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"Tell me and I will forget,
Show me and I will remember
Involve me and I will learn."
(Chinese proverb)

Mathematics has long become the language of science and technology, and is now increasingly penetrating into daily life and everyday language, is increasingly being introduced into areas that seem traditionally distant from it. Intensive mathematization of various fields human activity especially intensified with the rapid development of computers. Computerization of society, introduction of modern information technologies require mathematical literacy of a person in every workplace. This presupposes both specific mathematical knowledge and a certain style of thinking. In particular, learning trigonometry is an important aspect. The study of trigonometric functions is widely used in practice, in the study of many physical processes, in industry, and even in medicine. Students who will later in their professional activity will use mathematics, it is necessary to provide them with high mathematical preparation.

Trigonometry – component school course mathematics. Good knowledge and strong skills in trigonometry are evidence sufficient level mathematical culture, an indispensable condition successful study at the university of mathematics, physics, and a number of technical disciplines. However, a significant proportion of school graduates reveal from year to year very poor preparation in this important section of mathematics, as evidenced by the results of past years, since an analysis of the unified state exam showed that students make many mistakes when completing tasks in this particular section or do not take them at all for such tasks.

But even the Greeks, at the dawn of mankind, considered trigonometry the most important of the sciences, for geometry is the queen of mathematics, and trigonometry is the queen of geometry. Therefore, we, without disputing the ancient Greeks, will consider trigonometry one of the most important sections of the school course, and of all mathematical science in general.

Physics and geometry cannot do without trigonometry. The Unified State Exam cannot do without trigonometry. In Part B alone, questions on trigonometry are found in almost a third of the types of tasks. This is the simplest solution trigonometric equations in task B5, and work with trigonometric expressions in task B7, and the study of trigonometric functions in task B14, as well as task B12, in which there are formulas that describe physical phenomena and contain trigonometric functions. It is impossible not to note geometric tasks, in the solution of which the definitions of sine, cosine, tangent and cotangent are used acute angle right triangle, and basic trigonometric identities. And this is only part B! But there are also favorite trigonometric equations with selection of roots C1, and “not so favorite” geometric tasks C2 and C4.

How can students be trained on these topics? A large number of methods can be offered, but the most important thing is that the children do not have a feeling of fear and unnecessary anxiety, due to the huge variety of different tasks and formulas. And for this it is necessary to create a positive mood when solving these tasks. This presentation can be used for conducting classes with students and for speaking at seminars for mathematicians in preparation for the Unified State Exam. It offers some types of tasks and discusses their solutions.

Good training can be not only a simple solution of these tasks, but also the students’ own compilation of them. Depending on the preparation, these can be tests for working out limitations in solving trigonometric equations C1, and even the equations themselves.

To others active method is to conduct classes in the form mind games. One of the most convenient options, I think, is the “Custom Game” format. This game uniform, especially now with the use of computer presentations, can be used during test lessons, after studying topics, and in preparation for the Unified State Exam. The proposed work contains “Your own game. Solving trigonometric equations and inequalities.”

The result of the proposed work should be the successful solution of Unified State Examination tasks on the topic “Trigonometry”.

The video course “Get an A” includes all the topics necessary for successful passing the Unified State Exam in mathematics for 60-65 points. Completely all problems 1-13 Profile Unified State Examination mathematics. Also suitable for passing the Basic Unified State Examination in mathematics. If you want to pass the Unified State Exam with 90-100 points, you need to solve part 1 in 30 minutes and without mistakes!

Preparation course for the Unified State Exam for grades 10-11, as well as for teachers. Everything you need to solve Part 1 of the Unified State Exam in mathematics (the first 12 problems) and Problem 13 (trigonometry). And this is more than 70 points on the Unified State Exam, and neither a 100-point student nor a humanities student can do without them.

All the necessary theory. Quick solutions, pitfalls and secrets of the Unified State Exam. All current tasks of part 1 from the FIPI Task Bank have been analyzed. The course fully complies with the requirements of the Unified State Exam 2018.

The course contains 5 large topics, 2.5 hours each. Each topic is given from scratch, simply and clearly.

Hundreds of Unified State Exam tasks. Word problems and probability theory. Simple and easy to remember algorithms for solving problems. Geometry. Theory, reference material, analysis of all types of Unified State Examination tasks. Stereometry. Tricky solutions, useful cheat sheets, development of spatial imagination. Trigonometry from scratch to problem 13. Understanding instead of cramming. Visual explanation complex concepts. Algebra. Roots, powers and logarithms, function and derivative. Basis for solution complex tasks 2 parts of the Unified State Exam.

Educational and methodological manual
to prepare for the Unified State Exam in mathematics

TRIGONOMETRY IN THE USE IN MATHEMATICS

The purpose of this tutorial is to assistance to schoolchildren in preparing for the Unified State Examination in mathematics in the section “Trigonometry”.

IN textbook analysis is carried out and solutions are given to typical problems in trigonometry offered by the Moscow Institute open education in various control, diagnostic, training, demonstration and exam papers in mathematics for schoolchildren in grades 10 and 11.

