Number circle is a unit circle whose points correspond to certain real numbers.
A unit circle is a circle of radius 1.
General view of the number circle.
1) Its radius is taken as a unit of measurement.
2) The horizontal and vertical diameters divide the number circle into four quarters (see figure). They are respectively called the first, second, third and fourth quarter.
3) The horizontal diameter is denoted by AC, with A being the far right point.
The vertical diameter is designated BD, with B being the highest point.
Respectively:
the first quarter is the arc AB
second quarter – arc BC
third quarter – arc CD
fourth quarter – arc DA
4) The starting point of the number circle is point A.
Counting along the number circle can be done either clockwise or counterclockwise.
Counting from point A counterclockwise is called positive direction.
Counting from point A clockwise is called negative direction.
Number circle on the coordinate plane.
The center of the radius of the number circle corresponds to the origin (number 0).
The horizontal diameter corresponds to the axis x, vertical – axes y.
The starting point A of the number circle is on the axis x and has coordinates (1; 0).
Valuesx Andy in quarters of a number circle:
Basic values of the number circle:
Names and locations of the main points on the number circle:
How to remember number circle names.
There are several simple patterns that will help you easily remember the basic names of the number circle.
Before we begin, let us remind you: the counting is carried out in the positive direction, that is, from point A (2π) counterclockwise.
1) Let's start with the extreme points on the coordinate axes.
The starting point is 2π (the rightmost point on the axis X, equal to 1).
As you know, 2π is the circumference of a circle. This means that half a circle is 1π or π. Axis X divides the circle exactly in half. Accordingly, the leftmost point on the axis X equal to -1 is called π.
The highest point on the axis at, equal to 1, divides the upper semicircle in half. This means that if a semicircle is π, then half a semicircle is π/2.
At the same time, π/2 is also a quarter of a circle. Let's count three such quarters from the first to the third - and we will come to the lowest point on the axis at, equal to -1. But if it includes three quarters, then its name is 3π/2.
2) Now let's move on to the remaining points. Please note: all opposite points have the same numerator - and these are opposite points relative to the axis at, both relative to the center of the axes, and relative to the axis X. This will help us know their point values without cramming.
You only need to remember the meaning of the points of the first quarter: π/6, π/4 and π/3. And then we will “see” some patterns:
- Relative to the y axis at points of the second quarter, opposite to the points of the first quarter, the numbers in the numerators are 1 less than the size of the denominators. For example, take the point π/6. The point opposite to it relative to the axis at also has 6 in the denominator and 5 in the numerator (1 less). That is, the name of this point is: 5π/6. The point opposite π/4 also has 4 in the denominator and 3 in the numerator (1 less than 4) - that is, it is a 3π/4 point.
The point opposite π/3 also has 3 in the denominator, and 1 less in the numerator: 2π/3.
- Relative to the center of the coordinate axes everything is the other way around: the numbers in the numerators of opposite points (in the third quarter) are 1 greater than the value of the denominators. Let's take the point π/6 again. The point opposite to it relative to the center also has 6 in the denominator, and in the numerator the number is 1 larger - that is, it is 7π/6.
The point opposite the point π/4 also has 4 in the denominator, and in the numerator the number is 1 more: 5π/4.
The point opposite the point π/3 also has 3 in the denominator, and in the numerator the number is 1 more: 4π/3.
- Relative to axis X(fourth quarter) the matter is more complicated. Here you need to add to the value of the denominator a number that is 1 less - this sum will be equal to the numerical part of the numerator of the opposite point. Let's start again with π/6. Let's add to the denominator value equal to 6 a number that is 1 less than this number - that is, 5. We get: 6 + 5 = 11. This means that it is opposite to the axis X the point will have 6 in the denominator and 11 in the numerator - that is, 11π/6.
Point π/4. We add to the value of the denominator a number 1 less: 4 + 3 = 7. This means that it is opposite to the axis X the point has 4 in the denominator and 7 in the numerator - that is, 7π/4.
Point π/3. The denominator is 3. We add to 3 a smaller number by one - that is, 2. We get 5. This means that the point opposite to it has 5 in the numerator - and this is the point 5π/3.
3) Another pattern for the points of the midpoints of the quarters. It is clear that their denominator is 4. Let's pay attention to the numerators. The numerator of the middle of the first quarter is 1π (but it is not customary to write 1). The numerator of the middle of the second quarter is 3π. The numerator of the middle of the third quarter is 5π. The numerator of the middle of the fourth quarter is 7π. It turns out that the numerators of the middle quarters contain the first four odd numbers in ascending order:
(1)π, 3π, 5π, 7π.
This is also very simple. Since the midpoints of all quarters have 4 in the denominator, we already know their full names: π/4, 3π/4, 5π/4, 7π/4.
Features of the number circle. Comparison with the number line.
As you know, on the number line, each point corresponds to a single number. For example, if point A on a line is equal to 3, then it can no longer be equal to any other number.
It's different on the number circle because it's a circle. For example, in order to come from point A of a circle to point M, you can do it as if on a straight line (only passing an arc), or you can go around a whole circle, and then come to point M. Conclusion:
Let point M be equal to some number t. As we know, the circumference of a circle is 2π. This means that we can write a point on a circle t in two ways: t or t + 2π. These are equivalent values.
That is, t = t + 2π. The only difference is that in the first case you came to point M immediately without making a circle, and in the second case you made a circle, but ended up at the same point M. You can make two, three, or two hundred such circles . If we denote the number of circles by the letter k, then we get a new expression:
t = t + 2π k.
