An integral part of the Unified State Examination are trigonometric equations.

Unfortunately, there is no general unified method that can be followed to solve any equation involving trigonometric functions. Success here can only be ensured by good knowledge of formulas and the ability to see certain useful combinations, which can only be developed through practice.

The general goal is usually to transform the trigonometric expression involved in the equation so that the roots can be found from the so-called simplest equations:

To do this, you need to be able to use trigonometric formulas. It is useful to know and call them by “names”:

1. Formulas for double argument, triple argument:

сos 2x = cos 2 x – sin 2 x = 1 – 2 sin 2 x = 2 cos 2 x – 1;

sin 2x = 2 sin x cos x;

tg 2x = 2 tg x/1 – tg x;

ctg 2x = (ctg 2 x – 1)/2 ctg x;

sin 3x = 3 sin x – 4 sin 3 x;

cos 3x = 4 cos 3 x – 3 cos x;

tg 3x = (2 tg x – tg 3 x)/(1 – 3 tg 2 x);

ctg 3x = (ctg 3 x – 3ctg x)/(3ctg 2 x – 1);

2. Formulas for half argument or reduction of degree:

sin 2 x/2 = (1 – cos x)/2; cos 2 x/2 = (1 + cos x)/2;

tg 2 x = (1 – cos x)/(1 + cos x);

cot 2 x = (1 + cos x)/(1 – cos x);

3. Introduction of an auxiliary argument:

Let us consider the example of the equation a sin x + b cos x = c, namely, by determining the angle x from the conditions sin y = b/v(a 2 + b 2), cos y = a/v(a 2 + b 2), we we can reduce the equation under consideration to the simplest sin (x + y) = c/v(a 2 + b 2) the solutions of which can be written out without difficulty; thereby determining the solutions to the original equation.

4. Addition and subtraction formulas:

sin (a + b) = sin a cos b + cos a sin b;

sin (a – b) = sin a cos b – cos a sin b;

cos (a + b) = cos a cos b – sin a sin b;

cos (a – b) = cos a cos b + sin a sin b;

tg (a + b) = (tg a + tg b)/(1 – tg a tg b);

tg (a – b) = (tg a – tg b)/(1 + tg a tg b);

5. Universal trigonometric substitution:

sin a = 2 tan (a/2)/(1 + ( tg 2 (a/2));

cos a = (1 – tan 2 (a/2))/(1 + ( tg 2 (a/2));

tg a = 2 tg a/2/(1 – tg 2 (a/2));

6. Some important ratios:

sin x + sin 2x + sin 3x +…+ sin mx = (cos (x/2) -cos (2m + 1)x)/(2 sin (x/2));

cos x + cos 2x + cos 3x +…+ cos mx = (sin (2m+ 1)x/2 – sin (x/2))/(2 sin (x/2));

7. Formulas for converting the sum of trigonometric functions into a product:

sin a + sin b = 2 sin(a + b)/2 cos (a – b)/2;

cos a – cos b = -2 sin(a + b)/2 sin (b – a)/2;

tan a + tan b = sin (a + b)/(cos a cos b);

tan a – tan b = sin (a – b)/(cos a cos b).

And also reduction formulas.

During the solution process, one must especially carefully monitor the equivalence of the equations in order to prevent the loss of roots (for example, when reducing the left and right sides of the equation by a common factor), or the acquisition of extra roots (for example, when squaring both sides of the equation). In addition, it is necessary to control whether the receiving roots belong to the ODZ of the equation under consideration.

In all necessary cases (i.e. when unequal transformations were allowed), it is necessary to check. When solving equations, it is necessary to teach students to reduce them to certain types, usually starting with easy equations.

