Higher Unit 1

Outcome 2

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Higher MathsHigher Maths

www.mathsrevision.comwww.mathsrevision.com

What is a setFunction in various formats

Composite FunctionsExponential and Log Graphs

Connection between Radians and degrees & Exact valuesSolving Trig EquationsBasic Trig Identities

Graph TransformationsTrig Graphs

Inverse function

Mindmap

Exam Question Type

Outcome 2

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Higher

Sets & FunctionsSets & Functions

Notation & Terminology

SETS: A set is a collection of items which have some common property.

These items are called the members or elements of the set.

Sets can be described or listed using “curly bracket” notation.

Outcome 2

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N = {natural numbers}

= {1, 2, 3, 4, ……….}

W = {whole numbers} = {0, 1, 2, 3, ………..}Z = {integers} = {….-2, -1, 0, 1, 2, …..}

Q = {rational numbers}

This is the set of all numbers which can be written as fractions or ratios.

eg 5 = 5/1 -7 = -7/1 0.6 = 6/10 = 3/5

55% = 55/100 = 11/20 etc

We can describe numbers by the following sets:

Outcome 2

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R = {real numbers}This is all possible numbers. If we plotted values on a number line then each of the previous sets would leave gaps but the set of real numbers would give us a solid line.

We should also note that

N “fits inside” W

W “fits inside” Z

Z “fits inside” Q

Q “fits inside” R


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QZWN

When one set can fit inside another we say

that it is a subset of the other.

The members of R which are not inside Q are called irrational numbers. These cannot be expressed as

fractions and include , 2, 35 etc

R

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To show that a particular element/number belongs to a particular set we use the symbol

. eg 3 W but 0.9 Z

Examples

{ x W: x < 5 }= { 0, 1, 2, 3, 4 }

{ x Z: x -6 } = { -6, -5, -4, -3, -2, …….. }

{ x R: x2 = -4 } = { } or

This set has no elements and is called the empty set.


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Nat 5

What are Functions ?

Functions describe how one quantity

relates to another

Car Part

s

Assembly line

Cars

Defn: A function or mapping is a relationship between two sets in which each member of the first set is connected to exactly one member in the

second set.

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Nat 5

What are Functions ?

Functions describe how one quantity

relates to another

Dirty

Washing Machine

Clean

OutputInputyx

Functionf(x)

y = f(x)

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Defining a Functions

A function can be thought of as the relationship between

Set A (INPUT - the x-coordinate)

and

SET B the y-coordinate (Output) .

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Functions & MappingsFunctions & Mappings

A function can be though of as a black box

x - Coordinate

Input

Domain

Members (x - axis)Co-Domain

Members (y - axis)

Image

Range

Function

Output

y - Coordinatef(x) = x2+ 3x - 1

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Nat 5

Finding the Function

Find the output or input values for the functions below :

6

7

8

36

49

64

f(x) = x2

f: 0

f: 1

f:2

-1

3

7f(x) = 4x - 1

4 12

f(x) = 3x

5 15

6 18

Examples

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Functions & MappingFunctions & Mapping

Functions can be illustrated in three ways:

1) by a formula.

2) by arrow diagram.

3) by a graph (ie co-ordinate diagram).

ExampleSuppose that f: A B is defined by

f(x) = x2 + 3x where A = { -3, -2, -1, 0, 1}.FORMULA

then f(-3) = 0 , f(-2) = -2 , f(-1) = -2 , f(0) = 0 ,f(1) = 4

NB: B = {-2, 0, 4} = the range!

Outcome 2

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0

-3

-2

-1

0

1

0

-2

-2

0

4

A B

ARROW DIAGRAM

Functions & MappingFunctions & Mapping

f(-3) = 0

f(-2) = -2

f(-1) = -2

f(0) = 0

f(1) = 4

f(x)

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Functions & GraphsFunctions & Graphs

In a GRAPH we get :

NB: This graph consists of 5 separate points. It is not a solid curve.

Outcome 2

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Recognising FunctionsA B

a b c d

e

f

g

Not a function

two arrows leaving b!

A B

a bc d

e

f

g

YES


Outcome 2

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A B

a b c d

e f g

Not a function - d unused!

A B

a bc d

e f g h

YES

Outcome 2

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Recognising Functions from Graphs

If we have a function f: R R (R - real nos.) then every vertical line we could draw would cut

the graph exactly once!

