How to simplify expressions in trigonometry. Identical transformations of trigonometric expressions

Sections: Mathematics

Class: 11

Lesson 1

Subject: 11th grade (preparation for the Unified State Exam)

Simplifying trigonometric expressions.

The simplest solution trigonometric equations. (2 hours)

Goals:

Systematize, generalize, and expand students’ knowledge and skills related to the use of trigonometry formulas and solving simple trigonometric equations.

Equipment for the lesson:

Lesson structure:

Organizational moment
Testing on laptops. Discussion of results.
Simplifying trigonometric expressions
Solving simple trigonometric equations
Independent work.
Lesson summary. Explanation of homework assignment.

1. Organizational moment. (2 min.)

The teacher greets the audience, announces the topic of the lesson, reminds them that the task was previously given to repeat trigonometry formulas, and prepares students for testing.

2. Testing. (15 min + 3 min discussion)

The goal is to test knowledge trigonometric formulas and the ability to apply them. Each student has a laptop on their desk with a version of the test.

There can be any number of options, I will give an example of one of them:

I option.

Simplify expressions:

a) basic trigonometric identities

1. sin 2 3y + cos 2 3y + 1;

b) addition formulas

3. sin5x - sin3x;

c) converting a product into a sum

6. 2sin8y cos3y;

d) double angle formulas

7. 2sin5x cos5x;

e) formulas for half angles

e) formulas for triple angles

g) universal substitution

h) reduction in degree

16. cos 2 (3x/7);

Students see their answers on the laptop next to each formula.

The work is instantly checked by the computer. The results are displayed on a large screen for all to see.

Also, after finishing the work, the correct answers are shown on the students’ laptops. Each student sees where the mistake was made and what formulas he needs to repeat.

3. Simplification of trigonometric expressions. (25 min.)

The goal is to repeat, practice and consolidate the use of basic trigonometry formulas. Solving problems B7 from the Unified State Exam.

At this stage, it is advisable to divide the class into groups of strong students (work independently with subsequent testing) and weak students who work with the teacher.

Assignment for strong students (prepared in advance on a printed basis). The main emphasis is on reduction formulas and double angle, according to the Unified State Examination 2011.

Simplify expressions (for strong students):

At the same time, the teacher works with weak students, discussing and solving tasks on the screen under the students’ dictation.

Calculate:

5) sin(270º - α) + cos (270º + α)

Simplify:

It was time to discuss the results of the work of the strong group.

Answers appear on the screen, and also, using a video camera, the work of 5 different students(one task for each).

The weak group sees the condition and method of solution. Discussion and analysis are underway. With the use of technical means this happens quickly.

4. Solving simple trigonometric equations. (30 min.)

The goal is to repeat, systematize and generalize the solution of the simplest trigonometric equations and write down their roots. Solution of problem B3.

Any trigonometric equation, no matter how we solve it, leads to the simplest.

When completing the task, students should pay attention to writing down the roots of equations of special cases and general view and on the selection of roots in the last equation.

Solve equations:

Write down the smallest positive root as your answer.

5. Independent work (10 min.)

The goal is to test the acquired skills, identify problems, errors and ways to eliminate them.

Multi-level work is offered to the student's choice.

Option "3"

1) Find the value of the expression

2) Simplify the expression 1 - sin 2 3α - cos 2 3α

3) Solve the equation

Option for "4"

1) Find the value of the expression

2) Solve the equation Write down the smallest positive root in your answer.

Option for "5"

1) Find tanα if

2) Find the root of the equation Write down the smallest positive root as your answer.

6. Lesson summary (5 min.)

The teacher sums up the fact that during the lesson they repeated and reinforced trigonometric formulas and solving the simplest trigonometric equations.

Homework is assigned (prepared on a printed basis in advance) with a random check at the next lesson.

Solve equations:

10) In your answer, indicate the smallest positive root.

Lesson 2

Subject: 11th grade (preparation for the Unified State Exam)

Methods for solving trigonometric equations. Root selection. (2 hours)

Goals:

Generalize and systematize knowledge on solving trigonometric equations of various types.
Promote development mathematical thinking students, the ability to observe, compare, generalize, classify.
Encourage students to overcome difficulties in the process of mental activity, to self-control, and introspection of their activities.

Equipment for the lesson: KRMu, laptops for each student.

Lesson structure:

Organizational moment
Discussion of d/z and self. work from last lesson
Review of methods for solving trigonometric equations.
Solving trigonometric equations
Selection of roots in trigonometric equations.
Independent work.
Lesson summary. Homework.

1. Organizational moment (2 min.)

The teacher greets the audience, announces the topic of the lesson and the work plan.

2. a) Analysis homework(5 min.)

The goal is to check execution. One work is displayed on the screen using a video camera, the rest are selectively collected for teacher checking.

b) Analysis independent work(3 min.)

The goal is to analyze mistakes and indicate ways to overcome them.

Answers and solutions are on the screen; students have their work given out in advance. Analysis proceeds quickly.

3. Review of methods for solving trigonometric equations (5 min.)

The goal is to recall methods for solving trigonometric equations.

Ask students what methods for solving trigonometric equations they know. Emphasize that there are so-called basic (frequently used) methods:

variable replacement,
factorization,
homogeneous equations,

and there are applied methods:

using the formulas for converting a sum into a product and a product into a sum,
according to the formulas for reducing the degree,
universal trigonometric substitution
introduction of an auxiliary angle,
multiplication by some trigonometric function.

It should also be recalled that one equation can be solved in different ways.

4. Solving trigonometric equations (30 min.)

The goal is to generalize and consolidate knowledge and skills on this topic, to prepare for the C1 solution from the Unified State Exam.

I consider it advisable to solve equations for each method together with students.

