Material developer:
Matveeva Maria Viktorovna
mathematic teacher
GBOU Shi "Olympic Reserve"
Programmed lesson for grade 10 on the topic:
The concept of the arccosine. Equation of the form with os x = a.
As with solving ordinary equations, solving trigonometric equations comes down to the ability to solve simple equations.
Definition: An equation is called trigonometric if the unknown is under the sign of trigonometric functions.
The simplest trigonometric equations are: with os x = a, sinx = a, tgx = a.
Each of them has its own formula for solving. The only thing that needs to be clear remember- this is what happens when solving them infinite number of roots.
But you can find specific solutions.
In order to learn how to solve the first simple trigonometric equation, you need to get acquainted with such a concept as the arc cosine of a number.
It should be noted that number, for which the arc cosine is considered, belongs to the interval [-1; one].
Definition: The arc cosine of a [-one; 1] (denoted arccos a ) is such a number α , whose cosine is equal to a.I.e cos ( arccos a ) = a.
For example, arccos (-1) = π;as cos π= -1
arccos = , as cos =
Thus, the arc cosine is the inverse function of the cosine.
Write in a theoretical notebook: definition and examples.
Actually find the value arccos can be easily used painfully familiar to us table values of trigonometric functions.
When found arccos one must ask oneself the question, at what value cos equals ? And look at the table. Answer: "at 45 ° or in radian measure ».
It should be remembered that the value of the arccosine is usually written only in radians. Therefore, you should remember the correspondence between the degree and radian measure of angles.
If the number from which you want to find the arccosine is negative, then to find it you need to use the formula:
arccos (-a) \u003d π - arccos a.
For example, arccos (
= π
=
.
arccos (
= π
=
.
Write in a theoretical notebook: formula and examples.
Solve the tasks according to the textbook. 168 No. 568 - 570.
Solution of a trigonometric equation of the form cos x = a is reduced to using the formula:
x = ±
This formula can be illustrated in figure 68 p. 165 according to the textbook. Open the textbook.
The drawing shows that a point is marked on the cosine axis . A straight line drawn vertically through this point shows that the cosine for the values I and VI quarters match.
But how can we get these angles when we rotate the point? Yes, exactly in I quarters on the "+" corner, and in VI quarters on "-". Hence the sign "±". That is, with s and cos match.
Write in a theoretical notebook: a formula and a drawing from a textbook with explanations.
Let's analyze the solution of the trigonometric equation using an example:co s x = x = ±
(see value according to the table)x =±
Answer: x =±
Write in a theoretical notebook: the solution of the equation with explanations. Since an infinite number of roots is obtained, tasks are sometimes asked to find specific values of the roots, for example, belonging to the interval, that is I quarter or gap.
These tasks are very common in the exam. They can be found by substituting n specific numbers (highlighted to help you).
For example, consider the solution to our equation x =±
1.
Let be
n
=0
. Then x =±
±
, i.e. x 1
=
+
them 2
=
.
From this it can be seen that it turns out 45 ° and - 45 °. Of these two numbers, only one belongs to the interval, i.e. I quarters. Number only+
.
2.
Let be
n
=1
. Then x =±
±
,
i.e. x 1
=
+
them 2
=
,
X 1
=
=
them 2
=
=
From this it can be seen that x 1 \u003d 405 ° and x 2 \u003d 315 ° are obtained. So none of the numbers belong I a quarter, that is, an interval. Therefore, they cannot be written as an answer.
Write in a theoretical notebook: a way to find specific roots (belonging to a specific interval) of a trigonometric equation. For example 1 , solve the equationco s x = and find the roots that belong to the interval [ ]. First, what you need to do is just solve the equation using the formula and forget about the gap for a while.co s x = x = ±
(see value according to the table)x =±
Second, you need to decide on the quarter to which the roots should belong. this is the interval from 90° to 180°. So this II second quarter. Third, you need to substitute specific values n(highlighted to help you). - Let n=0.
For example 2 , solve the equationco s x = . co s x = x = ±
(see value according to the table, but there are no such values in the table, so calculate the value is not possible).Answer:x = ± Write out in a theoretical notebook: example 2 with explanations.If the cosine is equal to a negative number, it is necessary to use a different formula when solving the equation: x = ± ± Solve the tasks according to the textbook: p. 169 #571, 572. Equations are not always so simple, there are equations of varying degrees of complexity.For example, 3 . Solve the Equation2so s 3x = . co s 3x = ( you need to divide both sides of the equation by the number that comes before the cosine)3x = ± division sign can be written as a solid line s x = ,5 co s x = ,5 It is not possible to solve such an equation, since the value of the cosine is in the interval [-1; one].Answer: no solutions.Write in a theoretical notebook: examples with explanations. Solve the tasks according to the textbook: p. 169 #573.
Lesson type: setting a learning task.
Lesson Objectives:
educational: to systematize the knowledge of students about the methods for solving the simplest trigonometric equations, to consolidate the skills of working with a circle and a table.
