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An equality containing an unknown under the sign of a trigonometric function (`sin x, cos x, tan x` or` ctg x`) is called a trigonometric equation, and we will consider their formulas further.

The simplest equations are called `sin x = a, cos x = a, tg x = a, ctg x = a`, where` x` is the angle to be found, `a` is any number. Let's write down the root formulas for each of them.

1. Equation `sin x = a`.

For `| a |> 1` has no solutions.

For `| a | \ leq 1` has an infinite number of solutions.

Root formula: `x = (- 1) ^ n arcsin a + \ pi n, n \ in Z`

2. The equation `cos x = a`

For `| a |> 1` - as in the case with sine, solutions among real numbers does not have.

For `| a | \ leq 1` has an infinite number of solutions.

Root formula: `x = \ pm arccos a + 2 \ pi n, n \ in Z`

Special cases for sine and cosine in graphs.

3. The equation `tg x = a`

Has an infinite number of solutions for any values ​​of `a`.

Root formula: `x = arctan a + \ pi n, n \ in Z`

4. Equation `ctg x = a`

Also has an infinite number of solutions for any values ​​of `a`.

Root formula: `x = arcctg a + \ pi n, n \ in Z`

Formulas for roots of trigonometric equations in a table

For sine:
For cosine:
For tangent and cotangent:
Formulas for solving equations containing inverse trigonometric functions:

Methods for solving trigonometric equations

The solution to any trigonometric equation consists of two stages:

• using convert it to the simplest;
• solve the resulting simplest equation using the above written root formulas and tables.

Let's look at the examples of the main methods of solving.

Algebraic method.

In this method, variable replacement and substitution into equality is done.

Example. Solve the equation: `2cos ^ 2 (x + \ frac \ pi 6) -3sin (\ frac \ pi 3 - x) + 1 = 0`

`2cos ^ 2 (x + \ frac \ pi 6) -3cos (x + \ frac \ pi 6) + 1 = 0`,

we make the change: `cos (x + \ frac \ pi 6) = y`, then` 2y ^ 2-3y + 1 = 0`,

we find the roots: `y_1 = 1, y_2 = 1 / 2`, whence two cases follow:

1.` cos (x + \ frac \ pi 6) = 1`, `x + \ frac \ pi 6 = 2 \ pi n`,` x_1 = - \ frac \ pi 6 + 2 \ pi n`.

2.` cos (x + \ frac \ pi 6) = 1 / 2`, `x + \ frac \ pi 6 = \ pm arccos 1/2 + 2 \ pi n`,` x_2 = \ pm \ frac \ pi 3- \ frac \ pi 6 + 2 \ pi n`.

Answer: `x_1 = - \ frac \ pi 6 + 2 \ pi n`,` x_2 = \ pm \ frac \ pi 3- \ frac \ pi 6 + 2 \ pi n`.

Factorization.

Example. Solve the equation: `sin x + cos x = 1`.

Solution. Move all the terms of the equality to the left: `sin x + cos x-1 = 0`. Using, transform and factor the left side:

`sin x - 2sin ^ 2 x / 2 = 0`,

`2sin x / 2 cos x / 2-2sin ^ 2 x / 2 = 0`,

`2sin x / 2 (cos x / 2-sin x / 2) = 0`,

1. `sin x / 2 = 0`,` x / 2 = \ pi n`, `x_1 = 2 \ pi n`.
2. `cos x / 2-sin x / 2 = 0`,` tg x / 2 = 1`, `x / 2 = arctan 1+ \ pi n`,` x / 2 = \ pi / 4 + \ pi n` , `x_2 = \ pi / 2 + 2 \ pi n`.

Answer: `x_1 = 2 \ pi n`,` x_2 = \ pi / 2 + 2 \ pi n`.

Reduction to a homogeneous equation

First, you need to bring this trigonometric equation to one of two types:

`a sin x + b cos x = 0` (homogeneous equation of the first degree) or` a sin ^ 2 x + b sin x cos x + c cos ^ 2 x = 0` (homogeneous equation of the second degree).

Then divide both parts by `cos x \ ne 0` - for the first case, and by` cos ^ 2 x \ ne 0` - for the second. We obtain equations for `tg x`:` a tg x + b = 0` and `a tg ^ 2 x + b tg x + c = 0`, which need to be solved by known methods.

