How to solve a quadratic trigonometric equation. The simplest trigonometric equations
Line UMK G.K. Muravina. Algebra and beginnings mathematical analysis(10-11) (deep)
Line UMK G.K. Muravina, K.S. Muravina, O.V. Muravina. Algebra and the beginnings of mathematical analysis (10-11) (basic)
How to teach to solve trigonometric equations and inequalities: teaching methodology
The Russian Textbook corporation mathematics course, authored by Georgy Muravin and Olga Muravina, provides for a gradual transition to solving trigonometric equations and inequalities in grade 10, as well as continuing their study in grade 11. We present to your attention the stages of the transition to the topic with excerpts from the textbook "Algebra and the beginning of mathematical analysis" (advanced level).
1. Sine and cosine of any angle (propaedeutics for the study of trigonometric equations)
Job example. Find approximately the angles whose cosines are equal to 0.8.
Solution. The cosine is the abscissa of the corresponding point on the unit circle. All points with abscissas equal to 0.8 belong to a straight line parallel to the y-axis and passing through the point C(0.8; 0). This line intersects the unit circle at two points: P α ° and P β ° , symmetrical about the x-axis.
Using a protractor, we find that the angle α° approximately 37°. Means, general form rotation angles with end point P α°:
α° ≈ 37° + 360° n, where n- any integer.
By virtue of symmetry about the abscissa axis, the point P β ° - end point of rotation by an angle of –37°. So, for her the general form of the angles of rotation:
β° ≈ –37° + 360° n, where n- any integer.
Answer: 37° + 360° n, –37° + 360° n, where n- any integer.
Job example. Find the angles whose sines are equal to 0.5.
Solution. The sine is the ordinate of the corresponding point on the unit circle. All points with ordinates equal to 0.5 belong to a straight line parallel to the x-axis and passing through the point D(0; 0,5).
This line intersects the unit circle at two points: Pφ and Pπ–φ , symmetrical about the y-axis. In a right triangle OKPφ leg KPφ is half the hypotenuse OPφ , means,
General view of rotation angles with end point P φ :
where n- any integer. General view of rotation angles with end point P π–φ :
where n- any integer.
Answer: where n- any integer.
2. Tangent and cotangent of any angle (propaedeutics for the study of trigonometric equations)
Example 2
Job example. Find the general form of angles whose tangent is -1.2.
Solution. Mark a point on the tangent axis C with the ordinate equal to -1.2, and draw a straight line OC. Straight OC intersects the unit circle at points P α ° and Pβ° - ends of the same diameter. The angles corresponding to these points differ from each other by an integer number of half-turns, i.e. 180° n (n is an integer). Using a protractor, we find that the angle P α° OP 0 is -50°. This means that the general form of the angles, the tangent of which is -1.2, is as follows: -50 ° + 180 ° n (n- integer)
Answer:-50° + 180° n, n∈ Z.
Using the sine and cosine of the angles 30°, 45° and 60°, it is easy to find their tangents and cotangents. For example,
The listed angles are quite common in different problems, so it is useful to remember the values of the tangent and cotangent of these angles.
3. The simplest trigonometric equations
Designations are introduced: arcsin α, arccos α, arctg α, arcctg α. It is not recommended to hurry with the introduction of the combined formula. Two series of roots are much more convenient to write down, especially when it is necessary to select roots over an interval.
When studying the topic "the simplest trigonometric equations", the equations are most often reduced to squares.
4. Reduction formulas
The reduction formulas are identities, i.e. they are true for any allowed values φ . Analyzing the resulting table, you can see that:
1) the sign on the right side of the formula coincides with the sign of the reducible function in the corresponding quarter, if we assume φ sharp corner;
2) the name is changed only by the functions of the angles and
φ + 2π n |
||||
5. Properties and Graph of a Function y= sin x
The simplest trigonometric inequalities are solved either on a graph or on a circle. When solving a trigonometric inequality on a circle, it is important not to confuse which point to indicate first.
