Practical work on the topic of inverse trigonometric functions. "inverse trigonometric functions" - Document

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Assignment: Create a test “Inverse trigonometric functions”

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Independent work No. 14 (2 hours)

On the topic: “Stretching and compression along coordinate axes”

Target: systematization and consolidation of the acquired theoretical knowledge and practical skills of students;

Assignment: Abstract on the topic: “Extension and compression along the coordinate axes”

Literature: A.G. Mordkovich “Algebra and the beginnings of mathematical analysis” 10th grade

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Independent work No. 15 (1 hour)

On the topic: “Stretching and compression along coordinate axes”

Target: formation of independent thinking, ability for self-development, self-improvement and self-realization

Assignment: presentation: “Extension and compression along coordinate axes”

Literature: A.G. Mordkovich “Algebra and the beginnings of mathematical analysis” 10th grade

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Independent work No. 16 (2 hours)

On the topic: “Inverse trigonometric functions, their properties and graphs”

Target: systematization and consolidation of acquired theoretical knowledge and practical skills of students

Task completion form: research work.

Literature: A.G. Mordkovich “Algebra and the beginnings of mathematical analysis” 10th grade

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On topic: “Half argument formulas”

Goal: deepening and expanding theoretical knowledge

Assignment: Write a message on the topic “Formulas of half an argument.” Create a reference table for trigonometry formulas

Literature: A.G. Mordkovich “Algebra and the beginnings of mathematical analysis” 10th grade

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Lessons 32-33. Inverse trigonometric functions

Target: consider inverse trigonometric functions and their use for writing solutions to trigonometric equations.

I. Communicating the topic and purpose of the lessons

II. Learning new material

1. Inverse trigonometric functions

Let's begin our discussion of this topic with the following example.

Example 1

Let's solve the equation: a) sin x = 1/2; b) sin x = a.

a) On the ordinate axis we plot the value 1/2 and construct the angles x 1 and x2, for which sin x = 1/2. In this case x1 + x2 = π, whence x2 = π – x 1 . Using the table of values ​​of trigonometric functions, we find the value x1 = π/6, thenLet's take into account the periodicity of the sine function and write down the solutions to this equation:where k ∈ Z.

b) Obviously, the algorithm for solving the equation sin x = a is the same as in the previous paragraph. Of course, now the value a is plotted along the ordinate axis. There is a need to somehow designate the angle x1. We agreed to denote this angle with the symbol arcsin A. Then the solutions to this equation can be written in the formThese two formulas can be combined into one: at the same time

The remaining inverse trigonometric functions are introduced in a similar way.

Very often it is necessary to determine the magnitude of an angle from the known value of its trigonometric function. Such a problem is multivalued - there are countless angles whose trigonometric functions are equal to the same value. Therefore, based on the monotonicity of trigonometric functions, the following inverse trigonometric functions are introduced to uniquely determine angles.

Arcsine of the number a (arcsin , whose sine is equal to a, i.e.

Arc cosine of a number a(arccos a) is an angle a from the interval whose cosine is equal to a, i.e.

Arctangent of a number a(arctg a) - such an angle a from the intervalwhose tangent is equal to a, i.e.tg a = a.

Arccotangent of a number a(arcctg a) is an angle a from the interval (0; π), the cotangent of which is equal to a, i.e. ctg a = a.

Example 2

Let's find:

Taking into account the definitions of inverse trigonometric functions, we obtain:

Example 3

Let's calculate

Let angle a = arcsin 3/5, then by definition sin a = 3/5 and . Therefore, we need to find cos A. Using the basic trigonometric identity, we get:It is taken into account that cos a ≥ 0. So,

Let us give a number of more typical examples related to the definitions and basic properties of inverse trigonometric functions.

Example 4

Let's find the domain of definition of the function

In order for the function y to be defined, it is necessary to satisfy the inequalitywhich is equivalent to the system of inequalitiesThe solution to the first inequality is the interval x∈ (-∞; +∞), second - This interval and is a solution to the system of inequalities, and therefore the domain of definition of the function

Example 5

Let's find the area of ​​change of the function

Let's consider the behavior of the function z = 2x - x2 (see picture).

It is clear that z ∈ (-∞; 1]. Considering that the argument z the arc cotangent function changes within the specified limits, from the table data we obtain thatThus, the area of ​​change

Example 6

Let us prove that the function y = arctg x odd. LetThen tg a = -x or x = - tg a = tg (- a), and Therefore, - a = arctg x or a = - arctg X. Thus, we see thati.e. y(x) is an odd function.

