Topic:"Solution Methods trigonometric equations».

Lesson Objectives:

educational:

To form skills to distinguish types of trigonometric equations;

Deepening understanding of methods for solving trigonometric equations;

educational:

Upbringing cognitive interest to the educational process;

Formation of the ability to analyze the task;

developing:

To form the skill to analyze the situation with the subsequent choice of the most rational way out of it.

Equipment: poster with basic trigonometric formulas, computer, projector, screen.

Let's start the lesson by repeating the basic technique for solving any equation: reducing it to standard form. Through transformation linear equations reduce to the form ax \u003d in, square - to the form ax2+bx +c=0. In the case of trigonometric equations, it is necessary to reduce them to the simplest, of the form: sinx \u003d a, cosx \u003d a, tgx \u003d a, which can be easily solved.

First of all, of course, for this it is necessary to use the basic trigonometric formulas that are presented on the poster: addition formulas, formulas double angle, lowering the multiplicity of the equation. We already know how to solve such equations. Let's repeat some of them:

At the same time, there are equations, the solution of which requires knowledge of some special techniques.

The topic of our lesson is the consideration of these techniques and the systematization of methods for solving trigonometric equations.

Methods for solving trigonometric equations.

1. Convert to quadratic equation with respect to some trigonometric function, followed by a change of variable.

Let's consider each of listed methods on examples, but we will dwell on the last two in more detail, since we have already used the first two when solving equations.

1. Transformation to a quadratic equation with respect to any trigonometric function.

2. Solution of equations by the factorization method.

3. Solution of homogeneous equations.

Homogeneous equations of the first and second degree are called equations of the form:

respectively (a ≠ 0, b ≠ 0, c ≠ 0).

When solving homogeneous equations, both parts of the equation are divided term by term by cosx for (1) of the equation and by cos 2 x for (2). Such a division is possible, since sinx and cosx are not equal to zero at the same time - they turn to zero in different points. Consider examples of solving homogeneous equations of the first and second degree.

Remember this equation: when considering the next method - the introduction of an auxiliary argument, we will solve it in a different way.

4. Introduction of an auxiliary argument.

Consider the equation already solved by the previous method:

As you can see, the same result is obtained.

Let's look at another example:

In the examples considered, it was generally clear what the original equation needs to be divided into in order to introduce an auxiliary argument. But it may happen that it is not obvious which divisor to choose. There is a special technique for this, which we will now consider in general view. Let the equation be given:

Divide the equation by Square root from expression (3), we get:

asinx + bcosx = c ,

then a 2 + b 2 = 1 and hence a = sinx and b = cosx . Using the difference cosine formula, we obtain the simplest trigonometric equation:

which is easily solved.

Let's solve another equation:

We reduce the equation to one argument - 2 x using the double angle formulas and lowering the degree:

Similarly to the previous equations, using the sine formula of the sum, we get:

which is also easy to solve.

Decide for yourself by pre-defining the solution method:

The result of the lesson is to check the solution and evaluate the students.

Homework: p. 11, abstract, No. 164 (b, d), 167 (b, d), 169 (a, b), 174 (a, c).

Elementary trigonometric equations are equations of the form, where is one of the trigonometric functions: , .

Elementary trigonometric equations have infinitely many roots. For example, the equation is satisfied the following values: , etc. General formula by which all the roots of the equation are found, where, is:

Here it can take any integer values, each of them corresponds to a certain root of the equation; in this formula (as well as in other formulas by which elementary trigonometric equations are solved) is called parameter. They usually write it down, thereby emphasizing that the parameter can take any integer values.

The solutions of the equation, where, are found by the formula

The equation is solved by applying the formula

and the equation --- according to the formula

Let us especially note some special cases of elementary trigonometric equations, when the solution can be written without using general formulas:

When solving trigonometric equations important role plays the period of trigonometric functions. Therefore, we present two useful theorems:

Theorem If a --- basic period of the function, then the number is the main period of the function.

The periods of the functions and are said to be commensurate if there exist integers and what.

Theorem If a periodic functions and, have commensurate and, then they have general period, which is the period of the functions, .

The theorem says what is the period of the function, and not necessarily the main period. For example, the main period of the functions and is --- , and the main period of their product is --- .

Introducing an Auxiliary Argument

The standard way to transform expressions of the form is the following trick: let --- corner, given by the equalities, . For any and such an angle exists. In this way. If, or, in other cases.