After analyzing each typical task Similar problems are given for independent solution.

The necessary theoretical information used in solving problems can be found in the “Trigonometry” section of our “Handbook of Mathematics for Schoolchildren”.

With the main methods for solving trigonometric equations you can meet us in our educational manual"Solving trigonometric equations".

For schoolchildren in grades 10 and 11 who want to prepare well and pass Unified State Examination in mathematics or Russian language on high score, The educational center"Resolventa" conducts preparation courses for the Unified State Exam.

We also organize for schoolchildren

WITH demo options Unified State Exam published on the official information portal of the United State Exam, can be found at

A) Solve equation 2(\sin x-\cos x)=tgx-1.

b) \left[ \frac(3\pi )2;\,3\pi \right].

Solution

A) Opening the brackets and moving all the terms to the left side, we get the equation 1+2 \sin x-2 \cos x-tg x=0. Considering that \cos x \neq 0, the term 2 \sin x can be replaced by 2 tan x \cos x, we obtain the equation 1+2 tg x \cos x-2 \cos x-tg x=0,

1) which by grouping can be reduced to the form (1-tg x)(1-2 \cos x)=0. 1-tg x=0, tan x=1,

2) x=\frac\pi 4+\pi n, n \in \mathbb Z; 1-2 \cos x=0, \cos x=\frac12,

b) x=\pm \frac\pi 3+2\pi n, n \in \mathbb Z. Using the number circle, select the roots belonging to the interval

\left[ \frac(3\pi )2;\, 3\pi \right].

x_1=\frac\pi 4+2\pi =\frac(9\pi )4,

x_2=\frac\pi 3+2\pi =\frac(7\pi )3,

x_3=-\frac\pi 3+2\pi =\frac(5\pi )3.

A) Answer \frac\pi 4+\pi n,

b) \pm\frac\pi 3+2\pi n, n \in \mathbb Z; \frac(5\pi )3, \frac(7\pi )3,

\frac(9\pi )4.

A) Condition Solve the equation

b)(2\sin ^24x-3\cos 4x)\cdot \sqrt (tgx)=0. Indicate the roots of this equation that belong to the interval

Solution

A)\left(0;\,\frac(3\pi )2\right] ; ODZ:

\begin(cases) tgx\geqslant 0\\x\neq \frac\pi 2+\pi k,k \in \mathbb Z. \end(cases)

The original equation on the ODZ is equivalent to a set of equations

\left[\!\!\begin(array)(l) 2 \sin ^2 4x-3 \cos 4x=0,\\tg x=0. \end(array)\right. Let's solve the first equation. To do this we will make a replacement \cos 4x=t, t \in [-1; 1].

Then \sin^24x=1-t^2.

We get:

2(1-t^2)-3t=0, 2t^2+3t-2=0,

t_1=\frac12,

t_2=-2, t_2\notin [-1; 1].

\cos 4x=\frac12,

4x=\pm\frac\pi 3+2\pi n,

x=\pm \frac\pi (12)+\frac(\pi n)2, n \in \mathbb Z.

Let's solve the second equation.

tg x=0,\, x=\pi k, k \in \mathbb Z.

Using the unit circle, we find solutions that satisfy the ODZ. The “+” sign marks the 1st and 3rd quarters, in which tg x>0. We get: x=\pi k, k \in \mathbb Z;

b) x=\frac\pi (12)+\pi n, n \in \mathbb Z; x=\frac(5\pi )(12)+\pi m, m \in \mathbb Z.

Let's find the roots belonging to the interval \left(0;\,\frac(3\pi )2\right]. x=\frac\pi (12), x=\frac(5\pi )(12); x=\pi ;

x_3=-\frac\pi 3+2\pi =\frac(5\pi )3.

A) x=\frac(13\pi )(12); x=\frac(17\pi )(12). \pi k, k \in \mathbb Z;

b) \frac\pi (12)+\pi n, n \in \mathbb Z; \frac(5\pi )(12)+\pi m, m \in \mathbb Z. \pi; \frac\pi (12); \frac(5\pi )(12);

\frac(13\pi )(12); \frac(17\pi )(12). Source: “Mathematics. Preparation for the Unified State Exam 2017.

\frac(9\pi )4.

A) Profile level \cos ^2x+\cos ^2\frac\pi 6=\cos ^22x+\sin ^2\frac\pi 3;

b) List all roots belonging to the interval \left(\frac(7\pi )2;\,\frac(9\pi )2\right].

Solution

A) Because \sin \frac\pi 3=\cos \frac\pi 6, That \sin ^2\frac\pi 3=\cos ^2\frac\pi 6, Means, given equation is equivalent to the equation \cos^2x=\cos ^22x, which, in turn, is equivalent to the equation \cos^2x-\cos ^2 2x=0.

But \cos ^2x-\cos ^22x= (\cos x-\cos 2x)\cdot (\cos x+\cos 2x) And

\cos 2x=2 \cos ^2 x-1, so the equation becomes

(\cos x-(2 \cos ^2 x-1))\,\cdot(\cos x+(2 \cos ^2 x-1))=0,

(2 \cos ^2 x-\cos x-1)\,\cdot (2 \cos ^2 x+\cos x-1)=0.

