How to remember points on the unit circle. Lesson "definition of sine and cosine on the unit circle" Unit circle and coordinates of points

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 derived by looking at 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 side, and the cosine is equal to the ratio of the adjacent side 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 designated these points as “east”, “north”, “west” and “south”, although they do not have established names.

• "East" corresponds to the point with coordinates (1; 0) .
• "North" corresponds to the point with coordinates (0; 1) .
• "West" corresponds to the point with coordinates (-1; 0) .
• "South" corresponds to the point with coordinates (0; -1) .
• This is similar to a regular graph, so there is no need to memorize these values, just 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 the coordinates x And y with the same absolute values, but with different signs. In other words, you can draw horizontal and vertical lines from the points of the first quadrant and label the points of intersection with the circle with the same coordinates, but at the same time leave space on the left for the correct sign ("+" or "-").

To determine the sign of the coordinates, use the rules of symmetry. There are several ways to determine where to place the "-" sign:

• Remember the basic rules for regular charts. Axis x negative on the left and positive on the right. Axis y negative below and positive above;
• start with 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 coordinate axes), if you move along the unit circle counterclockwise. 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)))).

I hope you have already read about the number circle and know why it is called a number circle, where the origin of coordinates is on it and which side is the positive direction. If not, then run! If, of course, you are going to find points on number circle.

We denote the numbers \(2π\), \(π\), \(\frac(π)(2)\), \(-\frac(π)(2)\), \(\frac(3π)(2 )\)

As you know from the previous article, the radius of the number circle is \(1\). This means that the circumference is equal to \(2π\) (calculated using the formula \(l=2πR\)). Taking this into account, we mark \(2π\) on the number circle. To mark this number, we need to go from \(0\) along the number circle to a distance equal to \(2π\) in the positive direction, and since the length of the circle is \(2π\), it turns out that we will do full turn. That is, the number \(2π\) and \(0\) correspond to the same point. Don't worry, multiple values ​​for one point are normal for a number circle.

Now let's denote the number \(π\) on the number circle. \(π\) is half of \(2π\). Thus, to mark this number and the corresponding point, you need to go half a circle from \(0\) in the positive direction.

Let's mark the point \(\frac(π)(2)\) . \(\frac(π)(2)\) is half of \(π\), therefore, to mark this number, you need to go from \(0\) in the positive direction a distance equal to half of \(π\), that is quarter circle.

Let us denote the points on the circle \(-\)\(\frac(π)(2)\) . We move the same distance as last time, but in a negative direction.

Let's put \(-π\). To do this, we will walk a distance equal to half a circle in the negative direction.

Now let's look at a more complicated example. Let's mark the number \(\frac(3π)(2)\) on the circle. To do this, we translate the fraction \(\frac(3)(2)\) into \(\frac(3)(2)\) \(=1\)\(\frac(1)(2)\), i.e. e. \(\frac(3π)(2)\) \(=π+\)\(\frac(π)(2)\) . This means we need from \(0\) to positive side walk a distance of half a circle and another quarter.

Task 1. Mark the points \(-2π\),\(-\)\(\frac(3π)(2)\) on the number circle.

We denote the numbers \(\frac(π)(4)\), \(\frac(π)(3)\), \(\frac(π)(6)\)

Above we found the values ​​at the points of intersection of the number circle with the \(x\) and \(y\) axes. Now let's determine the position intermediate points. First, let's plot the points \(\frac(π)(4)\) , \(\frac(π)(3)\) and \(\frac(π)(6)\) .
\(\frac(π)(4)\) is half of \(\frac(π)(2)\) (that is, \(\frac(π)(4)\) \(=\)\ (\frac(π)(2)\) \(:2)\) , so the distance \(\frac(π)(4)\) is half a quarter circle.

\(\frac(π)(4)\) is a third of \(π\) (in other words,\(\frac(π)(3)\) \(=π:3\)), so the distance \ (\frac(π)(3)\) is a third of the semicircle.

