C3

Chapter 1 Algebraic fractionsPage 11

Chapter 2

Functions

Page 22

Chapter 3Exponentials

and Logarithms

Page 33

Chapter 4Numerical methodsPage 44

Chapter 5Transforming

graphsPage 55

Chapter 6Trigonometry

Page 66

Chapter 7 Further Trig

Page 77

Chapter 8Differentiation

Page 88

2

Chapter 1 – Algebraic fractions When we simplify fractions we can factorise and cancel out common factors from top and bottom.

1220

= 4 X 34 X 5

= 35

x + 32x + 6

= 1 X (x + 3)2 X (x + 3)

= 12

x + 2x + 3

This can’t be simplified and we definitely can’t cancel just the ‘x’s!!!

x + 2x + 3

= 23 X

3

12x + 1

3x + 6 =

2 X (12x + 1)

2 X (3x + 6) =

x + 26x + 12

= …

and now we can continue as normal.

When you have a fraction on the top or bottom it is a good idea to multiply to get rid of it.

2x + 613x + 1

= 3 X (2x + 6)

3 X (13x + 1)

= 6x + 18x + 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.

12x + 413x + 5

= 6 X (1

2x + 4)

6 X (13x + 5)

= 3x + 242x + 30

4

x2 – 3xx2 - 9

= x X (x - 3)

(x + 3) X (x - 3) =

x(x + 3)

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)2 – (Something else)2

x2 + 5x + 4x2 + 8x + 16

= (x + 1) X (x + 4)(x + 4) X (x + 4)

= (x + 1)(x + 4)

Split quadratics into two brackets to see common factors you can cancel.

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.

35 X

57 =

37

ab X

ca =

cb

23 X

57 =

1021

xy X

xz =

x2

yz

x + 2x + 4

X x2 - 16

3x + 6 =

(x + 2)(x - 4)

X (x + 4)(x – 4)3(x + 2)

=

(x – 4)3

6

When we divide fractions we flip the second fraction and continue as normal.

x2 – 3xy2 + y

÷ xy + 1

= x2 – 3xy2 + y

x y + 1x

= …

23 ÷

57 =

23 X

75 = …

ab ÷

cd =

ab X

dc = …

One thing we can’t do is cancel before we flip the second fraction!

p2

r - 1 ÷ r + 1

p =

pr - 1

÷ r + 11

= …

7

When we add or take away fractions we must first make the bottoms the same.

23 +

45

X 5 X 3

= 1015

+ 1215

= 2215

We don’t always need to multiply by both numbers! 3

10 +

45

X 2

= 3

10 +

810

= 1310

ab +

cd

X d X b

= adbd

+ bcbd

= ad + bc

bd

We don’t always need to multiply by all the letters!

efg

+ hfk

X k X g

= ekfgk

+ ghfgk

= ek +gh

fgk

8

We can think about dividing numbers as “What number do I need to multiply the small number by to equal the big number?”

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 2 left over in my hand.

17 ÷ 5 What number do I multiply 5 by to equal 17?

Answer – I need to multiply 5 by 3 and then add on an extra 2 So 17 ÷ 5 = 3 remainder 2 and 17 = (3 X 5) + 2 are different ways of saying the same thing.

12 ÷ 4 What number do I need to multiply 4 by to equal 12? Answer = 3 because 3 X 4 = 12 So 12 ÷ 4 = 2 and 3 X 4 = 12 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?”

(x2 – 3x) ÷ x = (x – 3) What do I need to multiply x by to equal x2 – 3x? Answer = (x – 3) because x X (x – 3) = x2 – 3x So (x2 – 3x) ÷ x = (x – 3) and x X (x – 3) = x2 – 3x are different ways of saying the same thing.

(x2 – 3x) ÷ (x – 3) = x What do I need to multiply (x – 3) by to equal (x2 – 3x)? Answer = x because x X (x – 3) = x2 – 3x So (x2 – 3x) ÷ (x – 3) = x and x X (x – 3) = x2 – 3x are different ways of saying the same thing.

(x2 + 5x + 6) ÷ (x + 2) = (x + 3) What do I need to multiply (x + 2) by to equal (x2 + 5x + 6)? Answer = (x + 3) because (x + 2) X (x + 3) = x2 + 5x + 6 So (x2 + 5x + 6) ÷ (x + 2) = (x + 3) and (x + 2) X (x + 3) = x2 +5x + 6 are different ways of saying the same thing.

