Mathematics Extension 1 HSC study notes

Table of Contents

Written by Vincent Liu (2018)

Trigonometric Identities 2 ..........................................................................................................

Mathematical Induction 3 ...........................................................................................................

Co-ordinate Geometry (3 Unit) 4 ...............................................................................................

Polynomials 5 ...............................................................................................................................

Parametrics 8 ................................................................................................................................

Permutations and Combinations 10 ..........................................................................................

Circle Geometry 11 .....................................................................................................................

Inverse Functions and Inverse Trigonometric Functions 13 ...................................................

Integration (3 Unit) 15 .................................................................................................................

Applications of Calculus to the Physical World (3 Unit) 16 ....................................................

The Binomial Theorem 21..........................................................................................................

�1

Trigonometric Identities

Pythagorean

� � �

Angle Sum

� �

� �

� �

Double Angle

�

� OR � OR �

Sums to Products (usually must be proved before using)

� �

� �

t-formulae

If � and � then:

� � �

Transformations (Auxiliary method)

� �

� �

Where � and �

General Solutions

� � �

sin2 x + cos2 x = 1 tan2 x + 1 = sec2 x cot2 x + 1 = cosec2x

sin(x + y) = sin x cos y + cos x sin y sin(x − y) = sin x cos y − cos x sin y

cos(x + y) = cos x cos y − sin x sin y cos(x − y) = cos x cos y + sin x sin y

tan(x + y) =tan x + tan y

1 − tan x tan ytan(x − y) =

tan x − tan y1 + tan x tan y

sin 2x = 2 sin x cos x

cos 2x = cos2 x − sin2 x cos 2x = 1 − 2 sin2 x cos 2x = 2 cos2 x − 1

sin x + sin y = 2 sin ( x + y2 ) cos ( x − y

2 ) sin x − sin y = 2 cos ( x + y2 ) sin ( x − y

2 )cos x + cos y = 2 cos ( x + y

2 ) cos ( x − y2 ) cos x − cos y = − 2 sin ( x + y

2 ) sin ( x − y2 )

t = tan ( x2 ) x ≠ π, 3π, 5π . . .

sin x =2t

1 + t2cos x =

1 − t2

1 + t2tan x =

2t1 − t2

a sin x + b cos x = r sin(x + α) a cos x + b sin x = r cos(x − α)

a sin x − b cos x = r sin(x − α) a cos x − b sin x = r cos(x + α)

r = a2 + b2 α = tan−1 ( ba )

sin θ = sin aθ = nπ + (−1)n a

cos θ = cos aθ = 2nπ ± a

tan θ = tan aθ = nπ + a

�2

Mathematical Induction

To prove a statement or proposition � true by mathematical induction:

1. We first prove it true for a base case, usually the minimum value of �.

2. We then assume that � is true (that is, � is true for some � ).

3. Then, we prove that � is true, usually by using the assumption that � is true.

If � is true, then � is true, so therefore � must be true…

i.e. if � is true for some integer �, then it is also true for all subsequent integers.

Proving the statement for a base value completes the proof.

Divisibility notation

The convenient notation � means that � divides � (that is, � is divisible by � ).

S(n)

n

S(k) S(n) n = k

S(k + 1) S(k)

S(k) S(k + 1) S((k + 1) + 1) = S(k + 2)

S(k) k

m |n m n n m

�3

Co-ordinate Geometry (3 Unit)

Angle between two lines

For two lines with gradients � and � , the angle � between them is given by: � if � , or

� if �

Division of Intervals

The order in which the endpoints of the interval � and � are specified is important. There are two cases:

Internal Division

External Division

� , i.e. � � , i.e. �

Note: these three statements are equivalent: • external division in the ratio � • division in the ratio � • division in the ratio �

In either case, the point � that divides the line � in the ratio � is given by the formula

� �

When using the formula with external division, one of � and � must be negative (it does not matter which).

m1 m2 θ

90∘ m1m2 = − 1

tan θ =m1 − m2

1 + m1m2m1m2 ≠ − 1

A B

Ifmn

> 1 |m | > |n | Ifmn

< 1 |m | < |n |

m : n−m : nm : − n

P A B m : n

x =m x2 + n x1

m + ny =

my2 + ny1

m + n

m n

�4

x

y

θ1

θ2

tan θ1 = m1tan θ2 = m2

B(x2, y2)

A(x1, y1)

P(x, y)

m

n

P(x, y)

A(x1, y1)

m

B(x2, y2) n

B(x2, y2)

P(x, y)

n

m

A(x1, y1)

Polynomials

A polynomial is an algebraic expression of the form � for � .