Scheme for solving trigonometric equations

The main scheme that we will be guided by when solving trigonometric equations is as follows:

solution given equation comes down to a decision elementary equations. Solution tools --- transformations, factorizations, change of unknowns. The guiding principle is not to lose roots. This means that when moving to the next equation (equations), we are not afraid of the appearance of extra (extraneous) roots, but only take care that each the following equation our "chain" (or the set of equations in the case of branching) was a consequence of the previous one. One of possible methods selection of roots is a check. We note right away that in the case of trigonometric equations, the difficulties associated with the selection of roots, with verification, as a rule, increase sharply in comparison with algebraic equations. After all, it is necessary to check the series consisting of an infinite number members.

Special mention should be made of the change of unknowns in solving trigonometric equations. In most cases, after the necessary replacement, it turns out algebraic equation. Moreover, it is not uncommon for equations which, although they are trigonometric in appearance, in fact, they are not, because already after the first step --- replacements variables --- turn into algebraic ones, and the return to trigonometry occurs only at the stage of solving elementary trigonometric equations.

Let us remind you once again: the replacement of the unknown should be done as soon as possible, the equation obtained after the replacement must be solved to the end, including the stage of selecting the roots, and only then will it return to the original unknown.

One of the features of trigonometric equations is that the answer in many cases can be written different ways. Even for solving the equation, the answer can be written like this:

1) in the form of two series: , ;

2) in standard form, which is a union of the above series: , ;

3) since, then the answer can be written in the form, . (Further on, the presence of a parameter, or in the response record automatically means that this parameter takes all possible integer values. Exceptions will be specified.)

Obviously, the three listed cases do not exhaust all the possibilities for writing the answer to the equation under consideration (there are infinitely many of them).

For example, when equality is true. Therefore, in the first two cases, if, we can replace with.

Usually, the answer is written on the basis of paragraph 2. It is useful to remember the following recommendation: if the work does not end with the solution of the equation, it is still necessary to conduct a study, the selection of roots, then the most convenient form of recording is indicated in paragraph 1. (A similar recommendation should be given for the equation.)

Let's consider an example illustrating what has been said.

Example Solve the equation.

Solution. The most obvious is next path. This equation splits into two: i. Solving each of them and combining the answers obtained, we find.

Another way. Since, then, replacing and by the formulas of lowering the degree. After small transformations, we get where.

At first glance, none special benefits the second formula has none compared to the first. However, if we take, for example, it turns out that, i.e. the equation has a solution, while the first way leads us to the answer. "Seeing" and proving equality is not so easy.

In algebra classes, teachers say that there is a small (in fact, a very large) class of trigonometric equations that cannot be solved in standard ways- neither through factorization, nor through a change of variable, nor even through homogeneous terms. In this case, a fundamentally different approach comes into play - the method auxiliary corner.

What is this method and how to apply it? First, let's recall the formulas for the sum/difference sine and the sum/difference cosine:

\[\begin(align)& \sin \left(\alpha \pm \beta \right)=\sin \alpha \cos \beta \pm \cos \alpha \sin \beta \\& \cos \left(\ alpha \pm \beta \right)=\cos \alpha \cos \beta \mp \sin \alpha \sin \beta \\\end(align)\]

I think these formulas are well known to you - formulas are derived from them double argument, without which trigonometry is nowhere at all. But let's now consider a simple equation:

Divide both parts by 5:

Note that $((\left(\frac(3)(5) \right))^(2))+((\left(\frac(4)(5) \right))^(2))= 1$, which means that there is sure to be an angle $\alpha $ for which these numbers are cosine and sine, respectively. Therefore, our equation will be rewritten as follows:

\[\begin(align)& \cos \alpha \sin x+\sin \alpha \cos x=1 \\& \sin \left(\alpha +x \right)=1 \\\end(align)\]

And this is already easily solved, after which it remains only to find out why is equal to the angle$\alpha $. How to find out, as well as how to choose the right number to divide both sides of the equation (in this simple example we divided by 5) - about this in today's video tutorial:

Today we will analyze the solution of trigonometric equations, or rather, one and only trick, which is called the “auxiliary angle method”. Why this particular method? Simply because over the past two or three days, when I was working with students, whom I talked about solving trigonometric equations, and we analyzed, among other things, the auxiliary angle method, and all the students as one make the same mistake. But the method is generally simple and, moreover, it is one of the main techniques in trigonometry. Therefore, many trigonometric problems otherwise than by the auxiliary angle method they are not solved at all.