Example. Solve the equation: `2 sin ^ 2 x + sin x cos x - cos ^ 2 x = 1`.

Solution. Rewrite the right side as `1 = sin ^ 2 x + cos ^ 2 x`:

`2 sin ^ 2 x + sin x cos x - cos ^ 2 x =` `sin ^ 2 x + cos ^ 2 x`,

`2 sin ^ 2 x + sin x cos x - cos ^ 2 x -` `sin ^ 2 x - cos ^ 2 x = 0`

`sin ^ 2 x + sin x cos x - 2 cos ^ 2 x = 0`.

This is a homogeneous trigonometric equation of the second degree, we divide its left and right sides by `cos ^ 2 x \ ne 0`, we get:

`\ frac (sin ^ 2 x) (cos ^ 2 x) + \ frac (sin x cos x) (cos ^ 2 x) - \ frac (2 cos ^ 2 x) (cos ^ 2 x) = 0`

`tg ^ 2 x + tg x - 2 = 0`. We introduce the replacement `tg x = t`, as a result,` t ^ 2 + t - 2 = 0`. The roots of this equation are `t_1 = -2` and` t_2 = 1`. Then:

1. `tg x = -2`,` x_1 = arctg (-2) + \ pi n`, `n \ in Z`
2. `tg x = 1`,` x = arctan 1+ \ pi n`, `x_2 = \ pi / 4 + \ pi n`,` n \ in Z`.

Answer. `x_1 = arctg (-2) + \ pi n`,` n \ in Z`, `x_2 = \ pi / 4 + \ pi n`,` n \ in Z`.

Go to half corner

Example. Solve the equation: `11 sin x - 2 cos x = 10`.

Solution. Apply the double angle formulas, as a result: `22 sin (x / 2) cos (x / 2) -`` 2 cos ^ 2 x / 2 + 2 sin ^ 2 x / 2 =` `10 sin ^ 2 x / 2 +10 cos ^ 2 x / 2`

`4 tg ^ 2 x / 2 - 11 tg x / 2 + 6 = 0`

Applying the above algebraic method, we get:

1. `tg x / 2 = 2`,` x_1 = 2 arctan 2 + 2 \ pi n`, `n \ in Z`,
2. `tg x / 2 = 3 / 4`,` x_2 = arctan 3/4 + 2 \ pi n`, `n \ in Z`.

Answer. `x_1 = 2 arctan 2 + 2 \ pi n, n \ in Z`,` x_2 = arctan 3/4 + 2 \ pi n`, `n \ in Z`.

Introducing an auxiliary angle

In the trigonometric equation `a sin x + b cos x = c`, where a, b, c are coefficients, and x is a variable, we divide both sides by` sqrt (a ^ 2 + b ^ 2) `:

`\ frac a (sqrt (a ^ 2 + b ^ 2)) sin x +` `\ frac b (sqrt (a ^ 2 + b ^ 2)) cos x = '' \ frac c (sqrt (a ^ 2 + b ^ 2)) `.

The coefficients on the left side have the properties of sine and cosine, namely, the sum of their squares is equal to 1 and their absolute values ​​are not greater than 1. We denote them as follows: `\ frac a (sqrt (a ^ 2 + b ^ 2)) = cos \ varphi` , `\ frac b (sqrt (a ^ 2 + b ^ 2)) = sin \ varphi`,` \ frac c (sqrt (a ^ 2 + b ^ 2)) = C`, then:

`cos \ varphi sin x + sin \ varphi cos x = C`.

Let's take a closer look at the following example:

Example. Solve the equation: `3 sin x + 4 cos x = 2`.

Solution. Divide both sides of the equality by `sqrt (3 ^ 2 + 4 ^ 2)`, we get:

`\ frac (3 sin x) (sqrt (3 ^ 2 + 4 ^ 2)) +` `\ frac (4 cos x) (sqrt (3 ^ 2 + 4 ^ 2)) = '' \ frac 2 (sqrt (3 ^ 2 + 4 ^ 2)) `

`3/5 sin x + 4/5 cos x = 2 / 5`.