Educational: to continue work on the formation of creative intellectual abilities of students through the use of various methods for solving trigonometric equations.
Educational: to develop the skills of collective mental activity, mutual support and acceptance of a point of view different from one's own.
During the classes
1. The situation of success.
Solve the equation: cosx=1; cox=0; cosx = -1.
2. Situation, gap” between knowledge and ignorance.
Solve the equation: cosx=½; cosx=a.
Discussion.
3. Statement of the learning task.
How to solve this type of equation?
1) What is the abscissa of the unit circle point obtained by rotating the point (1; 0) around the origin by an angle equal to: ?
2). What is equal to: 
?
Answer:
3). What is equal to: .
Answer:
;
;
(1)
.
Teacher's words: mathematicians called words, inversely cos “the word arccosine (arccos). The arccosine of a number is a number whose cosine is equal to a:
arccosa=α if cosα=a and 0≤α≤π.
4). Write equality (1) using the symbol arccos .
5). Solve equations: cosx=½, cosx=α.
Answer: x=arccos½, x=arccosa.
6). Name the angles of rotation of the point (1; 0) of the unit circle with the abscissa equal to ½.
Answer: the abscissa is ½ when the point is rotated through an angle equal to π / 3 and -π / 3.
i.e. cosx=½ for x=±arccos½
cosx=a when x=±arccosa.
7). What are the abscissas of the points obtained by turning the point (1; 0) through the angles: π/3+2π; π/3+6π; -π/3+4π; -π/3+8π; π/3+2πn; -π/3+2πn.
Answer: the abscissa is ½, and cosx=½ for x=±arccos½+2πn,.
cosx=a for x=±arccosa+2πn,.
eight). Conclusion: equation cosx=a
1) has roots if ≤1,
2) has no roots if >1.
nine). Lesson summary:
a) For what values of a and α does the equality arccosа=α make sense?
b) What is called the arc cosine of the number a?
c) For what values of a does the equation cosx=a have roots?
d) The formula for finding the roots of the equation cosx=a.
Synopsis lesson 1 (Shelest S.V.)
Subject name: Algebra and the beginnings of mathematical analysis Class 10 UMC: Algebra and the beginnings of mathematical analysis A.G. Mordkovich, P.V. Semenov, 2015 Level of study (basic) Lesson topic: The concept of the arc cosine and solving the equation cost = a The total number of hours allotted for studying the topic: 2 Place of the lesson in the system of lessons on the topic: 1 Lesson goal: introduce the concept of the arc cosine, derive a formula for finding of the roots of the equation cos t \u003d a Lesson objectives 1. Educational: a) to form the ability to calculate the arc cosine; b) to teach how to apply the formula when solving the simplest trigonometric equations; 2. Developing: a) develop the ability to briefly, logically, consistently express thoughts and judgments; b) develop the ability to argue their statements; 3. Educational: a) to teach the skills of planning activities, work at an optimal pace, b) to cultivate the ability to correctly assess one's capabilities, the results of educational activities, develop communication skills; Equipment: handout, unit circle model. Course of the lesson 1. Organizational moment. 2. Actualization of students' knowledge. Oral counting (tasks on the board) 1. Calculate. a) Calculate the values: cos ; cos ; cos. b) Calculate the values: cos ; cos; cos 2. Name several angles whose cosine is equal to: a) 0; b) c) d) –1. 3. Checking homework (3-4min) (3 students prepare in advance on the board solutions to equations using a unit circle) 1 student cos t \u003d t \u003d + 2πk, where kZ (explanation is carried out along a unit circle) Answer: t \u003d + 2πk, where kZ. 2 student cos t = 1.5, has no solution because -1≤а≤1 Answer: no solutions. cos t = 1, t = 2πk, where kZ. Answer: t = 2πk, where kZ. 3 student cos t = 0, t = + πk, k; Answer: t = + πk, k; cos t \u003d -1, t \u003d π + 2πk, k. Answer: t = π + 2πk, k. 4. Study of new material Statement of the problem problem. After the students remember the principle of solving equations of the form cos t \u003d a, invite them to solve the equation of the form on the blackboard, write on the main board next to the example cos t \u003d 0.5, all other students listen (the example and the unit circle are written in advance) Saying the algorithm solving the simplest trigonometric equation, the student solves the equation using the unit circle. t = t1 +2πk, t = t2 +2πk, where k., because t1= - t2, then t = ± t1 +2πk, where k. What kind of number t1 is is still unknown, it is only clear that t1. Faced with such a situation, mathematicians realized that they needed to come up with a way to describe it in