Then either 2 \cos ^2 x-\cos x-1=0, or 2 \cos ^2 x+\cos x-1=0.

Solving the first equation as quadratic equation relative to \cos x, we get:

(\cos x)_(1,2)=\frac(1\pm\sqrt 9)4=\frac(1\pm3)4. Therefore either \cos x=1 or \cos x=-\frac12. If \cos x=1, then x=2k\pi , k \in \mathbb Z. If \cos x=-\frac12, That x=\pm \frac(2\pi )3+2s\pi , s \in \mathbb Z.

Similarly, solving the second equation, we get either \cos x=-1 or \cos x=\frac12. If \cos x=-1, then the roots x=\pi +2m\pi , m \in \mathbb Z. If \cos x=\frac12, That x=\pm \frac\pi 3+2n\pi , n \in \mathbb Z.

Let's combine the solutions obtained:

x=m\pi , m \in \mathbb Z; x=\pm \frac\pi 3 +s\pi , s \in \mathbb Z.

b) Let's select the roots that fall within a given interval using a number circle.

We get: x_1 =\frac(11\pi )3, x_2=4\pi , x_3 =\frac(13\pi )3.

x_3=-\frac\pi 3+2\pi =\frac(5\pi )3.

A) m\pi, m\in \mathbb Z; \pm \frac\pi 3 +s\pi , s \in \mathbb Z;

b) \frac(11\pi )3, 4\pi , \frac(13\pi )3.

Source: “Mathematics. Preparation for the Unified State Exam 2017. Profile level." Ed. F. F. Lysenko, S. Yu. Kulabukhova.

\frac(9\pi )4.

A) Condition 10\cos ^2\frac x2=\frac(11+5ctg\left(\dfrac(3\pi )2-x\right) )(1+tgx).

b) Indicate the roots of this equation that belong to the interval \left(-2\pi ; -\frac(3\pi )2\right).

Solution

A) 1. According to the reduction formula, ctg\left(\frac(3\pi )2-x\right) =tgx. The domain of definition of the equation will be such values ​​of x such that \cos x \neq 0 and tan x \neq -1. Let's transform the equation using the cosine formula double angle 2 \cos ^2 \frac x2=1+\cos x. We get the equation:

5(1+\cos x) =\frac(11+5tgx)(1+tgx). notice, that \frac(11+5tgx)(1+tgx)= \frac(5(1+tgx)+6)(1+tgx)= 5+\frac(6)(1+tgx), so the equation becomes: 5+5 \cos x=5 +\frac(6)(1+tgx). From here \cos x =\frac(\dfrac65)(1+tgx),

\cos x+\sin x =\frac65. 2. Transform \sin x+\cos x using the reduction formula and the sum of cosines formula: \sin x=\cos \left(\frac\pi 2-x\right), \cos x+\sin x= 2\cos \frac\pi 4\cos \left(x-\frac\pi 4\right)= \sqrt 2\cos \left(x-\frac\pi 4\right) = \frac65.

From here \cos \left(x-\frac\pi 4\right) =\frac(3\sqrt 2)5. Means, x-\frac\pi 4= arc\cos \frac(3\sqrt 2)5+2\pi k, k \in \mathbb Z,

or x-\frac\pi 4= -arc\cos \frac(3\sqrt 2)5+2\pi t, t \in \mathbb Z.

That's why x=\frac\pi 4+arc\cos \frac(3\sqrt 2)5+2\pi k,k \in \mathbb Z,

or x =\frac\pi 4-arc\cos \frac(3\sqrt 2)5+2\pi t,t \in \mathbb Z.

The found values ​​of x belong to the domain of definition.

b) Let us first find out where the roots of the equation fall at k=0 and t=0. These will be numbers accordingly a=\frac\pi 4+arccos \frac(3\sqrt 2)5 And

b=\frac\pi 4-arccos \frac(3\sqrt 2)5.

1. Let us prove the auxiliary inequality:<\frac{3\sqrt 2}2<1.

\frac(\sqrt 2)(2) Really,<\frac{6\sqrt2}{10}=\frac{3\sqrt2}{5}.

\frac(\sqrt 2)(2)=\frac(5\sqrt 2)(10) Note also that<1^2=1, \left(\frac(3\sqrt 2)5\right) ^2=\frac(18)(25) Means<1.

\frac(3\sqrt 2)5 (1) 2. From inequalities

By the arccosine property we get:

0

From here arccos 1<\frac\pi 4+arc\cos \frac{3\sqrt 2}5<\frac\pi 4+\frac\pi 4,

0<\frac\pi 4+arccos \frac{3\sqrt 2}5<\frac\pi 2,

0

\frac\pi 4+0 Likewise,

-\frac\pi 4<\frac\pi 4-arccos \frac{3\sqrt 2}5< 0=\frac\pi 4-\frac\pi 4<\frac\pi 2,

0

\frac\pi 4

For k=-1 and t=-1 we obtain the roots of the equation a-2\pi and b-2\pi. \Bigg(a-2\pi =-\frac74\pi +arccos \frac(3\sqrt 2)5,\, b-2\pi =-\frac74\pi -arccos \frac(3\sqrt 2)5\Bigg). Wherein

-2\pi 2\pi This means that these roots belong to the given interval

\left(-2\pi , -\frac(3\pi )2\right).

