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Ti liu tp hun ging dy mn Ton bng ting Anh

1. Mathematical English Thut ng ton hc ting Anh

Chu Thu Hon - Trng PT Chuyn ngoi ng , i hc ngoi ng,HQG H ni

2. Mt s vn trong vic son bi v ging dy mn Ton bng ting Anh

Nhm bin son ti liu tp hun mn Ton

3. The sine rule and the cosine rule Cc cng thc sin v cos

T Ngc Tr B gio dc

Loi bi ging: bi ging trung hc ph thng

4. Trig Derivative o hm hm lng gic

T Ngc Tr B gio dc

Loi bi ging: bi ging trung hc ph thng

5. Equation of Circle Phng trnh ng trn

Nguyn c Thng Trng PT Amsterdam H Ni

Loi bi ging: bi ging trung hc ph thng

6. Parametric equation of a line Phng trnh tham s ca ng thng (chuyn th bi ging ting Vit sang ting Anh)

Nguyn c Thng Trng PT Amsterdam H Ni

Loi bi ging: bi ging trung hc ph thng

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7. Geometric Sequences Dy cp s nhn

Trn Thanh Tun i hc Khoa hc T nhin H Ni

Loi bi ging: A-level

8. Application of Differentiation: Related Rates ng dng ca php tnh vi phn: Cc tc bin thin ph thuc nhau

Trn Thanh Tun i hc Khoa hc T nhin H Ni

Loi bi ging: A-level

9. Permutations and Combinations - Hon v v T hp

Chu Thu Hon - Trng PT Chuyn ngoi ng , i hc ngoi ng,HQG H ni

Loi bi ging: SAT

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LI NI U

Vi mc ch l trong khong 8 bui tp hun, cc thy c s son c bi ging, v bc u c th ging bi c bng ting Anh, nn b ti liu tp hun ny s trnh by nhng hng dn ht sc c bn vi cc bi son mu c ni dung khng kh a s cc thy c s khng gp vn kh khn g trong ni dung, m ch tp trung vo cc phng php son bi ging v cch thc ging bi trn lp. c th ging bi trn lp, u tin cc thy c s phi son gio n bi ging . Cng vic ny kh l tng t vi vic son bi ging bng ting Vit, ch khc l cc thy c s phi son bng ting Anh. V vy, bi vit u tin trong ti liu tp hun l mt bng lit k cc thut ng Ton hc bng ting Anh kh l c bn v y . Nhng thut ng ny s l nhng t vng chuyn mn Ton cn thit cho cng vic ging dy ca cc thy c. Tip theo , bi vit th hai, nhm bin son trnh by nhng hng dn c bn v cn thit cc thy c c th son mt bi ging v cc bc trnh by bi ging trn lp. Bi vit ny s cung cp cho cc thy c nhng hng dn kh l chi tit cc thy c c th p dng v thc hnh c ngay. Phn cn li ca ti liu gm by bi son mu, trong bn bi c son theo cch chuyn th t bi son ting Vit sang ting Anh, hai bi c son theo gio trnh ging dy A-level, v mt bi c son theo gio trnh ging dy SAT. By bi son mu ny s cung cp cho cc thy c nhng v d cc thy c bc u c th thc hnh son nhng bi ging ca mnh.

Nhm bin son hy vng nhng ti liu ny s gip ch c cc thy c mt phn trong vic ging dy Ton bng ting Anh. Ti liu c bin son trong mt thi gian khng di nn s c khng t thiu st, rt mong cc thy c gp nhm bin son c th chnh sa thnh mt ti liu tt hn.

Nhm bin son ti liu tp hun mn Ton.

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MATHEMATICAL ENGLISH

By CHU THU HON

GV Trng PT Chuyn ngoi ng, i hc ngoi ng, HQG H ni

Arithmetic

0 zero 10 Ten 20 twenty 1 one 11 Eleven 30 thirty 2 two 12 Twelve 40 forty 3 three 13 thirteen 50 fifty 4 four 14 fourteen 60 sixty 5 five 15 fifteen 70 seventy 6 six 16 sixteen 80 eighty 7 seven 17 seventeen 90 ninety 8 eight 18 eighteen 100 one hundred 9 nine 19 nineteen 1000 one thousand

-245 minus two hundred and forty-five 22 731 twenty-two thousand seven hundred and

thirty-one 1 000 000 one million

56 000 000 fifty-six million 1 000 000 000 one billion [US usage, now universal] 7 000 000 000 seven billion [US usage, now universal]

1 000 000 000 000 one trillion [US usage, now universal] 3 000 000 000 000 three trillion [US usage, now universal]

Fractions [= Rational Numbers]

12

one half 38

three eighths

13

one third 269

twenty-six ninths

5

14

one quarter [= one fourth] 5

34 minus five thirty-fourths

15

one fifth 327

two and three sevenths

117

minus one seventeenth

Real Numbers

-0.067 minus nought point zero six seven 81.59 eighty-one point five nine

62.3 10 minus two point three times ten to the six [= -2 300 000 minus two million three hundred thousand]

34 10 four times ten to the minus three [ = 0.004=4/1000 four thousandths

[ 3.14159...] pi [pronounced as pie] [ 2.71828...]e e [base of the natural logarithm]

Complex Numbers

i I 3 4i three plus four i 1 2i one minus two i

1 2 1 2i i the complex conjugate of one minus two i equals one plus two i

The real part and the imaginary part of 3 4i are equal, respectively, to 3 and 4.

Basic arithmetic operations

Addition: 3 5 8 three plus five equals [ = is equal to] eight

Subtraction: 3 5 2 three minus five equals [ = ] minus two

Multiplication: 3 5 15 three times five equals [ = ]

6

fifteen Division: 3 / 5 0.6 three divided by five equals [ =

] zero point six

2 3 6 1 5 Two minus three in brackets times six plus one equals minus five

1 3 12 4 3

One minus three over two plus four equals minus one third 4! [ 1 2 3 4] four factorial

Exponentiation, Roots

25 5 5 25 five squared 35 5 5 5 125 five cubed 45 5 5 5 5 625 five to the (power of) four

15 1 5 0.2 five to the minus one 25 21 5 0.04 five to the minus two

3 1.73205... the square root of three 3 64 4 the cube root of sixty four 5 32 2 the fifth root of thirty two In the complex domain the notation n a is ambiguous, since any non-zero complex number has n -th roots. For example, 4 4 has four possible values: 1 i (with all possible combinations of signs).

2 21 2 one plus two, all to the power of two plus two

1ie e to the (power of) pi i equals minus one

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Divisibility

The multiples of a positive integer a are the numbers , 2 ,3 , 4 ,...a a a a . If b is a multiple of a , we also say that a divides b , or that a is a

divisor of b (notation: |a b ). This is equivalent to ba

being an integer.

Division with remainder

If ,a b are arbitrary positive integers, we can divide b by a , in general, only with a remainder. For example, 7 lies between the following two conseentive multiples of 3:

7 12 3 6 7 3 3 9. 7 2 3 1 23 3

In general, if qa is the largest multiple of a which is less than or equal to b , then

, 0,1,..., 1b qa r r a

The integer .,q resp r is the quotient (resp the remainder) of the division of b by a .

Euclids algorithm

The algorithm computes the greatest common divisor (notation: ,a b = gcd ,a b ) of two positive integers ,a b

It proceeds by replacing the pair ,a b (say, with a b ) by r ; a where r is the remainder of the division of b by a . This procedure, which preserves the gcd, is repeated until we arrive at 0r .

Example. Compute gcd(12, 44).

44 3 12 812 1 8 48 2 4 0

gcd(12,44) = gcd(8,12) = gcd(0,4) = 4.

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This calculation allows us to write the fraction 4412

in its lowest terms,

and also as a continued fraction:

44 44 4 11 13 112 12 4 3 12

If gcd ,a b =1, we say that a and b are relatively prime.

add (v) /d/ cng

algorithm (n)

Euclids algorithm

/lrm/

/jukld/

thut ton

thut ton Euclid

bracket (n) /brkt/ du ngoc

Left bracket

right bracket

curly bracket

/left/

/rat/

/kli/

du ngoc tri

du ngoc phi

du ngoc {}

denominator (n) /dnmnet(r)/ mu s

difference (n) /dfrns/ hiu

divide (v) /dvad/ chia

divisibility (n) /dvzblti/ tnh chia ht

Divisor (n) /dvaz(r)/ s chia

exponent (n ) /kspnnt/ s m

factorial (n) /fktril/ giai tha

fraction (n)

continued fraction

/frkn/

/kntnjud /

phn s

phn s lin tc

gcd [= greatest common divisor]

c s chung ln nht

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lcm [= least common multiple]

bi s chung nh nht

infinity (n) /nfnti/ v cc, v tn

Iterate (v) /tret/ ly nguyn hm

iteration (n) /tren/ nguyn hm

multiple (n) /mltpl/ bi s

multiply (v) /mltpla/ nhn

number (n)

even number (n)

odd number (n)

/nmb(r) /

/ivn/

/d/

s

s chn

s l

numerator (n) /njumret(r)/ t s

pair (n)

pairwise

/pe(r)/

/pe(r) waz/

cp

tng i, tng cp

power (n) /pa(r)/ ly tha

product (n) /prdkt/ tch

quotient (n) /kwnt/ thng s

ratio (n) /rei/ t s

rational

irrational (a)

/rnl/

/rnl/

hu t

v t

relatively prime (n) /reltvli/ - /pram/ s nguyn t cng nhau

remainder (n) /rmend(r)/ d, s d

root (n) /rut/ cn, nghim

sum (n) /sm/ tng s

subtract (v) /sbtrkt/ tr

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Algebra

Algebraic Expressions

2A a Capital a equals small a squared a x y a equals x plus y b x y b equals x minus y

c x y z c equals x times y times z c x y z c equals x y z

x y z xy x plus y in brackets z plus x y 2 3 5x y z x squared plus y cubed plus z to the (power of)

five n n nx y z x to the n plus y to the n equals z to the n

3mx y x minus y in brackets to the (power of) three m x minus y , all to the (power of) three m

2 3x y Two to the x times three to the y 2ax bx c a x squared plus b x plus c

3x y The square root of x plus the cube root of y n x y The n -th root of x plus y

a bc d

a plus b over c minus d

nm

(the binomial coefficient ) n over m

Indices

0x x zero; x nought

1 ix y x one plus y i

i jR (capital) R (subscript) i j ; (capital) R lower i j ;

ki jM (capital) M upper k lower i j;

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(capital) M superscript k subscript i j

0

n iii

a x sum of a I x to the i for i from nought [= zero] to n

sum over i (ranging) from zero to n of a i (times) x to the i

1 mmb

product of b m for m from one to infinity; product over m (ranging) from one to infinity of b m

1

ni j j kj

a b sum of a I j times b j k for j from one to n;

sum over j (ranging) from one to n of a i j times b j k

0

n i n ii

nx y

i

sum of n over i x to the i y to the n minus i for i

from nought [= zero] to n.

