Video lesson "Trigonometric functions of a numeric argument" presents visual material to ensure clarity when explaining the topic in class. During the demonstration, the principle of forming the value of trigonometric functions from a number is considered, a number of examples are described that teach how to calculate the values of trigonometric functions from a number. By using this manual It is easier to develop skills in solving relevant problems and to achieve memorization of the material. Using the manual increases the effectiveness of the lesson and helps quickly achieve learning goals.
At the beginning of the lesson, the title of the topic is shown. Then the task is to find the corresponding cosine to some numeric argument. It is noted that this task The solution is simple and can be clearly demonstrated. The screen displays a unit circle with its center at the origin. It is noted that the point of intersection of the circle with the positive semi-axis of the abscissa axis is located at point A(1;0). An example of point M is given, which represents the argument t=π/3. This point noted on unit circle, and from it a perpendicular descends to the abscissa axis. The found abscissa of the point is the cosine of cos t. IN in this case the abscissa of the point will be x=1/2. Therefore cos t=1/2.
Summarizing the facts considered, it is noted that it makes sense to talk about the function s=cos t. It is noted that students already have some knowledge about this function. Some values have been calculated cosine cos 0=1, cos π/2=0, cos π/3=1/2. Also related to this function are the functions s=sin t, s=tg t, s=ctg t. It is noted that they have a common name for all - trigonometric functions.
Important relationships that are used in solving problems with trigonometric functions: basic identity sin 2 t+ cos 2 t=1, expression of tangent and cotangent through sine and cosine tg t=sin t/cos t, where t≠π/2+πk for kϵZ, ctg t= cos t/sin t, where t≠πk for kϵZ, as well as the ratio of tangent to cotangent tg t·ctg t=1 where t≠πk/2 for kϵZ.
Next, we propose to consider the proof of the relation 1+ tg 2 t=1/ cos 2 t, with t≠π/2+πk for kϵZ. To prove the identity, it is necessary to represent tan 2 t in the form of a ratio of sine and cosine, and then reduce the terms on the left side to common denominator 1+ tan 2 t=1+sin 2 t/cos 2 t = (sin 2 t+cos 2 t)/ cos 2 t. Using the basic trigonometric identity, we obtain 1 in the numerator, that is, the final expression 1/ cos 2 t. Q.E.D.
The identity 1+ cot 2 t=1/ sin 2 t is proved in a similar way, for t≠πk for kϵZ. Just as in the previous proof, the cotangent is replaced by the corresponding ratio of cosine and sine, and both terms on the left side are reduced to a common denominator 1+ cot 2 t=1+ cos 2 t/sin 2 t= (sin 2 t+cos 2 t)/sin 2 t. After applying the main trigonometric identity to the numerator we get 1/ sin 2 t. This is the expression we are looking for.
The solution of examples in which the acquired knowledge is applied is considered. In the first task, you need to find the values of cost, tgt, ctgt, if the sine of the number sint=4/5 is known, and t belongs to the interval π/2< t<π. Для нахождения косинуса в данном примере рекомендуется использовать тождество sin 2 t+ cos 2 t=1, из которого следует cos 2 t=1-sin 2 t. Зная значение синуса, можно найти косинус cos 2 t=1-(4/5) 2 =9/25. То есть значение косинуса cost=3/5 и cost=-3/5. В условии указано, что аргумент принадлежит второй четверти координатной плоскости. В этой четверти значение косинуса отрицательное. С учетом данного ограничения находим cost=-3/5. Для нахождения тангенса числа пользуемся его определением tgt= sint/cost. Подставив известные значения синуса и косинуса, получаем tgt=4/5:(-3/5)=-4/3. Чтобы найти значение котангенса, также используется определение котангенса ctgt= cost/sint. Подставив известные значения синуса и косинуса в отношение, получаем ctgt=(-3/5):4/5=-3/4.
