What do I need to be able to do?

Y13 – Chapter 1 Algebraic Methods

By the end of this chapter you should be able to:

• Use proof by contradiction to prove true statements Key words:

• Add, subtract, multiply and divide two or more algebraic fractions • Contradiction – a disagreement between two statements
which means that both cannot be true.
• Convert an expression with linear factors in the denominator into
partial fractions • Coefficient – A number used to multiply by a variable
• Convert an expression with repeated linear fractions in the • Improper algebraic fraction – One whose numerator has a
degree equal to or larger than the denominator. It must be
denominator into partial fractions
converted to a mixed fraction before you can express it in
• Divide algebraic expressions partial fractions
• Convert an improper fraction into partial fraction form

Partial Fractions Pure Maths Year 2

Sometimes it can be useful to split a single algebraic fraction into two
or more partial fractions. Proof by Contradiction
7𝑥𝑥−13 2 5
Eg: (𝑥𝑥−3)(𝑥𝑥+1) = 𝑥𝑥−3 + 𝑥𝑥+1
To prove by contradiction you start by assuming that the
statement is false. You then use logical steps until you contradict
When solving partial fractions, you start by setting your function equal yourself by leading to something that is impossible. You can then
to the unknown fractions you are trying to find. There are 3 different conclude that your assumption was incorrect and that the
layouts which depend on the starting function original statement was true.

1) All linear terms in the denominator Eg: Prove by contradiction that √2 is irrational
7𝑥𝑥 − 13 𝐴𝐴 𝐵𝐵
= + 𝑎𝑎
(𝑥𝑥 − 3)(𝑥𝑥 + 1) (𝑥𝑥 − 3) (𝑥𝑥 + 1) Assumption: √2 is rational, therefore √2 can be written as 𝑏𝑏
𝑎𝑎
where a and b are in their lowest form and that 𝑏𝑏 is in its
2) A repeated term in the denominator:
3𝑥𝑥 2 + 7𝑥𝑥 − 12 𝐴𝐴 𝐵𝐵 𝐶𝐶 lowest terms
= + +
(𝑥𝑥 − 5)(𝑥𝑥 + 2)2 (𝑥𝑥 − 5) (𝑥𝑥 + 2) (𝑥𝑥 + 2)2
𝑎𝑎 2
⸫2= 𝑏𝑏
3) *Improper fractions: 𝑎𝑎 2
3𝑥𝑥 2 − 3𝑥𝑥 − 2 𝐵𝐵 𝐶𝐶 2 = 2
𝑏𝑏
= 𝐴𝐴 + + 2
⸫ 2𝑏𝑏 = 𝑎𝑎 2
(𝑥𝑥 − 1)(𝑥𝑥 − 2) (𝑥𝑥 − 1) (𝑥𝑥 − 2)
This means that a2 is even which means that a is even.
Steps to solve: If a is even then it can be expressed as 2k
1) Sett your functions equal to the correct unknown fraction as
above ⸫ 𝑎𝑎2 = 2𝑏𝑏 2
2) Add the fractions using a common denominator (this should be the (2𝑘𝑘)2 = 2𝑏𝑏 2
same as the original denominator) 4𝑘𝑘 2 = 2𝑏𝑏 2
3) Set the numerators as equal 2𝑘𝑘 2 = 𝑏𝑏 2
4) Substitute values for x that wil , in turn, make each bracket zero
and/or equate coefficients to create enough equations to find This means that b2 is even which means that b is even.
the values of A, B, C etc
Conclusion: If a and b are both even then the have a common
𝑎𝑎
*NB: You can either use algebraic division or the relationship F(x)= Q(x) factor of 2 so 𝑏𝑏 cannot be a fraction in its lowest terms which
x divisor + remainder to convert an improper fraction into a mixed is a contradiction. This means that the original assumption is not
fraction correct and therefore √2 is irrational.
What do I need to be able to do? Y13 – Chapter 2 Functions and Graphs
By the end of this chapter you should be able to:
Key words:
• Understand and use the modulus function • Modulus – the absolute value or modulus of a real number x,
• Understand mappings and functions, and use domain and range denoted │x│, is the non-negative value of x without regard
• Combine two or more functions to get a composite function to its sign. For example, the absolute value of 3 is 3, and the
• Know how to find the inverse of a function both graphically and absolute value of −3 is also 3.
algebraically • Composite function – A function made of other functions,
• Sketch the graphs of the modulus function where the output of one is the input to the other
• Apply a combination of transformations to a curve • Inverse function – An inverse function is a function that
• Transform a modulus function undoes the action of another function

The Modulus Function Pure Maths Year 2

To sketch the graph of y = │ax + b│, sketch y = ax + b and then Composite Functions
reflect any section of the graph that is below the x-axis in the x-axis
Always apply the inside function first.