Let's denote `3/5 = cos \ varphi`,` 4/5 = sin \ varphi`. Since `sin \ varphi> 0`,` cos \ varphi> 0`, then we take `\ varphi = arcsin 4 / 5` as an auxiliary angle. Then we write our equality in the form:

`cos \ varphi sin x + sin \ varphi cos x = 2 / 5`

Applying the formula for the sum of the angles for the sine, we write our equality in the following form:

`sin (x + \ varphi) = 2 / 5`,

`x + \ varphi = (- 1) ^ n arcsin 2/5 + \ pi n`,` n \ in Z`,

`x = (- 1) ^ n arcsin 2/5-` `arcsin 4/5 + \ pi n`,` n \ in Z`.

Answer. `x = (- 1) ^ n arcsin 2/5-` `arcsin 4/5 + \ pi n`,` n \ in Z`.

Fractional-rational trigonometric equations

These are equalities with fractions with trigonometric functions in the numerators and denominators.

Example. Solve the equation. `\ frac (sin x) (1 + cos x) = 1-cos x`.

Solution. Multiply and divide the right side of the equality by `(1 + cos x)`. As a result, we get:

`\ frac (sin x) (1 + cos x) =` `\ frac ((1-cos x) (1 + cos x)) (1 + cos x)`

`\ frac (sin x) (1 + cos x) =` `\ frac (1-cos ^ 2 x) (1 + cos x)`

`\ frac (sin x) (1 + cos x) =` `\ frac (sin ^ 2 x) (1 + cos x)`

`\ frac (sin x) (1 + cos x) -`` \ frac (sin ^ 2 x) (1 + cos x) = 0`

`\ frac (sin x-sin ^ 2 x) (1 + cos x) = 0`

Considering that the denominator cannot be equal to zero, we get `1 + cos x \ ne 0`,` cos x \ ne -1`, `x \ ne \ pi + 2 \ pi n, n \ in Z`.

Equate the numerator of the fraction to zero: `sin x-sin ^ 2 x = 0`,` sin x (1-sin x) = 0`. Then `sin x = 0` or` 1-sin x = 0`.

1. `sin x = 0`,` x = \ pi n`, `n \ in Z`
2. `1-sin x = 0`,` sin x = -1`, `x = \ pi / 2 + 2 \ pi n, n \ in Z`.

Considering that `x \ ne \ pi + 2 \ pi n, n \ in Z`, the solutions are` x = 2 \ pi n, n \ in Z` and `x = \ pi / 2 + 2 \ pi n` , `n \ in Z`.

Answer. `x = 2 \ pi n`,` n \ in Z`, `x = \ pi / 2 + 2 \ pi n`,` n \ in Z`.

Trigonometry, and trigonometric equations in particular, are used in almost all areas of geometry, physics, engineering. The study begins in grade 10, there are definitely tasks for the exam, so try to remember all the formulas of trigonometric equations - they will definitely come in handy!

However, you don't even need to memorize them, the main thing is to understand the essence and be able to deduce them. It's not as difficult as it sounds. See for yourself by watching the video.

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 - any number.

And here are the formulas with which you can immediately write down the solutions of these simple equations.

For sine:

For cosine:

х = ± arccos a + 2π n, n ∈ Z

For tangent:

x = arctan 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. Moreover, everything!) Nothing at all. However, the number of errors on this topic is simply off scale. Especially if the example deviates slightly from the template. Why?

Yes, because a lot of people write down these letters, not understanding their meaning at all! With caution he writes down, no matter how something happens ...) This must be dealt with. Trigonometry for humans, or humans for trigonometry after all !?)

Shall we figure it out?

One angle will be equal to arccos a, second: -arccos a.

And it will always work that way. For any but.

If you don't believe me, hover your mouse over the picture, or tap the picture on the tablet.) I changed the number but to some negative. Anyway, we got one corner arccos a, second: -arccos a.

Therefore, the answer can always be written in the form of two series of roots:

x 1 = arccos a + 2π n, n ∈ Z

х 2 = - arccos a + 2π n, n ∈ Z

We combine these two series into one:

x = ± arccos a + 2π n, n ∈ Z

And all the cases. Got 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 notation of two series of responses, you and the task "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 in a businesslike manner, and break it down into two separate answers, everything is decided.) Actually, this is why we understand. What, how and where.

In the simplest trigonometric equation

sinx = a

also two series of roots are obtained. Always. And these two series can be recorded too one line. Only this line will be more cunning:

х = (-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 a series of roots. And that's it!

Let's check the mathematicians? And then you never know ...)