Example 7

Let us express through all inverse trigonometric functions

Let It's obvious that Then since

Let's introduce the angle Because That

Similarly therefore And

So,

Example 8

Let's build a graph of the function y = cos(arcsin x).

Let us denote a = arcsin x, then Let's take into account that x = sin a and y = cos a, i.e. x 2 + y2 = 1, and restrictions on x (x∈ [-1; 1]) and y (y ≥ 0). Then the graph of the function y = cos(arcsin x) is a semicircle.

Example 9

Let's build a graph of the function y = arccos (cos x ).

Since the cos function x changes on the interval [-1; 1], then the function y is defined on the entire numerical axis and varies on the segment . Let's keep in mind that y = arccos(cosx) = x on the segment; the function y is even and periodic with period 2π. Considering that the function has these properties cos x Now it's easy to create a graph.

Let us note some useful equalities:

Example 10

Let's find the smallest and largest values ​​of the function Let's denote Then Let's get the function This function has a minimum at the point z = π/4, and it is equal to The greatest value of the function is achieved at the point z = -π/2, and it is equal Thus, and

Example 11

Let's solve the equation

Let's take into account that Then the equation looks like:or where By definition of arctangent we get:

2. Solving simple trigonometric equations

Similar to example 1, you can obtain solutions to the simplest trigonometric equations.

Example 12

Let's solve the equation

Since the sine function is odd, we write the equation in the formSolutions to this equation:where do we find it from?

Example 13

Let's solve the equation

Using the given formula, we write down the solutions to the equation:and we'll find

Note that in special cases (a = 0; ±1) when solving the equations sin x = a and cos x = and it’s easier and more convenient to use not general formulas, but to write down solutions based on the unit circle:

for the equation sin x = 1 solution

for the equation sin x = 0 solutions x = π k;

for the equation sin x = -1 solution

for the cos equation x = 1 solution x = 2π k ;

for the equation cos x = 0 solutions

for the equation cos x = -1 solution

Example 14

Let's solve the equation

Since in this example there is a special case of the equation, we will write the solution using the appropriate formula:where can we find it from?

III. Control questions (frontal survey)

1. Define and list the main properties of inverse trigonometric functions.

2. Give graphs of inverse trigonometric functions.

3. Solving simple trigonometric equations.

IV. Lesson assignment

§ 15, No. 3 (a, b); 4 (c, d); 7(a); 8(a); 12 (b); 13(a); 15 (c); 16(a); 18 (a, b); 19 (c); 21;

§ 16, No. 4 (a, b); 7(a); 8 (b); 16 (a, b); 18(a); 19 (c, d);

§ 17, No. 3 (a, b); 4 (c, d); 5 (a, b); 7 (c, d); 9 (b); 10 (a, c).

V. Homework

§ 15, No. 3 (c, d); 4 (a, b); 7 (c); 8 (b); 12(a); 13(b); 15 (g); 16 (b); 18 (c, d); 19 (g); 22;

§ 16, No. 4 (c, d); 7 (b); 8(a); 16 (c, d); 18 (b); 19 (a, b);

§ 17, No. 3 (c, d); 4 (a, b); 5 (c, d); 7 (a, b); 9 (d); 10 (b, d).

VI. Creative tasks

1. Find the domain of the function:

Answers:

2. Find the range of the function:

Answers:

3. Graph the function:

VII. Summing up the lessons

Federal Agency for Education of the Russian Federation

State Educational Institution of Higher Professional Education "Mari State University"

Department of Mathematics and MPM

Coursework

Inverse trigonometric functions

Completed:

student

33 JNF groups

Yashmetova L. N.

Scientific supervisor:

Ph.D. associate professor

Borodina M.V.

Yoshkar-Ola

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

Chapter I. Definition of inverse trigonometric functions.

1.1. Function y =arcsin x……………………………………………………........4

1.2. Function y =arccos x…………………………………………………….......5

1.3. Function y =arctg x………………………………………………………….6

1.4. Function y =arcctg x…………………………………………………….......7

Chapter II. Solving equations with inverse trigonometric functions.

Basic relations for inverse trigonometric functions....8

Solving equations containing inverse trigonometric functions………………………………………………………………………………..11

Calculating the values ​​of inverse trigonometric functions............21

Conclusion………………………………………………………………………………….25

List of references……………………………………………………………...26

Introduction

In many problems, there is a need to find not only the values ​​of trigonometric functions from a given angle, but also, conversely, an angle or arc from a given value of some trigonometric function.

Problems with inverse trigonometric functions are contained in the USE tasks (especially many in parts B and C). For example, in Part B of the Unified State Exam it was required to use the value of the sine (cosine) to find the corresponding value of the tangent or to calculate the value of an expression containing tabulated values ​​of inverse trigonometric functions. Regarding this type of tasks, we note that such tasks in school textbooks are not enough to develop a strong skill in their implementation.