10

Divide x3 + x2 – 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.

x2 + 4x + 12 (x – 3) x3 + x2 + 0x - 7 x3 - 3x2 4x2 + 0x 4x2 - 12x

12x - 7 12x - 36 + 29

We know we have reached the remainder because the power of the 29 is less than the thing we are dividing by.

----------------------------------------------------------

(x3 + x2 - 7) ÷ (x – 3) = x2 + 4x + 12 remainder 29 ----------------------------------------------------------- (x3 + x2 - 7) = (x2 + 4x + 12) X (x – 3) + 29 ÷ (x – 3) ÷ (x – 3) ÷ (x – 3)

x3 + x2 – 7(x – 3)

= (x2 + 4x + 12) + 29(x – 3)

These are all different ways of saying the same thing.

11

Divide x3 + x2 – 7 by x – 3 using the Remainder theorem.

Substitute x = 3 to make everything but the D disappear. x3 + x2 – 7 = (Ax2 + Bx + C)(x – 3) + D 27 + 9 – 7 = something X zero + D 29 = D ------------------------------------------------------------------------------------------------ Let x = 0 to make all the ‘x’ things disappear. x3 + x2 – 7 = (Ax2 + Bx + C)(x – 3) + D 0 + 0 - 7 = (0 + 0 + C)( 0 – 3) + D -7 = -3C + D 12 = C ------------------------------------------------------------------------------------------------ Compare coefficients of x3 x3 + x2 – 7 = (Ax2 + Bx + C)(x – 3) + D 1 = A ------------------------------------------------------------------------------------------------ Compare coefficients of x2 x3 + x2 – 7 = (Ax2 + Bx + C)(x – 3) + D 1 = -3 + B 4 = B ------------------------------------------------------------------------------------------------

So we get exactly the same answer as before x3 + x2 – 7 = (1x2 + 4x + 12)(x – 3) + 29

12

Chapter 2 - 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) = 2x - 1 or as a graph.

-11-10

-9-8-7-6-5-4-3-2-10123456789

10

-6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6

y = 2x - 1

-11-10

-9-8-7-6-5-4-3-2-10123456789

10

-6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6

f(x) = 2x - 1

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 2 into, go to 2 on the side, draw straight up to the line then straight across and you can see that this particular rule turns 2 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.

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(2) means what is the output when we put 2 through the f rule? f(2) = 3 X 2 + 1 = 7 g(-1) = 1 – (-1) = 2 x is the input and f(x) is the output.

rule

x

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

14

Some mappings are many to one like f(x) = sin x or f(x) = x2, there are lots of numbers we can put in and get the same answer out.

12

sin x

30°

150°

390°

x2

+5

-5

25

-1

-0.5

0

0.5

1

0 90 180 270 360 450

y = sin x

-5

0

5

10

15

20

25

-5 -4 -3 -2 -1 0 1 2 3 4 5

f(x) = x2

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 -4.

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.

√x

9

3 and -3

16

Some mappings may not be able to give an output for

certain inputs. For example, the mapping f(x) = 1x has

no output for the number zero.

1x

0

?

-5

-4

-3

-2

-1

0

1

2

3

4

5

-5 -4 -3 -2 -1 0 1 2 3 4 5

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.

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) = x2, domain {x є R, 0 < x} f(x) = x2, 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.

rule Domain = the numbers that we can put in

Notice that the word domain has IN at the end.

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) = x2 , we can only get numbers out that are greater than or equal to zero, so the range is { x є R, 0 ≤ x}.

-5

0

5

10

15

20

25

-5 -4 -3 -2 -1 0 1 2 3 4 5

f(x) = x2

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 }.

-5

-4

-3

-2

-1

0

1

2

3

4

5

-5 -4 -3 -2 -1 0 1 2 3 4 5

y = 1/x

-1

-0.5

0

0.5

1

0 90 180 270 360 450

y = sin 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}.

