trigonometry unit english for students of mathematics no2

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THAI NGUYEN UNIVERSITY OF EDUCATION MATHEMATICS

------TRIGONOMETRY

Supervisors: PhD. TRAN NGUYEN AN

Unit: English for students of mathematics (NO2)

Thai nguyen, April 2023

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Trigonometry is the branch of mathematics concerned with specificfunctions of angles and their application to calculations. There aresix functions of an angle commonly used in trigonometry. Theirnames and abbreviations are sine (sin), cosine (cos), tangent (tan), cotangent(cot). These four trigonometric functions in relation to a right triangle. Forexample, the triangle contains an angle A, and the ratio of the side opposite toA and the side opposite to the right angle (the hypotenuse) is called the sine ofA, or sin A; the other trigonometry functions are defined similarly. Thesefunctions are properties of the angle A independent of the size of the triangle,and calculated values were tabulated for many angles before computers madetrigonometry tables obsolete. Trigonometric functions are used in obtainingunknown angles and distances from known or measured angles in geometricfigures.

Trigonometry developed from a need to compute angles and distances in suchfields as astronomy, mapmaking, surveying, and artillery range finding.Problems involving angles and distances in one plane are covered in planetrigonometry. Applications to similar problems in more than one plane ofthree-dimensional space are considered in spherical

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TRIGONOMETRIC FUNCTIONS AND TRIGONOMETRIC EQUATIONS

 TRIGONOMETRIC FUNCTIONS:

1. DEFINITION

First of all, we recall the table of trig values of special ares.

 Sine and cosine functions

a) Sine functions

- In Grade 10, you knew that you could set each real number x corresponding toa unique point M on a trig circle with the measure of arc AM equal to x (rad).The y-coordiate M is completely determined. That value is sin x.

value of x on the x-axis and that of sinx on the y-axis, we get figure

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The rule of setting each real number x corresponding to the real number sinx

sin:sinx y x

Is called a sine function, denoted by y=sin x.The domain of a sine function is 

b) Cosine functions

The rule of setting each real number x corresponding to the real number sinx

cos:cosx y x

Is called a sine function, denoted by y=cosx.The domain of a cosine function is 

 Tangent and cotangent function

a) Tangent functions

The tangent function is defined by the formula

sin(cos 0)cos

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2. VARIATIONS AND GRAPHS OF TRIGONOMETRIC FUNCTIONS.a, Function y=sinx

VARIATIONS AND GRAPH OF FUNCTION Y=SIN X OVER INTERGER



0;



Consider real numbers x x where 1 2, 0 12

2x x 

  and x x1 2thus sin x <sin12

x .

Now x x belong to 3 4, 2; 

  and x x3 4 but sinx3 sinx 4

Therefore, fructions y=sinx increases on 0;2

  

  and decreases on 2; 

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+, is defintion with every x  and 1 cos  x1

+, is an even function;

+, is a periodic ffunction with period 2

With every x  we have equation sin x 2 cosx

  

Next, by translating the graph of function y = sinx through vector u ( ;0)2 

( over a unit of leghth equal to 2

to the left, parallel to the x-axis), we obtainthe graph of function y = cosx ( Figure 6)

From the graph of function y = cos x in figure 6, we infer :

Function y= cosx increases on interval



,0



and decereases on interval



0,



.

The Variation table:

x-π 0 πy = cos x

The graphs of functions y=cos x and y=sin x are called sinusoids.

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is an odd function

is a periodic function with period π.

A, The variation and graph of function y=tan x on half open interval 0;2 

 

From the geometric representation of tan x (Fig. 7a), with

1, 2 0; , 11, 22, 1 tan ,12 tan 2

x x  AMx AM x AT x AT  x

 

The variation table:X

To draw the graph of function y= tanx on half- open interval 0;2

Follow the steps below:

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First, calculate values of function y = tan x at some special points such as x = 0,

, the nearer the graph of function

y=tanx moves to line x 2

 . Function y=cot x

By definition, we find fuction y=cot x;Has domain D=R‚



k k Z, 



Is an odd function;

Is a periodic function with period π

We are going to consider the variation and the graph of function y=cot x oninterval (0; ) and then we will infer the graph of this function on D.

a) The variation and graph of function y=cot x on interval (0;π)

For two numbers x and 1 x such that 0 <2 x <1 x < π, we have 0 <2 x x2 1 π.