6. Properties and function graph y= cos x
The task of plotting a function graph y= cos x can be reduced to constructing a graph of the function y= sin x. Indeed, since function graph y= cos x can be obtained from the graph of the function y= sin x shift of the latter along the x-axis to the left by
7. Properties and graphs of functions y=tg x and y=ctg x
Function scope y=tg x includes all numbers except numbers of the form where n ∈ Z. As with the construction of a sinusoid, first we will try to get a graph of the function y = tg x in between
At the left end of this interval, the tangent is zero, and when approaching the right end, the values of the tangent increase indefinitely. Graphically, it looks like the graph of a function y =tg x clings to the straight line, leaving with it unlimitedly upwards.
8. Relationships between trigonometric functions of the same argument
Equality and express relations between trigonometric functions of the same argument φ. With their help, knowing the sine and cosine of a certain angle, you can find its tangent and cotangent. From these equalities it is easy to get that the tangent and cotangent are related by the following equality.
tan φ ctg φ = 1
There are other dependencies between trigonometric functions.
Unit circle equation centered at the origin x2 + y2= 1 connects the abscissa and ordinate of any point of this circle.
Basic trigonometric identity
cos 2 φ + sin 2 φ = 1
9. Sine and cosine of the sum and difference of two angles
Sum cosine formula
cos (α + β) = cos α cos β – sin α sin β
Difference cosine formula
cos (α - β) = cos α cos β + sin α sin β
Difference sine formula
sin (α - β) = sin α cos β - cos α sin β
The sine formula of the sum
sin (α + β) = sin α cos β + cos α sin β
10. Tangent of the sum and tangent of the difference of two angles
Sum tangent formula
Difference tangent formula
The textbook is included in the teaching materials for mathematics for grades 10–11, studying the subject in basic level. The theoretical material is divided into mandatory and optional, the system of tasks is differentiated by the level of complexity, each paragraph of the chapter ends control questions and assignments, and each chapter is a home control work. The textbook includes project topics and links to Internet resources.
11. Trigonometric functions of a double angle
Double angle tangent formula
cos2α = 1 – 2sin 2 α cos2α = 2cos 2 α – 1
Job example. solve the equation
Solution.
13. Solution of trigonometric equations
In most cases, the initial equation in the process of solution is reduced to the simplest trigonometric equations. However, there is no single solution method for trigonometric equations. In each case, success depends on knowledge trigonometric formulas and from the ability to choose the right ones. At the same time, the abundance of different formulas sometimes makes this choice rather difficult.
Equations that reduce to squares
Job example. Solve equation 2 cos 2 x+ 3 sin x = 0
Solution. With the help of the main trigonometric identity this equation can be reduced to a quadratic one with respect to sin x:
2cos 2 x+3sin x= 0, 2(1 - sin 2 x) + 3sin x = 0,
2-2sin2 x+3sin x= 0.2sin2 x– 3sin x – 2 = 0
Let's introduce a new variable y= sin x, then the equation will take the form: 2 y 2 – 3y – 2 = 0.
The roots of this equation y 1 = 2, y 2 = –0,5.
Back to Variable x and we get the simplest trigonometric equations:
1) sin x= 2 - this equation has no roots, since sin x < 2 при любом значении x;
2) sin x = –0,5,
Answer:
Homogeneous trigonometric equations
Job example. Solve the equation 2sin 2 x– 3sin x cos x– 5 cos 2 x = 0.
Solution. Consider two cases:
1) cos x= 0 and 2) cos x ≠ 0.
Case 1. If cos x= 0, then the equation takes the form 2sin 2 x= 0, whence sin x= 0. But this equality does not satisfy the condition cos x= 0, because for no x cosine and sine do not vanish at the same time.
Case 2. If cos x≠ 0, then we can divide the equation by cos 2 x “Algebra and the beginning of mathematical analysis. Grade 10 ”, like many other publications, is available on the LECTA platform. To do this, use the offer.
