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Higher
Mathematics
HSN24400
Course Revi on Notes
si
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2. Higher Mathematics Course Revision Notes
Contents
Unit 1 – Mathematics 1
Straight Lines 1
Functions and Graphs 2
Differentiation 5
Sequences 6
Unit 2 – Mathematics 2
Polynomials and Quadratics 7
Integration 8
Trigonometry 9
Circles 12
Unit 3 – Mathematics 3
Vectors 12
Further Calculus 15
Exponentials and Logarithms 16
The Wave Function 18
hsn.uk.net - ii - HSN24400
3. Higher Mathematics Course Revision Notes
Straight Lines
Distance Formula
stance = ( x2 − x1 ) + ( y2 − y1 ) between poi ( x1, y1 ) and ( x2 , y2 )
2 2
Di nts
Gradients
y2 − y1
m= between the poi ( x1, y1 ) and ( x2 , y2 ) where x1 ≠ x2
nts
x 2 − x1
Posi ve gradi
ti ents, negati gradi
ve ents, zero gradients, undef ned gradi
i ents
eg x = 2
eg y = 4
Li wi the sam e gradi are parallel
nes th ent
eg The li parallel to 2 y + 3 x = 5
ne
has gradi m = − 3 si
ent 2 nce 2 y + 3 x = 5
2 y = −3 x + 5
5
y = − 3 x + 2 (m ust be i the f
2 n orm y = mx + c )
Perpendicular li have gradi
nes ents such that m × mperp. = −1
eg i m = 2 then mperp. = − 2
f 3
3
m = tan
y
i the angle that the li m akes wi
s ne th
positive direction the posi ve di on of the x-axi
ti recti s
x
Equation of a Straight Line
The li passi through ( a, b ) wi gradi m has equati
ne ng th ent on:
y − b = m( x − a )
Medians
B x1 + x 2 y1 + y2
M i the m i
s dpoi of AC, i M =
nt e ,
2 2
BM i not usually perpendi
s cular to AC, so m1 × m2 = −1
A C cannot be used
M
To work out the gradi of BM, use the gradi f ula
ent ent orm
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4. Higher Mathematics Course Revision Notes
Altitudes
B
D i not usually the m i
s dpoi of AC
nt
BD i perpendi
s cular t AC, so m1 × m2 = −1 can be used t
o o
A C work out the gradi of BD
ent
D
Perpendicular Bisectors
D
CD passes through m i
dpoi of AC
nt
A B
C CD i perpendi
s cular t AB, so m1 × m2 = −1 can be used t
o o
C B f nd the gradi of CD
i ent
Perpendicular bisectors do not necessari have to appear
ly
D wi n a tri
thi angle – they can occur wi st ght li
th rai nes
A
Functions and Graphs
Composite Functions
Example
If f ( x ) = x 2 − 2 and g ( x ) = 1 , f nd a f ula f
x i orm or
(a) h ( x ) = f ( g ( x ) )
(b) k ( x ) = g ( f ( x ) )
and state a suitable dom ai f each.
n or
(a) h ( x ) = f ( g ( x ) ) (b) k ( x ) = g ( f ( x ) )
= f (1)
x = g( x 2 − 2)
=(1) −2
2 1
x =
x −2
2
1
= −2 Dom ai { x : x ∈ , x ≠ ± 2 }
n:
x2
Dom ai { x : x ∈ , x ≠ 0}
n:
You wi probably only be asked f a dom ai i the f
ll or n f uncti i
on nvolved a f on or an
racti
even root. Rem em ber that i a f on the denom i
n racti nator cannot be zero and any
num ber bei square rooted cannot be negati
ng ve
eg f ( x ) = x + 1 could have dom ai { x : x ∈ , x ≥ −1}
n:
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5. Higher Mathematics Course Revision Notes
Graphs of Inverses
To draw the graph of an inverse functi ref
on, lect the graph of the functi i the
on n
li y = x
ne
y y=x
y = g( x )
y = g−1 ( x )
O x
Exponential and Logarithmic Functions
Exponential Logarithmic
y y y y = loga x
y = ax, a >1 y = ax, 0 < a <1
( a ,1)
O x
1 (1, a ) 1 (1, a ) 1
O x O x
Trigonometric Functions
y = si x
n y = cos x y = tan x
y y y
1 1
O x O x O x
–1 –1
180° 360° 180° 360°
180° 360°
Peri = 360°
od Peri = 360°
od Peri = 180°
od
tude = 1
Am pli tude = 1
Am pli Am pli
tude i undef ned
s i
Graph Transformations
The next page shows the ef ect of transf ati on the two graphs shown below.