mathematical language. Therefore, a new symbol arccos a was introduced for consideration, which reads: arccosine a. Let's write down the topic of today's lesson: “The arccosine of the number a. Solving the equations cos t \u003d a ”- Today in the lesson we will study the concept of the arccosine of the number a, learn how to calculate and apply it when solving the simplest trigonometric equations. Arcus in Latin means arc, compare with the word arch. The symbol arcсosа, introduced by mathematicians, contains the sign (arc), сosа - a reminder of the original function Open the textbook on page 89 and read the definition of the arccosine (students open the textbook and read the definition from the book, highlighting the main thing) 5. Consolidation and development of the concept of the arccosine of the number a and the algorithm for its calculation (frontal work with the class) Teacher Student So, when calculating the arccosine of the number a, what question should you ask yourself? What is the cosine of a? Using the definition you have learned, find the value of the expression arccos ();arccos() arccos() arccos () = arccos() = arccos() = All values of a belong to the interval from -1 to 0. What quarter do the values of the arccosine a belong to? The values of arccosа belong to the segment from 0 to But how to calculate the value of arccos(–а)? Let's turn to the textbook and find the formula by which the value of arccos (-a) is calculated (read and highlight the formula). Compute: arccos(-); arccos(-); arccos(-); arccos (-)= arccos(-) = arccos(-) = All values(-a) belong to the segment from -1 to 0. Which quarter do the values of arccos(-a) belong to? Write down the reference material (slide 6) The values of arcсos(-а) belong to the segment from to π Students write the formula in a notebook. Consolidation and development of the concept of the arccosine of the number a and the algorithm for its calculation (frontal work with the class) Task Find the value of the expression: a) arccos ()-arccos (-)++arcos1 b) 2arccos 0 + 3 arccos 1 –arcos (-) 5. Independent work (with subsequent self-examination) 2 people work at the blackboard on their own, the rest work in notebooks, then check the correctness of the execution. Those who worked with homework write on pieces of paper at the blackboard, then hand them in for verification Teacher Student Let's return to the equation cos t =. which solved... Knowing the concepts of the arc cosine, we can now write the answer to the solution of this equation as follows. cos t =. t = ±arccos + 2πk, where kZ . Answer: t = ±arccos + 2πk, where kZ We solved the equation in two ways: using the unit circle and using the formula. Write down the solution for the teacher in a notebook So, let's write down the reference material and select it by solving the equation cos t = a, where a. t = ± arccos a + 2πk, k. Answer: t \u003d ± arccos a + 2πk, k. Write down in a notebook a model for solving the equation for the teacher 6. Consolidation of the studied material (13 min) No. 15.5 (b, d), 15.6 (a, b). (2 students work individually at the blackboard) 1 student: a) cos t = ; b) cos t = -; 2 account: a) cos t = ; b) cos t = . (pay attention to this example when evaluating a number) Solve equation: #15.5(b,d) b) cos t = . d) cos t = ; 15.6 (а, b) a) cos t =1; (pay attention to the answer and highlight special cases) b) cos t = - 7. Summing up the lesson (reflection). (3-4 min) (oral frontal work with the class) Teacher Student What new concepts did you learn in the lesson? We have learned a new concept of arccosine a. What new way to solve the simplest trigonometric equations did we consider in the lesson? Using formulas Once again, carefully review the reference material we have recorded. Close your notebooks, take a test on the desks, each option and fill in the blanks. You have 3 minutes for this work (peer review) (after 3 minutes of work, students change sheets of paper and check the correctness, the answers are projected onto an interactive whiteboard) (missing test points are highlighted in black) Perform the test Now you have identified gaps in your knowledge, and I ask pay attention to this at home. 8. Homework (differentiated) (1 min) We studied the educational material of the compulsory level and solved the tasks of the B level of testing in the USE format, at the same time you were asked to solve trigonometric equations that are reduced to the simplest §16, No. 15.3, 15.4,15.5 (in ,d), 15.6(c,d), *15.12
Download: Algebra 10kl - Summary of lesson 1 (Shelest S.V.).docxlesson 2 (Shelest S.V.)