To find fg(x) do g(x) first then substitute your answer into

f(x) to find the answer

Eg: f(x) = x2 and g(x) = x + 1

a) Find fg(2) b) Find gf(x)

g(2) = 2+1 = 3 f(x) = x2
f(3) = 32 = 9 g(x2) = x2 + 1

When solving modulus equations algebraically you consider the positive

and negative argument (the function inside the modulus) separately The Inverse Function
Eg:
Solve │2x – 1│= 5 The inverse of a function performs the opposite operation
to the original function. Inverse functions only exist for one-
2x – 1 = 5 to-one functions.
2x = 6 The inverse of a function f(x) is written as f-1(x)
x=3 The graphs of y = f(x) and y = f-1(x) are reflections of each
other in the line y = x
-(2x – 1) = 5 The domain of f(x) is the range of f-1(x)
-2x + 1 = 5 The range of f(x) is the domain of f-1(x)
-2x = 4
x=-2 To find the inverse function: f(x) = x2 – 3 find f-1(x)
1) Write it as y = 1) y = x2 – 3
2) Swap x and y 2) x = y2 – 3
Functions and Mappings 3) Rearrange to make y the subject 3) √(x + 3) = y
A mapping is a function if each input has a distinct output. Functions can 4) Replace y with f-1(x) 4) f-1(x) = √(x + 3)
either be one-to-one or many-to-one y= x2 – 3 y= x

one-to-one many-to-one not a function

y = √(x + 3)
What do I need to be able to do? Y13 – Chapter 3 Sequences and Series
By the end of this chapter you should be able to:
Key words:
• Find the nth term of an arithmetic sequence and a geometric • Sequence – A list of numbers or objects in a special order
sequence • Series – The sum of terms in a sequence
• Prove and use the formula for the sum of the first n terms of an • Arithmetic sequence – A sequence made by adding the
arithmetic series same value each time
• Prove and use the formula for the sum of a finite geometric • Geometric sequence – A sequence made by multiplying by
series the same value each time. There is a common ratio between
• Prove and use the formula for the sum to infinity of a convergent consecutive terms
geometric series • Arithmetic series– the sum of the terms of an arithmetic
• Use sigma notation sequence
• Generate sequences from recurrence relations • Geometric series – the sum of the terms in a geometric
• Model real life situations sequence
• Common ratio – The amount we multiply by each time in a
geometric sequence
Arithmetic Sequences and Series Pure • Converging sequence/series – A sequence/series

The formula for the nth term of an arithmetic sequence

Maths converges when it keeps getting closer and closer to a certain
value
is: Year • Divergent series – does not settle towards a certain value.
𝑢𝑢𝑛𝑛 = 𝑎𝑎 + 𝑛𝑛 − 1 𝑑𝑑 When a series diverges it goes off to infinity, minus infinity, or
𝑢𝑢𝑛𝑛 is the nth term
2 up and down without settling towards some value.
𝑎𝑎 is the first term
𝑑𝑑 is the common difference
Geometric Sequences and Series
The formula for the sum of the first n terms of an
arithmetic series is: The formula for the nth term of a geometric sequence is:
𝑛𝑛 𝑢𝑢𝑛𝑛 = 𝑎𝑎𝑟𝑟 𝑛𝑛−1
𝑆𝑆𝑛𝑛 = 2𝑎𝑎 + 𝑛𝑛 − 1 𝑑𝑑 𝑢𝑢𝑛𝑛 is the nth term
2
It can also be written as: 𝑎𝑎 is the first term
𝑛𝑛 𝑟𝑟 is the common ratio
𝑆𝑆𝑛𝑛 = (𝑎𝑎 + 𝑙𝑙)
2
𝑎𝑎 is the first term The formula for the sum of the first n terms of a geometric series is:
𝑑𝑑 is the common difference 𝑎𝑎(1 − 𝑟𝑟 𝑛𝑛 )
𝑙𝑙 is the last term 𝑆𝑆𝑛𝑛 = , 𝑟𝑟 ≠ 1
1 − 𝑟𝑟
It can also be written as:
Sigma Notation 𝑎𝑎(𝑟𝑟 𝑛𝑛 − 1)
𝑆𝑆𝑛𝑛 = , 𝑟𝑟 ≠ 1
𝑟𝑟 − 1
Σ means “the sum of”. You write on the top and bottom 𝑎𝑎 is the first term
to show which terms you are summing. 𝑟𝑟 is the common ratio
Eg:
5
Sum to Infinity
�(2𝑟𝑟 − 3) = −1 + 1 + 3 + 5 + 7
𝑟𝑟=1 As n tends to infinity, the sum of a geometric series is called the sum to
Substitute r=1, r=2, r=3, r=4 and r=5 into
the expression in brackets to find the 5 infinity.
This tells you that you are
summing the expression in
terms in this arithmetic series A geometric series is convergent only when │r│< 1, where r is the common
brackets with r=1 up to r=5 ratio

Recurrence Relations The formula for the sum to infinity of a convergent series is:
𝑎𝑎
𝑆𝑆∞ =
The next term in the sequence is the function of the 1 − 𝑟𝑟
previous term 𝑎𝑎 is the first term
𝑢𝑢𝑛𝑛+1 = 𝑓𝑓(𝑢𝑢𝑛𝑛 ) 𝑟𝑟 is the common ratio
What do I need to be able to do? Y13 – Chapter 4 Binomial Expansion
By the end of this chapter you should be able to:
• Expand (1+x)n for any rational constant n and determine the Key words:
range of values of x for which the expansion is valid • Infinite series – The sum of infinite terms that follow a rule
• Expand (a+bx)n for any rational constant n and determine the
range of values of x for which the expansion is valid
• Use partial fractions to expand fractional expressions Pure Maths Year 2

The Binomial Expansion

(a + b)n when n is a fraction or a negative number (i.e. NOT a positive integer) there wil be an infinite number of terms. This means that
the binomial expansion can only be used when -1 < x < 1.

