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Higher
Mathematics
HSN24400
Course Revision Notes
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Higher Mathematics
Course Revision Notes
Contents
Unit 1 – Mathematics 1
Straight Lines
Functions and Graphs
Differentiation
Sequences
1
2
5
6
Unit 2 – Mathematics 2
Polynomials and Quadratics
Integration
Trigonometry
Circles
7
8
9
12
Unit 3 – Mathematics 3
Vectors
Further Calculus
Exponentials and Logarithms
The Wave Function
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15
16
18
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HSN24400
Higher Mathematics
Course Revision Notes
Straight Lines
Distance Formula
( x2 − x1 )
Distance =
2
+ ( y2 − y1 ) between points ( x1, y1 ) and ( x2 , y2 )
2
Gradients
m=
y2 − y1
between the points ( x1, y1 ) and ( x2 , y2 ) where x1 ≠ x2
x 2 − x1
Positive gradients, negative gradients, zero gradients, undefined gradients
eg y = 4
eg x = 2
Lines with the sam e gradient are parallel
eg The line parallel to 2 y + 3 x = 5
has gradient m = − 32 since 2 y + 3 x = 5
2 y = −3 x + 5
y = − 32 x + 25 (m ust be in the form y = mx + c )
Perpendicular lines have gradients such that m × mperp. = −1
eg if m = 23 then mperp. = − 23
m = tan
y
positive direction
x
is the angle that the line m akes with
the positive direction of the x-axis
Equation of a Straight Line
The line passing through ( a, b ) with gradient m has equation:
y − b = m( x − a )
Medians
B
A
M
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x1 + x 2 y1 + y2
,
2
2
BM is not usually perpendicular to AC, so m1 × m2 = −1
cannot be used
To work out the gradient of BM, use the gradient form ula
M is the m idpoint of AC, ie M =
C
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HSN24400
Higher Mathematics
Course Revision Notes
Altitudes
B
A
D is not usually the m idpoint of AC
C
D
BD is perpendicular to AC, so m1 × m2 = −1 can be used to
work out the gradient of BD
Perpendicular Bisectors
D
B
A
C
B
C
D
CD passes through m idpoint of AC
CD is perpendicular to AB, so m1 × m2 = −1 can be used to
find the gradient of CD
Perpendicular bisectors do not necessarily have to appear
within a triangle – they can occur with straight lines
A
Functions and Graphs
Composite Functions
Example
If f ( x ) = x 2 − 2 and g ( x ) = 1x , find a form ula for
(a) h ( x ) = f ( g ( x ) )
(b) k ( x ) = g ( f ( x ) )
and state a suitable dom ain for each.
(a) h ( x ) = f ( g ( x ) )
= f ( 1x )
(b) k ( x ) = g ( f ( x ) )
= g( x 2 − 2)
= ( 1x ) − 2
2
=
1
−2
x2
=
1
x −2
2
Dom ain: { x : x ∈ , x ≠ ± 2 }
Dom ain: { x : x ∈ , x ≠ 0}
You will probably only be asked for a dom ain if the function involved a fraction or an
even root. Rem em ber that in a fraction the denom inator cannot be zero and any
num ber being square rooted cannot be negative
eg f ( x ) = x + 1 could have dom ain: { x : x ∈ , x ≥ −1}
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HSN24400
Higher Mathematics
Course Revision Notes
Graphs of Inverses
To draw the graph of an inverse function, reflect the graph of the function in the
line y = x
y=x
y
y = g( x )
y = g−1 ( x )
x
O
Exponential and Logarithmic Functions
Exponential
y
Logarithmic
y
y = ax, a >1
y = loga x
y
y = ax, 0 < a <1
( a ,1)
(1, a )
1
x
O
O
(1, a )
1
x
1
x
O
Trigonometric Functions
y = sin x
y
y
1
1
O
–1
x
180°
y = tan x
y = cos x
y
x
O
–1
180°
360°
x
O
360°
180°
360°
Period = 360°
Period = 360°
Period = 180°
Am plitude = 1
Am plitude = 1
Am plitude is undefined
Graph Transformations
The next page shows the effect of transform ations on the two graphs shown below.