Let's get acquainted with methods for solving equations:

1. Reduction to the form ax 2 + bx + c = 0

2. Homogeneity of equations.

3. Factorization.

4. Reduction to the form a 2 + b 2 + c 2 = 0

5. Replacement of variables.

6. Reducing the equation to an equation with one variable.

7. Evaluation of the left and right parts.

8. Gaze method.

9. Introduction of an auxiliary angle.

10. “Divide and conquer” method.

Let's look at examples:

1. Solve the equation: sin x + cos 2 x = 1/4.

Solution: Solve by reducing to a quadratic equation. Let's express cos 2 x through sin 2 x

sin x + 1 – sin 2 x = 1/4

4 sin 2 x – 4 sin x – 3 = 0

sin x = -1/2, sin x = 3/2 (does not satisfy the condition x€[-1;1]),

those. x = (-1) k+1 arcsin 1/2 + k, k€z,

Answer: (-1) k+1 /6 + k, k€z.

2. Solve the equation: 2 tg x cos x +1 = 2 cos x + tg x,

solve by factorization method

2 tg x cos x – 2 cos x + 1 – tg x = 0, where x /2 + k, k€z,

2 cos x (tg x – 1) – (tg x – 1) = 0

(2 cos x – 1) (tg x – 1) = 0

2 cos x – 1 = 0 or tg x – 1 = 0

cos x = 1/2, tgx = 1,

i.e. x = ± /3 + 2k, k€z, x = /4 + m, m€z.

Answer: ± /3 + 2k, k€z, /4 + m, m€z.

3. Solve the equation: sin 2 x – 3 sin x cos x + 2 cos 2 x = 0.

Solution: sin 2 x – 3 sin x cos x + 2 cos 2 x = 0 homogeneous equation of 2nd degree. Since cos x = 0 is not a root of this equation, we divide the left and right sides by cos 2 x. As a result, we arrive at a quadratic equation for tan x

tg 2 x – 3 tg x + 2 = 0,

tg x = 1 and tg x = 2,

whence x = /4 + m, m€z,

x = arctan 2 + k, k€z.

Answer: /4 + m, m€z, arctan 2 + k, k€z.

4. Solve the equation: cos (10x + 12) + 42 sin (5x + 6) = 4.

Solution: Method for introducing a new variable

Let 5x + 6 = y, then cos 2y + 4 2 sin y = 4

1 – 2 sin 2 y + 4 2 sin y – 4 = 0

sin y = t, where t€[-1;1]

2t 2 – 4 2t + 3 = 0

t = 2/2 and t = 3 2/2 (does not satisfy the condition t€[-1;1])

sin (5x + 6) = 2/2,

5x + 6 = (-1) k /4 + k, k€z,

x = (-1) k /20 – 6/5 + k/5, k€z.

Answer: (-1) k?/20 – 6/5 + ?k/5, k€z.

5. Solve the equation: (sin x – cos y) 2 + 40x 2 = 0

Solution: We use a 2 + b 2 + c 2 = 0, true if a = 0, b = 0, c = 0. Equality is possible if sin x – cos y = 0, and 40x = 0 from here:

x = 0, and sin 0 – cos y = 0, therefore, x = 0, and cos y = 0, hence: x = 0, and y = /2 + k, k€z, it is also possible to write (0; / 2 + k) k€z.

Answer: (0; /2 + k) k€z.

6. Solve the equation: sin 2 x + cos 4 x – 2 sin x + 1 = 0

Solution: Rearrange the equation and apply the divide and conquer method

(sin 2 x – 2 sin x +1) + cos 4 x = 0;

(sin x – 1) 2 + cos 4 x = 0; this is possible if

(sin x – 1) 2 = 0, and cos 4 x = 0, hence:

sin x – 1 = 0, and cos x = 0,

sin x = 1, and cos x = 0, therefore

x = /2 + k, k€z

Answer: /2 + k, k€z.

7. Solve the equation: sin 5x + sin x = 2 + cos 2 x.

Solution: We apply the method of estimating the left and right sides and the boundedness of the cos and sin functions.

