Outcome 2 Higher Higher Maths What is a set Function in various formats Composite Functions Exponential and Log Graphs Connection between Radians and degrees & Exact values Solving Trig Equations Basic Trig Identities Graph Transformations Trig Graphs Inverse function Mindmap Exam Question Type

Outcome 2 Higher 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.

Outcome 2 Higher Sets & Functions 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 / = 6 / 10 = 3 / 5 55% = 55 / 100 = 11 / 20 etc We can describe numbers by the following sets:

Outcome 2 Higher 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 Sets & Functions

Outcome 2 Higher Sets & Functions 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, 3  5 etc R

Outcome 2 Higher 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: x 2 = -4 }= { } or  This set has no elements and is called the empty set. Sets & Functions

Nat 5 What are Functions ? Functions describe how one quantity relates to another Car Parts 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.

Nat 5 What are Functions ? Functions describe how one quantity relates to another Dirty Washing Machine Clean OutputInput yx Function f(x) y = f(x)

Nat 5 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).

Outcome 2 Higher Functions & 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 - Coordinate f(x) = x 2 + 3x - 1

Nat 5 Finding the Function Find the output or input values for the functions below : f(x) = x 2 f: 0 f: 1 f: f(x) = 4x f(x) = 3x Examples

Outcome 2 Higher Functions & Mapping Functions can be illustrated in three ways: 1) by a formula. 2) by arrow diagram. 3) by a graph (ie co-ordinate diagram). Example Suppose that f: A  B is defined by f(x) = x 2 + 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 Higher A B ARROW DIAGRAM Functions & Mapping f(-3) = 0 f(-2) = -2 f(-1) = -2 f(0) = 0 f(1) = 4 f(x)

Outcome 2 Higher Functions & Graphs In a GRAPH we get : NB: This graph consists of 5 separate points. It is not a solid curve.

Outcome 2 Higher Recognising Functions ABAB abcdabcd efgefg Not a function two arrows leaving b! ABAB abcdabcd e f ge f g YES Functions & Graphs

Outcome 2 Higher Functions & Graphs A B abcdabcd efgefg Not a function - d unused! ABAB abcdabcd efghefgh YES

Outcome 2 Higher Functions & Graphs 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!

Outcome 2 Higher Function & Graphs x Y Function !!

Outcome 2 Higher x Y Not a function !! Cuts graph more than once ! Function & Graphs x must map to one value of y

Outcome 2 Higher Functions & Graphs X Y Not a function !! Cuts graph more than once!

Outcome 2 Higher X Y Function !! Functions & Graphs

Nat 5 The standard way to represent a function is by a formula. Function Notation Example f(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 =5 a + 4

Nat 5 For the function h(x) = 10 – x 2. Calculate h(1), h(-3) and h(5) h(1) = Examples h(-3) = h(5) = h(x) = 10 – x 2  Function Notation 10 – 1 2 = 9 10 – (-3) 2 = 10 – 9 = 1 10 – 5 2 =10 – 25 = -15

Nat 5 For the function g(x) = x 2 + x Calculate g(0), g(3) and g(2a) g(0) = Examples g(3) = g(2a) = g(x) = x 2 + x  Function Notation = = 12 (2a) 2 +2a =4a 2 + 2a

Nat 5 Outcome 1 Higher Inverse Functions A Inverse function is simply a function in reverse Input Function Output f(x) = x 2 + 3x - 1 InputOutput f -1 (x) = ?

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

Nat 5 Inverse Function Find the inverse function given f(x) = x 2 Example Remember f(x) is simply the y-coordinate y = x 2 Using Changing the subject rearrange into x = x = √y Rewrite replacing y with x. This is the inverse function f -1 (x) = √x

Nat 5 Inverse Function Find the inverse function given f(x) = 4x - 1 Example Remember f(x) is simply the y-coordinate y = 4x - 1 Using Changing the subject rearrange into x = x = Rewrite replacing y with x. This is the inverse function f -1 (x) = y x + 1 4

Outcome 2 Higher 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 Functions

Outcome 2 Higher Composite Functions ABCABC x y z f g h We can say that h(x) = g(f(x)) “function of a function” DEMO