f orm ons
( 2, 2 ) y = g ( x )
y y
1
O x
–1 0 3 x –1
180° 360°
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6. Higher Mathematics Course Revision Notes
Function Ef ect
f Ef ect on f ( x )
f Ef ect on si x °
f n
f ( x ) + a Shi ts the graph
f y y = g( x ) + 1 y y = si x ° + 1
n
a up the ( 2, 3 ) • 2
y-axis
1
• • ( 3,1)
( −1,1) 0 x
0 x 180° 360°
f ( x + a ) Shi ts the graph
f y y = g ( x + 1) y y = si ( x − 90 ) °
n
−a along the •
(1, 2 ) 1
x-axis
0 x
–2 0 2 x –1
180° 360°
− f (x) Reflects the y y y = − si x °
n
y = − g( x )
graph i the
n 0 1
x-axis –1 3 x
0 x
• ( 2, − 2 ) –1
180° 360°
f (−x ) Ref lects the ( −2, 2 ) y y = g( −x ) y = si ( − x ° )
n y
graph i the
n •
1
y-axi s
0 x
–3 0 1 x –1
−360° −180°
kf ( x ) Scales the graph y ( 2, 4 ) y = 2 g ( x ) y y = 1 si x °
2 n
vertically •
1
Stretches if 2
k >1 −1
0 x
x 2
Com presses if –1 0 3
k <1 180° 360°
f ( kx ) Scales the graph y (1, 2 ) y = g(2x ) y y = si 2 x °
n
horizontally •
1
Com presses if
k >1 0 x
Stretches if x
−1 0 3 –1
k <1 2 2 180° 360°
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7. Higher Mathematics Course Revision Notes
Differentiation
Differentiating
If f ( x ) = ax n then f ′ ( x ) = anx n −1
Bef you di f
ore f erenti all brackets should be m ulti ed out, and there should be no
ate, pli
f ons wi an x term i the denom i
racti th n nator (bottom li f exam ple:
ne), or
3 1 −1
1
= 1 x −2 = 3x −2 =x 2
3x 2 3 x2 x
Equations of Tangents
Tangents are strai li
ght nes, theref to f nd the equati of a tangent, you need a poi
ore i on nt
on the li and i gradi to substi
ne ts ent tute i y − b = m ( x − a )
nto
You wi always be gi one coordi
ll ven nate of the poi whi the tangent touches
nt ch
Fi the other coordi
nd nate by solvi the equati of the curve
ng on
nd ent f erenti ng then substi ng i the x-coordi
Fi the gradi by di f ati tuti n nate of the point
Example
Fi the equati of the tangent to the graph of y = x 3 at the poi where x = 9 .
nd on nt
y = x3 y = x3 At x = 9, m = 3 × 9 2
1
y − b = m(x − a)
2
= 93 = x2
3
=3 9 y − 27 = 9 ( x − 9 )
2
2
= 3 ×3 2 y − 54 = 9 x − 81
= 33 dy 3 1
= 2 x2 2
2 y = 9 x − 27
= 27 dx =9
2
( 9, 27 )
dy
Stationary poi occur at poi where
nts nts =0
dx
You m ust j f the nature of turni poi or poi of i lecti
usti y ng nts nts nf on
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8. Higher Mathematics Course Revision Notes
Graphs of Derived Functions
Quadratic Cubic Quartic
y y y
0
0 0
– +
+ –
+ – + –
0 +
x x x
0 0
y y y
+ +
+
x – – x
– + +
x
–
Linear Quadratic Cubic
Optimisation
These types of questions are usually practi problem s whi i
cal ch nvolve m axi um or
m
m i m um areas or volum es
ni
Rem em ber you m ust show that a m axi um or m i m um exi
m ni sts
Sequences
Linear Recurrence Relations
A linear recurrence relati i i the f
on s n orm un+1 = aun + b . Also be aware that thi m ay be
s
written as un = aun−1 + b
b
If −1 < a < 1 then a li i l =
m t exi You m ust state thi whenever you use the li i
sts. s m t
1− a
f ula
orm
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9. Higher Mathematics Course Revision Notes
Polynomials and Quadratics
Polynomials
The degree of a polynom i i the value of the hi
al s ghest power, eg 3 x 4 + 3 has degree 4
Syntheti di si (nested f ) can be used to f
c vi on orm actori polynom i
se als
Example
4 x 3 − 7 x 2 + 11
Find .