\(\frac(π)(6)\) is half of \(\frac(π)(3)\) (after all, \(\frac(π)(6)\) \(=\)\(\frac (π)(3)\) \(:2\)) so the distance \(\frac(π)(6)\) is half of the distance \(\frac(π)(3)\) .

This is how they are located relative to each other:

Comment: Location of points with value \(0\), \(\frac(π)(2)\) ,\(π\), \(\frac(3π)(2)\) , \(\frac(π)( 4)\) , \(\frac(π)(3)\) , \(\frac(π)(6)\) it’s better to just remember. Without them, the number circle, like a computer without a monitor, seems to be a useful thing, but is extremely inconvenient to use.

The different distances on the circle are clearly shown:

We denote the numbers \(\frac(7π)(6)\), \(-\frac(4π)(3)\), \(\frac(7π)(4)\)

Let us denote the point on the circle \(\frac(7π)(6)\) , to do this we perform the following transformations: \(\frac(7π)(6)\) \(=\)\(\frac(6π + π)( 6)\) \(=\)\(\frac(6π)(6)\) \(+\)\(\frac(π)(6)\) \(=π+\)\(\frac( π)(6)\) . From this we can see that from zero in the positive direction we need to travel a distance \(π\), and then another \(\frac(π)(6)\) .

Let's mark the point \(-\)\(\frac(4π)(3)\) on the circle. Transform: \(-\)\(\frac(4π)(3)\) \(=-\)\(\frac(3π)(3)\) \(-\)\(\frac(π)( 3)\) \(=-π-\)\(\frac(π)(3)\) . This means that from \(0\) you need to go in the negative direction the distance \(π\) and also \(\frac(π)(3)\) .

Let's plot the point \(\frac(7π)(4)\) , to do this we transform \(\frac(7π)(4)\) \(=\)\(\frac(8π-π)(4)\) \ (=\)\(\frac(8π)(4)\) \(-\)\(\frac(π)(4)\) \(=2π-\)\(\frac(π)(4) \) . This means that in order to place a point with the value \(\frac(7π)(4)\), you need to go from the point with the value \(2π\) in the negative direction to a distance \(\frac(π)(4)\) .

Task 2. Mark the points \(-\)\(\frac(π)(6)\) ,\(-\)\(\frac(π)(4)\) ,\(-\)\(\frac) on the number circle (π)(3)\) ,\(\frac(5π)(4)\) ,\(-\)\(\frac(7π)(6)\) ,\(\frac(11π)(6) \) , \(\frac(2π)(3)\) ,\(-\)\(\frac(3π)(4)\) .

We denote the numbers \(10π\), \(-3π\), \(\frac(7π)(2)\) ,\(\frac(16π)(3)\), \(-\frac(21π)( 2)\), \(-\frac(29π)(6)\)

Let us write \(10π\) in the form \(5 \cdot 2π\). Recall that \(2π\) is the distance equal to length circles, so to mark the point \(10π\), you need to go from zero to a distance equal to \(5\) circles. It is not difficult to guess that we will find ourselves again at point \(0\), just make five revolutions.

From this example we can conclude:

Numbers with a difference of \(2πn\), where \(n∈Z\) (that is, \(n\) is any integer) correspond to the same point.

That is, to put a number with a value greater than \(2π\) (or less than \(-2π\)), you need to extract from it an even number \(π\) (\(2π\), \(8π\), \(-10π\)...) and discard. Thus, we will remove “empty revolutions” from the numbers that do not affect the position of the point.

Another conclusion:

The point to which \(0\) corresponds also corresponds to all even quantities \(π\) (\(±2π\),\(±4π\),\(±6π\)…).

Now let's apply \(-3π\) to the circle. \(-3π=-π-2π\), which means \(-3π\) and \(–π\) are in the same place on the circle (since they differ by an “empty turn” in \(-2π\)).

By the way, all odd \(π\) will also be there.

The point to which \(π\) corresponds also corresponds to all odd quantities \(π\) (\(±π\),\(±3π\),\(±5π\)…).