The student dictates the solution, the teacher writes it down on the tablet, and the whole process is displayed on the screen. This will allow you to quickly and effectively recall previously covered material in your memory.

Solve equations:

1) replacing the variable 6cos 2 x + 5sinx - 7 = 0

2) factorization 3cos(x/3) + 4cos 2 (x/3) = 0

3) homogeneous sin equations 2 x + 3cos 2 x - 2sin2x = 0

4) converting the sum into a product cos5x + cos7x = cos(π + 6x)

5) converting the product into the sum 2sinx sin2x + cos3x = 0

6) reduction of the degree sin2x - sin 2 2x + sin 2 3x = 0.5

7) universal trigonometric substitution sinx + 5cosx + 5 = 0.

When solving this equation, it should be noted that using this method leads to a narrowing of the definition range, since sine and cosine are replaced by tg(x/2). Therefore, before writing out the answer, you need to check whether the numbers from the set π + 2πn, n Z are horses of this equation.

8) introduction of an auxiliary angle √3sinx + cosx - √2 = 0

9) multiplication by some trigonometric function cosx cos2x cos4x = 1/8.

5. Selection of roots of trigonometric equations (20 min.)

Since in conditions of fierce competition when entering universities, solving the first part of the exam alone is not enough, most students should pay attention to the tasks of the second part (C1, C2, C3).

Therefore, the goal of this stage of the lesson is to remember previously studied material and prepare to solve problem C1 from the Unified State Exam 2011.

There are trigonometric equations in which you need to select roots when writing out the answer. This is due to some restrictions, for example: the denominator of the fraction is not equal to zero, the expression under the even root is non-negative, the expression under the logarithm sign is positive, etc.

Such equations are considered equations of increased complexity and in version of the Unified State Exam are in the second part, namely C1.

Solve the equation:

A fraction is equal to zero if then by using unit circle let's select the roots (see Figure 1)

Figure 1.

we get x = π + 2πn, n Z

Answer: π + 2πn, n Z

On the screen, the selection of roots is shown on a circle in a color image.

The product is equal to zero when at least one of the factors is equal to zero, and the arc does not lose its meaning. Then

Using the unit circle, we select the roots (see Figure 2)

Figure 2.

Let's go to the system:

In the first equation of the system we make the replacement log 2 (sinx) = y, we then obtain the equation , let's return to the system

using the unit circle we select the roots (see Figure 5),

Figure 5.

6. Independent work (15 min.)

The goal is to consolidate and check the assimilation of the material, identify errors, and outline ways to correct them.

The work is offered in three versions, prepared in advance on a printed basis, for students to choose from.

You can solve equations in any way.

Option "3"

Solve equations:

1) 2sin 2 x + sinx - 1 = 0

2) sin2x = √3cosx

Option for "4"

Solve equations:

1) cos2x = 11sinx - 5

2) (2sinx + √3)log 8 (cosx) = 0

Option for "5"

Solve equations:

1) 2sinx - 3cosx = 2

7. Lesson summary, homework (5 min.)

The teacher summarizes the lesson and once again draws attention to the fact that a trigonometric equation can be solved in several ways. Most best way to achieve a quick result, it is the one that is best learned by a particular student.

When preparing for the exam, you need to systematically repeat formulas and methods for solving equations.

Homework (prepared in advance on a printed basis) is distributed and the methods for solving some equations are commented on.

Solve equations:

1) cosx + cos5x = cos3x + cos7x

2) 5sin(x/6) - cos(x/3) + 3 = 0

3) 4sin 2 x + sin2x = 3

4) sin 2 x + sin 2 2x - sin 2 3x - sin 2 4x = 0

5) cos3x cos6x = cos4x cos7x

6) 4sinx - 6cosx = 1

7) 3sin2x + 4 cos2x = 5

8)cosx cos2x cos4x cos8x = (1/8)cos15x

9) (2sin 2 x - sinx)log 3 (2cos 2 x + cosx) = 0

10) (2cos 2 x - √3cosx)log 7 (-tgx) = 0

11)

IN identity transformations trigonometric expressions The following algebraic techniques can be used: adding and subtracting identical terms; putting the common factor out of brackets; multiplication and division by the same quantity; application of abbreviated multiplication formulas; allocation full square; decomposition quadratic trinomial by multipliers; introduction of new variables to simplify transformations.

When converting trigonometric expressions that contain fractions, you can use the properties of proportion, reducing fractions, or reducing fractions to a common denominator. In addition, you can use isolating the whole part of the fraction, multiplying the numerator and denominator of the fraction by the same size, and also, if possible, take into account the homogeneity of the numerator or denominator. If necessary, you can represent a fraction as the sum or difference of several simpler fractions.

In addition, when applying all the necessary methods for converting trigonometric expressions, it is necessary to constantly take into account the range of permissible values of the expressions being converted.

Let's look at a few examples.

Example 1.

Calculate A = (sin (2x – π) cos (3π – x) + sin (2x – 9π/2) cos (x + π/2)) 2 + (cos (x – π/2) cos ( 2x – 7π/2) +
+ sin (3π/2 – x) sin (2x –5π/2)) 2

Solution.

From the reduction formulas it follows:

sin (2x – π) = -sin 2x; cos (3π – x) = -cos x;

sin (2x – 9π/2) = -cos 2x; cos (x + π/2) = -sin x;

cos (x – π/2) = sin x; cos (2x – 7π/2) = -sin 2x;

sin (3π/2 – x) = -cos x; sin (2x – 5π/2) = -cos 2x.

Whence, by virtue of the formulas for adding arguments and the main trigonometric identity, we get

A = (sin 2x cos x + cos 2x sin x) 2 + (-sin x sin 2x + cos x cos 2x) 2 = sin 2 (2x + x) + cos 2 (x + 2x) =
= sin 2 3x + cos 2 3x = 1

Answer: 1.