In the previous lesson, the solution (without any formulas) of a trigonometric equation with a sine was analyzed in detail:

The answer produced 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 an unfinished answer.) The student must know that arcsin 0.5 = π / 6. A complete answer would be:

x = (-1) n π / 6+ π n, n ∈ Z

Here arises interest Ask... Reply via x 1; x 2 (that's the right answer!) and through the lonely NS (and this is the correct answer!) - the same thing, or not? We'll find out now.)

Substitute in response with x 1 meaning n = 0; one; 2; and so on, we count, we get a series of roots:

x 1 = π / 6; 13π / 6; 25π / 6 etc.

With the same substitution in the answer with x 2 , we get:

x 2 = 5π / 6; 17π / 6; 29π / 6 etc.

Now we substitute the values n (0; 1; 2; 3; 4 ...) into the general formula for a lonely NS ... That is, we raise minus one to zero, then to the first, second, etc. And, of course, we substitute 0 in the second term; one; 2 3; 4, etc. And we count. We get the series:

x = π / 6; 5π / 6; 13π / 6; 17π / 6; 25π / 6 etc.

That's all you can see.) The general formula gives us exactly the same results, as the two answers separately. Only all at once, in order. The mathematicians were not fooled.)

Formulas for solving trigonometric equations with tangent and cotangent can also be checked. But we will not.) They are so simple.

I have described 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 short record of the answers. For this brevity, I had to insert plus / minus in the cosine solution and (-1) n in 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 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 write down the answer in two series, or solve the equation / inequality along the trigonometric circle. Then these inserts disappear and life becomes easier.)

You can summarize.

There are ready-made answer formulas for solving the simplest trigonometric equations. Four pieces. They are good for instantly recording the solution to an equation. For example, you need to solve the equations:

sinx = 0.3

Easily: х = (-1) n arcsin 0,3 + π n, n ∈ Z

cosx = 0.2

No problem: х = ± arccos 0,2 + 2π n, n ∈ Z

tgx = 1.2

Easily: x = arctan 1,2 + π n, n ∈ Z

ctgx = 3.7

One left: x = arcctg3,7 + π n, 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 the puddle.) The correct answer: no solutions. Do you understand why? Read what the arccosine is. In addition, if the tabular values ​​of sine, cosine, tangent, cotangent are on the right side of the original equation, - 1; 0; √3; 1/2; √3/2 etc. - the answer through the arches will be unfinished. Arches must be translated into radians.

And if you come across inequality like

then the answer is:

х πn, n ∈ Z

there is a rare nonsense, yes ...) Here it is necessary to decide on the trigonometric circle. What we will do in the relevant topic.

For those who have 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 alarming combat environment, even academically hardened nerds often get confused about where πn, And where 2π n. Here's a simple trick. In of all formulas worth πn. Except for the only formula with inverse cosine. It stands there 2πn. Two pien. Keyword - two. The same formula contains two sign at the beginning. Plus and minus. Here and there - two.

So if you wrote two sign in front of the inverse cosine, it is easier to remember what the end will be two pien. And even the opposite happens. Skip man sign ± , gets to the end, writes it right two pien, and it will come to its senses. 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. Instant validation testing. Learning - with interest!)

you can get acquainted with functions and derivatives.

Methods for solving trigonometric equations.

Solving a trigonometric equation consists of two stages: equation transformation to get its simplest view (see above) and solutionobtained simplest trigonometric equation. There are seven basic methods for solving trigonometric equations.

1. Algebraic method.

(variable substitution and substitution method).

2. Factoring.

PRI me R 1. Solve the equation: sin x+ cos x = 1 .

Solution. Move all the terms of the equation to the left:

Sin x+ cos x – 1 = 0 ,

We transform and factorize the expression in

Left side of the equation:

PRI me R 2. Solve the equation: cos 2 x+ sin x Cos x = 1.

SOLUTION cos 2 x+ sin x Cos x– sin 2 x- cos 2 x = 0 ,

Sin x Cos x– sin 2 x = 0 ,

Sin x(Cos x– sin x ) = 0 ,

PRI me R 3. Solve the equation: cos 2 x- cos 8 x+ cos 6 x = 1.

