Chapter 1 Algebraic fractions Page 11 Chapter 2 Functions Page 22. Chapter 8 Differentiation Page 88. Chapter 3 Exponentials and Logarithms Page 33

Transcription

1 Chapter 8 Differentiation Page 88 Chapter 1 Algebraic fractions Page 11 Chapter Functions Page Chapter 7 Further Trig Page 77 C3 Chapter 3 Exponentials and Logarithms Page 33 Chapter 6 Trigonometry Page 66 Chapter 5 Transforming graphs Page 55 Chapter 4 Numerical methods Page 44

2 Chapter 1 Algebraic fractions When we simplify fractions we can factorise and cancel out common factors from top and bottom. 1 0 = 4 X 3 4 X 5 = 3 5 x + 3 x + 6 = 1 X (x + 3) X (x + 3) = 1 x + x + 3 This can t be simplified and we definitely can t cancel just the x s!!! x + x + 3 = X 3

3 When you have a fraction on the top or bottom it is a good idea to multiply to get rid of it. 1 x + 1 3x + 6 = X ( 1 x + 1) X (3x + 6) = x + 6x + 1 = and now we can continue as normal. x x + 1 = 3 X (x + 6) 3 X ( 1 3 x + 1) = 6x + 18 x + 3 = and again we can continue as normal. If we have two fractions then it is quicker to multiply by a number that will get rid of both of them at the same time. 1 x x + 5 = 6 X ( 1 x + 4) 6 X ( 1 3 x + 5) = 3x + 4 x

4 Try and split expressions into as many factors as possible so you can see easily what you can cancel. Remember the difference of two squares if you see (Something) (Something else) x 3x x - 9 = x X (x - 3) (x + 3) X (x - 3) = x (x + 3) Split quadratics into two brackets to see common factors you can cancel. x + 5x + 4 x + 8x + 16 = (x + 1) X (x + 4) (x + 4) X (x + 4) = (x + 1) (x + 4) 4

5 When we multiply fractions we times the top by the top and the bottom by the bottom. Multiplying out brackets should be avoided whenever possible because it makes it a lot harder to spot common things on top and bottom. 3 X 5 7 = 10 1 x y X x z = x yz 3 5 X 5 7 = 3 7 a b X c a = c b x + X x - 16 x + 4 3x + 6 = (x + ) (x - 4) X (x + 4)(x 4) 3(x + ) = (x 4) 3 5

6 When we divide fractions we flip the second fraction and continue as normal = 3 X 7 5 = a b c d = a b X d c = x 3x y + y x y + 1 = x 3x y + y x y + 1 x = One thing we can t do is cancel before we flip the second fraction! p r - 1 r + 1 p r + 1 = = p r

7 When we add or take away fractions we must first make the bottoms the same. We don t always need to multiply by both numbers! X 5 X = X = 15 = We = don t always need 10 to multiply by all the letters! a b + c d e fg + h fk X d X b X k X g = ad bd + bc bd = ek fgk + gh fgk = ad + bc bd = ek +gh fgk 7

8 We can think about dividing numbers as What number do I need to multiply the small number by to equal the big number? 1 4 What number do I need to multiply 4 by to equal 1? Answer = 3 because 3 X 4 = 1 So 1 4 = and 3 X 4 = 1 are different ways of saying the same thing. We might also have some left over after we have shared the number out. We call this the remainder. If I share 17 marbles between 5 people, each person gets 3 marbles and I have left over in my hand What number do I multiply 5 by to equal 17? Answer I need to multiply 5 by 3 and then add on an extra 8 So 17 5 = 3 remainder and 17 = (3 X 5) + are different ways of saying the same thing.

