Gr 12 Maths
BOOKWORK
(an extract from Gr 12 Maths 2 in 1)
All the proofs you need to know!
Paper 1:
Series
(a maximum of 6 marks)
page 1
Paper 2:
Geometry (6 theorems) and
Trigonometry (3 proofs)
(a maximum of 12 marks)
page 1 to 3
Be sure to study the geometry and trigonometry summaries on
page 4 to 6.
Calculator instructions
page 7
BOOKWORK: PROOFS
PAPER 1
PAPER 2
Circle Geometry Theorems
Arithmetic and Geometric Series
Sn , the sum of n terms
1
Arithmetic Series:
Sn = a + (a + d) + (a + 2d) + . . . + Tn
. . . n terms
& Sn = Tn + (Tn – d) + (Tn – 2d) + . . . + a
. . . n terms
∴ 2 Sn = (a + Tn) + (a + Tn) + (a + Tn) + . . . (a + Tn)
The line segment drawn from the centre of a circle,
perpendicular to a chord, bisects the chord.
Given : ?O with OP ⊥ AB
To prove : AP = PB
Construction : Join OA and OB
s
Proof :
In Δ OPA & OPB
(1) OA = OB . . . radii
(2) Pˆ 1 = Pˆ 2 (= 90º) . . . given
(3) OP is common
â ΔOAP ≡ ΔOBP . . . SøS
â AP = PB, i.e. OP bisects chord AB
. . . n terms
∴ 2 Sn = n (a + Tn)
∴ Sn = n (a + Tn) 2
Tn = a + (n – 1)d
But
∴ Sn =
n
[ a + a + (n – 1)d ]
2
2
∴ Sn = n [ 2a + (n – 1)d ] Sn = a + ar + ar
% r ) ∴ r Sn =
ar + ar
2
+ ar
+ ar
– – — : ∴ Sn – r Sn = a – ar
3
3
+ . . . . . . + ar
n–2
+ ......
+ ar
+ ar
n –1
n–1
+ ar
... –
n
ˆ at the centre and APB
ˆ at the circumference.
?O, arc AB subtending AOB
P
12
x y
Then  = x . . . radii equal
â Oˆ 1 = 2x . . . exterior ø of ΔAOP
... —
Similarly, if Pˆ 2 = y, then Oˆ 2 = 2y
ˆ = 2x + 2y = 2(x + y) = 2 APB
ˆ
â AOB
the 'middle bit' falls away.
n
B
P
ˆ = 2APB
ˆ
To prove : AOB
Construction : Join PO and produce it to Q.
Pˆ 1 = x
Proof : Let
Geometric Series:
2
1 2
A
The angle which an arc of a circle subtends at the centre is
double the angle it subtends at any point on the circumference.
Given :
2
O
O
1 2
x
y
A
Q
B
n
∴ Sn (1 – r) = a(1 – r )
∴ Sn =
a(1 − r n )
1− r
. . . for r < 1
or
×
( − 1)
a(r n − 1)
: Sn =
( − 1)
r −1
3
. . . for r > 1
To prove : Â + Ĉ = 180º & B̂ + D̂ = 180º
Construction : Join BO and DO.
− (a − b)
b − a
a − b
=
=
d − c
c − d
− (c − d)
Proof :
 =
Let
Then
â Ĉ =
The derivation of the future and present value formulae, Fv and Pv, using the
sum formula of a geometric sequence, must be understood, but is not examinable.
See Topic 5 (Financial Maths) in the Gr 12 Maths 2 in 1 study guide.
â Â + Ĉ =
& â B̂ + D̂ =
1
x
2
O
D
1
B
. . . ø at centre = 2 %
ø at circumference
s
360º – 2x . . . ø about point O
1
(360º – 2x) = 180º – x . . . ø at centre = 2 %
2
ø at circumference
180º
s
180º
. . . sum of the ø of a
quadrilateral = 360º
ˆ = 2x
O
1
â Oˆ 2 =
Annuities:
Copyright © The Answer
x
Given : ?O and cyclic quadrilateral ABCD
Choose either depending on the value of r, whichever gives you a positive denominator.
Notice that :
A
The opposite angles of a cyclic quadrilateral are supplementary.