Using division, any polynomial can be expressed in the form � , where the polynomial � is the dividend, � is the divisor, � is the quotient and � is the remainder.

Polynomial Long Division

Polynomial long division occurs like normal long division.

e.g. �

1. Choose a term that can be multiplied with the first term of the divisor to give the first term of the dividend. This is the first term of the quotient.

�

2. Multiply the rest of the terms of the divisor by the term in step 1. and write down the result directly underneath.

�

3. Perform a subtraction.

�

4. Bring down the next term(s) (the number of terms you end up with must be the same number of terms in the divisor) and repeat.

�

Note: When dividing polynomials, the degree of the remainder is always less than that of the divisor.

anxn + an−1xn−1 + an−2xn−2 + . . . + a0 n ∈ ℤ+

P(x) ≡ D(x)Q(x) + R(x)P(x) D(x) Q(x) R(x)

x + 1 |2x3 + 3x2 + 2x + 1

2x2

x + 1 |2x3 + 3x2 + 2x + 12x3

2x2

x + 1 |2x3 + 3x2 + 2x + 12x3 + 2x2

2x2

x + 1 |2x3 + 3x2 + 2x + 12x3 + 2x2

x2

2x2

x + 1 |2x3 + 3x2 + 2x + 12x3 + 2x2

x2 + 2x

�5

Remainder Theorem

If a polynomial � is divided by � , the remainder is � .

More generally, if a polynomial � is divided by � , the remainder is � .

Proof: We use the form � with � . Since � has degree one, � must be of degree zero (that is, a constant).

�

Factor Theorem

It follows from the remainder theorem that � is a factor to � if and only if � .

This technique can be used to quickly factor polynomials by testing values of � .

More generally, � is a factor to � if and only if � .

Graphing Factorised Polynomials

The graph of a factorised polynomial can be sketched easily by inspecting its factors.

If a polynomial has a factor � , then its graph has a curve of degree � at � .

The exact nature of a curve can be determined by testing if the polynomial is positive or negative at a point on the curve.

e.g. �

Testing points: �

P(x) (x − α) P(α)

P(x) (bx − α) P ( αb )

P(x) ≡ D(x)Q(x) + R(x) D(x) = bx − αD(x) R(x)

P(x) = (bx − α)Q(x) + r

∴ P ( αb ) = 0 ⋅ Q(x) + r

P ( αb ) = r

(x − α) P(x) P(α) = 0

α

(bx − α) P(x) P ( αb ) = 0

(x − α)r r x = α

P(x) = (x − 1)(x − 2)2(x − 3)3

x = 0, y > 0

x = 32 , y < 0

x = 52 , y < 0

x = 4, y > 0

�6

x

y

degree 1

x

y

degree 2, 4, 6…

x

y

degree 3, 5, 7…

y

x1 2 3

y = P(x)y

x1 2 3

y = P(x)

Roots and Coefficients of a Polynomial

Approximation of roots

Suppose a polynomial function � has unknown roots that we wish to approximate. We have two methods.

An important observation: if � and � have different signs, then a root must lie between � and � .

Half-interval method

From the above observation, the first approximation using this method is given by � .

To apply the method again, choose � as the first point.

Then, the second point is whichever of � and � has a different sign to � .

Newton’s method

If � is an approximation, a better approximation � is given by

�

Newton’s method does not work when: 1. � is a stationary point, i.e. � 2. � is closer to another root, in which case � approximates

that root instead

In some cases, Newton’s method will oscillate or end up moving towards another root instead.