Outcome 2 Higher Composite Functions f(2)=3 x 2 – 2 =4 g(4)= =17 f(5)=5x3-2 =13 Example 1 Suppose that f(x) = 3x - 2 and g(x) = x 2 +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)= =677 g(2)= =5 f(1)=3x1 - 2 =1 g(5)= =26

Outcome 2 Higher Suppose that f(x) = 3x - 2 and g(x) = x 2 +1 Find formulae for (a) g(f(x)) (b) f(g(x)). (a) g(f(x)) =( ) 2 + 1= 9x x + 5 (b) f(g(x)) =3( ) - 2= 3x CHECK g(f(2)) =9 x x 2 + 5= = 17 f(g(2)) =3 x = 13 NB: g(f(x))  f(g(x)) in general. Composite Functions 3x - 2x 2 +1

Outcome 2 Higher Let h(x) = x - 3, g(x) = x and k(x) = g(h(x)). If k(x) = 8 then find the value(s) of x. k(x) = g(h(x)) = ( ) = x 2 - 6x + 13 Put x 2 - 6x + 13 = 8 then x 2 - 6x + 5 = 0 or (x - 5)(x - 1) = 0 So x = 1 or x = 5 Composite Functions x - 3

Outcome 2 Higher Choosing a Suitable Domain (i) Suppose f(x) = 1. x Clearly x  0 So x 2  4 So x  -2 or 2 Hence domain = {x  R: x  -2 or 2 } Composite Functions

Outcome 2 Higher (ii) Suppose that g(x) =  (x 2 + 2x - 8) We need (x 2 + 2x - 8)  0 Suppose (x 2 + 2x - 8) = 0 Then (x + 4)(x - 2) = 0 So x = -4 or x = 2 So domain = { x  R: x  -4 or x  2 } Composite Functions Sketch graph -42

Graphs & Functions Higher The functions f and g are defined on a suitable domain by a) Find an expression for b) Factorise a) Difference of 2 squares Simplify b)

Graphs & Functions Higher 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. a) b)

Graphs & Functions Higher a) Find b) If find in its simplest form. a) b)

Graphs & Functions Higher 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. a) b)

Outcome 2 Higher A function in the form f(x) = a x where a > 0, a ≠ 1 is called an exponential function to base a. Exponential (to the power of) Graphs Exponential Functions Consider f(x) = 2 x x f(x) 11 / 8 ¼ ½

Outcome 2 Higher The graph of y = 2 x (0,1) (1,2) Major Points (i) y = 2 x 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 Higher ie y x 1 / 8 ¼ ½ 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 = log 2 x “the logarithm to base 2 of x” or “log base 2 of x” Log Graphs

Outcome 2 Higher The graph of y = log 2 x (1,0) (2,1) Major Points (i) y = log 2 x 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 Higher The graph of y = a x always passes through (0,1) & (1,a) It looks like.. x Y y = a x (0,1) (1,a) Exponential (to the power of) Graphs

Outcome 2 Higher The graph of y = log a x always passes through (1,0) & (a,1) It looks like.. x Y y = log a x (1,0) (a,1) Log Graphs

Outcome 2 Higher 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) x 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(90 o,0) A(45 o,0.5) C(135 o,-0.5) B(90 o,1) A(45 o,1.5) C(135 o,0.5)

f(x) x 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 NOTE: Always state any coordinates given on f(x) on f(x ± k) graph Demo

f(x) x Transformation -f(x) (x, y)  (x, -y) Mapping f(x) Flip in x-axis Flip in x-axis

Transformation -f(x) Keypoints y = -f(x) Flips original f(x) graph in the x-axis y-coordinate changes sign NOTE: Always state any coordinates given on f(x) on -f(x) graph Demo

- f(x) A(-1,0)B(1,0) C(0,1)

- f(x) B(90 o,0) A(45 o,0.5) C(135 o,-0.5) A(45 o,-0.5) C(135 o,0.5)

f(x) x Transformation f(-x) (x, y)  (-x, y) Mapping f(x) Flip in y-axis Flip in y-axis

