Mark on the unit circle the points corresponding to the number. trigonometric circle. Detailed theory with examples. Guidelines for the study of the discipline and the performance of tests for students of correspondence courses All specialties

If you place the unit number circle on coordinate plane, then you can find coordinates for its points. The numerical circle is positioned so that its center coincides with the origin of the plane, i.e., the point O (0; 0).

Usually, on a unit number circle, points are marked corresponding to the origin on the circle

  • quarters - 0 or 2π, π/2, π, (2π)/3,
  • middle quarters - π/4, (3π)/4, (5π)/4, (7π)/4,
  • third quarters - π/6, π/3, (2π)/3, (5π)/6, (7π)/6, (4π)/3, (5π)/3, (11π)/6.

On the coordinate plane, with the above arrangement of the unit circle on it, one can find the coordinates corresponding to these points of the circle.

It is very easy to find the coordinates of the ends of the quarters. At point 0 of the circle, the x-coordinate is 1, and y is 0. We can write A (0) = A (1; 0).

The end of the first quarter will be located on the positive y-axis. Therefore, B (π/2) = B (0; 1).

The end of the second quarter is on the negative abscissa: C (π) = C (-1; 0).

End of the third quarter: D ((2π)/3) = D (0; -1).

But how to find the coordinates of the midpoints of quarters? To do this, build a right triangle. Its hypotenuse is a segment from the center of the circle (or the origin) to the midpoint of the quarter circle. This is the radius of the circle. Since the circle is unit, the hypotenuse is equal to 1. Next, a perpendicular is drawn from a point on the circle to any axis. Let it be to the x-axis. It turns out a right-angled triangle, the lengths of the legs of which are the x and y coordinates of the point of the circle.

A quarter circle is 90º. And half a quarter is 45º. Since the hypotenuse is drawn to the point of the middle of the quarter, the angle between the hypotenuse and the leg coming out of the origin is 45º. But the sum of the angles of any triangle is 180º. Therefore, the angle between the hypotenuse and the other leg also remains 45º. It turns out an isosceles right triangle.

From the Pythagorean theorem we obtain the equation x 2 + y 2 = 1 2 . Since x = y and 1 2 = 1, the equation simplifies to x 2 + x 2 = 1. Solving it, we get x = √1 = 1/√2 = √2/2.

Thus, the coordinates of the point M 1 (π/4) = M 1 (√2/2; √2/2).

In the coordinates of the points of the midpoints of other quarters, only the signs will change, and the modules of values ​​will remain the same, since the right-angled triangle will only turn over. We get:
M 2 ((3π)/4) = M 2 (-√2/2; √2/2)
M 3 ((5π)/4) = M 3 (-√2/2; -√2/2)
M 4 ((7π)/4) = M 4 (√2/2; -√2/2)

When determining the coordinates of the third parts of the quarters of the circle, a right triangle is also built. If we take the point π/6 and draw a perpendicular to the x-axis, then the angle between the hypotenuse and the leg lying on the x-axis will be 30º. It is known that the leg lying opposite an angle of 30º is equal to half the hypotenuse. So we have found the y coordinate, it is equal to ½.

Knowing the lengths of the hypotenuse and one of the legs, by the Pythagorean theorem we find the other leg:
x 2 + (½) 2 = 1 2
x 2 \u003d 1 - ¼ \u003d ¾
x = √3/2

Thus T 1 (π/6) = T 1 (√3/2; ½).

For the point of the second third of the first quarter (π / 3), it is better to draw a perpendicular to the axis to the y axis. Then the angle at the origin will also be 30º. Here, the x coordinate will already be equal to ½, and y, respectively, √3/2: T 2 (π/3) = T 2 (½; √3/2).