That. The purpose of the course work is to consider inverse trigonometric functions and their properties, and learn how to solve problems with inverse trigonometric functions.

To achieve the goal, we will need to solve the following tasks:

Study the theoretical foundations of inverse trigonometric functions,

Show the application of theoretical knowledge in practice.

ChapterI. Definition of inverse trigonometric functions

1.1. Function y =arcsinx

Consider the function,
. (1)

In this interval the function is monotonic (increases from -1 to 1), therefore, there is an inverse function

,
. (2)

Each given value at(sine value) from the interval [-1,1] corresponds to one well-defined value X(arc magnitude) from the interval
. Moving on to the generally accepted notation, we get

Where
. (3)

This is the analytical specification of the function inverse to function (1). Function (3) is called arcsine argument . The graph of this function is a curve symmetrical to the graph of the function, where , relative to the bisector of the I and III coordinate angles.

Let us present the properties of the function, where .

Property 1. Function value change area: .

Property 2. The function is odd, i.e.

Property 3. The function, where , has a single root
.

Property 4. If, then
; If , That.

Property 5. The function is monotonic: as the argument increases from -1 to 1, the value of the function increases from
to
.

1.2. Functiony = arWithcosx

Consider the function
, . (4)

In this interval the function is monotonic (decreases from +1 to -1), which means that there is an inverse function for it

, , (5)

those. each value (cosine values) from the interval [-1,1] corresponds to one well-defined value (arc values) from the interval . Moving on to the generally accepted notation, we get

, . (6)

This is the analytical specification of the function inverse to function (4). Function (6) is called arc cosine argument X. The graph of this function can be constructed based on the properties of graphs of mutually inverse functions.

The function , where , has the following properties.

Property 1. Function value change area:
.

Property 2. Quantities
And
related by the relation

Property 3. The function has a single root
.

Property 4. The function does not accept negative values.

Property 5. The function is monotonic: as the argument increases from -1 to +1, the function values ​​decrease from to 0.

1.3. Functiony = arctgx

Consider the function
,
. (7)

Note that this function is defined for all values ​​lying strictly within the interval from to ; at the ends of this interval it does not exist, since the values

- tangent break points.

In between
the function is monotonic (increases from -
to
), therefore, for function (1) there is an inverse function:

,
, (8)

those. each given value (tangent value) from the interval
corresponds to one very specific value (arc size) from the interval .

Moving on to the generally accepted notation, we get

,
. (9)

This is the analytical specification of the inverse function (7). Function (9) is called arctangent argument X. Note that when
function value
, and when

, i.e. the graph of the function has two asymptotes:
And.

The function , , has the following properties.

Property 1. Range of change of function values
.

Property 2. The function is odd, i.e. .

Property 3. The function has a single root.

Property 4. If
, That

; If , That
.

Property 5. The function is monotonic: as the argument increases from to, the function value increases from to +.

1.4. Functiony = arcctgx

Consider the function
,
. (10)

This function is defined for all values ​​lying within the range from 0 to ; at the ends of this interval it does not exist, since the values ​​and are the breakpoints of the cotangent. In the interval (0,) the function is monotonic (decreases from to), therefore, for function (1) there is an inverse function

, (11)

those. to each given value (cotangent value) from the interval (
) corresponds to one well-defined value (arc size) from the interval (0,). Moving on to generally accepted notations, we obtain the following relation: Abstract >> Mathematics trigonometric functions. TO reverse trigonometric functions usually referred to as six functions: arcsine...

The final work on the topic “Inverse trigonometric functions. Problems containing inverse trigonometric functions” was completed in advanced training courses.

Contains brief theoretical material, detailed examples and tasks for independent solution for each section.

The work is addressed to high school students and teachers.

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GRADUATE THESIS

TOPIC:

“INVERSE TRIGONOMETRIC FUNCTIONS.

PROBLEMS CONTAINING INVERSE TRIGONOMETRIC FUNCTIONS"

Completed:

math teacher

Municipal educational institution secondary school No. 5, Lermontov

GORBACHENKO V.I.

Pyatigorsk 2011

INVERSE TRIGONOMETRIC FUNCTIONS.

PROBLEMS CONTAINING INVERSE TRIGONOMETRIC FUNCTIONS

1. BRIEF THEORETICAL INFORMATION

1.1. Solutions to the simplest equations containing inverse trigonometric functions:

Table 1.

1.2. Solving simple inequalities involving inverse trigonometric functions

Table 2.