For other values ​​of k and t, the roots of the equation do not belong to the given interval. Indeed, if k\geqslant 1 and t\geqslant 1, then the roots are greater than 2\pi.

x_3=-\frac\pi 3+2\pi =\frac(5\pi )3.

A) If k\leqslant -2 and t\leqslant -2, then the roots are smaller

b) -\frac(7\pi )2.

Source: “Mathematics. Preparation for the Unified State Exam 2017. Profile level." Ed. F. F. Lysenko, S. Yu. Kulabukhova.

\frac(9\pi )4.

A) Condition \frac\pi4\pm arccos\frac(3\sqrt2)5+2\pi k, k\in\mathbb Z;

b)-\frac(7\pi)4\pm arccos\frac(3\sqrt2)5.

Solution

A)\sin \left(\frac\pi 2+x\right) =\sin (-2x).

Find all the roots of this equation that belong to the interval ;

Let's transform the equation:

\cos x =-\sin 2x,

\cos x+2 \sin x \cos x=0,

\cos x(1+2 \sin x)=0,

\cos x=0,

x =\frac\pi 2+\pi n, n\in \mathbb Z;

1+2 \sin x=0,

b)\sin x=-\frac12,

x=(-1)^(k+1)\cdot \frac\pi 6+\pi k, k \in \mathbb Z. We find the roots belonging to the segment using the unit circle.

x_3=-\frac\pi 3+2\pi =\frac(5\pi )3.

A) The indicated interval contains a single number \frac\pi 2.

b) We find the roots belonging to the segment using the unit circle.

Source: “Mathematics. Preparation for the Unified State Exam 2017. Profile level." Ed. F. F. Lysenko, S. Yu. Kulabukhova.

\frac(9\pi )4.

(-1)^(k+1)\cdot \frac\pi 6+\pi k, k \in \mathbb Z; is not included in the DZ.

Means, \sin x \neq 1. Divide both sides of the equation by a factor (\sin x-1), different from zero. We get the equation \frac 1(1+\cos 2x)=\frac 1(1+\cos (\pi +x)), or equation 1+\cos 2x=1+\cos (\pi +x). Applying the reduction formula on the left side and the reduction formula on the right, we obtain the equation \cos x=t, Where -1 \leqslant t \leqslant 1 reduce to square: 2t^2+t-1=0, whose roots t_1=-1 a=\frac\pi 4+arccos \frac(3\sqrt 2)5 t_2=\frac12. Returning to the variable x, we get \cos x = \frac12 or \cos x=-1, where x=\frac \pi 3+2\pi m, m \in \mathbb Z, x=-\frac \pi 3+2\pi n, n \in \mathbb Z, x=\pi +2\pi k, k \in \mathbb Z.

b) Let's solve inequalities

1) -\frac(3\pi )2 \leqslant \frac(\pi )3+2\pi m \leqslant -\frac \pi 2 ,

2) -\frac(3\pi )2 \leqslant -\frac \pi 3+2\pi n \leqslant -\frac \pi (2,)

3) -\frac(3\pi )2 \leqslant \pi+2\pi k \leqslant -\frac \pi 2 , m, n, k \in \mathbb Z.

1) -\frac(3\pi )2 \leqslant \frac(\pi )3+2\pi m \leqslant -\frac \pi 2 , -\frac32\leqslant \frac13+2m \leqslant -\frac12 -\frac(11)6 \leqslant 2m\leqslant -\frac56 , -\frac(11)(12) \leqslant m \leqslant -\frac5(12).

\left [-\frac(11)(12);-\frac5(12)\right].

2) -\frac (3\pi) 2 \leqslant -\frac(\pi )3+2\pi n \leqslant -\frac(\pi )(2), -\frac32 \leqslant -\frac13 +2n \leqslant -\frac12 , -\frac76 \leqslant 2n \leqslant -\frac1(6), -\frac7(12) \leqslant n \leqslant -\frac1(12).

There are no integers in the range \left[ -\frac7(12) ; -\frac1(12)\right].

3) -\frac(3\pi )2 \leqslant \pi +2\pi k\leqslant -\frac(\pi )2, -\frac32 \leqslant 1+2k\leqslant -\frac12, -\frac52 \leqslant 2k \leqslant -\frac32, -\frac54 \leqslant k \leqslant -\frac34.

This inequality is satisfied by k=-1, then x=-\pi.

x_3=-\frac\pi 3+2\pi =\frac(5\pi )3.

A) \frac \pi 3+2\pi m; -\frac \pi 3+2\pi n; \pi +2\pi k, m, n, k \in \mathbb Z;

b) -\pi .

\(\blacktriangleright\) Consider a rectangular coordinate system and in it a circle with unit radius and center at the origin.