– 1 sin 5x 1, and -1 sin x 1

0 + 2 2 + cos 2 x 1 + 2

2 2 + cos 2 x 3

sin 5x + sin x 2, and 2 + cos 2 x 2

2 sin 5x + sin x 2, i.e.

sin 5x + sin x 2,

we have the left side is 2, and the right side is 2,

equality is possible if they are both equal to 2.

cos 2 x = 0, and sin 5x + sin x = 2, therefore

x = /2 + k, k€z (be sure to check).

Answer: /2 + k, k€z.

8. Solve the equation: cos x + cos 2x + cos 3x + cos 4x = 0.

Solution: Solve using the factorization method. We group the terms on the left side into pairs.

(In this case, any method of grouping leads to the goal.) We use the formula cos a+cos b=2 cos (a + b)/2 cos (a – b)/2.

2 cos 3/2х cos x/2 + 2 cos 7/2х cos x/2 = 0,

cos x/2 (cos 3/2x + cos 7/2x) = 0,

2 cos 5/2x cos x/2 cos x = 0,

Three cases arise:

Answer: + 2k, /5 + 2/5k, /2 + k, k€z.

Let us note that the second case includes the first. (If in the second case we take k = 4 + 5, we get + 2n). Therefore, it is impossible to say which is more correct, but in any case the answer will look “more cultured and beautiful”: x 1 = /5 + 2/5k, x 2 = /2 + k, k€z. (Again, a typical situation leading to various forms of recording the answer). The first answer is also correct.

The equation considered illustrates a very typical solution scheme - factoring the equation through pairwise grouping and using the formulas:

sin a + sin b = 2 sin (a + b)/2 cos (a – b)/2;

sin a – sin b = 2 cos (a + b)/2 sin (a – b)/2;

cos a + cos b = 2 cos (a + b)/2 cos (a – b)/2;

cos a – cos b = -2 sin (a + b)/2 sin (b – a)/2.

The problem of selecting roots, sifting out unnecessary roots when solving trigonometric equations is very specific and usually turns out to be more complex than it was for algebraic equations. Let us present solutions to equations illustrating typical cases of the appearance of extra (extraneous) roots and methods of “combat” them.

Extra roots may appear due to the fact that during the solution process the domain of definition of the equations was expanded. Let's give examples.

9. Solve the equation: (sin 4x – sin 2x – cos 3x + 2sin x -1)/(2sin 2x – 3) = 0.

Solution: Let's equate the numerator to zero (in this case, the domain of definition of the equation is expanded - x values ​​are added, turning the denominator to zero) and try to factorize it. We have:

2 cos 3x sin x – cos 3x + 2sin x – 1 = 0,

(cos 3x + 1) (2 sin x – 1) = 0.

We get two equations:

cos 3x + 1 = 0, x = /3 + 2/3k.

Let's see which k suits us. First of all, we note that the left side of our equation is a periodic function with period 2. Therefore, it is enough to find a solution to the equation that satisfies the condition 0 x< 2 (один раз “обойти” круг), затем к найденным значениям прибавить 2k.

Inequality 0 x< 2 удовлетворяют три числа: /3, 5/3.

The first is not suitable, since sin 2/3 = 3/2, the denominator goes to zero.

The answer for the first case: x 1 = + 2k, x 2 = 5/3 + 2k (you can x 2 = – /3 + 2k), k€z.

Let's find a solution to this equation that satisfies the condition 0 x< 2. Их два: /6, 5/6. Подходит второе значение.

Answer: + 2k, 5/3 + 2k, 5/6 + 2k, k€z.

10. Find the roots of the equations: v(cos 2x + sin 3x) = v2 cos x.

The solution to this equation breaks down into two stages:

1) solving an equation obtained from a given equation by squaring both of its parts;

2) selection of those roots that satisfy the condition cos x 0. In this case (as in the case of algebraic equations) there is no need to worry about the condition cos 2x + sin 3x 0. All values ​​of k that satisfy the squared equation satisfy this condition.

