Tgx a are special cases. Trigonometric equations. How to solve trigonometric equations? Basic quantities of trigonometry

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

The simplest equations are `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 the root formulas for each of them.

1. Equation `sin x=a`.

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

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

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

2. Equation `cos x=a`

For `|a|>1` - as in the case of the sine, there are no solutions among real numbers.

With `|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. Equation `tg x=a`

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

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

4. Equation `ctg x=a`

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

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

Formulas for the roots of trigonometric equations in the table

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

Methods for solving trigonometric equations

The solution of any trigonometric equation consists of two stages:

• using to convert it to the simplest;
• solve the resulting simple equation using the above formulas for the roots and tables.

Let's consider the main methods of solution using examples.

algebraic method.

In this method, the replacement of a variable and its 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`,

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

we find the roots: `y_1=1, y_2=1/2`, from which 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 to the left all terms of equality: `sin x+cos x-1=0`. Using , we transform and factorize 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=arctg 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 forms:

`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 split both parts by `cos x \ne 0` for the first case, and by `cos^2 x \ne 0` for the second. We get equations for `tg x`: `a tg x+b=0` and `a tg^2 x + b tg x +c =0`, which must be solved using known methods.

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

Solution. Let's write 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, dividing 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`. Let's 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=arctg 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. Applying the double angle formulas, the result is: `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 algebraic method described above, we obtain:

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

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

Introduction of 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 parts 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 modulus is not greater than 1. 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. Dividing both sides of the equation 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`.

Denote `3/5 = cos \varphi` , `4/5=sin \varphi`. Since `sin \varphi>0`, `cos \varphi>0`, 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 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, in the numerators and denominators of which there are trigonometric functions.

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

Solution. Multiply and divide the right side of the equation 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`

Given that the denominator cannot be 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`.

Given 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, and engineering. The study begins in the 10th grade, there are always tasks for the exam, so try to remember all the formulas of trigonometric equations - they will definitely come in handy for you!

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

The concepts of sine, cosine, tangent and cotangent are the main categories of trigonometry - a branch of mathematics, and are inextricably linked with the definition of an angle. Possession of this mathematical science requires memorization and understanding of formulas and theorems, as well as developed spatial thinking. That is why trigonometric calculations often cause difficulties for schoolchildren and students. To overcome them, you should become more familiar with trigonometric functions and formulas.

Concepts in trigonometry

To understand the basic concepts of trigonometry, you must first decide what a right triangle and an angle in a circle are, and why all basic trigonometric calculations are associated with them. A triangle in which one of the angles is 90 degrees is a right triangle. Historically, this figure was often used by people in architecture, navigation, art, astronomy. Accordingly, studying and analyzing the properties of this figure, people came to the calculation of the corresponding ratios of its parameters.

The main categories associated with right triangles are the hypotenuse and the legs. The hypotenuse is the side of a triangle that is opposite the right angle. The legs, respectively, are the other two sides. The sum of the angles of any triangle is always 180 degrees.

Spherical trigonometry is a section of trigonometry that is not studied at school, but in applied sciences such as astronomy and geodesy, scientists use it. A feature of a triangle in spherical trigonometry is that it always has a sum of angles greater than 180 degrees.

Angles of a triangle

In a right triangle, the sine of an angle is the ratio of the leg opposite the desired angle to the hypotenuse of the triangle. Accordingly, the cosine is the ratio of the adjacent leg and the hypotenuse. Both of these values ​​always have a value less than one, since the hypotenuse is always longer than the leg.

The tangent of an angle is a value equal to the ratio of the opposite leg to the adjacent leg of the desired angle, or sine to cosine. The cotangent, in turn, is the ratio of the adjacent leg of the desired angle to the opposite cactet. The cotangent of an angle can also be obtained by dividing the unit by the value of the tangent.

unit circle

A unit circle in geometry is a circle whose radius is equal to one. Such a circle is constructed in the Cartesian coordinate system, with the center of the circle coinciding with the point of origin, and the initial position of the radius vector is determined by the positive direction of the X axis (abscissa axis). Each point of the circle has two coordinates: XX and YY, that is, the coordinates of the abscissa and ordinate. Selecting any point on the circle in the XX plane, and dropping the perpendicular from it to the abscissa axis, we get a right triangle formed by a radius to the selected point (let us denote it by the letter C), a perpendicular drawn to the X axis (the intersection point is denoted by the letter G), and a segment the abscissa axis between the origin (the point is denoted by the letter A) and the intersection point G. The resulting triangle ACG is a right triangle inscribed in a circle, where AG is the hypotenuse, and AC and GC are the legs. The angle between the radius of the circle AC and the segment of the abscissa axis with the designation AG, we define as α (alpha). So, cos α = AG/AC. Given that AC is the radius of the unit circle, and it is equal to one, it turns out that cos α=AG. Similarly, sin α=CG.