Now let's denote the number \(\frac(7π)(2)\) . As usual, we transform: \(\frac(7π)(2)\) \(=\)\(\frac(6π)(2)\) \(+\)\(\frac(π)(2)\ ) \(=3π+\)\(\frac(π)(2)\) \(=2π+π+\)\(\frac(π)(2)\) . We discard two pi, and it turns out that to designate the number \(\frac(7π)(2)\) you need to go from zero in the positive direction to a distance equal to \(π+\)\(\frac(π)(2)\ ) (i.e. half a circle and another quarter).

Solution:

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

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

Fig.9 Fig.10

Problem No. 2

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

N
.

Solution:

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

Fig.11

Exercises for independent solution:

1°. On the unit circle, construct a point obtained by rotating the point (1;0) by a given angle:

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

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

a) 3π; b) -
; c) 540°;

d) 810°; d)
, k – integer; e)
.

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

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

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

A)
; b)
; c) 4.5π; d) - 7π.

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

A)
; b)
; V)
; G)
.

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

A)
; b)
;

V)
; G)
.

DEFINITION OF SINE, COSINE OF ANGLE

Fig.12

In these definitions the angle α can be expressed in either degrees or radians. For example, when rotating a point (1;0) by an angle, i.e. angle 90°, the result is point (0;1). 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 No. 1

Find sin (- π) and cos (- π).

Solution:

Point (1;0) when rotated by an angle – π will go to point (-1; 0) (Fig. 13), therefore, sin (- π) = 0, cos (- π) = - 1.

Fig.13

Task No. 2

Solve the equation sin x = 0.

Solution:

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

x = πk., k
.

Answer: x = πk., k

(Z is a designation for a set of integers, read “k belongs to Z”).

Using similar reasoning, we can obtain the following solutions to 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 ​​of sine, cosine, tangent and cotangent.

Task No. 1

Calculate: 4sin +
cos - tg.

Solution:

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 meaning of the expression:

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

V)
; 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 meaning of the 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 with α = ; d) cos + sin at α = .

5*. Solve the equation:

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

Signs of sine, cosine and tangent

Let the point move along the unit circle counterclockwise, 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 No. 1

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

1) ; 2) 745°; 3)
.

Solution:

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

2) Since 745° = 2 ٠360° + 25°, then turning the point (1;0) by an angle of 745° corresponds to a point located at 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, the point is obtained third quarters. Therefore sin

Exercises for independent solution :

1°. In which quarter is the point obtained by rotating the point P(1;0) by an angle? α, If:

A) α = ; b) α = - ; V) α = ;Document

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>> Number circle

While studying the algebra course for grades 7-9, we have so far dealt with algebraic functions, i.e. functions defined analytically by expressions in which algebraic operations on numbers and variables were used (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. We will get acquainted with the first representatives of the class of non-algebraic functions - trigonometric functions - 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 that you have not yet encountered, but you are very familiar with the number line. Recall that the number line is a straight line on which the starting point O, the scale (unit segment) and the positive direction are given. We can compare any real number with a point on a line and vice versa.

How to find the corresponding point M on a line using the number x? The number 0 corresponds to the starting point O. If x > 0, then, moving along a straight line from point 0 in the positive direction, you need to pass 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 You have to move not only in a straight line. Quite often, movement along circle. Here concrete example. Let us consider the stadium running track to be a circle (in fact, it is, of course, not a circle, but remember, as sports commentators usually say: “the runner has run a circle”, “there is half a circle left to run before the finish”, etc.), its length is 400 m. The start is marked - point A (Fig. 97). The runner from point A moves around the circle counterclockwise. Where will he be in 200 m? in 400 m? in 800 m? in 1500 m? Where should he draw the finish line if he is running a marathon distance of 42 km 195 m?

After 200 m, he will be at point C, diametrically opposite to point A (200 m is the length of half the treadmill, i.e. the length of half a 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. 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 just have to deal with the marathon. After running 105 laps, the athlete will cover a distance of 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. You, of course, understand the convention of the last example. No one runs a marathon distance around the stadium, the maximum is 10,000 m, i.e. 25 laps.