Subject name: Algebra and the beginnings of mathematical analysis Class 10 UMC: Algebra and the beginnings of mathematical analysis A.G. Mordkovich, P.V. Semenov, 2015 Level of education (basic) Theme of the lesson: Solving the equation cost = a Total number of hours allotted for studying the topic: 2 Place of the lesson in the system of lessons on the topic: 2 Purpose of the lesson: to consolidate the ability to calculate the arc cosine; to form the ability to solve equations of the form cos t = a. Lesson objectives 1. Teaching: a) teach how to apply the formula when solving the simplest trigonometric equations; consider special cases of such equations. 2. Developing: a) develop the ability to briefly, logically, consistently express thoughts and judgments; b) develop the ability to argue their statements; 3. Educational: a) to teach the skills of planning activities, work at an optimal pace, b) to cultivate the ability to correctly assess one's capabilities, the results of educational activities, develop communication skills; Equipment: unit circle model. Course of the lesson 1. Organizational moment. 2. Actualization of students' knowledge. oral work. 1. Does the expression make sense. 2. Calculate. 3. Explanation of new material. Almost all of the new material was covered in the previous lesson. In this lesson, it is necessary to focus on solving equations of the form cos t \u003d a, consider special cases of such equations. First, you need to update the knowledge of students about the arc cosine and solving the equations cos t = a. Students should remember the formula derived in the previous lesson for solving these equations, which is put on the board: Then you should consider special cases of solving the equation cos t = a. Make appropriate notes on the board and in notebooks: 4. Formation of skills and abilities. 1. No. 15.5 (a; c), No. 15.6 (b; d). 2. No. 15.7. Solution: By solving several equations of the form cos t = a, where students often stop following the value of a. Therefore, it is useful to periodically offer them to solve such equations in which a) b) c) cos t = -1.1; no solutions since -1.1< –1. г) cos t = 2,04. нет решений, так как 2,04 >1. 3. No. 15.12 (a). Solution: cos t = 1 Answer: 4. No. 15.4 (a; b). Solution: a) if k = 0, then (in) (not included) if k = 1, then (not included) (included) Answer: b) if k = 0, then (not included) (not included) if k = 1, then (included) (not included) if k = 2, then (not included) (included) Answer: 5. No. 15.15 (d). Solution: if k = 0, then (included) (included) if k = 1, then (not included) (not included) if k = –1, then (included) (not included) Answer: 6. No. 15.17. You can ask students to complete additional tasks of an increased level of complexity. 7.* No. 15.18 (a; b). Solution: These inequalities are solved using a number circle. The main difficulty is to correctly determine the angles that correspond to the numbers and 8.* No. 15.19 (a). Decision: Let's make a replacement cos t = x and solve the inequality: We get: Taking into account the range of values of the function cos t, we get the inequality: Answer: 5. The results of the lesson. Questions to students: - What is called the arc cosine of the number a? – How to calculate arccos (–a)? - What is the formula for the roots of the equation cos t = a. – Formulate an algorithm for solving the simplest trigonometric inequalities. Homework: No. 15.5 (b; d), No. 15.6 (a; c), No. 15.12 (b), No. 15.15 (b; c). Additionally: No. 15.18 (c; d), No. 15.19 (d).
Download: Algebra 10kl - lesson 2 (Shelest S.V.).docxlesson 1 (Bakeyeva I. R.)
Subject name: Algebra and the beginning of analysis Grade: 10 TMC: Mordkovich A.G. Algebra and beginning of mathematical analysis. 10-11 grades. At 2 o'clock Textbook and task book for students of educational institutions (basic level). -M.: Mnemozina, 2011. Level of education: basic. Lesson topic: The concept of arc cosine. The total number of hours allocated for the study of the topic: 10 hours. The place of the lesson in the system of lessons on the topic: 1 lesson. Lesson objectives: to introduce the concept of the arc cosine, to form the ability to calculate the arc cosine; derive a formula for finding the roots of the equation cos t = a. 1) Educational objectives of the lesson: to create organizational and content conditions for the formation of students' skills to find the arc cosine of a number, the arc cosine of a negative number, compare the values of the arc cosine 2) Developing lesson objectives: to promote the formation of skills to apply the acquired knowledge in a new situation, develop logical thinking, mathematical speech. create conditions for the development of cognitive activity of students, cognitive interest in the subject; to develop an intellectual, reflective culture; develop the skills of independent activity of students; develop self-control skills; 3) Educational tasks of the lesson: to develop mobility, communication skills. foster a culture of mental work; to cultivate the ability to analyze the results of their own activities; ensure the humanistic nature of education. Expected results: 1. Be able to find the arc cosine of a number, the arc cosine of a negative number. 2. Be able to compare the values of the arc cosine. Technical support of the lesson: computer, projector, screen. Additional methodological and didactic support for the lesson: presentation in PowerPoint Lesson plan: I. Organizational moment. Self-determination for activities, setting goals and objectives of the lesson. II. oral work. 1. Calculate. 2. Name several angles whose cosine is equal to: a) 0; b) c) d) –1. III. Explanation of new material. The explanation is carried out according to the paragraph of the textbook in several stages. 1. Actualization of knowledge. You should repeat the method of solving equations of the form cos t = a using a number circle. Consider a textbook example that shows the solution to the equation cos t = . 