Next, we consider the solution to a similar problem in which the tangent tgt = -8/15 is known, and the argument is limited to the values 3π/2 To find the value of the sine, we use the definition of tangent tgt= sint/cost. From it we find sint= tgt·cost=(-8/15)·(15/17)=-8/17. Knowing that cotangent is the inverse function of tangent, we find ctgt=1/(-8/15)=-15/8. The video lesson “Trigonometric functions of a numerical argument” is used to increase the effectiveness of a mathematics lesson at school. During distance learning, this material can be used as a visual aid for developing skills in solving problems that involve trigonometric functions of a number. To acquire these skills, the student may be advised to independently examine visual material. TEXT DECODING: The topic of the lesson is “Trigonometric functions of a numerical argument.” Any real number t can be associated with a uniquely defined number cos t. To do this you need to do the following: 1) position the number circle on the coordinate plane so that the center of the circle coincides with the origin of coordinates, and the starting point A of the circle falls at point (1;0); 2) find a point on the circle that corresponds to the number t; 3) find the abscissa of this point. This is cos t. Therefore, we will talk about the function s = cos t (es equals cosine te), where t is any real number. We already got some idea of this function: All these functions are called trigonometric functions of the numerical argument t. From the definitions of sine, cosine, tangent and cotangent, some relationships follow: 1) sin 2 t + cos 2 t = 1 (sine square te plus cosine square te equals one) 2)tgt = for t ≠ + πk, kϵZ (tangent te is equal to the ratio of sine te to cosine te with te not equal to pi by two plus pi ka, ka belongs to zet) 3) ctgt = for t ≠ πk, kϵZ (cotangent te is equal to the ratio of cosine te to sine te when te is not equal to pi ka, ka belongs to zet). 4) tgt ∙ ctgt = 1 for t ≠ , kϵZ (the product of tangent te by cotangent te is equal to one when te is not equal to peak ka, divided by two, ka belongs to zet) Let us prove two more important formulas: One plus tangent squared te is equal to the ratio of one to cosine squared te when te is not equal to pi by two plus pi ka. Proof. Let us reduce the expression one plus tangent squared te to the common denominator cosine squared te. We get in the numerator the sum of the squares of the cosine te and sine te, which is equal to one. And the denominator remains the square of the cosine te. The sum of unity and the square of the cotangent te is equal to the ratio of unity to the square of the sine te when te is not equal to pi ka. Proof. The expression one plus cotangent squared te, similarly, we bring to a common denominator and apply the first relation. Let's look at examples. EXAMPLE 1. Find cost, tgt, ctgt if sint = and< t < π.(если синус тэ равен четырем пятым и тэ из промежутка от пи на два до пи) Solution. From the first relation we find the cosine squared te is equal to one minus sine squared te: cos 2 t = 1 - sin 2 t. This means that cos 2 t = 1 -() 2 = (cosine square te is equal to nine twenty-fifths), that is, cost = (cosine te is equal to three fifths) or cost = - (cosine te is equal to minus three fifths). By condition, the argument t belongs to the second quarter, and in it cos t< 0 (косинус тэ отрицательный). This means that the cosine te is equal to minus three-fifths, cost = - . Let's calculate tangent te: tgt = = ׃ (-)= - ;(tangent te is equal to the ratio of sine te to cosine te, and therefore four-fifths to minus three-fifths and equal to minus four-thirds) Accordingly, we calculate (the cotangent of the number te. since the cotangent te is equal to the ratio of the cosine of te to the sine of te,) ctgt = = - . (cotangent te is equal to minus three-fourths). Answer: cost = - , tgt= - ; ctgt = - . (we fill in the answer as we solve it) EXAMPLE 2. It is known that tgt = - and< t < 2π(тангенс тэ равен минус восемь пятнадцатых и тэ принадлежит промежутку от трех пи на два до двух пи). Найти значения cost, sint, ctgt. Solution. Let’s use this relationship and substitute the value into this formula to obtain: 1 + (-) 2 = (one per cosine square te is equal to the sum of one and the square minus eight fifteenths). From here we find cos 2 t = (cosine square te is equal to two hundred twenty-five two hundred eighty-ninths). This means cost = (cosine te is fifteen seventeenths) or cost = . By condition, the argument t belongs to the fourth quarter, where cost>0. Therefore cost = .(cosenus te is fifteen seventeenths) Let's find the value of the argument sine te. Since from the relation (show the relation tgt = for t ≠ + πk, kϵZ) sine te is equal to the product of tangent te by cosine te, then substituting the value of the argument te..tangent te is equal to minus eight fifteenths.. by condition, and cosine te is equal to solved earlier, we get sint = tgt ∙ cost = (-) ∙ = - , (sine te is equal to minus eight seventeenths) ctgt = = - . (since cotangent te is the reciprocal of tangent, which means cotangent te is equal to minus fifteen eighteenths) Trigonometric functions of a numeric argument.