y
y
( 2, 2 ) y = g ( x )
1
–1 0
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3
O
–1
x
Page 3
x
180°
360°
HSN24400
Higher Mathematics
Function
Course Revision Notes
Effect on f ( x )
Effect
f ( x ) + a Shifts the graph
a up the
y-axis
y
Effect on sinx °
y = g( x ) + 1
( 2, 3 ) •
2
• ( 3,1)
•
( −1,1
)
0
f ( x + a ) Shifts the graph
−a along the
x-axis
x
y = g ( x + 1)
y
•
0
1
0
x
360°
180°
y = sin ( x − 90 ) °
y
1
(1, 2 )
–2
y = sin x ° + 1
y
2
x
x
0
–1
360°
180°
− f (x)
y
Reflects the
graph in the
x-axis
0
–1
3
•
f (−x )
Reflects the
graph in the
y-axis
y
( −2, 2 )
Scales the graph
vertically
Stretches if
k >1
Com presses if
k <1
f ( kx )
Scales the graph
horizontally
Com presses if
k >1
Stretches if
k <1
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x
1
x
( 2, − 2 )
0
–1
y = g( −x )
y = sin ( − x ° )
180°
360°
y
•
1
0
–3
kf ( x )
y = − sin x °
y
y = − g( x )
x
1
−360°
y ( 2, 4 ) y = 2 g ( x )
−180°
0 x
–1
y = 12 sin x °
y
•
1
2
–1 0
3
y (1, 2 )
x
x
0
180°
y = g(2x )
•
− 12 0
− 12
y
360°
y = sin 2 x °
1
3
2
Page 4
x
0
–1
x
180°
360°
HSN24400
Higher Mathematics
Course Revision Notes
Differentiation
Differentiating
If f ( x ) = ax n then f ′ ( x ) = anx n −1
Before you differentiate, all brackets should be m ultiplied out, and there should be no
fractions with an x term in the denom inator (bottom line), for exam ple:
3
1
− 12
−2
1
−2
=
3x
=
x
1
= x
x2
x
3x 2 3
Equations of Tangents
Tangents are straight lines, therefore to find the equation of a tangent, you need a point
on the line and its gradient to substitute into y − b = m ( x − a )
You will always be given one coordinate of the point which the tangent touches
Find the other coordinate by solving the equation of the curve
Find the gradient by differentiating then substituting in the x-coordinate of the point
Example
Find the equation of the tangent to the graph of y = x 3 at the point where x = 9 .
y − b = m(x − a)
1
At x = 9, m = 32 × 9 2
y = x3
y = x3
= 93
= x2
= 32 9
= 33
= 27
dy 3 12
= x
dx 2
= 32 × 3
3
y − 27 = 92 ( x − 9 )
2 y − 54 = 9 x − 81
= 92
( 9, 27 )
2 y = 9 x − 27
dy
=0
dx
You m ust justify the nature of turning points or points of inflection
Stationary points occur at points where
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HSN24400
Higher Mathematics
Course Revision Notes
Graphs of Derived Functions
Quadratic
Cubic
y
Quartic
y
y
0
+
–
+
0
–
x
+
+
x
+
+
+
–
+
–
Linear
–
y
+
x
+
0
y
–
–
x
0
y
0
0
–
x
x
Quadratic
Cubic
Optimisation
These types of questions are usually practical problem s which involve m axim um or
m inim um areas or volum es
Rem em ber you m ust show that a m axim um or m inim um exists
Sequences
Linear Recurrence Relations
A linear recurrence relation is in the form un+1 = aun + b . Also be aware that this m ay be
written as un = aun−1 + b
If −1 < a < 1 then a lim it l =
b
exists. You m ust state this whenever you use the lim it
1− a
form ula
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HSN24400
Higher Mathematics
Course Revision Notes
Polynomials and Quadratics
Polynomials
The degree of a polynom ial is the value of the highest power, eg 3 x 4 + 3 has degree 4
Synthetic division (nested form ) can be used to factorise polynom ials
Example
4 x 3 − 7 x 2 + 11
Find
.