For other third quarter points, the signs and order of coordinate values ​​will change. All points that are closer to the x-axis will have a modulo value of the x-coordinate equal to √3/2. Those points that are closer to the y-axis will have a modulo y value equal to √3/2.
T 3 ((2π)/3) = T 3 (-½; √3/2)
T 4 ((5π)/6) = T 4 (-√3/2; ½)
T 5 ((7π)/6) = T 5 (-√3/2; -½)
T 6 ((4π)/3) = T 6 (-½; -√3/2)
T 7 ((5π)/3) = T 7 (½; -√3/2)
T 8 ((11π)/6) = T 8 (√3/2; -½)

5. TRIGONOMETRIC FUNCTIONS OF ANY ARGUMENT

§ 20. UNIT CIRCLE

948. What is the relationship between the arc length of a unit circle and its radian measure?

949. Construct points on the unit circle corresponding to the numbers: 0; one; 2; 3; 4; 5; .... Can any of these points coincide? Why?

950. The numbers are given by the formula α = 1 / 2 k, where k= 0; ±1; ±2; ....
Construct on the number axis and on the unit circle the points corresponding to these numbers. How many such points will be on the number axis and how many on the unit circle?

951. Mark on the unit circle and on the numerical axis the points corresponding to the numbers:
1) α = π k, k= 0; ±1, ±2, ...;
2) α = π / 2 (2k + 1), k= 0; ± 1; ±2; ...;
3) α = π k / 6 , k= 0; ±1; ±2; ... .
How many such points are on the number axis and how many are on the unit circle?

952. How will the points corresponding to the numbers be located on the number axis and on the unit circle:
1) a and - a; 2) a and a± π; 3) a+ π and a- π; 4) a and a+ 2π k, k= 0; ±1; ±2; ...?

953. What does it consist of fundamental difference between the representation of numbers by the points of the numerical axis and their representation by the points of the unit circle?

954. 1) Find the smallest non-negative numbers corresponding to the intersection points of the unit circle: a) with the coordinate axes; b) with bisectors of coordinate angles.

2) In each case, write general formula numbers corresponding to the specified points of the unit circle.

955. Knowing that a is one of the numbers corresponding to a given point on the unit circle, find:
1) all numbers corresponding to a given point;
2) all numbers corresponding to the point of the unit circle symmetrical to the given one:
a) relative to the abscissa axis; b) relative to the y-axis; c) relative to the origin of coordinates.
Solve the problem by taking a = 0; π / 2; one ; 2; π / 6; - π / 4 .

956. Find a condition that the numbers satisfy a corresponding to:
1) points of the 1st quarter of the unit circle;
2) points of the 2nd quarter of the unit circle;
3) points of the 3rd quarter of the unit circle;
4) points of the 4th quarter of the unit circle.

957. Vertex A of the regular octagon ABCDEFKL inscribed in the unit circle has coordinates (1; 0) (Fig. 39).

1) Determine the coordinates of the remaining vertices of the octagon.
2) Compose a general formula for arcs of a unit circle ending:
a) at points A, C, E and K; b) at points B, D, F and L; c) at points A, B, C, D, E, F, K and L.

958. 1) On the unit circle, construct a point whose ordinate is 0.5. How many points on the unit circle have a given ordinate? How are these points located relative to the y-axis.

2) Measure with a protractor (with an accuracy of 1 °) the smallest absolute arc, the end of which has an ordinate equal to 0.5, and draw up a general formula for arcs of a unit circle ending at points with an ordinate of 0.5.

959. Solve problem 958, taking the ordinate at equal to:

1) - 0,5; 2) 0 4; 3) 0,5√3 .

960. 1) On the unit circle, construct a point whose abscissa is 0.5. How many points on the unit circle have a given abscissa? How are these points located relative to the x-axis?

2) Measure with a protractor (with an accuracy of 1 °) the smallest positive arc, the end of which has an abscissa equal to 0.5, and draw up a general formula for arcs of a unit circle ending at points with an abscissa of 0.5.

961. Solve problem 960, taking the abscissa X equal to:

1) - 2 / 3 ; 2) 0,4; 3) 0,5√2 .

962. Determine the coordinates of the ends of the arcs of the unit circle given by the formula ( k= 0; ±1; ±2; ...):

1) α = 30°(2 k+ 1); 2) α = π k / 3 .