Example 2.

Convert the expression M = cos α + cos (α + β) · cos γ + cos β – sin (α + β) · sin γ + cos γ into a product.

Solution.

From formulas for adding arguments and formulas for converting sums trigonometric functions into the product after appropriate grouping we have

M = (cos (α + β) cos γ – sin (α + β) sin γ) + cos α + (cos β + cos γ) =

2cos ((β + γ)/2) cos ((β – γ)/2) + (cos α + cos (α + β + γ)) =

2cos ((β + γ)/2) cos ((β – γ)/2) + 2cos (α + (β + γ)/2) cos ((β + γ)/2)) =

2cos ((β + γ)/2) (cos ((β – γ)/2) + cos (α + (β + γ)/2)) =

2cos ((β + γ)/2) 2cos ((β – γ)/2 + α + (β + γ)/2)/2) cos ((β – γ)/2) – (α + ( β + γ)/2)/2) =

4cos ((β + γ)/2) cos ((α +β)/2) cos ((α + γ)/2).

Answer: M = 4cos ((α + β)/2) · cos ((α + γ)/2) · cos ((β + γ)/2).

Example 3.

Show that the expression A = cos 2 (x + π/6) – cos (x + π/6) cos (x – π/6) + cos 2 (x – π/6) takes one for all x from R and the same meaning. Find this value.

Solution.

Here are two ways to solve this problem. Applying the first method, by isolating a complete square and using the corresponding basic trigonometric formulas, we obtain

A = (cos (x + π/6) – cos (x – π/6)) 2 + cos (x – π/6) cos (x – π/6) =

4sin 2 x sin 2 π/6 + 1/2(cos 2x + cos π/3) =

Sin 2 x + 1/2 · cos 2x + 1/4 = 1/2 · (1 – cos 2x) + 1/2 · cos 2x + 1/4 = 3/4.

Solving the problem in the second way, consider A as a function of x from R and calculate its derivative. After transformations we get

А´ = -2cos (x + π/6) sin (x + π/6) + (sin (x + π/6) cos (x – π/6) + cos (x + π/6) sin (x + π/6)) – 2cos (x – π/6) sin (x – π/6) =

Sin 2(x + π/6) + sin ((x + π/6) + (x – π/6)) – sin 2(x – π/6) =

Sin 2x – (sin (2x + π/3) + sin (2x – π/3)) =

Sin 2x – 2sin 2x · cos π/3 = sin 2x – sin 2x ≡ 0.

Hence, due to the criterion of constancy of a function differentiable on an interval, we conclude that

A(x) ≡ (0) = cos 2 π/6 - cos 2 π/6 + cos 2 π/6 = (√3/2) 2 = 3/4, x € R.

Answer: A = 3/4 for x € R.

The main techniques for proving trigonometric identities are:

A) reducing the left side of the identity to the right through appropriate transformations;
b) reducing the right side of the identity to the left;
V) reducing the right and left sides of the identity to the same form;
G) reducing to zero the difference between the left and right sides of the identity being proved.

Example 4.

Check that cos 3x = -4cos x · cos (x + π/3) · cos (x + 2π/3).

Solution.

Transforming the right-hand side of this identity using the corresponding trigonometric formulas, we have

4cos x cos (x + π/3) cos (x + 2π/3) =

2cos x (cos ((x + π/3) + (x + 2π/3)) + cos ((x + π/3) – (x + 2π/3))) =

2cos x (cos (2x + π) + cos π/3) =

2cos x · cos 2x - cos x = (cos 3x + cos x) – cos x = cos 3x.

The right side of the identity is reduced to the left.

Example 5.

Prove that sin 2 α + sin 2 β + sin 2 γ – 2cos α · cos β · cos γ = 2 if α, β, γ are the interior angles of some triangle.

Solution.

Considering that α, β, γ are the interior angles of some triangle, we obtain that

α + β + γ = π and, therefore, γ = π – α – β.

sin 2 α + sin 2 β + sin 2 γ – 2cos α · cos β · cos γ =

Sin 2 α + sin 2 β + sin 2 (π – α – β) – 2cos α · cos β · cos (π – α – β) =

Sin 2 α + sin 2 β + sin 2 (α + β) + (cos (α + β) + cos (α – β) · (cos (α + β) =

Sin 2 α + sin 2 β + (sin 2 (α + β) + cos 2 (α + β)) + cos (α – β) (cos (α + β) =

1/2 · (1 – cos 2α) + ½ · (1 – cos 2β) + 1 + 1/2 · (cos 2α + cos 2β) = 2.

The original equality has been proven.

Example 6.

Prove that in order for one of the angles α, β, γ of the triangle to be equal to 60°, it is necessary and sufficient that sin 3α + sin 3β + sin 3γ = 0.

Solution.

The condition of this problem involves proving both necessity and sufficiency.

First let's prove necessity.

It can be shown that

sin 3α + sin 3β + sin 3γ = -4cos (3α/2) cos (3β/2) cos (3γ/2).

Hence, taking into account that cos (3/2 60°) = cos 90° = 0, we obtain that if one of the angles α, β or γ is equal to 60°, then

cos (3α/2) cos (3β/2) cos (3γ/2) = 0 and, therefore, sin 3α + sin 3β + sin 3γ = 0.

Let's prove now adequacy the specified condition.

If sin 3α + sin 3β + sin 3γ = 0, then cos (3α/2) cos (3β/2) cos (3γ/2) = 0, and therefore

either cos (3α/2) = 0, or cos (3β/2) = 0, or cos (3γ/2) = 0.

Hence,

or 3α/2 = π/2 + πk, i.e. α = π/3 + 2πk/3,

or 3β/2 = π/2 + πk, i.e. β = π/3 + 2πk/3,

or 3γ/2 = π/2 + πk,

those. γ = π/3 + 2πk/3, where k ϵ Z.

