Download Chapter 1 Algebraic fractions Page 11 Chapter 2 Functions Page 22

Chapter 1
Algebraic
fractions
Page 11
Chapter 8
Chapter 2
Differentiation
Functions
Page 88
Page 22
C3
Chapter 7
Further Trig
Page 77
Chapter 3
Exponentials
and
Logarithms
Page 33
Chapter 4
Chapter 6
Numerical
methods
Trigonometry
Page 66
Chapter 5
Transforming
graphs
Page 55
Page 44
Chapter 1 – Algebraic fractions
When we simplify fractions we can factorise and cancel
out common factors from top and bottom.
12
20
4X3
=
x+3
2x + 6
4X5
=
3
5
1 X (x + 3)
=
2 X (x + 3)
=
1
2
x+2
x+3
This can’t be simplified and we definitely can’t cancel
just the ‘x’s!!!
x+2
x+3
2
=
2
3
X
When you have a fraction on the top or bottom it is a
good idea to multiply to get rid of it.
1
x
2
+1
3x + 6
=
1
2
2 X ( x + 1)
2 X (3x + 6)
=
x+2
6x + 12
=…
and now we can continue as normal.
2x + 6
1
x
3
+1
=
3 X (2x + 6)
1
3
3 X ( 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
2
1
x
3
+4
+5
=
1
2
1
( x
3
6 X ( x + 4)
6X
+ 5)
=
3x + 24
2x + 30
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 – 3x
x2 - 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.
x2 + 5x + 4
2
x + 8x + 16
4
=
(x + 1) X (x + 4)
(x + 4) X (x + 4)
=
(x + 1)
(x + 4)
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.
2
3
3
5
X
X
5
=
7
5
=
7
x+2
x+4
X
10
x
21
y
3
a
7
b
x2 - 16
3x + 6
=
(x + 2)
(x - 4)
X
X
X
x
z
c
a
=
=
x2
yz
c
b
(x + 4)(x – 4)
3(x + 2)
=
(x – 4)
3
5
When we divide fractions we flip the second fraction
and continue as normal.
2
3
a
b
÷
÷
5
7
c
d
x2 – 3x
y2 + y
=
=
÷
2
3
a
b
X
X
x
y+1
7
5
d
c
=
=…
=…
x2 – 3x
y2 + y
x
y+1
x
= …
One thing we can’t do is cancel before we flip the
second fraction!
p2
r-1
6
÷
r+1
p
=
p
r-1
÷
r+1
1
=
…
When we add or take away fractions we must first
make the bottoms the same.
We don’t always need
2
3
+
=
numbers!
5
X5
=
to multiply by both
4
3
X3
10
15
+
10
4
+
5
12
X2
15
22
=
15
3
+
10
8
10
13
We
= don’t always need
10
to multiply by all the
letters!
a
b
+
Xd
=
=
e
c
fg
d
Xk
Xb
ad
bd
+
ad + bc
bd
+
bc
bd
=
=
ek
fgk
h
fk
Xg
+
gh
fgk
ek +gh
fgk
7
We can think about dividing numbers as
“What number do I need to multiply the small
number by to equal the big number?”
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.
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
8
different ways of saying the same thing.
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.
9
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.
(x – 3)
x2
+
4x
+
12
x3
+
x2
+
0x
x3
-
3x2
+
0x
-
12x
4x2
2
4x
- 7
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) =
÷ (x – 3)
(x2 + 4x + 12) X (x – 3) + 29
÷ (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.
10
Divide x3 + x2 – 7 by x – 3 using the Remainder theorem.