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The simplest trigonometric equations are the equations
Cos(x)=a, sin(x)=a, tg(x)=a, ctg(x)=a
Equation cos(x) = a
Explanation and rationale
- The roots of the equation cosx = a. When | a | > 1 the equation has no roots because | cosx |< 1 для любого x (прямая y = а при а >1 or at a< -1 не пересекает график функцииy = cosx).
Let | a |< 1. Тогда прямая у = а пересекает график функции
y = cos x. On the interval, the function y = cos x decreases from 1 to -1. But a decreasing function takes each of its values only at one point of its domain of definition, therefore the equation cos x \u003d a has only one root on this interval, which, by definition of the arc cosine, is: x 1 \u003d arccos a (and for this root cos x \u003d a).
Cosine - even function, so on the interval [-p; 0] the equation cos x = and also has only one root - the number opposite to x 1, that is
x 2 = -arccos a.
Thus, on the interval [-n; n] (length 2n) the equation cos x = a for | a |< 1 имеет только корни x = ±arccos а.
The function y = cos x is periodic with a period of 2n, so all other roots differ from those found by 2np (n € Z). We get the following formula for the roots of the equation cos x = a when
x = ± arccos a + 2n, n £ Z.
- Particular cases of solving the equation cosx = a.
It is useful to remember the special notation for the roots of the equation cos x = a when
a \u003d 0, a \u003d -1, a \u003d 1, which can be easily obtained using the unit circle as a guide.
Since the cosine is equal to the abscissa of the corresponding point on the unit circle, we get that cos x = 0 if and only if the corresponding point on the unit circle is point A or point B.
Similarly, cos x = 1 if and only if the corresponding point of the unit circle is the point C, therefore,
x = 2πp, k € Z.
Also cos x \u003d -1 if and only if the corresponding point of the unit circle is the point D, thus x \u003d n + 2n,
Equation sin(x) = a
Explanation and rationale
- The roots of the equation sinx = a. When | a | > 1 the equation has no roots because | sinx |< 1 для любого x (прямая y = а на рисунке при а >1 or at a< -1 не пересекает график функции y = sinx).
The simplest trigonometric equations are usually solved by formulas. Let me remind you that the following trigonometric equations are called the simplest:
sinx = a
cosx = a
tgx = a
ctgx = a
x is the angle to be found,
a is any number.
And here are the formulas with which you can immediately write down the solutions of these simplest equations.
For sinus:
For cosine:
x = ± arccos a + 2π n, n ∈ Z
For tangent:
x = arctg a + π n, n ∈ Z
For cotangent:
x = arcctg a + π n, n ∈ Z
Actually, this is the theoretical part of solving the simplest trigonometric equations. And, the whole!) Nothing at all. However, the number of errors on this topic just rolls over. Especially, with a slight deviation of the example from the template. Why?
Yes, because a lot of people write down these letters, without understanding their meaning at all! With apprehension he writes down, no matter how something happens ...) This needs to be dealt with. Trigonometry for people, or people for trigonometry, after all!?)
Let's figure it out?
One angle will be equal to arccos a, second: -arccos a.
And that's how it will always work. For any a.
If you don't believe me, hover your mouse over the picture, or touch the picture on the tablet.) I changed the number a to some negative. Anyway, we got one corner arccos a, second: -arccos a.
Therefore, the answer can always be written as two series of roots:
x 1 = arccos a + 2π n, n ∈ Z
x 2 = - arccos a + 2π n, n ∈ Z
We combine these two series into one:
x= ± arccos a + 2π n, n ∈ Z
And all things. We have obtained a general formula for solving the simplest trigonometric equation with cosine.
If you understand that this is not some kind of super-scientific wisdom, but just an abbreviated record of two series of answers, you and tasks "C" will be on the shoulder. With inequalities, with the selection of roots from a given interval ... There, the answer with plus / minus does not roll. And if you treat the answer businesslike, and break it into two separate answers, everything is decided.) Actually, for this we understand. What, how and where.