Quadratic Cubic Quartic

�αβγ =−da

�a x3 + bx2 + cx + d = 0

�αβγδ =ea

�α + β + γ + δ =−ba

�α + β =−ba

�αβ + αγ + βγ =ca

�αβγ + αβδ + βδγ + αδγ =−da

�a x4 + bx3 + cx2 + d x + e = 0 �a x2 + bx + c = 0

�α + β + γ =−ba

�αβ + αγ + αδ + βγ + βδ + γδ =ca

�αβ =ca

f (x)

f (x0) f (x1) x0 x1

f ( x0 + x1

2 )x0 + x1

2x0 x1

x0 + x1

2

x1 x2

x2 = x1 −f (x1)f ′�(x1)

x1 f ′�(x1) = 0x1 x2

�7

α x1x2

tan θ = f ʹ(x1)tan θ = f(x1) ______ (x1 - x2)Equating these gives Newton’s method.

θx

y

Here, when Newton’s method is applied on x1 to approximate α, the new approximation x2 is too far away. Applying Newton’s method again causes an approximation of β instead.

α x1

x2

x3

β x

y

Parametrics

A parametric equation is a set of two equations each describing � and � in terms of a parameter. Parametric equations can be converted to Cartesian equations by solving and eliminating the parameter.

Parametric representation of a parabola

The Cartesian form of a parabola with the vertex at the origin is � .

The corresponding parametric equation is � .

Note: The gradient of the tangent at a point is equal to the parameter: �

Equation of the chord

The gradient of the chord is � (since � ).

Using point-gradient form gives

�

If the chord is focal, it passes through the focus � . Substituting this point into the equation gives a condition for a focal chord: � .

Equation of the tangent

By obtaining the gradient and then using the point-gradient formula, we get:

Parametric: �

Cartesian: �

Equation of the normal

By obtaining the gradient of the tangent, taking the negative reciprocal and using the point-gradient formula, we get:

Parametric: �

Cartesian: �

x y

x2 = 4ay

{x = 2aty = at2

d yd x

=d y /dtd x /dt

=2at2a

= t

aq2 − ap2

2aq − 2ap=

p + q2

a ≠ 0

y = ( p + q2 ) x − apq

(0, a)pq = − 1

y = px − ap2

x x1 = 2a(y − y1)

x + py = ap3 + 2ap

y − y1 = −2ax1

(x − x1)

�8

x

y

P(2ap, ap2) Q(2aq, aq2)

x

y

P(2ap, ap2)

x

y

P(2ap, ap2)

Intersection of tangents

By simultaneously solving two parametric equations of the tangent with parameters � and �,

�

� and �

The intersection of two tangents is always outside the parabola.

Intersection of normals

By simultaneously solving two parametric equations of the normal with parameters � and �,

�

� and �

The intersection of two normals is always inside the parabola.

Chord of contact

The chord of contact is the line drawn between two points of contact on the parabola, i.e. points whose tangents intersect.

Observe that the equations for both tangents are:

�

Both these equations are of the form

�

A line can be unambiguously defined by two points — in this case, the two points of contact.

Since this equation represents a line and is satisfied by both points of contact, we deduce that it must therefore be the equation for the chord of contact.

p q

{y = px − ap2

y = qx − aq2

x = a(p + q) y = apq

p q

{x + py = ap3 + 2apx + q y = aq3 + 2aq

x = − pq(p + q) y = a(p2 + pq + q2 + 2)

{x1x0 = 2a(y1 + y0)x2x0 = 2a(y2 + y0)

x x0 = 2a(y + y0)

�9

x

y

Q(2aq, aq2)P(2ap, ap2)

x

y

P(2ap, ap2)Q(2aq, aq2)

x

y

(x2, y2)(x1, y1)

(x0, y0)

Permutations and Combinations

Permutations are arrangements where order is important i.e. � is different from � . Combinations are arrangements where order is not important i.e. � is the same as � . We introduce the factorial notation: � .

Multiplication Principle

If one operation can be performed in � different ways and afterwards, a second operation can be performed in � different ways and so on, the number of ways of performing the operations are � .

Addition Principle

If two operations are mutually exclusive, the number of arrangements for each can be added together to achieve the number of arrangements for either.

Circular Permutations

For every arrangement in a circle, there are � arrangements in a line which would be equivalent (e.g. for

� we have � ). Thus, the number of ways to arrange � people in a circle is � .

Permutations with Repetition

When arranging � things with � amount of one thing, � amount of another thing and so on, the number of

permutations is � .