Matrices

column (n)

column vector /klm/

/vekt(r)/

ct

vect ct

determinant (n) /dtmnnt/ nh thc

index (n)

(pl. indices) /ndeks/

( /ndsiz/)

s m

Matrix (n)

matrix entry (pl. entries)

m n matrix [ m by n matrix]

/ metrks/

/entri/

ma trn

h s ma trn

ma trn m n

row (n)

row vector /r/

/vekt(r)/

hng

vect hng

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Inequalities

x y x is greater than y x y x is greater (than) or equal to y x y x is smaller than y x y x is smaller (than) or equal to y

0x x is positive 0x x is positive or zero; x is non-negative 0x x is negative 0x x is negative or zero

Polynomial equations

A polynomial equation of degree 1n with complex coefficients

10 1 0... 0 0n n nf x a x a x a a

has n complex solutions (= roots), provided that they are counted with multiplicities.

For example, a quadratic equation 2ax 0 ( 0)bx c a

can be solved by completing the square , i.e. , by rewriting the L.H.S as

2ons tana x c t another constant.

This leads to an equivalent equation 2 2 4 ,

2 4b b aca xa a

whose solutions are

1,2 2bx

a

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where 22 2 1 24b ac a x x is the discriminant of the original equation. More precisely,

2 1 2ax .bx c a x x x x

If all coefficients , ,a b c are real, then the sign of plays a crucial role

If 0 , then 1 2 2x x b a is a double root;

If 0 , then 1 2x x are both real;

If 0 , then 1 2x x are complex conjugates of each other (and non-real).

coefficient (n) /kfnt/ h s Degree (n) /dri/ , cp bc discriminant bit s, bit thc Equation /kwen/ phng trnh L.H.S. [= left hand side] R.H.S. [= right hand side]

v tri v phi

polynomial (adj) /plinomil/ a thc polynomial (n) phng trnh i s Provided that /prvadd/ ga s root (n) /smpl/ cn, nghim simple root /dbl/ nghim n double root /trpl/ nghim kp Triple root /mltpl/ nghim bi ba multiple root nghim bi solution(n) /slun/ nghim, li gii,

php gii solve (v) /slv/ gii

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Congruences

Two integers ,a b are congruent modulo a positive integer m if they have the same remainder when divided by m (equivaqlently, if their difference a b is a multiple of m ).

(mod )a b m a is congruent to b modulo m

( )a b m

Some people use the following, slightly horrible, notation: .a b m

Fermats Little Theorem. If p is a prime number and a is an integer, then modpa a p . In other words, pa a is always divisible by p .

Chinese Remainder Theorem. If 1 2, ,..., km m m are pairwise relatively prime integers, then the system of congruences

1 1mod ... modk kx a m x a m

has a unique solution modulo 1 2, ,..., km m m , for any integers 1 2, ,..., ka a a .

Geometry

Lines and angles

line AB (n) /lain/ ng thng AB

ray AB (n) /rei/ tia AB

line segment AB (n) /lain 'segmnt/ on thng AB

length of segment AB (n)

/le v 'segmnt/ chiu di on thng AB

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AB XY means the same thing as AB = XY

/mi:nz seim i z/

ng ngha vi, tng t nh

angle(n) /'gl/ gc

vertex(n) /v:teks/ nh

acute angle /'kju:t 'gl/ gc nhn

right angle /'rait 'gl/ gc vung

obtuse angle /b'tju:s 'gl/ gc t

straight angle /stret 'gl/ gc bt

mA = mB we can write A B: congruent

/kgrunt/ tng ng,bng

supplementary (a) /splmntri/ ph

complementary(a) /,kmpli'mentri/ b

vertical angle(n) /'v:tikl 'gl/ gc i nh

bisect(v) /baisekt/ chia i

midpoint(n) /midpint/ trung im

perpendicular(a) /p:pn'dikjul/ vung gc

parallel(a) /'prlel/ song song

transversal(n)(a) /trnzv:sl/ ng ngang,ngang

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Triangle exterior angle(n) /eks'tiri 'gl/ gc ngoi

scalene triangle(n) /skeili:n traigl/ tam gic thng

isosceles triangle(n) /aissili:z traigl/ tam gic cn

equilateral

triangle(n)

/i:kwiltrl

traigl/

tam gic u

acute triangle(n) /'kju:t traigl/ tam gic nhn

Obtuse triangle (n) /b'tju:s traigl/ tam gic t

right triangle(n) /'rait traigl/ tam gic vung

hypotenuse(n) /hai'ptinju:z/ cnh huyn

leg(n) /leg/ cnh gc vung

Pythagorean

theorem and

corollaries

/paigrin 'irm

nd k'rlris/

nh l pythagore v

h qu

perimeter (n) /primit/ chu vi

triangle inequality(n) /traigl ,ini:'kwliti/ bt ng thc tam gic

height(n) /hait/ chiu cao

altitude(n) /ltitju:d/ chiu cao

Similar triangles(n) /simil traigls/ tam gic ng dng

ratio of similitude(n) /reiiou v

similitju:d/

t s ng dng

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Quadrilaterals and other polygons /kwdriltrls nd r pligns/ t gic v cc a

gic khc polygon(n)

/plign/ a gic

side (n)

/said/ Cnh

vertex (n)

/v:teks/ nh

vertices

s nhiu ca vertex

diagonal(n)

/daignl/ ng cho

quadrilateral(n)

/kwdriltrl/ t gic

regular polygon (n)

/'rgjul plign/ a gic u

exterior angle (n)

/eks'tiri 'gl/ gc ngoi

parallelogram(n) /prlelgrm/ hnh bnh hnh base (n) /beis/ y height (n) /hait/ chiu cao opposite sides are parallel: AB // CD and AD // BC

/'pzit saids : 'prlel/

cc cnh i din song song

opposite sides are congruent: AB CD and AD BC

/'pzit saids : kgrunt/

cc cnh i din tng ng/bng nhau

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opposite angles are congruent: A C and B D

/'pzit 'gls :kgrunt/

cc gc i din tng ng /bng nhau

two congruent triangles: ABC ACD

/tu: kgrunt traigls/:

hai tam gic tng ng/bng nhau

rectangle (n) /rektgl/ hnh ch nht length (n) /le/ chiu di width (n) /wd/ chiu rng

rhombus (n) /rmbs/ hnh thoi Square (n) /skwe/ hnh vung trapezoid(n) /trpizid/ hnh thang isosceles trapezoid(n)

/aissili:z trpizid/ hnh thang cn

perimeter (n)

/primit/ chu vi

Circles

circle (n) /'s:kl/ ng trn

center (n) /'sent/ tm

radius (n) /reidis/ bn knh diameter(n) /dai'mit/ ng knh chord (n) /krd/ dy cung circumference(n) /s:kmfrns/ chu vi ng trn semicircle(n)

/semis:kl/ na ng trn, bn nguyt

arc (n) /rk/ cung

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intercept (v) /'intsept/ chn central angle (n) /sentrl 'gl/ gc tm

Solid and coordinate geometry

rectangular solid =

box(n)

/rek'tgjul/ /'slid/ Hnh hp ch nht

face (n) /feis/ mt, b mt

edge (n) /ed/ Cnh

length (n) /le/ Chiu di, di

width (n) /wid / Chiu rng

height (n) /hait/ Chiu cao

cube (n) /kju:b/ Hnh lp phng

volume (n) /'vljum/ Khi, th tch

cubic unit (n) /'kju:bik/ /'ju:nit/ n v lp phng

surface area (n) /'s:fis/ /'eri/ Din tch b mt

diagonal (n) /dai'gnl/ ng cho

cylinder(n) /'silind/ Hnh tr

circle (n) /'s:kl/ ng trn

prism (n) /prism/ Lng tr

two congruent

parallel bases

/kgrunt/ /'prlel/

hai y song song

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right prism (n) /rait/ /prism/ Lng tr ng

cone (n) /koun/ Hnh nn

slant height (n) /sl:nt/ /hait/ ng sinh

circumference(n) /s'kmfrns/ Chu vi

Lateral surface area

(n)

/'ltrl/ /'s:fis/

/'eri/

Din tch xung

quang

pyramid(n) /'pirmid/ Hnh chp

polygon(n) /'plign/ a gic

Square(n) /skwe/ Bnh phng, hnh

vung

triangle (n) /traigl/ Hnh tam gic

Sphere(n) /sfi/ Hnh cu, mt cu

Radius(n) /reidis/ Bn knh

radii(n) /'reidjai/ S nhiu ca bn

knh

x-axis(n) /eks/ /'ksis/ Trc honh

y-axis (n) /wai/ /'ksis/ Trc tung

origin(n) /'ridin/ Gc ta

quadrant(n) /'kwdrnt/ Gc phn t, cung

phn t

x-coordinate (n) /eks/ /kou':dnit/ Honh

y-coordinate (n) /wait/ /kou':dnit/ Tung

horizontal line (n) /,hri'zntl/ /lain/ ng thng song

song trc honh

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vertical line (n) /'v:tikl/ /lain/ ng thng song

song trc tung

midpoint of a

segment (n)

/'segmnt/ Trung im ca

on thng

slope (n)=gradient /sloup/ H s gc

equation of line (n) /i'kwein/ /v, v/

/lain/

Phng trnh

ng thng

y-intercept (n) /'intsept/ Giao vi trc tung

parabola /p'rbl/ Parabol

axis of symmetry /'ksis/ /v, v/

/'simitri/

Trc i xng

vertex or turning

point

/'v:teks/ - /'t:ni/

/pint/

nh

y = ax2 + bx + c, a >

0

The parabola opens

upward and the

vertex is the lowest

point on the

parabola

y = ax2 + bx + c, a <

0

The parabola opens

downward and the

vertex is the highest

point on the parabola

Ellipse(n) /ilips/ elip

hyperbola(n) /haip:bl/ hyperbol

hexagon (n) /heksgn/ hnh su cnh

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parallelogram (n) /prlelgrm/ hnh bnh hnh pass through (v) /p:s ru:/ i qua pentagon (n) /pentgn/ hnh ng gic plane /plein/ mt phng (n),

phng (adj) point (n) /pint/ im (regular) polygon (n) (/'rgjul/) /plign/ a gic (u) (regular) polyhedron (n)

(/'rgjul/) /plihi:drn/(pl. polyhedra) (s nhiu. /'pli'hedr/)

khi nhiu mt (u)

projection (n) /prdekn/ hnh chiu central projection (n)

/sentrl prdekn/ php chiu qua tm

orthogonal projection (n)

/:gnl prdekn/

php chiu trc giao

parallel projection (n) /'prlel prdekn/ php chiu song song

quadrilateral (n) /kwdriltrl/ t gic radius (pl. radii) (n) /reidis/ bn knh rectangle (n) /rektgl/ hnh ch nht rectangular (adj) /rek'tgjul/ thuc hnh ch

nht, vung gc rotation (n) /routein/ php quay side (n) /said/ cnh slope (n) /sloup/ dc Sphere (n) /sfi/ hnh cu, mt cu Square /skwe/ bnh phng(n),

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vung(adj) surface (n) /srfs/ b mt tangent to (v) /'tndnt/ tip tuyn vi tangent line (n) /'tndnt lain/ ng tip tuyn tangent hyper (plane) (n)

/'tndnt haip/ siu phng tip xc

tetrahedron (n) /tetrhi:drn/ khi t din triangle (n) /traigl/ hnh tam gic equilateral triangle (n) /i:kwiltrl

traigl/ tam gic u

isosceles triangle (n) /aissili:z traigl/ tam gic cn right-angled triangle (n)

/raitld traigl/ tam gic vung

vertex (n) /v:teks/ nh

Eulers Formula Let P be a convex polyhedron. Eulers formula asserts that

2,V E F

V = the number of vertices of P ,

E = the number of edges of P ,

F = the number of faces of P . Exercise. Use this formula to classify regular polyhedra (there are precisely five of them: tetrahedron, cube, octahedron, dodecahedron and icosahedron). For example, an icosahedron has 20 faces, 30 edges and 12 vertices. Each face is an isosceles triangle, each edge belongs to two faces and there are 5 faces meeting at each vertex. The midpoints of its faces form a dual regular polyhedron, in this case a dodecahedron, which

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has 12 faces (regular pentagons), 30 edges and 20 vertices (each of them belonging to 3 faces).