21

Composite functions are when we do one rule after the other.

fg(2) means do g first on 2 then do f on the result. Notice that we do them opposite to the direction they are written. f(x) = x2 g(x) = 2x + 1 fg(2) = f(5) = 25 Notice that gf(3) gives us a different result gf(2) = g(4) = 9 so it does matter which order we do them in. f2(x) means the same as ff(x) or do the f rule twice. ff(2) = f(4) = 16

f

g

22

For a generalized version… f(x) = x2 g(x) = 2x + 1 fg(x) = f(2x + 1) = (2x + 1)2 So the function fg(x) = (2x + 1)2 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(x2) = 2x2 + 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

y x -2 0.25 -1 0.5 0 1 1 2 2 4 3 8 4 16

y x -2 0.11 -1 0.33 0 1 1 3 2 9 3 27 4 81

Chapter 3 – e and Ln We can draw the graphs of y = 2x and y = 3x . Graphs like these are called exponential functions.

0

5

10

15

20

-3 -2 -1 0 1 2 3 4 5

y = 2x

0

20

40

60

-3 -2 -1 0 1 2 3 4 5

y = 3x

Notice that they both cut the y axis at (0,1) because anything to the power of zero is one, but y = 3x gets steeper a lot quicker. The gradient of y = 2x at this point is 0.7. The gradient of y = 3x at this point is 1.1. There is a number between 2 and 3 for which the gradient at (0,1) is exactly one. We call this number e.

24

y x -2 0.14 -1 0.37 0 1 1 2.7 2 7.4 3 20 4 55

e is an irrational number like π, i.e. it goes on forever. However it is approximately equal to 2.718.

e ≈ 2.718

The graph of y = ex is called the exponential function rather than an exponential function.

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 which it approaches but never reaches.

0

5

10

15

20

25

-4 -3 -2 -1 0 1 2 3 4

y = ex

25

y = e2x y = e2x is the square of ex (because (ex)2 = e2x from the rules of indices) so everything happens quicker, i.e. it is steeper than ex and drops to zero quicker.

0

5

10

15

20

25

-4 -3 -2 -1 0 1 2 3 4

y = e2x

26

y = e½x With y = e½x everything happens slower so it is shallower than y = ex and drops to zero slower.

0

5

10

15

20

25

-4 -3 -2 -1 0 1 2 3 4

y = e½x

27

y = ex + 2 Everything jumps up 2 so it cuts through at (0, 3) and the asymptote is at 2 now.

0

5

10

15

20

25

-4 -3 -2 -1 0 1 2 3 4

y = ex + 2

28

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

0

5

10

15

20

25

-4 -3 -2 -1 0 1 2 3 4

y = 5ex

29

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

0

1

2

3

4

5

6

7

8

-3 -2 -1 0 1 2 3 4 5

y = e-x

This is a reflection of the curve y = ex in the y axis.

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 = 16000 X 1 (because anything to the zero is 1) P = 16000

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 = 9704.49

31

Example – The number of people infected with a disease varies according to the formula N = 300 - 100e-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 = 300 – 100e-0.5X0 N = 300 – 100 X 1 (because anything to the zero is 1) N = 200

32

Ln x is the shorthand that mathematicians use for Loge x. It is the inverse of ex which means that if you e a number and then Ln it, it will be back to the original number.

ex

Ln x

2

2

Remember from Chapter 2 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.

33

You can put any number you like into ex, 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 ex is the Real numbers and the range is the positive Real numbers. Domain – (x ε R) Range – (y ε R, y > 0)

ex

Real numbers

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)

34

Ln x

Positive Real numbers but

not zero

Real numbers

Example e2x+3 = 4 2x + 3 = Ln 4 2x = Ln 4 – 3

x = Ln 4 - 3

2

Ln both sides, e and Ln cancel

each other out.

Example 2Ln x + 1 = 4 2Ln x = 3

Ln x = 32

x = e1.5

e both sides, Ln and e cancel

each other out.

Note that we are not allowed to

bring the 3 inside the Ln and

calculate Ln 4-3 = Ln 1

35

Chapter 4 – Numerical methods We can solve equations like 2x + 1 = 7 and x2 + 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.