Angle in \(1^\circ\)- this is the central angle that rests on an arc whose length is equal to \(\dfrac1(360)\) the length of the entire circle.

\(\blacktriangleright\) We will consider angles on the circle in which the vertex is in the center of the circle, and one side always coincides with the positive direction of the \(Ox\) axis (highlighted in red in the figure).
The corners are marked in this way \(45^\circ,\ 180^\circ,\ 240^\circ\):

Note that the angle \(0^\circ\) is an angle whose both sides coincide with the positive direction of the axis \(Ox\) .

The point at which the second side of such an angle \(\alpha\) intersects the circle will be called \(P_(\alpha)\) .
The position of the point \(P_(0)\) will be called the initial position.

Thus, we can say that we rotate in a circle from the initial position \(P_0\) to the position \(P_(\alpha)\) by an angle \(\alpha\) .

\(\blacktriangleright\) A counterclockwise rotation in a circle is a positive rotation. A clockwise rotation is a negative rotation.

For example, in the figure the corners are marked \(-45^\circ, -90^\circ, -160^\circ\):

\(\blacktriangleright\) Consider the point \(P_(30^\circ)\) on a circle. In order to rotate in a circle from the initial position to the point \(P_(30^\circ)\), you need to rotate through the angle \(30^\circ\) (orange). If we make a full revolution (that is, by \(360^\circ\) ) and another turn by \(30^\circ\) , then we will again get to this point, although we have already made a turn by an angle \(390^\circ=360^\circ+30^\circ\)(blue). We can also get to this point by making a turn to \(-330^\circ\) (green), to \(750^\circ=360^\circ+360^\circ+30^\circ\) etc.

Thus, each point on the circle corresponds to an infinite number of angles, and these angles differ from each other by an integer number of full revolutions ( \(n\cdot360^\circ, n\in\mathbb(Z)\)).
For example, the angle \(30^\circ\) is \(360^\circ\) greater than the angle \(-330^\circ\) and \(2\cdot 360^\circ\) less than angle \(750^\circ\) .

All angles located at the point \(P_(30^\circ)\) can be written in the form: \(\alpha=30^\circ+n\cdot 360^\circ, \n\in\mathbb(Z)\).

\(\blacktriangleright\) Angle in \(1\) radians- this is the central angle that rests on an arc whose length is equal to the radius of the circle:

Because the length of the entire circle with radius \(R\) is equal to \(2\pi R\) , and in degree measure - \(360^\circ\), then we have \(360^\circ=2\pi \cdot 1\textbf(rad)\), where \ This is the basic formula with which you can convert degrees to radians and vice versa.

Example 1. Find the radian measure of the angle \(60^\circ\) .

Because \(180^\circ = \pi \Rightarrow 1^\circ = \dfrac(\pi)(180) \Rightarrow 60^\circ=\dfrac(\pi)3\)

Example 2. Find the degree measure of the angle \(\dfrac34 \pi\) .

Because \(\pi=180^\circ \Rightarrow \dfrac34 \pi=\dfrac34 \cdot 180^\circ=135^\circ\).

Usually they write, for example, not \(\dfrac(\pi)4 \text( rad)\), but simply \(\dfrac(\pi)4\) (i.e. the unit of measurement “rad” is omitted). Please note that the designation of degrees when writing an angle don't lower. Thus, by writing “the angle is equal to \(1\)” we mean that “the angle is equal to \(1\) radians”, and not “the angle is equal to \(1\) degrees”.

Because \(\pi \thickapprox 3.14 \Rightarrow 180^\circ \thickapprox 3.14 \textbf(rad) \Rightarrow 1 \textbf(rad) \thickapprox 57^\circ\).
Such an approximate substitution cannot be made in problems, but knowing what \(1\) radians in degrees is approximately equal to often helps in solving some problems. For example, in this way it is easier to find an angle of \(5\) radians on a circle: it is approximately equal to \(285^\circ\) .

\(\blacktriangleright\) From the course of planimetry (geometry on a plane) we know that for angles \(0<\alpha< 90^\circ\) определены синус, косинус, тангенс и котангенс следующим образом:
if given a right triangle with sides \(a, b, c\) and angle \(\alpha\), then:

Because any angles are defined on the unit circle \(\alpha\in(-\infty;+\infty)\), then you need to determine the sine, cosine, tangent and cotangent for any angle.
Consider the unit circle and on it the angle \(\alpha\) and the corresponding point \(P_(\alpha)\) :

Let us lower the perpendicular \(P_(\alpha)K\) from the point \(P_(\alpha)\) to the axis \(Ox\) . We get a right triangle \(\triangle OP_(\alpha)K\) from which we have: \[\sin\alpha=\dfrac(P_(\alpha)K)(P_(\alpha)O) \qquad \cos \alpha=\dfrac(OK)(P_(\alpha)O)\] Note that the segment \(OK\) is nothing more than the abscissa \(x_(\alpha)\) of the point \(P_(\alpha)\) , and the segment \(P_(\alpha)K\) is the ordinate \(y_(\alpha)\) . Note also that since we took the unit circle, then \(P_(\alpha)O=1\) is its radius.
Thus, \[\sin\alpha=y_(\alpha), \qquad \cos \alpha=x_(\alpha)\]

Thus, if the point \(P_(\alpha)\) had coordinates \((x_(\alpha)\,;y_(\alpha))\), then through the corresponding angle its coordinates can be rewritten as \((\ cos\alpha\,;\sin\alpha)\) .