In the simplest trigonometric equation
sinx = a
also get two series of roots. Is always. And these two series can also be recorded one line. Only this line will be smarter:
x = (-1) n arcsin a + π n, n ∈ Z
But the essence remains the same. Mathematicians simply constructed a formula to make one instead of two records of series of roots. And that's it!
Let's check the mathematicians? And that's not enough...)
In the previous lesson, the solution (without any formulas) of the trigonometric equation with a sine was analyzed in detail:
The answer turned out to be two series of roots:
x 1 = π /6 + 2π n, n ∈ Z
x 2 = 5π /6 + 2π n, n ∈ Z
If we solve the same equation using the formula, we get the answer:
x = (-1) n arcsin 0.5 + π n, n ∈ Z
Actually, this is a half-finished answer.) The student must know that arcsin 0.5 = π /6. The full answer would be:
x = (-1) n π /6+ πn, n ∈ Z
Here an interesting question arises. Reply via x 1; x 2 (this is the correct answer!) and through the lonely X (and this is the correct answer!) - the same thing, or not? Let's find out now.)
Substitute in response with x 1 values n =0; one; 2; etc., we consider, we get a series of roots:
x 1 \u003d π / 6; 13π/6; 25π/6 and so on.
With the same substitution in response to x 2 , we get:
x 2 \u003d 5π / 6; 17π/6; 29π/6 and so on.
And now we substitute the values n (0; 1; 2; 3; 4...) into the general formula for the lonely X . That is, we raise minus one to the zero power, then to the first, second, and so on. And, of course, we substitute 0 into the second term; one; 2 3; 4 etc. And we think. We get a series:
x = π/6; 5π/6; 13π/6; 17π/6; 25π/6 and so on.
That's all you can see.) The general formula gives us exactly the same results which are the two answers separately. All at once, in order. Mathematicians did not deceive.)
Formulas for solving trigonometric equations with tangent and cotangent can also be checked. But let's not.) They are so unpretentious.
I painted all this substitution and verification on purpose. It is important to understand one simple thing here: there are formulas for solving elementary trigonometric equations, just a summary of the answers. For this brevity, I had to insert plus/minus into the cosine solution and (-1) n into the sine solution.
These inserts do not interfere in any way in tasks where you just need to write down the answer to an elementary equation. But if you need to solve an inequality, or then you need to do something with the answer: select roots on an interval, check for ODZ, etc., these inserts can easily unsettle a person.
And what to do? Yes, either paint the answer in two series, or solve the equation / inequality in a trigonometric circle. Then these inserts disappear and life becomes easier.)
You can sum up.
To solve the simplest trigonometric equations, there are ready-made answer formulas. Four pieces. They are good for instantly writing the solution to an equation. For example, you need to solve the equations:
sinx = 0.3
Easily: x = (-1) n arcsin 0.3 + π n, n ∈ Z
cosx = 0.2
No problem: x = ± arccos 0.2 + 2π n, n ∈ Z
tgx = 1.2
Easily: x = arctg 1,2 + πn, n ∈ Z
ctgx = 3.7
One left: x= arcctg3,7 + πn, n ∈ Z
cos x = 1.8
If you, shining with knowledge, instantly write the answer:
x= ± arccos 1.8 + 2π n, n ∈ Z
then you already shine, this ... that ... from a puddle.) The correct answer is: there are no solutions. Don't understand why? Read what an arccosine is. In addition, if on the right side of the original equation there are tabular values \u200b\u200bof sine, cosine, tangent, cotangent, - 1; 0; √3; 1/2; √3/2 etc. - the answer through the arches will be unfinished. Arches must be converted to radians.
And if you already come across an inequality, like
then the answer is:
x πn, n ∈ Z
there is a rare nonsense, yes ...) Here it is necessary to decide on a trigonometric circle. What we will do in the corresponding topic.
For those who heroically read up to these lines. I just can't help but appreciate your titanic efforts. you a bonus.)