963. Express in one formula the following series of angles ( k= 0; ±1; ±2; ...):

1) α 1 = 180° k+ 120° and α 2 = 180° k+ 30°;

2) α 1 = π k + π / 6 and α 2 = π k - π / 3 ;

3) α 1 = 90° k and α 2 = 45° (2 k + 1);

4) α 1 = π k and α 2 = π / 3 (3k± 1);

5) α 1 = 120° k± 15° and α 2 = 120° k± 45°;

6) α 1 = π k; α 2 = 2π k ± π / 3 and α 3 \u003d 2l k± 2π / 3 ;

7) α 1 = 180° k+ 140°; α 2 = 180° k+ 80° and α 3 = 180° k+ 20°;

8) α 1 \u003d 180 ° k + (-1)k 60° and α 2 = 180° k - (-1)k 60°.

964. Eliminate repeating corners in the following formulas (k= 0-±1; ±2; ...):

1) α 1 = 90° k and α 2 = 60° k+ 30°;

2) α 1 = π k / 2 and α 2 = π k / 5 ;

3) α 1 = 1/4 π k and α 2 = 1 / 2 π k± 1 / 4 π;

4) α 1 = π (2 k+ 1) - π / 6 and α 2 = 2 / 5 π k+ 1 / 30 π;

5) α 1 \u003d 72 ° k+ 36° and α 2 = 120° k+ 60°.



Coordinates x points lying on the circle are equal to cos(θ), and the coordinates y correspond to sin(θ), where θ is the magnitude of the angle.

  • If you find it difficult to remember this rule, just remember that in the pair (cos; sin) "the sine comes last".
  • This rule can be deduced by considering right triangles and the definition of data trigonometric functions(the sine of the angle is equal to the ratio of the length of the opposite, and the cosine of the adjacent leg to the hypotenuse).

Write down the coordinates of four points on the circle. A "unit circle" is a circle whose radius is equal to one. Use this to determine the coordinates x and y at four points of intersection of the coordinate axes with the circle. Above, for clarity, we have designated these points as "east", "north", "west" and "south", although they do not have established names.

  • "East" corresponds to a point with coordinates (1; 0) .
  • "North" corresponds to a point with coordinates (0; 1) .
  • "West" corresponds to a point with coordinates (-1; 0) .
  • "South" corresponds to a point with coordinates (0; -1) .
  • This is similar to a normal graph, so there is no need to memorize these values, it is enough to remember the basic principle.

  • Remember the coordinates of the points in the first quadrant. The first quadrant is located in the upper right part of the circle, where the coordinates x and y take positive values. These are the only coordinates you need to remember:

    Draw straight lines and determine the coordinates of the points of their intersection with the circle. If you draw straight horizontal and vertical lines from the points of one quadrant, the second points of intersection of these lines with the circle will have coordinates x and y with the same absolute values ​​but different signs. In other words, you can draw horizontal and vertical lines from the points of the first quadrant and sign the intersection points with the circle with the same coordinates, but at the same time leave room for the correct sign ("+" or "-") on the left.

  • Use symmetry rules to determine the sign of coordinates. There are several ways to determine where to put the "-" sign:

    • remember the basic rules for regular charts. Axis x negative on the left and positive on the right. Axis y negative from below and positive from above;
    • start from the first quadrant and draw lines to other points. If the line crosses the axis y, coordinate x will change its sign. If the line crosses the axis x, the sign of the coordinate will change y;
    • remember that in the first quadrant all functions are positive, in the second quadrant only the sine is positive, in the third quadrant only the tangent is positive, and in the fourth quadrant only the cosine is positive;
    • whichever method you use, you should get (+,+) in the first quadrant, (-,+) in the second, (-,-) in the third, and (+,-) in the fourth.