From the fact that α, β, γ are the angles of a triangle, we have

0 < α < π, 0 < β < π, 0 < γ < π.

Therefore, for α = π/3 + 2πk/3 or β = π/3 + 2πk/3 or

γ = π/3 + 2πk/3 of all kϵZ only k = 0 is suitable.

It follows that either α = π/3 = 60°, or β = π/3 = 60°, or γ = π/3 = 60°.

The statement has been proven.

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Sections: Mathematics

Class: 11

Lesson 1

Subject: 11th grade (preparation for the Unified State Exam)

Simplifying trigonometric expressions.

Solving simple trigonometric equations. (2 hours)

Goals:

Systematize, generalize, and expand students’ knowledge and skills related to the use of trigonometry formulas and solving simple trigonometric equations.

Equipment for the lesson:

Lesson structure:

Organizational moment
Testing on laptops. Discussion of results.
Simplifying trigonometric expressions
Solving simple trigonometric equations
Independent work.
Lesson summary. Explanation of homework assignment.

1. Organizational moment. (2 min.)

The teacher greets the audience, announces the topic of the lesson, reminds them that the task was previously given to repeat trigonometry formulas, and prepares students for testing.

2. Testing. (15 min + 3 min discussion)

The goal is to test knowledge of trigonometric formulas and the ability to apply them. Each student has a laptop on their desk with a version of the test.

There can be any number of options, I will give an example of one of them:

I option.

Simplify expressions:

a) basic trigonometric identities

1. sin 2 3y + cos 2 3y + 1;

b) addition formulas

3. sin5x - sin3x;

c) converting a product into a sum

6. 2sin8y cos3y;

d) double angle formulas

7. 2sin5x cos5x;

e) formulas for half angles

e) formulas for triple angles

g) universal substitution

h) reduction in degree

16. cos 2 (3x/7);

Students see their answers on the laptop next to each formula.

The work is instantly checked by the computer. The results are displayed on a large screen for all to see.

Also, after finishing the work, the correct answers are shown on the students’ laptops. Each student sees where the mistake was made and what formulas he needs to repeat.

3. Simplification of trigonometric expressions. (25 min.)

The goal is to repeat, practice and consolidate the use of basic trigonometry formulas. Solving problems B7 from the Unified State Exam.

At this stage, it is advisable to divide the class into groups of strong students (work independently with subsequent testing) and weak students who work with the teacher.

Assignment for strong students (prepared in advance on a printed basis). The main emphasis is on the formulas of reduction and double angle, according to the Unified State Exam 2011.

Simplify expressions (for strong students):

At the same time, the teacher works with weak students, discussing and solving tasks on the screen under the students’ dictation.

Calculate:

5) sin(270º - α) + cos (270º + α)

Simplify:

It was time to discuss the results of the work of the strong group.

The answers appear on the screen, and also, using a video camera, the work of 5 different students is displayed (one task for each).

The weak group sees the condition and method of solution. Discussion and analysis are underway. With the use of technical means this happens quickly.

4. Solving simple trigonometric equations. (30 min.)

The goal is to repeat, systematize and generalize the solution of the simplest trigonometric equations and write down their roots. Solution of problem B3.

Any trigonometric equation, no matter how we solve it, leads to the simplest.

When completing the task, students should pay attention to writing the roots of equations of special cases and general form and to selecting the roots in the last equation.

Solve equations:

Write down the smallest positive root as your answer.

5. Independent work (10 min.)

The goal is to test the acquired skills, identify problems, errors and ways to eliminate them.

Multi-level work is offered to the student's choice.

Option "3"

1) Find the value of the expression

2) Simplify the expression 1 - sin 2 3α - cos 2 3α

3) Solve the equation

Option for "4"

1) Find the value of the expression

2) Solve the equation Write down the smallest positive root in your answer.

Option for "5"

1) Find tanα if

2) Find the root of the equation Write down the smallest positive root as your answer.

6. Lesson summary (5 min.)

The teacher sums up the fact that during the lesson they repeated and reinforced trigonometric formulas and solving the simplest trigonometric equations.

Homework is assigned (prepared on a printed basis in advance) with a random check at the next lesson.

Solve equations:

10) In your answer, indicate the smallest positive root.

Lesson 2

Subject: 11th grade (preparation for the Unified State Exam)

Methods for solving trigonometric equations. Root selection. (2 hours)

Goals:

Generalize and systematize knowledge on solving trigonometric equations of various types.
To promote the development of students’ mathematical thinking, the ability to observe, compare, generalize, and classify.
Encourage students to overcome difficulties in the process of mental activity, to self-control, and introspection of their activities.

Equipment for the lesson: KRMu, laptops for each student.

Lesson structure:

Organizational moment
Discussion of d/z and self. work from last lesson
Review of methods for solving trigonometric equations.
Solving trigonometric equations
Selection of roots in trigonometric equations.
Independent work.
Lesson summary. Homework.

1. Organizational moment (2 min.)

The teacher greets the audience, announces the topic of the lesson and the work plan.

2. a) Analysis of homework (5 min.)

The goal is to check execution. One work is displayed on the screen using a video camera, the rest are selectively collected for teacher checking.

b) Analysis of independent work (3 min.)

The goal is to analyze mistakes and indicate ways to overcome them.

Answers and solutions are on the screen; students have their work given out in advance. Analysis proceeds quickly.

3. Review of methods for solving trigonometric equations (5 min.)

The goal is to recall methods for solving trigonometric equations.

Ask students what methods for solving trigonometric equations they know. Emphasize that there are so-called basic (frequently used) methods:

variable replacement,
factorization,
homogeneous equations,

and there are applied methods:

using the formulas for converting a sum into a product and a product into a sum,
according to the formulas for reducing the degree,
universal trigonometric substitution
introduction of an auxiliary angle,
multiplication by some trigonometric function.