x +2
–2
4
–7
0
11
4
–8
–15
30
30
–60
–49
Rem em ber to put in 0
if there is no term
4 x 3 − 7 x 2 + 11
= 4 x 2 − 15 x + 30 rem ainder − 49
x +2
ie 4 x 3 − 7 x 2 + 11 = ( x + 2 ) ( 4 x 2 − 15 x + 30 ) − 49
If the divisor is a factor then the rem ainder is zero
If the rem ainder is zero then the divisor is a factor
Completing the Square
The x 2 term m ust have a coefficient of one. If it does not, you m ust take out a com m on
factor from the x 2 and x term , but not the constant
In the form y = a ( x + p ) + q the turning point of the graph is ( − p, q )
2
Example
Write 3 x 2 − 12 x + 7 in the form a(x + p)2 + q .
3 x 2 − 12 x + 7
= 3( x 2 − 4x ) + 7
= 3 ( x 2 − 4 x + ( −2 )2 − ( −2 )2 ) + 7
= 3 ( ( x − 2 )2 − 4 ) + 7
= 3 ( x − 2 )2 − 12 + 7
= 3 ( x − 2 )2 − 5
Note that in this exam ple, the graph is ∪ -shaped since the x 2 coefficient is positive; and
the turning point is ( 2, − 5 ) .
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HSN24400
Higher Mathematics
Course Revision Notes
The Discriminant
The discrim inant is part of the quadratic form ula and can be used to indicate how m any
roots a quadratic has. For the quadratic ax 2 + bx + c :
If b 2 − 4 ac > 0 , the roots are real and unequal (distinct)
2 roots
If b 2 − 4 ac = 0 , the roots are real and equal (ie repeated roots)
1 root
If b 2 − 4 ac < 0 , the roots are not real; they do not exist
no roots
The discrim inant can also be used to calculate the num ber of intersections between a
line and a curve. To use it, you m ust first equate them and set equal to zero, before
using the discrim inant
Rem em ber if b 2 − 4 ac = 0 , the line is a tangent
Integration
Integrating
ax n dx =
ax n +1
+c
n +1
As with differentiation, all brackets m ust be m ultiplied out, and there m ust be no
fractions with an x term in the denom inator
Examples
1. Find
dx
8
x5
dx
8
=
.
x5
2.
1
8
= x
x5
− 58
x 2 + 5x 7
dx = x −2 ( x 2 + 5 x 7 ) dx
2
x
= x 0 + 5 x 5 dx
dx
dx
= 1 + 5 x 5 dx
3
x8
= 3 +c
= x + 56 x 6 + c
8
3
= 83 x 8 + c
= 38 8 x 3 + c
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x 2 + 5x 7
dx
x2
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HSN24400
Higher Mathematics
Course Revision Notes
The Area under a Curve
If F ( x ) is the integral of f ( x ) , then
y
b
a
f ( x ) dx = F (b ) − F ( a )
y = f (x)
a
x
b
Rem em ber that areas split by the x-axis m ust be calculated separately and any negative
signs ignored; these just show that the area is under the axis.