x +2
–2 4 –7 0 11 Rem em ber to put i 0
n
i there i no term
f s
–8 30 –60
4 –15 30 –49
4 x 3 − 7 x 2 + 11
= 4 x 2 − 15 x + 30 rem ainder − 49
x +2
i 4 x 3 − 7 x 2 + 11 = ( x + 2 ) ( 4 x 2 − 15 x + 30 ) − 49
e
If the di sor i a f
vi s actor then the rem ainder i zero
s
If the rem ainder i zero then the di sor i a f
s vi s actor
Completing the Square
The x 2 term m ust have a coef i ent of one. If i does not, you m ust take out a com m on
f ci t
2
factor from the x and x term , but not the constant
orm y = a ( x + p ) + q the turni poi of the graph i ( − p, q )
2
In the f ng nt s
Example
Wri 3 x 2 − 12 x + 7 i the f
te n orm 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 i thi exam ple, the graph i ∪ -shaped si the x 2 coef i ent i posi ve; and
n s s nce f ci s ti
the turni poi i ( 2, − 5 ) .
ng nt s
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10. Higher Mathematics Course Revision Notes
The Discriminant
The di m i
scri nant i part of the quadrati f ula and can be used to i cate how m any
s c orm ndi
roots a quadrati has. For the quadrati ax + bx + c :
c c 2
If b 2 − 4 ac > 0 , the roots are real and unequal (di nct)
sti 2 roots
If b 2 − 4 ac = 0 , the roots are real and equal (i repeated roots)
e 1 root
If b 2 − 4 ac < 0 , the roots are not real; they do not exist no roots
The di m i
scri nant can also be used to calculate the num ber of intersecti between a
ons
li and a curve. To use i you m ust f rst equate them and set equal to zero, bef
ne t, i ore
usi the di m i
ng scri nant
Rem em ber i b 2 − 4 ac = 0 , the li i a tangent
f ne s
Integration
Integrating
ax n +1
ax n dx = +c
n +1
As wi di f
th f erenti on, all brackets m ust be m ulti ed out, and there m ust be no
ati pli
f ons wi an x term i the denom i
racti th n nator
Examples
dx x 2 + 5x 7
1. Find . 2. dx
8
x5 x2
dx 1 x 2 + 5x 7
= dx 2
dx = x −2 ( x 2 + 5 x 7 ) dx
8
x5 8
x5 x
= x
−5
8
dx = x 0 + 5 x 5 dx
x8
3
= 1 + 5 x 5 dx
= 3 +c
8 = x + 5 x6 + c
6
3
= 8 x8 + c
3
= 3 8 x3 + c
8
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11. Higher Mathematics Course Revision Notes
The Area under a Curve
b
If F ( x ) i the i
s ntegral of f ( x ) , then f ( x ) dx = F (b ) − F ( a )
a
y
y = f (x)
a b x
Rem em ber that areas spli by the x-axi m ust be calculated separately and any negati
t s ve
si i
gns gnored; these j show that the area i under the axi
ust s s.