Linear Algebra

basis (n) /beisis/ (pl. bases) c s, nn, y, c s

change of basis /tend/ /beisis/ bilinear form (n) /bailini:/ /f:m/ dng song tuyn

tnh coordinate (n) /kou':dneit/ ta (non-)degenerate /didenreit/ ng bin dimension (n) /dmnn,

damnn/ chiu, kch thc, kh, c

codimension (n) s i chiu finite dimension (n) /fanat/ /dmnn,

damnn/ hu hn chiu

infinite dimension (n) /infinit//dmnn, damnn/

v hn chiu

Image(n) /imind/ nh Isometry(n) /aismitri/ ng c kernel (n) /'k:nl/ nhn linear (a) /'lini/ tuyn tnh, ng

thng Origin(n) /'ridin/ Gc orthogonal; perpendicular

/:gnl/ /p:pn'dikjul/

vung gc, trc giao

(orthogonal) projection

/prdekn/ php chiu vung gc

quadratic form (n) /kw'drtik/ dng phng trnh bc hai

reflection (n) /riflekn/ s i xng represent (v) /,repr'zent/ biu din

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rotation (n) /routein/ php quay scalar (a) /skeil/ v hng scalar product (n) /prdkt/ tch v hng subspace (n) /sbspeis/ khng gian con (direct) sum (n) /sm/ tng skew-symmetric(a) phn i xng symmetric (a) /simetrik/ i xng vector (n) /'vekt/ vc-t vector space /speis/ khng gian vc-t vector subspace /sbspeis/ khng gian con

vc-t

Mathematical arguments Set theory

x A x is an element of A ; x lies in A ; x belongs to A ; x is in A . x A x is not an element of A ; x does not lie in A ; x does not belong to A ; x is not in A .

,x y A (both) x and y are elements of A ; ....lie in A ; ..... belong to A ; .... are in A .

,x y A (neither) x nor y is an element of A ; .... lies in A ; ...... belongs to A ; ... is in A . The empty set (= set with no elements). A = A is an empty set. A A is non-empty. A B the union of (the sets) A and B ; A union B . A B the intersection of (the sets) A and B ; A

intersection B . A B the product of (the sets) ) A and B ; A times B . A B A is disjoint from B ; the intersection of A and

B is empty.

26

,....x the set of all x such that ..... C the set of all complex numbers. Q the set of all rational numbers. R the set of all real numbers. A B contains those elements that belong to A or to B (or

to both). A B contains those elements that belong to both A and B . A B

contains the ordered pairs ( , )a b , where a (resp ..b ) belongs to A (resp..to B ).

...nn

A A A A

contains all ordered n-tuples of elements of A .

Belong to(v) [bil tu:] thuc v disjoint from(adj) [disdint frm] ri (nhau) element(n) [elimnt] phn t, yu t empty(a) [empti] trng, rng nonempty(a) [nn,empti] khng trng,

khng rng intersection(n) [,intsek()n] s tng giao inverse(n) [inv:s] s nghch o, s

kh nghch the inverse map to f [ inv:s mp tu: ef] nh x ngc ca

nh x f the inverse of f [ inv:s v ef] nh x ngc ca

nh x f map(n) [mp] nh x, bn , bn

phng n. pair(n) [pe(r)] cp, i ordered pair [:d(r) pe(r)] cp th t triple(n) [trip()l] bi ba quadruple(n) [kwdrup()l] gp bn, bi bn relation (n) [rilei()n] quan h, h thc

27

equivalence relation [ikwivlnt rilei()n] quan h tng ng

set (n) [set] tp hp finite set [fainait set] tp hu hn infinite set [infinit set] tp v hn union(n) [ju:nin] hp

Logic S T S or T . S T S and T . S T S implies T ; if S then T . S T S is equivalent to T ; S iff T . - S not S

.....x A for each [=for every] x in A ....... ....x A there exists [= there is] an x in A (such that).....

!x A .... there exists [= there is] a unique x in A (such that)..... x A .... there no x in A (such that).....

0 0x y 0x y if both x and y are positive, so is x y . x 2 2x no rational number has a square equal to two

,x y 2 / 3x y for every real number x there exists a rational number y such that the absolute value of x minus y is smaller than two thirds. Exercise. Read out the following statements. x A B ( )x A x B x A B ( )x A x B

x 2 0x x 2 0x y : x 2y x

28

Basic arguments It follows from .... that .... : T .suy ra .. We deduce from ... that ...: Ta suy ra t ..rng.. Conversely, .... implies that ....: Ngc li, .c ngha.. Equality (1) holds, by Proposition 2 : Theo mnh 2 , ng thc (1) ng. By definition, ...: Theo nh ngha The following statements are equivalent: Nhng pht biu sau l tng ng Thanks to ... the properties ... and ... of ... are equivalent to each other : Nh .nhng tnh chtl tng ng. ..... has the following properties: .c nhng tnh cht sau Theorem 1 hold unconditionally: nh l 1 c suy ra mt cch hin nhin This result is conditional on Axiom A : Kt qu ny c suy ra t tin A .... is an immediate consequence of Theorem 3: .l h qu trc tip t nh l 3. Note that .... is well- defined, since ....Ch rng .lun ng v .. As ... satifises .... formula (1) can be simplified as follows: vtha mn cng thc (1) c th c vit n gin nh sau . We conclude (the argument) by combining in equalities (2) and (3): T (1) v (2) ta suy ra iu phi chng minh (Let us) denote by X the set of all ..... K hiu X l tp hp Let X be the set of all .... Recall that ..... by assumption : Theo gi thit ta c .. It is enough to show that ....iu kin l . We are reduced to proving that ...Suy rat a cn chng minh rng.. The main idea is as follows : tng chnh l nh sau We argue by contradiction / Assume that ... exists : Gi s phn chng l :. The formal argument proceeds in several steps : kt lun c a ra t cc bc sau

29

Consider first the special case when ...xt trng hp c bit u tin The assumptions ... and ... are independent (of each other). Since ....Cc gi s l c lp nhau v .... which proves the required claim iu cn chng minh We use induction on n to show that ..... Ta s dng phng php chng minh quy np vi n On the other hand, ...mt mt. .... which mean that .... iu chng t rng In others word,....ni mt cch khc Argument (n) /':gjumnt/ Lp lun assume (v) /'sju:m/ Gi s assumption /'smpn/ S gi s Axiom (n) /'ksim/ tin Case (n) /keis/ cch special case /'spel keis / cch c bit Claim (n) /kleim/ li xc nhn (the following ) claim

Concept (n) /knsept/ khi nim Conclude(v) /knklud/ kt lun Conclusion (n) /knklun/ s kt lun Condition (n) /kn'dn/ iu kin a necessary and sufficient condition

/'nesseri/ /s'fint/

iu kin cn v

Conjecture (n) /kndekt/ s gi nh, gi s Consequence (n) /'knsikwns/ h qu, kt qu Consider (v) /knsid/ xt, ch n cho rng contradict (v) /kntrdikt/ mu thun vi, tri vi Contradiction (n) /,kntr'dikn/ s ph nh, s mu

thun Conversely (adv) /kn'v:sli/ ngc li

30

corollary (n) /k'rlri/ H qu Deduce (v) /didju:s/ suy ra Define(v) /di'fain/ nh ngha Well-defined /weldifaind/ c nh ngha Definition(n) /definin/ li nh ngha Equivalent(a) /ikwivlnt/ tng ng Establish(v) /stbl/ thit lp Example(n) /igza:mp()l/

v d

Exercise (n) /'ekssaiz/ bi tp Explain(v) /iks'plein/ gii thch Explanation(n) /,ekspl'nein/ s gii thch False(a) /f:ls/ sai Formal(a) /'f:ml/ hnh thc Hand(n) /hnd/ tay on one hand Mt mt on the other hand Mt khc iff [=if and only if] Khi v ch khi Imply(v) /im'plai/ bao hm; ko theo; c

h qu, c ngha Induction on(v) /In'dkn/ php quy np Lemma(n) /lem/ b Proof(n) /pru:f/ Bng chng Property(n) /'prpti/ tnh cht satisfy property P

/'stisfai/ /'prpti/

tha mn tnh cht

Proposition(n) /prpzn/ mnh Reasoning(n) /ri:zni/ s bin lun Reduce to(v) /ri'dju:s/ rt gn Remark(v) /ri'm:k/ ch , ch thch Require(v) /ri'kwai/ i hi, cn tm Result (v) /ri'zlt/ kt qu

31

s.t = such that Sao cho Statement(n) /'steitmnt/ cu lnh t.f.a.e = the following are equivalent

Tng ng vi

theorem (n) / 'irm/ nh l True (a) /tru:/ ng Truth (n) /tru:/ chn l wlog = without loss of generality

Khng mt tnh tng qut

Function Formulas/ Formulae

( )f x f of x ( , )g x y g of x (comma) y (2 ,3 )h x y h of two x (comma) three y

sin( )x sine x cos( )x cosine x tan( )x tan x arcsin( )x arc sine x arccos( )x arc cos x arctan( )x arc tan x sinh( )x hyperbolic sine x cosh( )x hyperbolic cosine x tanh( )x hyperbolic tan x

2sin( )x sine of x squared 2(sin )x sine squared of x ; sine x , all squared

4

1tan( )

xy

x plus one, all over over tan of y to the four

cos23x x three to the (power of) x minus cosine of two x 3 3exp( )x y exponential of x cubed plus y cubed

32

Intervals ( , )a b open interval ,a b ,a b closed interval ,a b ,a b halp open interval ,a b (open on the left, closed on

the right) [ , )a b halp open interval ,a b (open on the right, closed on

the left)

Derivatives

'f f dash; f prime; the first derivative of f ''f f double dash; f double prime; the second

derivative of f (3)f the third derivative of f ( )nf the n-th derivative of f

dydx

d y by d x; the derivative of y by x

2

2

d ydx

the second derivative of y by x; d squared y by d x squared

fx

the pattial derivative of f by x (with respect to x); partial d f by d x

2

2

fx

the second pattial derivative of f by x (with respect to x); partial d squared f by d x squared

f

nabla f; the gradient of f f

delta f

Example. The (total) differential of a function ( , , )f x y z in three real variables equal to

.f f fdf dx dy dzx y z

33

The gradient of f is the vector whose components are the partial derivatives of f with respect to the three variables:

( , , )f f ffx y z

.

The laplace operator acts on f by taking, the sum of the second partialderivatives with respect to the three variables

2 2 2

2 2 2

f f ffx y z

.