C
4
D
ˆ = APB
ˆ ,
A
1
To prove :
ˆ
= 90º
DAC
&
ˆ
DPA
= 90º
B
2
. . . diameter ⊥ tangent
2
. . . ø in semi-?
1
∴ Aˆ 1 + Aˆ 2 = 90º
and
6
1
P
Construction : Draw diameter AD and join DP.
Proof : We have
x
A
ˆ
ˆ
A
2 = P1
s
. . . ø in same seg.
EC
B
Q
E
F
B̂ = Ê & Ĉ = F̂
s
In Δ DPQ & ABC
. . . construction
(1) DP = AB
(2) DQ = AC . . . construction
(3)
D̂ = Â
C
But
stage 2 :
corresponding øs
s
corresponding ø equal
DP = AB and
. . . construction
â AB = AC
DE
DF
AD
= AE
DB
EC
stage 3 :
parallel lines
. . . proportion theorem
DQ = AC
A
. . . SøS
. . . given
The â PQ || EF . . .
focal
â DP = DQ
point
DE
DF
stage 1 :
congruency
. . . given
â ΔDPQ ≡ ΔABC
E
Given : ΔABC with DE || BC,
D & E on AB & AC respectively.
To prove :
AB
= AC = BC
DE
DF
EF
Proof :
A line parallel to one side of a triangle divides
the other two sides proportionally.
DB
C
To prove :
â 1̂ = B̂
= Ê
D
B
ΔABC & ΔDEF with  = D̂
A
i.e. DE || BC ² AD = AE
P 1
Given :
Remember :
Always give
REASONS for
each statement !
ˆ
∴ Aˆ 1 = Pˆ 2 ( = APB)
The Proportion Theorem
If two triangles are equiangular,
then their sides are proportional
and, therefore, they are similar.
Construction : Mark P & Q on DE & DF such that DP = AB & DQ = AC
C
Pˆ 1 + Pˆ 2 = 90º
But
D
A
ˆ ,
Given : Tangent AC and chord AB subtending APB
with C & P on opposite sides of AB.
5
The Similar Δs Theorem
The angle between a tangent to a circle and a chord drawn from the point of
contact is equal to the angle subtended by the chord in the alternate segment.
stage 4 :
proportions
Similarly, by marking S and T on DE and EF such that
Construction : Join DC & BE
1
AD . h
Area of ΔADE
= 2
= AD
Proof :
1
Area of ΔDBE
DB
DB . h
2
1
AE . h′
2
Similarly : Area of ΔADE = AE
1
Area of ΔEDC
EC
EC . h′
2
h′
D
B
SE = AB and ET = BC, it can be proved that :
h
E
AB
BC
=
DE
EF
â AB = AC = BC DE
C
DF
EF
â ΔABC and ΔDEF are similar.
s
Similar Δ
s
But :
ΔDBE = ΔEDC
and :
ΔADE is common
...
Δ are similar if : A : they are equiangular, and
B : their sides are proportional
on the same base DE ;
between || lines, DE & BC
In this proof, we show that A B
s
∴ The Δ are similar . . . Both conditions, A and B, apply
â Area of ΔADE = Area of ΔADE
Area of ΔDBE
Area of ΔEDC
AD
AE
â
=
DB
EC
The converse statement says : B A
s
∴ The Δ are similar
2
Copyright © The Answer
TRIGONOMETRY
3
The Cosine rule
Â
Area, Sine & Cosine Rules
1
A
a2 = b2 + c2 – 2bc cos A
b
c D
The Area rule
A
Ĉ acute
Area of ΔABC = 1 ab sin C
Construction :
Draw AD ⊥ BC
Proof :
1
Area of ΔABC = ah
2
But
Construction :
B
...
h
= sin C
b
2
C
D
a
2
a = BD + h
∴ a
...
. . . Pythagoras
2
2
= (c – AD) + h
2
2
2
= c – 2c . AD +
2
b
2
2
= b + c – 2c . AD
In ΔADC :
∴ h = b sin C
2
= c – 2c . AD + AD + h
B
C
a
2
. . . Pythagoras
...
...