The Area between two Curves
b
The area between the graphs of y = f ( x ) and y = g ( x ) i def ned as
s i a
f ( x ) − g ( x ) dx
y
y = g( x )
ven, f ( x ) and g ( x )
If the li i are not gi
m ts
should be equated to f nd a and b
i
x
a b y = f (x)
Trigonometry
Background Knowledge
You should know how to use all of the i orm ati below:
nf on
SOH CAH TOA
si x
n
tan x =
cos x
si 2 x + cos2 x = 1
n
a b c
The si rule:
ne = =
si A si B si C
n n n
b2 + c 2 − a2
The cosi rule: a = b + c − 2bc cos A or cos A =
ne 2 2 2
2bc
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12. Higher Mathematics Course Revision Notes
The area of a triangle, A = 1 ab si C
2 n
CAST diagram s
Exact values:
2 30°
2 45° 3
1
45° 60°
1 1
Radians
You should know how to convert between radi and degrees:
ans
360° = 2 90° = 2 45° = 4 ÷ 180 × →
Degrees Radians
180° = 60° = 3 30° = 6 ans × 180 ÷ →
Radi Degrees
5 × 180
eg 5 =
6 = 150°
6
Trigonometric Equations
Look at the restri ons on the dom ai eg 0 ≤ x ° < 360, or 0 ≤ x <
cti n,
Be aware of whether the answer i requi i degrees or radi
s red n ans
Rem em ber a CAST diagram whenever you are asked to “solve”
Examples
1. Solve 3si 2 x° = 1 where 0 ≤ x ° < 360 .
n
3si 2 x ° = 1
n
3 ( si x ° ) = 1
2
n
( si x ° )2 = 1
n 3
si x ° = ± 1
n 3 S A
x ° = si
n −1
(± 1 )
3 T C
x° = 35.3° x° = 180 − 35.3 x° = 180 + 35.3 x° = 360 − 35.3
= 144.7° = 215.3° = 324.7°
Soluti set = {35.3°, 144.7°, 215.3°, 324.7°}
on
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13. Higher Mathematics Course Revision Notes
2. Solve 2 si 2 x − 1 = 0 , 0 ≤ x < 2 .
n
2 si 2 x − 1 = 0
n
2si 2 x = 1
n
si 2 x = 1
n 2 S A
2 x = si −1 ( 1 )
n 2 T C
2 x ° = 30° 2 x ° = 180° − 30°
x ° = 15° 2 x ° = 150°
x ° = 75°
2 x ° = 360° + 30° 2 x ° = 360° +180° − 30°
2 x ° = 390° 2 x ° = 510°
x ° = 195° x ° = 255°
15
15° = 180 75
75° = 180 195
195° = 180 255
255° = 180
= 36
3 = 15
36 = 36
39 51
= 36
= 12 5
= 12 = 12
13 = 17
12
ons {
Soluti set = 12 , 12 , 12 , 17
5 13
12 }
Compound Angle Formulae
cos ( A ± B ) = cos A cos B si A si B
n n
si ( A ± B ) = si A cos B ± cos A si B
n n n
These are gi on the f ula sheet
ven orm
Double Angle Formulae
si 2 A = 2si A cos A
n n
cos 2 A = cos2 A − si 2 A
n
= 1 − 2 si 2 A
n
= 2 cos 2 A − 1
These are gi on the f ula sheet
ven orm
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14. Higher Mathematics Course Revision Notes
Circles
Equations of Circles
A ci wi centre ( a,b ) and radi r has the equati ( x − a )2 + ( y − b ) = r2
2
rcle th us on
Note that i a ci has centre ( 0, 0 ) then the equati i x 2 + y 2 = r2
f rcle on s
The equati can also be gi i the f
on ven n orm x 2 + y 2 + 2 gx + 2 fy + c = 0 where the centre
i ( − g,− f
s ) and the radi r = g2 + f 2 − c
us
You do not have to rem em ber any of these equations, si they are all gi i the
nce ven n
exam
stance f ula, d = ( x2 − x1 ) + ( y2 − y1 ) , si thi
2 2
You wi have to rem em ber the di
ll orm nce s
i not gi
s ven, and i f
s requently used i ci questi
n rcle ons
Intersection of a Line and a Circle