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

In principle, any circle can be considered as a numerical circle, 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 Let us agree to call arc AB the first quarter of the unit circle, arc BC the second quarter, arc CB the third quarter, arc DA the fourth quarter (Fig. 98). In this case, we are usually talking about an Open arc, i.e. about an arc without its ends (something like an interval on a number line).

Definition. A unit circle is given, and the starting point A is marked on it - the right end of the horizontal diameter (Fig. 98). Let's match each one 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 will describe a path along the circle with 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);

Let us associate point A with 0: A = A(0).

A unit circle with an established correspondence (between real numbers and points on 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 to find the corresponding points on the circle, you need to walk a path of a given length along the circle, moving from point A in the positive direction. Let us take into account that

The number 2 corresponds to point A, since, having passed along the circle a path of length 2, i.e. exactly one circle, we will again end up in starting point A So, A = A(2).
What's happened This means that moving from point A in a positive direction, you need to go through a whole circle.

Comment. When we are in 7th and 8th grades worked with the number line, then we agreed, for the sake of brevity, not to say “the point on the line corresponding to the number x,” but to say “point x.” We will adhere to exactly the same agreement when working with the number circle: “point f” - this means that we are talking about a point on the circle 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. Depositing arc AM (its length is -) from point A five times in the negative direction, we obtain point!, - the middle of arc BC. So,

Comment. Notice some of the 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 number). But both are designated 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 intended in the designation (arc, arc length, segment or segment length).

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

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

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

Please note that on both layouts we could assign other “names” to the given points.
Have you noticed that in all the analyzed examples of arc lengths
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 a circle or its quarter into equal parts, we get arcs whose lengths are expressed in fractions of the number and. Do you think it is possible to find a point E on the unit circle such that the length of the arc AE is equal to 1? Let's figure it out:

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

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

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

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

In fact, 2n is the length of the numerical (unit) circle, and the integer |th| 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 = -7, then this means that we are doing seven (| k | = | -71 = 7) rounds of the circle in the negative direction. But if we are at point M(1), then, having also completed | to | full circles around the circle, we will again find ourselves at point M.

A.G. Mordkovich Algebra 10th grade

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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. On the unit circle, construct points corresponding to the numbers: 0; 1; 2; 3; 4; 5; .... Could any of these points coincide? Why?

950. Numbers are given by the formula α = 1 / 2 k, Where k= 0; ±1; ±2; ....
Construct points on the number line and on the unit circle that correspond to these numbers. How many such points will there be on the number line and how many on the unit circle?

951. Mark the points on the unit circle and on the number axis that correspond 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 there on the number line and how many on the unit circle?

952. How are the points corresponding to numbers 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 is the fundamental difference between the representation of numbers by points on the number axis and their representation by points on the unit circle?

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

2) In each case, write a general formula for the numbers corresponding to the indicated 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 a point on the unit circle symmetrical to the given one:
a) relative to the x-axis; b) relative to the ordinate axis; c) relative to the origin.
Solve the problem by accepting A = 0; π / 2 ; 1 ; 2 ; π / 6; - π / 4 .

956. Find the condition that the numbers satisfy A, corresponding:
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 a regular octagon ABCDEFKL inscribed in a unit circle has coordinates (1; 0) (Fig. 39).

1) Determine the coordinates of the remaining vertices of the octagon.
2) Create a general formula for arcs of the 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) Construct a point on the unit circle whose ordinate is 0.5. How many points on the unit circle have a given ordinate? How are these points located relative to the ordinate axis?

2) Measure with a protractor (with an accuracy of 1°) the smallest absolute value arc, the end of which has a ordinate equal to 0.5, and compose a general formula for arcs of the 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) Construct a point on the unit circle 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 unit circle arcs 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 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; α2 = 2π k ± π / 3 and α 3 = 2l k± 2π / 3 ;

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

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

964. Eliminate duplicate angles 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 = 72° k+ 36° and α 2 = 120° k+ 60°.