2x + 1 = 7

The root

is 3

x2 – 1x - 6 = 0

The roots

are +3

and -2

36

-10

-5

0

5

10

15

20

-5 -4 -3 -2 -1 0 1 2 3 4 5 6

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 x3 – 3x2 + 3x – 4 = 0 has a root between x = 2 and x = 3. when x = 2 x3 – 3x2 + 3x – 4 = 8 – 12 + 6 – 4 = -2 = Negative when x = 3 x3 – 3x2 + 3x – 4 = 27 – 27 + 9 – 4 = +5 = Positive One is positive and one is negative therefore there is a root between x = 2 and x = 3

38

Given that f(x) = ex 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 exsin x – 1 = Positive when x = 0.6 exsin x – 1 = Negative One is positive and one is negative therefore there is a root between x = 0.5 and x = 0.6

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 x0 into the formula, this will give us our second iteration x1, 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.

40

Show that the formula x2 – 5x - 3 = 0 can be written in the form x = √(5x+3) and use the iterative formula xn+1 = √5x n + 3 to find a root of this equation. Use x0 = 5. x2 – 5x - 3 = 0 x2 = 5x + 3 x = √5x+3 If x0 = 5 then x1 = √5X5 +3 = √28 = 5.29 x2 = √5X5.29 +3 = 5.42 x3 = √5X5.42 +3 = 5.48 …eventually x4 = 5.53 x5 = 5.53 The iterations have stopped changing so this is the root.

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.

42

y = lf(x)l The modulus of a entire function makes the bits below the x axis reflect in the x axis.

-8

-6

-4

-2

0

2

4

6

8

-3 -2 -1 0 1 2 3 4 5 6

y=l2x-3l

-8

-6

-4

-2

0

2

4

6

8

-3 -2 -1 0 1 2 3 4 5 6

y=2x-3

-15

-10

-5

0

5

10

15

20

-5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8

y=x^2 -3x -10

-15

-10

-5

0

5

10

15

20

-5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8

y=lx^2 -3x -10l

43

y = f(lxl)

The modulus of just the x bit makes the entire function

reflect in the y axis.

-6

-5

-4

-3

-2

-1

0

1

2

3

-5 -4 -3 -2 -1 0 1 2 3 4 5

y = x - 2

-6

-5

-4

-3

-2

-1

0

1

2

3

-5 -4 -3 -2 -1 0 1 2 3 4 5

y = lxl - 2

-20

-15

-10

-5

0

5

10

15

20

-5 -4 -3 -2 -1 0 1 2 3 4 5

y = 4x - x^3 -20

-15

-10

-5

0

5

10

15

20

-5 -4 -3 -2 -1 0 1 2 3 4 5

y = 4lxl - lxl^3

-20

-15

-10

-5

0

5

10

15

20

-5 -4 -3 -2 -1 0 1 2 3 4 5

y = 4lxl - lxl^3

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.

0

1

2

3

4

5

-2 -1 0 1 2 3 4

l2x-1.5l = 3

45

There will be two answers to l2x-1.5l = 3 Positive answer Negative answer +(2x – 1.5) = 3 -(2x – 1.5) = 3 2x – 1.5 = 3 -2x + 1.5 = 3 x = 2.25 x = -0.75

46

There will be two answers to l5x-2l = l2xl Positive answer Negative answer +(5x - 2) = 2x -(5x – 2) = 2x 5x – 2 = 2x -5x + 2 = 2x x = 2/3 x = 2/7 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.

47

0

5

10

15

20

25

30

-8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8

y = x²

0

5

10

15

20

25

30

-8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8

y = (x+3)²

f(x + 3) f(x+3) moves the whole graph left three place.

48

0

5

10

15

20

25

30

-6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6

y = x² + 3

0

5

10

15

20

25

30

-6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6

y = x²

f(x) + 3 f(x) + 3 moves the whole graph up three places

49

2f(x) 2f(x) stretches the graph from the x axis and makes it two times bigger in the up and down direction.

-2

-1

0

1

2

0 90 180 270 360

y = sin x

-2

-1

0

1

2

0 90 180 270 360

y = 2 sin x

50

f(2x) f(2x) squashes the graph towards the y axis so that twice as much happens in the same space.