It should also be recalled that one equation can be solved in different ways.

4. Solving trigonometric equations (30 min.)

The goal is to generalize and consolidate knowledge and skills on this topic, to prepare for the C1 solution from the Unified State Exam.

I consider it advisable to solve equations for each method together with students.

Solve equations:

1) replacing the variable 6cos 2 x + 5sinx - 7 = 0

2) factorization 3cos(x/3) + 4cos 2 (x/3) = 0

3) homogeneous equations sin 2 x + 3cos 2 x - 2sin2x = 0

4) converting the sum into a product cos5x + cos7x = cos(π + 6x)

5) converting the product into the sum 2sinx sin2x + cos3x = 0

6) reduction of the degree sin2x - sin 2 2x + sin 2 3x = 0.5

7) universal trigonometric substitution sinx + 5cosx + 5 = 0.

When solving this equation, it should be noted that the use of this method leads to a narrowing of the definition range, since sine and cosine are replaced by tg(x/2). Therefore, before writing out the answer, you need to check whether the numbers from the set π + 2πn, n Z are horses of this equation.

8) introduction of an auxiliary angle √3sinx + cosx - √2 = 0

9) multiplication by some trigonometric function cosx cos2x cos4x = 1/8.

5. Selection of roots of trigonometric equations (20 min.)

Therefore, the goal of this stage of the lesson is to remember previously studied material and prepare to solve problem C1 from the Unified State Exam 2011.

Such equations are considered equations of increased complexity and in the Unified State Exam version they are found in the second part, namely C1.

Solve the equation:

A fraction is equal to zero if then using the unit circle we will select the roots (see Figure 1)

Figure 1.

we get x = π + 2πn, n Z

Answer: π + 2πn, n Z

On the screen, the selection of roots is shown on a circle in a color image.

The product is equal to zero when at least one of the factors is equal to zero, and the arc does not lose its meaning. Then

Using the unit circle, we select the roots (see Figure 2)

Voronkova Olga Ivanovna

MBOU "Secondary school"

No. 18"

Engels, Saratov region.

Math teacher.

« Trigonometric expressions and their transformations"

Introduction…………………………………………………………………………………......3

Chapter 1 Classification of tasks on the use of transformations of trigonometric expressions ………………………….…………………...5

1.1. Calculation tasks values of trigonometric expressions……….5

1.2.Tasks on simplifying trigonometric expressions.... 7

1.3. Tasks for converting numerical trigonometric expressions.....7

1.4 Mixed type tasks…………………………………………………….....9

Chapter 2. Methodological aspects of organizing the final repetition of the topic “Transformation of trigonometric expressions”……………………………11

2.1 Thematic repetition in 10th grade………………………………………………………...11

Test 1………………………………………………………………………………..12

Test 2………………………………………………………………………………..13

Test 3………………………………………………………………………………..14

2.2 Final repetition in 11th grade………………………………………………………...15

Test 1………………………………………………………………………………..17

Test 2………………………………………………………………………………..17

Test 3………………………………………………………………………………..18

Conclusion.………………………………………………………………………………......19

List of references………………………………………………………..…….20

Introduction.

In today's conditions, the most important question is: “How can we help eliminate some of the gaps in students’ knowledge and warn them against possible mistakes on the Unified State Exam?” To solve this issue, it is necessary to achieve from students not a formal assimilation of program material, but its deep and conscious understanding, development of the speed of oral calculations and transformations, as well as the development of skills in solving simple problems “in the mind.” It is necessary to convince students that only if they have an active position, when studying mathematics, provided they acquire practical skills and abilities and their use, can they count on real success. It is necessary to use every opportunity to prepare for the Unified State Exam, including elective subjects in grades 10-11, and conduct regular reviews difficult tasks with students, choosing the most rational way to solve problems in lessons and additional classes.Positive result inareas of solving standard problems can be achieved if mathematics teachers, by creatinggood basic training of students, look for new ways to solve the problems that have opened up to us, actively experiment, apply modern educational technologies, methods, techniques that create favorable conditions for effective self-realization and self-determination of students in new social conditions.

Trigonometry – component school mathematics course. Good knowledge and strong skills in trigonometry are evidence of a sufficient level of mathematical culture, an indispensable condition for successfully studying mathematics, physics, and a number of technical fields at a university. disciplines.

Relevance of the work. A significant proportion of school graduates show from year to year very poor preparation in this important section of mathematics, as evidenced by the results of past years (percentage of completion in 2011 - 48.41%, 2012 - 51.05%), since the analysis of passing the unified state exam showed that students make many mistakes when completing tasks in this particular section or do not take on such tasks at all. In One state exam Trigonometry questions are found in almost three types of assignments. This includes the solution of the simplest trigonometric equations in task B5, and work with trigonometric expressions in task B7, and the study of trigonometric functions in task B14, as well as tasks B12, in which there are formulas describing physical phenomena and containing trigonometric functions. And this is only part of the tasks B! But there are also favorite trigonometric equations with selection of C1 roots, and “not so favorite” geometric tasks C2 and C4.

Purpose of the work. Analyze Unified State Exam material tasks B7, devoted to transformations of trigonometric expressions and classify tasks according to the form of their presentation in tests.

The work consists of two chapters, introduction and conclusion. The introduction emphasizes the relevance of the work. The first chapter provides a classification of tasks for using transformations of trigonometric expressions into test tasks Unified State Exam (2012).

The second chapter examines the organization of repetition of the topic “Transformation of trigonometric expressions” in grades 10 and 11 and tests on this topic are developed.

The list of references includes 17 sources.