The Area between two Curves
The area between the graphs of y = f ( x ) and y = g ( x ) is defined as
b
a
f ( x ) − g ( x ) dx
y
y = g( x )
If the lim its are not given, f ( x ) and g ( x )
should be equated to find a and b
a
x
y = f (x)
b
Trigonometry
Background Knowledge
You should know how to use all of the inform ation below:
SOH CAH TOA
tan x =
sin x
cos x
sin 2 x + cos2 x = 1
The sine rule:
a
b
c
=
=
sin A sin B sin C
b2 + c 2 − a2
The cosine rule: a = b + c − 2bc cos A or cos A =
2bc
2
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2
2
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HSN24400
Higher Mathematics
Course Revision Notes
The area of a triangle, A = 12 ab sin C
CAST diagram s
Exact values:
2
45°
30°
2
1
3
60°
45°
1
1
Radians
You should know how to convert between radians and degrees:
90° = 2
45° = 4
÷ 180 × → Radians
Degrees 
60° = 3
5 × 180
eg 56 =
= 150°
6
30° = 6
× 180 ÷ → Degrees
Radians 
360° = 2
180° =
Trigonometric Equations
Look at the restrictions on the dom ain, eg 0 ≤ x ° < 360, or 0 ≤ x <
Be aware of whether the answer is required in degrees or radians
Rem em ber a CAST diagram whenever you are asked to “solve”
Examples
1. Solve 3sin 2 x° = 1 where 0 ≤ x ° < 360 .
3sin 2 x ° = 1
3 ( sin x ° ) = 1
2
( sin x ° )2 = 13
sin x ° = ± 13
x ° = sin
x° = 35.3°
−1
( ± 13 )
S
A
T C
x° = 180 − 35.3
x° = 180 + 35.3
x° = 360 − 35.3
= 144.7°
= 215.3°
= 324.7°
Solution set = {35.3°, 144.7°, 215.3°, 324.7°}
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HSN24400
Higher Mathematics
Course Revision Notes
2. Solve 2 sin 2 x − 1 = 0 , 0 ≤ x < 2 .
2 sin 2 x − 1 = 0
2sin 2 x = 1
sin 2 x = 12
S
2 x = sin −1 ( 12 )
A
T C
2 x ° = 30°
2 x ° = 180° − 30°
2 x ° = 150°
x ° = 15°
x ° = 75°
2 x ° = 360° + 30°
2 x ° = 360° +180° − 30°
2 x ° = 390°
2 x ° = 510°
x ° = 195°
x ° = 255°
75
75° = 180
15
15° = 180
195
195° = 180
255
255° = 180
3
= 36
= 15
36
39
= 36
51
= 36
= 12
5
= 12
13
= 12
= 17
12
{
5 , 13 , 17
Solutions set = 12 , 12
12 12
}
Compound Angle Formulae
cos ( A ± B ) = cos A cos B
sin A sin B
sin ( A ± B ) = sin A cos B ± cos A sin B
These are given on the form ula sheet
Double Angle Formulae
sin 2 A = 2sin A cos A
cos 2 A = cos2 A − sin 2 A
= 1 − 2 sin 2 A
= 2 cos 2 A − 1
These are given on the form ula sheet
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HSN24400
Higher Mathematics
Course Revision Notes
Circles
Equations of Circles
A circle with centre ( a,b ) and radius r has the equation ( x − a )2 + ( y − b ) = r2
2
Note that if a circle has centre ( 0, 0 ) then the equation is x 2 + y 2 = r2
The equation can also be given in the form x 2 + y 2 + 2 gx + 2 fy + c = 0 where the centre
is ( − g,− f
)
and the radius r = g2 + f 2 − c
You do not have to rem em ber any of these equations, since they are all given in the
exam
You will have to rem em ber the distance form ula, d =
is not given, and is frequently used in circle questions
( x2 − x1 )
2
+ ( y2 − y1 ) , since this
2
Intersection of a Line and a Circle
two intersections
one intersection (tangency)
no intersections
Rem em ber, a tangent and a line from the centre of a circle will m eet at right angles,
which m eans that m1 × m2 = −1 can be used
Vectors
Basic Facts
A vector is a quantity with both m agnitude (size) and direction
A vector is nam ed either by using a directed line segm ent (eg AB ) or a bold letter
(eg u written u)
A vector m ay also be defined in term s of i, j and k, the unit vectors in three
perpendicular directions:
1
i= 0
0
0
j= 1
0
0
k= 0
1
a
The m agnitude of vector AB = b is defined as AB = a 2 + b 2
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HSN24400
Higher Mathematics
Course Revision Notes
b1
a1 ± b1
a1
a2 ± b2 = a2 ± b2
a3
b3
a3 ± b3
ka1
a1
k a2 = ka2 where k is a scalar
a3
ka3
0
Z ero vector: 0
0
OA is called the position vector of the point A relative to the origin, written a
AB = b − a where a and b are the position vectors of A and B
If AB = k BC where k is a scalar, then AB is parallel to BC . Since B is com m on to both
AB and k BC , then A, B and C are collinear
Dividing Vectors in a Ratio
The point P can also be worked out from first principles, or
U sing the section form ula. If P divides AB in the ratio m : n , then:
n
m
p=
a+
b where p is the position vector OP
m+n
m+n
Example
P is the point ( −2, 4, − 1) and R is the point ( 8, − 1,19 ) . Point T divides PR in the ratio
2:3. Work out the coordinates of point T.
