember that π radians =180∘ and that π=3.14 to two decimal places, so
1 radian=3.14180∘=57.3∘to three significant figures,
These are the sine, cosine, tangent, cosecant, secant and cotangent of θ.
Each ratio has a single value for any one acute angle, since all right-angled triangles containing θ are similar, and the values are available from a calculator.
Length of an Arc and Area of a Sector
The circumference of a circle of radius r is 2πr and its area is πr2, and these formulae can be used to derive further results.
Consider an arc which subtends an angle θ at the centre of the circle, where θ is measured in radians. From the definition of a radian, the arc which subtends 1 radian at the centre has length r, so an arc which subtends θ radians at the centre has length rθ.
The area of a sector containing an angle of θ radians at the centre can be found by regarding the sector as a fraction of the circle. The ratio of the area of the sector to the area of the circle is equal to the ratio of the angle θ contained in the sector to the angle 2π contained in the whole circle, that is
πr2area of sector=2πθ,soarea of sector=21r2θ.Figure 5.4. A sector AOB containing the angle θ at the centre O.
Hence if an arc AB subtends an angle of θ radians at the centre O of a circle of radius r,
A chord AB divides a circle of radius 2 m into two segments. If AB subtends an angle of 60∘ at the centre O of the circle, find the area of the minor segment.
The triangle AOB has base OA=2 and height 2sin60∘, so
area of triangle AOB=21×2×2sin60∘=1.732m2.
Since 60∘=3π radians,
area of sector AOB=21×22×3π=32π=2.094m2.
So the area of the minor segment, shaded in Figure 5.5, is (2.094−1.732)m2=0.362m2.
Figure 5.5. The minor segment cut off by a chord subtending 60∘ at the centre.
Two discs of radii 3 cm and 4 cm are laid on a table with their centres 5 cm apart. Find the perimeter of the figure-eight shape so formed.
Let the discs have their centres at A and B and cross at C. Since 52=32+42, the triangle ABC is right-angled at C. Let α be the angle CAB and β the angle CBA. Then
The perimeter is made up of an arc subtending 2π−2α radians in the circle of radius 3 cm and an arc subtending 2π−2β radians in the circle of radius 4 cm. Hence, using length of arc =rθ,
perimeter=3(6.283−1.854)+4(6.283−1.287)=33.3 cm.
Figure 5.6. The figure-eight formed by the two discs, with its perimeter drawn heavily.
The moon subtends an angle of 31′ at the earth, and its distance from the earth is 382100 km. Find the diameter of the moon in kilometres.
An arc AB of length 5 cm is marked on a circle of radius 3 cm. Find the area of the sector bounded by this arc and the radii to A and B.
A chord AB of length 5.2 cm subtends an angle of 120∘ at the centre of a circle. Calculate the length of the arc AB, the area of the sector containing the angle of 120∘, and the area of the minor segment cut off by AB.
The Ratios of a General Angle
Since we now regard an angle as the measure of the rotation of a line about a fixed point, the size of an angle is unlimited, because the line can keep on rotating indefinitely. The six trigonometric ratios, however, have so far been given a meaning only for acute angles, since each is defined by an angle in a right-angled triangle. To use them for angles of any size they must be defined in a more general way.
The system of reference in which a general angle is measured is very similar to that used for polar coordinates. The point about which the line OP rotates is the pole or origin O, and the position from which the angle is measured is the initial line, the x-axis. An angle formed when the line rotates anticlockwise is positive, while clockwise rotation gives a negative angle. The pair of Cartesian axes divides the plane into four quadrants, numbered 1, 2, 3 and 4 as in Figure 5.7.
Figure 5.7. The four quadrants, and the positive and negative senses of rotation.
As the line OP rotates, P moves round the first quadrant, where both its coordinates are positive. As OP moves into the second quadrant its x-coordinate becomes negative. In the third quadrant both coordinates of P are negative, and in the fourth quadrant the x-coordinate is positive and the y-coordinate negative. The length r of OP, the radius vector, is always taken to be positive.
If the line OP has rotated through the angle θ from the positive x-axis, and P is the point (x,y) with OP=r, then
sinθ=ry,cosθ=rx,tanθ=xy,
and cosecθ=yr, secθ=xr, cotθ=yx, each ratio being defined whenever its denominator is not zero.
For an acute angle these agree with the ratios of the right-angled triangle formed by O, P and the foot of the perpendicular from P to the x-axis.
The numerical values of the ratios can be found as follows. From any position of P, the perpendicular from P meeting the x-axis at Q forms a right-angled triangle OPQ. The angle POQ formed in this way is always acute, whatever the value of θ, and is called the associated acute angle, α. For a particular value of θ, α is the difference between θ and 180∘ or 360∘, or a further multiple of 180∘ for larger angles. For example
θθ=160∘:=275∘:αα=180∘−160∘=20∘,=360∘−275∘=85∘,θθ=67π:=517∘:αα=67π−π=6π,=540∘−517∘=23∘.Figure 5.8. The associated acute angle α when P is in each of the four quadrants.
In the triangle OPQ the lengths PQ and OQ are rsinα and rcosα, so the ratios of θ have the numerical values of the ratios of α, and their signs depend on the signs of x and y, that is on the quadrant into which P has rotated.
In the first quadrant all six ratios are positive, and since θ is acute their values are those of θ itself. If OP has rotated through more than a complete revolution, α=θ−360∘.
In the second quadrant α=180∘−θ. The sine ratio ry is positive, the cosine ratio rx is negative and the tangent ratio xy is negative, so sinθ=+sinα, cosθ=−cosα and tanθ=−tanα.
In the third quadrant α=θ−180∘. The tangent ratio is positive while the sine and cosine ratios are both negative, so sinθ=−sinα, cosθ=−cosα and tanθ=+tanα.
In the fourth quadrant α=360∘−θ. The cosine ratio is positive but the sine and tangent ratios are negative, so sinθ=−sinα, cosθ=+cosα and tanθ=−tanα.
These results are summarised in the quadrant diagrams of Figure 5.9. The second shows in each quadrant the ratios which are positive there: all of them in the first, the sine in the second, the tangent in the third and the cosine in the fourth. The quadrant rule, together with the value of the associated acute angle, gives the value of any trigonometric ratio of any angle.
Figure 5.9. Quadrant diagrams for the signs of the sine, cosine and tangent.
Negative Angles
If OP rotates clockwise, so that P moves through the quadrants in the reverse order, 4th, 3rd, 2nd, 1st, then θ is negative. Every position of OP can be reached either by anticlockwise or by clockwise rotation, and so corresponds to two different values of θ, one positive and one negative. For example, θ=+120∘ and θ=−240∘ give the same position, with α=60∘ in both cases, so the angles +120∘ and −240∘ have the same trigonometric ratios. P is in the second quadrant, where only the sine ratio is positive, hence
sin120∘cos120∘tan120∘=sin(−240∘)=+sin60∘,=cos(−240∘)=−cos60∘,=tan(−240∘)=−tan60∘.Figure 5.10. The same position of OP reached by rotations of +120∘ and −240∘.
Turning through −θ instead of θ reflects P in the x-axis, which changes the sign of y and leaves x unchanged. So for every angle
θ is in a quadrant where the cosine ratio is positive and the tangent ratio is negative, that is in the fourth quadrant, where α=360∘−θ. Now cosα=0.866 gives α=30∘, so θ=330∘. In the fourth quadrant the sine ratio is negative, therefore
Given that tanθ=1 and −2π⩽θ⩽2π, give four possible values of θ.
The tangent ratio is positive in the first and third quadrants. When tanα=1, α=4π, or 45∘. The range of values of θ is specified in radians, so the solution should also be given in radians:
θ=4π,45π,−43πor−47π.
In this example we are solving a simple trigonometric equation.
An angle θ has an associated acute angle of 53∘. Find sinθ.