2. Statement of the problematic problem. After students remember the principle of solving equations of the form cos t \u003d a, invite them to solve an equation of the form cos t \u003d. Then, according to the paragraph of the textbook, introduce the concept of arc cosine. When determining the arc cosine, the attention of students should be drawn to the fact that the angle is taken from the interval. Explain: if this fact is not taken into account, then the arc cosine will take on infinitely many values. 3. Solution of the equation cos t = a. Derive a formula for solving the equation cos t = a, put it on the board. 4. Finding arccos(-a). Very often, students make mistakes when calculating the arccosine of a negative number. These errors are of two types. For example, when calculating arcos, students get (further confused with the arcsine) or (remember the parity of the function y \u003d cos x). To avoid these errors, an appeal to the definition of the arc cosine and to the numerical circle will help. According to the definition, the arccosine is in the interval, so it cannot be negative. It can be seen on the number circle that arcos After finding the values of several arccosines of negative numbers, an entry is made on the board: IV. Formation of skills and abilities. In this lesson, the focus should be on finding the arc cosines. The solution of the equations cos t = a can be postponed until the next lesson. 1. No. 15.1, No. 15.2, No. 15.3 (b, d). 2. Calculate. 3. No. 15.8 (a). Decision: 4. No. 15.10. It is very important that students realize what is the range of valid values of the expression arccos a. Only then can you move on to the next number. 5. No. 15.9 (a, d). Solution: a) arccos x Obviously, b) arccos (3 - 2x) 1 ≤ x ≤ 2 Answer: . 6. No. 15.11. Solution: tg (arccos 0.1 + arccos (- 0.1) + x) = tg x. Using the formula arccos a + arccos (- a) = π, we transform the left side of the equation: tg (arccos 0.1 + arccos (- 0.1) + x) = tg (π + x) = tg x. Proven. In a class with a high level of preparation, you can additionally offer several tasks of an increased level of complexity. 7.* No. 15.16. Solution: a) y \u003d arccos x + arccos (- x). It is quite common for students to simply transform a given expression, forgetting about the scope and range of the function. We have: arccos x + arccos (- x) = π, and D (y): and E (y): . Then the graph will look like this: b) y = cos(arccos x). It is obvious that cos(arccos x) = x, but again we must not forget about the domain of definition and the range of values of the original function. 8.* No. 15.21 (a), No. 15.22 (a). Decision: No. 15.21 (a). Tasks of this type often cause difficulties for students. For their implementation, a conscious understanding of the definition of the arc cosine is necessary. To help students find a way to solve this problem, you can reason as follows: - arccos is some angle, and we need to find the sine of this angle; – let arccos= α, according to the definition, this is an angle from the interval such that cosα = – since ˃ 0, then α lies in the first quarter; – we know that cos α = and 0 ˂ α ˂ , but we need to find sin α; - the task was reduced to finding the sine of a certain angle, if the cosine of this angle is known. For clarity, the solution to this problem can be formulated as follows: - ? We have: So, sin α = . Answer: . No. 15.22 (a). We argue in the same way as in the previous task. - ? Hence, tan α = - . Answer: - . V. The results of the lesson. Questions for students: - How to solve the equation cos t \u003d a using a number circle? What is the arccosine of a number called? - Why is the interval included in the definition of the arccosine? – How to calculate arccos (–a)? – What is the range of valid values of the expression arccos a? VI. Reflection. "10 points" Evaluate the work in the lesson on a 10-point scale from the position: "I" 0________10 "We" 0________10 "Business" 0________10 Homework: No. 15.3 (a, c), No. 15.4, No. 15.8 (b), No. 15.9 (b, c) Additionally: No. 15.21 (b), No. 15.22 (b).
Download: Algebra 10kl - lesson 1 (Bakeeva I. R.).docxlesson 2 (Bakeyeva I. R.)
Subject name: Algebra and the beginning of analysis Grade: 10 TMC: Mordkovich A.G. Algebra and beginning of mathematical analysis. 10-11 grades. At 2 o'clock Textbook and task book for students of educational institutions (basic level). -M.: Mnemozina, 2011. Level of education: basic. Lesson topic: Solving the equation cos t = a. The total number of hours allocated for the study of the topic: 10 hours. The place of the lesson in the system of lessons
THE RUSSIAN FEDERATION
YAMAL-NENETS AUTONOMOUS DISTRICT
DEPARTMENT OF EDUCATION OF THE ADMINISTRATION OF THE CITY OF NOYABRSK
MUNICIPAL EDUCATIONAL INSTITUTION
SECONDARY EDUCATIONAL SCHOOL №7
OF THE MUNICIPALITY OF THE CITY OF NOYABRSK"
Methodical development
algebra lesson (grade 10)
Topic: “Arc cosine of a.
Solving the equations cos x = a "
Mathematic teacher,
Noyabrsk
2009 Lesson of studying new material and primary consolidation of knowledge.
An open lesson on algebra and the beginnings of analysis in grade 10.
Lesson topic: Arccosine of number a. Solution of the equations cos x = a.
Lesson Objectives:
- Tutorials:
a) introduce the concept of the arc cosine of a;
b) develop the skill of calculating the arcsine of the number a;
c) derive the formula for the roots of the simplest trigonometric equations, the formula cos x = a;
d) teach how to apply the formula when solving the simplest trigonometric equations;
e) study a special case of the solutiontrigonometric equations for and equal to 0, -1, 1.
- Developing:
a) develop the ability to briefly, logically, consistently express thoughts and judgments;
b) develop the ability to argue their statements;
c) develop the ability to classify, compare, analyze and draw conclusions.
3.Educational:
a) to teach the skills of planning activities, working at an optimal pace,
b) to cultivate the ability to correctly assess their capabilities, the results of educational activities, develop communication skills;
c) educate diligence and purposefulness.
Equipment: computer, interactive whiteboard, handouts, cards on the reflection of educational activities (for each student), a poster with a unit circle.
Blackboard writing:
Each student has the right:
- Know more than the teacher and defend your hypotheses.
During the classes:
- Organizing time(2 minutes)
Teacher: Hello guys.