This basically means that every x-value has one, and only one, corresponding y-value!

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Function & Graphs Function & Graphs

x

YFunction !

!

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x

YNot a

function !!

Cuts graph

more than once !

Function & GraphsFunction & Graphs

x must map to

one value of y

Outcome 2

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Higher


X

Y Not a function !!

Cuts graph

more than once!

Outcome 2

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Higher

X

YFunction !

!


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Nat 5

The standard way to represent a function

is by a formula.

Function Notation

Examplef(x) = x + 4

We read this as “f of x equals x + 4”

or

“the function of x is x + 4

f(1) = 5 is the value of f at 1

f(a) = a + 4 is the value of f at a

1 + 4 = 5

a + 4

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Nat 5

For the function h(x) = 10 – x2.

Calculate h(1) , h(-3) and h(5)

h(1) =

Examples

h(-3) = h(5) =

h(x) = 10 – x2

Function Notation

10 – 12 = 9

10 – (-3)2 =

10 – 9 = 1

10 – 52 =

10 – 25 = -15

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Nat 5

For the function g(x) = x2 + x

Calculate g(0) , g(3) and g(2a)

g(0) =

Examples

g(3) = g(2a) =

g(x) = x2 + x

Function Notation

02 + 0 =

0

32 + 3 =

12

(2a)2 +2a =

4a2 + 2a

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Nat 5 Outcome 1

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Higher

Inverse FunctionsInverse Functions

A Inverse function is simply a function in reverse

Input

Function

Outputf(x) = x2+ 3x - 1

InputOutputf-1(x) = ?

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Nat 5

Inverse Function

Find the inverse function given

f(x) = 3x

Example

Remember

f(x) is simply the

y-coordinate

y = 3x

Using Changing the subject

rearrange into

x =

x =y

3

Rewrite replacing y with

x.

This is the inverse function

f-1(x) =x

3

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Nat 5

Inverse Function


f(x) = x2

Example

Remember

f(x) is simply the

y-coordinate

y = x2


rearrange into

x =

x = √y


x.


f-1(x) = √x

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Nat 5

Inverse Function


f(x) = 4x - 1

Example

Remember

f(x) is simply the

y-coordinate

y = 4x - 1


rearrange into

x =

x =


x.


f-1(x) =

y + 1

4

x + 1

4

Outcome 2

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COMPOSITION OF FUNCTIONS

( or functions of functions )

Suppose that f and g are functions where

f:A B and g:B C

with f(x) = y and g(y) = z

where x A, y B and z C.

Suppose that h is a third function where

h:A C with h(x) = z .

Composite FunctionsComposite Functions

Outcome 2

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A B C

x y zf g

h

We can say that h(x) = g(f(x))

“function of a function”

DEMO

Outcome 2

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Composite FunctionsComposite Functionsf(2)=3x2 – 2 =4

g(4)=42 + 1 =17

f(5)=5x3-2 =13Example 1

Suppose that f(x) = 3x - 2 and g(x) = x2 +1

(a) g( f(2) ) = g(4) = 17

(b) f( g (2) ) = f(5) = 13

(c) f( f(1) ) = f(1) = 1

(d) g( g(5) ) = g(26)= 677

f(1)=3x1 - 2 =1

g(26)=262

+ 1 =677

g(2)=22 + 1 =5

f(1)=3x1 - 2 =1

g(5)=52 + 1 =26

Outcome 2

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Higher

Suppose that f(x) = 3x - 2 and g(x) = x2 +1

Find formulae for (a) g(f(x)) (b) f(g(x)).

(a) g(f(x)) = ( )2 + 1

= 9x2 - 12x + 5

(b) f(g(x)) = 3( ) - 2= 3x2 + 1

CHECK

g(f(2)) = 9 x 22 - 12 x 2 + 5

= 36 - 24 + 5= 17

f(g(2)) = 3 x 22 + 1= 13

NB: g(f(x)) f(g(x)) in general.