Trigonometric functions of numeric argumentt are functions of the form y= cos t, Using these formulas, through the known value of one trigonometric function, you can find the unknown values of other trigonometric functions. Explanations. 1) Take the formula cos 2 t + sin 2 t = 1 and use it to derive a new formula. To do this, divide both sides of the formula by cos 2 t (for t ≠ 0, that is, t ≠ π/2 + π k). So: cos 2 t sin 2 t 1 The first term is equal to 1. We know that the ratio of sine to conis is tangent, which means the second term is equal to tg 2 t. As a result, we get a new (and already known to you) formula: 2) Now divide cos 2 t + sin 2 t = 1 by sin 2 t (for t ≠ π k): cos 2 t sin 2 t 1 The ratio of cosine to sine is the cotangent. Means: sin 2 t 1 sin 2 t cos 2 t + sin 2 t 1 In the same way, you can easily find the sum of one and the square of the cotangent, as well as many other identities. Trigonometric functions of angular argument.
In functionsat =
cost,
at =
sint,
at =
tgt,
at =
ctgt variablet can be more than just a numeric argument. It can also be considered a measure of the angle - that is, the angular argument. Using the number circle and coordinate system, you can easily find the sine, cosine, tangent, and cotangent of any angle. To do this, two important conditions must be met: 2) one of the sides of the angle must be a positive axis beam x. In this case, the ordinate of the point at which the circle and the second side of the angle intersect is the sine of this angle, and the abscissa of this point is the cosine of this angle. Explanation. Let's draw an angle, one side of which is the positive ray of the axis x, and the second side comes out from the origin of the coordinate axis (and from the center of the circle) at an angle of 30º (see figure). Then the point of intersection of the second side with the circle corresponds to π/6. We know the ordinate and abscissa of this point. They are also the cosine and sine of our angle: √3 1 And knowing the sine and cosine of an angle, you can easily find its tangent and cotangent. Thus, the number circle, located in a coordinate system, is a convenient way to find the sine, cosine, tangent, or cotangent of an angle. But there is an easier way. You don’t have to draw a circle and a coordinate system. You can use simple and convenient formulas: Example: find the sine and cosine of an angle equal to 60º. Solution : π 60 π √3 π 1 Explanation: we found out that the sine and cosine of an angle of 60º correspond to the values of a point on a circle π/3. Next, we simply find the values of this point in the table - and thus solve our example. The table of sines and cosines of the main points of the number circle is in the previous section and on the “Tables” page. Attention! Slide previews are for informational purposes only and may not represent all the features of the presentation. If you are interested in this work, please download the full version. Lesson objectives: Lesson type: training. Lesson type: lesson on skills and abilities. Form of study: group Type of groups:
group sitting together. Students of different levels of training, awareness of a given subject, compatible students, which allows them to complement and enrich each other. Equipment: board; chalk; table "Trigonometer"; route sheets; cards with letters (A, B, C.) for completing the test; plates with crew names; score sheets; tables with names of stages of the journey; magnets, multimedia complex. Students sit in groups: 4 groups of 5-6 people. Each group is a crew of a car with names corresponding to the names of trigonometric functions, led by a steering wheel. Each crew is given a route sheet and a goal is determined: to complete the given route successfully, without errors. The lesson is accompanied by a presentation. The teacher informs the topic of the lesson, the purpose of the lesson, the course of the lesson, the work plan of the groups, the role of the helmsmen. Teacher's opening remarks: –
Guys! Write down the number and topic of the lesson: “Trigonometric functions of a numerical argument.” Today in class we will learn: To do this you need to know: It has long been known that one head is good, but two are better, so today you work in groups. It is also known that the one who walks will master the road. But we live in an age of speed and time is precious, which means we can say this: “The road will be mastered by those who drive,” so today our lesson will be held in the form of a game “Mathematical Rally.” Each group is a vehicle crew, led by a steering wheel. Purpose of the game: The name of the crews corresponds to the make of the car you are driving. The crews and their helmsmen are introduced: The motto of the race: “Hurry up slowly!” You have to run through a “mathematical terrain” with many obstacles. Route sheets were issued to each crew. Crews who know definitions