When n is a fraction or a negative number the following form of the binomial expansion should be used:
𝑛𝑛 𝑛𝑛 − 1 2 𝑛𝑛 𝑛𝑛 − 1 𝑛𝑛 − 2 3 𝑛𝑛 𝑟𝑟
(1 + 𝑥𝑥)𝑛𝑛 = 1 + 𝑛𝑛𝑛𝑛 + 𝑥𝑥 + 𝑥𝑥 + ⋯ + 𝑥𝑥 + ⋯ , ( 𝑥𝑥 < 1, 𝑛𝑛 ∈ ℝ)
2! 3! 𝑟𝑟

The expansion is valid when 𝑥𝑥 < 1

1
The expansion of (1 + bx)n is valid for 𝑏𝑏𝑏𝑏 < 1 or 𝑥𝑥 < 𝑏𝑏

We can use the expansion of (1 + x)n to expand (a + bx)n by taking out a factor of an out of the expression
𝑏𝑏 𝑏𝑏
(𝑎𝑎 + 𝑏𝑏𝑏𝑏)𝑛𝑛 = (𝑎𝑎 1 + 𝑥𝑥 )𝑛𝑛 = 𝑎𝑎𝑛𝑛 (1 + 𝑥𝑥)𝑛𝑛
𝑎𝑎 𝑎𝑎

𝑏𝑏 𝑎𝑎
The expansion of (𝑎𝑎 + 𝑏𝑏𝑏𝑏)𝑛𝑛 is valid for 𝑎𝑎
𝑥𝑥 < 1 or 𝑥𝑥 < 𝑏𝑏

Partial Fractions
We can use partial fractions to simplify the expansions of more difficult expressions
E.g.
4−5𝑥𝑥
a) Express (1+𝑥𝑥)(2−𝑥𝑥) as partial fractions
4−5𝑥𝑥 7𝑥𝑥 11 2 25 3
b) Hence show that the cubic approximation of (1+𝑥𝑥)(2−𝑥𝑥) is 2 − 2
+ 4
𝑥𝑥 − 8
𝑥𝑥
c) State the range of values of x for which the expansion is valid

4−5𝑥𝑥 𝐴𝐴 𝐵𝐵 4−5𝑥𝑥 3 2
a) (1+𝑥𝑥)(2−𝑥𝑥) ≡ 1+𝑥𝑥 + 2−𝑥𝑥 b) (1+𝑥𝑥)(2−𝑥𝑥) = 1+𝑥𝑥 − 2−𝑥𝑥
𝐴𝐴 2−𝑥𝑥 +𝐵𝐵(1+𝑥𝑥) = 3(1 + 𝑥𝑥)−1 − 2(2 − 𝑥𝑥)−1
≡ 1+𝑥𝑥 (2−𝑥𝑥)
4 − 5𝑥𝑥 ≡ 𝐴𝐴 2 − 𝑥𝑥 + 𝐵𝐵(1 + 𝑥𝑥) The expansion of 3(1 + 𝑥𝑥)−1 = 3 − 3𝑥𝑥 + 3𝑥𝑥 2 − 3𝑥𝑥 3 + ⋯
𝑥𝑥 𝑥𝑥 2 𝑥𝑥 3
The expansion of 2(2 − 𝑥𝑥)−1 = 1 + 2 + + +⋯
Substitute x = 2: 4 8
4 − 10 = 𝐴𝐴 × 0 + 𝐵𝐵 × 3
4−5𝑥𝑥 𝑥𝑥 𝑥𝑥 2 𝑥𝑥 3
B = −2 Hence (1+𝑥𝑥)(2−𝑥𝑥) = 3 − 3𝑥𝑥 + 3𝑥𝑥 2 − 3𝑥𝑥 3 − 1 + 2 + +
4 8
7 11 2 25 3
= 2 − 2 𝑥𝑥 + 𝑥𝑥 − 𝑥𝑥
Substitute x = -1: 4 8
4 + 5 = 𝐴𝐴 × 3 + 𝐵𝐵 × 0 3 2
A=3 c) 1+𝑥𝑥 is valid if 𝑥𝑥 < 1 2−𝑥𝑥
is valid if 𝑥𝑥 < 2
So the expansion is valid when 𝑥𝑥 < 1
What do I need to be able to do? Y13 – Chapter 5 Radians
By the end of this chapter you should be able to:
• Convert between degrees and radians Key words:
• Know exact values of angles measured in radians • Radian – The angle made by taking the radius and
• Find arc length using radians wrapping it round the circle
• Find areas of sectors and segments using radians • Arc length – The distance along part of the
• Solve trigonometric equations circumference of a circle, or of any curve
• Use small angle approximations • Sector – the area between two radiuses and the
connecting arc of a circle
• Segment – The smallest part of a circle made when it is
cut by a line
Pure Maths Year 2

Converting between degrees and radians Arc lengths, Sectors and Segments

2π radians = 360° When working in radians:

π radians = 180°
1 radian = 180/𝜋𝜋 Arc length = 𝑟𝑟𝜃𝜃
× 𝜋𝜋 ÷ 180
1
Area of sector = 𝑟𝑟 2 𝜃𝜃
Degrees Radians 2