  • Check if you made a mistake. Below is full list coordinates of "special" points (except for four points on the coordinate axes), if you move counterclockwise along the unit circle. Remember that to determine all these values, it is enough to remember the coordinates of the points only in the first quadrant:

    • first quadrant :( 3 2 , 1 2 (\displaystyle (\frac (\sqrt (3))(2)),(\frac (1)(2)))); (2 2 , 2 2 (\displaystyle (\frac (\sqrt (2))(2)),(\frac (\sqrt (2))(2)))); (1 2 , 3 2 (\displaystyle (\frac (1)(2)),(\frac (\sqrt (3))(2))));
    • second quadrant :( − 1 2 , 3 2 (\displaystyle -(\frac (1)(2)),(\frac (\sqrt (3))(2)))); (− 2 2 , 2 2 (\displaystyle -(\frac (\sqrt (2))(2)),(\frac (\sqrt (2))(2)))); (− 3 2 , 1 2 (\displaystyle -(\frac (\sqrt (3))(2)),(\frac (1)(2))));
    • third quadrant :( − 3 2 , − 1 2 (\displaystyle -(\frac (\sqrt (3))(2)),-(\frac (1)(2)))); (− 2 2 , − 2 2 (\displaystyle -(\frac (\sqrt (2))(2)),-(\frac (\sqrt (2))(2)))); (− 1 2 , − 3 2 (\displaystyle -(\frac (1)(2)),-(\frac (\sqrt (3))(2))));
    • fourth quadrant :( 1 2 , − 3 2 (\displaystyle (\frac (1)(2)),-(\frac (\sqrt (3))(2)))); (2 2 , − 2 2 (\displaystyle (\frac (\sqrt (2))(2)),-(\frac (\sqrt (2))(2)))); (3 2 , − 1 2 (\displaystyle (\frac (\sqrt (3))(2)),-(\frac (1)(2)))).

    • >> Number circle


    While studying the algebra course of grades 7-9, we have so far dealt with algebraic functions, i.e. functions given analytically by expressions, in the notation of which were used algebraic operations over numbers and a variable (addition, subtraction, multiplication, division, exponentiation, extraction square root). But mathematical models of real situations are often associated with functions of a different type, not algebraic. With the first representatives of the class of non-algebraic functions - trigonometric functions - we will get acquainted in this chapter. You will study trigonometric functions and other types of non-algebraic functions (exponential and logarithmic) in more detail in high school.
    To introduce trigonometric functions, we need a new mathematical model- a number circle, which you have not yet met, but are well acquainted with the number line. Recall that a number line is a line on which the starting point O, the scale (single segment) and the positive direction are given. We can associate any real number with a point on a straight line and vice versa.

    How to find the corresponding point M on the line given the number x? The number 0 corresponds to the starting point O. If x > 0, then, moving in a straight line from the point 0 in the positive direction, you need to go n^th length x; the end of this path will be the desired point M(x). If x< 0, то, двигаясь по прямой из точки О в отрицательном направлении, нужно пройти путь 1*1; конец этого пути и будет искомой точкой М(х). Число х - координата точки М.

    And how did we solve the inverse problem, i.e. how did you find the x-coordinate of a given point M on the number line? We found the length of the segment OM and took it with the sign "+" or * - "depending on which side of the point O the point M is located on the straight line.

    But in real life move not only in a straight line. Quite often, movement is considered circles. Here is a specific example. We will consider the stadium treadmill as a circle (in fact, it is, of course, not a circle, but remember how sports commentators usually say: “the runner ran a circle”, “there is half a circle left to run to the finish line”, etc.), its length is 400 m. The start is marked - point A (Fig. 97). The runner from point A moves in a counterclockwise circle. Where will he be in 200 meters? after 400 m? after 800 m? after 1500 m? And where to draw the finish line if he runs a marathon distance of 42 km 195 m?

    After 200 m, he will be at point C, diametrically opposite point A (200 m is the length of half the treadmill, i.e. the length of half the circle). After running 400 m (i.e. “one lap”, as the athletes say), he will return to point A. After running 800 m (i.e. “two laps”), he will again be at point A. And what is 1500 m ? This is "three circles" (1200 m) plus another 300 m, i.e. 3

    Treadmill - the finish of this distance will be at point 2) (Fig. 97).

    We have to deal with the marathon. After running 105 laps, the athlete will overcome the distance 105-400 = 42,000 m, i.e. 42 km. There are 195 m left to the finish line, which is 5 m less than half the circumference. This means that the finish of the marathon distance will be at point M, located near point C (Fig. 97).