In addition, knowing these data, it is possible to determine the coordinate of point C on the circle, since cos α=AG, and sin α=CG, which means that point C has the given coordinates (cos α; sin α). Knowing that the tangent is equal to the ratio of the sine to the cosine, we can determine that tg α \u003d y / x, and ctg α \u003d x / y. Considering angles in a negative coordinate system, one can calculate that the sine and cosine values ​​of some angles can be negative.

Calculations and basic formulas

Values ​​of trigonometric functions

Having considered the essence of trigonometric functions through the unit circle, we can derive the values ​​of these functions for some angles. The values ​​are listed in the table below.

The simplest trigonometric identities

Equations in which there is an unknown value under the sign of the trigonometric function are called trigonometric. Identities with the value sin x = α, k is any integer:

1. sin x = 0, x = πk.
2. 2. sin x \u003d 1, x \u003d π / 2 + 2πk.
3. sin x \u003d -1, x \u003d -π / 2 + 2πk.
4. sin x = a, |a| > 1, no solutions.
5. sin x = a, |a| ≦ 1, x = (-1)^k * arcsin α + πk.

Identities with the value cos x = a, where k is any integer:

1. cos x = 0, x = π/2 + πk.
2. cos x = 1, x = 2πk.
3. cos x \u003d -1, x \u003d π + 2πk.
4. cos x = a, |a| > 1, no solutions.
5. cos x = a, |a| ≦ 1, х = ±arccos α + 2πk.

Identities with the value tg x = a, where k is any integer:

1. tg x = 0, x = π/2 + πk.
2. tg x \u003d a, x \u003d arctg α + πk.

Identities with value ctg x = a, where k is any integer:

1. ctg x = 0, x = π/2 + πk.
2. ctg x \u003d a, x \u003d arcctg α + πk.

Cast formulas

This category of constant formulas denotes methods by which you can go from trigonometric functions of the form to functions of the argument, that is, convert the sine, cosine, tangent and cotangent of an angle of any value to the corresponding indicators of the angle of the interval from 0 to 90 degrees for greater convenience of calculations.

The formulas for reducing functions for the sine of an angle look like this:

• sin(900 - α) = α;
• sin(900 + α) = cos α;
• sin(1800 - α) = sin α;
• sin(1800 + α) = -sin α;
• sin(2700 - α) = -cos α;
• sin(2700 + α) = -cos α;
• sin(3600 - α) = -sin α;
• sin(3600 + α) = sin α.

For the cosine of an angle:

• cos(900 - α) = sin α;
• cos(900 + α) = -sin α;
• cos(1800 - α) = -cos α;
• cos(1800 + α) = -cos α;
• cos(2700 - α) = -sin α;
• cos(2700 + α) = sin α;
• cos(3600 - α) = cos α;
• cos(3600 + α) = cos α.

The use of the above formulas is possible subject to two rules. First, if the angle can be represented as a value (π/2 ± a) or (3π/2 ± a), the value of the function changes:

• from sin to cos;
• from cos to sin;
• from tg to ctg;
• from ctg to tg.

The value of the function remains unchanged if the angle can be represented as (π ± a) or (2π ± a).

Secondly, the sign of the reduced function does not change: if it was initially positive, it remains so. The same is true for negative functions.

Addition Formulas

These formulas express the values ​​of the sine, cosine, tangent, and cotangent of the sum and difference of two rotation angles in terms of their trigonometric functions. Angles are usually denoted as α and β.

The formulas look like this:

1. sin(α ± β) = sin α * cos β ± cos α * sin.
2. cos(α ± β) = cos α * cos β ∓ sin α * sin.
3. tan(α ± β) = (tan α ± tan β) / (1 ∓ tan α * tan β).
4. ctg(α ± β) = (-1 ± ctg α * ctg β) / (ctg α ± ctg β).

These formulas are valid for any angles α and β.

Double and triple angle formulas

The trigonometric formulas of a double and triple angle are formulas that relate the functions of the angles 2α and 3α, respectively, to the trigonometric functions of the angle α. Derived from addition formulas:

1. sin2α = 2sinα*cosα.
2. cos2α = 1 - 2sin^2α.
3. tg2α = 2tgα / (1 - tg^2 α).
4. sin3α = 3sinα - 4sin^3α.
5. cos3α = 4cos^3α - 3cosα.
6. tg3α = (3tgα - tg^3 α) / (1-tg^2 α).

