Higher Outcome 2 Higher Unit 1 What is a set Recognising a Function in various formats Composite...

Higher Outcome 2

Higher Unit 1Higher Unit 1

What is a setRecognising a Function in various formatsComposite FunctionsExponential and Log Graphs

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

Exam Type Questions

Graph TransformationsTrig Graphs

Higher Outcome 2

Sets & 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.

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eg {colours in traffic lights}

eg {square nos. less than 30}

DESCRIPTION LIST

NB: Each of the above sets is finite because we can list every member

= {red, amber, green}

= { 0, 1, 4, 9, 16, 25}

Sets & FunctionsSets & Functions

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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:

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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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If the first set is A and the second B then we often write

f: A B

Functions & Mappings

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.

The members of set A are usually referred to as the domain of the function (basically the starting values or even x-values) while the corresponding values or images come from set B and are called the range of the function (these are like y-values).

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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!

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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.

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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


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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

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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

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X

Y Not a function !!

Cuts graph

more than once!

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X

YFunction !

!


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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

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ieA B C

x y zf g

h

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

“function of a function”

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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

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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)) = g(3x-2)

= (3x-2)2 + 1

= 9x2 - 12x + 5

(b) f(g(x)) = f(x2 + 1) = 3(x2 + 1) - 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 As in Ex1

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


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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))

= g(x - 3)

= (x - 3)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

CHECK g(h(5)) = g(2)= 22 + 4= 8


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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 }


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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 }


Check values below -4 , between -4 and 2, then above 2

x = -5

(-5 + 4)(-5 - 2) = positive

x = 0

(0 + 4)(0 - 2) =

negative

x = 3

(3 + 4)(3 - 2) =

positive

-4 2

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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

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The graph is like

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

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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

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The graph is likey =

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

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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

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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

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Graph Transformations

We will use the TI – 83 to 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 !

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y = f(x) y = x2

y = x2-3

Mathematically

y = f(x) ± k

moves f(x) up or down

Depending on the value of k

+ k move up

- k move down

y = x2+ 1

y = f(x) ± k

Graph of f(x) ± k Transformations

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Graph of -f(x ± k) Transformations

y = f(x) y = x2 y = (x-1)2

y = (x+2)2

y = f(x ± k)

Mathematically

y = f(x ± k)

moves f(x) to the left or right

depending on the value of k

-k move right

+ k move left

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y = f(x) y = x2

y = -x2

Mathematically

y = –f(x)

reflected f(x) in the x - axis

y = -f(x)

Graph of -f(x) Transformations

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y = f(x) y = 2x + 3y = -(2x + 3)

Mathematically

y = –f(x)


y = -f(x)


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y = f(x) y = x3

y = -x3

Mathematically

y = –f(x)


y = -f(x)


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Graph of f(-x) Transformations

y = f(x) y = x + 2

y = -x + 2

Mathematically

y = f(-x)

reflected f(x) in the y - axis

y = f(-x)

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

Mathematically

y = f(-x)

reflected f(x) in the y - axis

y = f(-x)

Graph of f(-x) Transformations

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Graph of k f(x) Transformations

y = f(x) y = x2-1

y = 4(x2-1)

Mathematically

y = k f(x)

Multiply y coordinate by a factor of k

k > 1 (stretch in y-axis direction)

0 < k < 1 (squash in y-axis direction)

y = 0.25(x2-1)

y = k f(x)

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y = f(x) y = x2-1

y = -4(x2-1)

Mathematically

y = -k f(x)

k = -1 reflect graph in x-axis

k < -1 reflect f(x) in x-axis & multiply by a factor k (stretch in y-axis direction)

0 < k < -1 reflect f(x) in x-axis multiply by a factor k (squash in y-axis direction)

y = -0.25(x2-1)

y = k f(x)

Graph of -k f(x) Transformations

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

y = (2x-2)2

Mathematically

y = f(kx)

Multiply x – coordinates by 1/k

k > 1 squashes by a factor of 1/k in the x-axis direction

k < 1 stretches by a factor of 1/k in the x-axis direction

y = (0.5x-2)2

y = f(kx)

Graph of f(kx) Transformations

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

y = (-2x-2)2

Mathematically

y = f(-kx)

k = -1 reflect in y-axis

k < -1 reflect & squashes by factor of 1/k in x direction

-1 < k > 0 reflect & stretches factor of 1/k in x direction

y = (-0.5x-2)2

y = f(-kx)

Graph of f(-kx) Transformations

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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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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

Higher Outcome 1

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.

y = cos(x - 50)

Trig Graphs

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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

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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

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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

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Radians

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

arc length radius

rθ

a

θ- theta

θ(radians) = ar

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If the arc length = the radiusθ

r

a = r

θ (radians) = r/r = 1

If we now take a semi-circle

θ (180o)

r

aHere a = ½ of circumference= ½ of πd

= πr

So θ (radians) = πr /r = π

Radians

Higher Outcome 1

Converting

degrees radians

÷180 then X π

÷ πthen x 180

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Since we have a semi-circle the angle must be 180o.