Thus cot x1 cot x = 2

cos cossin sin

x  x

=

sin cos cos sinsin sin

x x

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x xx x

Figure 10 represents the graph of function y=cot x on open interval (0; )

b) Graph of function y= cot x on D

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The graph of function y=cot x on D as represented in Figure 11.

The range of function y=cot x is open interval



  ;



I.Some basic vocabulary: M t sốố t v ng c b nộ ừ ự ơ ả

Ordinal

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8 Oriented circles/ _ rient d s k l /ɔː ɪ ˈ ɜː ə Đường tròn đ nh hị ướng

34 Positive real number/ p z. .t v r l ˈ ɒ ə ɪ ɪə

n mb r/

ˈ ʌ ə

Sốế th c dự ương

40 Anti-Clockwise/ ænti kl kwa z/ˈ ˈ ɒ ɪ Ngược chiêều kim đh

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43 Decrease/d kri s/ɪˈ ː S gi m điự ả

47 Starting point/ st t ŋ p nt/ˈ ɑː ɪ ˌ ɔɪ Đi m đâềuể

54 Opposite orientation/ p z t rien teˈɒ ə ɪ ɔː ˈ ɪʃə n/Chiêều ngượ ạc l i

55 One orientation/w nʌ ˌɔːrien teˈ ɪʃən/M t chiêềuộ

Đường tròn lượng giác

81 Arc of semi-circle/ k v semi s k l/ɑː ɒ ˈ ˌ ɜː ə Cung n a đử ường tròn

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Comment: Thus, to reduce the above expressions, we use a descending formulabased on the main idea of turning it into a whole.

(Bài t p 1: rút g n bi u th c lậ ọ ể ứ ượng giác

Rút g n bi u th c: A = cos10x + 2cosọ ể ứ 24x + 6cos3x.cosx - cosx - 8cosx.cos33x.Gi iả

b c d a trên ý tậ ự ưởng ch đ o là biêến đ i nó vêề d ng t ng.)ủ ạ ổ ạ ổ

Exercise 2: Simplify expressions:

sin x cos xbB

tan x tan  x 

2222

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Thus, to reduce the above expressions, we only need to use the relationshipbetween special angles.

sin x cos xbB

tan x tan  x 

cosx .

Bài t p 3ậ

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2. Calculate the value of trigonometric expressions. (Tính giá tr cácị

bi u th c lể ứ ượng giác)

From the given hypothesis (usually the value of an angle or a trigonometricvalue) the orientation transforms the expression to a form where only thegiven value of the hypothesis is present. Need to pay attention to theapplicable conditions (if any)

(T gi thiêết đêề cho (thừ ả ường là giá tr c a góc hay m t giá tr lị ủ ộ ị ượng giác)đ nh hị ướng biêến đ i bi u th c vêề d ng ch xuâết hi n giá tr đã cho c a giổ ể ứ ạ ỉ ệ ị ủ ả

thiêết đ tính. Câền chú ý điêều ki n áp d ng (nêếu có))ể ệ ụ

Exercise 4:Solution:

Let cos x = 1. Calculate the value of the expression

22 1

3P sin x cos x  cosx

we have,

22 12 114(1 cos ) cos 4 3()

3 3

.

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Exercise 5: Problems in triangles

1.Show that :

(i) tan 48° tan 23° tan 42° tan 67° = 1(ii) cos 38° cos 52° – sin 38° sin 52° = 0

Solution:

(i) tan 48° tan 23° tan 42° tan 67°

= tan (90° - 42°) tan (90° - 67°) tan 42° tan 67°= cot 42° cot 67° tan 42° tan 67°

= (cot 42° tan 42°) (cot 67° tan 67°) = 1×1 = 1(ii) cos 38° cos 52° - sin 38° sin 52°

= cos (90° - 52°) cos (90°-38°) - sin 38° sin 52°= sin 52° sin 38° - sin 38° sin 52° = 0

2. If tan A = cot B, prove that A + B = 90°

tan A = cot B

⇒ tan A = tan (90° - B)⇒ A = 90° - B⇒ A + B = 90°

E4. Prove a trigonometric expression that does not depend on x

Use basic trigonometric systems.Use properties of trigonometric values.+ Use memorable equality constants.