and trigonometric formulas will be able to overcome obstacles. During the run, each helmsman guides the crew, assisting, and assessing the contribution of each crew member to overcome the route in the form of “pros” and “cons” on the score sheet. For each correct answer the group receives a “+” and an incorrect answer “-”. You have to overcome the following stages of the journey: Stage I. SDA (traffic rules). And so off we go! 1) In each crew, the helmsmen distribute tickets with theoretical questions to each crew member: 2) Collect the “scattered” formulas. There is a table on the secret board (see below). The crews must align the formulas. Each team writes the answer on the board in the form of a line of corresponding letters (in pairs). Answer: ab, vg, de, hedgehog, zi, yk. Oral work: test. On the secret board it is written: task: simplify the expression. The answer options are written next to them. Crews determine the correct answers in 1 minute. and pick up the corresponding set of letters. Answer: C V A. The crews have 3 minutes for a meeting to decide the task, and then the crew representatives write the decision on the board. When the crew representatives finish writing down the solution to the first task, all students (together with the teacher) check the correctness and rationality of the solutions and write them down in a notebook. The helmsmen evaluate the contribution of each crew member using the “+” and “–” signs on the evaluation sheets. Tasks from the textbook: –
Your car has broken down. Your car needs to be repaired. Statements are given for each crew, but there are mistakes in them. Find these mistakes and explain why they were made. The statements use trigonometric functions that correspond to the make of your car. You are tired and need to rest. While the crew is resting, the helmsmen sum up preliminary results: they count the “pros” and “cons” of the crew members and the crew as a whole. For students: 3 or more “+” – score “5”; For crews:“+” and “-” cancel each other out. Only the remaining characters are counted. Guess the charade. From the numbers you take my first syllable, The word "trigonometry" (from the Greek words "trigonon" - triangle and "metreo" - measure) means "measurement of triangles." The emergence of trigonometry is associated with the development of geography and astronomy - the science of the movement of celestial bodies, the structure and development of the Universe. As a result of the astronomical observations made, the need arose to determine the position of the luminaries, calculate distances and angles. Since some distances, for example, from the Earth to other planets, could not be measured directly, scientists began to develop techniques for finding relationships between the sides and angles of a triangle, in which two vertices are located on the earth, and the third is a planet or star. Such relationships can be derived by studying various triangles and their properties. This is why astronomical calculations led to the solution (i.e., finding the elements) of the triangle. This is what trigonometry does. The beginnings of trigonometry were discovered in ancient Babylon. Babylonian scientists were able to predict solar and lunar eclipses. Some information of a trigonometric nature is found in ancient monuments of other ancient peoples. To successfully cross the finish line, all you have to do is strain yourself and make a “sprint.” It is very important in trigonometry to be able to quickly determine the values of sin t, cost, tgt, ctg t, where 0 ≤ t ≤ . Close textbooks. Crews alternately name the values of the functions sin t, cost, tgt, ctg t if: Results of the game. The helmsmen hand over evaluation sheets. The crew that became the champion of the “Mathematical Rally” is determined and the work of the remaining groups is characterized. Next are the names of those who received grades “5” and “4”. Lesson summary. - Guys! What did you learn in class today? (simplify trigonometric expressions; find values of trigonometric functions). What do you need to know for this? – I think you understand that you need to know the formulas well in order to apply them correctly. You also realized that trigonometry is a very important part of mathematics, as it is used in other sciences: astronomy, geography, physics, etc. Homework: Whatever real number t is taken, it can be associated with a uniquely defined number sin t. True, the matching rule is quite complex; as we saw above, it is as follows. To find the value of sin t using the number t, you need: 1) position the number circle in the coordinate plane so that the center of the circle coincides with the origin of coordinates, and the starting point A of the circle falls at point (1; 0); 2) find a point on the circle corresponding to the number t; 3) find the ordinate of this point. This ordinate is sin t. In