Transformation f(-x) Keypoints y = f(-x) Flips original f(x) graph in the y-axis x-coordinate changes sign NOTE: Always state any coordinates given on f(x) on f(-x) graph Demo

f(-x) B(0,0) C’(-1,1) A’(1,-1) A(-1,-1) C (1,1)

f(x) x Transformation kf(x) (x, y)  (x, ky) Mapping f(x) Stretch in y-axis 2f(x) 0.5f(x) Compress in y-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 NOTE: Always state any coordinates given on f(x) on kf(x) graph Demo

f(x) x Transformation f(kx) (x, y)  (1/kx, y) Mapping f(x) Compress in x-axis f(2x) f(0.5x) Stretch in x-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 NOTE: Always state any coordinates given on f(x) on f(kx) graph Demo

Outcome 2 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 Explain the effect the following have (d)f(x ± k) (e)kf(x) (f)f(kx) Name : (-1,-3) (1,3) f(x + 1) + 2 -f(x) - 2 (1,3) (-1,-3) (1,3) (-1,-3)

(1,-2) (-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 Explain the effect the following have (a)-f(x)flip in x-axis (b)f(-x) flip in y-axis (c)f(x) ± k move up or down Explain the effect the following have (d)f(x ± k) move left or right (e)kf(x)stretch / compress in y direction (e)f(kx) stretch / compress in x direction Name : (-1,-3) (1,3) (-2,-1) f(x + 1) + 2 -f(x) - 2 (-2,0) (1,3) (0,-6) (-1,-3) (1,3) (1,7) (-1,-5) (2,2) (-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 = 2 f (- x ) Graphs & Functions Higher a) Reflect across the y axis b) Now scale by 2 in the y direction

Graphs & Functions Higher 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. 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

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 a) b) Solve: 10 - f(x) is positive for -1 < x < 5

Graphs & Functions Higher 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. Graph is moved 2 units to the left, and 4 units up t.p.’s are: (1,4) (-1,8)

Outcome 3 Higher Trig Graphs The same transformation rules apply to the basic trig graphs. NB: If f(x) =sinx  then3f(x) = 3sinx  andf(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 !

Outcome 3 Higher 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 o 180 o 270 o 360 o DEMO

Outcome 3 Higher 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 o 90 o 180 o 270 o 360 o DEMO

Outcome 3 Higher 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  o 180 o 270 o 360 o DEMO

Outcome 3 Higher 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  o 180 o 270 o 360 o DEMO

Outcome 3 Higher Trig Graphs Sketch the graph of y = 2sin3x  If sinx  = f(x) then 2sin3x  = 2f(3x) So squash the sinx  graph to 1 / 3 size horizontally and also double its height. y = 2sin3x  90 o o 180 o 270 o DEMO

created by Mr. Lafferty Trig Graph o 180 o 270 o 360 o Write down equations for graphs shown ? Combinations Higher y = 0.5sin2x o y = 2sin4x o - 1 Write down the equations in the form f(x) for the graphs shown? y = 0.5f(2x) y = 2f(4x) - 1

DEMO created by Mr. Lafferty Trig Graphs o 180 o 270 o 360 o Combinations y = cos2x o + 1 y = -2cos2x o - 1 Higher Write down the equations for the graphs shown? Write down the equations in the form f(x) for the graphs shown? y = f(2x) + 1 y = -2f(2x) - 1

Outcome 3 Higher Radians Radian measure is an alternative to degrees and is based upon the ratio of arc Length radius r θ L θ- theta (angle at the centre) So, full circle360 o  2π radians

Radians Copy Table 360 o  2π180 o  π 90 o  π 2 45 o  π 4 30 o  π 6 60 o  π o  3π3π o  3π3π o  5π5π o  2π2π o  5π5π o  7π7π o  4π4π o  7π7π o  300 o  5π5π 3 11π 6 Demo

Outcome 3 Higher Converting degrees radians ÷180 then X π ÷ π then x 180 For any values

Outcome 3 Higher Ex172 o = 72 / 180 X π =2π / 5 Ex2330 o = 330 / 180 X π = 11 π / 6 Ex32π / 9 =2π / 9 ÷ π x 180 o = 2 / 9 X 180 o =40 o Ex4 23π/ 18 =23π / 18 ÷ π x 180 o = 23 / 18 X 180 o =230 o Converting