Today at the lesson we will learn(Slide 1)
a) briefly, logically, consistently express thoughts and judgments;
b) to argue statements;
c) compare, analyze and draw conclusions;
d) evaluate the results of their educational activities.
We remember that every student, as always, has the right to:
- Express your opinion and be heard;
- Independently plan home self-study;
- Know more than a teacher and defend your hypotheses(writing on the board)
2.Updating knowledge(3-4 min)
Oral counting (tasks are projected onto an interactive screen(Slide 2)
Teacher | Student |
Points of the unit circle, , belong to which quarter? | Points of the unit circle, , belong to 1 quarter? |
The cosine of which angle is a positive value? Conclusion: The cosine of an acute angle is a positive value. | If the angle belongs to the 1st quadrant |
2. Calculate values: cos ; cos ; cos
Teacher | Student |
Points of the unit circle, , belong to which quarter? | Points of the unit circle, , belong to 2 quarters. |
The cosine of which angle is negative? Conclusion: The cosine of an obtuse angle is negative | If the angle belongs to 2 quarters |
2. The cosine of which angle is; 0; ; one; ; - ; - , if ?
3. Checking homework(3-4min) (3 students prepare in advance on the board solutions to equations using a unit circle)
1 student
t = +2πk , where k Z ( explanation is on a unit circle)
Answer: t = +2πk , where k Z .
2 student
- cos t = 1.5,
Doesn't have a solution. -1≤a≤1
Answer: no solution.
- cos t = 1,
T = 2πk, where k Z.
Answer: t = 2πk, where k Z.
3 student
- cos t = 0,
t = + πk, k ;
Answer: t = + πk, k ;
- cos t = -1,
t = π + 2πk, k .
Answer: t = π + 2πk, k .
4. Learning new material(13-15 min)
Teacher | Student |
Now let's solve the equation cos t = . | on the board writes on the main board next to the example cos t = , all other students listen (the example and the unit circle are written in advance) Speaking the algorithm for solving the simplest trigonometric equation, the student solves the equation using a unit circle. t = t 1 +2πk, t = t 2 +2πk, where k Z, because t 1= - t 2, then t = ± t 1 +2πk, where k Z, |
Is this entry answer solution to the equation? | This entry is not the answer to the solution of the equation, because the values are not defined t1. |
Teacher: What is this number t 1 , is still unknown, it is only clear that t 1 . Faced with such a situation, mathematicians realized that they needed to come up with a way to describe it in mathematical language. Therefore, a new symbol arccos was introduced for consideration a , which reads: arc cosine a .
Let's write down the topic of today's lesson: “The arccosine of the number a. Solution of the equations cos t = a "(Slide 3.4)
Teacher | Student |
So, when calculating the arccosine of the number a, what question should you ask yourself? | What is the cosine of a? |
Applying the learned definition, find the value of the expression arccos(); arccos() arccos() (Slide 5) | arccos() = arccos() = arccos() = |
All values of a belong to the segment from -1 to 0. What quarter do the values of the arc cosine a belong to? | Arccosa values belong to the interval from 0 to |
But how to calculate the value of arccos (-a)? Let's turn to the textbook and find the formula by which the value of arccos (-a) (read and highlight the formula).(Slide 6) Calculate: arccos(-); arccos(-); arccos(-); (Slide 6) | arccos(-)= arccos(-) = arccos(-) = |
All values (-a) belong to the segment from -1 to 0. Which quarter do the values of arccos (-a) ? Record reference material (slide 6) | Arcсos(-а) values belong to the segment from up to π Students write the formula in their notebook. |
Consolidation and development of the concept of the arc cosine of the number a and the algorithm for its calculation (frontal work with the class)
Computing on a slide on an interactive whiteboard
Exercise |
Find the value of an expression:(Slide 7) a) arccos () - arccos (-) + + arcos1 |
b) 2arccos 0 + 3 arccos 1 - arcos (-) (Slide 8) |
5. Independent work (with subsequent self-examination)(Slide 9)
2 people work at the blackboard on their own, the rest work in notebooks, then check the correctness of the execution. Those who worked with homework write at the blackboard onleaflets, then hand them over for verification
Teacher | Student |
Let's return to the equation cos t =. which solved... Knowing the concepts of the arc cosine, we can now write the answer to the solution of this equation as follows. cos t = . t = ±arccos + 2πk , where k Z . Answer: t = ±arccos + 2πk , where k Z We solved the equation in two ways: using the unit circle and using the formula. | Write down the solution for the teacher in a notebook |
So, let's write down the reference material and select it by solving the equation(Slide 10) cos t = a, where a . t \u003d ± arccos a + 2πk, k. Answer: t \u003d ± arccos a + 2πk, k. | Write down in a notebook a model for solving the equation for the teacher |
6. Consolidation of the studied material(13min)
No. 15.5 (b, d), 15.6 (a, b).