fact, we are talking about the function u = sin t, where t is any real number. All these functions are called trigonometric functions of the numerical argument t. There are a number of relations that connect the values of various trigonometric functions; we have already obtained some of these relations: sin 2 t+cos 2 t = 1 From the last two formulas it is easy to obtain a relationship connecting tg t and ctg t: All of these formulas are used in cases where, knowing the value of a trigonometric function, it is necessary to calculate the values of other trigonometric functions. The terms “sine”, “cosine”, “tangent” and “cotangent” were actually familiar, however, they were still used in a slightly different interpretation: in geometry and physics they considered sine, cosine, tangent and cotangent at the head(but not numbers, as was in the previous paragraphs). From geometry it is known that the sine (cosine) of an acute angle is the ratio of the legs of a right triangle to its hypotenuse, and the tangent (cotangent) of an angle is the ratio of the legs of a right triangle. A different approach to the concepts of sine, cosine, tangent and cotangent was developed in the previous paragraphs. In fact, these approaches are interrelated. Let's take an angle with degree measure b o and place it in the “numeric circle in a rectangular coordinate system” model as shown in Fig. 14 the apex of the angle is compatible with the center circles (with the origin of the coordinate system), and one side of the corner is compatible with the positive ray of the x-axis. Full stop intersection of the second side of the angle with denote by the circle the letter M. Ordina- Fig. 14 b o, and the abscissa of this point is the cosine of the angle b o. To find the sine or cosine of an angle b o it is not at all necessary to do these very complex constructions every time. It is enough to note that the arc AM makes up the same part of the length of the number circle as the angle b o makes from the corner of 360°. If the length of the arc AM is denoted by the letter t, we get: Thus, For example, It is believed that 30° is a degree measure of an angle, and a radian measure of the same angle: 30° = rad. At all: In particular, I’m glad where, in turn, we get it from. So what is 1 radian? There are various measures of length of segments: centimeters, meters, yards, etc. There are also various measures to indicate the magnitude of angles. We consider the central angles of the unit circle. An angle of 1° is the central angle subtended by an arc that is part of a circle. An angle of 1 radian is the central angle subtended by an arc of length 1, i.e. on an arc whose length is equal to the radius of the circle. From the formula, we find that 1 rad = 57.3°. When considering the function u = sin t (or any other trigonometric function), we can consider the independent variable t to be a numerical argument, as was the case in the previous paragraphs, but we can also consider this variable to be a measure of the angle, i.e. corner argument. Therefore, when talking about a trigonometric function, in a certain sense it makes no difference to consider it a function of a numerical or angular argument. Definition1: The numerical function given by the formula y=sin x is called sine. This curve is called - sine wave. Properties of the function y=sin x 2. Function value range: E(y)=[-1; 1] 3. Parity function: y=sin x – odd,. 4. Periodicity: sin(x+2πn)=sin x, where n is an integer. This function takes on the same values after a certain period. This property of a function is called frequency. The interval is the period of the function. For the function y=sin x the period is 2π. The function y=sin x is periodic, with period Т=2πn, n is an integer. The smallest positive period is T=2π. Mathematically, this can be written as follows: sin(x+2πn)=sin x, where n is an integer. Definition2: The numerical function given by the formula y=cosx is called cosine. Properties of the function y=cos x 1. Function domain: D(y)=R 2. Function value area: E(y)=[-1;1] 3. Parity function: y=cos x – even. 4. Periodicity: cos(x+2πn)=cos x, where n is an integer. The function y=cos x is periodic, with period Т=2π. Definition 3: The numerical function given by the formula y=tan x is called tangent. Properties of the function y=tg x 1. Domain of the function: D(y) - all real numbers except π/2+πk, k – integer. Because at these points the tangent is not defined. 2. Function range: E(y)=R. 3. Parity function: y=tg x – odd. 4. Periodicity: tg(x+πk)=tg x, where k is an integer. The function y=tg x is periodic with period π. Definition 4: The numerical function given by the formula y=ctg x is called cotangent. Properties of the function y=ctg x 1. Domain of definition of the function: D(y) - all real numbers except πk, k is an integer. Because at these points the cotangent is not defined.