Chapter 1. Classification of tasks using transformations of trigonometric expressions.

In accordance with the standard of secondary (complete) education and the requirements for the level of preparation of students, the requirements codifier includes tasks on knowledge of the basics of trigonometry.

Learning the basics of trigonometry will be most effective when:

positive motivation will be provided for students to repeat previously learned material;

V educational process a person-centered approach will be implemented;

a system of tasks will be used that helps expand, deepen, and systematize students’ knowledge;

Advanced pedagogical technologies will be used.

Having analyzed the literature and Internet resources on preparation for the Unified State Exam, we have proposed one of the possible classifications of tasks B7 (KIM Unified State Exam 2012-trigonometry): calculation tasksvalues of trigonometric expressions; assignments forconverting numerical trigonometric expressions; tasks for converting literal trigonometric expressions; mixed type tasks.

1.1. Calculation tasks meanings of trigonometric expressions.

One of the most common types of simple trigonometry problems is calculating the values of trigonometric functions from the value of one of them:

a) Use of the basic trigonometric identity and its consequences.

Example 1 . Find if
And
.

Solution.
,
,

Because , That
.

Answer.

Example 2 . Find
, If
And .

Solution.
,
,
.

Because , That
.

Answer. .

b) Using double angle formulas.

Example 3 . Find
, If
.

Solution. , .

Answer.
.

Example 4 . Find the meaning of the expression
.

Solution. .

Answer.
.

1. Find , If

And

. Answer. -0.2

2. Find , If
And
. Answer. 0.4

3. Find

, If . Answer. -12.884. Find

, If

. Answer. -0.845. Find the meaning of the expression:

. Answer. 66. Find the meaning of the expression

.Answer. -19

1.2.Tasks on simplifying trigonometric expressions. Reduction formulas should be well understood by students, as they will find further application in geometry, physics and other related disciplines.

Example 5 . Simplify Expressions
.

Solution. .

Answer.
.

Tasks for independent solution:

1. Simplify the expression

. Answer. 0.62. Find

, If

And. Answer. 10.563. Find the meaning of the expression

, If

. Answer. 2

1.3. Tasks for converting numerical trigonometric expressions.

When practicing the skills of tasks for converting numerical trigonometric expressions, you should pay attention to knowledge of the table of values of trigonometric functions, the properties of parity and the periodicity of trigonometric functions.

a) Using exact values of trigonometric functions for some angles.

Example 6 . Calculate
.

Solution.
.

Answer.
.

b) Using parity properties trigonometric functions.

Example 7 . Calculate
.

Solution. .

Answer.

V) Using periodicity propertiestrigonometric functions.

Example 8 . Find the meaning of the expression
.

Solution. .

Answer.
.

Tasks for independent solution:

1. Find the meaning of the expression

. Answer. -40.52. Find the meaning of the expression

. Answer. 17

3. Find the meaning of the expression
. Answer. 6

. Answer. -24

Answer. -64

1.4 Mixed type tasks.

The certification test form has very significant features, so it is important to pay attention to tasks related to the use of several trigonometric formulas at the same time.

Example 9. Find
, If
.

Solution.
.

Answer.
.

Example 10 . Find
, If
And
.

Solution. .

Because , That
.

Answer.
.

Example 11. Find
, If .

Solution. , ,
,
,
,
,
.

Answer.

Example 12. Calculate
.

Solution. .

Answer.
.

Example 13. Find the meaning of the expression
, If
.

Solution. .

Answer.
.

Tasks for independent solution:

1. Find

, If

. Answer. -1.75
2. Find

, If

. Answer. 33. Find

, If .Answer. 0.254. Find the meaning of the expression

, If

. Answer. 0.35. Find the meaning of the expression

, If

. Answer. 5

Chapter 2. Methodological aspects of organizing the final repetition of the topic “Transformation of trigonometric expressions.”

One of the most important issues that contribute to the further improvement of academic performance and the achievement of deep and lasting knowledge among students is the issue of repeating previously covered material. Practice shows that in 10th grade it is more expedient to organize thematic repetition; in 11th grade - final repetition.

2.1. Thematic revision in 10th grade.

In the process of working on mathematical material especially great value acquires repetition of each completed topic or entire section of the course.

With thematic repetition, students' knowledge on a topic is systematized at the final stage of its completion or after a certain break.

For thematic repetition, special lessons are allocated, in which the material of one particular topic is concentrated and generalized.

Repetition in the lesson is carried out through conversation with the wide involvement of students in this conversation. After this, students are given the task to repeat a certain topic and are warned that test work will be carried out.

A test on a topic should include all its main questions. After completing the work, characteristic errors are analyzed and repetition is organized to eliminate them.

For thematic repetition lessons, we offer developed assessment work in the form of tests on the topic "Transformation of trigonometric expressions."

Test No. 1

Test No. 2

Test No. 3

Answer table

Test

2.2. Final review in 11th grade.

Final repetition is carried out at the final stage of studying the main issues of the mathematics course and is carried out in logical connection with the study educational material for this section or the course as a whole.

The final repetition of educational material pursues the following goals:

1. Activation of the entire material training course to clarify its logical structure and build a system within subject and inter-subject connections.

2. Deepening and, if possible, expanding students’ knowledge on the main issues of the course in the process of repetition.

In the context of the obligatory examination in mathematics for all graduates, the gradual introduction of the Unified State Exam forces teachers to take a new approach to preparing and conducting lessons, taking into account the need to ensure that all schoolchildren master the educational material on basic level, as well as the opportunity for motivated students interested in obtaining high scores for admission to a university to dynamically advance in mastering the material at an advanced and high level.