The Jacobian matrix of a pair of function ( , )g x y , ( , )h x y in two real variables is the 2 2 matrix whose entries are the partial derivatives of g and h , respectively, with respect to the two variables.

g gx xh hx y

.

Intergrals

( )f x dx integral of f of x d x 2

b

a

t dt integral from a to b of t squared d t

( , )S

h x y dxdy double integral over S of h x y d x d y

Differential equations An ordinary (resp.. a partial) differential equation, abbreviated as ODE (resp ..PDE) is an equation involving an unknown function f of one (resp.. more than one) varial together with its derivatives (resp.. partial derivatives). Its order is the maximal order of derivatives that appear in the equation. The equation is linear if f and its derivatives appear linearly, other wise it is non-linear.

34

' 0f xf first order linear ODE '' sin( ) 0f f second order non-linear ODE

2 2( ) ( ) 1 0f fx y x yx y

first order linear PDE

The classial linear PDE is arising from physics invole the Laplace operator

2 2 2

x y z

0f the Laplace equation f f the Helmholtz equation

ygx

the heat equation

2

2

ygx

the wave equation

Above, , ,x y z are the standard coordinates on a suit able domain U in

3 , f is the time variable, ( , , )f f x y z and ( , , , )g g x y z t . In addition, the function f (resp ..

g ) is required to satisfy suit able boundary conditions (resp .. initial conditions) on the boundary of U (resp .. for 0t ) act (v) /kt/ tc ng action (n) /kn/ tc ng bound (n) /'baund/ gii hn bounded (adj) /baundid/ b chn bounded above (adj) /baundid 'bv/ b chn trn bounded below (adj) /baundid bi'lou/ b chn di unbounded (adj) /nbaundid/ khng gii hn comma (n) /km/ du phy concave function (n) /'knkeiv fkn/ hm lm condition (n) /kn'dn/ iu kin

35

boundary condition (n) /'baundri kn'dn/ iu kin bin initial condition (n) /i'nil kn'dn/ iu kin ban u constant (n) /'knstnt/ i lng (s)

khng i, hng s

constant (adj) /'knstnt/ bt bin, khng thay i

constant function (n) /'knstnt fkn/ hm hng Non-constant (adj) /nn 'knstnt/ kh bin, thay i Non-constant function (n)

/nn 'knstnt fkn/

hm kh bin

continuous (adj) /kn'tinjus/ lin tc, st nhau continuous function (n) /kn'tinjus fkn/ hm lin tc Convex function (n) /knvks fkn/ hm li decrease (n) / 'di:kri:s/ s gim st decrease (v) / 'di:kri:s/ lm gim st decreasing function (n) / 'di:kri:si fkn/ hm nghch bin strictly decreasing function

/striktli 'di:kri:si fkn/

Hm gim mt cch nghim ngt

derivative (n) /di'rivtiv/ o hm second derivative (n) /sknd di'rivtiv/ o hm bc 2 n -th derivative (n) o hm bc n partial derivative (n) /prl di'rivtiv/ o hm tng

phn differential (n) /dif'renl/ vi phn differential form (n) /dif'renl f:m/ dng vi phn differentiable function (n)

/difrenibl fkn/

hm kh vi

twice differentiable function (n)

/twas difrenibl fkn/

hm kh vi bc 2

n -times continuously differentiable function

/en-taimskn'tinjusli difrenibl

hm kh vi bc n

36

fkn/ domain (n) /dmein/ min equation (n) /i'kwein/ phng trnh the heat equation (n) /hi:t i'kwein/ phng trnh nhit the wave equation (n) /wev i'kwein/ phng trnh sng function (n) /fkn/ hm, hm s graph (n) /gra:f/ th increase( n) /'kri:s/ s tng thm increase (v) /n'kri:s/ tng ln increasing function (n) /n'kri:si fkn/ hm ng bin strictly increasing function

/striktli n'kri:si fkn/

Hm tng mt cch nghim ngt

integral (n) /'intigrl/ tch phn integral (adj) /'intigrl/ nguyn closed integral (n) /klouzd 'intigrl/ tch phn ng open integral (n) /'oupn 'intigrl/ tch phn m linear (adj) /'lini/ tuyn tnh, thng Non-linear (adj) /nn 'lini/ khng thng maximum (adj) /mksimm/ cc i, cc ,

ti a global maximum (n) /gloubl mksimm/ cc i ton din local maximum (n) /'loukl mksimm/ cc i cc b minimum (n) /'mnmm/ cc tiu global minimum (n) /gloubl 'mnmm/ cc tiu ton din local minimum (n) /'loukl 'mnmm/ cc tiu cc b monotone function (n) /mntoun

fkn/ hm n iu

operator (n) /preit/ ton t , php ton otherwise / waz / cch khc, khc Partial(a) / pl / ring, phn ring Sign(n) / san / du, du hiu Value(n) / vlju / gi tr

37

complex-valued function

/ kmpleks / /vlju/

hm ly gi tr phc, hm phc

Real- value function

/ rl/ hm ly gi tr thc, hm thc

Variable(n) / veribl/ bin s, bin thin, bin i

complex variable / kmpleks / / veribl/

bin phc

Real variable / rl/ / veribl/ bin thc function in three variables

/ veribl/ hm ba bin

with respect to [=w.r.t.] /w rspekt t / i vi

This is all Greek to me

Small Greek letters used in mathematics

alpha Beta Gamma delta epsilon Zeta Eta theta iota Kappa Lamda muy nu Xi Omicron , pi rho Sigma Tau upsilon , phi Chi Psi omega

Capital Greek letters used in mathematics

Beta Gamma Delta Theta Lamda Xi Pi Sigma Upsilon Phi Psi Omega

38

Sequences, Series Convergence criteria

By definition, an infinite series of complex number 1

nn

a

converges ( to a complex numbers) if the sequence of partial sums 1n ns a a has a finite limit( equal to s ) otherwise it diverges.

The simplest convergence criteria are based on the following two facts.

Fact 1. If 1

nn

a

is convergent, so is

1n

na

( in this case we say that the

series 1

nn

a

is absolutely convergent).

Fact 2. If 0 n na b for all suflicently large n and if 1

nn

b

converges,

so does 1

nn

a

.

Taking nnb r and using the fact that the geometric series 1

n

nr

of

ratio r is convergent iff 1r , we deduce from Fact 2 the following statements. The ratio test (DAlembert). If there exists 0 1r such that, for all

sufficiently large n , 1n na r a then 1

nn

a

is (absolutely) convergent.

The root test( Cauchy). If there exists 0 1r such that, for all

sufficiently large n , n na r , the 1

nn

a

is (absolutely) convergent.

What is the sum 1 2 3 equal to?

At first glance, the answer is easy and not particularly interesting as the partial sums

1, 1+2=3, 1+2+3=6, 1+2+3+4=10,...

39

tend toward plus infinity, we have 1+2+3+...= +

It turns out that something much more interesting is going on behind the scenes. In fact, there are several ways of regularising this divergent series and they all lead to the following surprising answer:

( the regularized value of) 11 2 312

.

How is the possible? Let us pretend that the infinite sums a = 1 + 2 + 3 + 4 +... b = 1 2 + 3 4 +... c = 1 1 + 1 1 +...

all make sense. What can we say about their values? Firstly, adding e to itself yields.

1 1 1 1 ...11 1 1 ... .2

1 0 0 0 ... 1

cc c

c c

Secondly, computing 2 1 1 1 1 ... ...c c c c c c by adding the infinitely many rows in the following table

1 1 1 1 ...1 1 1 ...

1 1 ...

ccc

we obtain 2 14

b c . Alternatively, adding b to itself gives

1 2 3 4 ...11 2 3 ... .

2 41 1 1 1 ...

bcb b

b b c

Finally, we can relate a to b , by adding up the following two rows:

40

1 2 3 4 ... 1 134 4 8 ... 4 12a

a b aa

Exercise. Using the same method, compute the sum 2 2 2 21 2 3 4 .

1lim ( ) 2x

f x

the limit of f of x as x tends to one is equal to two

approach (n) / prt/ s gn ng, php xp x, cch tip cn

close (v) / klz/ ng kn

arbitrarily close to btrri klz tu:/ ng mt cch ty

Compare(v) / kmpe(r)/ so snh

Comparison (n) / kmprsn/ s so snh

Converge (v) / knvd/ hi t, ng quy

Convergence (n) / knvd/ s hi t, tnh ng quy

criterion (pl. criteria) / kratrin/ tiu chun

Diverge(v) / davd/ phn k, lch

Divergence(n) / davd/ s phn k, tnh phn k

Infinite(a) / nfnt/ v hn, v cc, v s

Infinity(n) / nfnti / v s, v cc, v hn, v tn

minus infinity / mans nfnti / m v cc

Plus infinity / pls nfnti / cng v cc

41

Large(a) / ld / ln, rng

Large enough / ld nf / ln

sufficiently large / sfnt ld / ln

Limit(n) / lmt / gii hn

tend to a limit / tend tu e lmt / tin ti gii hn

neighbo(u)rhood(n) / nebhd / ln cn

sequence (n) / sikwns / dy

bounded sequence / band sikwns / dy b chn

convergent sequence / knvd sikwns /

dy hi t

divergent sequence / davd sikwns /

dy phn k

unbounded sequence / nbandd sikwns /

dy khng b chn

Arithmetic sequence Cp s cng

Geometric sequence Cp s nhn

Series(n) /sriz / chui

absolutely convergent series

/ bslutli sriz / chui hi t tuyt i

geometric series / dimetrk sriz /

chui hnh hc

Sum(n) / sm / tng

partial sum /pl sm / tng ring

42

Prime Numbers

An integer 1a is a prime (number) if it cannot be written as a product of two integers , 1a b . If on the contrary, n ab for integers

, 1a b , we say that n is a composite number. The list of primes begins as follows:

2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 47, 53, 59, 61,

Note the presence of several twin primes (pairs of primes of the form , 2p p ) in this sequence:

11, 13 17,19 29, 31 41,43 59,61

Two fundamental properties of primes with proofs were already contained in Euclids Elements:

Proposition 1. There are infinitely many primes.

Proposition 2. Every integer 1n can be written in a unique way (up to reordering of the factors) as a product of primes.

Recall the proof of Proposition 1: given any finite set of primes 1,..., jp p , we must show that there is a prime p different from cach

ip . Set 1... jM p p ; the integer 1 2N M is divisible by at least one prime p (namely, the smallest divisor of N greater than 1). If p was equal to p , for some to ip for some 1,...,i j , then it would divide both N and ( )i iM p M p , hence also 1N M ,which is impossible. This contradiction implies that 1, ..., jp p p , concluding the proof.

The beauty of this argument lies in the fact that we do not need to know in advance any single prime, since the proof works even for

0j : in this case 2N (as the empty product M is equal to 1, by definition) and 2p .

43

It is easy to adapt this argument proof in order to show that there are infinitely many primes of the form 4 3n (resp.. 6 5n ). It is slightly more difficult, but still elementary, to do the same for the primes of the form 4 1n (resp.. 6 1n ).

Questions About Prime Numbers

Q1. Given a large integer n (say, with several hundred decimal digits), is it possible to decide whether or not n is a prime?

Yes, there are algorithms for primality testing which are reasonably fast both in theory (the Agrawal-Kayal-Saxena test) and in practise ( the Miller-Rabin test).

