Ch1 Algebra and functions. Ch 2 Sine and Cosine rule. Ch 10 Integration. Ch 9. Ch 3 Exponentials and Logarithms. Trigonometric.

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1 Ch1 Algebra and functions Ch 10 Integration Ch 2 Sine and Cosine rule Ch 9 Trigonometric Identities Ch 3 Exponentials and Logarithms C2 Ch 8 Differentiation Ch 4 Coordinate geometry Ch 7 Trigonometric functions Ch 5 The binomial expansion Ch 6 Radians

2 Chapter 1 Algebra and functions egwgwgw 2

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4 Chapter 2 Cosine and Sine rule egwgwgw 4

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6 Chapter 3 Exponentials and logarithms egwgwgw 6

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8 Chapter 4 Coordinate geometry in the xy plane egwgwgw 8

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11 Chapter 5 - The Binomial Theorem The binomial theorem tells us how to expand brackets like (3-2x) 11 quickly. If we have some brackets to a power then there are three ways we can expand them. If n (the power) is quite small then we can just expand them as normal. (a + b) 2 = (a + b)(a + b) = 1a 2 + 2a 1 b 1 + 1b 2 (a + b) 3 = (a + b)(a + b)(a + b) = 1a 3 + 3a 2 b 1 + 3a 1 b 2 + b 3 (a + b) 4 = (a + b)(a + b)(a + b)(a + b) = 1a 4 + 4a 3 b 1 + 6a 2 b 2 + 4a 1 b 3 + b 4 After this it gets a little time consuming so we start to use 11

12 Pascal s triangle If we look at the first few expansions then we start to notice a pattern in the coefficients (the numbers in front of the x, x 2, x 3, etc). (a + b) 2 = 1a 2 + 2a 1 b 1 + 1b 2 (a + b) 3 = 1a 3 + 3a 2 b 1 + 3a 1 b 2 + 1b 3 (a + b) 4 = 1a 4 + 4a 3 b 1 + 6a 2 b 2 + 4a 1 b 3 + 1b 4 The coefficients when n (the power) is four are the same as the numbers on the fifth row of Pascal s triangle. Pascal s triangle Each number is equal to the two above it added together. This predicts that (a + b) 5 = 1a 5 + 5a 4 b a 3 b a 2 b 3 + 5a 1 b 4 + 1b 5 12

13 Expand (2x + 3y) Pascal s triangle says (a + b) 4 = 1a 4 + 4a 3 b 1 + 6a 2 b 2 + 4a 1 b 3 + 1b 4 (2x + 3y) 4 = 1.(2x) (2x) 3.(3y) (2x) 2 (3y) (2x) 1 (3y) (3y) 4 = 16x x 3.3y + 6.4x 2.9y x 1.27y y 4 81y 4 = 16x x 3 y + 216x 2 y xy

14 The Binomial theorem For brackets with powers much bigger than this like (3 + 2x) 17 we start to use the Binomial theorem. (1 + x) n = 1 + nx + n(n 1) 1 X 2 x 2 + n(n 1)(n 2) 1 X 2 X 3 x (1 + x) 7 = 1 + 7x +. 7 X 6 1 X 2 x2 + 7 X 6 X 5 1 X 2 X 3 x3 +. = 1 + 7x + 21x x

15 (1 + x) n = 1 + nx + n(n 1) 1 X 2 x 2 + n(n 1)(n 2) 1 X 2 X 3 x (1 + 2x) 9 = X (2x) + 9 X 8 1 X 2 (2x)2 + 9 X 8 X 7 1 X 2 X 3 (2x)3 +. = x + 36 X 4x X 8x 3 +. = x + 144x x 3 +. (1 + x 2 )10 = X ( x 2 ) + 10 X 9 1 X 2 (x 2 ) X 9 X 8 1 X 2 X 3 (x 2 ) = 1 + 5x + 45 x x x ( x3 8 ) +... = 1 + 5x x x

16 This formula only works if the number is 1. If the number isn t 1 then we need to make it 1 by bringing it out of the bracket. (1 + x) n = 1 + nx + n(n 1) 1 X 2 x 2 + n(n 1)(n 2) 1 X 2 X 3 x (2 + x) 6 = [2(1 + x 2 )]6 = 2 6 x (1 + x 2 )6 = 64 x (1 + 6 X ( x 2 ) + 6 X 5 1 X 2 (x 2 )2 + 6 X 5 X 4 1 X 2 X 3 (x 2 )3 +. = 64 x (1 + 3x + 15 x x x ( x3 8 ) +... = 64 x (1 + 3x x x = x x x