Therefore, now, for a start, we will consider a couple of simple tasks, and then we will move on to more serious tasks. However, all of these, one way or another, will require us to use the auxiliary angle method, the essence of which I will tell in the first construction.

Solving simple trigonometric problems

Example #1

\[\cos 2x=\sqrt(3)\sin 2x-1\]

Let's change our expression a bit:

\[\cos 2x-\sqrt(3)\sin 2x=-1\left| \left(-1 \right) \right.\]

\[\sqrt(3)\cdot \sin 2x-\cos 2x=1\]

How are we going to solve it? Standard reception is to expand $\sin 2x$ and $\cos 2x$ using the double angle formulas, and then rewrite the unit as $((\sin )^(2))x((\cos )^(2))x $, get homogeneous equation, bring it to tangents and solve. However, this is a long and tedious path that requires a lot of calculations.

I suggest you think about this. We have $\sin$ and $\cos$. Recall the formula for the cosine and sine of the sum and difference:

\[\sin \left(\alpha \pm \beta \right)=\sin \alpha \cos \beta \pm \cos \alpha \sin \beta \]

\[\cos \left(\alpha +\beta \right)=\cos \alpha \cos \beta -\sin \alpha \sin \beta \]

\[\cos \left(\alpha -\beta \right)=\cos a\cos \beta +\sin \alpha \sin \beta \]

Let's go back to our example. Let's reduce everything to the sine of the difference. But first, the equation needs to be slightly transformed. Let's find the coefficient:

$\sqrt(l)$ is the same factor by which both parts of the equation must be divided so that numbers appear in front of the sine and cosine, which themselves are sines and cosines. Let's split:

\[\frac(\sqrt(3))(2)\cdot \sin 2x-\frac(1)(2)\cdot \cos 2x=\frac(1)(2)\]

Let's look at what we got on the left: is there such $\sin $ and $\cos $ such that $\cos \alpha =\frac(\sqrt(3))(2)$, and $\sin \alpha =\frac(1)(2)$? Obviously there is: $\alpha =\frac(\text( )\!\!\pi\!\!\text( ))(6)$. Therefore, we can rewrite our expression as follows:

\[\cos \frac(\text( )\!\!\pi\!\!\text( ))(\text(6))\cdot \sin 2x-\sin \frac(\text( )\! \!\pi\!\!\text( ))(\text(6))\cdot \cos 2x=\frac(1)(2)\]

\[\sin 2x\cdot \cos \frac(\text( )\!\!\pi\!\!\text( ))(\text(6))-\cos 2x\cdot \sin \frac(\ text( )\!\!\pi\!\!\text( ))(\text(6))=\frac(1)(2)\]

Now we have the formula for the sine of the difference. We can write like this:

\[\sin \left(2x-\frac(\text( )\!\!\pi\!\!\text( ))(\text(6)) \right)=\frac(1)(2) \]

Before us is the simplest classical trigonometric construction. Let me remind you:

This is what we write for our specific expression:

\[\left[ \begin(align)& 2x-\frac(\text( )\!\!\pi\!\!\text( ))(6)=\frac(\text( )\!\! \pi\!\!\text( ))(6)=2\text( )\!\!\pi\!\!\text( )n \\& 2x-\frac(\text( )\!\ !\pi\!\!\text( ))(\text(6))=\text( )\!\!\pi\!\!\text( )-\frac(\text( )\!\! \pi\!\!\text( ))(\text(6))+2\text( )\!\!\pi\!\!\text( )n \\\end(align) \right.\ ]

\[\left[ \begin(align)& 2x=\frac(\text( )\!\!\pi\!\!\text( ))(3)+2\text( )\!\!\pi \!\!\text( )n \\& 2x=\text( )\!\!\pi\!\!\text( )+2\text( )\!\!\pi\!\!\text ( )n \\\end(align) \right.\]

\[\left[ \begin(align)& x=\frac(\text( )\!\!\pi\!\!\text( ))(6)+\text( )\!\!\pi\ !\!\text( )n \\& x=\frac(\text( )\!\!\pi\!\!\text( ))(2)+\text( )\!\!\pi\ !\!\text( )n \\\end(align) \right.\]

Nuances of the solution

So, what should you do if you come across a similar example:

1. Modify the design if necessary.
2. Find the correction factor, take the root from it and divide both parts of the example by it.
3. We look at what values ​​\u200b\u200bof the sine and cosine are obtained from numbers.
4. We decompose the equation according to the formulas of the sine or cosine of the difference or sum.
5. We solve the simplest trigonometric equation.