The notation �

� represents the number of permutations where � places are filled with � objects. �

The notation �

� represents the number of combinations where � places are filled with � objects. �

or alternatively, the binomial notation � may be used.

The Stars and Bars method

When distributing � items across � places, permutations with repetition may be used to simulate the scenario by using dividers � to create places and stars (or similar objects) to represent items.

e.g. � simulates distributing 7 items across 4 places, the dividers creating places.

A BC CBAA BC CBA

n! = (n)(n − 1)(n − 2) . . . (1)

mn m × n × . . .

n

n = 4 A BCD = BCDA = . . . nn!n

n p qn!

p! q!

nPr

nPr r n nPr =n!

(n − r)!

nCr

nCr r n nCr =n!

(n − r)!r!nCr = (n

r)

n m|

⋆ | ⋆ ⋆ ⋆ | ⋆ ⋆ | ⋆

�10

Circle Geometry

Theorems

�11

θθ

θ

O

θ

O

θθ

O

O

θ

2θ

θ

180° – θ

θ

θ

θ

Equal angles at the centre stand on equal chords.

(and converse)

The perpendicular from the centre of a circle to a chord bisects the chord.

The line from the centre to the midpoint of a chord is perpendicular to the chord

(converse)

Equal chords in equal circles are equidistant from centres.

(and converse)

Opposite angles of a cyclic quad. are supplementary.

(converse to prove concyclic points)

The angle in a semicircle is a right angle.

(converse to prove diameter)

The angle at the centre is twice the angle at the circumference

subtended by the same arc.

Angles in the same segment are equal.

If an interval subtends equal angles at two points on the same side of it, the interval’s end points and the two points are concyclic.

The angle between a tangent and the chord is equal to the angle in

the alternate segment. (converse to prove tangent)

O

General Tips for Solving Circle Geometry Problems

• draw the most general form of a shape (e.g. if drawing a triangle, draw a scalene one, not isoceles)

• rotate the diagram to get a new perspective

• highlight cyclic quadrilaterals

• construct any common chords or tangents

�12

A

B

C

D

X

X

A

B

C

D

X

A

B

C

O Oʹ

O

Tangents to a circle from an external point are equal.

The tangent to a circle is perpendicular to the radius

drawn to the point of contact.

(and converse)

When two circles intersect, the line through their centres bisects the common chord at

right angles.

The products of the intercepts of two intersecting

chords are equal. AX · XB = CX · XD

The square of the tangent is equal to the product of the

intercepts of the secant. CX2 = AX · BX

The products of the intercepts of two secants to a

circle are equal. AX · BX = CX · DX

O

Oʹ

T

When two circles touch, the line joining their centres

passes through the point of contact.

Inverse Functions and Inverse Trigonometric Functions

An inverse function (denoted � ) is a function that “reverses” another.

If � , then � , and � .

To find the inverse, change � to � and � to � in the equation. e.g. � .

When taking the inverse of a function, its domain and range are switched.

Graphing Inverse Functions

The graph of � is the graph of � mirrored across the line � .

This means that the graph of � intersects the graph of � at � .

Restricting the Domain

If � is NOT a one-to-one function (that is, it does not pass the “horizontal line test”), then its inverse will not actually be a function (it will not pass the “vertical line test”).

In this case, we can restrict the domain of � such that it will have an inverse function. The domain is usually chosen to preserve as much of the original function as possible, preferably including � .

The Inverse Trigonometric Functions

By restricting the domain of the trigonometric functions, we may take their inverse.

The inverse sine function � is obtained by restricting the domain of � to � .

The inverse cosine function � is obtained by restricting the domain of � to � .

The inverse tangent function � is obtained by restricting the domain of � to � .

� � �

Note: � and � are both odd functions.

f −1(x)

f (x) = y f −1(y) = x f ( f −1(x)) = x = f −1( f (x))

f (x) x x f −1(x)f (x) = log x + 1 ⟹ x = log f −1(x) + 1 ⟹ f −1x = ex−1

y = f −1(x) y = f (x) y = x

y = f −1(x) y = f (x) y = x

f (x)

f (x)x = 0

sin−1 x sin x −π2

≤ x ≤π2

cos−1 x cos x 0 ≤ x ≤ π

tan−1 x tan x −π2

< x <π2

y = sin−1 x y = cos−1 x y = tan−1 x

sin−1 x tan−1 x

�13

x

y

y = f (x)

y = f -1(x)

y = x

x

y

1

-1

π—2

π–— 2

0

x

y

1

π—2

-1

π

y

0

π—2

π–— 2

Calculus with Inverse Trigonometric Functions

Proof: By using the property � ,

�

� (since � for � )

�

A similar procedure may be followed for � and � .