During final revision lessons, you can consider the following tasks:

Example 1 . Calculate the value of the expression.Solution. =

= =

=0,5. Answer. 0.5. Example 2. Specify the largest integer value that the expression can accept

Solution. Because
can take any value, belonging to the segment[–1; 1], then
takes any value of the segment [–0.4; 0.4], therefore . The expression has one integer value – the number 4.

Answer: 4 Example 3 . Simplify the expression

Solution: Let's use the formula for factoring the sum of cubes: . We have

We have:
.

Answer: 1

Example 4. Calculate
.

Solution. .

Answer: 0.28

For final revision lessons, we offer developed tests on the topic “Transformation of trigonometric expressions.”

Enter the largest integer not exceeding 1

Conclusion.

Having worked through the appropriate methodological literature on this topic, we can conclude that the ability and skills to solve problems related to trigonometric transformations V school course mathematics is very important.

In the course of the work done, a classification of tasks B7 was carried out. The trigonometric formulas most often used in CMMs in 2012 are considered. Examples of tasks with solutions are given. Differentiated tests have been developed to organize repetition and systematize knowledge in preparation for the Unified State Exam.

It is advisable to continue the work begun by considering solving the simplest trigonometric equations in task B5, studying trigonometric functions in task B14, tasks B12, which contain formulas that describe physical phenomena and contain trigonometric functions.

In conclusion, I would like to note that the effectiveness passing the Unified State Exam is largely determined by how effectively the training process is organized at all levels of education, with all categories of students. And if we are able to instill in students independence, responsibility and readiness to continue learning throughout their entire lives, then we will not only fulfill the order of the state and society, but also increase our own self-esteem.

Repetition of educational material requires the teacher creative work. He must provide a clear connection between types of repetition and implement a deeply thought-out system of repetition. Mastering the art of organizing repetition is the task of the teacher. The strength of students’ knowledge largely depends on its solution.

Literature.

Vygodsky Ya.Ya., Handbook of elementary mathematics. -M.: Nauka, 1970.

Problems of increased difficulty in algebra and basic analysis: Textbook for grades 10-11 high school/ B.M. Ivlev, A.M. Abramov, Yu.P. Dudnitsyn, S.I. Schwartzburd. – M.: Education, 1990.

Application of basic trigonometric formulas to transform expressions (10th grade) //Festival pedagogical ideas. 2012-2013.

Koryanov A.G. , Prokofiev A.A. We prepare good and excellent students for the Unified State Exam. - M.: Pedagogical University“First of September”, 2012.- 103 p.

Kuznetsova E.N. Simplifying trigonometric expressions. Solving trigonometric equations using various methods (preparation for the Unified State Exam). 11th grade. 2012-2013.

Kulanin E. D. 3000 competitive problems in mathematics. 4th edition, correct. and additional – M.: Rolf, 2000.

Mordkovich A.G. Methodological problems of studying trigonometry in secondary school// Mathematics at school. 2002. No. 6.

Pichurin L.F. About trigonometry and not only about it: -M. Enlightenment, 1985

Reshetnikov N.N. Trigonometry at school: -M. : Pedagogical University “First of September”, 2006, lx 1.

Shabunin M.I., Prokofiev A.A. Mathematics. Algebra. Beginnings of mathematical analysis. Profile level: textbook for grade 10 - M.: BINOM. Knowledge Laboratory, 2007.

Educational portal for preparing for the Unified State Exam.

Preparing for the Unified State Exam in Mathematics “Oh, this trigonometry! http://festival.1september.ru/articles/621971/

Project "Math? Easy!!!" http://www.resolventa.ru/

The video lesson “Simplifying Trigonometric Expressions” is designed to develop students' skills in solving trigonometric problems using basic trigonometric identities. During the video lesson, types of trigonometric identities and examples of solving problems using them are discussed. By using visual aids, it is easier for the teacher to achieve the lesson objectives. Vivid presentation of the material helps to remember important points. The use of animation effects and voice-over allows you to completely replace the teacher at the stage of explaining the material. Thus, by using this visual aid in mathematics lessons, the teacher can increase the effectiveness of teaching.

At the beginning of the video lesson, its topic is announced. Then we recall the trigonometric identities studied earlier. The screen displays the equalities sin 2 t+cos 2 t=1, tg t=sin t/cos t, where t≠π/2+πk for kϵZ, ctg t=cos t/sin t, correct for t≠πk, where kϵZ, tg t· ctg t=1, for t≠πk/2, where kϵZ, called the basic trigonometric identities. It is noted that these identities are often used in solving problems where it is necessary to prove equality or simplify an expression.

Below we consider examples of the application of these identities in solving problems. First, it is proposed to consider solving problems of simplifying expressions. In example 1, it is necessary to simplify the expression cos 2 t- cos 4 t+ sin 4 t. To solve the example, first take the common factor cos 2 t out of brackets. As a result of this transformation in parentheses, the expression 1- cos 2 t is obtained, the value of which from the main identity of trigonometry is equal to sin 2 t. After transforming the expression, it is obvious that it is possible to remove another common factor sin 2 t from brackets, after which the expression takes the form sin 2 t(sin 2 t+cos 2 t). From the same basic identity we derive the value of the expression in brackets equal to 1. As a result of simplification, we obtain cos 2 t- cos 4 t+ sin 4 t= sin 2 t.

In example 2, the expression cost/(1- sint)+ cost/(1+ sint) needs to be simplified. Since the numerator of both fractions contains the expression cost, it can be taken out of brackets as a common factor. Then the fractions in brackets are reduced to a common denominator by multiplying (1- sint)(1+ sint). After bringing similar terms the numerator remains 2, and the denominator remains 1 - sin 2 t. On the right side of the screen, the basic trigonometric identity sin 2 t+cos 2 t=1 is recalled. Using it, we find the denominator of the fraction cos 2 t. After reducing the fraction, we obtain a simplified form of the expression cost/(1- sint)+ cost/(1+ sint)=2/cost.