Definition: 1. The sine of the angle \(\alpha\) is the ordinate of the point \(P_(\alpha)\) corresponding to this angle on the unit circle.

2. The cosine of the angle \(\alpha\) is the abscissa of the point \(P_(\alpha)\) corresponding to this angle on the unit circle.

Therefore, the axis \(Oy\) is called the axis of sines, the axis \(Ox\) is called the axis of cosines.

\(\blacktriangleright\) The circle can be divided into \(4\) quarters, as shown in the figure.


Because in the \(I\) quarter both the abscissa and ordinates of all points are positive, then the cosines and sines of all angles from this quarter are also positive.
Because in the \(II\) quarter, the ordinates of all points are positive and the abscissas are negative, then the cosines of all angles from this quarter are negative, and the sines are positive.
Similarly, you can determine the sign of the sine and cosine for the remaining quarters.

Example 3. Since, for example, the points \(P_(\frac(\pi)(6))\) and \(P_(-\frac(11\pi)6)\) coincide, then their coordinates are equal, i.e. \(\sin\dfrac(\pi)6=\sin \left(-\dfrac(11\pi)6\right),\ \cos \dfrac(\pi)6=\cos \left(-\dfrac( 11\pi)6\right)\).

Example 4. Consider the points \(P_(\alpha)\) and \(P_(\pi-\alpha)\) . Let for convenience let \(0<\alpha<\dfrac{\pi}2\) .

Let's draw perpendiculars to the axis \(Ox\) : \(OK\) and \(OK_1\) . Triangles \(OKP_(\alpha)\) and \(OK_1P_(\pi-\alpha)\) are equal in hypotenuse and angle ( \(\angle P_(\alpha)OK=\angle P_(\pi-\alpha)OK_1=\alpha\)). Hence,\(OK=OK_1, KP_(\alpha)=K_1P_(\pi-\alpha)\) . Because point coordinates \(P_(\alpha)=(OK;KP_(\alpha))=(\cos\alpha\,;\sin\alpha)\), and the points \(P_(\pi-\alpha)=(-OK_1;K_1P_(\pi-\alpha))=(\cos(\pi-\alpha)\,;\sin(\pi-\alpha))\)

, hence, \[\cos(\pi-\alpha)=-\cos\alpha, \qquad \sin(\pi-\alpha)=\sin\alpha\]: In this way other formulas called

reduction formulas

\[(\large(\begin(array)(l|r) \hline \sin(\pi-\alpha)=\sin\alpha & \cos(\pi-\alpha)=-\cos\alpha\\ \sin(\pi+\alpha)=-\sin\alpha & \cos(\pi+\alpha)=-\cos\alpha\\ \sin(2\pi\pm\alpha)=\pm\sin\alpha & \cos (2\pi\pm\alpha)=\cos\alpha\\ \sin \left(\dfrac(\pi)2\pm\alpha\right)=\cos\alpha & \cos\left(\dfrac (\pi)2\pm\alpha\right)=\pm\sin\alpha\\ \hline \end(array)))\]
Using these formulas, you can find the sine or cosine of any angle, reducing this value to the sine or cosine of the angle from the \(I\) quarter.

Table of sines, cosines, tangents and cotangents of angles from the first quarter:

\[(\large(\begin(array)(|c|c|c|c|c|c|) \hline &&&&&\\[-17pt] & \quad 0 \quad (0^ \circ)& \quad \dfrac(\pi)6 \quad (30^\circ) & \quad \dfrac(\pi)4 \quad (45^\circ) & \quad \dfrac(\pi)3 \quad (60^\circ )& \quad \dfrac(\pi)2 \quad (90^\circ) \\ &&&&&\\[-17pt] \hline \sin & 0 &\frac12&\frac(\sqrt2)2&\frac(\sqrt3) 2&1\\ \hline \cos &1&\frac(\sqrt3)2&\frac(\sqrt2)2&\frac12&0\\ \hline \mathrm(tg) &0 &\frac(\sqrt3)3&1&\sqrt3&\infty\\ \hline \mathrm(ctg) &\infty &\sqrt3&1&\frac(\sqrt3)3&0\\ \hline \end(array)))\] Note that these values ​​were displayed in the section “Geometry on a plane (planimetry). Part II” in the topic “Initial information about sine, cosine, tangent and cotangent”.

Example 5. Find \(\sin(\dfrac(3\pi)4)\) .

Thus, Let's transform the angle:.