1.3. Some identities for inverse trigonometric functions

From the definition of inverse trigonometric functions, the identities follow

, (1)

, (2)

, (3)

, (4)

Moreover, the identities

, (5)

, (6)

, (7)

, (8)

Identities relating unlike inverse trigonometric functions

(9)

(10)

2. EQUATIONS CONTAINING INVERSE TRIGONOMETRIC FUNCTIONS

2.1. Equations of the form etc.

Such equations are reduced to rational equations by substitution.

Example.

Solution.

Replacement ( ) reduces the equation to a quadratic equation whose roots.

Root 3 does not satisfy the condition.

Then we get the reverse substitution

Answer .

Tasks.

2.2. Equations of the form, Where - rational function.

To solve equations of this type it is necessary to put, solve the equation of the simplest formand do the reverse substitution.

Example.

Solution .

Let . Then

Answer . .

Tasks.

2.3. Equations containing either different arc functions or arc functions of different arguments.

If the equation includes expressions containing different arc functions, or these arc functions depend on different arguments, then the reduction of such equations to their algebraic consequence is usually carried out by calculating some trigonometric function on both sides of the equation. The resulting foreign roots are separated by inspection. If tangent or cotangent is chosen as a direct function, then solutions included in the domain of definition of these functions may be lost. Therefore, before calculating the value of the tangent or cotangent from both sides of the equation, you should make sure that there are no roots of the original equation among the points not included in the domain of definition of these functions.

Example.

Solution .

Let's reschedule to the right side and calculate the value of the sine from both sides of the equation

As a result of transformations we get

The roots of this equation

Let's check

When we have

Thus, is the root of the equation.

Substituting , note that the left side of the resulting relationship is positive, and the right side is negative. Thus,- extraneous root of the equation.

Answer. .

Tasks.

2.4. Equations containing inverse trigonometric functions of one argument.

Such equations can be reduced to the simplest using basic identities (1) – (10).

Example.

Solution.

Answer.

Tasks.

3. INEQUALITIES CONTAINING INVERSE TRIGONOMETRIC FUNCTIONS

3.1. The simplest inequalities.

The solution to the simplest inequalities is based on the application of the formulas in Table 2.

Example.

Solution.

Because , then the solution to the inequality is the interval.

Answer .

Tasks.

3.2. Inequalities of the form, - some rational function.

Inequalities of the form, is some rational function, and- one of the inverse trigonometric functions is solved in two stages - first, the inequality with respect to the unknown is solved, and then the simplest inequality containing the inverse trigonometric function.

Example.

Solution.

Let it be then

Solutions to inequalities

Returning to the original unknown, we find that the original inequality can be reduced to two simplest ones

Combining these solutions, we obtain solutions to the original inequality

Answer .

Tasks.

3.3. Inequalities containing either opposite arc functions or arc functions of different arguments.

It is convenient to solve inequalities connecting the values ​​of various inverse trigonometric functions or the values ​​of one trigonometric function calculated from different arguments by calculating the values ​​of some trigonometric function from both sides of the inequalities. It should be remembered that the resulting inequality will be equivalent to the original one only if the set of values ​​of the right and left sides of the original inequality belong to the same monotonicity interval of this trigonometric function.

Example.

Solution.

Multiple Valid Valuesincluded in the inequality:. At . Therefore, the valuesare not solutions to the inequality.

At both the right side and the left side of the inequality have values ​​belonging to the interval. Because in betweenthe sine function increases monotonically, then whenthe original inequality is equivalent

Solving the last inequality

Crossing with a gap, we get a solution

Answer.

Comment. Can be solved using

Tasks.

3.4. Inequality of the form, Where - one of the inverse trigonometric functions,- rational function.

Such inequalities are solved using the substitutionand reduction to the simplest inequality in Table 2.

Example.

Solution.

Let it be then

Let's do the reverse substitution and get the system

Answer .

Tasks.

Preparation for the Unified State Exam in mathematics

Experiment

Lesson 9. Inverse trigonometric functions.

Practice

Lesson summary

We will mainly need the ability to work with arc functions when solving trigonometric equations and inequalities.

The tasks that we will now consider are divided into two types: calculating the values ​​of inverse trigonometric functions and their transformations using basic properties.

Calculation of values ​​of arc functions

Let's start by calculating the values ​​of the arc functions.

Task No. 1. Calculate.

As we see, all the arguments of arc functions are positive and tabular, which means that we can restore the value of angles from the first part of the table of values ​​of trigonometric functions for angles from to . This range of angles is included in the range of values ​​of each of the arc functions, so we simply use the table, find the value of the trigonometric function in it and restore which angle it corresponds to.