Next, we consider examples of proofs of identities that use the acquired knowledge about the basic identities of trigonometry. In example 3, it is necessary to prove the identity (tg 2 t-sin 2 t)·ctg 2 t=sin 2 t. The right side of the screen displays three identities that will be needed for the proof - tg t·ctg t=1, ctg t=cos t/sin t and tg t=sin t/cos t with restrictions. To prove the identity, the brackets are first opened, after which a product is formed that reflects the expression of the main trigonometric identity tg t·ctg t=1. Then, according to the identity from the definition of cotangent, ctg 2 t is transformed. As a result of the transformations, the expression 1-cos 2 t is obtained. Using the main identity, we find the meaning of the expression. Thus, it has been proven that (tg 2 t-sin 2 t)·ctg 2 t=sin 2 t.

In example 4, you need to find the value of the expression tg 2 t+ctg 2 t if tg t+ctg t=6. To calculate the expression, first square the right and left sides of the equality (tg t+ctg t) 2 =6 2. The abbreviated multiplication formula is recalled on the right side of the screen. After opening the brackets on the left side of the expression, the sum tg 2 t+2· tg t·ctg t+ctg 2 t is formed, to transform which you can apply one of the trigonometric identities tg t·ctg t=1, the form of which is recalled on the right side of the screen. After the transformation, the equality tg 2 t+ctg 2 t=34 is obtained. The left side of the equality coincides with the condition of the problem, so the answer is 34. The problem is solved.

The video lesson “Simplification of trigonometric expressions” is recommended for use in traditional school lesson mathematics. The material will also be useful to the teacher implementing distance learning. In order to develop skills in solving trigonometric problems.

TEXT DECODING:

"Simplification of trigonometric expressions."

Equalities

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 with te 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)

are called basic trigonometric identities.

They are often used in simplifying and proving trigonometric expressions.

Let's look at examples of using these formulas to simplify trigonometric expressions.

EXAMPLE 1. Simplify the expression: cos 2 t - cos 4 t + sin 4 t. (expression a cosine squared te minus cosine of the fourth degree te plus sine of the fourth degree te).

Solution. cos 2 t - cos 4 t + sin 4 t = cos 2 t∙ (1 - cos 2 t) + sin 4 t =cos 2 t ∙ sin 2 t + sin 4 t = sin 2 t (cos 2 t + sin 2 t) = sin 2 t 1= sin 2 t

(we take out the common factor cosine square te, in brackets we get the difference between unity and the squared cosine te, which is equal to the squared sine te by the first identity. We get the sum of the fourth power sine te of the product cosine square te and sine square te. We take out the common factor sine square te outside the brackets, in brackets we get the sum of the squares of the cosine and sine, which is basically trigonometric identity equals one. As a result, we obtain the square of the sine te).

EXAMPLE 2. Simplify the expression: + .

(the expression be is the sum of two fractions in the numerator of the first cosine te in the denominator one minus sine te, in the numerator of the second cosine te in the denominator of the second one plus sine te).

(Let’s take the common factor cosine te out of brackets, and in brackets we bring it to a common denominator, which is the product of one minus sine te by one plus sine te.

In the numerator we get: one plus sine te plus one minus sine te, we present similar ones, the numerator is equal to two after bringing similar ones.

In the denominator, you can apply the abbreviated multiplication formula (difference of squares) and obtain the difference between unity and the square of the sine te, which, according to the basic trigonometric identity

equal to the square of the cosine te. After reducing by cosine te we get the final answer: two divided by cosine te).

Let's look at examples of using these formulas when proving trigonometric expressions.

EXAMPLE 3. Prove the identity (tg 2 t - sin 2 t) ∙ ctg 2 t = sin 2 t (the product of the difference between the squares of tangent te and sine te by the square of cotangent te is equal to the square of sine te).

Proof.

Let's transform the left side of the equality:

(tg 2 t - sin 2 t) ∙ ctg 2 t = tg 2 t ∙ ctg 2 t - sin 2 t ∙ ctg 2 t = 1 - sin 2 t ∙ ctg 2 t =1 - sin 2 t ∙ = 1 - cos 2 t = sin 2 t

(Let us open the parentheses; from the previously obtained relation it is known that the product of the squares of tangent te by cotangent te is equal to one. Recall that cotangent te equal to the ratio cosine te by sine te, which means the square of the cotangent is the ratio of the square of the cosine te by the square of the sine te.

After reduction by sine square te we obtain the difference between unity and cosine square te, which is equal to sine square te). Q.E.D.

EXAMPLE 4. Find the value of the expression tg 2 t + ctg 2 t if tgt + ctgt = 6.

(the sum of the squares of tangent te and cotangent te, if the sum of tangent and cotangent is six).

Solution. (tgt + ctgt) 2 = 6 2

tg 2 t + 2 ∙ tgt ∙ctgt + ctg 2 t = 36

tg 2 t + 2 + ctg 2 t = 36

tg 2 t + ctg 2 t = 36-2

tg 2 t + ctg 2 t = 34

Let's square both sides of the original equality:

(tgt + ctgt) 2 = 6 2 (the square of the sum of tangent te and cotangent te is equal to six squared). Let us recall the formula for abbreviated multiplication: The square of the sum of two quantities is equal to the square of the first plus twice the product of the first by the second plus the square of the second. (a+b) 2 =a 2 +2ab+b 2 We get tg 2 t + 2 ∙ tgt ∙ctgt + ctg 2 t = 36 (tangent squared te plus double the product of tangent te by cotangent te plus cotangent squared te equals thirty-six) .

Since the product of tangent te and cotangent te is equal to one, then tg 2 t + 2 + ctg 2 t = 36 (the sum of the squares of tangent te and cotangent te and two is equal to thirty-six),