SOLUTION cos 2 x+ cos 6 x= 1 + cos 8 x,

2 cos 4 x cos 2 x= 2 cos² 4 x ,

Cos 4 x · (cos 2 x- cos 4 x) = 0 ,

Cos 4 x 2 sin 3 x Sin x = 0 ,

one). cos 4 x= 0, 2). sin 3 x= 0, 3). sin x = 0 ,

3. Bringing to homogeneous equation. The equation called homogeneous from relationally sin and cos , if all of him members of the same degree with respect to sin and cos same angle... To solve a homogeneous equation, you need: but) move all its members to the left side; b) take all common factors out of brackets; in) equate all factors and parentheses to zero; G) the parentheses equated to zero give homogeneous equation of lesser degree, which should be divided by cos(or sin) in the senior degree; d) solve the resulting algebraic equation with respect totan . sin 2 x+ 4 sin x Cos x+ 5 cos 2 x = 2. SOLUTION. 3sin 2 x+ 4 sin x Cos x+ 5 cos 2 x= 2sin 2 x+ 2cos 2 x , Sin 2 x+ 4 sin x Cos x+ 3 cos 2 x = 0 , Tan 2 x+ 4 tan x + 3 = 0 , from here y 2 + 4y +3 = 0 , The roots of this equation are:y 1 = - 1, y 2 = - 3, hence 1) tan x= –1, 2) tan x = –3,

4. Move to half corner.

Let's consider this method with an example:

EXAMPLE Solve equation: 3 sin x- 5 cos x = 7.

SOLUTION.6 sin ( x/ 2) cos ( x/ 2) - 5 cos ² ( x/ 2) + 5 sin ² ( x/ 2) =

7 sin ² ( x/ 2) + 7 cos ² ( x/ 2) ,

2 sin² ( x/ 2) - 6 sin ( x/ 2) cos ( x/ 2) + 12 cos ² ( x/ 2) = 0 ,

tan ² ( x/ 2) - 3 tan ( x/ 2) + 6 = 0 ,

. . . . . . . . . .

5. Introduction of an auxiliary angle.

Consider an equation of the form:

a sin x + b cos x = c ,

Where a, b, c- coefficients;x- the unknown.

Now the coefficients of the equation have the properties of sine and cosine, namely: modulus (absolute value) of each of which no more than 1, and the sum of their squares is 1. Then we can denote them respectively how cos and sin (here - so-called auxiliary corner), andtake our equation

UMK line by G. K. Muravin. Algebra and the beginnings of mathematical analysis (10-11) (in-depth)

UMK line 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 methods

The course of mathematics of the Russian Textbook Corporation, by Georgy Muravin and Olga Muravina, provides for a gradual transition to solving trigonometric equations and inequalities in grade 10, as well as their continuation in grade 11. We present to your attention the stages of 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 to the study of trigonometric equations)

An example of a task. Find approximately the angles whose cosines are 0.8.

Solution. Cosine is the abscissa of the corresponding point unit circle... All points with abscissas equal to 0.8 belong to a straight line parallel to the ordinate 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 abscissa axis.

Using the protractor, we find that the angle α° approximately equal to 37 °. Means, general form rotation angles with end point P α°:

α ° ≈ 37 ° + 360 ° n, where n- any integer.

Due to symmetry about the abscissa axis, the point P β ° - the end point of rotation by an angle of –37 °. This means that the general view of the rotation angles for it is:

β ° ≈ –37 ° + 360 ° n, where n- any integer.

Answer: 37 ° + 360 ° n, –37 ° + 360 ° n, where n- any integer.

An example of a task. Find the angles with sines 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 abscissa axis and passing through the point D(0; 0,5).

This line intersects the unit circle at two points: Pφ and Pπ – φ, symmetric about the ordinate axis. IN right triangle OKPφ leg KPφ is equal to half the hypotenuse OPφ , means,

General view of the angles of rotation with the end point P φ :

where n- any integer. General view of the angles of rotation with the end point P π–φ :

where n- any integer.

Answer: where n- any integer.

2. Tangent and cotangent of any angle (propaedeutics to the study of trigonometric equations)

Example 2.

An example of a task. Find the general view of the angles whose tangent is –1.2.

Solution. Let's mark the point on the tangent axis C with 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 the protractor, we find that the angle P α° OP 0 is equal to –50 °. This means that the general view 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 angles listed 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

The designations are introduced: arcsin α, arccos α, arctan α, arcctan α. It is not recommended to rush into the introduction of a unified formula. It is much more convenient to record two series of roots, especially when it is necessary to select roots at 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, that is, they are true for any admissible 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 reduced function in the corresponding quarter, if we consider φ acute angle;

2) only the functions of the angles change the name and

	φ + 2π n

5. Properties and graph of the function y = sin x

The simplest trigonometric inequalities are solved either on a graph or on a circle. When solving trigonometric inequality on a circle, it is important not to confuse which point to indicate first.