Only α is given, so θ could be in any of the four quadrants. In quadrants 1 and 2, sinθ=sin53∘=0.7986, and in quadrants 3 and 4, sinθ=−sin53∘=−0.7986. Therefore sinθ=±0.7986.
Find the sine, cosine and tangent of 300∘ and of −160∘ in terms of the ratios of their associated acute angles, and evaluate them.
Within the range −360∘⩽θ⩽360∘, give all the values of θ for which cosθ=−0.5, and all those for which tanθ=1.2.
Find the smallest angle, positive or negative, for which cosθ=0.8 and sinθ is positive, and the smallest for which sinθ=−0.6 and tanθ is negative.
Solving Triangles
Triangles are involved in many practical measurements, in surveying for instance. A triangle has three sides and three angles, and its size and shape can be specified by suitable data, such as three sides, two angles and a side, or two sides and their included angle. The third angle is not an independent item: the angles of a triangle add up to 180∘, so if two of them are known the third follows directly. Two sides and a non-included angle can sometimes give two different triangles. From any data sufficient to define a triangle the remaining sides and angles can be calculated. This is called solving the triangle, and it uses one of the formulae that relate the sides and angles of a triangle. The two used most frequently are the sine rule and the cosine rule.
In a triangle ABC we use A, B and C to denote the angles at the vertices A, B and C, and a, b and c to denote the sides opposite these vertices.
The Sine Rule
In a triangle ABC,
sinAa=sinBb=sinCc.
We show this by taking O and R as the centre and the radius of the circle through A, B and C, drawing the diameter AD and joining DB, as in Figure 5.11. Then ∠ABD=90∘, the angle in a semicircle, so in the right-angled triangle ABD, whose hypotenuse AD is 2R,
c=2RsinD.
In diagram (i), C=D, since they are angles in the same segment. In diagram (ii), C=180∘−D, since they are opposite angles of a cyclic quadrilateral, and so sinC=sinD by the rule for the second quadrant. In both diagrams, therefore, sinC=sinD, and hence
c=2RsinC,that issinCc=2R.
Similarly sinBb=2R and sinAa=2R, so
sinAa=sinBb=sinCc=2R.Figure 5.11. The circle through A, B and C, with the diameter AD.
Any pair of these three equal ratios gives an equation containing two sides and two angles. So the sine rule can be used to solve a triangle in which we know either two sides and one angle, or two angles and one side, provided that one given side is opposite a given angle.
In a triangle ABC, A=73∘, B=49∘ and a=12.2 cm. Find b and c.
As a, A and B are known, b can be calculated using
sinBb=sinAa,b=sin73∘12.2sin49∘=9.63 cm.
Also C=180∘−73∘−49∘=58∘, so c can be calculated using
sinCc=sinAa,c=sin73∘12.2sin58∘=10.82 cm.
In this example the given data define one and only one triangle.
The Ambiguous Case
Sometimes, when two sides and one angle are specified, two different triangles can be found from the data. Suppose that we have to solve the triangle in which A=24∘, c=2.6 cm and a=1.1 cm. Knowing a, A and c we can find C using
sinAa=sinCc,sinC=1.12.6sin24∘=0.9614.
In a triangle, sinC=0.9614 gives C=74∘ or C=106∘, since the sine is positive in both the first and the second quadrants. We must check whether both are possible values for C.
If C=74∘, then A+C=98∘, which is less than 180∘. So 74∘ is a possible value for C, corresponding to B=82∘.
If C=106∘, then A+C=130∘, which is also less than 180∘. So 106∘ is a possible value too, corresponding to B=50∘.
So there are two possible triangles with the given data. This is known as the ambiguous case, and it is easily understood by attempting to construct the triangle from its specification: an arc of radius 1.1 about B cuts the line from A in two places. Each position of C corresponds to a different pair of values for B and b, and in each case the solution of the triangle is completed using sinAa=sinBb, which gives b=2.68 cm and b=2.07 cm.
Figure 5.12. The two positions C1 and C2 of the vertex C in the ambiguous case.
There are not always two possible triangles when one angle and two sides are given, as the next example shows.
If A=37∘, a=4.59 cm and c=2.1 cm, show that there is only one possible triangle ABC, and find its remaining angles.
Using sinAa=sinCc,
sinC=4.592.1sin37∘=0.2753,soC=16∘orC=164∘.
If C=16∘ and A=37∘, then A+C=53∘, which is less than 180∘, so 16∘ is a possible value for C, corresponding to B=127∘. If C=164∘, then A+C=201∘, which is more than 180∘, so 164∘ is not a possible value. Hence there is only one triangle defined by the given data, and its other angles are C=16∘ and B=127∘.
In each part the data refer to a triangle in the standard notation.
B=35∘, a=2.7 cm and b=5.1 cm; find A.
b=3.8 cm, A=25∘ and a=1.8 cm; find the possible values of C.
In a triangle PQR the angle PQR is 30∘ and the angle QPR is θ. Show that sinθ=2qp.
The Cosine Rule
The sine rule can be applied only when we know an angle opposite a given side. It cannot be used, for instance, when two sides and the included angle are given. Such cases are solved using the cosine rule,
a2=b2+c2−2bccosA.
We show this by placing the triangle ABC on Cartesian axes with A at the origin and AB along the positive x-axis, as in Figure 5.13. In diagram (i) the coordinates of B and C are (c,0) and (bcosA,bsinA). In diagram (ii), where A is obtuse, they are (c,0) and (−bcos(180∘−A),bsin(180∘−A)), which by the rule for the second quadrant is again (bcosA,bsinA). So by the length of the line joining two points,
where sin2A means (sinA)2. Since C is at a distance b from the origin, (bcosA)2+(bsinA)2=b2, and therefore
a2=b2+c2−2bccosA.
Similarly it can be shown that
b2=c2+a2−2cacosB,c2=a2+b2−2abcosC.Figure 5.13. The triangle ABC placed on axes, with A acute and with A obtuse.
The calculation involved in using the cosine rule is less straightforward than that required for the sine rule, so the cosine rule is used only when the sine rule is inapplicable.
In a triangle ABC, a=17.5 cm, b=8.4 cm and c=11.9 cm. Find the largest angle.
The longest side is opposite the largest angle, so we must find A. As a, b and c are given we use the cosine rule, rearranged so that A can be found conveniently:
Using cosA=2bcb2+c2−a2, show that A is acute if a2<b2+c2 and obtuse if a2>b2+c2, and check that taking A=90∘ gives the result of Pythagoras.
Find the angles of a triangle whose sides are in the ratio 2:3:4.
The Graphs of the Circular Functions
There is a single value for each trigonometric ratio of any angle, so the mappings θ↦sinθ, θ↦cosθ, and so on, are functions, and we can plot graphs showing how each trigonometric function behaves as θ varies. The graphs of the sine, cosine and tangent are particularly important.
The Sine Function
Figure 5.14. The graph of f(θ)=sinθ.
The graph of f:θ↦sinθ, θ∈R, shows that the sine function has the following characteristics.
It is continuous: its graph has no breaks.
Its range is −1⩽sinθ⩽1.
The shape of the graph from θ=0 to θ=2π is repeated for each further complete revolution.
A function whose graph repeats a pattern is periodic, or cyclic. The width of the repeating pattern, measured on the horizontal axis, is the period of the function: it is the smallest positive number p for which f(θ+p)=f(θ) for every θ.
So θ↦sinθ is a periodic function with a period of 2π, a maximum value of 1 and a minimum value of −1. A graph of this shape is known as a sine wave. The amplitude is half the distance between the maximum and minimum values; for sinθ its value is 1.
The Cosine Function
Figure 5.15. The graph of f(θ)=cosθ, with the sine curve dashed.
The characteristics of the graph of f:θ↦cosθ, θ∈R, are as follows.
It is continuous.
It lies entirely within the range −1⩽cosθ⩽1.