P
2:3
R
T
U sing the section form ula
The ratio is 2:3, so let m = 2 and n = 3
From first principles
n
m
p+
r
m+n
m+n
= 35 p + 25 r
t=
(
= 15 3 p + 2r
= 15
(
3t − 3 p = 2r − 2t
3t + 2t = 2r + 3 p
−6
16
12 + −2
−3
38
16
−6
5t = −2 + 12
38
−3
10
10
5t = 10
35
= 15 10
35
2
= 2
7
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)
3 t − p = 2 (r − t )
)
Therefore T is the point ( 2, 2, 7 ) .
PT 2
=
TR 3
3PT = 2TR
2
t= 2
7
Therefore T is the point ( 2, 2, 7 ) .
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HSN24400
Higher Mathematics
Course Revision Notes
The Scalar Product
The scalar product a.b = a b cos , where is the sm allest angle between a and b
Rem em ber that both vectors m ust point away from the angle, eg
a
b
b1
a1
If a = a2 and b = b2 then a.b = a1b1 + a2b2 + a3b3
a3
b3
cos =
ab +a b +a b
a.b
or cos = 1 1 2 2 3 3
ab
a b
If a and b are perpendicular then a.b = 0
If a.b = 0 then a and b are perpendicular
Example
8
4
If u = 0 and v = 0 , calculate the angle between the vectors u + v and u − v .
1
4
Let a = u + v
Let b = u − v
8
4
a= 0 + 0
1
4
8
4
b= 0 − 0
1
4
12
a= 0
5
4
b= 0
3
cos =
=
a.b
ab
(12 × 4 ) + ( 0 × 0 ) + ( 5 × 3 )
122 + 02 + 52 4 2 + 02 + 32
63
=
169 25
= cos −1
63
169 25
= 14·3°
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HSN24400
Higher Mathematics
Course Revision Notes
Further Calculus
Trigonometry
D ifferentiation
This is straightforward, since the form ulae are given on the form ula sheet:
f (x)
f ′( x )
sin ax
a cos ax
cos ax
− a sin ax
Integration
Again, the form ulae are provided in the paper:
f ( x ) dx
f (x)
sin ax
− 1a cos ax + c
cos ax
1 sin ax + c
a
Examples
1. Differentiate x 3 + cos3 x with respect to x.
(
)
d x 3 + cos3 x = 3 x 2 − 3sin3 x
dx
2. Find 4 x 3 + sin3 x dx .
4x 4 1
4 x + sin3 x dx =
− 3 cos3 x + c
4
= x 4 − 13 cos3 x + c
3
Chain Rule Differentiation
n −1
If f ( x ) = ( ax + b )n then f ′ ( x ) = n ( ax + b )
n −1
× a = an ( ax + b )
or
n −1
If f ( x ) = ( p ( x ) ) then f ′ ( x ) = n ( p ( x ) )
n
× p′ ( x )
“The power m ultiplies to the front, the bracket stays the sam e, the power lowers by one
and everything is m ultiplied by the differential of the bracket”
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HSN24400
Higher Mathematics
Course Revision Notes
Examples
1. G iven f ( x ) =
1
+ x − sin3 x , find f ′ ( x ) .