9 We can do this with expressions as well What do I need to multiply the small expression by to equal the big expression? (x 3x) x = (x 3) What do I need to multiply x by to equal x 3x? Answer = (x 3) because x X (x 3) = x 3x So (x 3x) x = (x 3) and x X (x 3) = x 3x are different ways of saying the same thing. (x 3x) (x 3) = x What do I need to multiply (x 3) by to equal (x 3x)? Answer = x because x X (x 3) = x 3x So (x 3x) (x 3) = x and x X (x 3) = x 3x are different ways of saying the same thing. (x + 5x + 6) (x + ) = (x + 3) What do I need to multiply (x + ) by to equal (x + 5x + 6)? Answer = (x + 3) because (x + ) X (x + 3) = x + 5x + 6 So (x + 5x + 6) (x + ) = (x + 3) and (x + ) X (x + 3) = x +5x + 6 are different ways of saying the same thing. 9

10 Divide x 3 + x 7 by long division. When we write out the division we need to include a zero x term as well because there are no x s in the original expression. x + 4x + 1 (x 3) x 3 + x + 0x - 7 x 3-3x 4x + 0x 4x - 1x 1x - 7 1x We know we have reached the remainder because the power of the 9 is less than the thing we are dividing by (x 3 + x - 7) (x 3) = x + 4x + 1 remainder (x 3 + x - 7) = (x + 4x + 1) X (x 3) + 9 (x 3) (x 3) (x 3) x 3 + x 7 (x 3) = (x + 4x + 1) + 9 (x 3) 10 These are all different ways of saying the same thing.

11 Divide x 3 + x 7 by x 3 using the Remainder theorem. Substitute x = 3 to make everything but the D disappear. x 3 + x 7 = (Ax + Bx + C)(x 3) + D = something X zero + D 9 = D Let x = 0 to make all the x things disappear. x 3 + x 7 = (Ax + Bx + C)(x 3) + D = ( C)( 0 3) + D -7 = -3C + D 1 = C Compare coefficients of x 3 x 3 + x 7 = (Ax + Bx + C)(x 3) + D 1 = A Compare coefficients of x x 3 + x 7 = (Ax + Bx + C)(x 3) + D 1 = -3 + B 4 = B So we get exactly the same answer as before x 3 + x 7 = (1x + 4x + 1)(x 3)

12 Chapter - Functions A mapping is a rule that turns one number into another number. It can be written in words take the number double it and take away 1 in function notation f(x) = x - 1 or as a graph. y = x f(x) = x We can have either y or f(x) up the side. This is just a picture of the rule. If you want to know what this rule turns into, go to on the side, draw straight up to the line then straight across and you can see that this particular rule turns into 3. Notice that with this particular function if we know our output we can work out what we put in. If we got 7 out we can go across to the line then straight down and we must have put 4 into our mapping. 1

13 We might use different letters for different rules so that we know which rule we are talking about at any time. f(x) = 3x + 1 g(x) = 1 x f() means what is the output when we put through the f rule? f() = 3 X + 1 = 7 g(-1) = 1 (-1) = x is the input and f(x) is the output. x rule f(x) If f(a) = 10 what is a? If we have put a certain number a into the f rule and the output is 10 what number did we put in? 3 X a + 1 = 10 a = 3 13

14 Some mappings are many to one like f(x) = sin x or f(x) = x, there are lots of numbers we can put in and get the same answer out sin x x y = sin x f(x) = x Notice that now if we know the output, we can no longer find our way back to the input. If we have an output of 16 then the input could have been either 4 or

15 We can also have a one to many mapping where one number in produces more than one number out, for example f(x) = x. 9 x 3 and -3 15

16 Some mappings may not be able to give an output for certain inputs. For example, the mapping f(x) = 1 x has no output for the number zero. 1 0? x y = 1/x

17 We are going to concentrate on a certain kind of mapping called a function. These are one to one mappings. Every input has one and only one output and every output has one and only one input. This allows us to find our way back to the input if we know our output or work out the reverse of the rule. 17

18 One way we can turn a mapping into a function is by restricting the numbers that we are going to put in or the domain. Up till now we have just assumed that the domain for all of these functions is just all of the Real numbers. Remember the Real numbers are all the whole numbers, fractions, decimals, surds, zero, in fact any kind of number that you can think of. But we can decide to confine the numbers that we are allowed to put in so that each input has only one output and every number that we put in does have an output. f(x) = x, domain {x є R, 0 < x} f(x) = x, x is a Real number and x is bigger than zero by restricting ourselves to only putting in numbers bigger that zero we no longer have the problem of getting the same number out for two different inputs so now this mapping is a function. Domain = the numbers that we can put in rule Notice that the word domain has IN at the end. 18