The first step leads us to the equation sin 3x = 1, from which x 1 = /6 + 2/3k.

Now we need to determine at what k cos (/6 + 2/3k) 0 will occur. To do this, it is enough to consider the values ​​0, 1, 2 for k, i.e. as usual, “go around the circle once,” since further the cosine values ​​will differ from those already considered by a multiple of 2.

Answer: /6 + 2k, 3/2/3 + 2k, 5/6 + 2k, k€z.

11. Solve the equation: sin 8 x – cos 5 x = 1.

The solution to this equation is based on the following simple consideration: if 0< a < 1 то a t убывает с ростом t.

This means sin 8 x sin 2 x, – cos 5 x cos 2 x;

Adding these inequalities term by term, we have:

sin 8 x – cos 5 x sin 2 x + cos 2 x = 1.

Therefore, the left side of this equation is equal to one if and only if two equalities are satisfied:

sin 8 x = sin 2 x, cos 5 x = cos 2 x,

those. sin x can take values ​​-1, 0

Answer: /2 + k, + 2k, k€z.

To complete the picture, consider another example.

12. Solve the equation: 4 cos 2 x – 4 cos 2 3x cos x + cos 2 3x = 0.

Solution: We will consider the left side of this equation as a quadratic trinomial with respect to cos x.

Let D be the discriminant of this trinomial:

1/4 D = 4 (cos 4 3x – cos 2 3x).

From the inequality D 0 it follows cos 2 3x 0 or cos 2 3x 1.

This means that two possibilities arise: cos 3x = 0 and cos 3x = ± 1.

If cos 3x = 0, then it follows from the equation that cos x = 0, whence x = /2 + k.

These values ​​of x satisfy the equation.

If cos 3x = 1, then from the equation cos x = 1/2 we find x = ± /3 + 2k. These values ​​also satisfy the equation.

Answer: /2 + k, /3 + 2k, k€z.

13. Solve the equation: sin 4 x + cos 4 x = 7/2 sin x cos x.

Solution: Transform the expression sin 4 x + cos 4 x, highlighting the perfect square: sin 4 x + cos 4 x = sin 4 x + 2 sin 2 x cos 2 x + cos 4 x – 2 sin 2 x cos 2 x = (sin 2 x + cos 2 x) 2 – 2 sin 2 x cos 2 x, whence sin 4 x + cos 4 x = 1 – 1/2 sin 2 2x. Using the resulting formula, we write the equation in the form

1-1/2 sin 2 2x = 7/4 sin 2x.

denoting sin 2х = t, -1 t 1,

we get the quadratic equation 2t 2 + 7t – 4 = 0,

solving which, we find t 1 = 1/2, t 2 = – 4

equation sin 2x = 1/2

2x = (- 1) k /6 + k, k€z, x = (- 1) k //12 + k /2, k€z.

One of the areas of mathematics that students struggle with the most is trigonometry. It is not surprising: in order to freely master this area of ​​​​knowledge, you need spatial thinking, the ability to find sines, cosines, tangents, cotangents using formulas, simplify expressions, and be able to use the number pi in calculations. In addition, you need to be able to use trigonometry when proving theorems, and this requires either a developed mathematical memory or the ability to derive complex logical chains.

Origins of trigonometry

Getting acquainted with this science should begin with the definition of sine, cosine and tangent of an angle, but first you need to understand what trigonometry does in general.

Historically, the main object of study in this branch of mathematical science was right triangles. The presence of an angle of 90 degrees makes it possible to carry out various operations that allow one to determine the values ​​of all parameters of the figure in question using two sides and one angle or two angles and one side. In the past, people noticed this pattern and began to actively use it in the construction of buildings, navigation, astronomy and even in art.