Transition from sum to product

Considering that 2sinx*cosy = sin(x+y) + sin(x-y), simplifying this formula, we obtain the identity sinα + sinβ = 2sin(α + β)/2 * cos(α − β)/2. Similarly, sinα - sinβ = 2sin(α - β)/2 * cos(α + β)/2; cosα + cosβ = 2cos(α + β)/2 * cos(α − β)/2; cosα - cosβ = 2sin(α + β)/2 * sin(α − β)/2; tgα + tgβ = sin(α + β) / cosα * cosβ; tgα - tgβ = sin(α - β) / cosα * cosβ; cosα + sinα = √2sin(π/4 ∓ α) = √2cos(π/4 ± α).

Transition from product to sum

These formulas follow from the identities for the transition of the sum to the product:

• sinα * sinβ = 1/2*;
• cosα * cosβ = 1/2*;
• sinα * cosβ = 1/2*.

Reduction formulas

In these identities, the square and cubic powers of the sine and cosine can be expressed in terms of the sine and cosine of the first power of a multiple angle:

• sin^2 α = (1 - cos2α)/2;
• cos^2α = (1 + cos2α)/2;
• sin^3 α = (3 * sinα - sin3α)/4;
• cos^3 α = (3 * cosα + cos3α)/4;
• sin^4 α = (3 - 4cos2α + cos4α)/8;
• cos^4 α = (3 + 4cos2α + cos4α)/8.

Universal substitution

The universal trigonometric substitution formulas express trigonometric functions in terms of the tangent of a half angle.

• sin x \u003d (2tgx / 2) * (1 + tg ^ 2 x / 2), while x \u003d π + 2πn;
• cos x = (1 - tg^2 x/2) / (1 + tg^2 x/2), where x = π + 2πn;
• tg x \u003d (2tgx / 2) / (1 - tg ^ 2 x / 2), where x \u003d π + 2πn;
• ctg x \u003d (1 - tg ^ 2 x / 2) / (2tgx / 2), while x \u003d π + 2πn.

Special cases

Particular cases of the simplest trigonometric equations are given below (k is any integer).

Private for sine:

sin x value	x value
0	pk
1	π/2 + 2πk
-1	-π/2 + 2πk
1/2	π/6 + 2πk or 5π/6 + 2πk
-1/2	-π/6 + 2πk or -5π/6 + 2πk
√2/2	π/4 + 2πk or 3π/4 + 2πk
-√2/2	-π/4 + 2πk or -3π/4 + 2πk
√3/2	π/3 + 2πk or 2π/3 + 2πk
-√3/2	-π/3 + 2πk or -2π/3 + 2πk

Cosine quotients:

cos x value	x value
0	π/2 + 2πk
1	2πk
-1	2 + 2πk
1/2	±π/3 + 2πk
-1/2	±2π/3 + 2πk
√2/2	±π/4 + 2πk
-√2/2	±3π/4 + 2πk
√3/2	±π/6 + 2πk
-√3/2	±5π/6 + 2πk

Private for tangent:

tg x value	x value
0	pk
1	π/4 + πk
-1	-π/4 + πk
√3/3	π/6 + πk
-√3/3	-π/6 + πk
√3	π/3 + πk
-√3	-π/3 + πk

Cotangent quotients:

ctg x value	x value
0	π/2 + πk
1	π/4 + πk
-1	-π/4 + πk
√3	π/6 + πk
-√3	-π/3 + πk
√3/3	π/3 + πk
-√3/3	-π/3 + πk

Theorems

Sine theorem

There are two versions of the theorem - simple and extended. Simple sine theorem: a/sin α = b/sin β = c/sin γ. In this case, a, b, c are the sides of the triangle, and α, β, γ are the opposite angles, respectively.

Extended sine theorem for an arbitrary triangle: a/sin α = b/sin β = c/sin γ = 2R. In this identity, R denotes the radius of the circle in which the given triangle is inscribed.

Cosine theorem

The identity is displayed in this way: a^2 = b^2 + c^2 - 2*b*c*cos α. In the formula, a, b, c are the sides of the triangle, and α is the angle opposite side a.

Tangent theorem

The formula expresses the relationship between the tangents of two angles, and the length of the sides opposite them. The sides are labeled a, b, c, and the corresponding opposite angles are α, β, γ. The formula of the tangent theorem: (a - b) / (a+b) = tg((α - β)/2) / tg((α + β)/2).