19 The Range tells us what numbers we could get out of our function. rule Range = the numbers that could come out For f(x) = x, we can only get numbers out that are greater than or equal to zero, so the range is { x є R, 0 x}. f(x) = x

20 For g(x) = 1/x, as long as agree not to put zero in we can get any number from minus infinity to plus infinity out so the range is { x є R }. y = 1/x For h(x) = sin x, we can only get number between -1 and +1 out so the range is { x є R, -1 x +1}. 1 y = sin x

21 Composite functions are when we do one rule after the other. fg() means do g first on then do f on the result. Notice that we do them opposite to the direction they are written. f(x) = x g(x) = x + 1 fg() = f(5) = 5 Notice that gf(3) gives us a different result gf() = g(4) = 9 so it does matter which order we do them in. f (x) means the same as ff(x) or do the f rule twice. ff() = f(4) = 16 g f 1

22 For a generalized version f(x) = x g(x) = x + 1 fg(x) = f(x + 1) = (x + 1) So the function fg(x) = (x + 1) This function has a domain {x є R}, because we can put anything we like in without there being a problem. The range will be {x є R, 0 x} because no matter what we put in we can only get numbers bigger than or equal to zero out gf(x) = g(x ) = x + 1 Again we can put anything we like in so the domain = {x є R} but we can only get numbers bigger than or equal to one out, so the range = {x є R, 1 x}.

23 Chapter 3 e and Ln We can draw the graphs of y = x and y = 3 x. Graphs like these are called exponential functions. y x y = x y x y = 3 x Notice that they both cut the y axis at (0,1) because anything to the power of zero is one, but y = 3 x gets steeper a lot quicker. The gradient of y = x at this point is 0.7. The gradient of y = 3x at this point is 1.1. There is a number between and 3 for which the gradient at (0,1) is exactly one. We call this number e. 3

24 e is an irrational number like π, i.e. it goes on forever. However it is approximately equal to.718. e.718 The graph of y = e x rather than an exponential function. is called the exponential function y x y = e x Notice that as x, y As usual when x = 0, y = 1, i.e. it cuts through the y axis at one. And as x -, y 0, i.e. it has an asymptote at zero 4 which it approaches but never reaches.

25 y = e x y = e x is the square of e x (because (e x ) = e x from the rules of indices) so everything happens quicker, i.e. it is steeper than e x and drops to zero quicker. y = e x

26 y = e ½x With y = e ½x everything happens slower so it is shallower than y = e x and drops to zero slower. y = e ½x

27 y = e x + Everything jumps up so it cuts through at (0, 3) and the asymptote is at now. y = e x

28 y = 5e x For this curve the asymptote is still zero but now it cuts through at (0, 5) y = 5e x

29 y = e -x This type of curve is sometimes called exponential decay and appears in the decay of radioactivity. y = e -x This is a reflection of the curve y = e x in the y axis. 9

30 Example The price of a used car can be represented by the formula P = 16000e -t/10 where P is the price in s and t is the age in years from new. Calculate a The price when new b The value at 5 years old c Sketch the graph of P against t and say what this suggests about the eventual price of the car a. To find the price when new we substitute t = 0 P = 16000e -0/10 P = X 1 (because anything to the zero is 1) P = b. To find the price when the car is five years old we substitute t = 5 P = 16000e -5/10 P = 16000e -0.5 P =

31 Example The number of people infected with a disease varies according to the formula N = e -0.5t where N is the number of people infected with the disease and t is the time in years after detection. Calculate a How many people were first diagnosed with the disease b Graph N against t and state what this shows about the long term prediction for how the disease will spread. a. When t = 0 N = e -0.5X0 N = X 1 (because anything to the zero is 1) N = 00 31

32 Ln x is the shorthand that mathematicians use for Log e x. It is the inverse of e x which means that if you e a number and then Ln it, it will be back to the original number. Ln x e x Remember from Chapter that the inverse of a function is always the reflection in the line y = x. Notice that as x 0, y - As x +, y + It cuts through the x axis at x = 1 Ln x does not exist for negative values of x. 3