Substitute x = 3 to make everything but the D disappear.
x3 +
x2 – 7
27 + 9 – 7
=
(Ax2 + Bx + C)(x – 3)
+D
=
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
=
(0 + 0 + C)( 0 – 3)
+D
-7
-7
=
-3C
12
=
C
+D
-----------------------------------------------------------------------------------------------Compare coefficients of x3
x3 + x2 – 7
(Ax2 + Bx + C)(x – 3)
=
1
=
+D
A
-----------------------------------------------------------------------------------------------Compare coefficients of x2
x3 + x2 – 7
(Ax2 + Bx + C)(x – 3)
=
1
=
-3 + B
4
=
B
+D
-----------------------------------------------------------------------------------------------So we get exactly the same answer as before
x3 + x2 – 7 = (1x2 + 4x + 12)(x – 3) + 29
11
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.
y = 2x - 1
f(x) = 2x - 1
10
9
8
7
6
5
4
3
2
1
0
-1
-6 -5 -4 -3 -2 -2
-1 0 1 2 3 4 5 6
-3
-4
-5
-6
-7
-8
-9
-10
-11
10
9
8
7
6
5
4
3
2
1
0
-1
-6 -5 -4 -3 -2 -2
-1 0 1 2 3 4 5 6
-3
-4
-5
-6
-7
-8
-9
-10
-11
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.
12
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.
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
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.
30°
150°
sin x
1
2
x2
25
390°
+5
-5
y = sin x
f(x) = x2
1
25
20
0.5
15
0
10
0
90
180
270
360
450
5
-0.5
0
-5 -4 -3 -2 -1 0
-5
-1
1
2
3
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.
14
4
5
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
Some mappings may not be able to give an output for
certain inputs. For example, the mapping f(x) =
no output for the number zero.
1
x
0
?
y = 1/x
5
4
3
2
1
0
-5
-4
-3
-2
-1 -1 0
-2
-3
-4
-5
16
1
2
3
4
5
1
x
has
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
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.
Domain = the
numbers that
rule
we can put in
Notice that the word domain has IN at the end.
18
The Range tells us what numbers we could get out of
our function.
Range = the
rule
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}.
f(x) = x2
25
20
15
10
5
0
-5
-4
-3
-2
-1 0
-5
1
2
3
4
5
19
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
5
4
3
2
1
0
-5
-4
-3
-2
-1-1 0
1
2
3
4
5
-2
-3
-4
-5
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}.
y = sin x
1
0.5
0
0
-0.5
-1
20
90
180
270
360
450
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
g
f
21
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}.
22
Chapter 3 – e and Ln
We can draw the graphs of y = 2x and y = 3x . Graphs
like these are called exponential functions.
y
-2
-1
0
1
2
3
4
x
0.25
0.5
1
2
4
8
16
y
-2
-1
0
1
2
3
4
x
0.11
0.33
1
3
9
27
81
y = 2x
20
15
10
5
0
-3
-2
-1
0
1
2
3
4
5
3
4
5
y = 3x
60
40
20
0
-3
-2
-1
0
1
2
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.
23
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.
y
-2
-1
0
1
2
3
4
x
0.14
0.37
1
2.7
7.4
20
55
y = ex
25
20
15
10
5
0
-4
-3
-2
-1
0
1
2
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
24
which it approaches but never reaches.
3
4
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.
y = e2x
25
20
15
10
5
0
-4
-3
-2
-1
0
1
2
3
4
25
y = e½x
With y = e½x everything happens slower so it is
shallower than y = ex and drops to zero slower.
y = e½x
25
20
15
10
5
0
-4
26
-3
-2
-1
0
1
2
3
4
y = ex + 2
Everything jumps up 2 so it cuts through at (0, 3) and
the asymptote is at 2 now.
y = ex + 2
25
20
15
10
5
0
-4
-3
-2
-1
0
1
2
3
4
27
y = 5ex
For this curve the asymptote is still zero but now it cuts
through at (0, 5)
y = 5ex
25
20
15
10
5
0
-4
28
-3
-2
-1
0
1
2
3
4
y = e-x
This type of curve is sometimes called exponential
decay and appears in the decay of radioactivity.
y = e-x
8
7
6
5
4
3
2
1
0
-3
-2
-1
0
1
2
3
4
5
This is a reflection of the curve y = ex in the y axis.
29
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
30
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
31
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.
2
ex
Ln x
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.
32
2
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)
Positive Real
Real numbers
ex
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
Positive Real
numbers but
not zero
Example
e2x+3 = 4
Ln x
Real numbers
Ln both sides, e and Ln cancel
each other out.