(2 students work individually at the blackboard)
1 account: a) cos t = ; b) cos t = - ;
2 account: a) cos t = ; b) cos t = . ( pay attention to this example when evaluating the number)
Solve the equation:
№15.5(b,d)
b) cos t = .
d) cos t = ;
15.6 (a,b)
a) cos t = 1; (pay attention to the answer and highlight special cases)
b) cos t = -
7. Summing up the lesson (reflection).(3-4min)
(oral frontal work with the class)
Teacher | Student |
What new concepts did you learn in class? | We have learned a new concept of arccosine a. |
What new way to solve the simplest trigonometric equations did we consider in the lesson? | Using formulas |
Once again, carefully review the reference material recorded by us. Close your notebooks, take a test on the desks, each option and fill in the blanks. You have 3 minutes for this work (peer review) (after 3 minutes of work, students change sheets of paper and check the correctness, the answers are projected onto an interactive whiteboard)(missing test points are highlighted in black) | Perform the test (Slide 11) |
Now you have identified gaps in your knowledge, and I ask you to pay attention to this at home. |
8. Homework (differentiated)(1min) (Slide 12)
Teacher: We studied the educational material of the compulsory level and solved the tasks of level B testing in the USE format, at the same time you were asked to solve trigonometric equations that are reduced to the simplest
§16, #15.3, 15.4,15.5(c,d), 15.6(c,d), *15.12
Preview:
Compute: a rc with os - arc with os + + a rc with os 1 =
Calculate: 2) 2 a rc with os 0 + 3 arc with os 1 - arc with os =
Independent work No. 15.1 (a, b, c), 15.2 (c, d)
cos t \u003d a, where a ϵ [-1; 1] t \u003d ± arc with os a + 2 π k, k ϵ Z Answer: ± arc with os a + 2π k, k ϵ Z No. 15.5 (b), 15.6 (b), 15.5(d), 15.6(a)
Option 1 Option 2 If a ϵ [-1; 1], then arc c os a is such a number from the segment [ 0; π] whose cosine is equal to a. if in ϵ [-1;0], then arc with os in ϵ if a ¢ [-1;1], then the equation cos t = a has no solutions if cos t = 1, then t = 2π k , k ϵ Z ; if a ϵ , then ar with cos a ϵ if a ϵ , then ar with cos (-a)= π- ar with cos and if cos t = 0, then t = + π k , k ϵ Z ; if a ϵ [-1;1], then the equation cos t = a has solutions t = ± arc with os a + 2π k , k ϵ Z
Homework §16, #15.3, 15.4, 15.5(c,d), 15.6(c,d), *15.12
Thank you for the lesson
If | a | 1, then the equation cos t = a has no real roots
Special cases if cos t = 1 , then t = 2 π k , k ϵ Z if cos t = -1 , then t = π + 2 π k , k ϵ Z if cos t = 0 , then t = + π k , k ϵ Z
In continuation of the previous topic, which looked at examples of solving trigonometric functions, this video tutorial introduces students to the inverse cosine and the solution of the equation cos t = a.
An example of solving the equation cos t =1/4 is considered. Using the number circle, we find the points with the coordinate x = 1/4, mark these points on the graph as M (t 1) and N (t 2).
The graph shows that t 1 is the length of AM, and t 2 is the length of AN. In another way, we can say that t 1 \u003d arccos 1/4; t 2 \u003d - arccos 1/4. Solution of the equation t = ± arccos ¼ + 2πk.
Thus, arccos 1/4 is the number (length of AM) whose cosine is 1/4. This number belongs to the segment from 0 to π/2, i.e. first quarter of the circle.
Further, the solution of the equation cos t = - 1/4 is considered. By analogy with the previous example, t \u003d ± arccos (-1/4 + 2πk. We can say that arccos (-1/4 is the number (length of the arc AM), whose cosine is - ¼ and this number belongs to the second quarter circle, i.e. i.e. a segment from π/2 to π.
Based on two examples, the definition of the arc cosine is given: if the modulus a is less than or equal to 1, then arccos and is such a number from the segment from 0 to π, the cosine of which is equal to a. Then the expression cos t \u003d a with modulus a less than or equal to 1 can look like t \u003d ± arccos a + 2πk. Below are the values of t at cos t = 0; cos t = 1; cos t = - 1.
The author gives an example 1. Find the solution of the expression arccos. We indicate that the given value of arccos is equal to t, therefore, cos t is equal to this value, where t belongs to the segment from 0 to π. Using the table of values, we find that cos t corresponds to the value t =π/6. Find the corresponding cosine value, where π/6 belongs to the segment from 0 to π.
Let's analyze example 2. Calculate the arccos of a negative number. Let us assume that the arccos of this number is, therefore, cos t is equal to this number, where t belongs to the segment from 0 to π. According to the table of values, we will see what value corresponds to cos t, this is t \u003d 5π / 6. Those. cos 5π/6 is minus the square root of three divided by two, where 5π/6 belongs to the segment from 0 to π.