Bonus:
When writing formulas in an anxious combat situation, even hardened nerds often get confused where pn, And where 2πn. Here's a simple trick for you. In all formulas pn. Except for the only formula with arc cosine. It stands there 2πn. Two pien. Keyword - two. In the same single formula are two sign at the beginning. Plus and minus. Here and there - two.
So if you wrote two sign in front of the arc cosine, it is easier to remember what will happen at the end two pien. And vice versa happens. Skip the man sign ± , get to the end, write correctly two pien, yes, and catch it. Ahead of something two sign! The person will return to the beginning, but he will correct the mistake! Like this.)
If you like this site...
By the way, I have a couple more interesting sites for you.)
You can practice solving examples and find out your level. Testing with instant verification. Learning - with interest!)
you can get acquainted with functions and derivatives.
Class: 10
"Equations will exist forever."
A. Einstein
Lesson Objectives:
- Educational:
- deepening understanding of methods for solving trigonometric equations;
- to form skills to distinguish, correctly select ways to solve trigonometric equations.
- Educational:
- education of cognitive interest in the educational process;
- formation of the ability to analyze the task;
- contribute to the improvement of the psychological climate in the classroom.
- Educational:
- to promote the development of the skill of self-acquisition of knowledge;
- encourage students to argue their point of view;
Equipment: poster with basic trigonometric formulas, computer, projector, screen.
1 lesson
I. Actualization of basic knowledge
Orally solve the equations:
1) cosx = 1;
2) 2 cosx = 1;
3) cosx = –;
4) sin2x = 0;
5) sinx = -;
6) sinx = ;
7) tgx = ;
8) cos 2 x - sin 2 x \u003d 0
1) x = 2k;
2) x = ± + 2k;
3) x = ± + 2k;
4) x = k;
5) x \u003d (-1) + k;
6) x \u003d (-1) + 2k;
7) x = + k;
8) x = + k; to Z.
II. Learning new material
- Today we will consider more complex trigonometric equations. Consider 10 ways to solve them. Then there will be two lessons to consolidate, and the next lesson will be a test. At the “To the Lesson” stand, tasks are posted that are similar to those that will be on the test work, they must be solved before the test work. (The day before, before the test work, hang out the solutions to these tasks on the stand).
So, we turn to the consideration of methods for solving trigonometric equations. Some of these methods will probably seem difficult to you, while others will be easy, because. you already know some methods of solving equations.
Four students in the class received an individual task: to understand and show you 4 ways to solve trigonometric equations.
(Speaking students prepared slides in advance. The rest of the students in the class write down the main steps in solving equations in a notebook.)
1 student: 1 way. Solving equations by factoring
sin 4x = 3 cos 2x
To solve the equation, we use the formula for the sine of a double angle sin 2 \u003d 2 sin cos
2 sin 2x cos 2x - 3 cos 2x = 0,
cos 2x (2 sin 2x - 3) = 0. The product of these factors is equal to zero if at least one of the factors is equal to zero.
2x = + k, k Z or sin 2x = 1.5 - no solutions, because | sin| one
x = + k; to Z.
Answer: x = + k, k Z.
2 student. 2 way. Solving equations by converting the sum or difference of trigonometric functions into a product
cos 3x + sin 2x - sin 4x = 0.
To solve the equation, we use the formula sin–sin = 2 sin cos
cos 3x + 2 sin cos = 0,
cos 3x - 2 sin x cos 3x \u003d 0,
cos 3x (1 - 2 sinx) = 0. The resulting equation is equivalent to the combination of two equations:
The set of solutions of the second equation is completely included in the set of solutions of the first equation. Means
Answer:
3 student. 3 way. Solving equations by converting the product of trigonometric functions into a sum
sin 5x cos 3x = sin 6x cos2x.