A)

b)

V)

G)

Answer. .

Task No. 2. Calculate

.

In this example we already see negative arguments. A typical mistake in this case is to simply remove the minus from under the function and simply reduce the task to the previous one. However, this cannot be done in all cases. Let us remember how in the theoretical part of the lesson we discussed the parity of all arc functions. The odd ones are arcsine and arctangent, i.e., the minus is taken out of them, and arccosine and arccotangent are functions of a general form; to simplify the minus in the argument, they have special formulas. After calculation, to avoid errors, we check that the result is within the range of values.

When the function arguments are simplified to positive form, we write out the corresponding angle values ​​from the table.

The question may arise: why not write down the value of the angle corresponding, for example, directly from the table? Firstly, because the table before is harder to remember than before, and secondly, because there are no negative values ​​of the sine in it, and negative values ​​of the tangent will give the wrong angle according to the table. It is better to have a universal approach to a solution than to get confused by many different approaches.

Task No. 3. Calculate.

a) A typical mistake in this case is to start taking out a minus and simplify something. The first thing to notice is that the arcsine argument is not in the scope of

Therefore, this entry has no meaning, and the arcsine cannot be calculated.

b) The standard mistake in this case is that they confuse the values ​​of the argument and the function and give the answer. This is not true! Of course, the thought arises that in the table the value corresponds to cosine, but in this case, what is confused is that arc functions are calculated not from angles, but from the values ​​of trigonometric functions. That is, not .

In addition, since we have found out what exactly the argument of the arc cosine is, it is necessary to check that it is included in the domain of definition. To do this, let us remember that , i.e., which means arccosine does not make sense and cannot be calculated.

By the way, for example, the expression makes sense, because , but since the value of the cosine equal is not tabular, it is impossible to calculate the arc cosine using the table.

Answer. The expressions don't make sense.

In this example, we do not consider arctangent and arccotangent, since their domain of definition is not limited and the function values ​​will be for any arguments.

Task No. 4. Calculate .

In essence, the task comes down to the very first one, we just need to separately calculate the values ​​of the two functions, and then substitute them into the original expression.

The arctangent argument is tabular and the result belongs to the range of values.

The arccosine argument is not tabular, but this should not scare us, because no matter what the arccosine is equal to, its value when multiplied by zero will result in zero. There is one important note left: it is necessary to check whether the arccosine argument belongs to the domain of definition, since if this is not the case, then the entire expression will not make sense, regardless of the fact that it contains multiplication by zero. But, therefore, we can say that it makes sense and we get zero in the answer.

Let us give another example in which it is necessary to be able to calculate one arc function, knowing the value of another.

Problem #5. Calculate if it is known that .

It may seem that it is necessary to first calculate the value of x from the indicated equation, and then substitute it into the desired expression, i.e., into the inverse tangent, but this is not necessary.

Let us remember the formula by which these functions are related to each other:

And let’s express from it what we need:

To be sure, you can check that the result lies in the arc cotangent range.

Transformations of arc functions using their basic properties

Now let's move on to a series of tasks in which we will have to use transformations of arc functions using their basic properties.

Problem #6. Calculate .

To solve, we will use the basic properties of the indicated arc functions, only making sure to check the corresponding restrictions.

A)

b) .

Answer. A) ; b) .

Problem No. 7. Calculate.

A typical mistake in this case is to immediately write 4 in response. As we indicated in the previous example, to use the basic properties of arc functions, it is necessary to check the corresponding restrictions on their argument. We are dealing with the property:

at

But . The main thing at this stage of the decision is not to think that the specified expression does not make sense and cannot be calculated. After all, we can reduce the four, which is the argument of the tangent, by subtracting the period of the tangent, and this will not affect the value of the expression. Having done these steps, we will have a chance to reduce the argument so that it falls within the specified range.

Because since, therefore, , because .

Problem No. 8. Calculate.

In the above example, we are dealing with an expression that is similar to the basic property of the arcsine, but only it contains cofunctions. It must be reduced to the form sine from arcsine or cosine from arccosine. Since it is easier to transform direct trigonometric functions than inverse ones, let’s move from sine to cosine using the “trigonometric unit” formula.

As we already know:

In our case, in the role. Let us first calculate for convenience .

Before substituting it into the formula, let’s find out its sign, i.e., the sign of the original sine. We must calculate the sine from the arccosine value, whatever this value is, we know that it lies in the range. This range corresponds to the angles of the first and second quarters, in which the sine is positive (check this yourself using a trigonometric circle).

In today's practical lesson we looked at the calculation and transformation of expressions containing inverse trigonometric functions

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