Differentiation (General form)

� �

� �

� �

Integration (General form)

� �

� �

d yd x

=1dxdy

y = sin−1 x

x = sin y (for −π2

< y <π2 )

d xd y

= cos y

= 1 − x2

cos y ≥ 0 −π2

< y <π2

∴d yd x

=1

1 − x2

cos−1 x tan−1 x

dd x

sin−1 x =1

1 − x2

dd x

sin−1 f (x) =f ′�(x)

1 − f (x)2

dd x

cos−1 x = −1

1 − x2

dd x

cos−1 f (x) = −f ′�(x)

1 − f (x)2

dd x

tan−1 x =1

1 + x2

dd x

tan−1 f (x) =f ′�(x)

1 + f (x)2

∫1

1 − x2= sin−1 x + C

= − cos−1 x + C

∫1

a2 − b2x2=

1b

sin−1 bxa

+ C

= −1b

cos−1 bxa

+ C

∫1

1 + x2= tan−1 x + C ∫

1a2 + b2x2

=1

abtan−1 bx

a+ C

�14

Integration (3 Unit)

Integration by Substitution

Indefinite Integrals

Suppose we have an integral in the form � .

We may perform a substitution � , which gives � (or � ).

Substituting this result gives:

�

After integrating, we can undo the substitution � to obtain the result in terms of �.

Definite Integrals

If the integral has bounds, they must also be dealt with during the substitution process.

Because definite integrals are to be evaluated numerically, substituting back for � is not necessary.

�

� , so �

� (given that � : if � , then � , etc.)

There are two types of substitutions that can be performed — either � or � .

When performing a substitution from one variable to another, all instances of the original variable must be replaced with the new variable. The differential e.g. � must also be accounted for.

I = ∫ f (g(x)) ⋅ g′ �(x) d x

u = g(x)dud x

= g′�(x) du = g′�(x) d x

I = ∫ f (u)dud x

d x

= ∫ f (u) du

u = g(x) x

x

I = ∫a

bf (g(x)) ⋅ g′�(x) d x

u = g(x) du = g′�(x) d x

I = ∫x=a

x=bf (g(x)) ⋅ g′�(x) d x

= ∫u=g(a)

u=g(b)f (u) du

= ∫g(a)

g(b)f (u) du

u = g(x) x = a u = g(a)

u = g(x) x = g(u)

d x

�15

Applications of Calculus to the Physical World (3 Unit)

Related Rates

Suppose we have two quantities � and � each defined as a function of time �. If given the rate of change of one of the quantities and a relationship between the two quantities (e.g. having � in terms of �), we can calculate the rate of change of the other quantity.

If � , � and we relate the two variables � , then we can use the chain rule to obtain:

� and � (note that � .)

Exponential Growth and Decay

Basic Exponential Growth and Decay

If the rate of change of a quantity � over time is proportional to the quantity, then � for some �.

This implies that:

� (for constants � and � where � is the value of � at � .)

Proof: �

�

If � is positive, then � is increasing. If � is negative, then � is decreasing.

To find the values of the constants, substitute any given conditions into the equation.

Modified Exponential Growth and Decay

If the rate of change of a quantity � over time is proportional to � for some � , then � .

This implies that:

� (for constants �, � and � .)

Proof: As before, replacing � with � .

For negative �, � . (i.e. � is the value that � approaches as time passes.)

x y ty x

y = g(t) x = h(t) y = f (x)

d ydt

=d yd x

⋅d xdt

d xdt

=d xd y

⋅d ydt

d xd y

=1dydx

PdPdt

= k P k

P = P0ekt k P0 P0 P t = 0

dPdt

= k P

ddP

t =1

k P

t =1k

log P + C

P = ekt−kC = (e−kC) ekt

k P k P

P P − N NdPdt

= k (P − N )

P = N + Aekt k N A

P P − N

k limt→∞

P = N N P

�16

Motion

Displacement, velocity and acceleration can be expressed as functions of time.