Q2. Is it possible to find concrete large primes?

Searching for huge prime numbers usually involves numbers of special form, such as the Mersenne numbers 2 1nnM (if nM is a prime, n is necessarily also a prime). The point is that there is a simple test (the Lucas-Lehmer criterion) for deciding whether nM is a prime or not.

In practive, if we wish to generate a prime with several hundred decimal digits, it is computationally feasible to pick a number randomly and then apply a primality testing algorithm to numbers in its vicinity (having first eliminated those which are divisible by small primes).

Q3. Given a large integer n , is it possible to make explicit the factorisation of n into a product of primes? For example,

3999999 3 7 11 13 37.

At present, no (unless n has special form). It is an open question whether a fast prime factorisation algorithm exists (such an algorithm is known for a hypothetical quantum computer).

Q4. Are there infinitely many primes of special form?

44

According to Dirichlets theorem on primes in arithmetic progressions, there are infinitely many primes of the form an b , for fixed integers , 1a b without a common factor.

It is unknown whether there are infinitely many primes of the form 2 1n (or, more generally, of the form ( )f n , where ( )f n is a polynomial of degree deg( ) 1f ).

Similarly, it is unknown whether there are infinitely many primes of the form 2 1n (the Mersenne primes) or 2 1n (the Fermat primes).

Q5. Is there anything interesting about primes that one can actually prove?

Green and Tao have recently shown that there are arbitrarily long arithmetic progressions consisting entirely of primes.

Digit(n) / ddt / ch s, hng s

prime number (n) / pram nmb(r) / S nguyn t

Twin primes / twn nmb(r) / S nguyn t cng nhau

progression / prren / Chui

arithmetic progression

/ rmtk prren /

Chui s cng

geometric progression

/ dimetrk prren /

Chui s nhn

45

Probability and Randomness Probability theory attempts to describe in quantitative terms

various random events For example, if we roll a die, we expect each of the six possible

outcomes to occur with the same probability, namely 16

(this should

de true for a fair die; professional gamblers would prefer to use loaded dice, instead).

The following basic rules are easy to remember. Assume that an event A (resp., B ) occurs with probability p (resp., q ). Rule 1. If A and B are independent, then the probability of both A and B occurring is equal to pq . For example, if we roll the die twice in a row, the probability that

we get twice 6 points is equal to 1 1 16 6 36 .

Rule 2. If A and B are mutually exclusive (= they can never occur together), then the probability that A or B occurs is equal to p q .

For example, if we roll the die once, the probability that we get 5

or 6 points is equal to 1 1 16 6 3 .

It tunrs out that human intuition is not very good at estimating probabilities. Here are three classical examples. Example 1. The winner of a regular TV show can win a car hidden behind one of three doors. The winner makes preliminary choice of one of the doors (the first door). The show moderator then opens one of the remaining two doors behind which there is no car (the second door). Should the winner open the initially chosen first door, or the remaining third door? Example 2. The probability that two randomly chosen people have

birthday on the same day of the year is equal to 1365

(we disregard

the occasional existence of February 29). Given 2n randomly

46

chosen people, what is the probability nP that at least two of them have birthday on the same day of the year? What is the smallest value

of n for which 12n

P ?

Example 3. 100 letters should have been put into 100 addressed envelopes, but the letters got mixed up and were put into the envelops completely randomly. What is the probability that no (resp., exactly one) letter is in the correct envelope? See the next page for answers.

coin (n) / kn / tin bng kim loi toss [=flip] a coin / ts/ /kn / nm, gieo ng tin

bng kim loi die (pl. Dice) / da/ qun sc sc fair [= unbiased] die / fe(r) da/ qun sc sc cn

bng biased [= loaded] die / bast da/ qun sc sc khng

i xng roll [= throw] a die / rl da/ ln mt qun sc

sc Head(n) / hed/ u, phn trn, phn

trc, mc Probability(n) / prbblti/ xc sut Random(a) / rndm/ ngu nhin randomly chosen / rndmli tzn/ chn ngu nhin Tail(n) / tel/ ui, phn d with respect to [= w.r.t.] / w rspekt tu/ i vi Permutation(n) / pmjuten/ Hon v, chnh hp Combination(n) / kmbnen/ t hp Basic(a) / besk/ C s Count (v) / kant/ m

47

Fundamental(a) / fndmentl/ C s Principle(n) / prnspl/ Nguyn l, nguyn

tc Row(n) / r/ Dy Adjacent(a) / desnt/ Lin tip, lin k Integer(n) / ntd(r)/ S nguyn Arrange(v) / rend/ Sp xp Arrangement(n) / rendmnt/ S sp xp, s

chnh hp Remain(v) / rmen/ Cn li Restriction(n) / rstrkn/ Gii hn Without restriction / wat rstrkn/ Khng c gii hn Mean (n) /mi:n/ Trung bnh

Arithmetic mean (n) /'rimtik/ /mi:n/ Trung bnh cng

Median (n) /'mi:djn/ Trung v

Mode (n) /moud/ Mod

Range (n) /reind/ min gi tr

Average (n) /'vrid/ Trung bnh

Answer to Example 1. The third door. The probability that the car is

behind the first (resp,. the second) door is equal to 13

(resp,. zero); the

probability that it is behind the third one is, therefore, equal to 1 21 03 3

.

48

Answer to Example 2. Say, we have n people birthdays on the days 1 2, ,..., nD D D . We compute 1 nP , namely, the probability that all the

days iD are distinct. There are 365 possibilitices for each iD . Given

1D , only 364 possible values of 2D are distinct from 1D . Given distinct 1 2,D D , only 363 possible values of 3D are distinct from

1 2,D D . Similarly, given distinct 1 2 1, ,..., nD D D , only 365 (n-1) values of nD are distinct from 1 2 1, ,..., nD D D . As a result.

365 1364 3631365 365 365

1 2 11 1 1 1365 365 365

n

n

nP

nP

.

One computes that 22 0.476...P and 23 0.507...P In other words, it is more likely than not that a group of 23 randomly chosen people will contain two people who share the same birthday! Answer to Example 3. Assume that there are N letters and N envelopes (with 10N ). The probability mp , that there will be exactly m letters in the correct envelopes is equal to

1 1 1 1 1 1

! 0! 1! 2! 3! !mp

m N m

(where ! 1 2m m and 0! 1 , as usual). For small values of m (with respect to N ), mp is very close to the infinite sum

11 1 1 1 1 1 1! 0! 1! 2! 3! ! !

m

mq em e m m

.

which is the probability occurring in the Poisson distribution, and which does not depend on the (large) number of envelopes. In particular, both 0p and 1p are very close to

0 11 0.368...q qe

, which implies that the probability that there will

be at most one letter in the correct envelope is greater than 73%!

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depend on (v) / dpend n/ ph thuc (to be) independent of / ndpendnt v/ c lp ca Correspondence(n) / krspndns / php(s) tng

ng Transcendental(a) / trnsendentl/ siu vit

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MT S VN TRONG VIC SON BI V GING DY MN TON BNG TING ANH

Nhm bin son ti liu tp hun mn Ton

Tm tt

Bo co trnh by cc kinh nghim c nhn ca nhm bin son ti liu tp hun trong vic ging dy mn Ton bng ting Anh. Cc thnh vin ca nhm l nhng ging vin i hc, cc gio vin trung hc c nhiu kinh nghim ging dy mn Ton bng ting Anh trn lp cng nh tham gia luyn thi cc chng trnh thi Ton vo cc trng i hc nc ngoi (v d nh SAT, A-level,). Cc kinh nghim bao gm vic chun b bi ging, cch thc ging bi, tng tc vi sinh vin v cc kh khn trong vic ging bi bng ting Anh. i tng m bo co ny ch yu hng n l nhng gio vin trung hc cha tng, hoc c rt t kinh nghim trong vic ging dy mn Ton bng ting Anh. Do vy, bo co ch tp trung vo nhng kinh nghim rt c bn.

1. M u

Vic ging dy bn thn n l mt ngh thut. Mt ngi thy dy gii khng nhng cn c trnh kin thc chuyn mn gii m cn cn mt phng php truyn t, ging dy tt, hp l, hp dn c th truyn ti nhiu nht lng kin thc ti hc sinh v gip hc sinh nhn c nhiu nht kin thc c th. thc hin c iu ny, a s cc thy c phi tri qua mt thi gian ng lp tm ra cho mnh mt phng php, phong

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cch ging dy ph hp nht cho ring mnh. Tuy mi thy c c mt phng php ging dy ring nhng u phi tun theo mt s nguyn tc c bn trong vic son v trnh by bi ging. Ging dy bng ting Anh cng tng t nh vy. thc hin c cc thy c cng cn phi nm mt s nguyn tc c bn ri t pht trin ring cho mnh mt phng php ph hp nht i vi bn thn. V vy, bo co ny trnh by mt s kinh nghim, nguyn tc trong vic son v trnh by bi ging bng ting Anh.

Cc kinh nghim trong bo co s c trnh by thnh ba phn. Phn u lin quan n vic son gio n. Phn th hai trnh by mt s nguyn tc c bn v cc kinh nghim c nhn trong vic trnh by bi ging trn lp. Phn ny s trnh by cc bc c bn cn thc hin trong mt bi ging v s cung cp cho cc thy c cc cu trc ngn ng v cc v d gip cc thy c c th t tin ging bi bng ting Anh trn lp. Cui cng, bo co s trnh by mt s kinh nghim chia s ni chung trong cng vic ging dy Ton bng ting Anh.

2. Thit k v chun b bi ging

Thit k bi ging l khu rt quan trng trong ging dy. Khi ging dy mn Ton bng ting Vit th chng trnh mn hc c chun ha v cc gio vin s dy theo chng trnh chun ha ny. Tuy nhin, vic ging dy Ton bng ting Anh hin nay mi ch c thc hin th im ti mt s trng, nn mt chng trnh chun hin ti cha c xy dng. Do , cc gio vin hin nay c xu hng l chn mt chng trnh ging dy ca mt trng no ging dy mn Ton bng ting Anh dng lm chng trnh ging dy cho lp ca mnh. Vic lm ny ni chung l kh thun li v d dng v cc chng trnh ging dy Ton ca cc trng nc ngoi c xy dng v chnh sa trong thi gian di nn kh l y v tt. Tuy nhin, vi mc tiu l a vic ging dy bng ting Anh vo mn Ton hc sinh va pht trin c ngoi ng li va m bo c chng trnh mn hc theo chun ca quc gia, bi v hc sinh vn phi thi tt nghip v thi i hc theo chng trnh ca Vit Nam, cc thy c s phi son bi ging ca mnh vi ni dung khng khc nhiu so vi

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chng trnh ang ging dy m vn phi a vo c vn ngn ng v rn cho hc sinh t duy theo kiu Anh/M. Vic lm ny s l rt kh khn cho nhng thy c mi lm quen vi vic ging dy ny. Tuy nhin, son c mt bi ging tt th khng phi l mt vic khng th thc hin c nu cc thy c thc hin cc cng vic cn thit.

Vic dy hc Ton bng ting Anh hin nay c nhiu mc v hnh thc khc nhau. C th ni l c ba hnh thc ang c vn dng ch yu hin nay:

Hnh thc

Mc Mc tiu dy hc

Yu cu v ting Anh

Hot ng ging dy

1 Dy Ton bng ting Anh di hnh thc Cu Lc B.

Mc tiu gip gio vin v hc sinh cng tip cn vi vic s dng ting Anh trong mn Ton.