1
× 180 ÷ 𝜋𝜋 Area of a segment = 𝑟𝑟 2 (𝜃𝜃 − sin 𝜃𝜃)
2

Small Angle Approximations

Solving Trigonometric Equations
When ϴ is small and measured in radians:
This works the same way as solving trigonometric equations 𝑦𝑦 = 𝜃𝜃

in degrees. sin ϴ ≈ ϴ
𝑦𝑦 = 𝑠𝑠𝑠𝑠𝑠𝑠𝜃𝜃
π/2
π–ϴ
S A
ϴ ϴ
π 0/2π
ϴ ϴ 𝑦𝑦 = 𝜃𝜃
T C tan ϴ ≈ ϴ 𝑦𝑦 = 𝑡𝑡𝑡𝑡𝑡𝑡𝜃𝜃
π+ϴ 2π– ϴ
3π/2
sin 𝜃𝜃 = sin(𝜋𝜋 − 𝜃𝜃)
cos 𝜃𝜃 = cos(2𝜋𝜋 − 𝜃𝜃)
tan 𝜃𝜃 = tan(𝜋𝜋 + 𝜃𝜃) ϴ2
cos ϴ ≈ 1 −
2
−sin 𝜃𝜃 = sin 𝜋𝜋 + 𝜃𝜃 = sin(2𝜋𝜋 − 𝜃𝜃)
− cos 𝜃𝜃 = cos(𝜋𝜋 − 𝜃𝜃) = cos(𝜋𝜋 + 𝜃𝜃) 𝑦𝑦 = 𝑐𝑐𝑐𝑐𝑐𝑐𝜃𝜃
−tan 𝜃𝜃 = tan 𝜋𝜋 − 𝜃𝜃 = tan(2𝜋𝜋 − 𝜃𝜃)
𝜃𝜃 2
𝑦𝑦 = 1 −
2
What do I need to be able to do? Y13 – Chapter 6 Trigonometric
By the end of this chapter you should be able to:
Functions
• Understand secant, cosecant and cotangent and their
relationship to cosine, sine and tangent Key words:
• Understand the graphs of secant, cosecant and cotangent • Cosecant – In a right angled triangle, the cosecant of an
• Simplify expressions, prove identities and solve equations involving angle is: The length of the hypotenuse divided by the length
secant, cosecant and cotangent of the side opposite the angle
• Understand and use inverse trigonometric functions • Secant– In a right angled triangle, the secant of an angle
is: The length of the hypotenuse divided by the length of
the adjacent side.
Pure Maths Year 2 • Cotangent – In a right angled triangle, the cotangent of
an angle is: The length of the adjacent side divided by the
length of the side opposite the angle
Trig Identities
1 + 𝑡𝑡𝑡𝑡𝑡𝑡2 𝑥𝑥 ≡ 𝑠𝑠𝑠𝑠𝑠𝑠 2 𝑥𝑥 Secant, Cosecant and Cotangent
1 + 𝑐𝑐𝑐𝑐𝑐𝑐 2 𝑥𝑥 ≡ 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 2 𝑥𝑥
1
sec 𝑥𝑥 =
Graphs of Inverse Functions cos 𝑥𝑥
The graph has symmetry in the y-axis, a period of 360/2 π
It has vertical asymptotes at all of the values for which cos(x)=0
The inverse of sin(x) is arcsin(x)
The domain of y=sec(x) is x∈ ℝ
The domain is -1 ≤x ≤1
In degrees:
The range is -π/2 ≤ arcsin(x) ≤ π/2 x≠ 90, 270, 450.. (any odd multiple of 90)
In radians:
x≠ π/2, 3π/2, 5π/2.. (any odd multiple of π/2)
The range of y=sec(x) is y≤-1 or y≥1

1
𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 𝑥𝑥 =
sin 𝑥𝑥
The inverse of cos(x) is arccos(x) The graph has symmetry in the y-axis, a period of 360/2 π
The domain is -1 ≤x ≤1 It has vertical asymptotes at all of the values for which cos(x)=0
The range is 0 ≤ arcsin(x) ≤ π
The domain of y=sec(x) is x∈ ℝ
In degrees:
x≠ 90, 270, 450.. (any odd multiple of 90)
In radians:
x≠ π/2, 3π/2, 5π/2.. (any odd multiple of π/2)
The range of y=sec(x) is y≤-1 or y≥1

1
The inverse of tan(x) is arctan(x) 𝑐𝑐𝑐𝑐𝑐𝑐 𝑥𝑥 =
𝑡𝑡𝑡𝑡𝑡𝑡 𝑥𝑥
The domain is x∈ ℝ The graph has vertical asymptotes at all of the values for which
The range is -π/2 ≤ arctan(x) ≤ π/2 tan(x)=0

The domain of y=cot(x) is x∈ ℝ

In degrees:
x≠ 0, 180, 360.. (any odd multiple of 180)
In radians:
x≠ 0, π, 2π.. (any multiple of π)
The range of y=sec(x) is y∈ ℝ
What do I need to be able to do? Y13 – Chapter 7 Trigonometry and
By the end of this chapter you should be able to: Modelling
• Prove and use the addition formulae
• Understand and use the double angle formulae Addition Formulae
• Solve trigonometric equations using addition and double angle
formulae
• Write expressions in the form Rcos(ϴ±α) and Rsin(ϴ±α) Sometimes are known as the compound angle
• Prove trigonometric identities formulae
• Model real life situations
𝑠𝑠𝑠𝑠𝑠𝑠 𝐴𝐴 + 𝐵𝐵 ≡ 𝑠𝑠𝑠𝑠𝑠𝑠A𝑐𝑐𝑐𝑐𝑐𝑐B + 𝑐𝑐𝑐𝑐𝑐𝑐A𝑠𝑠𝑠𝑠𝑠𝑠B
Proof of the Addition Formulae 𝑠𝑠𝑠𝑠𝑠𝑠 𝐴𝐴 − 𝐵𝐵 ≡ 𝑠𝑠𝑠𝑠𝑠𝑠A𝑐𝑐𝑐𝑐𝑐𝑐B − 𝑐𝑐𝑐𝑐𝑐𝑐A𝑠𝑠𝑠𝑠𝑠𝑠B