3x - 2 x2 +1

Outcome 2

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Higher

Let h(x) = x - 3 , g(x) = x2 + 4 and k(x) = g(h(x)). If k(x) = 8 then find the value(s) of x.

k(x) = g(h(x))

= ( )2 + 4

= x2 - 6x + 13

Put x2 - 6x + 13 = 8

then x2 - 6x + 5 = 0

or (x - 5)(x - 1) = 0

So x = 1 or x = 5


x - 3

Outcome 2

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Higher

Choosing a Suitable Domain

(i) Suppose f(x) = 1 . x2 - 4

Clearly x2 - 4 0

So x2 4

So x -2 or 2

Hence domain = {xR: x -2 or 2 }


Outcome 2

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(ii) Suppose that g(x) = (x2 + 2x - 8)

We need (x2 + 2x - 8) 0

Suppose (x2 + 2x - 8) = 0

Then (x + 4)(x - 2) = 0

So x = -4 or x = 2

So domain = { xR: x -4 or x 2 }

Composite FunctionsComposite FunctionsSketch graph

-4 2

Graphs & Functions Higher

The functions f and g are defined on a suitable domain by

a) Find an expression for b) Factorise

2 2( ) 1 and ( ) 2f x x g x x

( ( ))f g x ( ( ))f g x

a) 22 2( ( )) ( 2) 2 1f g x f x x

2 22 1 2 1x x Difference of 2 squares

Simplify 2 23 1x x

b)


Functions and are defined on suitable domains.

a) Find an expression for h(x) where h(x) = f(g(x)).

b) Write down any restrictions on the domain of h.

1( )

4f x

x

( ) 2 3g x x

( ( )) (2 3)f g x f x a)1

2 3 4x

1

( )2 1

h xx

b) 2 1 0x 1

2x


3( ) 3 ( ) , 0x

f x x and g x x

a) Find

b) If find in its simplest form.

( ) where ( ) ( ( ))p x p x f g x

3

3( ) , 3

xq x x

( ( ))p q x

3( ) ( ( ))x

p x f g x f

a) 33

x 3 3x

x

3( 1)x

x

b)

33 1

3333

3

( ( )) x

xx

p q x p

9 3

33 3x x

9 3(3 ) 3

3 3

x x

x

3 3

3 3

x x

x

x

Graphs & Functions HigherFunctions f and g are defined on the set of real numbers by

a) Find formulae for

i) ii)

b) The function h is defined by

Show that and sketch the graph of h.

2( ) 1 and ( )f x x g x x

( ( ))f g x ( ( ))g f x

( ) ( ( )) ( ( ))h x f g x g f x

2( ) 2 2h x x x

a)

b)

2 2( ( )) ( ) 1f g x f x x 2( ( )) ( 1) 1g f x g x x

22( ) 1 1h x x x 2 2( ) 1 2 1h x x x x 22 2x x

Outcome 2

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Higher

A function in the form f(x) = ax where a > 0, a ≠ 1

is called an exponential function to base a .

Exponential (to the power of) Graphs

Exponential Functions

Consider f(x) = 2x

x -3 -2 -1 0 1 2 3

f(x) 1 1/8 ¼ ½ 1 2 4 8

Outcome 2

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The graph of

y = 2x

(0,1)(1,2)

Major Points

(i) y = 2x passes through the points (0,1) & (1,2)

(ii) As x ∞ y ∞ however as x -∞ y 0 .(iii) The graph shows a GROWTH function.

Graph

Outcome 2

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ie

y -3 -2 -1 0 1 2 3

x 1/8 ¼ ½ 1 2 4 8

To obtain y from x we must ask the question

“What power of 2 gives us…?”

This is not practical to write in a formula so we say

y = log2x“the logarithm to base 2 of x”

or “log base 2 of x”

Log Graphs

Outcome 2

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The graph of

y = log2x (1,0)

(2,1)

Major Points

(i) y = log2x passes through the points (1,0) & (2,1) .(ii) As x ∞ y ∞ but at a very slow rate

and as x 0 y -∞ .