-1

0

1

0 90 180 270 360

y = sin x

-1

0

1

0 90 180 270 360

y = sin 2x

51

Chapter 6 - Trigonometry

cosec x = 1

sin x

sec x = 1

cos x

cot x = 1

tan x

52

cosec 37 = 1

sin 37 =

10.602

= 1.66

sec 142 = 1

cos 142 =

1- 0.788

= - 1.27

cot (-63) = 1

tan (-63) =

1- 1.96

= - 0.510

Notice that we can get negative answers as well as putting in angles that are negative or greater than ninety degrees.

53

54

-1

-0.5

0

0.5

1

0 30 60 90 120 150 180 210 240 270 300 330 360

sin x

-3

-2.5

-2

-1.5

-1

-0.5

0

0.5

1

1.5

2

2.5

3

0 30 60 90 120 150 180 210 240 270 300 330 360

cosec x = 1/sin x

55

When x = 90°, sin x = 1, so 1sin x

= 11 = 1.

As we head away from the top of the curve sin x gets

smaller and smaller so 1sin x

gets bigger and bigger and

heads off towards infinity.

When we get to x = 0, sin x = 0, we can’t work out 10

so cosec is undefined at this point (and at x = 180 and 360, in fact there will be an asymptote every 180 degrees.

56

-1

-0.5

0

0.5

1

0 30 60 90 120 150 180 210 240 270 300 330 360

cos x

-3

-2

-1

0

1

2

3

0 30 60 90 120 150 180 210 240 270 300 330 360

sec x = 1/cos x

57

When x = 0°, cos x = 1, so 1cos x

= 11 = 1.

As we head away from the top of the curve cos x gets

smaller and smaller so 1cos x

gets bigger and bigger and

heads off towards infinity.

When we get to x = 90, cos x = 0, we can’t work out 10

so cosec is undefined at this point (and at x = 270 and 450, in fact there will be an asymptote every 180 degrees.

58

-3

-2

-1

0

1

2

3

0 30 60 90 120 150 180 210 240 270 300 330 360

tan x

-3

-2

-1

0

1

2

3

0 30 60 90 120 150 180 210 240 270 300 330 360

cot x = 1/tan x

59

When tan x is very small, cot x = 1tan x

is very big.

When x = 45°, tan x = 1, so 1tan x

= 11 = 1.

As we head towards x = 90, tan x gets bigger and

bigger and cot x = 1tan x

gets smaller and smaller. At

exactly x = 90 tan x is undefined and then becomes negative.

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 Θsin Θ

X 1

cos Θ =

1

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 +

1sin Θ

) =

sin Θ cos Θ ( sin Θ + cos Θsin Θ cos Θ

) =

sin Θ + cos Θ

62

Simplify cot x cosec x sec2 x + cosec2 x

= cos3 x

--------------------------------------------------------------

cot x cosec x

sec2 x + cosec2 x =

cos xsin x X 1

sin x1

cos2 x + 1sin2 x

=

cos xsin2 x

sin2 x + cos2 xsin2 x cos2 x

=

cos xsin2 x

1sin2 x cos2 x

=

cos xsin2 x

÷ 1sin2 x cos2 x

=

cos xsin2 x

X sin2 x cos2 x

1 =

cos 3 x

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+B) = tan A – tan B

1 + tan A tan B

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

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 + sin B cos A

cos A cos B cos A cos B cos A cos B – sin A sin B

cos A cos B

tan (A+B) = sin Acos A + sin B

cos B

1 - sin A sin B cos A cos B

= tan A + tan B 1 - tan A tan B

66

Use the formula for sin(A-B) to find the exact value of sin 15°. -------------------------------------------------------------- Remember that

sin 45° = 1√2

= √22

cos 45° = 1√2

= √22

sin 30° = 12 cos 30° = √3

2

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°

= √22

X √32

- 12 X √2

2

= √64

- √24

= √6 - √24

67

Given that sin A = − 35 and 180<A<270 and cos B = − 12

13 and B is

obtuse find the value of cos(A-B). --------------------------------------------------------------------- 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 = − 35 , by drawing a triangle and using Pythagoras we

can see that the value of cos A is 45 but will the sign be negative

or positive? If A is between 180 and 270 then it is in the third quadrant and

cos is negative in the third quadrant so cos A = − 45

---------------------------------------------------------------------

Similarly if cos B = − 1213

then by drawing a triangle and using

Pythagoras we can see that the value of sin B will be 513

but will it

be positive or negative? As B is obtuse it is in the second quadrant, sin is positive in the

second quadrant so sin B = + 513

--------------------------------------------------------------------- cos(A-B) = cos A cos B + sin A sin B