G cos(α+ E
β) 𝑐𝑐𝑐𝑐𝑐𝑐 𝐴𝐴 + 𝐵𝐵 ≡ 𝑐𝑐𝑐𝑐𝑐𝑐A𝑐𝑐𝑐𝑐𝑐𝑐B − 𝑠𝑠𝑠𝑠𝑠𝑠A𝑠𝑠𝑠𝑠𝑠𝑠B
α 𝑐𝑐𝑐𝑐𝑐𝑐 𝐴𝐴 − 𝐵𝐵 ≡ 𝑐𝑐𝑐𝑐𝑐𝑐A𝑐𝑐𝑐𝑐𝑐𝑐B + 𝑠𝑠𝑠𝑠𝑠𝑠A𝑠𝑠𝑠𝑠𝑠𝑠B
cos(α)sin(

sin(α)sin( 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 + 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡

𝑡𝑡𝑡𝑡𝑡𝑡 𝐴𝐴 + 𝐵𝐵 ≡
sin(α+β)

β) 1 − 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡
C
β)

𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 − 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡
sin(α)cosβ)

𝑡𝑡𝑡𝑡𝑡𝑡 𝐴𝐴 − 𝐵𝐵 ≡
1 + 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡

β Simplifying acos(x)± bcos(x)

α
A D B
Sometimes known as the harmonic form. You can
cos(α)cos(β) write expressions in the form acos(x) ± bcos(x) as a
Using the properties of sine and cosine we can label the diagram as
above.
function of sine or cosine only.

Using triangle ADE: 𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝜃𝜃 ± 𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏 can be written as either:

DE= sin(α+β)
AD = cos (α+β) 𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅 𝑥𝑥 ± 𝛼𝛼 where R>0 and 0< 𝛼𝛼<90
DE = DF + FE
𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅 𝑥𝑥 ± 𝛽𝛽 where R>0 and 0< 𝛽𝛽<90
⸫ sin(α+β) = sin(α)cos(β) + cos(α)sin(β)

AD = AB – DB Where 𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅 = 𝑎𝑎 and 𝑅𝑅𝑅𝑅𝑖𝑖𝑖𝑖𝛼𝛼 = 𝑏𝑏 and

⸫ cos(α+β) = cos(α)cos(β) – sin(α)sin(β) 𝑅𝑅 = 𝑎𝑎2 + 𝑏𝑏 2

Double Angle Formulae

You can use the addition formulae to derive the following double angle formulae: Pure Maths
𝑠𝑠𝑠𝑠𝑠𝑠 2𝐴𝐴 ≡ 2𝑠𝑠𝑠𝑠𝑠𝑠A𝑐𝑐𝑐𝑐𝑐𝑐A

𝑐𝑐𝑐𝑐𝑐𝑐 2𝐴𝐴 ≡ 𝑐𝑐𝑐𝑐𝑐𝑐 2 𝐴𝐴 − 𝑠𝑠𝑠𝑠𝑠𝑠2 𝐴𝐴 ≡ 2𝑐𝑐𝑐𝑐𝑐𝑐 2 𝐴𝐴 − 1 ≡ 1 − 2𝑠𝑠𝑠𝑠𝑠𝑠2 𝐴𝐴

Year 2
2𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡
𝑡𝑡𝑡𝑡𝑡𝑡 2𝐴𝐴 ≡
1 − 𝑡𝑡𝑡𝑡𝑡𝑡2 𝐴𝐴
What do I need to be able to do? Y13 – Chapter 8 Parametric Equations
By the end of this chapter you should be able to:
Pure Key words:
• Convert parametric equations into Cartesian form
• Understand and use parametric equations of curves and sketch Maths • Cartesian equations – Gives a direct
relationship between x and y
parametric curves Year • Parametric equations – Uses a third
• Solve problems involving parametric equations
• Use parametric equations in modelling 2 variable (usually t or ϴ) to define x and y

Sketching parametric equations of Converting between parametric and cartesian

curves equations

When sketching a parametric equation, sub in values Combine the two equations by rearranging one of them to
of t to find x and y values and then sketch as make t the subject and then substitute into the other
normal! equation.
OR
Sketch the curve defined by 𝒙𝒙=𝟐𝟐𝒕𝒕 and 𝒚𝒚=−𝒕𝒕𝟐𝟐 Rearrange both equations to make t the subject and then
between 𝒕𝒕=−𝟑𝟑 𝒂𝒂𝒏𝒏𝒅𝒅 𝟑𝟑. equate the two equations
T -3 -2 -1 0 1 2 3
Eg: Convert the following parametric equations into cartesian
X -6 -4 -2 0 2 4 6
form
Y -9 -4 -1 0 -1 -4 -9

x = t + 3 y = 2t2 x = t + 3 y = 2t2

x=t+3t=x–3 x=t+3t=x–3

y = 2(x – 3)2 OR y = 2t2  t = √(y/2)

√(y/2) = x – 3
Calculus with parametric equations* y/2 = (x – 3)2
* This section actually appears in your text book in chapters 9 and 11 y = 2 (x – 3)2

Differentiation: If your parametric equations contains trigonometric functions,

If x = f(t) and y = g(t) first find an identity that connects then rearrange the
Then: parametric equations so that you can substitute into the
𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑
= ÷ identity
𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑

Integration:
If x = f(t) and y = g(t) Domain and range
Then:
𝑑𝑑𝑑𝑑
� 𝑦𝑦 𝑑𝑑𝑑𝑑 = � 𝑦𝑦 𝑑𝑑𝑑𝑑 For parametric equations x = p(t) and y = q(t) with Cartesian
𝑑𝑑𝑑𝑑 equation y = f(x)
Remember to adjust limits if you are using definite • The domain of f(x) is the range of p(t)
integration • The range of f(x) is the range of q(t)
What do I need to be able to do? Y13 – Chapter 9 Differentiation
By the end of this chapter you should be able to: Pure
Key words:
• Differentiate trigonometric functions • Concave – Curves inwards Maths
• Differentiate exponentials and logarithms • Convex – Curves outwards Year 2
• Use the chain, product and quotient rules
• Use the second derivative to describe a function’s behaviour
• Solve problems involving connected rates of change Chain, Product and Quotient Rules
• Construct differential equations
• Differentiate parametric equations (see parametric equations
sheet)
Chain Rule:
𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑
= ×
𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑
Differentiating Trigonometric functions
Where y is a function of u and u is another function of x
𝑑𝑑𝑑𝑑
If 𝑦𝑦 = sin 𝑘𝑘𝑘𝑘, then 𝑑𝑑𝑑𝑑 = 𝑘𝑘 cos 𝑘𝑘𝑘𝑘 In function notation:
𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑
If 𝑦𝑦 = cos 𝑘𝑘𝑘𝑘, then 𝑑𝑑𝑑𝑑 = −𝑘𝑘 sin 𝑘𝑘𝑘𝑘 If y = (f(x))n then 𝑑𝑑𝑑𝑑 = 𝑛𝑛(𝑓𝑓 𝑥𝑥 )𝑛𝑛−1 𝑓𝑓𝑓(𝑥𝑥)
𝑑𝑑𝑑𝑑
If 𝑦𝑦 = tan 𝑘𝑘𝑘𝑘, then 𝑑𝑑𝑑𝑑 = 𝑘𝑘 sec 2 𝑘𝑘𝑘𝑘
𝑑𝑑𝑑𝑑
If 𝑦𝑦 =
𝑑𝑑𝑑𝑑
cosec 𝑘𝑘𝑘𝑘, then 𝑑𝑑𝑑𝑑 = −𝑘𝑘 cosec 𝑘𝑘𝑘𝑘 cot 𝑘𝑘𝑘𝑘 If y = f(g(x)) then 𝑑𝑑𝑑𝑑 = 𝑓𝑓 ′ 𝑔𝑔 𝑥𝑥 𝑔𝑔𝑔(𝑥𝑥)
𝑑𝑑𝑑𝑑
If 𝑦𝑦 = sec 𝑘𝑘𝑘𝑘, then 𝑑𝑑𝑑𝑑 = −𝑘𝑘 sec 𝑘𝑘𝑘𝑘 tan 𝑘𝑘𝑘𝑘
𝑑𝑑𝑑𝑑 Product Rule:
If 𝑦𝑦 = cot 𝑘𝑘𝑘𝑘, then 𝑑𝑑𝑑𝑑 = −𝑘𝑘 𝑐𝑐𝑐𝑐sec 2 𝑘𝑘𝑘𝑘
𝑑𝑑𝑑𝑑 1
If 𝑦𝑦 = arcsin 𝑥𝑥, then 𝑑𝑑𝑑𝑑 = If y = uv where u and v are functions of x, then:
1−𝑥𝑥 2
𝑑𝑑𝑑𝑑 1 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑
If 𝑦𝑦 = arccos 𝑥𝑥, then 𝑑𝑑𝑑𝑑 = − = 𝑢𝑢 + 𝑣𝑣
1−𝑥𝑥 2 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑
𝑑𝑑𝑑𝑑 1
If 𝑦𝑦 = arctan 𝑥𝑥, then 𝑑𝑑𝑑𝑑 = 1+𝑥𝑥2
In function notation:
If f(x) = g(x)h(x) then:
Implicit Differentiation 𝑓𝑓 ′ 𝑥𝑥 = 𝑔𝑔 𝑥𝑥 ℎ′ 𝑥𝑥 + ℎ 𝑥𝑥 𝑔𝑔𝑔(𝑥𝑥)

Use when equations are difficult to rearrange into the form Quotient Rule:
y = f(x)
𝑑𝑑 𝑛𝑛 𝑑𝑑𝑑𝑑 If y = u/v where u and v are functions of x, then:
𝑦𝑦 = 𝑛𝑛𝑦𝑦 𝑛𝑛−1
𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑
𝑑𝑑 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 𝑣𝑣 𝑑𝑑𝑑𝑑 − 𝑢𝑢 𝑑𝑑𝑑𝑑
𝑥𝑥𝑥𝑥 = 𝑥𝑥 + 𝑦𝑦 =
𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 𝑣𝑣 2