    Comment. Of course, you understand the convention of the last example. Nobody runs the marathon distance around the stadium, the maximum is 10,000 m, i.e. 25 circles.

    You can run or walk a path of any length along the stadium's running track. This means that any positive number corresponds to some point - the “finish of the distance”. Moreover, any negative number can be associated with a circle point: you just need to make the athlete run in the opposite direction, i.e. start from point A not in the opposite direction, but in the clockwise direction. Then the stadium running track can be considered as a numerical circle.

    In principle, any circle can be considered as a numerical one, but in mathematics it was agreed to use a unit circle for this purpose - a circle with a radius of 1. This will be our "treadmill". The length b of a circle with radius K is calculated by the formula The length of a half circle is n, and the length of a quarter circle is AB, BC, SB, DA in Fig. 98 - equal We agree to call the arc AB the first quarter of the unit circle, the arc BC - the second quarter, the arc CB - the third quarter, the arc DA - the fourth quarter (Fig. 98). At the same time, usually we are talking about the Open Arc, i.e. about an arc without its ends (something like an interval on a number line).


    Definition. A unit circle is given, the starting point A is marked on it - the right end of the horizontal diameter (Fig. 98). Let's match each real number I point of the circle according to the following rule:

    1) if x > 0, then, moving from point A in a counterclockwise direction (the positive direction of going around the circle), we describe a path along the circle with a length and the end point M of this path will be the desired point: M = M (x);

    2) if x< 0, то, двигаясь из точки А в направлении по часовой стрелке (отрицательное направление обхода окружности), опишем по окружности путь длиной и |; конечная точка М этого пути и будет искомой точкой: М = М(1);

    0 we assign point A: A = A(0).

    A unit circle with an established correspondence (between real numbers and points of the circle) will be called a number circle.
    Example 1 Find on the number circle
    Since the first six of the given seven numbers are positive, then in order to find the corresponding points on the circle, you need to go along the circle a path of a given length, moving from point A in a positive direction. At the same time, we take into account that


    Point A corresponds to the number 2, since, having passed a path of length 2 along the circle, i.e. exactly one circle, we will again fall into starting point A So, A = A(2).
    What So, moving from point A in a positive direction, you need to go through a whole circle.

    Comment. When we're in 7th or 8th grade worked with the number line, we agreed, for the sake of brevity, not to say "the point of the line corresponding to the number x", but to say "the point x". We will adhere to exactly the same agreement when working with a numerical circle: "point f" - this means that we are talking about a circle point that corresponds to the number
    Example 2
    Dividing the first quarter AB into three equal parts by points K and P, we get:

    Example 3 Find points on the number circle that correspond to numbers
    We will make constructions using Fig. 99. Postponing the arc AM (its length is equal to -) from the point A five times in the negative direction, we get the point!, - the middle of the arc BC. So,

    Comment. Notice some liberties we take in using mathematical language. It is clear that the arc AK and the length of the arc AK are different things (the first concept is geometric figure, and the second concept is a number). But both are denoted the same way: AK. Moreover, if points A and K are connected by a segment, then both the resulting segment and its length are denoted in the same way: AK. It is usually clear from the context what meaning is attached to the designation (arc, arc length, segment or segment length).

    Therefore, two layouts of the number circle will be very useful to us.

    FIRST LAYOUT
    Each of the four quarters of the numerical circle is divided into two equal parts, and their “names” are written near each of the eight available points (Fig. 100).

    SECOND LAYOUT Each of the four quarters of the numerical circle is divided into three equal parts, and their “names” are written near each of the twelve points available (Fig. 101).