33 You can put any number you like into e x, positive or negative and get out a sensible answer but you only ever get positive numbers out the other end (but never zero). This means that the domain of e x is the Real numbers and the range is the positive Real numbers. Domain (x ε R) Range (y ε R, y > 0) Real numbers e x Positive Real numbers but not zero Because the domain and range of inverse functions are always the same but they switch places, the domain of Ln x is the positive Real numbers and the range is the Real numbers. Domain (x ε R, x>0) Range (y ε R) 33

34 Positive Real numbers but not zero Ln x Real numbers Example e x+3 = 4 Ln both sides, e and Ln cancel each other out. x + 3 = Ln 4 x = Ln 4 3 x = Ln 4-3 Note that we are not allowed to bring the 3 inside the Ln and calculate Ln 4-3 = Ln 1 Example Ln x + 1 = 4 Ln x = 3 Ln x = 3 x = e 1.5 e both sides, Ln and e cancel each other out. 34

35 Chapter 4 Numerical methods We can solve equations like x + 1 = 7 and x + x 5 = 0 through a process, however, in real life the equations that we need to solve are often not so easy to work out. The roots of an equation are the values that we need to put in to make it true. The root is 3 x + 1 = 7 The roots are +3 and - x 1x - 6 = 0 35

36 When we did trial and improvement at GCSE we wanted to know if our guess made the expression too big or too small. Once we had trapped it between one guess that made it too big and another that made it too small we knew that the answer lay between these two answers. At A level we normally rearrange so that the equation is equal to zero so then if we find one number that makes the expression positive and another that makes it negative then we know the root must lie between these two numbers Notice that there may be more than one root so the signs might go from positive to negative and back again.

37 Show that the x 3 3x + 3x 4 = 0 has a root between x = and x = 3. when x = x 3 3x + 3x 4 = = - = Negative when x = 3 x 3 3x + 3x 4 = = +5 = Positive One is positive and one is negative therefore there is a root between x = and x = 3 37

38 Given that f(x) = e x sin x 1, show that the equation f(x) = 0 has a root x = r where r lies in the interval 0.5 < r < 0.6. The most important thing to remember with this type of question is that x is in radians, and your calculator must be in radians mode! when x = 0.5 e x sin x 1 = Positive when x = 0.6 e x sin x 1 = Negative One is positive and one is negative therefore there is a root between x = 0.5 and x =

39 Once we have found where the interval where the root lies we can use an iterative process to get a more and more accurate answer to the equation. We need to rewrite the equation so that it is in the form x = something, it doesn t matter if we also have x s on the other side. Then we put our first iteration x 0 into the formula, this will give us our second iteration x 1, we put this into the formula and so on. Each iteration will get closer and closer to the correct answer and eventually will stop moving altogether. 39

40 Show that the formula x written in the form 5x - 3 = 0 can be x = (5x+3) and use the iterative formula x n+1 = 5x n + 3 to find a root of this equation. Use x 0 = 5. x 5x - 3 = 0 x = 5x + 3 x = 5x+3 If x 0 = 5 then x 1 = 5X5 +3 = 8 = 5.9 x = 5X = 5.4 x 3 = 5X = 5.48 eventually x 4 = 5.53 x 5 = 5.53 The iterations have stopped changing so this is the root. 40

41 Chapter 5 Transforming graphs The modulus of something is the positive value of it. l-5l = 5 l-7l = 7 l9l = 9 It turns negative numbers into positive and leaves positive numbers unchanged. 41

42 y = lf(x)l The modulus of a entire function makes the bits below the x axis reflect in the x axis y=x y=lx-3l y=x^ -3x y=lx^ -3x -10l

43 y = f(lxl) The modulus of just the x bit makes the entire function reflect in the y axis. y = x y = lxl y = 4lxl - lxl^ y = 4x - x^3 y = 4lxl - lxl^

44 Solving equations Equations involving modulus will usually have double the amount of solutions that they would normally have. We can find the solutions in two ways. We can draw two graphs on top of one another and see where they cross or we can solve them using algebra. 5 lx-1.5l =

45 There will be two answers to lx-1.5l = 3 Positive answer Negative answer +(x 1.5) = 3 -(x 1.5) = 3 x 1.5 = 3 -x = 3 x =.5 x =