In function notation:
Second Derivatives If f(x) = g(x)/h(x) then:
ℎ 𝑥𝑥 𝑔𝑔′ 𝑥𝑥 − 𝑔𝑔 𝑥𝑥 ℎ′(𝑥𝑥)
𝑓𝑓 ′ 𝑥𝑥 =
The function f(x) is concave on a given interval if and only if (ℎ(𝑥𝑥))2
f’’(x) ≤ 0 for every value of x in the value in that interval
Differentiating Exponential and Logarithms
The function f(x) is convex on a given interval if and only if
f’’(x) ≥ 0 for every value of x in that interval If 𝑦𝑦 = 𝑒𝑒 𝑘𝑘𝑘𝑘 , then
𝑑𝑑𝑑𝑑
= 𝑘𝑘𝑒𝑒 𝑘𝑘𝑘𝑘
𝑑𝑑𝑑𝑑
𝑑𝑑𝑑𝑑 1
If 𝑦𝑦 = ln𝑥𝑥, then =
𝑑𝑑𝑑𝑑 𝑥𝑥
A point of inflection is a point at which f’’(x) changes sign 𝑑𝑑𝑑𝑑
If 𝑦𝑦 = 𝑎𝑎𝑘𝑘𝑘𝑘 , where k is a real constant and a>0, then =
𝑑𝑑𝑑𝑑
𝑎𝑎𝑘𝑘𝑘𝑘 𝑘𝑘ln𝑎𝑎
What do I need to be able to do? Y13 – Chapter 10 Numerical Methods
By the end of this chapter you should be able to:
Key words:
• Locate roots of f(x)=0 by considering sign changes • Root – Where a function equals zero
• Use iteration to find an approximation to the equation f(x)=0 • Continuous function – The function does not ‘jump’ from
• Use the Newton Raphson method to find approximations to the one value to another. If the graph of a function has a
solutions of equations in the form f(x)=0 vertical asymptote between two points then the function is
• Use numerical methods to solve problems in context not continuous in the interval between the two points.

Locating Roots
You can sometimes show the existence of a root within a given interval by showing that the function changes sign in
that interval.

Eg Show that f(x) = x3 – 4x2 + 3x + 1 has a root between x=1.4 and x=1.5

f(1.4) = (1.4)3 – 4(1.4)2 + 3(1.4) + 1 = 0.104

f(1.5) = (1.5)3 – 4(1.5)2 + 3(1.5) + 1 = -0.125

The function is continuous and there is a change of sign between 1.4 and 1.5 so there is at least one root in this
interval

Watch out! The change of sign doesn’t mean exactly one root and the absence of a sign change does not necessarily
mean that a root does not exist

a b

a b

a b

There are multiple roots within the There are multiple roots within the There is a vertical asymptote within
interval [a, b] – odd number of roots interval [a, b] but a sign change the interval [a, b] – a sign change
does not occur – even number of occurs but no roots
roots

Iteration The Newton-Raphson Method

To solve an equation in the form f(x) = 0 using iteration first rearrange An alternative way to find roots using
f(x)=0 into the form x=g(x) and then use the iterative formula: differentiation
𝑥𝑥𝑛𝑛+1 = 𝑔𝑔(𝑥𝑥𝑛𝑛 ) 𝑓𝑓(𝑥𝑥𝑛𝑛 )
𝑥𝑥𝑛𝑛+1 = 𝑥𝑥𝑛𝑛 −
𝑓𝑓𝑓(𝑥𝑥𝑛𝑛 )
Some iterations will converge and some will diverge
Successful iterations can be shown on staircase or cobweb diagrams Pure Maths Year 2
 What do I need to be able to do? Y13 – Chapter 11 Integration
By the end of this chapter you should be able to:
• Integrate standard mathematical function
Key words:
• Use trigonometric identities in integration • Differential equation – An equation with a function and
one or more of its derivatives
• Use the reverse chain rule
• Integrate functions by substitution, by parts and using partial
fractions Standard integrals
• Find the are under a curve using integration
• Use the trapezium rule 𝑥𝑥 𝑛𝑛+1
• Solve differential equations and model with differential equations � 𝑥𝑥 𝑛𝑛 𝑑𝑑𝑑𝑑 = + 𝑐𝑐; 𝑛𝑛 ≠ −1
𝑛𝑛 + 1

Reverse Chain Rule � 𝑒𝑒 𝑥𝑥 𝑑𝑑𝑑𝑑 = 𝑒𝑒 𝑥𝑥 + 𝑐𝑐

1
Functions in the form f(ax+b): � 𝑑𝑑𝑑𝑑 = ln 𝑥𝑥 + 𝑐𝑐
𝑥𝑥
1
� 𝑓𝑓 ′ 𝑎𝑎𝑎𝑎 + 𝑏𝑏 𝑑𝑑𝑑𝑑 = 𝑓𝑓 𝑎𝑎𝑎𝑎 + 𝑏𝑏 + 𝑐𝑐
𝑎𝑎 � cos 𝑥𝑥 𝑑𝑑𝑑𝑑 = sin 𝑥𝑥 + 𝑐𝑐
𝑓𝑓𝑓(𝑥𝑥)
To integrate expressions in the form∫ 𝑘𝑘 𝑓𝑓(𝑥𝑥)
𝑑𝑑𝑑𝑑, try ln|f(x)|and
� sin 𝑥𝑥 𝑑𝑑𝑑𝑑 = − cos 𝑥𝑥 + 𝑐𝑐
differentiate to check and adjust the constant as necessary

To integrate functions in the form ∫ 𝑘𝑘𝑘𝑘𝑘(𝑥𝑥)(𝑓𝑓(𝑥𝑥))𝑛𝑛 𝑑𝑑𝑑𝑑, try � sec 2 𝑥𝑥 𝑑𝑑𝑑𝑑 = tan 𝑥𝑥 + 𝑐𝑐
(f(x))n+1 and differentiate to check and adjust the constant as necessary
� cosec 𝑥𝑥 cot 𝑥𝑥 𝑑𝑑𝑑𝑑 = − cosec 𝑥𝑥 + 𝑐𝑐
Partial Fractions and Integration by substitution
� cosec 2 𝑥𝑥 𝑑𝑑𝑑𝑑 = − cot 𝑥𝑥 + 𝑐𝑐
You can integrate by substitution by choosing a function (𝑢𝑢(𝑥𝑥)) (which
can be differentiated) to help you to integrate a tricky function. You
need to substitute all 𝑥𝑥′𝑠𝑠 (including d𝑥𝑥) with terms involving 𝑢𝑢. � sec 𝑥𝑥 tan 𝑥𝑥 𝑑𝑑𝑑𝑑 = sec 𝑥𝑥 + 𝑐𝑐

