TRIGONOMETRY
MEMORIZE THE VALUES
0
45
30
sin
0
cos
1
2
tan
o
g
1
make
table
0
30
sin
F
as
F
Trick
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to
90
60
undefined
60
90
if
F
1
F
F
EE
45
E
Ej's Igined
GRAPHS
I
1
do
sin
0
COS
I
3
182
1
1
7211270
1
316
1
P
ts
tan
1
1
in
sty
a
EQUATION
SOLVING
Imipletiescfnieptpafutw.iq
study
in A Levels
90
Ytre
180
IIE
38
270
Step1 Take inverse ignoring all
and find α
Step2 DecideQuadrant
4 signs
2 180 α
α
and then apply
this
1
180 2
Sin
x
2 360
360
2
sñ
x 180 2
α
30
N
210,330
2182703
2 360 α
2 360 30
x 330
α
tank
2
x
o
360
Quadrant 2 and4
T
tan
2 180 2
60
α
N
x
2 360 60
x 300
120 300
2
505k
052
03
4
Quadrant 2,4
2
05
3
66.42
s
x 180 α
2 180 2
T
180
X
180
α
360 α
180 66.42
113.58
180 66.42 246.42
A
This is some part of a
DIFFERENTIATION
complete concept
in A Levels
you will study
gradient
DIFFERENTIATION
g
CURVE TANGENT
oftangent
GRADIENT OF
BASIC RULES
21
1
22
2
52
5
331
3
1
2
1 5,1 gm
2
0
3
0
MAIN RULE
STEP 1
BRING POWER BACK AND MULTIPLY
STEP 2
SUBTRACT ONE
FROM
POWER
EXAMPLES
a
y
41
142
12 2
1214123
48 3
2
y
gdy
qty
3
x
4
3
4 3
0
Y
1b
523
3R
1
1522
If
it
12
3 2 k
513 k
STATIONARY POINT
5
122
bx
0
12
NATURE
AND
TURWING POINT
MAX
day
o gives
stationary
Min
dhy
you
points
decides whether it
dk
a
or
maximum
is a minimum
nature
nature turning point
Min
4
y
Max
2
DIFF
F
dya
to
statilnatypothts
Min
1
2
Oman
y 23 x2
1
3
2
22
322 22 1
0
1h
x
3227022 1
0
32
2
1
32 RTI
1 2 1
1
K 1 32
K 1
2
y
y
23 x2 x
13
I
1
0
0
1
0
32
0
1
0
1
we already know
this one
112 41
I
l
I
1
x 0 Y 0
4
id
t.gg minimum
Minimum
orig Y
Y K
K 71
Y 2 x 427
1 2K x u 7
110
2101
18 1
11
2
Horizontal
K
17
STEM
AND LEAF
get rid of repeats
we
1 17
1.21
1.31
1 18
127
1 32
1 43
1.47
1 19
1 23
1.37
1.49
must be singledigit
mustbeinteger
Leaf
stem
11
Median
DIAGRAMS
7
lastsignificant figuregenerally
8
12
2 05ᵗʰ
13
2 709ᵗʰ
14
7 9
n
lowerQuartile
upperduartile 3
term.fi
f
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120
thtem o.sttemT
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jhtem 3 Ejten 9nstent
InterquartileRange
UQ
1.4
L0
1.2
0.2
EF
10
4,4
12
0 4,508h
13
2,4
14
15
2 5012ᵗʰ
16
0
LO
n
n
124breathan
8
2
ten
Median
3
1214means
7
11
UQ
key
9
1
152
ten
ten 3
15,11
15,11
Interquartile
tem 8thterm
Median
term 4ᵗʰterm
LQ
term 12ᵗʰterm
UQ
range
v0
LO
145
115
125
115
145
30
CONDITIONAL PROBABILITY
WILL HAVE TWO EVENTS
YOU
A
FIRST
B SECOND
GIVEN THAT SECOND EVENT HAS ALREADYHAPPENED
FIND PROBABILITY OF FIRST EVENT
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pantage
I
5
After condition
iEe
T.EE
Year
that there was a hailstorm
to
to
that day
now treediagram has
branches available
only two
Findyour favouvable branch
Plevent
Favourable
Total
I
88
70 5
pantage
2
To