\(\dfrac(3\pi)4=\dfrac(4\pi-\pi)(4)=\pi-\dfrac(\pi)4\)

\(\sin(\dfrac(3\pi)4)=\sin\left(\pi-\dfrac(\pi)4\right)=\sin\dfrac(\pi)4=\dfrac(\sqrt2) 2\)\(\blacktriangleright\) To make it easier to remember and use reduction formulas, you can follow the following rule. Case 1. \[\cos(n\cdot \pi\pm \alpha)=\bigodot \cos\alpha\]

The sign of an angle can be found by determining which quadrant it is in. Using this rule, we assume that the angle \(\alpha\) is in the \(I\) quadrant.

Case 2. If the angle can be represented in the form , where \(n\in\mathbb(N)\) , then \[\sin(n\cdot \pi+\dfrac(\pi)2\pm \alpha)=\bigodot \cos\alpha\] where in place of \(\bigodot\) is the sign of the sine of the angle \(n\cdot \pi\pm \alpha\) . \[\cos(n\cdot \pi+\dfrac(\pi)2\pm \alpha)=\bigodot \sin\alpha\] where in place of \(\bigodot\) is the sign of the cosine of the angle \(n\cdot \pi\pm \alpha\) .

The sign is determined in the same way as in the case of \(1\) .

Note that in the first case the function remains unchanged, and in the second case it changes (they say that the function changes to a cofunction).

Example 6. Find \(\sin \dfrac(13\pi)(3)\) .

Example 5. \(\dfrac(13\pi)(3)=\dfrac(12\pi+\pi)(3)=4\pi+\dfrac(\pi)3\), and the points \(\sin \dfrac(13\pi)(3)=\sin \left(4\pi+\dfrac(\pi)3\right)=\sin\dfrac(\pi)3=\dfrac(\sqrt3) 2\)

Example 7. Find \(\cos \dfrac(17\pi)(6)\) .

Example 5. \(\dfrac(17\pi)(6)=\dfrac(18\pi-\pi)(6)=3\pi-\dfrac(\pi)6\), and the points \(\cos \dfrac(17\pi)(6)=\cos \left(3\pi-\dfrac(\pi)6\right)=-\cos\dfrac(\pi)6=-\dfrac( \sqrt3)2\)

\(\blacktriangleright\) Range of sine and cosine values.
Because the coordinates \(x_(\alpha)\) and \(y_(\alpha)\) of any point \(P_(\alpha)\) on the unit circle are in the range from \(-1\) to \(1\) , and \(\cos\alpha\) and \(\sin\alpha\) are the abscissa and ordinate, respectively, of this point, then \[(\large(-1\leq \cos\alpha\leq 1 ,\qquad -1\leq\sin\alpha\leq 1))\]

From a right triangle according to the Pythagorean theorem we have: \(x^2_(\alpha)+y^2_(\alpha)=1^2\)
Because \(x_(\alpha)=\cos\alpha,\ y_(\alpha)=\sin\alpha \Rightarrow\) \[(\large(\sin^2\alpha+\cos^2\alpha=1)) - \textbf(basic trigonometric identity (GTT))\]

\(\blacktriangleright\) Tangent and cotangent.

Because \(\mathrm(tg)\,\alpha=\dfrac(\sin\alpha)(\cos\alpha), \cos\alpha\ne 0\)

\(\mathrm(ctg)\,\alpha=\dfrac(\cos\alpha)(\sin\alpha), \sin\alpha\ne 0\), That:

1) \((\large(\mathrm(tg)\,\alpha\cdot \mathrm(ctg)\,\alpha=1, \cos\alpha\ne 0, \sin\alpha \ne 0))\)

2) tangent and cotangent are positive in \(I\) and \(III\) quarters and negative in \(II\) and \(IV\) quarters.

3) the range of values ​​of tangent and cotangent - all real numbers, i.e. \(\mathrm(tg)\,\alpha\in\mathbb(R), \ \mathrm(ctg)\,\alpha\in\mathbb(R)\)

4) reduction formulas are also defined for tangent and cotangent.

\(\sin(\dfrac(3\pi)4)=\sin\left(\pi-\dfrac(\pi)4\right)=\sin\dfrac(\pi)4=\dfrac(\sqrt2) 2\) \[\mathrm(tg)\,(n\cdot \pi\pm \alpha)=\bigodot \mathrm(tg)\,\alpha\] where in place of \(\bigodot\) is the sign of the tangent of the angle \(n\cdot \pi\pm \alpha\) (\(\cos\alpha\ne 0\) ). \[\mathrm(ctg)\,(n\cdot \pi\pm \alpha)=\bigodot \mathrm(ctg)\,\alpha\] where in place of \(\bigodot\) is the sign of the angle cotangent \(n\cdot \pi\pm \alpha\) (\(\sin\alpha\ne 0\) ).

Case 2. If the angle can be represented as \(n\cdot \pi+\dfrac(\pi)2\pm\alpha\), where \(n\in\mathbb(N)\) , then \[\mathrm(tg)\,(n\cdot \pi+\dfrac(\pi)2\pm \alpha)=\bigodot \mathrm(ctg)\,\alpha\] where in place \(\bigodot\) there is a sign of the tangent of the angle \(n\cdot \pi\pm \alpha\) (\(\sin\alpha\ne 0\) ). \[\mathrm(ctg)\,(n\cdot \pi+\dfrac(\pi)2\pm \alpha)=\bigodot \mathrm(tg)\,\alpha\] where in place of \(\bigodot\) is the sign of the angle cotangent \(n\cdot \pi\pm \alpha\) (\(\cos\alpha\ne 0\) ).