Initial stage

Initially, people talked about the relationship between angles and sides solely using the example of right triangles. Then special formulas were discovered that made it possible to expand the boundaries of use in everyday life of this branch of mathematics.

The study of trigonometry in school today begins with right triangles, after which students use the acquired knowledge in physics and solving abstract trigonometric equations, which begin in high school.

Spherical trigonometry

Later, when science reached the next level of development, formulas with sine, cosine, tangent, cotangent began to be used in spherical geometry, where different rules apply, and the sum of the angles in a triangle is always more than 180 degrees. This section is not studied in school, but it is necessary to know about its existence at least because the earth’s surface, and the surface of any other planet, is convex, which means that any surface marking will be “arc-shaped” in three-dimensional space.

Take the globe and the thread. Attach the thread to any two points on the globe so that it is taut. Please note - it has taken on the shape of an arc. Spherical geometry deals with such forms, which is used in geodesy, astronomy and other theoretical and applied fields.

Right triangle

Having learned a little about the ways of using trigonometry, let's return to basic trigonometry in order to further understand what sine, cosine, tangent are, what calculations can be performed with their help and what formulas to use.

The first step is to understand the concepts related to a right triangle. First, the hypotenuse is the side opposite the 90 degree angle. It is the longest. We remember that according to the Pythagorean theorem, its numerical value is equal to the root of the sum of the squares of the other two sides.

For example, if the two sides are 3 and 4 centimeters respectively, the length of the hypotenuse will be 5 centimeters. By the way, the ancient Egyptians knew about this about four and a half thousand years ago.

The two remaining sides, which form a right angle, are called legs. In addition, we must remember that the sum of the angles in a triangle in a rectangular coordinate system is equal to 180 degrees.

Definition

Finally, with a firm understanding of the geometric basis, one can turn to the definition of sine, cosine and tangent of an angle.

The sine of an angle is the ratio of the opposite leg (i.e., the side opposite the desired angle) to the hypotenuse. The cosine of an angle is the ratio of the adjacent side to the hypotenuse.

Remember that neither sine nor cosine can be greater than one! Why? Because the hypotenuse is by default the longest. No matter how long the leg is, it will be shorter than the hypotenuse, which means their ratio will always be less than one. Thus, if in your answer to a problem you get a sine or cosine with a value greater than 1, look for an error in the calculations or reasoning. This answer is clearly incorrect.

Finally, the tangent of an angle is the ratio of the opposite side to the adjacent side. Dividing the sine by the cosine will give the same result. Look: according to the formula, we divide the length of the side by the hypotenuse, then divide by the length of the second side and multiply by the hypotenuse. Thus, we get the same relationship as in the definition of tangent.

Cotangent, accordingly, is the ratio of the side adjacent to the corner to the opposite side. We get the same result by dividing one by the tangent.

So, we have looked at the definitions of what sine, cosine, tangent and cotangent are, and we can move on to formulas.

The simplest formulas

In trigonometry you cannot do without formulas - how to find sine, cosine, tangent, cotangent without them? But this is exactly what is required when solving problems.

The first formula you need to know when starting to study trigonometry says that the sum of the squares of the sine and cosine of an angle is equal to one. This formula is a direct consequence of the Pythagorean theorem, but it saves time if you need to know the size of the angle rather than the side.

Many students cannot remember the second formula, which is also very popular when solving school problems: the sum of one and the square of the tangent of an angle is equal to one divided by the square of the cosine of the angle. Take a closer look: this is the same statement as in the first formula, only both sides of the identity were divided by the square of the cosine. It turns out that a simple mathematical operation makes the trigonometric formula completely unrecognizable. Remember: knowing what sine, cosine, tangent and cotangent are, transformation rules and several basic formulas, you can at any time derive the required more complex formulas on a piece of paper.

Formulas for double angles and addition of arguments

Two more formulas that you need to learn are related to the values ​​of sine and cosine for the sum and difference of angles. They are presented in the figure below. Please note that in the first case, sine and cosine are multiplied both times, and in the second, the pairwise product of sine and cosine is added.