Tp trung vo vic lm quen vi cc thut ng v cu trc cu hay s dng trong ton hc c bn.

c, dch, cc hot ng nhm ghi nh thut ng v cu trc cu.

2 Dy Ton bng ting Anh p ng nhu cu hc sinh thi chng ch SAT ca M.

Gip hc sinh c hiu v s dng ng thut ng ton trong ting anh gii chnh xc nht.

Cc thut ng v cu trc cu theo chuyn ca ton hc ph thng. i hi dch v hiu ngha chnh xc trong cc ng cnh.

c, dch, thng k thut ng v cu trc cu theo chuyn mc cao.

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3 Dy Ton bng ting Anh nh mt mn hc.

Gio vin v hc sinh cng s dng ngn ng ting Anh trao i v chim lnh tri thc Ton hc m bo hiu ng v t yu cu ca BGD v ni dung khoa hc.

Gio vin v hc sinh phi c kh nng giao tip bng ting Anh v trnh by kin bng ting Anh.

Cu trc bi ging nh mt gi dy ton ting vit, kt hp vi cc hot ng nhm cng c ngn ng v kin thc.

Sau y l cc bc tham kho son bi ging thuc hnh thc th ba trong bng trn.

Bc 1: Tm c ni dung bi ging tng t bng ting Anh lc ra cc thut ng chnh, cu trc cu chnh thng hay s dng trong ni dung .

y, ti liu tt nht tham kho l cc gio trnh chun ca cc nc Anh, M, c, Singapore. Cc gio trnh ny s cung cp cc thut ng chun v cc nh ngha chun v mt ting Anh a vo ging dy.

Bc 2: Kt hp ni dung ca SGK ting Vit v ti liu ting Anh bin son ti liu ging dy.

Vi yu cu hc sinh phi m bo cc kin thc mn hc tham d c cc k thi trong nc nn ngi dy phi mt nhiu cng sc hn trong vic t chc cc hot ng hp l nhm gip hc sinh ghi nh v s dng c thut ng bng ting Anh v lm quen c cc dng bi tp trong cc k thi ca Vit Nam. Sch ting Anh cung cp thut ng, cu trc ting Anh chun. Sch ting Vit cung cp h thng bi tp nhm p ng nhu cu thi c. Kt qu ca vic son bi l cc sn phm nh sau (mt s hoc tt c ty vo tnh cht bi hc: l thuyt, luyn tp, thc hnh)

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i. H thng thut ng v cu trc cu s dng trong bi hc.

ii. H thng bi tp c hiu, in khuyt cc on vn bng ting Anh.

iii. Gio n theo cc bc ln lp v thi gian cho mi bc ln lp. Trong cn phi a ra c cc h thng l thuyt chun v ngn ng ly t cc gio trnh chun. Bn cnh l h thng cc cu hi dn dt ngi hc chim lnh tri thc mi.

iv. H thng bi tp dng c bn vn dng cc kin thc nn tng ca bi hc.

v. H thng bi tp nng cao p ng vic thi c. vi. H thng tr chi kim tra v cng c ngn ng. vii. Giao nhim v cho bi hc k tip.

Mt s bi son ging mu theo cc hnh thc dy hc khc nhau cc thy c c th xem trong ti liu tp hun.

3. Ging bi trn lp v vn ngn ng

Nhng kinh nghim trong phn ny s gip cc thy c c mt bi ging trn lp bng ting Anh tt. Tt nhin, trc khi ln bc ging th cc thy c phi nm r ni dung m mnh sp ging dy v bit c tt c cc thut ng chuyn mn bng ting Anh trong bi ging. Do phn ny s tp trung vo cc kinh nghim lin quan n cu trc, trnh t bi ging v a ra cc cch trnh by bng ting Anh tng ng.

Mt bi ging tt s ph thuc vo mt s yu t nh l ni dung ca bi ging, cch din t v tm l ca gio vin. Sau khi chun b ni dung bi ging tt, chng ta phi bit cch truyn t bi ging mt cch c hiu qu nht. Vic ny ph thuc kh nhiu vo tm l ca chng ta khi ng trn bc ging. Nu ch ging dy bng ting Vit th vn tm l khng phi l mt vn ln i vi cc thy c, l cc gio vin c rt nhiu kinh nghim ng trn bc ging. Tuy nhin, vic ging dy bng ting Anh li l mt vn khc. Vi nhng gio vin tr m khng c o to v lm vic nc ngoi th chc chn s khng t tin vi pht

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m ting Anh ca mnh, iu ny dn n s mt t tin khi ng trn bc ging. i vi cc thy c c tui th vn ngn ng li cng tr nn kh khn hn. Ngay bn thn nhng thy c mc d c o to ti nc ngoi nhng nhng ngy u ng trn bc ging cng khng t tin lm v kh nng ni ting Anh ca mnh. Nhng iu ny hon ton c th c khc phc nu chng ta thc hin mt s vic sau (da trn mt s kinh nghim c nhn). Th nht l phi chun b bi ging tht k, thc hnh cch trnh by nhng cu bng ting Anh lin quan n bi ging trc. Da vo cc bi ging video mu ca gio vin nc ngoi tng t nh bi ging ca chng ta, chng ta c th sa li cch pht m mt s t theo chun ca M. Khi chun b k bc ny, lc ging bi chng ta ban u hy ni chm v rnh mch. Khng nn ging bi nhanh qu v khi hc sinh c th s khng hiu mt s thut ng chuyn mn bng ting Anh. Gii thch cn k cc khi nim v ban u nn dng cc cu trc ng php v t ng n gin. Cch ging bi chm ny s gip hc sinh hiu bi v nm c cc t mi chuyn mn d hn, do s c hiu qu hn.

Phn tip theo s trnh by cc kinh nghim ni ting Anh trong mt bi ging. Trnh t ca mt bi ging c th c chia ra theo th t nh sau:

- Gio vin vo lp, n nh lp.

- Gii thiu bi hc, cu trc bi hc.

- i vo cc phn v cc hot ng chi tit.

- Tng kt li bi hc, giao bi tp v thng bo v ni dung bi hc tip theo.

Sau khi n nh lp, cc thy c s gii thiu ni dung ca bi hc hm . Vi phn gii thiu ni dung chnh ca bi hc, cc thy c c th dng cc cu v d nh sau:

- Today, we are going to study

- Our topic today is

- What I want to talk about today is

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- We are going to discuss

- Today I am going to focus on

- Today, I want to give you some background on ..

Tip n cc thy c s gii thiu cu trc ca bi hc, cc thy c c th dng cc cch din t sau:

- First well look at and then well look at

- Im going to cover and then

- Well discuss a few examples of/types of

V d 1

Good morning. Its nice to see you all. It looks like you are ready for the lesson, so lets get started. In todays lesson, Ill be talking about one of the special sequence called geometric progression. First, well look at some examples about G.P which is the short for geometric progression, then well learn about properties of G.P, and finally well solve some real-life problems relating to G.P.

V d 2

Hi everyone. Please take your seats so we can get started. Great. In todays lesson, we are going to look at some rules in trigonometric functions. More specifically, well be looking at the cosine rule and the sine rule. In the first half of the lesson, we will prove these two rules and then we will use these rules to solve some problems in the second half. Now, to help understand the proofs of these two rules, I want to review some important properties of the trig functions.

V d 3

Hi there, everyone. Its ten oclock, so lets go ahead and get started. What I want to talk about this morning is the parametric equation of line. Now, why do I want to talk about the parametric equation of line? Well, for some complicated curves, it is not easy to find their Cartesian equations. To

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overcome this difficulty, luckily we could express the variables, says, x and y, as functions of some parameters. In this way, the curve is said to be defined parametrically. Alright, the lesson consists of three parts. First, we will . Second, we will . And finally we will.

V d 4

Greetings everyone. This morning we have an interesting topic. Were going to discuss the derivative of trig functions. Thats right, how to find the derivative of trig function and how to apply it to the problems. Are you ready? All right. First, well look at a couple of examples and then well go into the detail of

Sau khi gii thiu ni dung v cu trc ca mn hc, cc thy c s i vo ging bi v thc hin cc ni dung v hot ng chi tit. Trong phn ny, cc thy c s phi dng rt nhiu cc thut ng, cu trc ting Anh trong ton hc. din t c cc cu trc ton hc bng ting Anh, cc thy c nn xem k phn Mathematical English trong ti liu tp hun. Phn tip theo sau y s trnh by mt s cu trc cu hay dng kt ni cc phn nh trong phn ni dung bi ging.

Sau khi gii thiu cu trc bi ging, cc thy c bt u i vo phn u tin ca bi ging bng cch din t v d nh sau:

- First lets look at

- Let me start with

Sau cc thy c i vo phn ni dung ca phn u tin . Sau khi ht /phn th nht, cc thy c chuyn tip sang phn tip theo bng cch c th ni:

- Next, lets talk about

- Now lets move on to

- Now, we are ready for (able to) .

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- With what we have discussed, we now have all necessary information to solve

- Now that weve talked about ________, lets talk about

- Thats enough about _______. Lets go on to

V d

Now, let me start with an interesting example about the interest paid by the bank into fix deposit accounts. The example is ________. To solve this problem, we must use formula of the general term of a geometric progression. So, lets go on to the first section which is about the definition of G.P.

Khi mun nh ngha mt i lng no , cc thy c c th ni:

- That is,

- In other words

- X, meaning

- By X, I mean

- What do I mean by X? Well, I mean

- Let me define that:

- The definition of that is

Khi cn gii thch mt vn hay mt thut ng mi m cc thy c a ra, cc thy c c th dng

- Let me explain

- What I mean is

- Let me show you what I mean

- Lets look at how this works

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V d

Today, we are going to explore one of the applications of differentiation in the max-min problems. In this type of problem, well look specifically at the end points and the critical points of the function. Let me explain what I mean by critical points. By critical points, I mean the undefined points and the stationary points the points at which the derivative of the function equals zero.

Trong bi ging, khi a ra mt kt qu hay kin no , cc thy c i khi phi a ra nhng lp lun hay l do ca mnh, hay n gin l bo v kin . Cc thy c c th ni:

- Let me tell you why

- Let me give you an example

- The reason is

- This is because

- I think

Khi a ra mt v d, chng ta c th din t bi

- For example,

- Here are some examples:

- Take X, for example

- For instance,

- , such as,

- Let me give you an example

V d

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There are some types of sequence that are useful and are used frequently in life, such as the arithmetic sequence, or geometric sequence. In the next lesson, we will study these two kinds of sequence.

Khi t tn cho mt bin no , chng ta hay dng cu trc: Let be/denote.

- Let x be the distance from A to B.

- Let y denote the price of product X.

Khi pht biu mt nh lut (law), nh l (theorem) chng ta hay dng cu trc s dng ng t states (pht biu), v d nh:

- Let a, b, and c be the length of the legs of a triangle opposite angles A, B, and C. Then the law of cosine states that

a2 = b2 + c2 2bc cos A

b2 = a2 + c2 2ac cos B

c2 = a2 + b2 2ab cos C.

- The Pythagorean Theorem states that for a right triangle with legs a, b, and hypotenuse c,

a2 + b2 = c2 .