AD
= cos A
b
∴ AD = b cos A
1
∴ Area of ΔABC =
ab sin C
2
h
Draw CD ⊥ BA
2
Proof :
b
h
2
in :
acute :
in :
∴ a2 = b2 + c2 – 2bc cos A
2
The Sine rule
= sin C
sin A
= sin B
a
b
Construction :
Proof :
B̂ acute
Always construct the height from a vertex not involved in the formula :
D
i.e. To prove :
b
c
h
B
Draw CD ⊥ AB
In ΔADC :
h
= sin A
b
∴ h = b sin A
In ΔBDC :
...
a
C
∴
...
Copyright © The Answer
sin A
sin B
=
, draw a height from C, not A or B.
a
b
To prove :
a = b + c – 2bc cos A, draw a height from B or C, not A.
30º
sin A
sin B
=
a
b
Similarly, by drawing a perpendicular from B, one can prove :
∴
To prove :
Remember :
& : ∴ b sin A = a sin B
÷ ab)
1
ab sin C, draw a height from A or B, not C.
2
Area =
2
2
150º
2
II
sin (180º – θ) = sin θ
θ
h
= sin B
a
∴ h = a sin B
Equating
A
180º – θ
sin 150º = sin 30º
cos 150º = – cos 30º
cos (180º – θ) = – cos θ
y
x
x + y = 180º sin x = sin y
cos x = – cos y
Formulae for Compound Angles
The derivation of the formula : cos (A – B) = cos A cos B + sin A sin B
will not be examined. (However, it is always good to understand how a formula arises ! ).
There is an easy guide to this derivation in Topic 6 in the Gr 12 Maths 2 in 1 Study guide.
You do need to know how to derive all the other formulae from this formula for cos(A – B).
See examples of this, also found in Topic 6 in the Gr 12 Maths 2 in 1 Study guide.
sin A
sin C
=
a
c
sin A
sin B
sin C
=
=
a
c
b
3
3 ways to prove that a quadrilateral is cyclic
GROUPING OF CIRCLE GEOMETRY THEOREMS
I
The grey arrows indicate how
various theorems are used to
prove subsequent ones.
The 'Centre' group
We use 3 converse theorem statements to prove that a quadrilateral is cyclic.
converse of 'ø's in same segment' theorem
Prove that one side
subtends equal angles
at the two other points
(on the same side).
x
2x
diameter
2x
x
x
Equal
radii !
II The 'No Centre' group
Equal
chords !
x1
z1
y2
B
Prove that a pair of
opposite angles
is supplementary.
Equal chords subtend
equal angles
and, vice versa,
equal angles are subtended
by equal chords.
Either prove :
 + Ĉ = 180º
A
or B̂ + D̂ = 180º
D
C
converse of 'exterior ø of c.q.' theorem
Prove that an exterior angle
of a quadrilateral is equal
to the interior opposite angle.
Either prove :
5
2
7
1̂ = 2̂
4
8 3
6
1
x
There are 3 ways
to prove that
a quad. is a
cyclic quad.
Either prove :
x1 = x2
or y1 = y2
or z1 = z2
or p1 = p2
p1 x
2
z2
p2
converse of 'ø's of c.q.' theorem
III The 'Cyclic Quadrilateral' group
y
y1
or 3̂ = 4̂
or 5̂ = 6̂
or 7̂ = 8̂
2 ways to prove that a line is a tangent to a ?
We use 2 converse theorem statements to prove that a line is a tangent to a ?.
x + y = 180º
converse of 'tangent ⊥ radius' theorem
IV The 'Tangent' group
O
Prove that the line is perpendicular
to a radius at a point where the
radius meets the circumference.
Prove that OP ⊥ AB ;
then AB will be a tangent
A
P
B
converse of 'tan-chord' theorem
Equal
tangents !
Prove that when the line is
drawn through the end point of
a chord, it makes with the chord
an angle equal to an angle in
the alternate segment.
There are 2 ways to prove that
a line is a tangent to a ?.
4
A
y
x
B
Prove that
x = y;
then AB
will be
a tangent
Note :
It is a good idea to
draw a feint circle
(as shown) to see
this converse
theorem clearly.
Copyright © The Answer
QUADRILATERALS
- pathways of definitions, areas and properties - A Summary
A Rectangle
All you need to know!