46 There will be two answers to l5x-l = lxl Positive answer Negative answer +(5x - ) = x -(5x ) = x 5x = x -5x + = x x = /3 x = /7 Notice that when we take the modulus of we only need to put a plus or minus in front of one of the brackets, if we put them on both we will end up with the same answer both times. 46

47 f(x + 3) f(x+3) moves the whole graph left three place. y = x² y = (x+3)²

48 f(x) + 3 f(x) + 3 moves the whole graph up three places y = x² y = x²

49 f(x) f(x) stretches the graph from the x axis and makes it two times bigger in the up and down direction. y = sin x y = sin x

50 f(x) f(x) squashes the graph towards the y axis so that twice as much happens in the same space. 1 y = sin x y = sin x

51 Chapter 6 - Trigonometry cosec x = 1 sin x sec x = 1 cos x 1 cot x = tan x 51

52 cosec 37 = 1 sin 37 = = 1.66 sec 14 = 1 cos 14 = = cot (-63) = 1 tan (-63) = = Notice that we can get negative answers as well as putting in angles that are negative or greater than ninety degrees. 5

53 53

54 1 sin x cosec x = 1/sin x

55 When x = 90, sin x = 1, so 1 sin x = 1 1 = 1. As we head away from the top of the curve sin x gets smaller and smaller so 1 sin x heads off towards infinity. gets bigger and bigger and When we get to x = 0, sin x = 0, we can t work out 1 0 so cosec is undefined at this point (and at x = 180 and 360, in fact there will be an asymptote every 180 degrees. 55

56 1 cos x sec x = 1/cos x

57 When x = 0, cos x = 1, so 1 cos x = 1 1 = 1. As we head away from the top of the curve cos x gets smaller and smaller so 1 cos x heads off towards infinity. gets bigger and bigger and When we get to x = 90, cos x = 0, we can t work out 1 0 so cosec is undefined at this point (and at x = 70 and 450, in fact there will be an asymptote every 180 degrees. 57

58 3 tan x cot x = 1/tan x

59 When tan x is very small, cot x = 1 tan x is very big. When x = 45, tan x = 1, so 1 tan x = 1 1 = 1. As we head towards x = 90, tan x gets bigger and bigger and cot x = 1 tan x gets smaller and smaller. At exactly x = 90 tan x is undefined and then becomes negative. 59

60 We can simplify expressions with sec, cosec and cot in. Simplify sin Θ cot Θ sec Θ sin Θ cot Θ sec Θ = sin Θ X cos Θ sin Θ X 1 cos Θ = 1 60

61 Simplify sin Θ cos Θ (sec Θ + cosec Θ) sin Θ cos Θ (sec Θ + cosec Θ) = sin Θ cos Θ ( 1 cos Θ + 1 sin Θ ) = sin Θ cos Θ ( sin Θ + cos Θ sin Θ cos Θ ) = sin Θ + cos Θ 61

62 Simplify cot x cosec x sec x + cosec x = cos3 x cot x cosec x sec x + cosec x = cos x sin x X 1 sin x 1 cos x + 1 sin x = cos x sin x sin x + cos x sin x cos x = cos x sin x 1 sin x cos x = cos x 1 sin x sin x cos x = cos x sin x X sin x cos x 1 = cos 3 x 6

63 sin(a+b) = sin A cos B + sin B cos A cos(a+b) = cos A cos B sin A sin B tan(a+b) = tan A + tan B 1 - tan A tan B sin(a-b) = sin A cos B - sin B cos A cos(a-b) = cos A cos B + sin A sin B tan A tan B tan(a+b) = 1 + tan A tan B 63

64 Using the formula for sin(a+b) derive the formula for sin(a-b) sin(a+b) = sin A cos B + sin B cos A If we replace B with (-B) sin(a+(-b)) = sin A cos (-B) + sin (-B) cos A now the cos (-B) is just the same as cos B but sin (- B) is the negative of sin B so sin(a-b) = sin A cos B sin B cos A as required 64