In this regard, attentive students are likely to have two questions.

What prevents us from writing $\sin $ and $\cos $ at the stage of finding the correction factor? — We are hindered by the basic trigonometric identity. The fact is that the resulting $\sin $ and $\cos $, like any other with the same argument, should add up to exactly “one” when squaring. In the process of solving, you need to be very careful not to lose the “deuce” in front of the “X”.

The auxiliary angle method is a tool that helps reduce an "ugly" equation to a perfectly adequate and "beautiful" one.

Example #2

\[\sqrt(3)\sin 2x+2((\sin )^(2))x-1=2\cos x\]

We see that we have $((\sin )^(2))x$, so let's use the reduction calculations. However, before using them, let's get them out. To do this, remember how to find the cosine of a double angle:

\[\cos 2x=((\cos )^(2))x-((\sin )^(2))x=2((\cos )^(2))x-1=1-2(( \sin )^(2))x\]

If we write $\cos 2x$ in the third variant, we get:

\[\cos 2x=1-2((\sin )^(2))x\]

\[((\sin )^(2))x=\frac(1-((\cos )^(2))x)(x)\]

I will write separately:

\[((\sin )^(2))x=\frac(1-\cos 2x)(2)\]

The same can be done for $((\cos )^(2))x$:

\[((\cos )^(2))x=\frac(1+\cos 2x)(2)\]

We need only the first calculations. Let's get to work on the task:

\[\sqrt(3)\cdot \sin 2x+2\cdot \frac(1-\cos 2x)(2)-1=2\cos x\]

\[\sqrt(3)\cdot \sin 2x+1-\cos 2x-1=2\cos x\]

\[\sqrt(3)\cdot \sin 2x-\cos 2x=2\cos x\]

Now we use the calculations of the cosine of the difference. But first, let's calculate the correction $l$:

Let's rewrite with this fact in mind:

\[\frac(\sqrt(3))(2)\cdot \sin 2x-\frac(1)(2)\cdot \cos 2x=\cos x\]

In this case, we can write that $\frac(\sqrt(3))(2)=\frac(\text( )\!\!\pi\!\!\text( ))(3)$, and $\frac(1)(2)=\cos \frac(\text( )\!\!\pi\!\!\text( ))(3)$. Let's rewrite:

\[\sin \frac(\text( )\!\!\pi\!\!\text( ))(\text(3))\cdot \sin 2x-\cos \frac(\text( )\! \!\pi\!\!\text( ))(\text(3))\cdot \cos 2x=\cos x\]

\[-\cos \left(\frac(\text( )\!\!\pi\!\!\text( ))(\text(3))+2x \right)=\cos x\]

Let's put the "minus" in the bracket in a tricky way. To do this, note the following:

\[\cos \left(\frac(\text( )\!\!\pi\!\!\text( ))(\text(3))+2x \right)=\cos \left(\text( )\!\!\pi\!\!\text( )-\text( )\!\!\pi\!\!\text( +)\frac(\text( )\!\!\pi\! \!\text( ))(\text(3))+2x \right)=\]

\[=\cos \left(\text( )\!\!\pi\!\!\text( )-\frac(2\text( )\!\!\pi\!\!\text( )) (3)+2x \right)=\cos \left(\text( )\!\!\pi\!\!\text( )+\varphi \right)=-\cos \varphi \]

We return to our expression and remember that in the role of $\varphi $ we have the expression $-\frac(2\text( )\!\!\pi\!\!\text( ))(3)+2x$. Therefore, we write:

\[-\left(-\cos \left(-\frac(2\text( )\!\!\pi\!\!\text( ))(3)+2x \right) \right)=\cos x\]

\[\cos \left(2x-\frac(2\text( )\!\!\pi\!\!\text( ))(3) \right)=\cos x\]

To solve a similar problem, you need to remember the following:

\[\cos \alpha =\cos \beta \]

\[\left[ \begin(align)& \alpha =\beta +2\text( )\!\!\pi\!\!\text( )n \\& \alpha =-\beta +2\text ( )\!\!\pi\!\!\text( )n \\\end(align) \right.\]