It is periodic with a period of 2π.
It has the same shape as the sine graph, but is displaced a distance 2π to the left on the horizontal axis. Such a displacement is known as a phase difference, or phase shift.
So θ↦cosθ is a cyclic function with period 2π and values from −1 to 1.
The Tangent Function
Figure 5.16. The graph of f(θ)=tanθ.
The behaviour of the tangent function f:θ↦tanθ is different from that of the sine and cosine functions in several respects.
It is not continuous, being undefined when θ=±2π,±23π,±25π,…, where the lines drawn dashed in Figure 5.16 are asymptotes to the curve.
The range of possible values of tanθ is unlimited.
The tangent function is periodic, but its period is π, not 2π as for the sine and cosine.
Special Values
It is useful to note the angles whose trigonometric ratios have the values 0 and ±1, and the angles at which tanθ is undefined. Reference to the graphs shows that, for n∈Z,
sinθsinθsinθcosθcosθcosθtanθ=0=1=−1=0=1=−1=0when θ=…,−2π,−π,0,π,2π,3π,…,when θ=…,−23π,2π,25π,…,when θ=…,−2π,23π,27π,…,when θ=…,−2π,2π,23π,25π,…,when θ=…,−2π,0,2π,4π,…,when θ=…,−π,π,3π,5π,…,when θ=…,−π,0,π,2π,…,that is θ=nπ,that is θ=2nπ+2π,that is θ=2nπ−2π,that is θ=(2n+1)2π,that is θ=2nπ,that is θ=(2n+1)π,that is θ=nπ,
and tanθ is undefined, its graph going off to ±∞, when θ=(2n+1)2π.
Remark.
Throughout, 2n stands for any even integer and 2n+1 for any odd integer, provided that n∈Z.
The Reciprocal Ratios
The three ratios sinθ, cosθ and tanθ are used much more frequently than their reciprocals cosecθ, secθ and cotθ, but the reciprocal ratios must not be overlooked. The graph of f(θ)=cosecθ can be drawn without any table of values, simply by observing the graph of f(θ)=sinθ and using the following properties of any expression and its reciprocal.
As an expression approaches zero its reciprocal becomes numerically large without limit, and as an expression becomes numerically large without limit its reciprocal approaches zero.
The reciprocal of 1 is 1, and the reciprocal of −1 is −1.
Where an expression has a maximum value its reciprocal has a minimum value, and conversely.
Where an expression is increasing its reciprocal is decreasing, and conversely.
An expression and its reciprocal have the same sign.
These properties are reasonable enough to accept without detailed analysis at this stage.
Figure 5.17. The graph of f(θ)=cosecθ for −π⩽θ⩽2π, with the sine curve dashed.
In the same way the graphs of f(θ)=secθ and f(θ)=cotθ can be deduced from those of cosθ and tanθ.
Figure 5.18. The graphs of secθ and cotθ, each drawn over the curve it is the reciprocal of.
Inverse Circular Functions
The function f:x↦sinx is a many-one mapping for the domain x∈R, and so it does not have an inverse function. However, if the domain is redefined as −2π⩽x⩽2π, the function f:x↦sinx is a one-one mapping, and now it does have an inverse. This inverse sine function is denoted by arcsin or sin−1. Thus
if f:x↦sinx,−2π⩽x⩽2π,thenf−1:x↦arcsinx,−1⩽x⩽1.
For f:x↦sinx the input is an angle and the output is a number. So for the inverse function f−1:x↦arcsinx the input is a number and the output is an angle, and arcsinx means “the angle whose sine is x”.
Similarly, if f:x↦cosx, 0⩽x⩽π, then f−1 exists and is denoted by arccos or cos−1, where arccosx means “the angle whose cosine is x”:
if f:x↦cosx,0⩽x⩽π,thenf−1:x↦arccosx,−1⩽x⩽1.
Further, if f:x↦tanx for −2π<x<2π, the inverse function f−1 exists and is written arctan or tan−1, where arctanx means “the angle whose tangent is x”. The domain of x↦arctanx is x∈R. In the same way arccotx is the angle whose cotangent is x, and for positive x it is arctanx1.
Figure 5.19. The restricted sine, cosine and tangent functions and their inverses, reflected in the line y=x.
Remark (A warning about notation).
If the notation sin−1 is adopted, it is most important to appreciate that sin−1x is not the same as sinx1.
Common Trigonometric Ratios
The angles 30∘, 45∘ and 60∘, and the angles for which these are the associated acute angles, such as 150∘, 225∘ and 300∘, are used frequently, so their trigonometric ratios are well worth noting.
Consider first an equilateral triangle ABC which is bisected by the line AD. In the triangle BAD the angle B is 60∘, or 3π, since the triangle ABC is equilateral, and the angle at A is 30∘, or 6π, since the angle BAC is bisected. If AB=2 units then BD=1 unit, and AD=3 units by Pythagoras. Therefore
Now consider a triangle ABC in which AB=BC and the angle B is a right angle, so that the angles at A and C are each 45∘. If AB=BC=1 unit, then AC=2 units by Pythagoras. Therefore
If the sum of two acute angles is 90∘, or 2π, they are complementary, and each is the complement of the other.
Consider a right-angled triangle ABC containing the angles α and β, as in Figure 5.21. Then
sinα=ca=cosβ,cosα=cb=sinβ,tanα=ba=cotβ,cotα=ab=tanβ.Figure 5.21. A right-angled triangle containing two complementary angles α and β.
But α and β are complementary, so we have shown that the sine of an angle is the cosine of its complement, and the tangent of an angle is the cotangent of its complement. Because of this property the sine and cosine of an angle are called complementary ratios, and similarly the tangent and cotangent are complementary ratios.
The same relationships hold for angles of any size. Reflecting OP in the line y=x exchanges the coordinates x and y of P, and turns the angle θ into 2π−θ, so for every angle
An equation in which at least one term contains a trigonometric ratio is a trigonometric equation. Solving it means finding the angle or angles for which it is true.
Consider the simple equation sinθ=0. Referring to the graph of the sine function, we see that sinθ=0 when θ is any multiple of π, that is when θ=nπ, where n∈Z. The full, or general, solution of this equation is the infinite set of angles θ=nπ, or θ=180n∘.
Sometimes it is necessary to extract certain values of θ from the infinite set. Solving the equation sinθ=0 for −π⩽θ⩽π, for instance, gives the finite solution set θ=−π,0,π.
There are two basic approaches to finding the solution of a trigonometric equation. One of them was used above, and refers to the graph of the appropriate circular function; it is usually best for sines and cosines with the values ±1 and 0, and for tangents which are zero or undefined. Alternatively, the position of the rotating line OP in the appropriate quadrants can lead to a clear solution. In all cases the first step is to find the principal solution, which is the principal value, PV, of θ.
Principal Values
Figure 5.22. The intervals in which each value of the sine, cosine and tangent occurs exactly once.
From the graph of the sine function, every possible value of sinθ occurs once and only once in the interval −2π⩽θ⩽2π. So any equation sinθ=s, with −1⩽s⩽1, has one and only one solution in this interval, and this is the principal value of θ, in either the first or the fourth quadrant. For example, if sinθ=21 the principal solution is θ=6π, and if sinθ=−21 it is θ=−6π.
Every possible value of cosθ occurs once and only once in the interval 0⩽θ⩽π, so there is one and only one solution of cosθ=c in this interval. This is the principal value of θ, in either the first or the second quadrant. If cosθ=21 the principal solution is θ=3π, and if cosθ=−21 it is θ=32π.
Every possible value of tanθ occurs once and only once for angles in the interval −2π<θ<2π, so one and only one solution of tanθ=t is in this interval, and it is the principal value of θ, in the first or the fourth quadrant. If tanθ=1 the principal solution is θ=4π, and if tanθ=−1 it is θ=−4π.