= (− 45) X (− 12

13) + (− 3

5) X ( 5

13)

= 3365

68

sin 2A = 2 sin A cos A

cos 2A = cos2A - sin2A = 2 cos2A – 1 = 1 – 2 sin2 A

tan 2A = 2tan A

1 + tan2 A

69

Use the formula for cos(A+B) to derive the formula for cos 2A. ---------------------------------------------------------------- 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 2A = cos2 A – sin2 A as required

70

Show that cos 2A = 2cos2 A – 1 using cos 2A = cos2 A – sin2 A. -------------------------------------------------------------- Remember that sin2 A + cos2 A = 1 so sin2 A = 1 – cos2 A -------------------------------------------------------------- cos 2A = cos2 A – sin2 A = cos2 A – (1 – cos2 A) = 2cos2 A – 1 as required

71

Rewrite 2 sin 15° cos 15° as a single trigonometric ratio. -------------------------------------------------------------- 2 sin A cos A = sin 2A 2 sin 15° cos 15° = sin (2X15°)

= sin 30°

72

Given that cos x = ½ find the exact value of cos 2x. ----------------------------------------------------------------

cos 2x = 2cos2 x - 1 = 2(½)2 - 1 = -½

73

Given that cos x = 34 and 180<x<360 find the exact

value of sin 2x. -----------------------------------------------------------------------

sin 2x = 2 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 = − √7 4

----------------------------------------------------------------------- sin 2x = 2 sin x cos x

= 2 X − √7 4 X 3

4

74

By expanding sin(2A+A) show that sin 3A = 3 sin A – 4sin3 A ______________________________________________ Remember that

sin (A+B) = sin A cos B + sin B cos A sin 2A = 2 sin A cos A cos 2A = cos2 A – sin2 A

--------------------------------------------------------------- sin (2A+A) = sin 2A cos A + sin A cos 2A

= (2 sin A cos A)cos A + sin A (cos2 A –

sin2 A) = 2 sin A cos2 A + sin A cos2 A – sin3 A = 3 sin A cos2 A – sin3 A = 3 sin A (1 – sin2 A) – sin3 A = 3 sin A – 4 sin3 A

75

Prove the identity tan 2A = 2

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 2A = 2 tan A

1 – tan2 A

Divide everything by tan A

= 2

1tan A – tan A

= 2

cot A – tan A as required

76

Prove the identity tan 2A = 2

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 2A = 2 tan A

1 – tan2 A

Divide everything by tan A

= 2

1tan A – tan A

= 2

cot A – tan A as required

77

Given that x = 3 sin Θ and y = 3 – 4 cos 2Θ, eliminate Θ and express y in terms of x. ______________________________________________ Remember that cos 2Θ = 1 – 2 sin2 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 = x3

----------------------------------------------------------------

cos 2Θ = 3 - y4

1 – 2 sin2 Θ = 3 - y4

Substitute in the first equation

1 – 2 (x3)2 = 3 - y

4

and rearrange

y = 8(x3)2 - 1

78

Solve 3 cos 2x – cos x + 2 = 0 for 0° < x < 360°. ______________________________________________ Remember that cos 2x = 2 cos2 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 2x as we can’t have a quadratic that mixes up cos 2x and cos x.

---------------------------------------------------------------- 3 cos 2x – cos x + 2 = 0 3 (2 cos2 x – 1) – cos x + 2 = 0 6 cos2 x – 3 – cos x + 2 = 0

6 cos2 x – cos x – 1 = 0 A quadratic in cos x

(3 cos x + 1)(2 cos x – 1) = 0 3 cos x + 1 = 0 or 2 cos x – 1 = 0

cos x = − 13 cos x = 12

x = 109.5, 250.5 x = 60, 300 x = 60, 109.5, 250.5, 300

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 = 4 -------------------------------------------------------------- R sin α