Khi iu hnh tho lun nhm, cc thy c c th s dng cc cu sau:

- Is everybody ready to start?

- Lets start with question number 1.

- Nam, do you want to begin?

- Hung, what do you think about that?

- Has everyone had a chance to speak?

- Any other comments?

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- Thanks, everyone. Good discussion.

Trong lc iu hnh tho lun trong lp, khi mun a kin ca mnh, cc thy c c th s dng:

- In my opinion,

- I think/feel

- I noticed that

- I think it was interesting that

- is really important because

tng kt li mt phn hoc ton b bi, cc thy c c th dng

- So we have learned

- Lets wrap up what we have studied today

- Well, I have talked everything about

- Ok, I gave/explained you two examples with the solutions, now lets take a look at them again and point out the important facts.

Sau khi kt thc bi ging, chng ta nn thng bo cho hc sinh bit v ni dung bi k tip cc em c th chun b trc. Cc thy c c th dng

- Ok, thats all for today. Tomorrow, we will come back to this problem.

- Well, we have finished Chapter 3 today. In the next lesson, we will have a test for this chapter and we will move on to the Chapter 4 Integration.

Mt s ch thm

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Trong bui u tin gii thiu mn hc, hoc khi cc thy c dy mt topic cho mt lp mi th cc thy c c th thng bo trc vi c lp v vic liu c th hc sinh ngt li cc thy c t cu hi hoc yu cu ging k hn phn cc em cha hiu hay khng. y l iu m nhng ngi thuyt trnh trc mt s lng ln khn gi hay lm. Cc thy c c th quy c:

- During the lecture/talk/presentation, you can interrupt me when there is anything you dont understand by making a small hand gesture or raising your hand. I am very willing to answer you questions.

- I would like to finish my lecture/talk/presentation first, so all the questions or comments are welcome at the end of the lecture.

Trong qu trnh ging bi, mt trong nhng iu quan trng l quan st xem cc em hc sinh c hiu bi hay khng. Nu c mt vn cc thy c ang ging m c v cc em khng hiu, cc thy c c th dng li v hi Do you get it? hoc Got it?, hay l Do you get the point?. Khi ht mt phn no quan trng trong bi ging, cc thy c cng c th tm dng xem cc em hc sinh c cu hi no khng. Khi cc thy c n gin l dng li v hi Any questions? hoc n gin l Questions?.

Khi hc sinh t cu hi hoc trao i vi gio vin, nu cu hi l di hoc cc em kh din t bng ting Anh nn ni chm, cc thy c c th khuyn khch cc em tip tc bng cch ni nhng t nh Uh huh, Hmm, Ok, I see. Nu cha hiu cc cu hi ca cc em th c th hi Could you repeat that?.

Trc khi tr li cu hi chng ta nn ni Thats a good question! hoc I get your point.. Vic lm ny s khin hc sinh thy cu hi ca mnh c tn trng v s khuyn khch cc em hi thm.

Khi cc thy c t cu hi cho cc em v cc em a ra cu tr li, hoc khi cc em a ra cc nhn xt lin quan n ni dung mn hc, nu ng vi cu tr li ca cc em th chng ta c th ni I agree with you. hay Thats a good point, hay l Youre right.. Nu khng ng vi cu tr li th chng ta c th ni I dont agree with that , hoc l I disagree

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with you., hay I have a different idea/point of view/answer. v trnh by kin ca mnh.

i khi, chng ta s phi s dng nhng t m vo cc cu din t trn lp (small words or interjection) lm chm li tc truyn t vi mc ch lm cho hc sinh ch ti vn mi sp trnh by. V d nh:

- Well, Id like to say something here.

- Um, can I add something to that?

- So, could I say something here?

- Alright, lets go to the next problem.

4. Mt s kinh nghim chia s khc - Mt trong nhng vn ha ca vic hc ti M m chng ta nn hc

tp l s giao tip gia gio vin v hc sinh. Khc vi cch dy truyn thng mt chiu xy ra trong rt nhiu lp hc Vit Nam, chng ta nn t chc lp hc mt cch m v bnh ng hn. Chng ta c th thng nht vi hc sinh l c th ngt li gio vin hi bt c nhng vn g cha hiu hoc l hi sau mi tit. Ti th thch cch u tin hn v ti s khi hc sinh khng hiu mt vn g hoc khng hiu mt thut ng g th s lm nh hng n ton b bi ging. V vy ti cho php hc sinh ngt li ti v hi cc vn . Vic ny tt nhin c nh hng n bi ging nhng li ch ca n quan trng hn: hc sinh ho hng vi bi ging v khng kh trong lp hc si ni v khuyn khch hc sinh giao tip vi thy gio nhiu hn.

- Khc vi cc mn hc khc, trong gi dy l thuyt mn Ton, phn trng v bng en vn l s la chn tt nht. Hc sinh s ch hiu c nhng vn kh trong l thuyt Ton bng cch thng qua cc bin i ton hc c trnh by tng bc trn bng v khi vit chng vo v. Vi nhng vn quan trng v kh, i khi vic ging bng ting Vit cng khng phi d dng gip cc em hiu

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c mu cht vn , ch khng ni g n vic ging bng ting Anh. Do vy, nu cn thit cc thy c nn chuyn sang ging bng ting Vit nhng phn ny. Cc thy c ng do d lm vic ny, nht l khi ging bi m quan st thy a s hc sinh ca lp c v cha nm c vn ang c ging, bi v suy cho cng th chng ta vn t mc tiu ging dy ni dung Ton l quan trng nht.

- mt thy c bt u thc hin ging bi bng ting Anh t c cht lng tt th thy c phi tri qua vic son bi v thc hin ging dy t nht l mt nm lin tc. Do , trong nm u tin, cc thy c s cm thy mt gnh nng tm l v cc thy c phi n lc luyn tp nhiu. p lc s l nh nhau i vi mi mt bi hc mi. Tuy nhin, khi ging li bi hc ln th hai, hoc ln th ba sau mt thi gian, cc thy c s thy n tr nn d dng hn rt nhiu. V sau mt s ln, cc thy c s thy mnh c th ging bi mt cch t nhin gn nh l ging bi bng ting Vit.

- Cc bc ging bi bng ting Anh (cho gi dy l thuyt) Bc 1: Gii thiu thut ng bng ting Anh trong bi hc v cu trc ting anh thng c s dng thng qua hnh nh v on vn bn bng ting Anh v vn lin quan n bi hc, phn ny thng nn cho lm vic nhm 2 ngi ngi hc h tr nhau v tit kim thi gian. Vic lm ny s gip ngi hc ghi nh lu hn so vi vic ta gii thiu thut ng v ngha bng ting vit ngay khi bt u bi hc.

Bc 2: Gii thiu cc khi nim v tnh cht ton hc. Vic ny nn kp hp vi hnh nh nhiu nht c th ngi hc c th hiu vn m khng cn phi dch ra ting Vit. Nu kh nng ngi hc cha th thnh tho ngn ng th ngi dy nn tr trng vo h thng cu hi dn dt ngi hc tr li v tm ra c cu tr li. Qua vic t cu hi ngi hc cng s c im ta a ra cu tr li v qua ngi hc hc c cch t cu hi.

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Bc 3: a ra cc bi tp c bn v vic xc nh cc i tng trong nh ngha, tnh cht, cha i hi phi lp lun nhiu. Yu cu ngi hc a ra p n bng cc cu tr li y .

Bc 4: a ra cc bi tp nng cao dn theo yu kin thc ca vic kim tra v thi c. Bt u bng cc phiu tr li in khuyt ngi hc lm quen vi cch vit ngn ng.

Bc 5: Giao bi tp v nh v nhim v ca bui hc sau.

Trong thc t, ngi dy cn linh hot ty vo kh nng ngn ng ca hc sinh v ca chnh ngi dy c s ph hp, khng gy ra tnh trng nhm chn.

5. Kt lun Ni chung th vic ging bi bng ting Anh cng khng c g khc bit my so vi cch m chng ta ang ging bi bng ting Vit. Chng ta ch cn thay i v luyn tp mt cht l hon ton c th ging bi tt p ng c cc yu cu hin nay. Chng ti hy vng nhng kinh nghim c nhn c trnh by trong bo co ny s gip ch c cho cc thy c.

Ph lc

Trong mt gi ging dy bng ting Anh, thng thng s phi tri qua cc hot ng nh yu cu mi vic trn lp, t cu hi, bt u / kt thc bi hc, thc hin cc hot ng trong sch gio khoa v trn bng, iu khin lp hc, ng vin v khuyn khch hc sinh. Vi mi hot ng c th dng rt nhiu mu cu khc nhau, linh hot, ty tnh hung.

V d nh cu mnh lnh: - Close your books. - You say it, Mai. - Answer it, somebody.

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- Don't be quiet now. - Just sit down and be qiuet. - I want you to try exercise 1. Yu cu (tng t cu mnh lnh nhng dng ng iu thp hn): - Come here, please. - Would you like to write on the board? - Can/Could you say it again? - Do you mind repeating what you said? Gi v thuyt phc (Suggesting and persuading): - Let's start now. - What about if we translate these sentences? - You can leave question 1 out. - There is no need to translate everything. Cu hi (Questions): - Do you agree with A? - Can you all see? - Are you sure? - Do you really think so? Bt u bi hc (Beginning of the lesson): - Hurry up so that I can start the lesson. - Is everybody ready to start? - I think we can start now. - I'm waiting for you to be quiet. - What's the day today / What day is it today? Kt thc bi hc (End of lesson): - It's almost time to stop. - I make it almost time. We'll have to stop here. - All right, that's all for day. - We'll finish this next time. - We'll continue working on this chapter next time.

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- Please re-read this lesson for Monday's. - You were supposed to do this exercise for homework. - There will bw a test on this next Monday. - Remember your homework. - See you again on Monday. - Up you get. - Off / out you go. Khi gio vin gy ra sai st trong lp hc hoc c vic bn phi ra ngoi, c th xin li hc sinh bng cch: - Mind out of the way. - Could I get past? - I'll be back in the moment/ - Would you excuse me for a while? - I'm sorry, I didn't notice it. - I've made a mistake on the board. Cnh bo hc sinh khi cc em gp sai lm (tr li sai, thiu tn trng gio vin): - Be careful / Look out / Watch out. - Mind / watch the step. - You will be in detention next week. - I'll send you to see the headmaster . Hot ng trong sch gio khoa: - Give out the books, please. - Open your books at page 10. - Turn to page 10, please. - Has everybody got a book? / Does everybody have a book? - Books put (out with you books) / Books away (away with your books). - Take out books and open them at page 10. - Look at exercise 1 on page 10. - Turn back to the page 10. - Have a look at the dialog on page 10. - Stop working now.

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- Put your pens down. - Let's read the text aloud. - Do you understand everything? Lm vic nhm: - Work in twos / pairs. - Work together with your friend. - I want you go form groups. 4 pupils in each group. - Get into groups of 4. - Discuss it with your neighbor. Lm vic trn bng: - Come out to board, please. - Come out and write the word on the board. - Take a piece of chalk and write the sentence out. - Are these sentences on the board right? - Anything wrong with sentence one? - Everyone, look at the board, please. Trn y l mt s cu hc sinh thng dng trong gi hc, cc thy c tham kho thm.