Let's take a look at our example:

\[\left[ \begin(align)& 2x-\frac(2\text( )\!\!\pi\!\!\text( ))(3)=x+2\text( )\!\ !\pi\!\!\text( )n \\& 2x-\frac(2\text( )\!\!\pi\!\!\text( ))(3)=-x+2\text ( )\!\!\pi\!\!\text( )n \\\end(align) \right.\]

Let's calculate each of these equations:

And the second:

Let's write the final answer:

\[\left[ \begin(align)& x=\frac(2\text( )\!\!\pi\!\!\text( ))(3)+2\text( )\!\!\ pi\!\!\text( )n \\& x=\frac(2\text( )\!\!\pi\!\!\text( ))(9)+\frac(2\text( ) \!\!\pi\!\!\text( )n)(3) \\\end(align) \right.\]

Nuances of the solution

In fact, this expression is solved in many different ways, however, it is the auxiliary angle method that is in this case optimal. In addition, using the example of this design, I would like to draw your attention to several more interesting tricks and facts:

• Degree reduction formulas. These formulas do not need to be memorized, but you need to know how to derive them, which I told you about today.
• Solution of equations of the form $\cos \alpha =\cos \beta $.
• Adding "zero".

But that's not all. Until now, $\sin$ and $\cos$, which we output as an additional argument, we thought they should be positive. Therefore, now we will solve more complex problems.

Analysis of more complex problems

Example #1

\[\sin 3x+4((\sin )^(3))x+4\cos x=5\]

Let's transform the first term:

\[\sin 3x=\sin \left(2x+x \right)=\sin 2x\cdot \cos x+\cos 2x\cdot \sin x\]

\[=2\left(1-\cos 2x \right)\cdot \sin x\]

And now we substitute all this into our original construction:

\[\sin 2x\cos x+\cos 2x\sin x+2\sin x-2\cos x\sin x+4\cos x=5\]

\[\sin 2x\cos x-\operatorname(cosx)-cos2\sin x+2\sin x+4\cos x=5\]

\[\sin \left(2x-x \right)+2\sin x+4\cos x=5\]

Let's introduce our correction:

We write down:

\[\frac(3)(5)\sin x+\frac(4)(5)\cos x=1\]

$\alpha $ such that $\sin $ or $\cos $ would be equal to $\frac(3)(5)$ and $\frac(4)(5)$ in trigonometric table no. Therefore, let's just write and reduce the expression to the sine of the sum:

\[\sin x\cdot \cos \varphi +\cos x\cdot \sin \varphi =1\]

\[\sin \left(x+\varphi \right)=1\]

it special case, the simplest trigonometric construction:

It remains to find what $\varphi $ is equal to. This is where many students go wrong. The fact is that two requirements are imposed on $\varphi $:

\[\left\( \begin(align)& \cos \varphi =\frac(3)(5) \\& \sin \varphi =\frac(4)(5) \\\end(align) \right .\]

Let's draw a radar and see where these values ​​occur:

Returning to our expression, we write the following:

But this entry can be improved a bit. Because we know the following:

\[\alpha:\arcsin \alpha +\arccos \alpha =\frac(\text( )\!\!\pi\!\!\text( ))(\text(2)),\]

then in our case we can write it like this:

Example #2

This will require an even deeper understanding of the methods of solving standard tasks no trigonometry. But to solve this example, we also use the auxiliary angle method.\[\]

The first thing that catches your eye is that there are no degrees higher than the first, and therefore nothing can be decomposed according to the expansion formulas of degrees. Uses inverses:

Why did I spread $5$. Look here:

Unit according to the main trigonometric identity we can write as $((\sin )^(2))x+((\cos )^(2))x$:

What gives us such a record? The fact is that in the first bracket there is an exact square. Let's roll it up and get:

I propose to introduce a new variable:

\[\sin x+\cos x=t\]

In this case, we get the expression:

\[((t)_(1))=\frac(5+1)(4)=\frac(3)(2)\]

\[((t)_(2))=\frac(5-1)(4)=1\]

In total we get:

\[\left[ \begin(align)& \sin x+\cos x=\frac(3)(2) \\& \sin x+\cos x=1 \\\end(align) \right.\]

Of course, knowledgeable students will now say that such constructions are easily solved by reduction to a homogeneous one. However, we will solve each equation using the auxiliary angle method. To do this, we first calculate the correction $l$:

\[\sqrt(l)=\sqrt(2)\]