two intersections one intersection (tangency) no intersections
Rem em ber, a tangent and a li f
ne rom the centre of a circle wi m eet at ri angles,
ll ght
whi m eans that m1 × m2 = −1 can be used
ch
Vectors
Basic Facts
A vector i a quanti wi both m agni
s ty th tude (si and di on
ze) recti
A vector i nam ed ei
s ther by usi a di
ng rected li segm ent (eg AB ) or a bold letter
ne
(eg u written u)
A vector m ay also be def ned i term s of i, j and k, the uni vectors i three
i n t n
perpendicular di ons:
recti
1 0 0
i= 0 j= 1 k= 0
0 0 1
a
The m agnitude of vector AB = b i def ned as AB = a 2 + b 2
s i
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15. Higher Mathematics Course Revision Notes
a1 b1 a1 ± b1 a1 ka1 0
a2 ± b2 = a2 ± b2 k a2 = ka2 where k i a scalar
s Z ero vector: 0
a3 b3 a3 ± b3 a3 ka3 0
OA i called the posi on vector of the poi A relati to the ori n, wri
s ti nt ve gi tten a
AB = b − a where a and b are the posi on vectors of A and B
ti
If AB = k BC where k i a scalar, then AB i parallel to BC . Si B i com m on to both
s s nce s
AB and k BC , then A, B and C are collinear
Dividing Vectors in a Ratio
The poi P can also be worked out f
nt rom f rst pri ples, or
i nci
U si the secti f ula. If P di des AB i the rati m : n , then:
ng on orm vi n o
n m
p= a+ b where p i the posi on vector OP
s ti
m+n m+n
Example
P i the poi ( −2, 4, − 1) and R i the poi ( 8, − 1,19 ) . Poi T di des PR i the rati
s nt s nt nt vi n o
2:3. Work out the coordinates of poi T.
nt
2:3
P R
T
U si the secti f ul
ng on orm a From fi pri pl
rst nci es
The rati i 2:3, so let m = 2 and n = 3
o s PT 2
=
n m TR 3
t= p+ r
m+n m+n 3PT = 2TR
= 3 p + 2r
5 5 ( )
3 t − p = 2 (r − t )
(
= 1 3 p + 2r
5 ) 3t − 3 p = 2r − 2t
3t + 2t = 2r + 3 p
−6 16
=1 12 + −2 16 −6
5
−3 38 5t = −2 + 12
38 −3
10
= 1 10 10
5
35 5t = 10
35
2
= 2 2
7 t= 2
7
Theref T i the poi ( 2, 2, 7 ) .
ore s nt Theref T i the poi ( 2, 2, 7 ) .
ore s nt
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16. Higher Mathematics Course Revision Notes
The Scalar Product
The scalar product a.b = a b cos , where i the sm allest angle between a and b
s
Rem em ber that both vectors m ust poi away f
nt rom the angle, eg
a b
a1 b1
If a = a2 and b = b2 then a.b = a1b1 + a2b2 + a3b3
a3 b3
a.b ab +a b +a b
cos = 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 .
4 1
Let a = u + v Let b = u − v
8 4 8 4
a= 0 + 0 b= 0 − 0
4 1 4 1
12 4
a= 0 b= 0
5 3
a.b
cos =
ab
(12 × 4 ) + ( 0 × 0 ) + ( 5 × 3 )
=
122 + 02 + 52 4 2 + 02 + 32
63
=
169 25
63
= cos −1
169 25
= 14· °
3
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17. Higher Mathematics Course Revision Notes
Further Calculus
Trigonometry
Differenti on
ati
Thi i strai orward, si the f ulae are gi on the f ula sheet:
s s ghtf nce orm ven orm
f (x) f ′( x )
si ax
n a cos ax
cos ax − a si ax
n
Integration
Agai the f ulae are provi i the paper:
n, orm ded n
f (x) f ( x ) dx
si ax
n − 1 cos ax + c
a
cos ax 1 si ax + c
a n
Examples
f erenti x 3 + cos3 x wi respect to x.