These intervals are the ranges of the inverse functions, so the principal values of the solutions of sinθ=s, cosθ=c and tanθ=t are arcsins, arccosc and arctant.
Secondary Values
Having found the principal value of the solution of a trigonometric equation, we usually find a second angle with the same trigonometric ratio in the interval −π<θ⩽π. This solution lies in a different quadrant and is called the secondary value, SV, of θ, or the secondary solution of the equation.
If sinθ=21, the secondary solution is in the second quadrant, where the sine ratio is also positive, and it is θ=65π. If sinθ=−21, the secondary solution is in the third quadrant, where the sine ratio is also negative, and it is θ=−65π.
If cosθ=21, the secondary value is in the fourth quadrant and is θ=−3π. If cosθ=−21, the secondary value is in the third quadrant and is θ=−32π. For an equation of the form cosθ=c, SV=−PV.
If tanθ=1, the secondary solution is in the third quadrant and is θ=−43π. If tanθ=−1, the secondary value is in the second quadrant and is θ=43π.
Figure 5.23. Principal and secondary values for six equations.
Determine the principal solutions of sinθ=−23, cosθ=−21 and tanθ=−33.
Find the principal and secondary solutions of sinθ=21, cosθ=−23 and tanθ=−3.
By referring to the graphs of the appropriate circular functions, explain why the equations sinθ=−1 and cosθ=1 have no secondary solution.
Solutions in a Specified Range
In solving a trigonometric equation in a specified range we first find the principal angle and the secondary angle, except in those cases where there is no secondary angle. A quadrant diagram can then be drawn showing the two solution positions, and any angle measured from the positive x-axis to either of the solution positions is a solution of the equation.
Solve the equation sinθ=0.4 within the interval −360∘⩽θ⩽360∘.
The principal solution is θ=23.58∘. The secondary solution is in the second quadrant, since the sine ratio is positive in the first and second quadrants, and it is θ=156.42∘. Therefore the solutions within the specified interval are
θ=−336.42∘,−203.58∘,23.58∘,156.42∘.
Figure 5.24. The four solutions of sinθ=0.4 between −360∘ and 360∘.
Solve the equation tanθ=−31 in the interval 0⩽θ⩽2π.
If tanθ=−31, the principal solution is in the fourth quadrant and the secondary solution is in the second quadrant: the PV is −6π and the SV is 65π. Within the specified interval the solution set is
θ=65π,611π.
As here, the principal value is not always included in the solution set.
Find the angles in the interval −360∘⩽θ⩽0 which satisfy the equation cosθ=0.7.
Since cosθ is positive, the principal solution is in the first quadrant and the secondary solution is in the fourth: the PV is 45.57∘ and the SV is −45.57∘. In the interval −360∘⩽θ⩽0 the solution set is θ=−314.43∘,−45.57∘.
Solve, within the interval 0⩽θ⩽360∘, the equation sinθ+3sinθcosθ=0.
First the equation must be factorised:
sinθ(1+3cosθ)=0.
Therefore either sinθ=0 or cosθ=−31.
Referring to the sine graph, sinθ=0 gives θ=0,180∘,360∘.
For cosθ=−31 the principal solution is in the second quadrant and the secondary solution in the third: the PV is 109.47∘ and the SV is −109.47∘. Within the specified range these give θ=109.47∘,250.53∘.
So the complete solution set from 0 to 360∘ is θ=0,109.47∘,180∘,250.53∘,360∘.
Although the solutions of sinθ=0 could conveniently be expressed in radians, degrees are used because the range is specified in degrees. Units must not be mixed in any one example.
The general solution of a trigonometric equation is an expression which represents all the angles which satisfy the equation, that is an infinite set of angles.
In looking for a general solution we use the graphs of the circular functions, the period of each circular function, and the principal solution together with, except when the tangent ratio is involved, the secondary solution.
Consider the equation sinθ=s, where −1⩽s⩽1. The period 2π of the sine function is covered by the interval −π<θ⩽π, which includes both the PV and the SV of θ. So by adding or subtracting any multiple of 2π to either the PV or the SV we get another angle with the same sine. Thus the complete solution of sinθ=s is
θ=PV+2nπorθ=SV+2nπ,n∈Z,
or, in degrees, θ=PV+360n∘ or θ=SV+360n∘.
A similar situation arises for the equation cosθ=c, because both the PV and the SV of the cosine lie within one period, which is again 2π. So the complete solution of cosθ=c is also given by adding multiples of 2π to either the PV or the SV. Remembering that for cosines the PV and the SV are equal in value but opposite in sign, the general solution of cosθ=c can be given in the form
θ=±PV+2nπorθ=±PV+360n∘,n∈Z.
For the equation tanθ=t, only the principal value is included in the complete period −2π<θ<2π. All further angles with the same tangent are given by adding multiples of π, the period, to the PV:
Find the general solution set of the equation sinθ=21.
The principal value of θ for which sinθ=21 is 6π, and the secondary value, in the second quadrant, is 65π. So the general solution set includes
θ=6π+2nπandθ=65π+2nπ.
Remark.
There is a way of combining the two parts of the general solution of sinθ=s into one formula,
θ=(−1)nPV+nπ,n∈Z.
When n is even this is PV+2kπ, and when n is odd it is π−PV+2kπ, which is the secondary value plus a multiple of 2π. So when sinθ=21, θ=(−1)n6π+nπ. Either form of the general solution may be used.
Find the general solution of the equation 4sinθ(2tanθ+3)+6tanθ+9=0.
First we simplify and factorise the equation:
4sinθ(2tanθ+3)+3(2tanθ+3)=0,(4sinθ+3)(2tanθ+3)=0.
So either sinθ=−43 or tanθ=−23.
For sinθ=−43 the principal solution is θ=−48.59∘, and the secondary solution, in the third quadrant, is θ=−131.41∘. Hence the general solution is θ=−48.59∘+360n∘ or θ=−131.41∘+360n∘.
For tanθ=−23 the principal solution is the only one we need, and it is θ=−56.31∘. The general solution is then θ=−56.31∘+180n∘.
Combining these results, the general solution set of the given equation is
Find the general solution of each of the following equations.
sinθ=−23;
cosθ=0;
cosθ=0.371.
Multiple Angles
Equations are frequently met in which the angle involved is a multiple of θ, such as cos2θ=21 or tan3θ=−2. Such equations are solved by determining first the necessary values of the multiple angle and then, by division, the corresponding values of θ.
Find the angles within the range −180∘⩽θ⩽180∘ which satisfy the equation tan3θ=−2.
Let 3θ=φ, so that tanφ=−2. The principal value is in the fourth quadrant, φ=−63.43∘, and the secondary value is in the second quadrant, φ=116.57∘. Values of θ are required in the range −180∘⩽θ⩽180∘, and φ=3θ, so we need the values of φ in the range −540∘⩽φ⩽540∘. These are
Alternatively, quoting the general solution for φ gives φ=−63.43∘+180n∘, so that θ=−21.14∘+60n∘. Giving n the values −2,−1,0,1,2,3, which cover the required range, gives the same values of θ.
Find the solutions of the equation sin2θ=0.6 for values of θ between 0 and 360∘.
Let 2θ=φ, so that sinφ=0.6. The principal value of φ is 36.87∘, and the secondary value, in the second quadrant, is 143.13∘. The required range of values of θ is from 0 to 360∘, so the range of values of φ is from 0 to 180∘, which gives φ=36.87∘,143.13∘. Therefore θ=73.74∘,286.26∘.
Solve the equations tan2θ=1 and sin3θ=0.7 within the interval 0⩽θ⩽360∘.
Find the general solution of the equation cos2θ=0.63.