2x + 3 = Ln 4
2x = Ln 4 – 3
x=
Ln 4 - 3
2
Note that we are not allowed to
bring the 3 inside the Ln and
calculate Ln 4-3 = Ln 1
Example
2Ln x + 1 = 4
2Ln x = 3
Ln x =
3
2
x = e1.5
34
e both sides, Ln and e cancel
each other out.
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.
The root
is 3
2x + 1 = 7
The roots
are +3
x2 – 1x - 6 = 0
and -2
35
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.
20
15
10
5
0
-5
-4
-3
-2
-1
0
1
2
3
4
5
6
-5
-10
Notice that there may be more than one root so the
signs might go from positive to negative and back
36 again.
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
37
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
38
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.
39
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.
40
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
y = lf(x)l
The modulus of a entire function makes the bits
below the x axis reflect in the x axis.
y=2x-3
y=l2x-3l
8
8
6
6
4
4
2
2
0
0
-3 -2 -1
-2 0
1
2
3
4
5
6
-4
-4
-6
-6
-8
-8
y=x^2 -3x -10
1
2
3
4
5
6
y=lx^2 -3x -10l
20
20
15
15
10
10
5
5
0
0
-5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8
-5
42
-3 -2 -1
-2 0
-5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8
-5
-10
-10
-15
-15
y = f(lxl)
The modulus of just the x bit makes the entire function
reflect in the y axis.
y=x-2
y = lxl - 2
3
3
2
2
1
1
0
0
-5 -4 -3 -2 -1
-1 0
1
2
3
4
5
-5 -4 -3 -2 -1-1 0
y = 4lxl - lxl^3
20
1
2
3
4
515
-2
-2
10
-3
-3
5
-4
-4
0
-5
-5
-6
-6
-5 -4 -3 -2 -1
-5 0
1
-10
-15
y = 4lxl - lxl^3
y = 4x - x^3
20
20
15
15
10
10
5
5
0
0
-5 -4 -3 -2 -1
-5 0
1
2
3
4
5
-5 -4 -3 -2 -1
-5 0
-10
-10
-15
-15
-20
-20
1
2
3
-20
4
5
43
2
3
4
5
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.
l2x-1.5l = 3
5
4
3
2
1
0
-2
44
-1
0
1
2
3
4
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
x = 2.25
-2x + 1.5 = 3
x = -0.75
45
There will be two answers to l5x-2l = l2xl
Positive answer
Negative answer
+(5x - 2) = 2x
-(5x – 2) = 2x
5x – 2 = 2x
x = 2/3
-5x + 2 = 2x
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.
46
f(x + 3)
f(x+3) moves the whole graph left three place.
y = x²
y = (x+3)²
30
30
25
25
20
20
15
15
10
10
5
5
0
0
-8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8
-8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8
47
f(x) + 3
f(x) + 3 moves the whole graph up three places
y = x² + 3
30
30
25
25
20
20
15
15
10
10
5
5
0
0
-6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6
48
y = x²
-6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6
2f(x)
2f(x) stretches the graph from the x axis and makes it
two times bigger in the up and down direction.
y = sin x
2
1
0
0
90
180
270
360
270
360
-1
-2
y = 2 sin x
2
1
0
0
90
180
-1
-2
49
f(2x)
f(2x) squashes the graph towards the y axis so that
twice as much happens in the same space.
y = sin x
1
0
0
90
180
270
360
270
360
-1
y = sin 2x
1
0
0
-1
50
90
180
Chapter 6 - Trigonometry
cosec x =
sec x =
cot x =
1
sin x
1
cos x
1
tan x
51
cosec 37 =
sec 142 =
cot (-63) =
1
sin 37
1
cos 142
1
tan (-63)
=
=
=
1
0.602
= 1.66
1
- 0.788
1
- 1.96
= - 1.27
= - 0.510
Notice that we can get negative answers as well as
putting in angles that are negative or greater than
ninety degrees.