Divide everything by $\sqrt(2)$:

\[\left[ \begin(align)& \frac(\sqrt(2))(2)\sin x+\frac(\sqrt(2))(2)\cos x=\frac(3)(2\ sqrt(2)) \\& \frac(\sqrt(2))(2)\sin x+\frac(\sqrt(2))(2)\cos x=\frac(\sqrt(2))(2 ) \\\end(align) \right.\]

Let's reduce everything to $\cos$:

\[\cos x\cdot \cos \frac(\text( )\!\!\pi\!\!\text( ))(4)+\sin x\sin \frac(\text( )\!\ !\pi\!\!\text( ))(\text(4))\]

\[\left[ \begin(align)& \cos \left(x-\frac(\text( )\!\!\pi\!\!\text( ))(\text(4)) \right) =\frac(3)(2\sqrt(2)) \\& \cos \left(x-\frac(\text( )\!\!\pi\!\!\text( ))(4) \ right)=\frac(\sqrt(2))(2) \\\end(align) \right.\]

Let's take a look at each of these expressions.

The first equation has no roots, and irrationality in the denominator will help us to prove this fact. Note the following:

\[\sqrt(2)<1,5\]

\[\frac(3)(2\sqrt(2))>\frac(3)(3\cdot 1,5)=\frac(3)(3)=1\]

In total, we have clearly proved that it is required that $\cos \left(x-\frac(\text( )\!\!\pi\!\!\text( ))(4) \right)$ be is equal to the number, which is greater than "one" and, therefore, this construction has no roots.

Let's deal with the second:

Let's solve this design:

In principle, you can leave the answer like this, or you can paint it:

Important Points

In conclusion, I would like to once again draw your attention to the work with "ugly" arguments, i.e. when $\sin$ and $\cos$ are not table values. The problem is that if we say that in our equation $\frac(3)(5)$ is $\cos $ and $\frac(4)(5)$ is $\sin $, then in the end, after we decide the design, we need to take into account both of these requirements. We get a system of two equations. If we do not take this into account, we get the following situation. In this case, we will get two points and in place of $\varphi $ we will have two numbers: $\arcsin \frac(4)(5)$ and $-\arcsin \frac(4)(5)$, but the last one in no way satisfied. The same will happen with the point $\frac(3)(5)$.

This problem only occurs when we are talking about "ugly" arguments. When we have table values, then there is nothing.

I hope today's lesson helped you understand what the auxiliary angle method is and how to apply it with examples. different levels difficulties. But this is not the only lesson devoted to solving problems using the auxiliary angle method. So stay with us!

Lesson summary for grades 10-11

Topic 1 : Auxiliary argument input method. Derivation of formulas.

Goals:

Formation of knowledge of a new method for solving tasks in trigonometry, in which its application is possible or necessary;

Formation of skills to analyze the condition of the problem, compare and find differences;

Development of thinking, logic and validity of statements, the ability to draw conclusions and generalize;

Development of speech, enrichment and complication vocabulary, mastering the expressive properties of the language by students;

Formation of attitude to the subject, enthusiasm for knowledge, creation of conditions for a creative non-standard approach to mastering knowledge.

Required knowledge, skills and abilities:

Be able to derive trigonometric formulas and use them in further work;

Be able to solve or have an idea of ​​how to solve trigonometric tasks;

Know basic trigonometric formulas.

The level of preparedness of students for conscious perception:

Equipment: AWP, presentation with task conditions, solutions and necessary formulas, cards with tasks and answers.

Lesson structure:

1. Setting the goal of the lesson (2

Preparation for the study of new material (12 min).

Acquaintance with new material (15 min).

Primary comprehension and application of what has been learned (10 min).

Setting homework (3 min).

Summing up the lesson (3 min).

During the classes.

1. Setting the goal of the lesson.

Check the readiness of students and equipment for the lesson. It is advisable to prepare in advance homework on the board to discuss the solution. Note that the purpose of the lesson is to expand knowledge about the methods for solving some tasks in trigonometry and try your hand at mastering them.

2. Preparation for the study of new material.

Discuss homework: remember the basic trigonometric formulas, the values ​​​​of trigonometric functions for the simplest arguments. Review the homework assignment.

Formulas:

; ;

; ;

A task: Express the expression as a product.

Students are likely to offer next solution:

Because they know the formulas for converting the sum of trigonometric functions into a product.