The Equation cosA=cosB
This type of equation can be solved very neatly as follows. Let cosA=cosB=c, where −1⩽c⩽1. In general there are two solution positions for cosB=c, OP1 and OP2, and the set of angles represented by OP1 and OP2 is 2nπ±B. But we also know that cosA=c, so OP1 and OP2 together represent all possible values of A. Thus
cosA=cosBgivesA=2nπ±B,
that is, the values of A are the general solution set for B. The same argument for tangents and sines gives
tanA=tanB gives A=nπ+B,sinA=sinB gives A=2nπ+B or A=(2n+1)π−B.
Find the values in the range 0⩽θ⩽360∘ which satisfy the equation tan(3θ−40∘)=tanθ.
The general solution is
3θ−40∘=180n∘+θ,2θ=180n∘+40∘,θ=90n∘+20∘.
For 0⩽θ⩽360∘ we let n=0,1,2,3, giving θ=20∘,110∘,200∘,290∘.
This method can be used only for equations containing two terms involving the same trigonometric ratio. That situation can sometimes be arranged in an apparently unsuitable case, as in the next example.
Find the general solutions of the equations cos4θ=cos3θ and sin4θ=sin3θ.
Solve the equation cos(2θ+60∘)=cosθ, giving the values of θ from −180∘ to 180∘.
Graphs of Multiple and Compound Angles
Consider f:θ↦sin2θ. The following table gives pairs of corresponding values of θ and f(θ).
θ
0
4π
2π
43π
π
45π
23π
47π
2π
2θ
0
2π
π
23π
2π
25π
3π
27π
4π
f(θ)
0
1
0
−1
0
1
0
−1
0
Figure 5.25. The graph of f(θ)=sin2θ, with the graph of sinθ dashed.
The following characteristics can be observed.
The function sin2θ is cyclic, and its period is π, that is 21×2π.
Its range is −1⩽sin2θ⩽1.
Its shape is a sine wave.
Within the domain 0⩽θ⩽2π there are two complete cycles of the curve, compared with only one for the basic function θ↦sinθ: the complete cycle appears with twice the frequency.
When the same investigation is carried out on θ↦sin3θ, we find that the function is cyclic with period 32π, so that three complete cycles occur between 0 and 2π. In general the graph of θ↦sinkθ, for k>0, is a sine wave with period k2π and a frequency k times that of θ↦sinθ. We show this by noting that
sink(θ+k2π)=sin(kθ+2π)=sinkθ,
so that the graph repeats after a width of k2π, while as θ runs through any interval of that width, kθ runs through an interval of width 2π and so through one complete sine wave. Similar properties hold for θ↦coskθ, whose period is k2π, and for θ↦tankθ, whose period is kπ. For example, the period of tan2θ is 2π and its frequency is 2; the period of cos4θ is 2π and its frequency is 4; the period of sin2θ is 4π and its frequency is 21.
Figure 5.26. The graphs of tan2θ, cos4θ and sin21θ.
Now consider the function f(θ)=cos(θ−α). This is clearly a cosine function, but
f(θ)=0 when θ−α=2π,23π,…, that is when θ=2π+α,23π+α,…;
f(θ)=1 when θ−α=0,2π,4π,…, that is when θ=α,2π+α,4π+α,…;
f(θ)=−1 when θ−α=π,3π,5π,…, that is when θ=π+α,3π+α,5π+α,….
So the graph of f(θ)=cos(θ−4π), for example, is identical in shape to the graph of f(θ)=cosθ, but is in a position given by moving the standard cosine curve a horizontal distance 4π to the right. Similarly the graph of f(θ)=cos(θ+α) is given by moving the standard cosine curve a horizontal distance α to the left, as the graph of cos(θ+3π) shows.
Figure 5.27. The graphs of cos(θ−4π) and cos(θ+3π).
Now consider the function f(θ)=sin(2θ+α). Adding a constant angle moves a curve to the left, and in this case sin(2θ+α)=0 when θ=−2α, so the graph of this function is obtained by moving the graph of sin2θ a distance 2α to the left. For example, the graph of sin(2θ+2π) is that of sin2θ moved 4π to the left.
Similarly, if f(θ)=cos(3θ−4π), putting 3θ−4π=0 shows that the graph is obtained by moving the graph of cos3θ a distance 12π to the right. And if f(θ)=tan(2θ+4π), we have a tangent curve with a frequency of 21 which is moved a distance 2π to the left.
Figure 5.28. The graphs of sin(2θ+2π), cos(3θ−4π) and tan(21θ+4π).
An equation containing a compound angle is solved in the same way as one containing a multiple angle.
Sketch the graphs of sin4θ and sec2θ in the domain 0⩽θ⩽2π, and state the period and the frequency of each function.
Find the general solution of the equation cos(θ−6π)=−21.
Trigonometric Identities
Any one angle has six trigonometric ratios, and a particular value of one ratio applies to an infinite set of angles, so it is not surprising that relationships exist between the various circular functions. They are identities, true for every angle for which the ratios are defined, and they are very useful in the development of trigonometry.
Consider first the relationship between the sine, cosine and tangent of any angle. With P(x,y) as in Figure 5.29,
sinθ=OPy,cosθ=OPx,tanθ=xy.
But xy=OPy÷OPx, so for all angles
tanθ=cosθsinθ,and similarlycotθ=sinθcosθ.Figure 5.29. The right-angled triangle OPQ for any position of P.
The Pythagorean Identities
For any position of OP a right-angled triangle OPQ can be drawn, for which, by Pythagoras,
x2+y2=OP2.
Dividing throughout, in turn, by OP2, x2 and y2 gives
Again we have a quadratic equation, but because it has no simple factors we solve it by the formula:
tanθ=65±25+12=1.8471or−0.1805.
If tanθ=1.8471 the principal solution is θ=61.57∘, and if tanθ=−0.1805 it is θ=−10.23∘. The complete general solution is therefore
θ=180n∘+61.57∘orθ=180n∘−10.23∘.
Other applications of the standard identities include the derivation of further trigonometric relationships, the elimination of trigonometric terms from pairs of equations, and the calculation of the remaining trigonometric ratios of an angle for which only one ratio is known.
Because this relationship has yet to be established, we must not assume that it is true by using the complete identity in our working: the left- and right-hand sides must be kept apart throughout. Considering the left-hand side,
This is already a very simple form, but it is not obviously identical to the right-hand side, so this time we work independently on the right-hand side:
Solve the equation tanθ+cotθ=2 for angles in the range −180∘⩽θ⩽180∘.
Find the general solution of the equation 5cosθ−4sin2θ=2.
Show that cotθ+tanθ=secθcosecθ, and that 1+cosAsinA=sinA1−cosA.
Eliminate θ from the equations x=4secθ and y=5tanθ.
Compound Angle Identities
It is often useful to be able to express the trigonometric ratios of angles such as A+B or A−B in terms of the ratios of A and of B. At first sight it is dangerously easy to think, for instance, that sin(A+B) is sinA+sinB. That this is false can be seen by considering
So the sine function is not distributive, and similarly for the other trigonometric ratios. The correct expression is
sin(A+B)=sinAcosB+cosAsinB.
We show this geometrically when A and B are both acute, using Figure 5.30. The right-angled triangles OPQ and OQR contain the angles A and B, RT is perpendicular to OP, and QS is perpendicular to RT. Since QS is parallel to OP, the angle SQO is A, so the angle SQR is 90∘−A and the angle SRQ is equal to A. Since TS=PQ,
sin(A+B)=ORTR=ORTS+SR=ORPQ+ORSR=OQPQ⋅OROQ+QRSR⋅ORQR=sinAcosB+cosAsinB.Figure 5.30. The construction for sin(A+B) when A and B are acute.
Accepting at this stage that the formula is valid for all angles, it can be adapted to give the full set of compound angle identities.
Replacing B by −B, and using cos(−B)=cosB and sin(−B)=−sinB, gives sin(A−B)=sinAcosB−cosAsinB.