Outcome 3 Higher º º 33 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

Outcome 3 Higher x0º30º45º60º90º Sin xº Cos xº Tan xº  ½ ½ 33  3 2  Exact Values

Outcome 3 Higher Exact Values 11 45º 2 2 Consider the square with sides 1 unit 1 1 We are now in a position to calculate exact values for sin, cos and tan of 45 o

Outcome 3 Higher x0º30º45º60º90º Sin xº Cos xº Tan xº Undefined ½ ½ 33  3 2  Exact Values 1  2 1

Outcome 3 Higher Exact value table and quadrant rules. tan150 o (Q2 so neg) = - tan( ) o = - tan30 o = -1 / √3 cos300 o (Q4 so pos) = cos( ) o = cos60 o = 1 / 2 sin120 o (Q2 so pos) = sin( ) o = sin60 o = √ 3/2 tan300 o (Q4 so neg) = - tan( ) o = - tan60 o = - √ 3

Outcome 3 Higher Find the exact value of cos 2 ( 5π / 6 ) – sin 2 ( π / 6 ) cos( 5π / 6 ) = cos150 o (Q2 so neg) = cos( ) o = - cos30 o = - √3 / 2 sin( π / 6 )= sin30 o = 1 / 2 cos 2 ( 5π / 6 ) – sin 2 ( π / 6 )= (- √3 / 2 ) 2 – ( 1 / 2 ) 2 = ¾ - 1 / 4 = 1 / 2 Exact value table and quadrant rules.

Outcome 3 Higher Exact value table and quadrant rules. Prove thatsin( 2 π / 3 ) = tan ( 2 π / 3 ) cos ( 2 π / 3 ) sin( 2π / 3 ) = sin120 o = sin(180 – 120) o = sin60 o = √3 / 2 cos( 2 π / 3 ) = cos120 o tan( 2 π / 3 ) = tan120 o = cos(180 – 120) o = tan(180 – 120) o = - cos60 o = -tan60 o = - 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

Outcome 3 Higher created by Mr. Lafferty Solving Trig Equations All +veSin +ve Tan +ve Cos +ve 180 o - x o 180 o + x o 360 o - x o 1234

Outcome 3 Higher created by Mr. Lafferty Solving Trig Equations Example 1 Type 1: Solving the equation sin x o = 0.5 in the range 0 o to 360 o Graphically what are we trying to solve x o = sin -1 (0.5) x o = 30 o There is another solution x o = 150 o (180 o – 30 o = 150 o ) sin x o = (0.5) 1234 C A S T 0o0o 180 o 270 o 90 o

Outcome 3 Higher created by Mr. Lafferty Solving Trig Equations Example 2 : Solving the equation cos x o = 0 in the range 0 o to 360 o Graphically what are we trying to solve cos x o = x o = 51.3 o (360 o o = o ) x o = cos -1 (0.625) There is another solution 1234 x o = o C A S T 0o0o 180 o 270 o 90 o

Outcome 3 Higher created by Mr. Lafferty Solving Trig Equations Example 3 : Solving the equation tan x o – 2 = 0 in the range 0 o to 360 o Graphically what are we trying to solve tan x o = 2 x o = 63.4 o x = 180 o o = o x o = tan -1 (2) There is another solution 1234 C A S T 0o0o 180 o 270 o 90 o

Outcome 3 Higher created by Mr. Lafferty Solving Trig Equations Example 4 Type 2 : Solving the equation sin 2x o = 0 in the range 0 o to 360 o Graphically what are we trying to solve 2x o = sin -1 (0.6) 2x o =217 o, 323 o 577 o, 683 o sin 2x o = (-0.6) x o =108.5 o, o o, o C A S T 0o0o 180 o 270 o 90 o 2x o = 37 o ( always 1 st Q First) ÷2

Outcome 3 Higher created by Mr. Lafferty Solving Trig Equations Graphically what are we trying to solve sin (2x - 30 o ) = √3 ÷ 2 2x o - 30 o = 60 o, 120 o,420 o, 480 o sin (2x - 30 o ) = √3 x o =45 o, 75 o 225 o, 255 o 2x - 30 o = sin -1 (√3 ÷ 2) Example 5 Type 3 : Solving the equation 2sin (2x o - 30 o ) - √3 = 0 in the range 0 o to 360 o C A S T 0o0o 180 o 270 o 90 o 2x o = 90 o, 150 o,450 o, 510 o ÷2