R cos α = 4

3

sin α

cos α = 4

3

tan α = 4

3

α = 53.1° --------------------------------------------------------------

R cos α = 3 and R sin α = 4 Square both sides

R2 cos2 α = 9 and R2 sin2 α = 16 Add together

R2 cos2 α + R2 sin2 α = 9 + 16 R2 (cos2 α + sin2 α) = 25

Because cos2 α + sin2 α = 1 R2 = 25 R = 5 -------------------------------------------------------------- 3 sin x + 4 cos x = 5 sin (x – 53.1)

80

Express 7 cos Θ - 24 sin Θ in the form R cos(x + α). ____________________________________________ R cos (Θ + α) = 7 cos Θ – 24 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 = 24 --------------------------------------------------------------

R sin αR cos α

= 247

sin αcos α

= 247

tan α = 247

α = ° --------------------------------------------------------------

R cos α = 7 and R sin α = 24 Square both sides

R2 cos2 α = 49 and R2 sin2 α = 576 Add together

R2 cos2 α + R2 sin2 α = 49 + 576 R2 (cos2 α + sin2 α) = 625

Because cos2 α + sin2 α = 1

R2 = 625 R = 25 -------------------------------------------------------------- 7 cos Θ – 24 sin Θ = 25 cos (Θ

81

sin P + sin Q = 2 sin P + Q

2 cos

P - Q 2

sin P - sin Q = 2 cos P + Q

2 sin

P - Q 2

cos P + cos Q = 2 cos P + Q

2 cos

P - Q 2

cos P - cos Q = -2 sin P + Q

2 sin

P - Q 2

82

Use the formula for sin (A + B) and sin (A – B) to derive the result that

sin P + sin Q = 2 sin P + Q

2 cos

P - Q 2

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) = 2 sin A cos B -------------------------------------------------------------- Let A + B = P and A – B = Q

then A = P + Q

2 and B =

P - Q 2

so

so sin P + sin Q = 2 sin P + Q

2 cos

P - Q 2

as required

83

Use the formula for sin (A + B) and sin (A – B) to derive the result that

sin P + sin Q = 2 sin P + Q

2 cos

P - Q 2

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) = 2 sin A cos B -------------------------------------------------------------- Let A + B = P and A – B = Q

then A = P + Q

2 and B =

P - Q 2

so

so sin P + sin Q = 2 sin P + Q

2 cos

P - Q 2

as required

84

85

Chapter 8 – Differentiation

86

The Chain rule We can use the chain rule when we want to differentiate one thing that is inside another. For example (2x3 + 1)5 e2x Ln(4x7- 3) are all functions inside functions a bit like Russian dolls.

The Chain rule

dydx

= dydu

X dudx

Notice that the ‘du’s cancel a bit like fractions.

87

If y = (2x3 + 1)5 what is dydx

?

____________________________________________ If we let u = 2x3 + 1 then using the Chain rule.

y = u5 u = 2x3 + 1

dydu

= 5u4 dudx

= 6x2

--------------------------------------------------------------

dydx

= dydu

X dudx

dydx

= 5u4 X 6x2

dydx

= 6x2 X 5u4

dydx

= 30x2(2x3 + 1)4

88

In general if we have

y = (something)number

dydx

= n x something differentiated X

(something)n - 1

y = (3x + 2)7

dydx

= 7 x 3 X (3x + 2)6

dydx

= 21(3x + 2)6

Notice that the inside stays the same.

y = (4x3 + 6x)9

dydx

= 9 X (12x2 + 6) x (4x3 + 6x)8

dydx

= 9(12x2 + 6)(3x + 2)8

89

We can rearrange the Chain rule into another useful format

dydx

= 1

dxdy

this means we can work out dxdy

and then flip it to find

dydx

.

If x = y2 + 3y find dydx

_______________________________________

It’s too hard to find out dydx

so let’s differentiate both

sides with respect to y instead. dxdy

= 2y + 3

so dydx

= 1

2y + 3

This means that we now need to stick in the value of y if we want to find the gradient instead of x.

90

The Product rule We use the product rule when we want to differentiate two things that are multiplying each other. For example 4x2(3x + 1)4 x3e2x 3x5 Ln(2x+1) We call the first thing u and the second thing v.

The Product rule If y = u X v then

dydx

= vdudx

+ udvdx

‘the second one left alone times the first one differentiated plus the first one left alone times the second one differentiated’

91

If y = x2(3x2 + 4)5 find dydx

.