65 Using the formula for sin(a+b) and cos(a+b) derive the formula for tan(a+b) tan (A+B) = sin(a+b) cos(a+b) tan (A+B) = sin A cos B + sin B cos A cos A cos B sin A sin B Divide everything by cos A cos B. tan (A+B) = sin A cos B cos A cos B cos A cos B cos A cos B + sin B cos A cos A cos B sin A sin B cos A cos B tan (A+B) = sin A cos A + sin B cos B sin A sin B 1 - cos A cos B = tan A + tan B 1 - tan A tan B 65

66 Use the formula for sin(a-b) to find the exact value of sin Remember that sin 45 = 1 = cos 45 = 1 = sin 30 = 1 cos 30 = 3 and sin(a-b) = sin A cos B - sin B cos A sin 15 = sin (45 30 ) = sin 45 cos 30 sin 30 cos 45 = X 3-1 X = =

67 Given that sin A = 3 1 and 180<A<70 and cos B = 5 13 obtuse find the value of cos(a-b). and B is cos(a-b) = cos A cos B + sin A sin B so we need to find out cos A and sin B as well If sin A = 3 5, by drawing a triangle and using Pythagoras we can see that the value of cos A is 4 5 or positive? but will the sign be negative If A is between 180 and 70 then it is in the third quadrant and cos is negative in the third quadrant so cos A = Similarly if cos B = 1 13 then by drawing a triangle and using Pythagoras we can see that the value of sin B will be 5 but will it 13 be positive or negative? As B is obtuse it is in the second quadrant, sin is positive in the second quadrant so sin B = cos(a-b) = cos A cos B + sin A sin B = ( 4 1 ) X ( ) + ( 3 ) X ( 5 ) =

68 sin A = sin A cos A cos A = cos A - sin A = cos A 1 = 1 sin A tan A tan A = 1 + tan A 68

69 Use the formula for cos(a+b) to derive the formula for cos A cos(a+b) = cos A cos B sin A sin B If we let B = A cos(a+a) = cos A cos A sin A sin A cos A = cos A sin A as required 69

70 Show that cos A = cos A 1 using cos A = cos A sin A Remember that sin A + cos A = 1 so sin A = 1 cos A cos A = cos A sin A = cos A (1 cos A) = cos A 1 as required 70

71 Rewrite sin 15 cos 15 as a single trigonometric ratio sin A cos A = sin A sin 15 cos 15 = sin (X15 ) = sin 30 71

72 Given that cos x = ½ find the exact value of cos x cos x = cos x - 1 = (½) - 1 = -½ 7

73 Given that cos x = 3 4 and 180<x<360 find the exact value of sin x sin x = sin x cos x so we are going to have to find sin x as well If we draw a triangle then by Pythagoras the value of sin x = 7. 4 What sign is sin x going to be? cos x is positive and x is a reflex angle. This means that x is in the forth quadrant. In the forth quadrant sin is negative so sin x = sin x = sin x cos x = X 7 4 X

74 By expanding sin(a+a) show that sin 3A = 3 sin A 4sin 3 A Remember that sin (A+B) = sin A cos B + sin B cos A sin A = sin A cos A cos A = cos A sin A sin (A+A) = sin A cos A + sin A cos A = ( sin A cos A)cos A + sin A (cos A sin A) = sin A cos A + sin A cos A sin 3 A = 3 sin A cos A sin 3 A = 3 sin A (1 sin A) sin 3 A = 3 sin A 4 sin 3 A 74

75 Prove the identity tan A = cot A tan A Remember that to prove an identity you need to start on one side and work your way through to the other side. You can t work on both at the same time and meet in the middle tan A = tan A 1 tan A = 1 tan A tan A Divide everything by tan A = cot A tan A as required 75

76 Prove the identity tan A = cot A tan A Remember that to prove an identity you need to start on one side and work your way through to the other side. You can t work on both at the same time and meet in the middle tan A = tan A 1 tan A Divide everything by tan A = 1 tan A tan A = cot A tan A as required 76

77 Given that x = 3 sin Θ and y = 3 4 cos Θ, eliminate Θ and express y in terms of x. Remember that cos Θ = 1 sin Θ We want to rearrange the two equations so that we can substitute one into the other and get rid of the Θ Rearrange the first equation sin Θ = x cos Θ = 3 - y 4 1 sin Θ = 3 - y 4 Substitute in the first equation 1 ( x 3 ) = 3 - y 4 and rearrange y = 8( x 3 )