NB: x > 0

Graph

Outcome 2

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Higher

The graph of y = ax always passes through (0,1) & (1,a)

It looks like ..

x

Y

y = ax

(0,1)

(1,a)

Exponential (to the power of) Graphs

Outcome 2

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Higher

The graph of y = logax always passes through (1,0) & (a,1)

It looks like ..

x

Y

y = logax

(1,0)

(a,1)

Log Graphs

Outcome 2

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Graph Transformations

We will investigate f(x) graphs of the form

1. f(x) ± k

2. f(x ± k)

3. -f(x)

4. f(-x)

5. kf(x)

6. f(kx)

Each moves the

Graph of f(x) in a certain

way !

f(x)

0 2 4 6 8x

-2-4-6

2

4

6

-2

-4

-6

Transformation f(x) ± k

(x , y) (x , y ± k)

Mapping

f(x) + 5 f(x) - 3

f(x)

Transformation f(x) ± k

Keypoints

y = f(x) ± k

moves original f(x) graph vertically up or down

+ k move up

- k move down

Only y-coordinate changes

NOTE: Always state any coordinates given on f(x)

on f(x) ± k graph

Demo

f(x) - 2

A(-1,-2) B(1,-2)C(0,-3)

f(x) + 1

B(90o,0)

A(45o,0.5)

C(135o,-0.5)

B(90o,1)

A(45o,1.5)

C(135o,0.5)

f(x)

0 2 4 6 8x

-2-4-6

2

4

6

-2

-4

-6

Transformation f(x ± k)

(x, y) (x ± k , y)

Mapping

f(x - 2) f(x + 4)

f(x)

Transformation f(x ± k)

Keypoints

y = f(x ± k)

moves original f(x) graph horizontally left or right

+ k move left

- k move right

Only x-coordinate changes


on f(x ± k) graph

Demo

f(x)

0 2 4 6 8x

-2-4-6

2

4

6

-2

-4

-6

Transformation -f(x)

(x, y) (x , -y)

Mapping

f(x)

Flip inx-axis

Flip inx-axis

Transformation -f(x)

Keypoints

y = -f(x)

Flips original f(x) graph in the x-axis

y-coordinate changes sign


on -f(x) graph

Demo

- f(x)

A(-1,0) B(1,0)

C(0,1)

- f(x)

B(90o,0)

A(45o,0.5)

C(135o,-0.5)A(45o,-0.5)

C(135o,0.5)

f(x)

0 2 4 6 8x

-2-4-6

2

4

6

-2

-4

-6

Transformation f(-x)

(x, y) (-x , y)

Mapping

f(x)

Flip iny-axis

Flip iny-axis

Transformation f(-x)

Keypoints

y = f(-x)

Flips original f(x) graph in the y-axis

x-coordinate changes sign


on f(-x) graph

Demo

f(-x)

B(0,0)

C’(-1,1)

A’(1,-1)A(-1,-1)

C (1,1)

f(x)

0 2 4 6 8x

-2-4-6

2

4

6

-2

-4

-6

Transformation kf(x)

(x, y) (x , ky)

Mapping

f(x)

Stretch iny-axis

2f(x) 0.5f(x)

Compress iny-axis

Transformation kf(x)

Keypoints

y = kf(x)

Stretch / Compress original f(x) graph in the

y-axis direction

y-coordinate changes by a factor of k


on kf(x) graph

Demo

f(x)

0 2 4 6 8x

-2-4-6

2

4

6

-2

-4

-6

Transformation f(kx)

(x, y) (1/kx , y)

Mapping

f(x)

Compress inx-axis

f(2x) f(0.5x)

Stretch inx-axis

Transformation f(kx)

Keypoints

y = f(kx)

Stretch / Compress original f(x) graph in the

x-axis direction

x-coordinate changes by a factor of 1/k


on f(kx) graph

Demo

Outcome 2

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Higher

You need to be able to work with combinations

Combining Transformations

Demo

(1,3)

(-1,-3)

(1,3)

(-1,-3)

2f(x) + 1

f(0.5x) - 1

f(-x) + 1

-f(x + 1) - 3

Explain the effect the following have

(a)-f(x)

(b)f(-x)

(c)f(x) ± k


(d)f(x ± k)

(e)kf(x)

(f)f(kx)

Name :

(-1,-3)

(1,3)

(-1,-3)

(1,3)

f(x + 1) + 2

-f(x) - 2

(1,3)

(-1,-3)

(-1,-3)

(1,3)

(1,3)

(-1,-3)

(1,-2)

(1,3)(-1,4)

(-1,-3)(1,-5)

(1,3)

(-1,1)

(-1,-3)

(0,5)

2f(x) + 1

f(0.5x) - 1

f(-x) + 1

-f(x + 1) - 3


(a)-f(x) flip in x-axis

(b)f(-x) flip in y-axis

(c)f(x) ± k move up or down


(d)f(x ± k) move left or right

(e)kf(x) stretch / compressin y

direction

(e)f(kx) stretch / compress

in x direction

Name :