Replacing A by 2π−A in this identity, and using the complementary ratios, gives sin(2π−(A+B))=cosAcosB−sinAsinB, that is cos(A+B)=cosAcosB−sinAsinB.
Replacing B by −B in this identity gives cos(A−B)=cosAcosB+sinAsinB.
Dividing sin(A+B) by cos(A+B), and then dividing the numerator and the denominator by cosAcosB, gives tan(A+B)=1−tanAtanBtanA+tanB.
Replacing B by −B in this identity, with tan(−B)=−tanB, gives tan(A−B)=1+tanAtanBtanA−tanB.
the last after rationalising the denominator. The value of cos105∘ is negative, which is consistent with the cosine of an angle in the second quadrant. In each part there are alternative compound angles which could be used, such as 75∘=120∘−45∘, 105∘=150∘−45∘ and −15∘=30∘−45∘.
A is obtuse and sinA=53, and B is acute and sinB=1312. Without finding the values of A and B, evaluate cos(A+B) and tan(A−B).
In order to use the compound angle formulae we need cosA, cosB, tanA and tanB. These are most simply obtained by using Pythagoras in the appropriate right-angled triangles, with sides 3, 4, 5 and 5, 12, 13, remembering that A is obtuse:
and these alternative expressions can themselves be rearranged to give
2sin2A=1−cos2A,2cos2A=1+cos2A.
Complete familiarity with all the double angle formulae, including all the alternative forms of cos2A, is essential. They are probably the most useful of all the trigonometric identities for simplifying trigonometric functions.
Express as a single trigonometric ratio 2sin14∘cos14∘, 1−2sin240∘ and 1−tanx1+tanx.
Solve the equation cos2x=sinx for angles in the range 0∘⩽x⩽360∘, and state the general solution.
Show that cos3θ=4cos3θ−3cosθ.
Identities and the Inverse Functions
The double angle and compound angle identities are often useful in simplifying expressions, or solving equations, which contain inverse trigonometric functions.
Let arcsinx=θ, so that sinθ=x, and cosθ=1−x2 since −2π⩽θ⩽2π. Let arccos2x=φ, so that cosφ=2x, and sinφ=24−x2 since 0⩽φ⩽π. The given equation then becomes θ+φ=65π, so
sin(θ+φ)=21,sinθcosφ+cosθsinφ=21,
that is
2x2+21−x24−x2=21,1−x24−x2=1−x2.
Squaring both sides, and not cancelling 1−x2, which would lose solutions, gives
Since x>−1, both x and 1+x1−x are greater than −1, so α and β each lie between −4π and 2π, and α+β lies between −2π and π. The only angle in that range whose tangent is 1 is 4π. Thus
Solve the equation arctan(1+x)+arctan(1−x)=arctan2.
Simplify sin(2arctanx).
The Half Angle Identities
We already know that tan2A=1−tan2A2tanA. Dividing sin2A=2sinAcosA and cos2A=cos2A−sin2A by cos2A+sin2A, which is 1, and then dividing the numerator and the denominator by cos2A, gives also
sin2A=1+tan2A2tanA,cos2A=1+tan2A1−tan2A.
If we replace 2A by θ and use t to denote tan2θ, we have
tanθ=1−t22t,sinθ=1+t22t,cosθ=1+t21−t2.
These three identities allow all the trigonometric ratios of any one angle to be expressed in terms of a common variable t. In problems where none of the identities used so far can be applied, this group can be helpful.
Therefore either tan2θ=−31 or tan2θ=1. The range of values specified for θ is 0∘ to 360∘, so the range of values required for 2θ is 0∘ to 180∘. Within this range tan2θ=−31 gives 2θ=161.57∘, and tan2θ=1 gives 2θ=45∘. Thus θ=323.13∘,90∘.
Extra care is sometimes needed with this method, as a very similar equation shows. Consider
sinθ−cosθ=1.
Using t=tan2θ gives
1+t22t−1+t21−t2=1,2t−1+t2=1+t2.
The two t2 terms cancel, leaving 2t=2, so t=1. But t=tan2θ is not defined when 2θ is an odd multiple of 2π, that is when θ=(2n+1)π, and these angles have to be checked separately: sinπ−cosπ=0+1=1, so they are solutions as well. Hence tan2θ=1, or tan2θ is undefined, giving
When the half angle identities are used, t does not always represent tan2θ. For instance, in solving the equation sin4θ+tan2θ=0 we would use t=tan2θ.
The Expression acosθ+bsinθ
It is often useful to reduce acosθ+bsinθ to a single term such as rcos(θ−α). This is possible provided that we can find values of r and α for which
r(cosθcosα+sinθsinα)=acosθ+bsinθ
for every θ. Comparing the coefficients of cosθ and of sinθ,
rcosα=a,rsinα=b.
Squaring and adding gives r2=a2+b2, so r is equal to the length of the hypotenuse of the triangle containing α with sides a and b, and dividing gives tanα=ab, with α in the quadrant given by the signs of a and b. Thus
Express 3cosθ+4sinθ in the form rcos(θ−α), giving the values of r and α.
Let
r(cosθcosα+sinθsinα)=3cosθ+4sinθ,
so that rcosα=3 and rsinα=4. Thus r=5 and tanα=34, which gives α=53.13∘, and
3cosθ+4sinθ=5cos(θ−53.13∘).
It is sometimes more convenient to begin by comparing acosθ+bsinθ with rsin(θ+α), so that
r(sinθcosα+cosθsinα)=acosθ+bsinθ,rsinα=a,rcosα=b.
Then tanα=ba and r=a2+b2. The value of α is not the same as it was when we used rcos(θ−α). Further variations that could be used are rsin(θ−α) and rcos(θ+α). When using this method it is better to work from the basic comparison each time, as in the example above, than to quote values of r and α.
Express 5cosθ+12sinθ in the forms rcos(θ−α) and rsin(θ+α).
Express 3sinθ+4cosθ in the forms rsin(θ+α) and rcos(θ−α).
Express 3cosθ−sinθ in the form rcos(θ+α).
The Graph of acosθ+bsinθ
First consider the function f(θ)=kcosθ, where k>0, which has the following characteristics.
f(θ)=0 when θ=2π,23π,25π,….
f(θ)=k when θ=0,2π,4π,….
f(θ)=−k when θ=π,3π,5π,….
So the graph of this function is very similar to a standard cosine curve, but has maximum and minimum values ±k; we say that the curve has an amplitude of k. We also saw that the graph of cos(θ−α) is given by moving the graph of cosθ a distance α to the right. Combining these two modifications of a standard cosine curve, the graph of the function f(θ)=kcos(θ−α) can be sketched.
Figure 5.31. A cosine curve with amplitude 5, and a cosine curve with amplitude k moved a distance α to the right.
Consider now the function acosθ+bsinθ. At first sight its graph is not easy to visualise, but using acosθ+bsinθ=rcos(θ−α) we see that its graph is a cosine curve modified in two ways.
Its maximum and minimum values are ±r, that is its amplitude is r.
Its position is a distance α to the right of the standard curve.
Sketch the graph of the function 3cosθ+4sinθ from −180∘ to 180∘.
As in the last example, 3cosθ+4sinθ=rcos(θ−α) with r=5 and tanα=34, so α=53.13∘. Hence the graph is a cosine curve with an amplitude of 5 and a phase shift of 53.13∘ to the right.
Figure 5.32. The graph of 3cosθ+4sinθ, with cosθ dashed.
It is interesting to see how the same graph is produced if the alternative form 3cosθ+4sinθ=rsin(θ+α′) is used. With this approach r=5 and tanα′=43, so that 3cosθ+4sinθ=5sin(θ+α′), and the graph is a sine curve with an amplitude of 5 and a displacement of α′ to the left. But since tanα′=cotα, the angles α′ and α are complementary, that is α+α′=2π. We also know that a cosine curve is the same as a sine curve displaced 2π to the left. So a cosine curve moved a distance α to the right coincides with a sine curve moved a distance α′ to the left.