Exercise 6: If tan(A + B) = √3 and tan(A – B) = 1/√3 ; 0° < A + B ≤ 90°; A > B, find A and B.

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Solution:

tan(A + B) = √3tan(A + B) = tan 60°A + B = 60°….(i)And tan(A – B) = 1/√3tan(A – B) = tan 30°A – B = 30°….(ii)Adding (i) and (ii),A + B + A – B = 60° + 30°2A = 90°

A = 45°

Substituting A = 45° in (i),45° + B = 60°

A = 45°

Thay A = 45° vào (i), ta có

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45° + B = 60°B = 60° – 45° = 15°Do đó, A = 45° và B = 15°

Exercise 7: Given triangle . Prove that:

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Bài t pậ 7: Cho tam giác ABC.. Ch ng minh răềng:ứ

Exercise 8: Prove that the following expression does not depend on x:Solution:

 So P does not depend on x

Bài t p ậ 8: Ch ng minh rằằng bi u th c sau không ph thu c vào x:ứ ể ứ ụ ộ

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sin 6cos 3cos

(i) tan 48° tan 23° tan 42° tan 67°

= tan (90° - 42°) tan (90° - 67°) tan 42° tan 67°= cot 42° cot 67° tan 42° tan 67°

= (cot 42° tan 42°) (cot 67° tan 67°) = 1×1 = 1(ii) cos 38° cos 52° - sin 38° sin 52°

= cos (90° - 52°) cos (90°-38°) - sin 38° sin 52°= sin 52° sin 38° - sin 38° sin 52° = 0

Exercise 10: If tan A = cot B, prove that A + B = 90°.

tan A = cot B

⇒ tan A = tan (90° - B)⇒ A = 90° - B⇒ A + B = 90°

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Exercise 11. A 20 m long rope, which is tightly stretched and tied from the top of a vertical pole to the ground. Find the height of the pole, if the angle made by the rope with the ground level is 30° (see figure ).

⇒

12 20

⇒

⇒ AB = 10

The height of the pole is 10 m

Bài 11. M t s i dây dài 20 m độ ợ ược căng và bu c ch t t đâều m t chiêếc c t ộ ặ ừ ộ ộ

th ng đ ng xuốếng đâết. Tìm chiêều cao c a c t, nêếu góc t o b i s i dây v i ẳ ứ ủ ộ ạ ở ợ ớ

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⇒

12 20

⇒

In the right triangle PQR, Q is right angle.By Pythagoras theorem,

PR2 = PQ + QR22QR2 = (13) – (12)22= 169 – 144= 25QR = 5 cmtan P =

PQ =

cot R =

PQ =

So, tan P – cot R = (

12 ) –(

12) = 0

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Exercise 13: Given 15 cot A = 8, find sin A and sec A.

Let ΔABC be a right-angled triangle, right-angled at B.We know that cot A =

cos A =

Bài 13: Cho 15 cot A = 8, tìm sin A và cos A.Tr l i:ả ờ

Cho ∆ABC là tam giác vuống cân t i B.ạ

Ta biêết răềng cũi A =

(Cho trước)G i AB là 8 và BC là 15 v i k là sốế th c dọ ớ ự ương.Theo đ nh lý Pythagore ta đị ược

AC2 = AB + BC22

AC2 = (8) + (15)22

AC2 = 64 + 2252 2

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AC = 289AC = 17 sin A =

cos A =

Exercise 14: In the right angle in the figure below, angle Z is 50 degrees.The length of XZ is 5 units.

sin 50° = 0.766, cos 50° = 0.643, and tan 50° = 1.192.Approximately how many units long is XY?