(-1,-3)

(1,3)

(-1,-3)

(1,3)

(-2,-1)

f(x + 1) + 2

-f(x) - 2

(-2,0)

(1,3)

(0,-6)

(-1,-3)

(-1,-3)

(1,3)

(1,7)

(-1,-5)

(2,2)

(1,3)

(-2,-4)(-1,-3)

The diagram shows the graph of a function f.

f has a minimum turning point at (0, -3) and a

point of inflexion at (-4, 2).

a) sketch the graph of y = f(-x).

b) On the same diagram, sketch the graph of y = 2f(-x)


a) Reflect across the y axis

b) Now scale by 2 in the y direction-1 3 4

2

y = f(-x)

-3

y

x

4

y = 2f(-x)

-6


Part of the graph of is shown in the diagram.

On separate diagrams sketch the graph of

a) b)

Indicate on each graph the images of O, A, B, C, and D.

( )y f x

( 1)y f x 2 ( )y f x

a)

b)

graph moves to the left 1 unit

graph is reflected in the x axis

graph is then scaled 2 units in the y direction

(2, 1)

(2, -1)

(2, 1)

5

y=f(x)

y= -f(x)

y= 10 - f(x)

Graphs & Functions Higher =

a) On the same diagram sketch

i) the graph of

ii) the graph of

b) Find the range of values of x for

which is positive

2( ) 4 5f x x x

( )y f x

10 ( )y f x

10 ( )f x

2( 2) 1x

a)

b) Solve:210 ( 2) 1 0x

2( 2) 9x ( 2) 3x 1 or 5x

10 - f(x) is positive for -1 < x < 5


A sketch of the graph of y = f(x) where is shown.

The graph has a maximum at A (1,4) and a minimum at B(3, 0)

.

Sketch the graph of

Indicate the co-ordinates of the turning points. There is no need to

calculate the co-ordinates of the points of intersection with the axes.

3 2( ) 6 9f x x x x

( ) ( 2) 4g x f x

Graph is moved 2 units to the left, and 4 units up(3, 0) (1, 4)

(1, 4) ( 1, 8)

t.p.’s are:

(1,4)

(-1,8)

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Outcome 3Higher

Trig Graphs

The same transformation rules

apply to the basic trig graphs.

NB: If f(x) =sinx then 3f(x) = 3sinx

and f(5x) = sin5x

Think about sin replacing f !

Also if g(x) = cosx then g(x) – 4 = cosx – 4

and g(x + 90) = cos(x + 90) Think about cos replacing g !

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Outcome 3Higher

Sketch the graph of y = sinx - 2

If sinx = f(x) then sinx - 2 = f(x) - 2

So move the sinx graph 2 units down.

y = sinx - 2

Trig Graphs

1

-1

-2

-3

090o 180o 270o 360o

DEMO

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Outcome 3Higher

Sketch the graph of y = cos(x - 50)

If cosx = f(x) then cos(x - 50) = f(x - 50)So move the cosx graph 50 units right.

Trig Graphs

y = cos(x - 50)o

1

-1

-2

-3

0

50o

90o 180o 270o 360o

DEMO

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Outcome 3Higher

Trig Graphs

Sketch the graph of y = 3sinx

If sinx = f(x) then 3sinx = 3f(x)

So stretch the sinx graph 3 times vertically.

y = 3sinx

1

-1

-2

-3

0

2

3

90o 180o 270o 360o

DEMO

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Trig Graphs

Sketch the graph of y = cos4x

If cosx = f(x) then cos4x = f(4x)

So squash the cosx graph to 1/4 size horizontally

y = cos4x

1

-1

090o 180o 270o 360o

DEMO

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Trig Graphs

Sketch the graph of y = 2sin3xIf sinx = f(x) then 2sin3x = 2f(3x)So squash the sinx graph to 1/3 size horizontally and also double its height.

y = 2sin3x

90o

1

-1

-2

-3

0

2

3

360o180o 270o

DEMO

DEMO

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Trig Graph Trig Graph

1

2

3

-3

-2

-1

090o 180o 270o 360o

Write down equations for

graphs shown ?