We propose another solution to the problem: . Here, when solving, the formula for the cosine of the difference of two arguments was used, where is auxiliary. Note that in each of these methods, other similar formulas could be used.

3. Acquaintance with new material.

The question arises, where did the auxiliary argument come from?

To get an answer, consider common decision problem, we transform the expression into a product, where and are arbitrary non-zero numbers.

we introduce an additional angle (auxiliary argument), where , , then our expression will take the form:

Thus, we got the formula: .

If the angle is entered according to the formulas, then the expression will take the form and we will get a different form of the formula: .

We have derived formulas for the additional angle, which are called formulas of the auxiliary argument:

Formulas can also have a different form (it is necessary to pay attention to this Special attention and show with examples).

Note that in the simplest cases, the method of introducing an auxiliary argument is reduced to replacing numbers; ; ; ; one; trigonometric functions corresponding corners.

4. Primary comprehension and application of what has been learned .

To consolidate the material, it is proposed to consider a few more examples of tasks:

Express as a product of the expression:

It is advisable to analyze tasks 3 and 4 in the class (the analysis of tasks is present in the materials for classes). Tasks 1, 2 and 5 can be taken for independent decision(answers given).

To analyze the features of the conditions of typical tasks in which the considered solution method can be used, various methods can be used. Note that task 1. can be performed in various ways, and to complete tasks 2 - 5 it is more convenient to use the method of introducing an auxiliary angle

In the course of a frontal conversation, it should be discussed how these tasks are similar to the example considered at the beginning of the lesson, what are the differences, whether the proposed method can be applied to solve them and why its use is more convenient.

Similarity: in all the proposed examples, it is possible to apply the method of introducing an auxiliary argument, and this is a more convenient method that leads immediately to the result.

Difference: in the first example, a different approach is possible, and in all the others, a method of applying an auxiliary argument using not one, but several formulas is possible.

After discussing the tasks, you can invite the guys to solve the rest on their own at home.

5. Statement of homework.

At home, you are invited to carefully study the summary of the lesson and try to solve the following exercises.

Lesson topic: A method for introducing an auxiliary angle in solving trigonometric equations.

Actualization.

Teacher.

Guys! We got acquainted with various types of trigonometric equations and learned how to solve them. Today we will generalize the knowledge of methods for solving trigonometric equations various kinds. To do this, I ask you to work on the classification of the equations proposed to you (see equations No. 1-10 in the Appendix - at the end of the abstract in PDF form)

Fill in the table: indicate the type of equation, the method for solving it and match the numbers of the equations to the type to which they belong.

Students. Fill out the table.

Type of equation	Solution method	Equations
Protozoa	Root formulas	№1
Reducible to square	Variable replacement method	№2,3
Complex trigonometric view	Simplify to known form using trigonometry formulas	№4,5
Homogeneous first degree	Divide the equation term by term by the cosine of the variable	№6
Homogeneous second degree	Divide the equation term by term by the square of the cosine of the variable	№7

Problematization.

Filling in the table, students are faced with a problem. They cannot determine the type and method of solving three equations: No. 8,9,10.

Teacher. Have you managed to classify all equations according to the form and method of solution?

Students response. No, three equations could not be placed in the table.

Teacher. Why?

Students response. They don't look like known species. The solution method is not clear.

Goal setting.

Teacher. How, then, shall we formulate the purpose of our lesson?

Answer students. Define Discovered new type equations and find a method for solving them.

Teacher. Is it possible to formulate the topic of the lesson if we do not know the type of the discovered equations and the method for solving them?

Student response. No, but we can do it later, when we figure out what we are dealing with.

Activity planning.

Teacher. Let's plan our activities. Usually we define the type and then look for a method for solving trigonometric equations. In our current situation, is it possible to give a specific name to the type of equations discovered? And in general, do they belong to the same species?

Students response. It's hard to do.

Teacher. Then think, maybe something unites them, or are they similar to some type?

Students response. The left side of these equations is the same as the homogeneous ones, but their right side is not equal to zero. So, dividing by cosine will only complicate the solution.

Teacher. Maybe we'll start by looking for a solution method, and then determine the type of the equation? Which of the 3 equations do you think is the simplest?

Students answer but there is no consensus. Perhaps someone will guess that the coefficients in equation No. 8 should be expressed as the sine and cosine of the table angle. And then the class will determine the equation that can be solved first. If not, the teacher suggests considering additional equation (see equation No. 11 in the Appendix - at the end of the abstract in PDF form). In it, the coefficients are equal to the sine and cosine of a known angle, and students should notice this.