5) the tangent axis passes through the point \((1;0)\) parallel to the sine axis, and the positive direction of the tangent axis coincides with the positive direction of the sine axis;
the cotangent axis is through the point \((0;1)\) parallel to the cosine axis, and the positive direction of the cotangent axis coincides with the positive direction of the cosine axis.


We will give a proof of this fact using the example of the tangent axis.

\(\triangle OP_(\alpha)K \sim \triangle AOB \Rightarrow \dfrac(P_(\alpha)K)(OK)=\dfrac(BA)(OB) \Rightarrow \dfrac(\sin\alpha)( \cos\alpha)=\dfrac(BA)1 \Rightarrow BA=\mathrm(tg)\,\alpha\).

Thus, if the point \(P_(\alpha)\) is connected by a straight line to the center of the circle, then this straight line will intersect the tangent line at a point whose value is \(\mathrm(tg)\,\alpha\) .

6) the following formulas follow from the main trigonometric identity: \ The first formula is obtained by dividing the right and left sides of the OTT by \(\cos^2\alpha\), the second by dividing by \(\sin^2\alpha\) .

Please note that the tangent is not defined at angles where the cosine is zero (this is \(\alpha=\dfrac(\pi)2+\pi n, n\in\mathbb(Z)\));
cotangent is not defined at angles where the sine is zero (this is \(\alpha=\pi+\pi n, n\in\mathbb(Z)\)).

\(\blacktriangleright\) Evenness of cosine and oddness of sine, tangent, cotangent.

Recall that a function \(f(x)\) is called even if \(f(-x)=f(x)\) .

A function is called odd if \(f(-x)=-f(x)\) .

It can be seen from the circle that the cosine of the angle \(\alpha\) is equal to the cosine of the angle \(-\alpha\) for any values ​​of \(\alpha\) :

Thus, cosine is an even function, which means that the formula \[(\Large(\cos(-x)=\cos x))\] is true

It is clear from the circle that the sine of the angle \(\alpha\) is opposite to the sine of the angle \(-\alpha\) for any values ​​of \(\alpha\) :

Thus, sine is an odd function, which means the formula is correct \[(\Large(\sin(-x)=-\sin x))\]

Tangent and cotangent are also odd functions: \[(\Large(\mathrm(tg)\,(-x)=-\mathrm(tg)\,x))\] \[(\Large(\mathrm(ctg)\,(-x)=-\mathrm(ctg)\,x))\]

Because \(\mathrm(tg)\,(-x)=\dfrac(\sin (-x))(\cos(-x))=\dfrac(-\sin x)(\cos x)=-\mathrm (tg)\,x \qquad \mathrm(ctg)\,(-x)=\dfrac(\cos(-x))(\sin(-x))=-\mathrm(ctg)\,x\))

As practice shows, one of the most difficult sections of mathematics that schoolchildren encounter in the Unified State Examination is trigonometry. The science of aspect ratios in triangles begins to be learned in the 8th grade. Equations of this type contain a variable under the sign of trigonometric functions. Despite the fact that the simplest of them: \(sin x = a\) , \(cos x = a\) , \(tg x = a\) , \(ctg x = a\) - are familiar to almost every schoolchild, their implementation is often difficult.

In the Unified State Exam in mathematics at the profile level, a correctly solved trigonometry task is rated very highly. A student can receive up to 4 primary points for correctly completing a task from this section. To do this, looking for trigonometry cheat sheets for the Unified State Exam is almost pointless. The most reasonable solution is to prepare well for the exam.

How to do it?

To ensure that trigonometry in the Unified State Examination in mathematics does not scare you, use our portal when preparing. It's convenient, simple and effective. In this section of our educational portal, open to students in both Moscow and other cities, theoretical material and formulas on trigonometry for the Unified State Exam are presented in an accessible manner. Also, for all mathematical definitions, we have selected examples with a detailed description of the process of solving them.

After studying the theory in the “Trigonometry” section in preparation for the Unified State Exam, we recommend going to “Catalogues” so that the acquired knowledge is better assimilated. Here you can select problems on a topic of interest and view their solutions. Thus, repeating the theory of trigonometry in the Unified State Examination will be as effective as possible.

What do you need to know?

First of all, you need to learn the values ​​of \(sin\) , \(cos\) , \(tg\) , \(ctg\) acute angles from \(0°\) to \(90°\) . Also, when preparing for the Unified State Exam in Moscow, it is worth remembering the basic methods of solving trigonometry problems. It should be noted that when completing tasks, you must reduce the equation to its simplest form. You can do this as follows:

• factoring the equation;
• replacing a variable (reduction to algebraic equations);
• leading to a homogeneous equation;
• moving to the half corner;
• converting products into sums;
• by entering an auxiliary angle;
• using the universal substitution method.

In this case, most often the student has to use several of the listed methods during the solution.