y= sin t, y= tg t, y= ctg t.
--- + --- = ---
cos 2 t cos 2 t cos 2 t
--- + --- = ---, where t ≠ π k + π k, k– integer
sin 2 t sin 2 t sin 2 t
Knowing the basic principles of mathematics and having learned the basic formulas of trigonometry, you can easily derive most of the other trigonometric identities on your own. And this is even better than just memorizing them: what you learn by heart is quickly forgotten, but what you understand is remembered for a long time, if not forever. For example, it is not necessary to memorize what the sum of one and the square of the tangent is equal to. If you forgot, you can easily remember if you know the simplest thing: tangent is the ratio of sine to cosine. In addition, apply the simple rule of adding fractions with different denominators and get the result:
1 + tg 2 t = 1 + --- = - + --- = ------ = ---
cos 2 t 1 cos 2 t cos 2 t cos 2 t
1) the vertex of the angle must be the center of the circle, which is also the center of the coordinate axis;
--; --
2 2
sin 60º = sin --- = sin -- = --
180 3 2
cos 60º = cos -- = -
3 2
Back forward
During the classes
I. Organizational moment.
Stage II. Technical inspection.
Stage III. Cross-country race.
Stage IV. A sudden stop is an accident.
V stage. Halt.
Stage VI. Finish.
VII stage. Results.Stage I. SDA (traffic rules).
A
tg 2 t + 1
e
1
V
tg t
and
cos t / sin t, t ≠ k, kZ.
d
sin 2 t + cos 2 t
And
1/ sin 2 t, t ≠ k, kZ.
e
ctg t
To
1,t ≠ k / 2, kZ.
h
1 + ctg 2 t
G
sin t /cos t, t ≠ /2 + k, kZ.
th
tg t ∙ctg t
b
1/ cos 2 t, t ≠ /2 + k, kZ.
Stage II. Technical inspection.
№
Expression
Answer options
A
IN
WITH
1.
1 – cos 2 t
cos 2 t
- sin 2 t
sin 2 t
2.
sin 2 t – 1
cos 2 t
- cos 2 t
2 cos 2 t
3.
(cos t – 1)(1+ cos t)
-sin 2 t
(1+ cos t) 2
(cos t – 1) 2
Stage III. Cross-country race.
Stage IV. A sudden stop is an accident.
V stage. Halt.
2 “+” – rating “4”;
1 “+” – rating “3”.
The second is from the word “proud”.
And you will drive the third horses,
The fourth will be the bleating of a sheep.
My fifth syllable is the same as the first
The last letter in the alphabet is the sixth,
And if you guess everything correctly,
Then in mathematics you will get a section like this.
(Trigonometry)Stage VI. Finish.
VII stage. Results.
Trigonometric functions of a numeric argument. Properties and graphs of trigonometric functions.