____________________________________________

u = x2 v = (3x2 + 4)5 dudx

= 2x dvdx

= 30x(3x2 + 4)4

--------------------------------------------------------------

dydx

= vdudx

+ udvdx

dydx

= (3x2 + 4)5 X 2x + x2 X 30x(3x2 + 4)4

take the common factors of 2x and (3x2 + 4)4 down the front

dydx

= (2x) X (3x2 + 4)4 X (1 + 15x)

dydx

= 2x(1 + 15x)(3x2 + 4)4

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(2x + 1)(3x2 + 1)5

We call the thing on top u and the thing underneath v.

The Quotient rule

If y = uv

dydx

= vdu

dx – udvdx

v2

93

If y = 5x2

(3x -1) find dy

dx.

____________________________________________

Use the Quotient rule as it’s a fraction.

u = 5x2 v = 3x – 1 dudx

= 10x dvdx

= 3

--------------------------------------------------------------

dydx

= vdu

dx – udvdx

v2

dydx

= (3x-1))(10x) – (5x2)(3)

(3x-1)2

dydx

= (15x2 – 10x)

(3x-1)2

94

If y = ex then dydx

also = ex.

Differentiation of ex and esomething

If y = e3x then dydx

= 3e3x.

If y = ex² then dydx

= 2xex².

If y = esomething then

dydx

= (something differentiated) X esomething

Notice that the power doesn’t decrease by one like it

normally does.

95

We could do esomething questions the long way by using

the Chain rule, dydx

= dydu

X dudx

.

If y = e3x what is dydx

?

____________________________________________ If we let u = 3x then using the Chain rule.

y = eu u = 3x

dydu

= eu dudx

= 3

--------------------------------------------------------------

dydx

= dydu

X dudx

dydx

= eu X 3

dydx

= 3eu

dydx

= 3e3x

96

If y = Ln x then dydx

= 1x

Differentiation of Ln(x) and Ln(something)

If y = Ln(3x +1) then dydx

= 33x + 1

If y = Ln(2x3 +5x) then dydx

= 6x + 52x3 + 5x

If y = Ln(something)

dydx

= something differentiated

something left alone

97

Again we can differentiate Ln questions the long way by

using the chain rule dydx

= dydu

X dudx

.

If y = Ln(3x2 + 4x) what is dydx

?

____________________________________________ If we let u = 3x2 + 4x then using the Chain rule.

y = Ln(u) u = 3x2 + 4x

dydu

= 1u

dudx

= 6x + 4

--------------------------------------------------------------

dydx

= dydu

X dudx

dydx

= 1u X (6x + 4)

dydx

= 6x + 4

u

dydx

= 6x + 4

3x2 + 4x

98

If y = sin x then dydx

= cos x

Differentiation of Trig functions All of the following only work when x is in radians!

If y = cos x then dydx

= - sin x

If y = tan x then dydx

= sec2 x

If y = cosec x then dydx

= -cosec x cot x

If y = sec x then dydx

= sec x tan x

If y = cot x then dydx

= -cosec2 x

99

If y = sin 3x then find dydx

using the Chain rule.

-------------------------------------------------------------- y = sin u u = 3x

dydu

= cos u dudx

= 3

--------------------------------------------------------------

dydx

= dydu

X dudx

dydx

= cos u X 3

dydx

= 3 cos 3x

Notice that the inside, the 3x, has stayed the same.

In general if y = sin (something)

dydx

= something differentiated X cos (something)

100

If y = sin 5x

dydx

= 3 x cos 3x

If y = cos 7x

dydx

= 7 x - sin 7x

If y = tan 5x

dydx

= 5 x sec2 5x

If y = cosec 6x

dydx

= 6 X – cosec 6x cot 6x

101

If we write y = sin3 x this actually means y = (sin x)3

We can find dydx

using the Chain rule.

-------------------------------------------------------------- y = u3 u = sin x

dydu

= 3u2 dudx

= cos x

--------------------------------------------------------------

dydx

= dydu

X dudx

dydx

= 3u2 X cos x

dydx

= 3(sin x)2 X cos x

dydx

= 3 cos x(sin x)2

102