17 We can use the binomial theorem to calculate high powers of numbers, like (0.98) 10, quickly. (1 2x) 10 = 1 20x + 180x 2 960x Compare 1 2x and x = 0.98 x = 0.01 So if we substitute x = 0.01 into our expansion for (1 2x) 10 we get an approximation for (0.98) (1 2x) 10 = 1 20x + 180x 2 960x = 1 20(0.01) + 180(0.01) 2 960(0.01) 3 = x X = = and the real value of (0.98) 10 is so we get a very good approximation just using the first four terms of the expansion. We don t need to go any further than x 3 because (0.01) 4 is which is so small it is irrelevant. 17

18 We can also write the formula using combination and factorial notation. 9! (9 factorial) = 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1 Bob has seven friends but only three spaces in his car. How many different ways can he choose three people to give a lift to out of a group of seven? Arrangements where order doesn t matter are called combinations. We can calculate the numbers of ways to choose 3 people from a group of 7 people by using factorial and 7C 3 = 7! 4! 3! = 7 3 these are all just different ways of writing the same thing. nc r = n! (n r)! r! = n r 18

19 (1 + x) n = 1 + nx + n(n 1) 1 X 2 x 2 + n(n 1)(n 2) 1 X 2 X 3 x (1 + x) n = n C 0 + n C 1 x 1 + n C 2 x n C r x r nc n x n (1 + x) n = n 0 + n 1 x x xr nc n 7 3 xn These are just three different ways of writing the same thing. Notice that n C 1 (the number of ways of choosing 1 person from n) is n and n C n (the number of ways of choosing n people from n people) is 1. If you have 8 friends, how many different ways can you choose 1 of them? 8. If you have 8 friends, how many different ways can you choose 8 of them? 1. 19

20 Chapter 6 Radian measure and its applications ufuful 20

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22 Chapter 7 Geometric sequences and series Arithmetic sequences are adding (or taking away) number patterns. We always call the first number a and the number we add on each time d. Take the sequence with a = 4 and d = 3 1 st 2 nd 3 rd 4 th 4, 7, 10, 13, = 4, 4 + 1X3, 4 + 2X3, 4 + 3x3 = a, a + d, a + 2d, a + 3d Notice that the number of d s is always one less than the term number. In general the nth term is a + (n-1)xd so the 10 th term here would be 4 + 9X3 =

23 A Geometric sequence is a multiplying number pattern (though we could be multiplying by a fraction so the numbers are going down). We always call the first number a and the number we multiply by each time r. Take the sequence with a = 3 and r = 2 1 st 2 nd 3 rd 4 th 3, 6, 12, 24, = 3, 3 X 2 1, 3 X 2 2, 3 X 2 3, = a, a X r 1, a X r 2, a X r 3, Notice that the number of r s is always one less than the term number. In general the nth term is a X r n-1 so the 10 th term here would be 4 X 3 9 = 78, st 2 nd 3 rd 4 th n th a, ar, ar 2, ar 3, ar n-1 23

24 Notice that when we divide one term by another the a s cancel. This can be very useful when answering questions. 3rd term 2nd term = ar 2 = r ar 1 The second term of a geometric sequence is 2 and the fifth term is 54. Find the common ratio and the 10 th term th term = ar 4 = 54 2 nd term = ar 1 = ar 4 = ar r 3 = 27 r = 3 24

25 If we end up with r 2 = something the answer could be positive or negative. The second term of a geometric sequence is 5 and the fourth term is 20. Find the common ratio and the 10 th term ar 3 = 20 ar 1 = ar 3 = ar r 2 = 4 r = +2 or -2 25

26 We can use geometric sequences to solve interest questions. Ann invests an amount of money at a rate of interest of 4% per year. After 5 years it was worth 10,000. How much did she invest at the start? This is a geometric sequence with a = amount and r = After 5 years = 6 th term = ar 5 = 10,000 a X (1.04) 5 = 10,000 a X = 10,000 a = 10, a =

27 We can use the laws of logarithms to solve geometric sequence questions. What is the first term in the geometric progression 3, 6, 12, 24,... to exceed 1 million? This is a geometric sequence with a = 3 and r = 2 n th term = ar n-1 = 3 X 2 n X 2 n-1 > 1,000,000 log(3x2 n-1 ) > log 1,000,000 log 3 + log 2 n-1 > log 1,000,000 log 3 + (n-1)log 2 > log 1,000,000 (n-1)log 2 > log 1,000,000 log 3 n 1 > log 1,000,000 log 3 log 2 integer. n > n = 20 as n must be an 27