CombinationsHigher

y = 0.5sin2xo + 0.5

y = 2sin4xo- 1

Write down the equations in the form f(x) for the graphs shown?

y = 0.5f(2x) + 0.5

y = 2f(4x) - 1

DEMO

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Trig GraphsTrig Graphs

1

2

3

-3

-2

-1

090o 180o 270o 360o

Combinationsy = cos2xo + 1

y = -2cos2xo - 1Higher

Write down the equations for the graphs shown?

Write down the equations in the form f(x) for the graphs shown?

y = f(2x) + 1y = -2f(2x) - 1

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Outcome 3Higher

Radians

Radian measure is an alternative to degrees and is based upon the ratio of

arc Length radius

rθ

L

θ- theta

(angle at the centre)

Circumf erence 2 r

2 Circumference

r

So, full circle 360o 2π radians

RadiansCopy Table360o 2π180o π

90o π2

45o π4

30o π6

60o π3

270o 3π2

135o 3π4

150o 5π6

120o 2π3

225o 5π4

210o 7π6

240o 4π3

315o 7π4

330o

300o 5π3

11π6

Demo

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Outcome 3Higher

Converting

degrees radians

÷180then X π

÷ π then x

180

For any values

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Outcome 3Higher

Ex1 72o =72/180 X π = 2π /5

Ex2 330o =330/180 X π =11 π /6

Ex3 2π /9 =2π /9 ÷ π x 180o = 2/9 X 180o = 40o

Ex4 23π/18 = 23π /18 ÷ π x 180o

= 23/18 X 180o = 230o

Converting

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22

2

60º

60º60º160º

230º3

This triangle will provide exact values for

sin, cos and tan 30º and 60º

Exact Values

Some special values of Sin, Cos and Tan are useful left as fractions, We call these exact values

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Outcome 3Higher

x 0º 30º 45º 60º 90º

Sin xº

Cos xº

Tan xº

½

½

3

3

2

3

20

1

0

1

0

Exact Values

1

3

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Outcome 3Higher

Exact Values

1 1 45º

45º

2

Consider the square with sides 1 unit

11

We are now in a position to calculate exact values for sin, cos and tan of 45o

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Outcome 3Higher

x 0º 30º 45º 60º 90º

Sin xº

Cos xº

Tan xº Undefined

½

½

3

3

2

3

20

1

0

1

0

Exact Values

1

3

1 2

1 2

1

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Exact value table

and quadrant rules.

tan150o

(Q2 so neg)

= - tan(180 - 150) o

= - tan30o= -1/√3

cos300o

(Q4 so pos)

= cos(360 - 300) o

= cos60o

= 1/2

sin120o

(Q2 so pos)

= sin(180 - 120)

o

= sin60o= √ 3/2

tan300o

(Q4 so neg)

= - tan(360-300)o

= - tan60o= - √ 3

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Outcome 3Higher

Find the exact value of cos2(5π/6) – sin2(π/6)

cos(5π/6) =

cos150o

(Q2 so neg)

= cos(180 - 150)o

= - cos30o= - √3 /2

sin(π/6) = sin30o = 1/2

cos2(5π/6) – sin2(π/6) = (- √3 /2)2 – (1/2)2= ¾ - 1/4 = 1/2

Exact value table

and quadrant rules.

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Outcome 3Higher

Exact value table

and quadrant rules.

Prove that sin(2 π /3) = tan (2 π /3)

cos (2 π /3) sin(2π/3) = sin120o = sin(180 – 120)o=

sin60o= √3/2

cos(2 π /3) = cos120o

tan(2 π /3) = tan120o

= cos(180 – 120)o

= tan(180 – 120)o

= - cos60o

= -tan60o

= -1/2

= - √3

LHS =sin(2 π /3) cos (2 π /3)

= √3/2 ÷ -1/2 = √3/2 X -2

= - √3 = tan(2π/3) = RHS

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created by Mr. Laffertycreated by Mr. Lafferty

Solving Trig Equations Solving Trig Equations

All +veSin +ve

Tan +ve Cos +ve

180o - xo

180o + xo 360o - xo

1 2 3 4

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Solving Trig Equations

Example 1 Type 1:

Solving the equation sin xo = 0.5 in the range 0o to 360o

Graphically what are we

trying to solve

xo = sin-1(0.5)

xo = 30o

There is another solution

xo = 150o

(180o – 30o = 150o)

sin xo = (0.5)