6. Properties and graph of the function y= cos x

The task of plotting a function y= cos x can be reduced to plotting a function y = sin x... Indeed, since function graph y= cos x can be obtained from the function graph y= sin x by shifting the latter along the abscissa 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 in the case of plotting 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 presses against a straight line, leaving with it unlimitedly upward.

8. Dependencies between trigonometric functions of the same argument

Equality and express the 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 obtain that the tangent and cotangent are related to each other by the following equality.

tg φ ctg φ = 1

There are other dependencies between trigonometric functions.

Equation of the unit circle centered at the origin x 2 + y 2= 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

Cosine sum 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

Sine Difference Formula

sin (α - β) = sin α cos β - cos α sin β

Sine sum formula

sin (α + β) = sin α cos β + cos α sin β

10. The tangent of the sum and the tangent of the difference of two angles

Formula for tangent of sum

Difference tangent formula

The textbook is included in the teaching materials in mathematics for grades 10-11, studying the subject at the basic level. The theoretical material is divided into mandatory and optional, the system of tasks is differentiated according to the level of difficulty, each point of the chapter ends with control questions and tasks, and each chapter is home test work... The tutorial includes project topics and links to Internet resources.

11. Double angle trigonometric functions

Double angle tangent formula

cos2α = 1 - 2sin 2 α cos2α = 2cos 2 α - 1

An example of a task. Solve the equation

Solution.

13. Solving trigonometric equations

In most cases, the original equation in the process of solving is reduced to the simplest trigonometric equations. However, there is no single solution method for trigonometric equations. In each specific case, success depends on knowledge of trigonometric formulas and on the ability to choose the right ones from them. At the same time, the abundance of various formulas sometimes makes this choice quite difficult.

Equations Reducing to Squares

An example of a task. Solve the 2 cos 2 equation x+ 3 sin x = 0

Solution... Using the basic trigonometric identity, this equation can be reduced to square with respect to sin x:

2cos 2 x+ 3sin x= 0, 2 (1 - sin 2 x) + 3sin x = 0,

2 - 2sin 2 x+ 3sin x= 0, 2sin 2 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.

Returning to the 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

An example of a task. Solve the 2sin 2 equation x- 3sin x cos x- 5cos 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, since for no x cosine and sine do not vanish simultaneously.

Case 2. If cos x≠ 0, then the equation can be divided 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 concept of solving trigonometric equations.

• To solve a trigonometric equation, convert it to one or more basic trigonometric equations. Solving a trigonometric equation ultimately comes down to solving four basic trigonometric equations.

Solving basic trigonometric equations.

• There are 4 types of basic trigonometric equations:
• sin x = a; cos x = a
• tg x = a; ctg x = a
• Solving basic trigonometric equations involves considering different provisions"X" on the unit circle, and the use of a conversion table (or calculator).
• Example 1.sin x = 0.866. Using a conversion table (or calculator), you get the answer: x = π / 3. The unit circle gives another answer: 2π / 3. Remember: all trigonometric functions are periodic, that is, their values ​​are repeated. For example, the periodicity of sin x and cos x is 2πn, and the periodicity of tg x and ctg x is πn. Therefore, the answer is written as follows:
• x1 = π / 3 + 2πn; x2 = 2π / 3 + 2πn.
• Example 2.cos x = -1/2. Using a conversion table (or calculator), you get the answer: x = 2π / 3. The unit circle gives another answer: -2π / 3.
• x1 = 2π / 3 + 2π; x2 = -2π / 3 + 2π.
• Example 3.tg (x - π / 4) = 0.
• Answer: x = π / 4 + πn.
• Example 4. ctg 2x = 1.732.
• Answer: x = π / 12 + πn.

Transformations used to solve trigonometric equations.

• To transform trigonometric equations, algebraic transformations are used (factorization, reduction homogeneous members etc.) and trigonometric identities.
• Example 5. Using trigonometric identities, the equation sin x + sin 2x + sin 3x = 0 is transformed into the equation 4cos x * sin (3x / 2) * cos (x / 2) = 0. Thus, you need to solve the following basic trigonometric equations: cos x = 0; sin (3x / 2) = 0; cos (x / 2) = 0.