To solve the equation, we use the formula
Answer:
4 student. 4 way. Solving Equations Reducing to Quadratic Equations
3 sin x - 2 cos 2 x \u003d 0,
3 sin x - 2 (1 - sin 2 x) \u003d 0,
2 sin 2 x + 3 sin x - 2 = 0,
Let sin x = t, where | t |. Get quadratic equation 2t 2 + 3t - 2 = 0,
D = 9 + 16 = 25.
In this way . does not satisfy the condition | t |.
So sin x = . That's why .
Answer:
III. Consolidation of what was studied from the textbook by A. N. Kolmogorov
1. No. 164 (a), 167 (a) (quadratic equation)
2. No. 168 (a) (factorization)
3. No. 174 (a) (converting a sum to a product)
4. (convert product to sum)
(At the end of the lesson, show the solution of these equations on the screen for verification)
№ 164 (a)
2 sin 2 x + sin x - 1 = 0.
Let sin x = t, | t | 1. Then
2 t 2 + t - 1 = 0, t = - 1, t= . Where
Answer: - .
№ 167 (a)
3 tg 2 x + 2 tg x - 1 = 0.
Let tg x \u003d 1, then we get the equation 3 t 2 + 2 t - 1 \u003d 0.
Answer:
№ 168 (a)
Answer:
№ 174 (a)
Solve the equation:
Answer:
2 lesson (lesson-lecture)
IV. Learning new material(continuation)
- So, let's continue studying ways to solve trigonometric equations.
5 way. Solution of homogeneous trigonometric equations
Equations of the form a sin x + b cos x = 0, where a and b are some numbers, are called homogeneous equations of the first degree with respect to sin x or cos x.
Consider the equation
sin x – cos x = 0. Divide both sides of the equation by cos x. This can be done, the loss of the root will not occur, because. , if cos x = 0, then sin x = 0. But this contradicts the basic trigonometric identity sin 2 x + cos 2 x = 1.
Get tg x - 1 = 0.
tan x = 1,
Equations of the form a sin 2 x + bcos 2 x + c sin x cos x = 0 , where a, b, c some numbers are called homogeneous equations of the second degree with respect to sin x or cos x.
Consider the equation
sin 2 x - 3 sin x cos x + 2 cos 2 \u003d 0. We divide both parts of the equation by cos x, and the root will not be lost, because cos x = 0 is not the root of this equation.
tg 2x - 3tgx + 2 = 0.
Let tgx = t. D \u003d 9 - 8 \u003d 1.
Then Hence tg x = 2 or tg x = 1.
As a result x = arctg 2 + , x =
Answer: arctg 2 + ,
Consider another equation: 3 sin 2 x - 3 sin x cos x + 4 cos 2 x = 2.
We transform the right side of the equation in the form 2 = 2 1 = 2 (sin 2 x + cos 2 x). Then we get:
3sin 2 x – 3sin x cos x + 4cos 2 x = 2 (sin 2 x + cos 2 x),
3sin 2 x – 3sin x cos x + 4cos 2 x – 2sin 2 x – 2 cos 2 x = 0,
sin 2 x - 3sin x cos x + 2cos 2 x \u003d 0. (We got the 2nd equation, which we have already analyzed).
Answer: arctg 2 + k,
6 way. Solution of linear trigonometric equations
A linear trigonometric equation is an equation of the form a sin x + b cos x = c, where a, b, c are some numbers.
Consider the equation sin x + cos x= – 1.
Let's rewrite the equation in the form:
Considering that and, we get:
Answer:
7 way. Introduction of an additional argument
Expression a cos x + b sin x can be converted:
(we have already used this transformation when simplifying trigonometric expressions)
We introduce an additional argument - the angle is such that
Then
Consider the equation: 3 sinx + 4 cosx = 1. =
Homework: No. 164 -170 (c, d).
In this lesson, we'll look at main trigonometric functions, their properties and graphs, and also list main types of trigonometric equations and systems. In addition, we indicate general solutions of the simplest trigonometric equations and their special cases.
This lesson will help you prepare for one of the types of assignments. B5 and C1.