78 Solve 3 cos x cos x + = 0 for 0 < x < 360. Remember that cos x = cos x - 1 We are aiming for a quadratic equation in cos x which we are going to put into brackets and solve. We need to get start off by getting rid of the cos x as we can t have a quadratic that mixes up cos x and cos x cos x cos x + = 0 3 ( cos x 1) cos x + = 0 6 cos x 3 cos x + = 0 6 cos x cos x 1 = 0 A quadratic in cos x (3 cos x + 1)( cos x 1) = 0 3 cos x + 1 = 0 or cos x 1 = 0 cos x = 1 3 cos x = 1 x = 109.5, 50.5 x = 60, 300 x = 60, 109.5, 50.5,

79 Express 3 sin x + 4 cos x in the form R sin(x + α) R sin (x + α) = 3 sin x + 4 cos x Compare to formula multiplied by R R sin (x + α) = R sin x cos α + R sin α cos x R cos α = 3 and R sin α = R sin α = 4 R cos α 3 sin α = 4 cos α 3 tan α = 4 3 α = R cos α = 3 and R sin α = 4 Square both sides R cos α = 9 and R sin α = 16 Add together R cos α + R sin α = R (cos α + sin α) = 5 Because cos α + sin α = 1 R = 5 R = sin x + 4 cos x = 5 sin (x 53.1) 79

80 Express 7 cos Θ - 4 sin Θ in the form R cos(x + α). R cos (Θ + α) = 7 cos Θ 4 sin Θ Compare to formula multiplied by R R cos (Θ + α) = R cos Θ cos α R sin Θ sin α R cos α = 7 and R sin α = R sin α = 4 R cos α 7 sin α = 4 cos α 7 tan α = 4 7 α = R cos α = 7 and R sin α = 4 Square both sides R cos α = 49 and R sin α = 576 Add together R cos α + R sin α = R (cos α + sin α) = 65 Because cos α + sin α = 1 R = 65 R = cos Θ 4 sin Θ = 5 cos (Θ 80

81 sin P + sin Q = sin P + Q cos P - Q sin P - sin Q = cos P + Q sin P - Q cos P + cos Q = cos P + Q cos P - Q cos P - cos Q = - sin P + Q sin P - Q 81

82 Use the formula for sin (A + B) and sin (A B) to derive the result that sin P + sin Q = sin P + Q cos P - Q sin (A + B) = sin A cos B + sin B cos A sin (A B) = sin A cos B sin B cos A Add them together and sin (A + B) + sin (A B) = sin A cos B Let A + B = P and A B = Q then A = P + Q and B = P - Q so so sin P + sin Q = sin P + Q cos P - Q as required 8

83 Use the formula for sin (A + B) and sin (A B) to derive the result that sin P + sin Q = sin P + Q cos P - Q sin (A + B) = sin A cos B + sin B cos A sin (A B) = sin A cos B sin B cos A Add them together and sin (A + B) + sin (A B) = sin A cos B Let A + B = P and A B = Q then A = P + Q and B = P - Q so so sin P + sin Q = sin P + Q cos P - Q as required 83

84 84

85 Chapter 8 Differentiation 85

86 The Chain rule We can use the chain rule when we want to differentiate one thing that is inside another. For example (x 3 + 1) 5 e x Ln(4x 7-3) are all functions inside functions a bit like Russian dolls. The Chain rule = dx du X du dx Notice that the du s cancel a bit like fractions. 86

87 If y = (x 3 + 1) 5 what is dx? If we let u = x then using the Chain rule. y = u 5 u = x du = 5u4 du dx = 6x dx = du X du dx dx = 5u4 X 6x dx = 6x X 5u 4 dx = 30x (x 3 + 1) 4 87

88 In general if we have y = (something) number = n x something differentiated X dx (something) n - 1 y = (3x + ) 7 dx = 7 x 3 X (3x + )6 dx = 1(3x + )6 Notice that the inside stays the same. y = (4x 3 + 6x) 9 dx = 9 X (1x + 6) x (4x 3 + 6x) 8 dx = 9(1x + 6)(3x + ) 8 88