• Finding angles from known values ​​of functions.

  • Before learning methods for solving trigonometric equations, you need to learn how to find angles from known values ​​of functions. This can be done using a conversion table or calculator.
  • Example: cos x = 0.732. The calculator will give the answer x = 42.95 degrees. The unit circle will give additional angles, the cosine of which is also 0.732.

• Set the solution aside on the unit circle.

  • You can defer the solutions to the trigonometric equation on the unit circle. The solutions of the trigonometric equation on the unit circle are the vertices of a regular polygon.
  • Example: The solutions x = π / 3 + πn / 2 on the unit circle are the vertices of a square.
  • Example: The solutions x = π / 4 + πn / 3 on the unit circle represent the vertices of a regular hexagon.

• Methods for solving trigonometric equations.

  • If a given trig equation contains only one trig function, solve that equation as the basic trig equation. If a given equation includes two or more trigonometric functions, then there are 2 methods for solving such an equation (depending on the possibility of its transformation).
    • Method 1.
  • Convert this equation to an equation of the form: f (x) * g (x) * h (x) = 0, where f (x), g (x), h (x) are the basic trigonometric equations.
  • Example 6.2cos x + sin 2x = 0. (0< x < 2π)
  • Solution. Using the double angle formula sin 2x = 2 * sin x * cos x, replace sin 2x.
  • 2cos x + 2 * sin x * cos x = 2cos x * (sin x + 1) = 0. Now solve the two basic trigonometric equations: cos x = 0 and (sin x + 1) = 0.
  • Example 7.cos x + cos 2x + cos 3x = 0. (0< x < 2π)
  • Solution: Using trigonometric identities, transform this equation into an equation of the form: cos 2x (2cos x + 1) = 0. Now solve the two basic trigonometric equations: cos 2x = 0 and (2cos x + 1) = 0.
  • Example 8.sin x - sin 3x = cos 2x. (0< x < 2π)
  • Solution: Using trigonometric identities, transform this equation into an equation of the form: -cos 2x * (2sin x + 1) = 0. Now solve the two basic trigonometric equations: cos 2x = 0 and (2sin x + 1) = 0.
    • Method 2.
  • Convert the given trigonometric equation to an equation containing only one trigonometric function. Then replace this trigonometric function with some unknown, for example, t (sin x = t; cos x = t; cos 2x = t, tg x = t; tg (x / 2) = t, etc.).
  • Example 9.3sin ^ 2 x - 2cos ^ 2 x = 4sin x + 7 (0< x < 2π).
  • Solution. In this equation, replace (cos ^ 2 x) with (1 - sin ^ 2 x) (by identity). The transformed equation is:
  • 3sin ^ 2 x - 2 + 2sin ^ 2 x - 4sin x - 7 = 0. Replace sin x with t. Now the equation looks like: 5t ^ 2 - 4t - 9 = 0. This is quadratic equation with two roots: t1 = -1 and t2 = 9/5. The second root t2 does not satisfy the range of values ​​of the function (-1< sin x < 1). Теперь решите: t = sin х = -1; х = 3π/2.
  • Example 10.tg x + 2 tg ^ 2 x = ctg x + 2
  • Solution. Replace tg x with t. Rewrite the original equation as follows: (2t + 1) (t ^ 2 - 1) = 0. Now find t and then find x for t = tg x.

• Special trigonometric equations.

  • There are some special trigonometric equations that require specific transformations. Examples:
  • a * sin x + b * cos x = c; a (sin x + cos x) + b * cos x * sin x = c;
  • a * sin ^ 2 x + b * sin x * cos x + c * cos ^ 2 x = 0

• Periodicity of trigonometric functions.

  • As mentioned earlier, all trigonometric functions are periodic, that is, their values ​​are repeated after a certain period. Examples:
    • The period of the function f (x) = sin x is 2π.
    • The period of the function f (x) = tan x is equal to π.
    • The period of the function f (x) = sin 2x is π.
    • The period of the function f (x) = cos (x / 2) is 4π.
  • If the period is specified in the problem, calculate the value "x" within this period.
  • Note: Solving trigonometric equations is not an easy task and often leads to errors. So check your answers carefully. To do this, you can use a graphing calculator to plot a graph of this equation R (x) = 0. In such cases, the solutions will be presented in the form decimal fractions(that is, π is replaced by 3.14).