The teacher gives the order of activities. ( See equations in the Appendix - in PDF form at the end of the abstract).

1. Solve the first equation (№11), by replacing the coefficients with the values ​​of the sine and cosine of the known angle and applying the formula for the sine of the sum.
2. Try to convert other equations to the form of the first one and apply the same method. ( see Equation #8,9,12)
3. Generalize and extend the method to any coefficients and construct a general algorithm of actions (see Equation #10).
4. Apply the method to solve other equations of the same type. (see Equations Nos. 12,13, 14).

Implementation of the plan.

Teacher. Well, we've made a plan. Let's start implementing it.

At the blackboard, the student solves equation No. 11.

The second student solves the following equation No. 8, after dividing it by constant number and, thereby, reducing the situation to an already found solution.

The teacher suggests solving equations No. 9.12 on their own. Checks the correctness of the transformations and the set of solutions.

Teacher. Guys, how can you call the angle that appears instead of the coefficients of the equation and helps us reach a solution?

Students response. Additional. (Option: auxiliary).

Teacher. It is not always easy to find such an auxiliary angle. Is it possible to find it if the coefficients are not sine and cosine known corners? What identity must such coefficients satisfy if we want to represent them as the sine and cosine of the auxiliary angle?

Answer. Basic trigonometric identity.

Teacher. Well done! Correctly! So our task is to obtain coefficients such that the sum of their squares is equal to one! Try to come up with a number by which you need to divide the equation so that the condition we indicated is satisfied.

Students think and, perhaps, offer to divide everything by the square root of the sum of the squares of the coefficients of the equation. If not, the teacher leads them to this idea.

Teacher. It remains for us to choose which of the new coefficients to designate as the sine of the auxiliary angle, and which as the cosine. There are two options. The transition to the simplest equation with a sine or a cosine depends on the choice.

Students they offer a solution, and the teacher completes it, paying attention to the form of recording the reasoning and the answer. Solve Equation 10.

Teacher. Have we discovered a method for solving a new type of equation? What do we call this type?

Answer. We worked by the method of finding an auxiliary angle. Maybe the equations should be called equations that are solved using auxiliary angles?

Teacher. Yes, you certainly may. Can you think of a formula for them? It will be shorter.

Answer. Yes. Equations with coefficients A, B and C.

Teacher. Let's generalize the method for arbitrary coefficients.

The teacher discusses and writes on the board the formulas for the sine and cosine of the auxiliary angle for generalized coefficients. Then, with their help, he solves equations No. 13 and 14.

Teacher. Have we mastered the method well enough?

Answer. No. It is necessary to solve such equations and consolidate the ability to use the auxiliary angle method.

Teacher. How do we know that the method has been mastered?

Answer. If we solve several equations ourselves.

Teacher. Let's establish a qualitative scale for mastering the method.

Get acquainted with the characteristics of the levels and place them on a scale that reflects the level of mastery of this skill. Correlate the characteristic of the level and the score (from 0 to 3)

• I can solve equations with different coefficients
• Can't solve equations
• I can solve complex equations
• I can solve equations with tabular coefficients

Teacher.(After students answer) So, our rating scale is as follows:

By the same principle, we estimate independent work topic in the next lesson.

And now, please solve equations No. 1148 g, 1149 g, 1150 g and determine your level of assimilation of the topic.

Do not forget to complete the entries in the table and name the topic: "Introduction of an auxiliary angle when solving trigonometric equations."

Reflection of the way to achieve the goal.

Teacher. Guys, have we reached the goal of the lesson?

Student responses. Yes, we have learned to recognize a new type of equation.

We found a method for solving them using an auxiliary angle.

Learned to apply the method in practice.

Teacher. How did we act? How did we come to understand what we need to do?

Answer. We considered several special cases of equations with "recognizable" coefficients and extended this logic to any values ​​of A, B and C.

Teacher. This is an inductive way of thinking: we have derived a method from several cases and applied it in similar cases.

Perspective. Where can we apply this way of thinking? (student answers)

You did a good job today in class. At home, read the description of the auxiliary angle method in the textbook and solve No. 1148 (a, b, c), 1149 (a, b, c), 1150 (a, b, c). I hope that in the next lesson you will all be great at using this method when solving trigonometric equations.

Thanks for the lesson!