There are also formulas associated with double angle arguments. They are completely derived from the previous ones - as a practice, try to get them yourself by taking the alpha angle equal to the beta angle.

Finally, note that double angle formulas can be rearranged to reduce the power of sine, cosine, tangent alpha.

Theorems

The two main theorems in basic trigonometry are the sine theorem and the cosine theorem. With the help of these theorems, you can easily understand how to find the sine, cosine and tangent, and therefore the area of ​​the figure, and the size of each side, etc.

The sine theorem states that dividing the length of each side of a triangle by the opposite angle results in the same number. Moreover, this number will be equal to two radii of the circumscribed circle, that is, the circle containing all the points of a given triangle.

The cosine theorem generalizes the Pythagorean theorem, projecting it onto any triangles. It turns out that from the sum of the squares of the two sides, subtract their product multiplied by the double cosine of the adjacent angle - the resulting value will be equal to the square of the third side. Thus, the Pythagorean theorem turns out to be a special case of the cosine theorem.

Careless mistakes

Even knowing what sine, cosine and tangent are, it is easy to make a mistake due to absent-mindedness or an error in the simplest calculations. To avoid such mistakes, let's take a look at the most popular ones.

First, you should not convert fractions to decimals until you get the final result - you can leave the answer as a fraction unless otherwise stated in the conditions. Such a transformation cannot be called a mistake, but it should be remembered that at each stage of the problem new roots may appear, which, according to the author’s idea, should be reduced. In this case, you will waste your time on unnecessary mathematical operations. This is especially true for values ​​such as the root of three or the root of two, because they are found in problems at every step. The same goes for rounding “ugly” numbers.

Further, note that the cosine theorem applies to any triangle, but not the Pythagorean theorem! If you mistakenly forget to subtract twice the product of the sides multiplied by the cosine of the angle between them, you will not only get a completely wrong result, but you will also demonstrate a complete lack of understanding of the subject. This is worse than a careless mistake.

Thirdly, do not confuse the values ​​for angles of 30 and 60 degrees for sines, cosines, tangents, cotangents. Remember these values, because the sine of 30 degrees is equal to the cosine of 60, and vice versa. It is easy to confuse them, as a result of which you will inevitably get an erroneous result.

Application

Many students are in no hurry to start studying trigonometry because they do not understand its practical meaning. What is sine, cosine, tangent for an engineer or astronomer? These are concepts with which you can calculate the distance to distant stars, predict the fall of a meteorite, or send a research probe to another planet. Without them, it is impossible to build a building, design a car, calculate the load on a surface or the trajectory of an object. And these are just the most obvious examples! After all, trigonometry in one form or another is used everywhere, from music to medicine.

In conclusion

So you're sine, cosine, tangent. You can use them in calculations and successfully solve school problems.

The whole point of trigonometry comes down to the fact that using the known parameters of a triangle you need to calculate the unknowns. There are six parameters in total: the length of three sides and the size of three angles. The only difference in the tasks lies in the fact that different input data are given.

You now know how to find sine, cosine, tangent based on the known lengths of the legs or hypotenuse. Since these terms mean nothing more than a ratio, and a ratio is a fraction, the main goal of a trigonometry problem is to find the roots of an ordinary equation or system of equations. And here regular school mathematics will help you.

We will begin our study of trigonometry with the right triangle. Let's define what sine and cosine are, as well as tangent and cotangent of an acute angle. This is the basics of trigonometry.

Let us remind you that right angle is an angle equal to 90 degrees. In other words, half a turned angle.

Acute angle- less than 90 degrees.

Obtuse angle- greater than 90 degrees. In relation to such an angle, “obtuse” is not an insult, but a mathematical term :-)

Let's draw a right triangle. A right angle is usually denoted by . Please note that the side opposite the corner is indicated by the same letter, only small. Thus, the side opposite angle A is designated .