'Any' Quadrilateral
f
a
A Trapezium
b
m
A || with one right ø
2
h
Area = ℓ % b
1
d
DIAGONALS are EQUAL
a
s
Sum of the ø of
any quadrilateral = 360º
The Square
DEFINITION :
b
c
e
A Parallelogram
DEFINITION :
Quadrilateral with 1 pair of opposite sides ||
DEFINITION :
Quadrilateral with 2 pairs opposite sides ||
the 'ultimate' quadrilateral !
Area = s 2
Area = base % height
Area = Δ 1 + Δ 2
P
= 1 ah + 1 bh
2
2
Sum of the interior angles
= (a + b + c) + (d + e + f)
s
= 2 % 180º
. . . (2 Δ )
= 360º
Copyright © The Answer
D
B
|| ABCD = ABCQ + ΔQCD
rect. PBCQ = ABCQ + ΔPBA
where ΔQCD ≡ ΔPBA
Properties :
It's all been said 'before' !
Since a square is a rectangle,
a rhombus, a parallelogram,
a kite, . . . ALL the properties
of these quadrilaterals apply.
C
m
. . . SS90º
m
â || ABCD = rect. PBCQ (in area)
= BC % QC
A Kite
a
2
a
2
b
See how the properties
accumulate as we
move from left to right.
i.e. the first quad has
no special properties
and each successive
quadrilateral has all
preceding properties.
Q
A Rhombus
= 1 (a + b) . h
2
'Half the sum of the || sides
% the distance between them.'
The arrows indicate
various ‘ROUTES’
from ‘any’
quadrilateral to the
square (the ‘ultimate
quadrilateral’). These
routes, which combine
logic and fact, are
essential to use when
proving specific types
of quadrilaterals.
A
Properties :
2 pairs opposite sides equal
2 pairs opposite angles equal
& DIAGONALS BISECT ONE ANOTHER
Area
=
1
product of diagonals (as for a kite)
2
THE DIAGONALS
• bisect one another PERPENDICULARLY
• bisect the angles of the rhombus
• bisect the area of the rhombus
Given diagonals a and b
s
Area = 2Δ = 2 ⎛⎜ 1 b . a ⎞⎟ = ab
2⎠
m
A || with one pair of
adjacent sides equal
or
= base % height (as for a parallelogram)
DEFINITION :
Quadrilateral with 2 pairs of
consecutive sides equal
⎝2
DEFINITION :
2
'Half the product of the diagonals'
THE DIAGONALS
• cut perpendicularly
• the LONG DIAGONAL bisects:
the short diagonal, the opposite angles and the area of the kite
5
Note :
x
y y
x
x
x
y y
s
2x + 2y = 180º . . . ø of Δ, or
² x + y = 90º
co-int. øs suppl.
Quadrilaterals
play a
prominent role
in both Euclidian
and Analytical
Geometry right
through to
Grade 12 !
2
SUMMARY : TRIGONOMETRY (Gr. 12)
' SPECIAL LS
' THE RATIOS
sin 30E '
& their
' DEFINITIONS :
The ANSWER SERIES
y
r
O
H
x
r
A
H
y
x
3
1
2
tan 45E ' 1
2
1
and cos 60° '
2
O
A
tan θ '
y
x
& THEIR “FAMILIES” :
' SIGNS :
' GENERAL FORMS
' CRITICAL VALUES :
any ratio
' 90E - θ (an acute angle)
(see denominator ! )
' GRAPHS :
±
that SAME
ratio of
'
'
θ
sin(90E + θ)
cos(90E - θ) = sin θ
cos(90E + θ) = - sin θ
AREA '
SINE rule
1
ab sin C
2
1
2
And in RIGHT-ANGLED Δ S :
2
bh as well
2
' cos(A ± B) = cos A cos B
6
COSINE rule
'
c 2 ' a 2 % b 2 & 2ab cos C
' sin 2x = 2 sin x cos x
' sin(A ± B) = sin A cos B ± cos A sin B
& cos θ = 1 - sin θ
sin A
sin B
sin C
'
'
a
b
c
'
use regular trig. ratios & the theorem of Pythagoras !!!