Cotangent theorem

Associates the radius of a circle inscribed in a triangle with the length of its sides. If a, b, c are the sides of a triangle, and A, B, C, respectively, are their opposite angles, r is the radius of the inscribed circle, and p is the half-perimeter of the triangle, the following identities hold:

• ctg A/2 = (p-a)/r;
• ctg B/2 = (p-b)/r;
• ctg C/2 = (p-c)/r.

Applications

Trigonometry is not only a theoretical science associated with mathematical formulas. Its properties, theorems and rules are used in practice by various branches of human activity - astronomy, air and sea navigation, music theory, geodesy, chemistry, acoustics, optics, electronics, architecture, economics, mechanical engineering, measuring work, computer graphics, cartography, oceanography, and many others.

Sine, cosine, tangent and cotangent are the basic concepts of trigonometry, with which you can mathematically express the relationship between angles and lengths of sides in a triangle, and find the desired quantities through identities, theorems and rules.

Trigonometry, as a science, originated in the Ancient East. The first trigonometric ratios were developed by astronomers to create an accurate calendar and orientate by the stars. These calculations were related to spherical trigonometry, while in school course study the ratio of the sides and angle of a flat triangle.

Trigonometry is a branch of mathematics dealing with the properties of trigonometric functions and the relationship between the sides and angles of triangles.

During the heyday of culture and science in the 1st millennium AD, knowledge spread from the Ancient East to Greece. But the main discoveries of trigonometry are the merit of the men of the Arab Caliphate. In particular, the Turkmen scientist al-Marazvi introduced such functions as tangent and cotangent, compiled the first tables of values ​​for sines, tangents and cotangents. The concept of sine and cosine was introduced by Indian scientists. A lot of attention is devoted to trigonometry in the works of such great figures of antiquity as Euclid, Archimedes and Eratosthenes.

Basic quantities of trigonometry

The basic trigonometric functions of a numerical argument are sine, cosine, tangent, and cotangent. Each of them has its own graph: sine, cosine, tangent and cotangent.

The formulas for calculating the values ​​of these quantities are based on the Pythagorean theorem. It is better known to schoolchildren in the formulation: “Pythagorean pants, equal in all directions,” since the proof is given on the example of an isosceles right triangle.

Sine, cosine and other dependencies establish a relationship between acute angles and sides of any right triangle. We give formulas for calculating these quantities for angle A and trace the relationship of trigonometric functions:

As you can see, tg and ctg are inverse functions. If we represent leg a as the product of sin A and hypotenuse c, and leg b as cos A * c, then we get the following formulas for tangent and cotangent:

trigonometric circle

Graphically, the ratio of the mentioned quantities can be represented as follows:

The circle, in this case, represents all possible values ​​of the angle α - from 0° to 360°. As can be seen from the figure, each function takes a negative or positive value depending on the angle. For example, sin α will be with a “+” sign if α belongs to the I and II quarters of the circle, that is, it is in the range from 0 ° to 180 °. With α from 180° to 360° (III and IV quarters), sin α can only be a negative value.

Let's try to build trigonometric tables for specific angles and find out the meaning of the quantities.

The values ​​of α equal to 30°, 45°, 60°, 90°, 180° and so on are called special cases. The values ​​of trigonometric functions for them are calculated and presented in the form of special tables.

These angles were not chosen by chance. The designation π in the tables is for radians. Rad is the angle at which the length of a circular arc corresponds to its radius. This value was introduced in order to establish a universal relationship; when calculating in radians, the actual length of the radius in cm does not matter.

The angles in the tables for trigonometric functions correspond to radian values:

So, it is not difficult to guess that 2π is a full circle or 360°.

Properties of trigonometric functions: sine and cosine

In order to consider and compare the basic properties of sine and cosine, tangent and cotangent, it is necessary to draw their functions. This can be done in the form of a curve located in a two-dimensional coordinate system.

