trigonometry unit english for students of mathematics no2

Theirnames and abbreviations are sine sin, cosine cos, tangent tan, cotangentcot.. Forexample, the triangle contains an angle A, and the ratio of the side opposite toA and the side oppos

THAI NGUYEN UNIVERSITY OF EDUCATION MATHEMATICS

Unit: English for students of mathematics (NO2)

Thai nguyen, April 2023

Trigonometry is the branch of mathematics concerned with specific functions of angles and their application to calculations There are six functions of an angle commonly used in trigonometry Their names and abbreviations are sine (sin), cosine (cos), tangent (tan), cotangent (cot) These four trigonometric functions in relation to a right triangle For example, the triangle contains an angle A, and the ratio of the side opposite to

A and the side opposite to the right angle (the hypotenuse) is called the sine of

A, or sin A; the other trigonometry functions are defined similarly These functions are properties of the angle A independent of the size of the triangle, and calculated values were tabulated for many angles before computers made trigonometry tables obsolete Trigonometric functions are used in obtaining unknown angles and distances from known or measured angles in geometric figures.

T

Trigonometry developed from a need to compute angles and distances in such fields as astronomy, mapmaking, surveying, and artillery range finding Problems involving angles and distances in one plane are covered in plane trigonometry Applications to similar problems in more than one plane of three-dimensional space are considered in spherical

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 to

a 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.

esent ing the

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

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

sin:

sin

x y x

Is called a sine function, denoted by y=sin x.

The domain of a sine function is 

Is called a sine function, denoted by y=cosx.

The domain of a cosine function is 

 Tangent and cotangent function

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 1 2

  and x x3 4 but sinx3 sinx 4

Therefore, fructions y=sinx increases on 0;2

+, 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

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

Function y= cosx increases on interval ,0and decereases on interval

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

To draw the graph of function y= tanx on half- open interval 0;2

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

Is a periodic function with period π

We are going to consider the variation and the graph of function y=cot x on interval (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

1 2

1 2

cos cossin sin

Therefore, function y =cot x decreases on interval (0; )

The variation table:

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

b) Graph of function y= cot x on D

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

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

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

Comment: Thus, to reduce the above expressions, we use a descending formula based 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

A = cos10x + 1 + cos8x - cosx - 2(4cos33x - 3cos3x)cosx

= 2cos9x.cosx + 1 - cosx - 2cos9x.cosx = 1 - cosx.



Nh v y, đ rút g n các bi u th c trên chúng ta s d ng cống th c hư ậ ể ọ ể ứ ử ụ ứ ạ

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:

2

(cos

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

sinx sin x sin x

cosx cosx cosx

Solution:

We have in turn:

= 2sin3x.cos2x + sin3x = sin3x(2cos2x + 1) (first)

cosx + cos3x + cos5x = cosx + cos5x + cos3x

= 2cos3x.cos2x + cos3x = cos3x(1cos2x - 1) (2)

From (1) and (2) deduce: A = sin3xcos3x = tan3x

33

sinx

cosx

Bài t p 3ậ

A =

sinx sinx sinx

cosx cos x cos x

Gi iả

Ta lâền lượt có: sinx + sin3x + sin5x = sinx + sin5x + sin3x

= 2sin3x.cos2x + sin3x = sin3x(2cos2x + 1) (1)

cosx + cos3x + cos5x = cosx + cos5x + cos3x

= 2cos3x.cos2x + cos3x = cos3x(1cos2x - 1) (2)

T (1) và (2) suy ra: A =ừ sin3xcos3x = tan3x.

33

sinxcosx

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 trigonometric value) the orientation transforms the expression to a form where only the given value of the hypothesis is present Need to pay attention to the applicable 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

3 3

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°

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.

BB

BB

 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

sin 6cos 3cos

(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

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

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 ).

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

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

Solution:

Let ΔABC be a right-angled triangle, right-angled at B.

We know that cot A =

815

AB

BC

(Given) Let AB be 8 and BC will be 15 where k is a positive real number.

By Pythagoras theorem we get,

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

Ta biêết răềng cũi A =

815

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.

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

Chiêều dài c a XZ là 5 đ n v ủ ơ ị

sin 50° = 0,766, cos 50° = 0,643 và tan 50° = 1,192.

XY dài bao nhiêu đ n vơ ị ?

Gi i:ả Sin c a góc Z đủ ược tính băềng cách chia XY cho XZ.

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 many degrees 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 toán (ậ ừ ộ ủ ∠XYZ) t 180°.ừ

180° − 60° = 120°

Có hai góc còn 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.

T ng c a hai góc còn l i này, tamổ ủ ạ giác WXY vâềtm giác YZV băềng 120°.

Vì tam giác XYZ là cân nên các góc còn l i seỗ băềng nhau vì hai c nh XY và YZạ ạ

băềng nhau.

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

giác WXY và tam giác YZV.

sin 3A = cos (A – 26°); 3A is an acute angle

cos (90° – 3A) = cos (A – 26°)

⇒ A = 29°

7 problems on variation and graph of trigonometric functions

Exercise 17: Determine the value of x on the segment

3

;2

a, Get value equal to 0

b, Get value equal to 1

c, Get a positive value

d, Get negative value

Solution:

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

3

;2

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=

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 ạ

3

;2

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

3

;2