    Note that on both layouts we could given points assign other "names".
    Have you noticed that in all the examples analyzed, the lengths of the arcs
    expressed by some fractions of the number n? This is not surprising: after all, the length of a unit circle is 2n, and if we divide the circle or its quarter into equal parts, then we get arcs whose lengths are expressed as fractions of the number and. And what do you think, is it possible to find such a point E on the unit circle that the length of the arc AE will be equal to 1? Let's guess:

    Arguing in a similar way, we conclude that on the unit circle one can find both the point Eg, for which AE, = 1, and the point E2, for which AEg = 2, and the point E3, for which AE3 = 3, and the point E4, for which AE4 = 4, and point Eb, for which AEb = 5, and point E6, for which AE6 = 6. In fig. 102 (approximately) the corresponding points are marked (moreover, for orientation, each of the quarters of the unit circle is divided by dashes into three equal parts).


    Example 4 Find on the number circle the point corresponding to the number -7.

    We need, starting from the point A (0) and moving in a negative direction (in a clockwise direction), go around the circle path of length 7. If we go through one circle, we get (approximately) 6.28, which means we still need to go ( in the same direction) a path of length 0.72. What is this arc? Slightly less than half a quarter of a circle, i.e. its length is less than number -.

    So, a numerical circle, like a numerical straight line, each real number corresponds to one point (only, of course, it is easier to find it on a straight line than on a circle). But for a straight line, the opposite is also true: each point corresponds to a single number. For a numerical circle, such a statement is not true, we have repeatedly convinced ourselves of this above. For a number circle, the following statement is true.
    If the point M of the numerical circle corresponds to the number I, then it also corresponds to the number of the form I + 2k, where k is any integer (k e 2).

    Indeed, 2n is the length of the numerical (unit) circle, and the integer |d| can be considered as the number of complete rounds of the circle in one direction or another. If, for example, k = 3, then this means that we make three rounds of the circle in the positive direction; if k \u003d -7, then this means that we make seven (| k | \u003d | -71 \u003d 7) rounds of the circle in the negative direction. But if we are at the point M(1), then by doing more | to | full circles, we will again find ourselves at the point M.

    A.G. Mordkovich Algebra Grade 10

    Lesson content lesson summary support frame lesson presentation accelerative methods interactive technologies Practice tasks and exercises self-examination workshops, trainings, cases, quests homework discussion questions rhetorical questions from students Illustrations audio, video clips and multimedia photographs, pictures graphics, tables, schemes humor, anecdotes, jokes, comics parables, sayings, crossword puzzles, quotes Add-ons abstracts articles chips for inquisitive cheat sheets textbooks basic and additional glossary of terms other Improving textbooks and lessonscorrecting errors in the textbook updating a fragment in the textbook elements of innovation in the lesson replacing obsolete knowledge with new ones Only for teachers perfect lessons calendar plan for the year guidelines discussion programs Integrated Lessons

    Decision:

    1) Since 7π = 3٠2π + π , then turning by 7π produces the same point as turning by π, i.e. a point with coordinates (- 1; 0) is obtained. (fig.9)

    2) Since = -2π - , then turning on produces the same point as turning on - , i.e. a point with coordinates (0; 1) is obtained (Fig. 10)

    Fig.9 Fig.10

    Task #2

    Write down all the angles by which you need to rotate the point (1; 0) to get the point

    N
    .

    Decision:

    From right triangle AON (Fig. 11) it follows that the angle AON is , i.e. one of the possible rotation angles is . Therefore, all the angles by which the point (1;0) must be rotated in order to obtain the point are expressed as follows: + 2πk, where k is any integer.

    Fig.11

    Exercises for self-solving:

    1°. Construct a point on the unit circle obtained by rotating the point (1; 0) by a given angle:

    a) 4π; b) - 225°; in) - ; G) - ; e)
    ; e)
    .

    2°. Find the coordinates of the point obtained by rotating the point Р(1;0) by an angle:

    a) 3π; b) -
    ; c) 540°;

    d) 810°; e)
    , k is an integer; e)
    .

    3°. Determine the quarter in which the point is located, obtained by turning the point P (1; 0) by an angle:

    a) 1; b) 2.75; c) 3.16; d) 4.95.

    4*. On the unit circle, construct a point obtained by turning the point P (1; 0) through an angle:

    a)
    ; b)
    ; c) 4.5π; d) - 7π.