x2
1
f ( x ) = x −2 + x 2 − sin3 x
−1
f ′ ( x ) = −2 x −3 + 12 x 2 − 3cos3 x
2
1
=− 3 +
− 3cos3 x
x
2 x
2. G iven f ( x ) = ( 3 x 2 + 2 x + 1) , find f ′ ( x ) .
3
f ′ ( x ) = 3 ( 3 x 2 + 2 x + 1) × ( 6 x + 2 )
2
= 3 ( 6 x + 2 ) ( 3 x 2 + 2 x + 1)
2
3. Differentiate y = cos 2 x = ( cos x )2 with respect to x.
dy
= 2 ( cos x ) × ( − sin x )
dx
= −2cos x sin x
n
Integration of (ax + b)
( ax + b ) dx =
n
( ax + b )n+1
( n + 1) × a
+c
Example
Find ( 3 x + 5 )4 dx .
( 3 x + 5 ) dx =
4
( 3 x + 5 )5
5× 3
+c =
( 3 x + 5 )5
15
+c
It is possible for any type of ‘further calculus’ to be exam ined in the style of a standard
calculus question (eg optim isation, area under a curve, etc)
Exponentials and Logarithms
An exponential is a function in the form f ( x ) = a x
Logarithm s and exponentials are inverses
y = a x ⇔ loga y = x
On a calculator, log is log10 and ln is loge
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HSN24400
Higher Mathematics
Course Revision Notes
Laws of Logarithms
loga x + loga y = loga xy
(Squash)
loga x − loga y = loga xy
(Split)
loga x n = n loga x
(Fly)
Examples
1. Evaluate log2 4 + log2 6 − log2 3
log2 4 + log2 6 − log2 3
4×6
3
= log2
= log2 8
=3
(since 23 = 8)
2. Below is a diagram of part of the graph of y = ke 0.7 x
y
y = ke 0.7 x
3
0
x =1
x
(a) Find the value of k
(b) The line with equation x = 1 intersects at R. Find the coordinates of R.
(a) At ( 0, 3 ) , y = ke 0.7 x
3 = ke 0.7×0
3 = ke 0
k =3
(b) x = 1
y = 3e 0.7×1
= 6.04
So R is the point (1, 6.04 ) .
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HSN24400
Higher Mathematics
Course Revision Notes
The Wave Function
Example
1. Express
6 sin x ° − 2 cos x ° in the form k cos ( x − a ) ° where 0 ≤ a ° < 360 .
k cos ( x − a ) ° = k cos x ° cos a ° + k sin x ° sin a °
= k cos a ° cos x ° + k sin a ° sin x °
k cos a ° = − 2
k sin a ° = 6
(− 2 )
k=
2
+ 6
2
= 2+6
S A
= 8
T C
=2 2
k sin a °
k cos a °
6
=−
2
=− 3
tan a ° =
a ° = 180° − tan −1 ( 3 )
= 180° − 60°
= 120°
Therefore 6 sin x ° − 2 cos x ° = 2 2 cos ( x − 120 ) ° .
2. Express cos x − sin x in the form k sin ( x + ) where 0 ≤ < 2 .
k sin ( x + ) = k sin x cos + k cos x sin
= k cos sin x + k sin cos x
k cos = −1
k sin = 1
k = ( −1)2 + 12
= 2
S A
k sin
k cos
= −1
tan =
° = 180° − tan −1 (1)
T C
= 180° − 45°
= 135°
135
= 180
= 34
(
Therefore cos x − sin x = 2 sin x + 34
).
The m axim um value of an expression in the form k cos ( x ± a ) occurs when
cos ( x ± a ) = 1 ; and sin ( x ± a ) = 1 for k sin ( x ± a )
The m inim um value of an expression in the form k cos ( x ± a ) occurs when
cos ( x ± a ) = −1 ; and sin ( x ± a ) = −1 for k sin ( x ± a )
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