So any correct compound angle form of acosθ+bsinθ gives a quick method of sketching the graph of that function, and in particular of finding its maximum and minimum values, which for sine and cosine functions are also the greatest and least values.
The Equation acosθ+bsinθ=c
One way of solving an equation of this type, using the half angle formulae, has already been used. A compound angle form gives an alternative method. Applied to the equation sinθ+2cosθ=1, solved above with t, it gives
2cosθ+sinθ=r(cosθcosα+sinθsinα)=rcos(θ−α),
where rcosα=2 and rsinα=1, that is tanα=21 and r=5. Hence
5cos(θ−α)=1,cos(θ−α)=51,θ−α=360n∘±63.43∘,
from which θ=360n∘±63.43∘+α. But α=arctan21=26.57∘, so θ=360n∘+90∘ or θ=360n∘−36.87∘. Using the values of n which give θ between 0∘ and 360∘, that is n=0 and n=1, we have θ=90∘,323.13∘.
Find the general solution of the equation cosθ−3sinθ=1, first by using the half angle formulae and then by using a compound angle form.
With t=tan2θ, the equation becomes 1+t21−t2−23t=1, so 1−t2−23t=1+t2, that is 2t(t+3)=0. Hence either t=0 or t=−3, and the principal values of 2θ are 0 and −3π. So 2θ=nπ or 2θ=nπ−3π, and θ=2nπ or θ=2nπ−32π. When t is undefined, θ=(2n+1)π, the left-hand side is −1, so no solutions are lost.
Let cosθ−3sinθ=r(cosθcosα−sinθsinα)=rcos(θ+α), where rcosα=1 and rsinα=3. Then tanα=3, so α=3π and r=2, and the equation can be written 2cos(θ+3π)=1. The principal value of θ+3π is 3π, so θ+3π=2nπ±3π, and again θ=2nπ or θ=2nπ−32π.
Express 5sinθ+12cosθ in the form rsin(θ+α), giving the values of r and α. Show that 5sinθ+12cosθ+7⩽20, and find the minimum value of 5sinθ+12cosθ+7. Sketch the graph of the function 5sinθ+12cosθ1 for 0⩽θ⩽2π.
Let 5sinθ+12cosθ=r(sinθcosα+cosθsinα)=rsin(θ+α), so that rcosα=5 and rsinα=12, giving r=13 and tanα=512, α=67.38∘. Hence
5sinθ+12cosθ=13sin(θ+α).
But −1⩽sin(θ+α)⩽1, so −13⩽5sinθ+12cosθ⩽13, and adding 7 throughout gives
−6⩽5sinθ+12cosθ+7⩽20.
This shows that 5sinθ+12cosθ+7⩽20, and that the minimum value of 5sinθ+12cosθ+7 is −6.
Now 5sinθ+12cosθ=13sin(θ+α), so the graph of 5sinθ+12cosθ1 is also the graph of 131cosec(θ+α). Its general shape is a typical cosecant curve, except that its branches turn at the values 131 and −131, and its position is a distance α to the left of the standard curve.
Figure 5.33. The graph of 5sinθ+12cosθ1 for 0⩽θ⩽2π.
Using t=tan2θ, solve the equation 3cosθ+2sinθ=3, giving the values of θ from −180∘ to 180∘.
Find the maximum and minimum values of 7cosθ−24sinθ+3, and the values of θ between 0∘ and 360∘ at which they occur.
Find the general solution of the equation cosx+sinx=2.
The Factor Formulae
To factorise is to express in the form of a product. The set of identities called the factor formulae converts expressions such as sinA+sinB into a product. To derive them we use the compound angle group. Adding and subtracting
Identities (5) to (8) are best used when a sum or difference is to be expressed as a product, while identities (1) to (4) should be used when a given product is to be changed into a sum or difference. For example, to express sin6θ−sin4θ as a product we use (6):
sin6θ−sin4θ=2cos26θ+4θsin26θ−4θ=2cos5θsinθ.
But to express 2cos7θcos2θ as a sum we use (3):
2cos7θcos2θ=cos(7θ+2θ)+cos(7θ−2θ)=cos9θ+cos5θ.
When these identities are used regularly they are not too difficult to remember. Most people find it best to memorise them in words rather than as symbols. For example, (5) can be remembered as “the sum of two sines is twice the sine of the semi-sum times the cosine of the semi-difference”, and (1) as “twice sin cos is the sine of the sum plus the sine of the difference”. Identities (4) and (8) need special care because of the minus sign.
If A, B and C are the angles of a triangle, show that sinA+sinB+sinC=4cos2Acos2Bcos2C.
Now A+B+C=180∘, so 2A+B=90∘−2C, and therefore sin2A+B=cos2C and sin2C=cos2A+B. Considering the left-hand side, and using (5) and the double angle formula for sinC,
Solve the equation cos2x+cos4x=0, giving the values of x from 0∘ to 360∘.
If A, B and C are the angles of a triangle, show that cos(B+C)=−cosA and that cosA+cosB+cosC=1+4sin2Asin2Bsin2C.
Small Angles
A glance at the values of sinθ and tanθ when θ is a very small positive angle shows that these two ratios are almost equal. Further, if the small angle is measured in radians, both are found to be almost equal to θ. These relationships can be demonstrated as follows.
Consider a small angle θ, measured in radians, subtended by an arc AB at the centre O of a circle of radius r. The area of the sector OAB is 21r2θ. If AC is drawn perpendicular to OA, to cut OB produced at C, then OAC is a right-angled triangle with base r and height rtanθ, so its area is 21r2tanθ. Further, when the chord AB is drawn, an isosceles triangle OAB is formed with base r and height rsinθ, so its area is 21r2sinθ.
Figure 5.34. The triangle OAB, the sector OAB and the triangle OAC.
Now area of triangle OAB< area of sector OAB< area of triangle OAC, that is
21r2sinθ<21r2θ<21r2tanθ.
Dividing throughout by 21r2, which is positive, gives sinθ<θ<tanθ. But sinθ, θ and tanθ are all positive, since θ is a small positive angle, so we can divide throughout by any of them. Dividing by sinθ,
1<sinθθ<secθ.
For small values of θ, secθ→1 as θ→0. Hence as θ→0, sinθθ lies between 1 and a number which approaches 1, and we can say that sinθθ→1 as θ→0, a limit in the sense of the last lesson. Similarly, by dividing the first inequalities by tanθ, we get cosθ<tanθθ<1, which shows that tanθθ→1 as θ→0.
The same results are obtained if θ is a small negative angle, since sinθθ and tanθθ are unchanged when θ is replaced by −θ. These limiting values show that, for small values of θ,
sinθ≈θandtanθ≈θ.
So far we have not found an approximate value for cosθ when θ is small. To do this we use the double angle identity
cosθ=1−2sin22θ.
If θ is small then so is 2θ, and sin2θ≈2θ, so cosθ≈1−2(2θ)2=1−2θ2.
Thus, when θ is measured in radians,
θ→0limθsinθ=1andθ→0limθtanθ=1,
and for any small angle θ, measured in radians,
sinθ≈θ,tanθ≈θ,cosθ≈1−2θ2.
For angles in the range −0.105⩽θ⩽0.105, that is about −6∘⩽θ⩽6∘, each of these approximations differs from the true value by less than 0.0005.
If θ is small, find approximations for 1−cosθθsinθ and θsin4θ.
If θ is small enough for θ2 to be neglected, show that tan(4π+θ)≈1−θ1+θ.
Further Properties of Triangles
There is a great variety of relationships between the sides and angles of a triangle besides the sine and cosine rules, and some of the most useful can be derived from the identities above.
The Cotangent Formula
If D divides the side AB of a triangle ABC in the ratio m:n then, with the angles marked in Figure 5.35,
(m+n)cotθ=mcotα−ncotβ.