52
53
sin x
1
0.5
0
-0.5
0
30
60
90
120
150
180
210
240
270
300
330
360
270
300
330
360
-1
cosec x = 1/sin x
3
2.5
2
1.5
1
0.5
0
-0.5 0
-1
-1.5
-2
-2.5
-3
54
30
60
90
120
150
180
210
240
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
gets bigger and bigger and
heads off towards infinity.
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
cos x
1
0.5
0
0
30
60
90
120
150
180
210
240
270
300
330
360
270
300
330
360
-0.5
-1
sec x = 1/cos x
3
2
1
0
0
-1
-2
-3
56
30
60
90
120
150
180
210
240
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
gets bigger and bigger and
heads off towards infinity.
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 = 270 and
450, in fact there will be an asymptote every 180
degrees.
57
tan x
3
2
1
0
-1
0
30
60
90
120
150
180
210
240
270
300
330
360
270
300
330
360
-2
-3
cot x = 1/tan x
3
2
1
0
-1
-2
-3
58
0
30
60
90
120
150
180
210
240
When tan x is very small, cot x =
When x = 45°, tan x = 1, so
1
tan x
1
tan x
=
1
1
is very big.
= 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
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
1
60
cos Θ
sin Θ
X
=
1
cos Θ
=
Simplify sin Θ cos Θ (sec Θ + cosec Θ)
-------------------------------------------------------------sin Θ cos Θ (sec Θ + cosec Θ) =
sin Θ cos Θ (
1
cos Θ
sin Θ cos Θ (
sin Θ + cos Θ
+
1
sin Θ
sin Θ + cos Θ
sin Θ cos Θ
)
=
)
=
61
Simplify
cot x cosec x
2
2
sec x + cosec x
= cos3 x
-------------------------------------------------------------cot x cosec x
=
sec2 x + cosec2 x
cos x
1
X
sin x
sin x
1
1
+ 2
cos2 x
sin x
=
cos x
sin2 x
sin2 x + cos2 x
sin2 x cos2 x
=
cos x
sin2 x
1
sin2 x cos2 x
=
cos x
sin2 x
÷
cos x
X
sin2 x
cos
62
3
x
1
=
sin2 x cos2 x
sin2 x cos2 x
1
=
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
63
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
Using the formula for sin(A+B) and cos(A+B)
derive the formula for tan(A+B).
--------------------------------------------------------------
tan (A+B) =
tan (A+B) =
sin(A+B)
cos(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) =
tan (A+B) =
=
sin 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
cos A cos B
sin A
sin B
+
cos A
cos B
sin A sin B
1 cos A cos B
tan A + tan B
1 - tan A tan B
65
Use the formula for sin(A-B) to find the exact
value of sin 15°.
-------------------------------------------------------------Remember that
sin 45° =
sin 30° =
1
√2
=
1
√2
2
cos 45° =
cos 30° =
2
1
√2
=
√2
2
√3
2
and
sin(A-B) = sin A cos B - sin B cos A
-------------------------------------------------------------sin 15° =
66
sin (45°–30°)
=
sin 45° cos 30° – sin 30° cos 45°
=
√2
=
√6
=
√6 - √2
2
4
X
-
4
√3
2
√2
4
-
1
2
X
√2
2
Given that sin A = −
3
5
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 = −
3
5
, by drawing a triangle and using Pythagoras we
can see that the value of cos A is
4
5
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 = −
4
5
--------------------------------------------------------------------Similarly if cos B = −
12
13
then by drawing a triangle and using
Pythagoras we can see that the value of sin B will be
5
13
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 = +
5
13
--------------------------------------------------------------------cos(A-B) = cos A cos B + sin A sin B
4
12
3
5
5
13
5
13
= (− ) X (−
=
) + (− ) X ( )
33
65
67
sin 2A = 2 sin A cos A
cos 2A =
=
2 cos2A – 1
=
1 – 2 sin2 A
tan 2A =
68
cos2A - sin2A
2tan A
1 + tan2 A
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
69
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
70
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°
71
Given that cos x = ½ find the exact value of cos 2x.