If displacement � is a function of time �: �

Then velocity is its first derivative: �

And acceleration is its second derivative: �

Acceleration can be expressed in four useful forms.

Proof: � and �

When integrating acceleration or velocity, we can use the initial conditions given to us (e.g. when � , � ) to find the value of the constant of integration.

Taking the Square Root of Velocity

If we know that the velocity of an object doesn’t change signs, then it is either positive or negative.

This means that if � , then we may take either � or � (not both at once).

Usually, if the given initial conditions for velocity are either positive or negative, then only one of these equations can hold, so we may discard the other.

Choosing Which Form of Acceleration to Use

If acceleration is given as:

A function of time: use � or �

A function of displacement: use �

A function of velocity: check the initial conditions:

If they are � : use �

If they are � : use �

x t x = f (t)

v = ·x =d xdt

a = ··x =d2xdt2

=dvdt

= vdvd x

=d

d x ( 12

v2)

··x =dvdt

=dvd x

⋅d xdt

= vdvd x

vdvd x

=dvd x

⋅d ( 1

2 v2)dv

=d

d x ( 12

v2)t = 0

v = 0

v2 = f (x) v = f (x) v = − f (x)

··x =dvdt

d2xdt2

··x =d

d x ( 12

v2)

(t, v) ··x =dvdt

(x , v) ··x = vdvd x

�17

Simple Harmonic Motion (SHM)

In simple harmonic motion, acceleration is proportional to displacement, but acts in the opposite direction.

Simple harmonic motion can be modelled by a sine wave.

If the object starts from the origin:

Acceleration: � OR � Velocity: � OR � Displacement: � OR � (by the angle sum formula)

where � is the amplitude, � is the phase shift angle, and the period � is given by � .

(if the centre of motion is not the origin, replace � with � ).

Proof: Given that � ,

� �

Properties of SHM

Acceleration �

• reaches maximum magnitude � when � . • is zero when � .

Velocity �

• reaches maximum magnitude � when � . • is zero when �

·x = − an2 sin(nt + α) ··x = − n2x·x = an cos(nt + α) v2 = n2 (a2 − x2)x = a sin(nt + α) x = A sin(nt) + B cos(nt)

a α T T =2πn

x x − x0

··x = − n2x

dd x ( 1

2v2) = − n2x

12

v2 = −n2x2

2+ C1

Let C1 =n2a2

2for some constant a ,

∴ v2 = n2 (a2 − x2)

v2 = n2 (a2 − x2)d xdt

= n a2 − x2

dd x

(t) =1n ( 1

a2 − x2 )nt = sin−1 (x /a) + C2

∴ x = a sin (nt − C2)⟹ x = a sin(nt + α)

··x = − n2x

|a | x = ± ax = 0

v2 = n2 (a2 − x2)|v | x = 0

x = ± a

�18

t

x

O

| a | max,v = 0

| v | max,a = 0

a—a O

t

x

a

—a

OT

Choosing Which Form of SHM to Use

Which form of SHM to use depends on where the moving object is initially, at � .

Centre of motion Positive extreme Negative extreme � � �

Note: These formulae assume positive initial velocity. If initial velocity is negative, reverse the sign.

Otherwise:

• If the conditions are given are initial, use � , (i.e. for � ) � .

• If the conditions given are NOT initial, use � , � .

After picking an equation, substitute in the given conditions to find unknowns.

t = 0

x = a sin(nt) + b x = a cos(nt) + b x = − a cos(nt) + b

x = A sin(nt) + B cos(nt) + bt = 0 v = An cos(nt) − Bn sin(nt)

x = a sin(nt + α) + bv = an cos(nt + α)

�19

b + ab—a b b + ab—a b b + ab—a b

Projectile Motion

In the 3 unit course, several assumptions are made about projectile motion: There is no air resistance, gravity � is constant, and projectiles have negligible mass.

This leads to the two fundamental equations of projectile motion:

Fundamental Equations of Projectile Motion

� and � where � is the acceleration due to gravity.