We now get a simple connection between degrees and radians. π (radians) =

180oThis now gives us 2π = 360o

π /2 = 90o

π /3 = 60o

3π /2 = 270o

2π /3 = 120o

π /4 = 45o 3π /4 = 135o

π /6 = 30o 5π /6 = 150o

NB: Radians are usually expressed as fractional multiples of π.

Radians

Higher Outcome 1

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

Higher Outcome 1

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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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

Higher Outcome 1

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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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

1 2

1 2

1

Higher Outcome 1

Exact value table

and quadrant rules.

tan150o

(Q2 so neg)

= tan(180-30)

o

= -tan30o = -1/√3

cos300o

(Q4 so pos)

= cos(360-60)

o

= cos60o

= 1/2

sin120o

(Q2 so pos)

= sin(180-60) o = sin60o = √ 3/2

tan300o

(Q4 so neg)

= tan(360-60)o

= -tan60o = - √ 3

Higher Outcome 1

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

cos(5π/6) =

cos150o

(Q2 so neg)

= cos(180-30)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.

Higher Outcome 1

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 – 60)o=

sin60o

= √3/2

cos(2 π /3) = cos120o

tan(2 π /3) = tan120o

= cos(180 – 60)o

= tan(180 – 60)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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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

Higher Outcome 2


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

Higher Outcome 2


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

Higher Outcome 2


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

Higher Outcome 2

created by Mr. Laffertycreated by Mr. Lafferty

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 , 265o

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 , 530o .........

÷2

Higher Outcome 2

created by Mr. Laffertycreated by Mr. Lafferty


Example 5 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

Higher Outcome 2

Example 6 Type 5 : Solving the equation 3sin2x + 2sin x - 1 = 0 in the range 0o to 360o


3x – 1 = 0

xo = 19.5o and 160.5o

xo = 90o

Let x = sin x

We have 3x2 + 2x - 1 = 0

(3x – 1)(x + 1) = 0Factorise

x = 1/3

x – 1 = 0x = 1sin x =

1/3sin x = 1

C

AS

T0o180

o

270o

90o

3

2

2

C

AS

T0o180

o

270o

90o

3

2

2

Higher Outcome 1

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

Higher Outcome 1

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

Higher Outcome 1

= tan

sin2θ + cos2 θ = 1

sin2 θ = 1 - cos2

θ cos2 θ = 1 - sin2

θ

Simply rearranging we get two other forms

Trig Identities

Higher Outcome 1

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

Higher Outcome 1

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

Graphs & Functions

Strategies

Higher Maths

Click to start

Graphs & Functions Higher

Graphs & Functons

The following questions are on

Non-calculator questions will be indicated

Click to continue

You will need a pencil, paper, ruler and rubber.

Hint

QuitQuitPrevious Next

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

Hint



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

Hint



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

(2, 1)

(2, -1)

(2, 1)

5

y=f(x)

y= -f(x)

y= 10 - f(x)

Hint



a) Express in the form

b) On the same diagram sketch

i) the graph of

ii) the graph of

c) Find the range of values of x for

which is positive

2( ) 4 5f x x x 2( )x a b

( )y f x

10 ( )y f x

10 ( )f x

a) 2( 2) 4 5x 2( 2) 4 5x 2( 2) 1x

b)

c) Solve:210 ( 2) 1 0x

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

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

Hint



The graph of a function f intersects the x-axis at (–a, 0)

and (e, 0) as shown.

There is a point of inflexion at (0, b) and a maximum turning

point at (c, d).

Sketch the graph of the derived function f

m is + m is + m is -

f(x)

Hint



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

Hint



The diagram shows the graphs of two quadratic

functions

Both graphs have a minimum turning point at (3, 2).

Sketch the graph of

and on the same diagram

sketch the graph of

and( ) ( )y f x y g x

( )y f x

( )y g x

y=g(x)

y=f(x)

Hint



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

Hint



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

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

a) Find the co-ordinates of the turning point at A.

b) Hence, 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.

c) Write down the range of values of k for which g(x) = k has 3 real roots.

3 2( ) 6 9f x x x x

( ) ( 2) 4g x f x

a) Differentiate 2( ) 3 12 9f x x x for SP, f(x) = 0 1 3x or x

when x = 1 4y t.p. at A is: (1, 4)

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

(1, 4) ( 1, 8)

t.p.’s are:

c) For 3 real roots, line y = k has to cut graph at 3 points

from the graph, k 4

Hint



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

Hint



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

Hint



Functions 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.

c) Find the area enclosed between this graph and the x-axis.

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

c) Graph cuts x axis at 0 and 1 Now evaluate1 2

02 2x x dx

2unit

1

3Area

Hint



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)

You have completed all 13 questions in this section

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30° 45° 60°

sin

cos

tan 1

6

4

3

1

2

1

23

2

3

2

1

21

21

3 3

Table of exact values

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