Outcome 3 Higher created by Mr. Lafferty Solving Trig Equations Example 6 Type 4 : Solving the equation cos 2 x = 1 in the range 0 o to 360 o Graphically what are we trying to solve cos x o = ± 1 cos x o = 1 cos 2 x o = 1 x o = 0 o and 360 o C A S T 0o0o 180 o 270 o 90 o cos x o = -1 x o = 180 o

Outcome 3 Higher Example 7 Type 5 : Solving the equation 3sin 2 x + 2sin x - 1 = 0 in the range 0 o to 360 o Solving Trig Equations 3p – 1 = 0 x o = 19.5 o and o x o = 270 o Let p = sin x We have 3p 2 + 2p - 1 = 0 (3p – 1)(p + 1) = 0Factorise p = 1/3 p + 1 = 0 p = - 1 sin x = 1/3sin x = -1 C A S T 0o0o 180 o 270 o 90 o C A S T 0o0o 180 o 270 o 90 o

Graphs & Functions Higher Functions f and g are defined on suitable domains by and a)Find expressions for: i) ii) b)Solve a) b)

Graphs & Functions Higher 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 a) b) Now use exact values Repeat for ii) equation reduces to

Graphs & Functions Higher 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 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

Outcome 3 Higher An identity is a statement which is true for all values. eg 3x(x + 4) = 3x x eg(a + b)(a – b) = a 2 – b 2 Trig Identities (1)sin 2 θ + cos 2 θ = 1 (2)sin θ = tan θ cos θ θ ≠ an odd multiple of π/2 or 90°. Trig Identities

Outcome 3 Higher Reason θoθo c b a a 2 +b 2 = c 2 sin θ o = a/c cos θ o = b/c (1) sin 2 θ o + cos 2 θ o = Trig Identities

Outcome 3 Higher sin 2 θ + cos 2 θ = 1 sin 2 θ = 1 - cos 2 θ cos 2 θ = 1 - sin 2 θ Simply rearranging we get two other forms Trig Identities

Outcome 3 Higher Example1 sin θ = 5 / 13 where 0 < θ < π / 2 Find the exact values of cos θ and tan θ. cos 2 θ = 1 - sin 2 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 Socos θ = 12 / 13 tan θ = sinθ cos θ = 5 / 13 ÷ 12 / 13 = 5 / 13 X 13 / 12 tan θ = 5 / 12 Trig Identities

Outcome 3 Higher Given that cos θ = -2 / √ 5 where π< θ < 3 π / 2 Find sin θ and tan θ. sin 2 θ = 1 - cos 2 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 Higher Are you on Target ! Update you log book Make sure you complete and correct MOST of the Composite FunctionComposite Function questions in the past paper booklet. Make sure you complete and correct SOME of the TrigonometryTrigonometry questions in the past paper booklet.

f(x) Graphs & Functions y = -f(x) y = f(-x) y = f(x) ± k y = f(kx) Move vertically up or downs depending on k flip in y-axis flip in x-axis + - Stretch or compress vertically depending on k y = kf(x) Stretch or compress horizontally depending on k f(x) y = f(x ± k) Move horizontally left 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) = x g(x) = 1 x x Domain x-axis values Input Range y-axis values Output x x Restrictionx ≠ 0 (x – 2)(x + 2) ≠ 0 x ≠ 2x ≠ -2 g(f(x)) g(f(x)) = f(x) = x g(x) = 1 x x Domain x-axis values Input Range y-axis values Output f(g(x)) Restriction x 2 ≠ 0 1 x = Similar to composite Area Write down g(x) with brackets for x g(x) = 1 ( ) inside bracket put f(x) g(f(x)) = 1 x x x2x2 f(g(x)) = Write down f(x) with brackets for x f(x) = ( ) inside bracket put g(x) f(g(x)) = 1 x2x2 - 4

Functions & Graphs TYPE questions (Sometimes Quadratics) Sketching Graphs Composite Functions Steps : 1.Outside function stays the same EXCEPT replace x terms with a ( ) 2.Put inner function in bracket You need to learn basic movements Exam questions normally involve two movements Remember order BODMAS Restrictions : 1.Denominator NOT ALLOWED to be zero 2.CANNOT take the square root of a negative number