This relationship is known as the cotangent formula, and we show it by using the sine rule in the triangles ACD and BCD. In the triangle ACD the angle at D is 180∘−θ, so the angle A is θ−α, and
sinACD=sinαAD,CD=sinαADsin(θ−α).
In the triangle BCD the angle B is 180∘−θ−β, so sinB=sin(θ+β), and
If D is the midpoint of AB, this becomes 2cotθ=cotα−cotβ.
Figure 5.35. The angles in the cotangent formula, with AD:DB=m:n.
Other Relationships
The Projection Formula
In any triangle ABC,
c=bcosA+acosB.
When A and B are acute, the perpendicular from C to AB divides AB into two parts of lengths bcosA and acosB, as in Figure 5.36. If one of the angles is obtuse, say B, the foot of the perpendicular lies beyond B, and the part acosB is negative, so the formula still holds.
The Difference of Two Sides
In any triangle ABC,
a+ba−b=tan2A−Btan2C.
We show this by using the sine rule in the form sinAa=sinBb=k, say, so that a=ksinA and b=ksinB. Hence, by the factor formulae,
But A+B+C=π, so 2A+B=2π−2C, and cot2A+B=tan2C. So
a+ba−b=tan2A−Btan2C.
The Bisector of an Angle
In a triangle ABC the bisector of the angle A divides BC in the ratio c:b.
We show this by letting the bisector meet BC at D, with the angle ADB equal to θ. In the triangle ABD,
sin21ABD=sinθc,
and in the triangle ACD, where the angle at D is 180∘−θ,
sin21ADC=sin(180∘−θ)b=sinθb.
Hence
sinθsin21A=cBD=bDC,soBD:DC=c:b.Figure 5.36. The perpendicular from C divides AB into bcosA and acosB; the bisector AD divides BC in the ratio c:b.
Points Associated with a Triangle
The following geometric properties of a triangle should also be familiar.
The perpendicular bisectors of the sides of a triangle ABC meet at a point called the circumcentre, which is the centre of the circle through A, B and C, the circumcircle. Its radius is the R of the sine rule.
The bisectors of the angles A, B and C meet at a point called the incentre, which is the centre of the circle that touches all three sides, the inscribed circle.
The altitudes meet at a point H called the orthocentre.
The medians meet at a point G called the centroid.
Figure 5.37. The circumcentre, incentre, orthocentre and centroid of a triangle.
The Area of a Triangle
The area of a triangle can be found using any of the following.
Half the base times the perpendicular height.
21absinC, or the corresponding 21bcsinA and 21casinB. This follows from the first, since the height of A above the side CB, of length a, is bsinC.
s(s−a)(s−b)(s−c), where s=21(a+b+c), which we accept here.
In a surveying exercise, P and Q are two points on land which is inaccessible. To find the distance PQ, a line AB of length 200 metres is drawn so that P and Q are on opposite sides of AB. The following angles are measured:
∠ABP=60∘,∠ABQ=46∘,∠BAP=30∘,∠BAQ=67∘.
Find the distance PQ.
In the triangle APB, ∠APB=90∘, hence AP=200cos30∘=173.2 m. In the triangle ABQ, ∠AQB=67∘, hence
sin46∘AQ=sin67∘200,AQ=156.3 m.
Then in the triangle APQ, where ∠PAQ=30∘+67∘=97∘, the cosine rule gives
In a triangle ABC, BC=7.4 cm, AC=4.1 cm and ∠ACB=66∘. Calculate the other angles of the triangle, the area of the triangle, and the distance of the incentre I of the triangle from BC.
Using the cosine rule, c2=7.42+4.12−2(7.4)(4.1)cos66∘ gives c=6.85 cm. Then the sine rule, sinA7.4=sinB4.1=sin66∘6.85, gives A=80.8∘ and B=33.2∘; A is acute since a2<b2+c2.
The area of the triangle is 21absinC=21(7.4)(4.1)sin66∘=13.86cm2.
The incentre I is the centre of the inscribed circle, and is therefore at the same distance r from all three sides. Joining AI, BI and CI divides the triangle into three triangles, AIB, BIC and CIA, with areas 21cr, 21ar and 21br. So the total area of the triangle ABC is 21r(a+b+c)=9.175r. But this area is 13.86cm2, hence the distance of I from BC is r=1.5 cm.
Figure 5.39. The incentre I at the distance r from each side, dividing the triangle into three.
Calculate the area of the triangle whose sides are 11, 10 and 15.
In a triangle PQR, p=2.8 m and q=4.5 m. If the area of the triangle is 5.84m2, find the two possible values of r.
Show that, for any triangle ABC, abc=4ΔR, where Δ is the area of the triangle and R is the radius of its circumcircle.
ABC is a triangle and D is the point on BC for which BD=DC. The angle BAD is 20∘ and the angle CAD is 30∘. Find the angle ACB.
Exercises
Questions marked with an examining board are taken from past A-level papers: JMB is the Joint Matriculation Board, U of L the University of London, C Cambridge, and AEB the Associated Examining Board.
A chord of a circle subtends an angle of θ radians at the centre of the circle. If the area of the minor segment cut off by the chord is one sixth of the area of the circle, show that
P and Q are points on a circle of radius r, and the chord PQ subtends an angle of 2θ radians at the centre O. If A is the area enclosed by the minor arc PQ and the chord PQ, and B is the area enclosed by the arc PQ and the tangents to the circle at P and Q, show that
Three cylinders are placed in contact with each other with their axes parallel. The radii of the cylinders are 3 cm, 4 cm and 5 cm. An elastic band is stretched round the three cylinders so that the plane of the band is perpendicular to the axes of the cylinders. Calculate the length of the part of the band in contact with the largest cylinder. (U of L)
If sin(θ−α)=ksin(θ+α), find tanθ in terms of tanα and k, and so determine the possible values of θ between 0∘ and 360∘ when k=21 and α=150∘.
Show, without the use of a calculator, that x=10π satisfies the equation cos3x=sin2x. By expressing this equation in terms of sinx and cosx, show that sin10π is a root of the equation 4s2+2s−1=0. (C)
Find, to the nearest minute, the acute angle α for which 4cosθ−3sinθ=5cos(θ+α). Calculate the values of θ in the interval −180∘<θ<180∘ for which the function f(θ)=4cosθ−3sinθ−4 attains its greatest value, its least value and the value zero. (JMB)
Show that secx+tanx=tan(4π+2x), and deduce a similar expression for secx−tanx. Hence find in surd form the values of tan12π and tan125π. (AEB, 1975)
By expressing sec2x and tan2x in terms of tanx, or otherwise, solve the equation 2tanx+sec2x=2tan2x, giving all the solutions between −180∘ and 180∘. (U of L)
Express 3sinθ−cosθ in the form Rsin(θ−α), where R is positive. Find all the values of θ in the range 0∘⩽θ⩽360∘ which satisfy the equation 4sinθcosθ=3sinθ−cosθ. (JMB)
Express 7sinx−24cosx in the form Rsin(x−α), where R is positive and α is an acute angle. Hence, or otherwise, solve the equation 7sinx−24cosx=15 for 0∘⩽x⩽360∘.
Solve the simultaneous equations cosx+cosy=1 and secx+secy=4 for 0∘⩽x⩽180∘ and 0∘⩽y⩽180∘. (AEB, 1976)
A quadrilateral ABCD is right-angled at B and at D, and the angle DAB is 132∘. If DA=4 and AB=7, find the lengths of the diagonals of the quadrilateral and the radius of the inscribed circle of the triangle ABC. (U of L)
Three towns A, B and C are all at sea level. The bearings of the towns B and C from A, measured clockwise from north, are 36∘ and 247∘ respectively. If B is 120 km from A and C is 234 km from A, calculate the distance and the bearing of the town B from C. (AEB, 1972)
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