Preparation for the exam in mathematics
Experiment
Lesson 10 Trigonometric equations and their systems.
Theory
Lesson summary
We have already repeatedly used the term "trigonometric function". Back in the first lesson of this topic, we defined them with right triangle and the unit trigonometric circle. Using such methods of defining trigonometric functions, we can already conclude that for them one value of the argument (or angle) corresponds to exactly one value of the function, i.e. we have the right to call sine, cosine, tangent and cotangent exactly functions.
In this lesson, it's time to try to abstract from the previously discussed methods for calculating the values of trigonometric functions. Today we will move on to the usual algebraic approach to working with functions, we will consider their properties and draw graphs.
As for the properties of trigonometric functions, special attention should be paid to:
Domain of definition and range of values, since for sine and cosine there are restrictions on the range of values, and for tangent and cotangent there are restrictions on the range of definition;
The periodicity of all trigonometric functions, since we have already noted the presence of the smallest non-zero argument, the addition of which does not change the value of the function. Such an argument is called the period of the function and is denoted by the letter . For sine/cosine and tangent/cotangent, these periods are different.
Consider a function:
1) Domain of definition;
2) Range of values ;
3) The function is odd ;
Let's plot the function. In this case, it is convenient to start the construction from the image of the area, which limits the graph from above by the number 1 and from below by the number , which is associated with the range of the function. In addition, for plotting, it is useful to remember the values of the sines of several main table angles, for example, that This will allow you to build the first complete "wave" of the graph and then redraw it to the right and left, taking advantage of the fact that the picture will be repeated with an offset by a period, i.e. on the .
Now let's look at the function:
The main properties of this function:
1) Domain of definition;
2) Range of values ;
3) The function is even This implies the symmetry of the graph of the function with respect to the y-axis;
4) The function is not monotone throughout its domain of definition;
Let's plot the function. As in the construction of the sine, it is convenient to start with the image of the area that limits the graph from above by the number 1 and from below by the number , which is related to the range of the function. We will also plot the coordinates of several points on the graph, for which it is necessary to remember the cosine values of several main table angles, for example, using these points, we can build the first complete “wave” of the graph and then redraw it to the right and left, taking advantage of the fact that the picture will repeat with a period shift, i.e. on the .
Let's move on to the function:
The main properties of this function:
1) Domain of definition except for , where . We have already indicated in previous lessons that doesn't exist. This statement can be generalized by taking into account the period of the tangent;
2) The range of values, i.e. tangent values are not limited;
3) The function is odd ;
4) The function monotonically increases within its so-called tangent branches, which we will now see in the figure;
5) The function is periodic with a period
Let's plot the function. In this case, it is convenient to start the construction from the image of the vertical asymptotes of the graph at points that are not included in the domain of definition, i.e. etc. Next, we depict the branches of the tangent inside each of the strips formed by the asymptotes, pressing them to the left asymptote and to the right one. At the same time, do not forget that each branch is monotonically increasing. We depict all branches in the same way, because the function has a period equal to . This can be seen from the fact that each branch is obtained by shifting the neighboring one along the x-axis.
And we conclude with a look at the function:
The main properties of this function:
1) Domain of definition except for , where . According to the table of values of trigonometric functions, we already know that it does not exist. This statement can be generalized by taking into account the period of the cotangent;
2) The range of values, i.e. cotangent values are not limited;
3) The function is odd ;
4) The function monotonically decreases within its branches, which are similar to the tangent branches;
5) The function is periodic with a period
Let's plot the function. In this case, as for the tangent, it is convenient to start the construction from the image of the vertical asymptotes of the graph at points that are not included in the domain of definition, i.e. etc. Next, we depict the branches of the cotangent inside each of the strips formed by the asymptotes, pressing them to the left asymptote and to the right one. In this case, we take into account that each branch is monotonically decreasing. All branches, similarly to the tangent, are depicted in the same way, because the function has a period equal to .