28 Up till now we have been looking at geometric sequences which involves each term by itself. We are now going to look at geometric series which are the sum of the sequences. Sum of the first n terms of a geometric series = S n S n = a + ar + ar ar n-3 + ar n-2 + ar n-1 S n = a(1 - rn ) 1 - r Sometimes it is more convenient to write it as S n = a(rn 1) r 1 by multiplying top and bottom by 1. 28

29 Proof of formula We are going to - write out the sum of the first n terms - multiply both sides by r - take the bottom line away from the top line - put both sides into brackets - divide through S n = a + ar + ar ar n-3 + ar n-2 + ar n-1 rs n = ar + ar 2 + ar ar n-2 + ar n-1 + ar n S n rs n = a ar n S n rs n = a - ar n (1 r)s n = a(1 r n ) S n = a(1 - r n ) 1 - r 29

30 Find the sum of the 1 st 10 terms of the following series This is a Geometric Series with a = 2, r = 3 and n = 10. S n = a(1 - r n ) 1 - r Sum of the first 10 terms = S 10 S 10 = 2( ) 1-3 = 59,

31 Find the how many terms are in the following series This is a Geometric Series with a = 1024, r = 1 2 but what is n? nth term = ar n-1 nth term = 1 ar n-1 = X ( 1 2 )n-1 = 1 ( 1 2 )n-1 = Log( 1 2 )n-1 = (n-1)log( 1 2 ) = Log (n-1) = Log Log 1 2 n = 1 31

32 What is the least value of n such that the sum exceeds 2,000,000? This is a geometric series with a = 1 and r = 2 S n = a(r n 1) r a(r n 1) r - 1 > 2,000,000 1(2 n 1) 2-1 > 2,000,000 2 n - 1 > 2,000,001 2 n > 2,000,001 log 2 n > log 2,000,001 n log 2 > log 2,000,001 n > log 2,000,001 log 2 n > 20.9 n = 21 as n must be an integer. 32

33 Find (3 X 2 r ) = 3 x X X X 2 10 = 3 X ( ) This is 3 X a geometric series with a = 2, r = 2 and n = S n = a(r n 1) r - 1 = 2(2 10 1) 2-1 =

34 If we look at the series with a = 3 and r = If we look at the series or sum at certain points Sum of 1 st 5 terms = S 5 = Sum of 1 st 10 terms = S 10 = Sum of 1 st 20 terms = S 20 = As the number of terms we are adding goes up the sum goes up as well but not by a lot. In fact it will never actually get to 6, it will just get closer and closer and closer for ever. We say that it tends to 6 or it has a limit of 6 or the sum to infinity is 6 This series is convergent. Not all series converge, some diverge. If the series is going to converge then r (the number we multiply by) must be between 1 and + 1. If it isn t then the series will be divergent. 34

35 Sum to infinity of a geometric series S = a 1 - r for lrl < 1 Proof of formula Take a number less than 1, say 1 2 (½) 2 = 0.25 (½) 3 = (½) 4 = The bigger n gets the smaller (½) n gets. It gets nearer and nearer (but never actually gets to) zero. We say as n tends to infinity (½) n tends to zero. As n», (½) n» 0 And this is true for any r between 1 and + 1. So in the formula r n disappears, (1 0) = 1 and S n = S = a(1 - r n ) 1 - r a 1 - r becomes 35

36 Find the sum to infinity of the series This is an infinite geometric series with a = 120 and r = S = a 1 - r = =

37 Find the sum to infinity of the series p + 1 p p This is an infinite geometric series with a = 1 and r = 1 p S = a 1 - r = p (multiply top and bottom by p) p = p

38 The sum of the first 4 terms of a geometric series is 15 and the sum to infinity is 16. Find the possible values of r Sum of 1 st n terms Sum to infinity S n = a(1 - r n ) 1 - r S = a 1 - r Sum of the first 4 terms Sum to infinity a(1 r 4 ) 1 - r = 15 a 1 - r = If we divide the 1 st equation by the 2 nd we get rid of the a and the 1 r (remember to divide we can flip the 2 nd fraction and multiply). 1 r 4 = r 4 = r = or r =

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40 Chapter 8 Graphs of trigonometric functions ufufuli 40

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42 Chapter 9 Differentiation ufuful 42

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44 Chapter 10 Trigonometric identities and simple equations fuful 44

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46 Chapter 11 - Integration ufufu 46

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