---------------------------------------------------------------cos 2x
= 2cos2 x - 1
= 2(½)2 - 1
= -½
72
Given that cos x =
3
4
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
=2X−
3
√7
X
4
4
73
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
74
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
1
tan A
– tan A
2
cot A – tan A
as required
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
=
=
76
2
1
tan A
– tan A
2
cot A – tan A
as required
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 =
x
3
---------------------------------------------------------------cos 2Θ =
3-y
4
1 – 2 sin2 Θ =
3-y
4
Substitute in the first equation
x
3-y
3
4
1 – 2 ( )2 =
and rearrange
x
y = 8( )2 - 1
3
77
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
cos x = −
1
3
x = 109.5, 250.5
or
2 cos x – 1 = 0
cos x =
2
x = 60, 300
x = 60, 109.5, 250.5, 300
78
1
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
sin α
cos α
=
=
4
3
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)
79
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 α
R cos α
sin α
cos α
=
=
7
24
7
tan α =
α=
24
24
7
°
-------------------------------------------------------------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
80
sin P + sin Q = 2 sin
sin P - sin Q = 2 cos
P+Q
2
P+Q
cos P + cos Q = 2 cos
cos P - cos Q = -2 sin
2
P+Q
2
P+Q
2
P-Q
cos
sin
2
P-Q
cos
sin
2
P-Q
2
P-Q
2
81
Use the formula for sin (A + B) and sin (A – B) to
derive the result that
sin P + sin Q = 2 sin
P+Q
cos
2
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
then A =
so
82
P+Q
2
and
and
A–B=Q
B=
sin P + sin Q = 2 sin
P-Q
P+Q
2
2
cos
so
P-Q
2
as required
Use the formula for sin (A + B) and sin (A – B) to
derive the result that
sin P + sin Q = 2 sin
P+Q
cos
2
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
then A =
so
P+Q
2
and
and
A–B=Q
B=
sin P + sin Q = 2 sin
P-Q
P+Q
2
2
cos
so
P-Q
2
as required
83
84
Chapter 8 – Differentiation
85
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
dy
dx
=
dy
du
X
du
dx
Notice that the ‘du’s cancel a bit like fractions.
86
If y = (2x3 + 1)5 what is
dy
dx
?
____________________________________________
If we let u = 2x3 + 1 then using the Chain rule.
y =
dy
du
u5
u = 2x3 + 1
du
= 5u4
dx
= 6x2
-------------------------------------------------------------dy
dx
dy
dx
=
dy
du
X
du
dx
= 5u4 X 6x2
dy
dx
dy
dx
= 6x2 X 5u4
= 30x2(2x3 + 1)4
87
In general if we have
(something)number
y =
dy
=
dx
n
x
something
differentiated
(something)n - 1
y =
(3x + 2)7
dy
7 x 3 X (3x + 2)6
dx
dy
dx
=
21(3x + 2)6
=
Notice that the inside stays the same.
y =
dy
dx
dy
dx
88
=
=
(4x3 + 6x)9
9 X (12x2 + 6) x (4x3 + 6x)8
9(12x2 + 6)(3x + 2)8
X
We can rearrange the Chain rule into another useful
format
dy
dx
=
1
dx
dy
this means we can work out
dx
dy
and then flip it to find
dy
.
dx
If x = y2 + 3y find
dy
dx
_______________________________________
It’s too hard to find out
dy
dx
so let’s differentiate both
sides with respect to y instead.
dx
dy
= 2y + 3
so
dy
dx
=
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.
89
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
dy
dx
‘the
du
dx
=v
second
dv
dx
+u
one
left
alone
times
the
first
one
differentiated plus the first one left alone times the
second one differentiated’
90
If y = x2(3x2 + 4)5 find
dy
dx
.
____________________________________________
u = x2
du
dx
v = (3x2 + 4)5
dv
= 2x
dx
= 30x(3x2 + 4)4
-------------------------------------------------------------dy
dx
dy
dx
=v
du
dx
+u
dv
dx
= (3x2 + 4)5 X 2x
+
x2 X 30x(3x2 + 4)4
take the common factors of 2x and (3x2 + 4)4 down the
front
dy
dx
= (2x) X (3x2 + 4)4 X (1 + 15x)
dy
dx
= 2x(1 + 15x)(3x2 + 4)4
91
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
u
If y =
dy
dx
92
=
v
v
du
dx
–u
v2
dv
dx
5x2
If y =
(3x -1)
find
dy
dx
.