Separately, it should be noted that trigonometric functions with a complex argument may have a non-standard period. It's about about view functions:
They have the same period. And about functions:
They have the same period.
As you can see, to calculate a new period, the standard period is simply divided by the factor in the argument. It does not depend on other modifications of the function.
You can understand and understand in more detail where these formulas come from in the lesson about constructing and converting function graphs.
We have come to one of the most important parts of the topic "Trigonometry", which we will devote to solving trigonometric equations. The ability to solve such equations is important, for example, when describing oscillatory processes in physics. Let's imagine that you have driven a few laps on a kart in a sports car, solving a trigonometric equation will help determine how long you have already been participating in the race, depending on the position of the car on the track.
Let's write down the simplest trigonometric equation:
The solution of such an equation is the arguments, the sine of which is equal to. But we already know that because of the periodicity of the sine, there are an infinite number of such arguments. Thus, the solution of this equation will be, etc. The same applies to solving any other simple trigonometric equation, there will be an infinite number of them.
Trigonometric equations are divided into several basic types. Separately, one should dwell on the simplest, because. all the rest are reduced to them. There are four such equations (according to the number of basic trigonometric functions). For them, common solutions are known, they must be remembered.
The simplest trigonometric equations and their general solutions look like this:
Please note that the sine and cosine values must take into account the limitations known to us. If, for example, , then the equation has no solutions and this formula should not be applied.
In addition, these root formulas contain a parameter in the form of an arbitrary integer . AT school curriculum this is the only case when the solution of an equation without a parameter contains a parameter. This arbitrary integer shows that it is possible to write down an infinite number of roots of any of the indicated equations simply by substituting all the integers in turn.
You can get acquainted with the detailed receipt of these formulas by repeating the chapter “Trigonometric Equations” in the 10th grade algebra program.
Separately, it is necessary to pay attention to the solution of particular cases of the simplest equations with sine and cosine. These equations look like:
Formulas for finding should not be applied to them. common solutions. Such equations are most conveniently solved using a trigonometric circle, which gives a simpler result than general solution formulas.
For example, the solution to the equation is . Try to get this answer yourself and solve the rest of the indicated equations.
In addition to the most common type of trigonometric equations indicated, there are several more standard ones. We list them, taking into account those that we have already indicated:
1) Protozoa, for example, ;
2) Particular cases of the simplest equations, for example, ;
3) Complex Argument Equations, for example, ;
4) Equations reduced to their simplest form by taking out a common factor, for example, ;
5) Equations reduced to their simplest form by transforming trigonometric functions, for example, ;
6) Equations Reducible to the Simplest by Substitution, for example, ;
7) Homogeneous equations, for example, ;
8) Equations that are solved using the properties of functions, for example, . Don't be intimidated by the fact that this equation has two variables, it is solved at the same time;
As well as equations that are solved using various methods.
In addition to solving trigonometric equations, it is necessary to be able to solve their systems.
The most common types of systems are:
1) In which one of the equations is a power law, for example, ;
2) Systems of simple trigonometric equations, for example, .
In today's lesson, we looked at the basic trigonometric functions, their properties and graphs. And also got acquainted with the general formulas for solving the simplest trigonometric equations, indicated the main types of such equations and their systems.
In the practical part of the lesson, we will analyze the methods for solving trigonometric equations and their systems.
Box 1.Solution of special cases of the simplest trigonometric equations.
As we said in the main part of the lesson, special cases of trigonometric equations with sine and cosine of the form:
have more simple solutions than give formulas for general solutions.
For this, a trigonometric circle is used. Let us analyze the method for solving them using the equation as an example.
Draw a point on a trigonometric circle at which the cosine value is zero, which is also the coordinate along the x-axis. As you can see, there are two such points. Our task is to indicate what the angle that corresponds to these points on the circle is.
We start counting from the positive direction of the abscissa axis (cosine axis) and, when postponing the angle, we get to the first point shown, i.e. one solution would be this angle value. But we are still satisfied with the angle that corresponds to the second point. How to get into it?