You can also use partial fractions in order to integrate algebraic

fractions You can use trigonometric identities to replace expressions
that cannot be integrated with one that can
Differential Equations
Eg ∫ tan2 𝑥𝑥 𝑑𝑑𝑑𝑑
𝑑𝑑𝑑𝑑 1 + tan2 𝑥𝑥 ≡ sec 2 𝑥𝑥
When 𝑑𝑑𝑑𝑑 = 𝑓𝑓 𝑥𝑥 𝑔𝑔(𝑦𝑦) you can solve it by separating the variables So
as follows: tan2 𝑥𝑥 ≡ sec 2 𝑥𝑥 − 1
1 So
� 𝑑𝑑𝑑𝑑 = � 𝑓𝑓(𝑥𝑥) 𝑑𝑑𝑑𝑑
𝑔𝑔(𝑦𝑦) ∫ tan2 𝑥𝑥 𝑑𝑑𝑑𝑑 = ∫ sec 2 𝑥𝑥 − 1 𝑑𝑑𝑑𝑑 which we can
When you integrate to solve a differential equation you stil need to integrate using a standard integral
include a constant of integration – this wil give the general solution.
Integration by Parts
Area Bounded by Two Curves 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑
� 𝑢𝑢 𝑑𝑑𝑑𝑑 = 𝑢𝑢𝑢𝑢 − � 𝑣𝑣 𝑑𝑑𝑑𝑑
𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑
𝑏𝑏 𝑏𝑏 𝑏𝑏
𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴 𝑜𝑜𝑜𝑜 𝑅𝑅 = � 𝑓𝑓 𝑥𝑥 − 𝑔𝑔(𝑥𝑥) 𝑑𝑑𝑑𝑑 = � 𝑓𝑓(𝑥𝑥) 𝑑𝑑𝑑𝑑 − � 𝑔𝑔(𝑥𝑥) 𝑑𝑑𝑑𝑑
𝑎𝑎 𝑎𝑎 𝑎𝑎
Limit Notation
𝑏𝑏
The Trapezium Rule Pure 𝑏𝑏
𝑏𝑏
1 Maths � 𝑓𝑓(𝑥𝑥) 𝑑𝑑𝑑𝑑 = lim � 𝑓𝑓(𝑥𝑥)𝛿𝛿𝛿𝛿
𝛿𝛿𝛿𝛿→0
� 𝑦𝑦 𝑑𝑑𝑑𝑑 ≈ ℎ(𝑦𝑦0 + 2 𝑦𝑦1 + 𝑦𝑦2 … + 𝑦𝑦𝑛𝑛−1 + 𝑦𝑦𝑛𝑛 ) Year 2
𝑎𝑎 𝑎𝑎
𝑎𝑎 2
What do I need to be able to do? Y13 – Chapter 12 Vectors
By the end of this chapter you should be able to:
• Understand 3D Cartesian coordinates Key words:
• Use vectors in three dimensions • Coplanar vectors – Vectors in the same plane
• Use vectors to solve geometric problems
• Model 3D motion in mechanics with vectors
• Magnitude– The size of the vector

3D Coordinates 3D Vectors
z
y
Unit vectors along the x, y and z axes
are denoted by I, j and k respectively
0 x 1 0 0
𝑖𝑖 = 0 𝑗𝑗 = 1 𝑘𝑘 = 0
When visualising 3D coordinates, think of the x and y 0 0 1
axis drawn on a flat surface with the z axis sticking
up from the flat surface. For any 3D vector 𝑝𝑝𝒊𝒊 + 𝑞𝑞𝒋𝒋 + 𝑟𝑟𝒌𝒌 =
𝑝𝑝
The distance from the origin to the point (x, y, z) is: 𝑞𝑞
𝑥𝑥 2 + 𝑦𝑦 2 + 𝑧𝑧 2 𝑟𝑟

The distance between the points (x1, y1, z1) and (x2, Magnitude and Direction
y2, z2)
is: Vector a = 𝑝𝑝𝒊𝒊 + 𝑞𝑞𝒋𝒋 + 𝑟𝑟𝒌𝒌
(𝑥𝑥1 − 𝑥𝑥2 )2 + (𝑦𝑦1 − 𝑦𝑦2 )2 + (𝑧𝑧1 − 𝑧𝑧2 )2
Magnitude of vector a:
𝒂𝒂 = 𝑝𝑝2 + 𝑞𝑞 2 + 𝑟𝑟 2
Parallel Vectors in 3D
Direction of vector a:
If a, b, and c are 3D, non-coplanar vectors (not 𝑝𝑝
The angle with the x-axis: cos 𝜃𝜃𝑥𝑥 =
in the same plane) then you can compare |𝒂𝒂|
𝑞𝑞
coefficients on both sides of an equation: The angle with the y-axis: cos 𝜃𝜃𝑦𝑦 =
|𝒂𝒂|
𝑟𝑟
The angle with the z-axis: cos 𝜃𝜃𝑧𝑧 =
|𝒂𝒂|
Eg
If
𝑝𝑝𝒊𝒊 + 𝑞𝑞𝒋𝒋 + 𝑟𝑟𝒌𝒌 = 𝑢𝑢𝒊𝒊 + 𝑣𝑣𝒋𝒋 + 𝑤𝑤𝒌𝒌 Pure Maths Year 2
Then: p=u, q=v and r=w