Further, the author considers the theorem: for any a belonging to the segment from minus one to one, the equality arccos a + arccos (-a) = π is valid. In the proof, for definiteness, we assume that a > 0, then - a< 0. На окружности отметим arccos a, это длина АК, и arccos (- a), это длина TС. АК = ТС, т.к. они симметричны относительно вертикального диаметра окружности ТК. Следовательно, arccos a + arccos (- а) = АК + АТ = ТС + АТ =π. Из написанного равенства можно сделать вывод, что arccos (- а) = π- arccos a, где 0 ≤ а ≤ 1.
When a > 0, arccos a belongs to the first quarter of the circle (marked in the figure), and when a< 0, arccos a принадлежит II четверти.
Let's consider one more example. Solve the expression where cos t is a negative number. Let's write down what t is equal to in this case. Then we will find the value of the arc cosine, this is 3π / 4. Let's substitute the found value of arccos into the value of t and get that t = ± 3π/4+ 2πk.
Let us analyze the solution of the cos t inequality. To solve, we need to find points on the number circle where x is equal to the cosine value. These are points with values π/4 and - π/4. As you can see in the figure, the length of the arc MN is - π/4≤ t ≤π/4. So the answer of the inequality will be - π/4 + 2πk≤ t ≤ π/4+ 2πk.
TEXT INTERPRETATION:
Arccosine. Solution of the equation cost = a
Consider the solution of the equation cost = .
Considering that cos t is the abscissa of the point M(t) (em from te) of the numerical circle, we find points on the numerical circle with the abscissa
On the numerical circle, we mark the points M (t 1), N (t 2) - the points of intersection of the line x \u003d with this circle.
t 1 is the length of the arc AM, t 2 is the length of the arc AN, t 2 = - t 1.
When mathematicians first encountered this situation, they introduced the new symbol arccos
arccos (arc cosine of one fourth).
Then t 1 = arccos; t 2 \u003d - arccos
And then the roots of the cost = equation can be written in two formulas:
t = arccos + 2πk, t = - arccos + 2πk or t = arccos + 2πk.
What does arcos mean?
This number
(the length of the arc AM), the cosine of which is equal to one fourth and this number belongs to the first quarter, that is, the segment.
Now consider the equation
cost = - . Similarly to the solution of the previous equation, we write
t = arccos) + 2πk.
How to understand arccos(-)? This number
(the length of the arc AM), whose cosine is equal to minus one fourth and this number belongs to the second quarter, that is, the segment [; ].
Let's define the arccosine:
DEFINITION. Let | a | 1 (module a is less than or equal to one). The arc cosine a is such a number from the segment whose cosine is equal to a. (Fig. 1)
EXAMPLE 1. Calculate arccos. (Arc cosine root of three by two)
Decision. Let arccos = t. Then cost = and t [ ; ](te belongs to the segment from zero to pi). Recall the value of cos corresponds to
(Show table of values) So t = (pi times six) because cos = and . So arccos = .
arcos is the length of the arc, but the length of the arc of a circle is the t in the definition of cost
(Conventionally, we can say that the arccosine is the “angle value” that the point from M left the point A, if you remember, we entered the number t as part of the circumference, the radius equal to 1 (one), and then 2π- the entire circle is 360 ° , π- half circle =180°, ==60°)
EXAMPLE 2. Calculate arccos (- (arc cosine minus the root of three by two).
Decision. Let arccos(-) = t. Then cost = and t [ ; ](te belongs to the segment from zero to pi). Hence, t = (five pi by six), since cos = - and [; ]. So, arccos) = .
Let's prove THEOREM. For any a [; ](and from the segment from minus one to one) the equality arccosа + arccos(-а) = π(the sum of the arccosine a and the arccosine minus a is equal to pi) is fulfilled.
Proof. For definiteness, we will assume that a 0, then - a 0. On the number circle, mark arcos a (this is the length of the arc AK) and
arccos(-a) (this is the length of the arc AT) (see Fig. 2)
From the proved theorem it follows: arcos (-a) \u003d π - arcos a (the arc cosine minus a is equal to the difference between pi and the arc cosine a), where 0 a 1 (where a is greater than or equal to zero and less than or equal to one).
When a > 0, consider that arcos a belongs to the first quarter of the number circle.
When a< 0 считают, что arcosa belongs to the second quarter of the number circle.
EXAMPLE 3. Solve the equation cost = - .
Decision. Let's make a solution formula: t = arccos(-)+ 2πk.
Let's calculate the values of the arccosine: arccos(-) = π - arccos = π - = .
(According to the relation arccos(-) = π - arсcos arсcos , then substituting this value into the formula, we get that arccos(-) =) .
Let's substitute the found value into the solution formula t = arccos(-)+ 2πk and get the value of t: t = + 2πk.
EXAMPLE 4. Solve the cost inequality.
Decision. We know that cost is the abscissa of the point M(t) on the number circle. This means that you need to find such points M(t) on the number circle that satisfy the inequality x.
The line x = intersects the number circle at two points M and N.
The inequality x corresponds to the points of the open arc MN. Point M corresponds, and point N -
- (minus pi by four).
Hence, the kernel of the analytical representation of the arc MN is the inequality
T , and the analytic notation of the arc MN itself has the form