- Excuse me. May I come in?

- Can you help me do this exercise?

- Could you speak more slowly, please?

- Can you lend me a pen?

- Can you repeat that please? I didn't understand.

- Have you done your homework?

- Sorry. I don't understand that.

- How do you say 'mesa' in English?

- Sorry, I can't remember your name.

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- How do you spell 'table'?

- Can I share your book with you?

- What's the difference between 'do' and 'did'?

- Where's Angela today?

- I'm sorry, I've left my book at home.

- She's absent.

- Excuse me. I'm sorry I'm late.

- What page is it on?

- Can I leave a bit early today, please?

- Can you pass me that piece of paper, please?

- Can you explain that again, please?

- Do we have to work in pairs?

- I didn't have time to do my homework. I'm sorry.

- Who's going to start?

- It's time to go.

- Whose turn is it?

- See you next lesson.

- It's my turn now.

- Have a nice weekend.

- Sorry, can you say that again'!

- The same to you. Bye.

- Excuse me, that's my book.

- Have you finished? What do we have to do now?

70

The sine rule and the cosine rule

Vocabulary

Sine rule, Cosine rule (Sine law, Cosine law; or The law of sines, The law of cosines in American English);

Solving a triangle.

Prior Knowledge

Pupils should be confident in applying sine, cosine and tangent to problems in right-angled triangles, as covered in the previous lessons.

Lesson Objective

The sine rule, The cosine rule;

Use the sine rule, the cosine rule to solve triangles;

Use the sine rule, the cosine rule to solve problems

Apply the sine and cosine rules in solving different kinds of maths exercises.

Number(s) of 45-minute period: 2

Material: Geometry Grade 10 Textbook (pp. 46-60), Geometry Grade 10 Advanced Textbook (pp. 53-67), University Entrance Maths Exams, Internet

Warming up

Let be given a right-angled (right) triangle ABC with A=900. Denote by a, b and c the respective sides opposite to angle A, B and C as usual.

The trigonometric ratios of an angle are defined as

Sin A=? Tan A=?

Teacher: Ta Ngoc Tri, B.Sc. (Maths), M.Sc. (Maths), M.Sc. (Maths Edu.), Ph.D. (Maths)

71

Cos A=? Cot A=?

The Pythagoras' theorem is stated as a2 =b2 + c2

Let be given a triangle ABC (oblique). Denote by a, b and c the lengths of three sides of ABC, and by A, B and C the three interior angles of ABC corresponding to a, b, and c, respectively. We have

Starter 1

Illustrate what is in effect the sine rule, from a specific example. Draw the triangle on the board. Ask pupils for ideas on how they might find the length of side b.

Start by asking why we cannot use sin, cos or tan in this case. (They can only be used in right-angled triangles.) Then suggest pupils consider what happens if we make a right-angled triangle by dropping a perpendicular from vertex C. Now can pupils see how they might find the length of line AC?

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Eventually this should lead to the appreciation that b sin 50 = 5 sin 65 and, therefore, that

65sinb =

05sin5 .

Point out that the equal ratios are the lengths of the sides multiplied by the sines of the opposite angles. Is this always true? Investigate by forming a general triangle and thus deriving the sine rule.

Handout 1 (proof for the sine rule)

The circumcircle is a triangles circumscribed circle, the unique circle that passes through each of the triangles three vertices. The centre O of the circumcircle is called the circumcentre, and the circles radius R is called the circumradius. The complete sine rule is given by:

Aa

sin =

Bb

sin =

Cc

sin = 2R

It can be proved as follows. Let O be the centre of the circumcircle of triangle ABC and let M be the midpoint of BC.

Then MOC = 21 BOC = 21 (2BAC) = BAC.

Therefore, sin A = 2aMC

OC R and so

Aa

sin = 2R. The complete sine rule

then follows.

Handout

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Given a triangle ABC.

Then, we have the cosine rule as follows

a2 = b2 + c2 2bc cos A

b2 = a2 + c2 2ac cos B

c2 = a2 + b2 2ab cos C

Proof. Consider the following triangle:

By Pythagoras theorem we have h2 = c2 x2 and h2 = b2 (a x)2.

Solving simultaneously we obtain c2 x2 = b2 (a x)2 = b2 a2 + 2ax x2.

Therefore c2 = b2 a2 + 2ax.

But x = c cos B and so c2 = b2 a2 + 2ac cos B, or b2 = a2 + c2 2ac cos B.

From the symmetry of the triangle we can therefore deduce the three standard forms of the cosine rule:

a2 = b2 + c2 2bc cos A

b2 = a2 + c2 2ac cos B

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c2 = a2 + b2 2ab cos C

Summary of activities

Teacher Student

Warm up (see above)

Starter 1 (see above)

Introduce the sine rule to students

Explain and prove the rule

Introduce the cosine rule to students (Starter 2, in a handout)

Sketch some points about how to prove the rule

Discuss with teachers

Do with the teacher

Write the rule sine in the notebook

Work out the poof for the sine rule with the teacher

Write the rule cosine in the notebook

Note the ideas to prepare to work out the proof home

1 Hanoi University of Technology 2 International Mathematical Olympiads

75

Provide students with Worksheet 1(to solve triangles)

Provide students with Worksheet 2 (to solve problem with the application of the cosine rule, one question in HUT1 maths exam 2000).

Give out the answer for the exercise in Worksheet 2 (in a separate paper)

Worksheet 3 (to solve problem with the application of the cosine rule, Problem 4 IMO2 1979).

Summarise the lesson and provide students with homework

Work out the solutions and discuss with the teacher

Work out the solutions and discuss with the teacher

Discuss the problem with friends and the teacher

Worksheet 1

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Sine and Cosine Rule Questions

1. ABC is a triangle. AC = 19 cm, BC = 17 cm and angle BAC = 60

Not to scale

600

17 cm

19 cm

B

CA

Calculate the size of angle ABC.

2. PQR is a triangle. PR = 23 cm, PQ = 22 cm and angle QPR = 48

Not to scale

480

22 cm

23 cmP R

Q

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Calculate the length of QR.

Give your answer to an appropriate degree of accuracy.

Worksheet 2:

Exercise (Question VI in HUT entrance Maths exam 1995)

Let be given an equilateral triangle ABC with its circumcircle centered at O. Take a point M on the small chord AB. Given MA=1, MB=2. Calculate MC=?

Solution.

Remark that angle AMB=1200 . Using the cosine rule for the triangle AMB we have

AB2= MA2 + MB2 -2 MA MB cos AMB

= 12 + 22 - 2. 1. 2. cos 1200

= 7.

Denote MC=x, x>0. Obviously, AMC= 1200 the cosine rule once more time for the triangle AMC we get

AC2= MA2 + MC2 -2 MA MC cos AMC. Then,

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7= 12 + x2 -2.1.x. cos 600.

Or, x2- x-6= 0 and we get x=MC=3 since x>0.

Worksheet 3

Exercise (Problem 4, IMO1979)

Given a plane , a point P in the plane and a point Q not in the plane, find all points R in such that the ratio (QP + PR)/QR is a maximum.

Solution

Consider the points R on a circle center P. Let X be the foot of the perpendicular from Q to . Assume P is distinct from X, then we minimise QR (and hence maximise (QP + PR)/QR) for points R on the circle by taking R on the line PX. Moreover, R must lie on the same side of P as X. Hence if we allow R to vary over k, the points maximising (QP + PR)/QR must lie on the ray PX. Take S on the line PX on the opposite side of P from X so that PS = PQ. Then for points R on the ray PX we have (QP + PR)/QR = SR/QR = sin RQS/sin QSR. But sin QSR is fixed for points on the ray, so we maximise the ratio by taking angle RQS = 90. Thus there is a single point maximising the ratio.

If P = X, then we still require angle RQS = 90, but R is no longer restricted to a line, so it can be anywhere on a circle center P.

Homework

A. Do all exercises in Geometry Grade 10 textbook (pp. 59-60) B. Work out the worksheet below

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1. Given an isosceles triangle ABC in which its side lengths a, b and c satisfying

3 3 32b c a a

b c a

.

Prove that ABC is an equilateral triangle.

2. Assume that ABC with sides of a, b and c satisfying 4 4 4a b c . Prove that ABC is acute.

3. A triangle ABC has lengths of its sides given by a=x2+ x+1, b= 2x+1 and c=x2-1. Prove that ABC has one interior angle of 1200.

4. Given ABC with CM as a median, ACM= , BCM= . Prove that

a. sin sinsin sin

AB

b. Given , . Calculate A, B and C.

5. Solve the triangle ABC knowing that a=1, b=2 and c= 3 .

6. Given ABC with lengths of sides as a, b and c; area S. Its inscribed circle touches BC, CA and AB at A', B' and C', respectively. Assume that A'B'C' has lengths of sides as a', b' and c'; area S'. Prove that

a. ' ' 2sin sin sin2 2 2

a b C A Ba b

b. ' 2sin sin sin .2 2 2

S A B CS

7. A ruined tower is fenced off for safety reasons.

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To find the height of the tower Rashid stands at a point A and measures the angle of elevation as 18. He then walks 20 metres directly towards the base of the tower to point B where the angle of elevation is 31.

Calculate the height, h, of the tower.

8. ABCD is a quadrilateral.

AB = 7 cm, AD = 6 cm and BC = 9 cm. Angle ABC = 75 and angle ADC = 90

Calculate the perimeter of ABCD.

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9. In triangle ABC, AB = 11 cm, BC = 9 cm and CA = 10 cm.

Find the area of triangle ABC.

10.Two ships, A and B, leave port at 13 00 hours.

Ship A travels at a constant speed of 18 km per hour on a bearing of 070.

Ship B travels at a constant speed of 25 km per hour on a bearing of 152.

Calculate the distance between A and B at 14 00 hours.

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Trig Derivative

Vocabulary: Derivative, trigonometric functions: sine, cosine, tangent and cotangent

Lesson Objectives

Number(s) of 45-minute period: 1

Materials: Algebra & Analysis Grade 11 Textbook (pp. 163-167), University entrance maths exams, Internet.

Prior Knowledge

1. Find the derivatives of four basic trigonometric functions

2. Differentiate functions which involve trigonometric functions

3. Can use the chain, the product and Quotient rules when

using the trigonometric functions

Teacher: Ta Ngoc Tri, B.Sc. (Maths), M.Sc. (Maths), M.Sc. (Maths Edu.), Ph.D. (Maths)

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ANSWERS

1)

2)

3) 0 0 0 0

sin in sin in sin sin in sinlim lim lim lim 1.1 1sin sinx x x x

s x s x x s x xx x x x x

.

A. Evaluate each of the following limits

1) 2) 3) 0

sin sinlimx

xx

4) 0

1 coslimx

xx

B. Given f(x)=x(x-1)(x-2)...(x-2013). Calcualte f'(0)=?

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4)

22

0 0 0

2sin sin1 cos 1 12 2lim lim lim2 2

2x x x

x xx

xx x

.

LESSON PLAN:

Introduction:

Basic trigonometric functions include sin(x), cos(x), tan(x) and cot x. Knowledge of differentiation from first principles is required, along with competence in the use of trigonometric identities and limits. All functions involve the arbitrary variable x, with all differentiation performed with respect to x. Knowing the derivatives of sin(x) and cos(x), one can easily compute the d