Solution: The sine of angle Z is calculated by dividing XY by XZ.

sin Z =

XYXZ

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3.83 = XY. So line segment XY has length 3.83

Exercise 14 : Trong góc vuống trong hình dưới đây, góc Z là 50 đ .ộ

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0,766 × 5 = 5

× 50,766 × 5 = XY3,83 = XY .

V y XY dài 3,83 đ n v .ậ ơ ị

Exercise 15: ∠XYZ is an isosceles triangle, where XY is equal to YZ.Angle Y is 60° and points W, X, and Z are co-linear.

What is the measurement of ∠WXY?

Solution: We know that any straight line is 180°.

So, we need to subtract the degree of the angle stated in the problem (∠XYZ)from 180°.

180° − 60° = 120°

There are two remaining angles lying on the straight line.

One is WXY, and we will call the other one that extends past point Z ∠ ∠YZV.

The sum of these two remaining angles, WXY and YZV equals 120°.∠ ∠

Since XYZ is isosceles, the remaining angles will be equal to each other since∠

their two sides, XY and YZ are equal.

So, we divide the remaining degrees by two in order to find out how manydegrees there are in WXY and ∠ ∠YZV.

120°÷ 2 = 60°

Bài 15: tam giác XYZ là tam giác cân, trong đó XY băềng YZ.Góc Y băềng 60° và các đi m W, X, Z th ng hàng.ể ẳ

Phép đo c a tamủ giác WXY là gì?

Gi i:ả Ta biêết răềng bâết kỳ đường th ng nào cũng băềng 180°..ẳ

Vì v y, chúng ta câền tr đi đ c a góc đã nêu trong bài tốn (ậ ừ ộ ủ ∠XYZ) t 180°.ừ

180° − 60° = 120°

Có hai góc cịn l i năềm trên đạ ường th ng.ẳ

M t là ộ ∠WXY, và chúng ta seỗ g i cái còn l i kéo dài qua đi m Z là ọ ạ ể ∠YZV.

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T ng c a hai góc cịn l i này, tamổ ủ ạ giác WXY vâềtm giác YZV băềng 120°.Vì tam giác XYZ là cân nên các góc cịn l i seỗ băềng nhau vì hai c nh XY và YZạ ạ

băềng nhau.

Vì v y, ta chia các đ cịn l i cho hai đ tìm xem có bao nhiêu đ trong tamậ ộ ạ ể ộ

giác WXY và tam giác YZV.120°÷ 2 = 60°

Exercise 16: If sin 3A = cos (A – 26°), where 3A is an acute angle, find the value of A.

⇒ A =

⇒ A = 29°

Bài 16 Nếốu sin 3A = cos (A – 26°), trong đó 3A là góc nh n, hãy tìm giá : ọ

tr c a A.ị ủGi i:ả

Ta có,

Sin 3A = cos (A – 26°); 3A là góc nh nọ

cos (90° – 3A) = cos (A – 26°)

⇒ 90° – 3A = A – 26⇒ 3A + A = 90° + 26°⇒ 4A = 116°

⇒ A =

o

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⇒ A = 29°

7. problems on variation and graph of trigonometric functions

Exercise 17: Determine the value of x on the segment

2 

  so that the

function y=tan x;a, Get value equal to 0b, Get value equal to 1c, Get a positive valued, Get negative value

a, obsever the graph of the function y= tan x on the segment

2 



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a, tan x=0 at the value x= - ,0, 

(The point on the horizontal axis intersect the graph of the function y=tan x)+, Similar:

b, tan x=1 at the value x=

3 , ,54 4 4

  

7.Bài toán vếề s biếốn thiến và đốề th c a hàm sốố lự ị ủ ượ ng giác

Bài 17: Hãy xác đ nh giá tr c a x trên đo n ị ị ủ ạ3;

2 

L i gi i:ờ ả

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a, Quan sát đốề th hàm sốế y= tan x trên đo nị ạ3;

2 

  

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trigonometry unit english for students of mathematics no2

<b>THAI NGUYEN UNIVERSITY OF EDUCATION MATHEMATICS</b>

<i> </i>

<b>------TRIGONOMETRY</b>

<b>Unit: English for students of mathematics (NO2)</b>

<i>Thai nguyen, April 2023</i>





















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