    5*. Find the coordinates of the point obtained by turning the point P (1; 0) by an angle (k is an integer):

    a)
    ; b)
    ; in)
    ; G)
    .

    6*. Write down all the angles by which you need to rotate the point P (1; 0) to get a point with coordinates:

    a)
    ; b)
    ;

    in)
    ; G)
    .

    DEFINITION OF SINE, COSINE OF ANGLE

    Fig.12

    In these definitions, the angle α can be expressed in both degrees and radians. For example, when turning the point (1; 0) by the angle , i.e. the angle is 90°, the point (0;1) is obtained. Point ordinate ( 0 ;1 ) is equal to 1 , so sin = sin 90° = 1; the abscissa of this point is equal to 0 , so cos = cos 90° = 0

    Task #1

    Find sin (- π) and cos (- π).

    Decision:

    The point (1; 0) when turning through the angle - π will go to the point (-1; 0) (Fig. 13), therefore, sin (- π) \u003d 0, cos (- π) \u003d - 1.

    Fig.13

    Task #2

    Solve the equation sin x = 0.

    Decision:

    Solving the equation sin x \u003d 0 means finding all the angles whose sine zero. An ordinate equal to zero has two points of the unit circle (1; 0 )and (- 1; 0 ). These points are obtained from the point (1;0) by turning through the angles 0, π, 2π, 3π, etc., as well as through the angles - π, - 2π, - 3π, etc.. therefore, sin x = 0 for x = πk., where k is any integer i.e. the solution can be done like this:

    x = πk., k
    .

    Answer: x = πk., k

    (Z is the notation for the set of integers, read "k belongs to Z").

    Arguing similarly, we can obtain the following solutions of trigonometric equations:

    sinx

    x = + 2πk, k

    x = - +2πk., k

    x = +2πk., k

    x = 2πk., k

    x = π + 2 πk., k

    Here is a table of common values ​​for sine, cosine, tangent and cotangent.

    Task #1

    Calculate: 4sin +
    cos-tg.

    Decision:

    Using the table, we get

    4 sin + cos - tg = 4 ٠ + ٠ -1 = 2 + 1,5 = 2,5.

    :

    1°. Calculate:

    a) sin + sin; b) sin - cos π; c) sin 0 - cos 2π; d) sin3 - cos .

    2°. Find the value of an expression:

    a) 3 sin + 2 cos - tg; b)
    ;

    in)
    ; d) cos 0 - sin 3π.

    3°. Solve the equation:

    a) 2 sin x = 0; b) cos x = 0; c) cos x - 1 = 0; d) 1 – sin x = 0.

    4*. Find the value of an expression:

    a) 2 sin α +
    cos α at α = ; b) 0.5 cos α - sin α at α = 60°;

    c) sin 3 α - cos 2 α at α = ; d) cos + sin at α = .

    5*. Solve the equation:

    a) sinx \u003d - 1; b) cos x = 0; c) sin
    ; d) sin3 x = 0.

    Signs of sine, cosine and tangent

    Let the point move counterclockwise along the unit circle, then sinus positive in first and second coordinate quarters (Fig. 14); cosine positive in first and fourth coordinate quarters (Fig. 15); tangent and cotangent positive in first and third coordinate quarters (Fig. 16).

    Fig.14 Fig.15 Fig.16

    Task #1

    Find out the signs of the sine, cosine and tangent of an angle:

    1) ; 2) 745°; 3)
    .

    Decision:

    1) An angle corresponds to a point on the unit circle located in second quarters. Therefore, sin > 0, cos

    2) Since 745° = 2 ٠360° + 25° , then the rotation of the point (1; 0) by an angle of 745° corresponds to a point located in first quarters.

    Therefore sin 745° > 0, cos 745° > 0, tg 745° > 0.

    3) The point moves clockwise, therefore - π , then when the point (1; 0) is rotated by an angle, a point is obtained third quarters. Therefore sin

    Exercises for self-solving :

    1°. In what quarter is the point obtained by turning the point P (1; 0) through the angle α, if:

    a) α = ; b) α = - ; in) α = ;Document

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