89 We can rearrange the Chain rule into another useful format 1 dx = dx this means we can work out dx. dx and then flip it to find If x = y + 3y find dx It s too hard to find out so let s differentiate both dx sides with respect to y instead. dx = y + 3 so dx = 1 y + 3 This means that we now need to stick in the value of y if we want to find the gradient instead of x. 89

90 The Product rule We use the product rule when we want to differentiate two things that are multiplying each other. For example 4x (3x + 1) 4 x 3 e x 3x 5 Ln(x+1) We call the first thing u and the second thing v. The Product rule If y = u X v then dx = vdu dx + udv dx the second one left alone times the first one differentiated plus the first one left alone times the second one differentiated 90

91 If y = x (3x + 4) 5 find dx. u = x v = (3x + 4) 5 du dx = x dv dx = 30x(3x + 4) dx = vdu dx + udv dx dx = (3x + 4) 5 X x + x X 30x(3x + 4) 4 take the common factors of x and (3x + 4) 4 down the front dx = (x) X (3x + 4) 4 X (1 + 15x) dx = x(1 + 15x)(3x + 4) 4 91

92 The Quotient rule We use the product rule when we want to differentiate two things that are in a fraction. For example y = Ln(x + 1) (3x + 1) 5 We call the thing on top u and the thing underneath v. The Quotient rule If y = u v = v dx du dx udv dx v 9

93 If y = 5x (3x -1) find dx. Use the Quotient rule as it s a fraction. u = 5x v = 3x 1 du dx = 10x dv dx = v dx = du dx udv dx v dx = (3x-1))(10x) (5x )(3) (3x-1) dx = (15x 10x) (3x-1) 93

94 Differentiation of e x and e something If y = e x then dx also = ex. If y = e something then dx = (something differentiated) X esomething Notice that the power doesn t decrease by one like it normally does. If y = e 3x then dx = 3e3x. If y = e x² then dx = xex². 94

95 We could do e something questions the long way by using the Chain rule, dx = du X du dx. If y = e 3x what is dx? If we let u = 3x then using the Chain rule. y = e u u = 3x du = e u du dx = dx = X du du dx dx = eu X 3 dx = 3eu dx = 3e3x 95

96 Differentiation of Ln(x) and Ln(something) If y = Ln x then dx = 1 x If y = Ln(something) dx = something differentiated something left alone If y = Ln(3x +1) then dx = 3 3x + 1 If y = Ln(x 3 +5x) then dx = 6x + 5 x 3 + 5x 96

97 Again we can differentiate Ln questions the long way by using the chain rule dx = du X du dx. If y = Ln(3x + 4x) what is dx? If we let u = 3x + 4x then using the Chain rule. y = Ln(u) u = 3x + 4x du = 1 u du dx = 6x dx = du X du dx dx = 1 u X (6x + 4) dx = 6x + 4 u dx = 6x + 4 3x + 4x 97

98 Differentiation of Trig functions All of the following only work when x is in radians! If y = sin x then dx = cos x If y = cos x then dx = - sin x If y = tan x then dx = sec x If y = cosec x then dx = -cosec x cot x If y = sec x then dx = sec x tan x If y = cot x then dx = -cosec x 98

99 If y = sin 3x then find dx using the Chain rule y = sin u u = 3x du = cos u du dx = dx = X du du dx dx = cos u X 3 dx = 3 cos 3x Notice that the inside, the 3x, has stayed the same. In general if y = sin (something) dx = something differentiated X cos (something) 99

100 If y = sin 5x dx = 3 x cos 3x If y = cos 7x dx = 7 x - sin 7x If y = tan 5x dx = 5 x sec 5x If y = cosec 6x = 6 X cosec 6x cot 6x dx 100

101 x) 3 If we write y = sin 3 x this actually means y = (sin We can find dx using the Chain rule y = u 3 u = sin x du = du 3u dx = cos x dx = du X du dx dx = 3u X cos x dx = 3(sin x) X cos x = 3 cos x(sin x) dx 101

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