Resolution of Velocity

Velocity can be resolved into horizontal and vertical components, and vice versa. If the velocity is � and the angle of inclination is � at a point in time, then

� OR � and �

are the equations that give the velocity of an object, at that point in time.

Deriving Projectile Motion Equations by Integrating

Suppose that the projectile starts from the origin with velocity � and angle of inclination � . Then, by integrating and using initial conditions to find the constants of integration, we get:

Acceleration: � � Velocity: � �

Displacement: � �

(Cartesian): �

These equations lead to some important expressions:

Time of Flight: � (solve for � at � )

Horizontal Range: � (substitute the time of flight � into the equation for �)

Maximum Height: � (solve for � at � , substitute into equation for �)

g

··x = 0 ··y = − g g

v α

{·x = v cos α·y = v sin α v2 = ·x2 + ·y2 tan α =

·y·x

V α

··x = 0 ··y = − g

·x = V cos α ·y = − gt + V sin α

x = (V cos α) t y = −12

gt2 + (V sin α) t

y = x tan α − ( g sec2 α2V 2 ) x2

T =2V sin α

gt y = 0

R =V 2 sin 2α

gT x

H =V 2 sin2 α

2gt ·y = 0 y

�20

The velocity is always tangent to the path of the projectile.

. y

. x

v

α

The Binomial Theorem

Using the � notation for binomial coefficients,

The expansion of the binomial � is given by

�

OR

�

In this way, the � term of a binomial expansion is given by � .

Binomial Identities

�

� (generally, this is not important to know)

Pascal’s Triangle

Each row of Pascal’s triangle is made by summing adjacent numbers on the row before.

Pascal’s triangle can be used to quickly find the coefficients of a binomial expansion.

For example, to determine the coefficients of � we look at the corresponding row in Pascal’s triangle to find that the coefficients are 1, 2, 1 — i.e. � .

(nr) =

n!(n − r)! r!

(a + b)n

(a + b)n = (n0) an + (n

1) an−1b + (n2) an−2b2 + . . . + (n

k) an−kbk + . . . + (nn) bn

(a + b)n =n

∑k=0

(nk) an−kbk

(k + 1)th Tk+1 = (nk) an−kbk

(nk) = ( n

n − k)(n

k) = (n − 1k ) + (n − 1

k − 1)

(a + b)2

(a + b)2 = 1a2 + 2ab + 1b2

�21

1 1 1

1 2 1 1 3 3 1

1 4 6 4 1 …

(a + b)0 (a + b)1 (a + b)2 (a + b)3 (a + b)4

…

Finding the Greatest Coefficients/Terms

It’s easy to see that if � then � (given that both terms are positive, otherwise see below).

Therefore, to find out which coefficients/terms are greatest, we can solve � for �.

�

To get the greatest coefficient, discard the variable (e.g. �) and then solve � for �.

To get the greatest term, substitute the given value into the variable and then solve � for �.

If negative coefficients/terms are involved and only the magnitude is needed, use � instead.

Remember that � must be an integer, e.g. if � then � .

Equal Coefficients/Terms

If the final inequality involves an integer e.g. � , then there are equal coefficients/terms present.

i.e. for some constant �, if � when � , then � when � , that is, � .

Binomial Probability

Suppose an event has two outcomes: success (probability �) and failure (probability �).

Incidentally, means that � .

Then, in � independent trials, the probability of � successes is given by � .

This can be thought of as choosing which trials to fail/succeed and multiplying by the probability of the desired number of successes/failures happening.

Tk+1

Tk> 1 Tk+1 > Tk

Tk+1

Tk> 1 k

Tk+1

Tk=

(nk) an−kbk

( nk − 1) an−k+1bk−1

=n!

(n − k)!k!⋅

(n − k + 1)!(k − 1)!n!

⋅ba

=n − k + 1

k⋅

ba

xTk+1

Tk> 1 k

Tk+1

Tk> 1 k

Tk+1

Tk> 1

k k < 2.5 k = 2, 1

k < 5

r k < rTk+1

Tk> 1 k = r

Tk+1

Tk= 1 Tr+1 = Tr

p q

p + q = 1 ⟹ q = 1 − p

n r P(r) = (nr) qn−r pr

�22