The angle is denoted by the corresponding Greek letter.

Hypotenuse of a right triangle is the side opposite the right angle.

Legs- sides lying opposite acute angles.

The leg lying opposite the angle is called opposite(relative to angle). The other leg, which lies on one of the sides of the angle, is called adjacent.

Sinus The acute angle in a right triangle is the ratio of the opposite side to the hypotenuse:

Cosine acute angle in a right triangle - the ratio of the adjacent leg to the hypotenuse:

Tangent acute angle in a right triangle - the ratio of the opposite side to the adjacent:

Another (equivalent) definition: the tangent of an acute angle is the ratio of the sine of the angle to its cosine:

Cotangent acute angle in a right triangle - the ratio of the adjacent side to the opposite (or, which is the same, the ratio of cosine to sine):

Note the basic relationships for sine, cosine, tangent, and cotangent below. They will be useful to us when solving problems.

Let's prove some of them.

Okay, we have given definitions and written down formulas. But why do we still need sine, cosine, tangent and cotangent?

We know that the sum of the angles of any triangle is equal to.

We know the relationship between parties right triangle. This is the Pythagorean theorem: .

It turns out that knowing two angles in a triangle, you can find the third. Knowing the two sides of a right triangle, you can find the third. This means that the angles have their own ratio, and the sides have their own. But what should you do if in a right triangle you know one angle (except the right angle) and one side, but you need to find the other sides?

This is what people in the past encountered when making maps of the area and the starry sky. After all, it is not always possible to directly measure all sides of a triangle.

Sine, cosine and tangent - they are also called trigonometric angle functions- give relationships between parties And corners triangle. Knowing the angle, you can find all its trigonometric functions using special tables. And knowing the sines, cosines and tangents of the angles of a triangle and one of its sides, you can find the rest.

We will also draw a table of the values ​​of sine, cosine, tangent and cotangent for “good” angles from to.

Please note the two red dashes in the table. At appropriate angle values, tangent and cotangent do not exist.

Let's look at several trigonometry problems from the FIPI Task Bank.

1. In a triangle, the angle is , . Find .

The problem is solved in four seconds.

Since , .

2. In a triangle, the angle is , , . Find .

Let's find it using the Pythagorean theorem.

The problem is solved.

Often in problems there are triangles with angles and or with angles and. Remember the basic ratios for them by heart!

For a triangle with angles and the leg opposite the angle at is equal to half of the hypotenuse.

A triangle with angles and is isosceles. In it, the hypotenuse is times larger than the leg.

We looked at problems solving right triangles - that is, finding unknown sides or angles. But that's not all! There are many problems in the Unified State Examination in mathematics that involve sine, cosine, tangent or cotangent of an external angle of a triangle. More on this in the next article.

Table of values ​​of trigonometric functions

Note. This table of trigonometric function values ​​uses the √ sign to represent the square root. To indicate a fraction, use the symbol "/".

See also useful materials:

For determining the value of a trigonometric function, find it at the intersection of the line indicating the trigonometric function. For example, sine 30 degrees - we look for the column with the heading sin (sine) and find the intersection of this table column with the row “30 degrees”, at their intersection we read the result - one half. Similarly we find cosine 60 degrees, sine 60 degrees (once again, at the intersection of the sin column and the 60 degree line we find the value sin 60 = √3/2), etc. The values ​​of sines, cosines and tangents of other “popular” angles are found in the same way.

Sine pi, cosine pi, tangent pi and other angles in radians

The table below of cosines, sines and tangents is also suitable for finding the value of trigonometric functions whose argument is given in radians. To do this, use the second column of angle values. Thanks to this, you can convert the value of popular angles from degrees to radians. For example, let's find the angle of 60 degrees in the first line and read its value in radians under it. 60 degrees is equal to π/3 radians.