' COMPOUND LS
ˆ sin2 θ = 1 - cos2 θ
= cos θ
For non-right angled Δ S :
But remember : area of ∆=
sin 2 θ % cos 2 θ ' 1
90E + θ (an obtuse angle)
sin(90E - θ) = cos θ
' SOLUTION OF ΔS
sin θ
tan θ '
cos θ
=
' CO-RATIOS (sine and cosine)
' MINIMUM & MAXIMUM VALUES : (sin θ & cos θ)
' MUTUAL RELATIONSHIPS
180° - θ
180° + θ
360° - θ
-θ
K
sin A sin B
' cos 2x = cos2 x - sin2 x
2
= 1 - 2 sin x
2
= 2 cos x - 1
Copyright © The Answer
CALCULATOR INSTRUCTIONS
MEAN AND STANDARD DEVIATION
REGRESSION & CORRELATION
There are 3 phases for each calculator procedure :
The equation of the regression line
STEP 1 : how to get there
Brief instructions for the CASIO fx - 82ES
STEP 2 : how to enter the data
STEP 3 : how to find the mean and the standard deviation.
STEP 1 :
Press/Select : MODE ; 2 (STAT) ; 2 (A + Bx )
(and, in the next column, A, B and r)
Grouped Data/Frequency Tables
Ungrouped Data
Casio fx-82ES
Enter the x and y values, each followed by =.
Casio fx-82ES
You'll see :
[MODE] [2 : STAT] [1 : 1 - VAR]
Enter each value, followed by [=]
after the last value : [=] [AC]
1
2
3
X
…
…
…
FREQ
automatically
entered
as 1
,
,
[SHIFT] [STAT] [5 : VAR] [3 : x σ n] [=]
If this is not done, the mean and standard deviation
will be added as entries in the data table.
(These would then need to be deleted.)
Enter each value, followed by [M+]
To find the mean : [RCL] [ x ]
To find the S.D. :
[RCL] [σ x ]
FREQ
…
…
…
Press AC
STEP 3 :
Press
SHIFT
STAT (on '1') ; 7 (Reg) ; then :
For A, press 1
=
or
For B, press 2
=
To find B after A, press AC
and then do the whole of STEP 3.
to move to the top of the right-hand column.
Just like for A and B in the regression function :
Brief instructions for the CASIO fx - 82ES
To find the mean : [SHIFT] [STAT] [5 : VAR] [2 : x ] [=]
To find the S.D. :
You'll see :
Stat
Type in the correct frequencies followed by [=] each
,
time; after the last frequency : [=] [AC]
0.
Sharp EL-531WH
You'll see :
[MODE] [1 : STAT] - to get into stats mode . . .
then [0 : SD]
Enter midpoint of each interval (or each score if
provided), followed by [STO].
Then enter the frequency followed by [M+].
(Enter all the data like this.)
To find the S.D. :
STEP 1 :
Press/Select : MODE ; 2 (STAT) ; 2 (A + Bx )
[SHIFT] [STAT] [5 : VAR] [3 : x σ n] [=]
To find the mean : [RCL] [ x ]
Copyright © The Answer
1
2
3
X
…
…
…
The correlation coefficient
It is essential to use [AC] at the end of entering all the data.
[MODE] [1 : STAT] - to get into stats mode . . .
then [0 : SD]
[MODE] [2 : STAT] [1 : 1 - VAR]
[SHIFT] [SETUP] Scroll down (use arrow)
[3 : STAT] [1 : ON]
Use
Note for Casio fx-82ES
Sharp EL-531WH
You'll see :
Enter the midpoint of each interval
(or each score if provided) followed by [=], then . . .
After the last value : [=] - and not [AC] - then . . .
To find the mean : [SHIFT] [STAT] [5 : VAR] [2 : x ] [=]
To find the S.D. :
STEP 2 :
[RCL] [σ x ]
7
Stat
0.
STEP 2 :
Enter the x and y values, each followed by =.
Press AC
STEP 3 :
Press
SHIFT
STAT (on '1') ; 7 (Reg) ; then :
But, now : For r press 3
=
If you have found A and B first, press AC
before going back to the beginning of STEP 3.