____________________________________________
Use the Quotient rule as it’s a fraction.
u = 5x2
du
dx
v = 3x – 1
dv
= 10x
dx
=3
-------------------------------------------------------------dy
dx
dy
dx
dy
dx
=
=
=
du
dx
v
–u
v2
dv
dx
(3x-1))(10x) – (5x2)(3)
(3x-1)2
(15x2 – 10x)
(3x-1)2
93
Differentiation of ex and esomething
If y = ex then
dy
dx
also = ex.
If y = esomething then
dy
dx
= (something differentiated) X esomething
Notice that the power doesn’t decrease by one like it
normally does.
If y = e3x then
If y = ex² then
94
dy
dx
dy
dx
= 3e3x.
= 2xex².
We could do esomething questions the long way by using
dy
the Chain rule,
dx
If y = e3x what is
=
dy
dx
dy
du
X
du
dx
.
?
____________________________________________
If we let u = 3x then using the Chain rule.
eu
y =
dy
u = 3x
du
= eu
du
dx
=3
-------------------------------------------------------------dy
dx
dy
dx
dy
dx
dy
dx
=
dy
=
du
X
du
dx
eu X 3
= 3eu
= 3e3x
95
Differentiation
of
Ln(x)
Ln(something)
If y = Ln x then
dy
dx
=
1
x
If y = Ln(something)
dy
dx
=
something differentiated
something left alone
If y = Ln(3x +1) then
If y = Ln(2x3 +5x) then
96
dy
dx
dy
dx
=
=
3
3x + 1
6x + 5
2x3 + 5x
and
Again we can differentiate Ln questions the long way by
dy
using the chain rule
dx
=
dy
du
If y = Ln(3x2 + 4x) what is
X
du
dx
dy
dx
.
?
____________________________________________
If we let u = 3x2 + 4x then using the Chain rule.
y =
dy
Ln(u)
=
du
u=
1
du
u
dx
=
3x2 + 4x
6x + 4
-------------------------------------------------------------dy
dx
dy
dx
dy
dx
dy
dx
=
=
=
dy
=
du
1
u
X
du
dx
X (6x + 4)
6x + 4
u
6x + 4
3x2 + 4x
97
Differentiation of Trig functions
All of the following only work when x is in radians!
dy
If y = sin x then
If y = cos x then
dx
dy
dx
If y = tan x then
If y = cosec x then
If y = sec x then
dy
dx
dy
dx
If y = cot x then
98
dy
dx
= cos x
= - sin x
= sec2 x
= -cosec x cot x
= sec x tan x
dy
dx
= -cosec2 x
If y = sin 3x then find
dy
dx
using the Chain rule.
-------------------------------------------------------------y = sin u
dy
du
=
u = 3x
du
cos u
dx
=3
-------------------------------------------------------------dy
dx
dy
dx
=
dy
dx
=
dy
du
X
du
dx
cos u X 3
=
3 cos 3x
Notice that the inside, the 3x, has stayed the same.
In general if y = sin (something)
dy
dx
= something differentiated X cos (something)
99
If y = sin 5x
dy
dx
= 3 x cos 3x
If y = cos 7x
dy
dx
= 7 x - sin 7x
If y = tan 5x
dy
dx
= 5 x sec2 5x
If y = cosec 6x
dy
dx
100
= 6 X – cosec 6x cot 6x
If we write y = sin3 x this actually means y = (sin
x)3
We can find
dy
dx
using the Chain rule.
-------------------------------------------------------------y = u3
dy
du
u = sin x
du
= 3u2
dx
= cos x
-------------------------------------------------------------dy
dx
dy
dx
dy
dx
=
dy
du
X
du
dx
= 3u2 X cos x
= 3(sin x)2 X cos x
dy
dx
= 3 cos x(sin x)2
101
102