The number pi unambiguously expresses the dependence of the circumference on the degree measure of the angle. Thus, pi radians are equal to 180 degrees.

Any number expressed in terms of pi (radians) can be easily converted to degrees by replacing pi (π) with 180.

Examples:
1. Sine pi.
sin π = sin 180 = 0
thus, the sine of pi is the same as the sine of 180 degrees and it is equal to zero.

2. Cosine pi.
cos π = cos 180 = -1
thus, the cosine of pi is the same as the cosine of 180 degrees and it is equal to minus one.

3. Tangent pi
tg π = tg 180 = 0
thus, tangent pi is the same as tangent 180 degrees and it is equal to zero.

Table of sine, cosine, tangent values ​​for angles 0 - 360 degrees (common values)

If in the table of values ​​of trigonometric functions a dash is indicated instead of the function value (tangent (tg) 90 degrees, cotangent (ctg) 180 degrees), then for a given value of the degree measure of the angle the function does not have a specific value. If there is no dash, the cell is empty, which means we have not yet entered the required value. We are interested in what queries users come to us for and supplement the table with new values, despite the fact that current data on the values ​​of cosines, sines and tangents of the most common angle values ​​is quite sufficient to solve most problems.

Table of values ​​of trigonometric functions sin, cos, tg for the most popular angles
0, 15, 30, 45, 60, 90 ... 360 degrees
(numeric values ​​“as per Bradis tables”)

Sinus acute angle α of a right triangle is the ratio opposite leg to hypotenuse.
It is denoted as follows: sin α.

Cosine The acute angle α of a right triangle is the ratio of the adjacent leg to the hypotenuse.
It is designated as follows: cos α.

Tangent acute angle α is the ratio of the opposite side to the adjacent side.
It is designated as follows: tg α.

Cotangent acute angle α is the ratio of the adjacent side to the opposite side.
It is designated as follows: ctg α.

The sine, cosine, tangent and cotangent of an angle depend only on the size of the angle.

Rules:

Basic trigonometric identities in a right triangle:

(α – acute angle opposite to the leg b and adjacent to the leg a . Side With – hypotenuse. β – second acute angle).

As the acute angle increasessin α andtan α increase, andcos α decreases.

For any acute angle α:

sin (90° – α) = cos α

cos (90° – α) = sin α

Example-explanation:

Let in a right triangle ABC
AB = 6,
BC = 3,
angle A = 30º.

Let's find out the sine of angle A and the cosine of angle B.

Solution .

1) First, we find the value of angle B. Everything is simple here: since in a right triangle the sum of the acute angles is 90º, then angle B = 60º:

B = 90º – 30º = 60º.

2) Let's calculate sin A. We know that the sine is equal to the ratio of the opposite side to the hypotenuse. For angle A, the opposite side is side BC. So:

BC 3 1
sin A = -- = - = -
AB 6 2

3) Now let's calculate cos B. We know that the cosine is equal to the ratio of the adjacent leg to the hypotenuse. For angle B, the adjacent leg is the same side BC. This means that we again need to divide BC by AB - that is, perform the same actions as when calculating the sine of angle A:

BC 3 1
cos B = -- = - = -
AB 6 2

The result is:
sin A = cos B = 1/2.

sin 30º = cos 60º = 1/2.

It follows from this that in a right triangle, the sine of one acute angle is equal to the cosine of another acute angle - and vice versa. This is exactly what our two formulas mean:
sin (90° – α) = cos α
cos (90° – α) = sin α

Let's make sure of this again:

1) Let α = 60º. Substituting the value of α into the sine formula, we get:
sin (90º – 60º) = cos 60º.
sin 30º = cos 60º.

2) Let α = 30º. Substituting the value of α into the cosine formula, we get:
cos (90° – 30º) = sin 30º.
cos 60° = sin 30º.

(For more information about trigonometry, see the Algebra section)