1 2 3 4

C

AS

T0o180

o

270o

90o

3

2

2

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Outcome 3Higher



Example 2 :

Solving the equation cos xo - 0.625 = 0 in the range 0o to 360o


trying to solve

cos xo = 0.625

xo = 51.3o

(360o - 53.1o = 308.7o)

xo = cos -1 (0.625)


1 2 3 4xo = 308.7o

C

AS

T0o180

o

270o

90o

3

2

2

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Outcome 3Higher



Example 3 :

Solving the equation tan xo – 2 = 0 in the range 0o to 360o


trying to solve

tan xo = 2

xo = 63.4o

x = 180o + 63.4o = 243.4o

xo = tan -1(2)


1 2 3 4

C

AS

T0o180

o

270o

90o

3

2

2

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Example 4 Type 2 :

Solving the equation sin 2xo + 0.6 = 0 in the range 0o to 360o


trying to solve

2xo = sin-1(0.6)

2xo = 217o , 323o

577o , 683o ......

sin 2xo = (-0.6)

xo = 108.5o , 161.5o

288.5o , 341.5o

C

AS

T0o180

o

270o

90o

3

2

2

2xo = 37o ( always 1st Q First)

÷2

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Outcome 3Higher


Solving Trig EquationsGraphically what are we

trying to solve

sin (2x - 30o) = √3 ÷ 2

2xo - 30o = 60o , 120o ,420o , 480o .........

2sin (2x - 30o) = √3

xo = 45o , 75o

225o , 255o

2x - 30o = sin-1(√3 ÷ 2)

Example 5 Type 3 :

Solving the equation 2sin (2xo - 30o) - √3 = 0 in the range 0o to 360o

C

AS

T0o180

o

270o

90o

3

2

2

2xo = 90o , 150o ,450o , 510o .........

÷2

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Outcome 3Higher



Example 6 Type 4 :

Solving the equation cos2x = 1 in the range 0o to 360o


trying to solve

cos xo = ± 1

cos xo = 1

cos2 xo = 1

xo = 0o and 360o

C

AS

T0o180

o

270o

90o

3

2

2

cos xo = -1xo = 180o

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Outcome 3Higher

Example 7 Type 5 : Solving the equation 3sin2x + 2sin x - 1 = 0 in the range 0o to 360o


3p – 1 = 0

xo = 19.5o and 160.5o

xo = 270o

Let p = sin x

We have 3p2 + 2p - 1 = 0

(3p – 1)(p + 1) = 0Factorise

p = 1/3

p + 1 = 0p = - 1sin x =

1/3sin x = -1

C

AS

T0o180

o

270o

90o

3

2

2

C

AS

T0o180

o

270o

90o

3

2

2


Functions f and g are defined on suitable domains by and

a) Find expressions for:

i)

ii)

b) Solve

( ) sin( )f x x ( ) 2g x x

( ( ))f g x

( ( ))g f x

2 ( ( )) ( ( )) 0 360f g x g f x for x

( ( )) (2 )f g x f xa) sin 2x ( ( )) (sin )g f x g x 2sin x

b) 2sin 2 2sinx x sin 2 sin 0x x

2sin cos sin 0x x x sin (2cos 1) 0x x 1

or2

sin 0 cosx x 0 , 180 , 360x 60 , 300x


Functions

are defined on a suitable set of real numbers.

a) Find expressions for

b) i) Show that

ii) Find a similar expression for

and hence solve the equation

4and( ) sin , ( ) cos ( )f x x g x x h x x

( ( ))f h x ( ( ))g h x

1 1

2 2( ( )) sin cos f h x x x

( ( ))g h x

for( ( )) ( ( )) 1 0 2f h x g h x x

4( ( )) ( )f h x f x a) 4

sin( )x 4

( ( )) cos( )g h x x

sin cos4 4 4

sin( ) sin cos xx x b) Now use exact values

Repeat for ii)

equation reduces to2

sin 12

x 2 1sin

2 2x

3,

4 4x


The diagram shows a sketch of part of

the graph of a trigonometric function

whose equation is of the form

Determine the values of a, b and c

sin( )y a bx c

a is the amplitude: a = 4

b is the number of waves in 2 b = 2

c is where the wave is centred vertically c = 1

2a

1 in

2 in 2

1

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Outcome 3Higher

An identity is a statement which is true for all values.

eg 3x(x + 4) = 3x2 + 12x

eg (a + b)(a – b) = a2 – b2

Trig Identities

(1) sin2θ + cos2 θ = 1

(2) sin θ = tan θ cos θ θ ≠ an odd multiple of π/2 or 90°.