Consider a comparative table of properties for a sine wave and a cosine wave:

sinusoid	cosine wave
y = sin x	y = cos x
ODZ [-1; 1]	ODZ [-1; 1]
sin x = 0, for x = πk, where k ϵ Z	cos x = 0, for x = π/2 + πk, where k ϵ Z
sin x = 1, for x = π/2 + 2πk, where k ϵ Z	cos x = 1, for x = 2πk, where k ϵ Z
sin x = - 1, at x = 3π/2 + 2πk, where k ϵ Z	cos x = - 1, for x = π + 2πk, where k ϵ Z
sin (-x) = - sin x, i.e. odd function	cos (-x) = cos x, i.e. the function is even
	the function is periodic, the smallest period is 2π
sin x › 0, with x belonging to quarters I and II or from 0° to 180° (2πk, π + 2πk)	cos x › 0, with x belonging to quarters I and IV or from 270° to 90° (- π/2 + 2πk, π/2 + 2πk)
sin x ‹ 0, with x belonging to quarters III and IV or from 180° to 360° (π + 2πk, 2π + 2πk)	cos x ‹ 0, with x belonging to quarters II and III or from 90° to 270° (π/2 + 2πk, 3π/2 + 2πk)
increases on the interval [- π/2 + 2πk, π/2 + 2πk]	increases on the interval [-π + 2πk, 2πk]
decreases on the intervals [ π/2 + 2πk, 3π/2 + 2πk]	decreases in intervals
derivative (sin x)' = cos x	derivative (cos x)’ = - sin x

Determining whether a function is even or not is very simple. It is enough to imagine a trigonometric circle with signs of trigonometric quantities and mentally “fold” the graph relative to the OX axis. If the signs are the same, the function is even; otherwise, it is odd.

The introduction of radians and the enumeration of the main properties of the sinusoid and cosine wave allow us to bring the following pattern:

It is very easy to verify the correctness of the formula. For example, for x = π/2, the sine is equal to 1, as is the cosine of x = 0. Checking can be done by looking at tables or by tracing function curves for given values.

Properties of tangentoid and cotangentoid

The graphs of the tangent and cotangent functions differ significantly from the sinusoid and cosine wave. The values ​​tg and ctg are inverse to each other.

1. Y = tgx.
2. The tangent tends to the values ​​of y at x = π/2 + πk, but never reaches them.
3. The smallest positive period of the tangentoid is π.
4. Tg (- x) \u003d - tg x, i.e., the function is odd.
5. Tg x = 0, for x = πk.
6. The function is increasing.
7. Tg x › 0, for x ϵ (πk, π/2 + πk).
8. Tg x ‹ 0, for x ϵ (— π/2 + πk, πk).
9. Derivative (tg x)' = 1/cos 2 ⁡x .

Consider graphic image cotangentoids below.

The main properties of the cotangentoid:

1. Y = ctgx.
2. Unlike the sine and cosine functions, in the tangentoid Y can take on the values ​​of the set of all real numbers.
3. The cotangentoid tends to the values ​​of y at x = πk, but never reaches them.
4. The smallest positive period of the cotangentoid is π.
5. Ctg (- x) \u003d - ctg x, i.e., the function is odd.
6. Ctg x = 0, for x = π/2 + πk.
7. The function is decreasing.
8. Ctg x › 0, for x ϵ (πk, π/2 + πk).
9. Ctg x ‹ 0, for x ϵ (π/2 + πk, πk).
10. Derivative (ctg x)' = - 1/sin 2 ⁡x Fix

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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 the trigonometric equation ultimately comes down to solving the four basic trigonometric equations.

Solution of basic trigonometric equations.

• There are 4 types of basic trigonometric equations:
• sin x = a; cos x = a
• tan x = a; ctg x = a
• Solving basic trigonometric equations involves looking at the various x positions on the unit circle, as well as using 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. So the answer is written like this:
• 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 \u003d π / 4 + πn.
• Example 4. ctg 2x = 1.732.
• Answer: x \u003d π / 12 + πn.

Transformations used in solving 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 converted to the equation 4cos x*sin (3x/2)*cos (x/2) = 0. Thus, the following basic trigonometric equations need to be solved: cos x = 0; sin(3x/2) = 0; cos(x/2) = 0.

• Finding angles from known values ​​of functions.

  • Before learning how to solve 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 equal to 0.732.

• Set aside the solution on the unit circle.

  • You can put 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 the square.
  • Example: The solutions x = π/4 + πn/3 on the unit circle are the vertices of a regular hexagon.

• Methods for solving trigonometric equations.

  • If the given trigonometric equation contains only one trigonometric function, solve this equation as a basic trigonometric 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
  • Transform this equation into 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 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 two basic trigonometric equations: cos 2x = 0 and (2cos x + 1) = 0.
  • Example 8. sin x - sin 3x \u003d 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 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) (according to the identity). The transformed equation looks like:
  • 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 a quadratic equation with two roots: t1 = -1 and t2 = 9/5. The second root t2 does not satisfy the range 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.