1. Di f ate th
( )
d x 3 + cos3 x = 3 x 2 − 3si x
dx
n3
2. Find 4 x 3 + si x dx .
n3
4x 4 1
4 x + si x dx =
3
n3 − 3 cos3 x + c
4
= x 4 − 1 cos3 x + c
3
Chain Rule Differentiation
n −1 n −1
If f ( x ) = ( ax + b )n then f ′ ( x ) = n ( ax + b ) × a = an ( ax + b )
or
n −1
If f ( x ) = ( p ( x ) ) then f ′ ( x ) = n ( p ( x ) )
n
× p′ ( x )
“The power m ulti es to the f
pli ront, the bracket stays the sam e, the power lowers by one
and everythi i m ulti ed by the di f
ng s pli f erenti of the bracket”
al
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18. Higher Mathematics Course Revision Notes
Examples
1
1. G i f ( x ) =
ven + x − si x , f nd f ′ ( x ) .
n3 i
x2
1
f ( x ) = x −2 + x 2 − si x
n3
−1
f ′ ( x ) = −2 x −3 + 1 x 2 − 3cos3 x
2
2 1
=− 3 + − 3cos3 x
x 2 x
ven f ( x ) = ( 3 x 2 + 2 x + 1) , f nd f ′ ( x ) .
3
2. G i i
f ′ ( x ) = 3 ( 3 x 2 + 2 x + 1) × ( 6 x + 2 )
2
= 3 ( 6 x + 2 ) ( 3 x 2 + 2 x + 1)
2
f erenti y = cos 2 x = ( cos x )2 wi respect to x.
3. Di f ate th
dy
= 2 ( cos x ) × ( − si x )
n
dx
= −2cos x si x n
n
Integration of (ax + b)
( ax + b )n+1
( ax + b ) dx =
n
+c
( n + 1) × a
Example
Find ( 3 x + 5 )4 dx .
( 3 x + 5 )5 ( 3 x + 5 )5
( 3 x + 5 ) dx =
4
+c = +c
5× 3 15
It i possi f any type of ‘urther calculus’ to be exam i i the style of a standard
s ble or f ned n
calculus questi (eg opti i on, area under a curve, etc)
on m sati
Exponentials and Logarithms
An exponenti i a f
al s uncti i the f
on n orm f ( x ) = a x
Logari s and exponenti are i
thm als nverses
y = a x ⇔ loga y = x
On a calculator, log i log10 and ln i loge
s s
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19. Higher Mathematics Course Revision Notes
Laws of Logarithms
loga x + loga y = loga xy (Squash)
loga x − loga y = loga x
y (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
= log2
3
= log2 8
=3 (si 23 = 8)
nce
2. Below i a di
s agram of part of the graph of y = ke 0.7 x
y
y = ke 0.7 x
3
0 x =1 x
(a) Fi the value of k
nd
(b) The li wi equati x = 1 i
ne th on ntersects at R. Fi the coordi
nd nates 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 i the poi (1, 6.04 ) .
s nt
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20. Higher Mathematics Course Revision Notes
The Wave Function
Example
1. Express 6 si x ° − 2 cos x ° i the f
n n orm k cos ( x − a ) ° where 0 ≤ a ° < 360 .
k cos ( x − a ) ° = k cos x ° cos a ° + k si x ° si a °
n n
= k cos a ° cos x ° + k si a ° si x °
n n
k si a °
k cos a ° = − 2 (− 2 ) n
2 2
k= + 6 tan a ° =
k cos a °
k si a ° = 6
n
= 2+6 6
=−
S A = 8 2
T C =2 2 =− 3
a ° = 180° − tan −1 ( 3 )
= 180° − 60°
= 120°
Therefore 6 si x ° − 2 cos x ° = 2 2 cos ( x − 120 ) ° .
n
2. Express cos x − si x i the f
n n orm k si ( x + ) where 0 ≤ < 2 .
n
k si ( x + ) = k si x cos + k cos x si
n n n
= k cos si x + k si cos x
n n
k cos = −1 k = ( −1)2 + 12 k si
n
tan =
k si = 1
n = 2
k cos
= −1
S A
° = 180° − tan −1 (1)
T C
= 180° − 45°
= 135°
135
= 180
=3
4
Theref cos x − si x = 2 si x + 3
ore n n 4 ( ).
The m axi um value of an expressi i the f
m on n orm k cos ( x ± a ) occurs when
cos ( x ± a ) = 1 ; and si ( x ± a ) = 1 f k si ( x ± a )
n or n
The m i m um value of an expressi
ni on i the f
n orm k cos ( x ± a ) occurs when
cos ( x ± a ) = −1 ; and si ( x ± a ) = −1 f k si ( x ± a )
n or n
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