Trig Identities

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Reason

θo

c

b

a

a2 +b2 = c2

sinθo = a/c

cosθo = b/c

(1) sin2θo + cos2 θo =

Trig Identities

2 2

2 2

a bc c

2 2

2

a bc

2

2 1cc

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Outcome 3Higher

sin2θ + cos2 θ = 1

sin2 θ = 1 - cos2

θ cos2 θ = 1 - sin2

θ

Simply rearranging we get two other forms

Trig Identities

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Outcome 3Higher

Example1 sin θ = 5/13 where 0 < θ < π/2

Find the exact values of cos θ and tan θ .

cos2 θ = 1 - sin2 θ

= 1 – (5/13)2

= 1 – 25/169= 144/169

cos θ = √(144/169)

= 12/13 or -12/13

Since θ is between 0 < θ < π/2

then cos θ > 0

So cos θ = 12/13

tan θ = sinθ cos θ

= 5/13 ÷ 12/13

= 5/13 X 13/12

tan θ = 5/12

Trig Identities

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Outcome 3Higher

Given that cos θ = -2/ √ 5 where π< θ < 3 π /2Find sin θ and tan θ.

sin2 θ = 1 - cos2 θ

= 1 – (-2/ √

5 )2

= 1 – 4/5

= 1/5

sin θ = √(1/5)

= 1/ √ 5 or - 1/ √ 5

Since θ is between π< θ < 3 π /2

sinθ < 0

Hence sinθ = - 1/√5

tan θ = sinθ cos θ

= - 1/ √ 5 ÷ -2/ √

5 = - 1/ √ 5 X - √5 /2

Hence tan θ = 1/2

Trig Identities

Outcome 2

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Higher

Are you on Target !

• Update you log book

• Make sure you complete and correct MOST of the Composite Functionquestions in the past paper booklet.

• Make sure you complete and correct SOME of the Trigonometryquestions in the past paper booklet.

f(x)

f(x)

Graphs & Functions

y = -f(x)

y = f(-x)

y = f(x) ± k

y = f(kx)

Move verticallyup or downs

depending on k

flip iny-axis

flip inx-axis

+

- Stretch or compressvertically

depending on k

y = kf(x)

Stretch or compress

horizontally depending on k

f(x)

f(x)f(x)

f(x)y = f(x ± k)

Move horizontallyleft or right

depending on k

+-

Remember we can combine

these together !!

0 < k < 1 stretch

k > 1 compress

0 < k < 1 compress

k > 1 stretch

Composite Functions

A complex function made up of 2 or

more simpler functions

= +

f(x) = x2 - 4 g(x) = 1x

x

Domainx-axis valuesInput

Rangey-axis valuesOutput

x2 - 41

x2 - 4

Restriction x2 - 4 ≠ 0

(x – 2)(x + 2) ≠ 0

x ≠ 2 x ≠ -2

g(f(x)) g(f(x)) =

f(x) = x2 - 4g(x) = 1x

x

Domainx-axis valuesInput

Rangey-axis valuesOutput

f(g(x))

Restriction x2 ≠ 0

1

x

2- 4 =

Similar to composite

Area

Write down g(x) with brackets for x

g(x) =1

( )

inside bracket put f(x)

g(f(x)) =1

x2 - 4

1x

- 41x2

f(g(x)) =Write down f(x) with brackets for x

f(x) = ( )2 - 4

inside bracket put g(x)

f(g(x)) =1

x2- 4

Functions & Graphs

TYPE questions(Sometimes Quadratics)

SketchingGraphs

CompositeFunctions

Steps :

1.Outside function staysthe same EXCEPT replacex terms with a ( )

2. Put inner function in bracket

You need to learn basic movements

Exam questionsnormally involve two movements

Remember orderBODMAS

Restrictions :

1.Denominator NOT ALLOWEDto be zero

2.CANNOT take the square rootof a negative number

Documents

Higher Unit 1