Hence the formula:
Number circle equation
(the second equation is in the section “Sine, cosine, tangent, cotangent”):
x 2 + y 2 = 1 |
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In this article we will analyze in great detail the definition of the number circle, find out its main property and arrange the numbers 1,2,3, etc. Learn how to mark other numbers on a circle (including pi).
Number circle called a circle of unit radius whose points correspond , arranged according to the following rules:
1) The origin is at the extreme right point of the circle;
2) Counterclockwise - positive direction; clockwise – negative;
3) If we plot the distance \(t\) on the circle in the positive direction, then we will get to a point with the value \(t\);
4) If we plot the distance \(t\) on the circle in the negative direction, then we will get to a point with the value \(–t\).
Why is the circle called a number circle?
Because it has numbers on it. In this way, the circle is similar to the number axis - on the circle, like on the axis, there is a specific point for each number.
Why know what a number circle is?
Using the number circle, the values of sines, cosines, tangents and cotangents are determined. Therefore, to know trigonometry and pass the Unified State Exam with 60+ points, you must understand what a number circle is and how to place dots on it.
What do the words “...of unit radius...” mean in the definition?
This means that the radius of this circle is equal to \(1\). And if we construct such a circle with the center at the origin, then it will intersect with the axes at points \(1\) and \(-1\).
It doesn’t have to be drawn small; you can change the “size” of the divisions along the axes, then the picture will be larger (see below).
Why is the radius exactly one? This is more convenient, because in this case, when calculating the circumference using the formula \(l=2πR\), we get:
The length of the number circle is \(2π\) or approximately \(6.28\).
What does “...the points of which correspond to real numbers” mean?
As we said above, on the number circle for any real number there will definitely be its “place” - a point that corresponds to this number.
Why determine the origin and direction on the number circle?
The main purpose of the number circle is to uniquely determine its point for each number. But how can you determine where to put the point if you don’t know where to count from and where to move?
Here it is important not to confuse the origin on the coordinate line and on the number circle - these are two different reference systems! And also do not confuse \(1\) on the \(x\) axis and \(0\) on the circle - these are points on different objects.
Which points correspond to the numbers \(1\), \(2\), etc.?
Remember, we assumed that the number circle has a radius of \(1\)? This will be our unit segment (by analogy with the number axis), which we will plot on the circle.
To mark a point on the number circle corresponding to the number 1, you need to go from 0 to a distance equal to the radius in the positive direction.
To mark a point on the circle corresponding to the number \(2\), you need to travel a distance equal to two radii from the origin, so that \(3\) is a distance equal to three radii, etc.
When looking at this picture, you may have 2 questions:
1. What happens when the circle “ends” (i.e. we make a full revolution)?
Answer: let's go for the second round! And when the second one is over, we’ll go to the third one and so on. Therefore, an infinite number of numbers can be plotted on a circle.
2. Where will the negative numbers be?
Answer: right there! They can also be arranged, counting from zero the required number of radii, but now in a negative direction.
Unfortunately, it is difficult to denote integers on the number circle. This is due to the fact that the length of the number circle will not be equal to an integer: \(2π\). And at the most convenient places (at the points of intersection with the axes) there will also be fractions, not integers
Lesson and presentation on the topic: "Number circle on the coordinate plane"
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Manuals and simulators in the Integral online store for grade 10 from 1C
Algebraic problems with parameters, grades 9–11
We solve problems in geometry. Interactive construction tasks for grades 7-10
What we will study:
1. Definition.
2. Important coordinates of the number circle.
3. How to find the coordinate of the number circle?
4. Table of the main coordinates of the number circle.
5. Examples of problem solving.
Definition of the number circle on the coordinate plane
Let's place the number circle in the coordinate plane so that the center of the circle coincides with the origin of coordinates, and take its radius as a unit segment. The starting point of the number circle A is aligned with the point (1;0).Each point on the number circle has its own x and y coordinates in the coordinate plane, and:
1) for $x > 0$, $y > 0$ - in the first quarter;
2) for $x 0$ - in the second quarter;
3) for $x 4) for $x > 0$, $y
For any point $M(x; y)$ on the number circle the following inequalities are satisfied: $-1
Remember the equation of the number circle: $x^2 + y^2 = 1$.
It is important for us to learn how to find the coordinates of the points on the number circle presented in the figure. 
Let's find the coordinate of the point $\frac(π)(4)$
Point $M(\frac(π)(4))$ is the middle of the first quarter. Let us drop the perpendicular MR from point M to straight line OA and consider triangle OMP. Since arc AM is half of arc AB, then $∠MOP=45°$. This means that triangle OMP is an isosceles right triangle and $OP=MP$, i.e. at point M the abscissa and ordinate are equal: $x = y$.
Since the coordinates of the point $M(x;y)$ satisfy the equation of the number circle, then to find them you need to solve the system of equations:
$\begin (cases) x^2 + y^2 = 1,\\ x = y. \end (cases)$
Having solved this system, we obtain: $y = x =\frac(\sqrt(2))(2)$.
This means that the coordinates of the point M corresponding to the number $\frac(π)(4)$ will be $M(\frac(π)(4))=M(\frac(\sqrt(2))(2);\frac (\sqrt(2))(2))$.
The coordinates of the points presented in the previous figure are calculated in a similar way.
Coordinates of points on the number circle
Let's look at examples
Example 1.Find the coordinate of a point on the number circle: $P(45\frac(π)(4))$.
Solution:
$45\frac(π)(4) = (10 + \frac(5)(4)) * π = 10π +5\frac(π)(4) = 5\frac(π)(4) + 2π*5 $.
This means that the number $45\frac(π)(4)$ corresponds to the same point on the number circle as the number $\frac(5π)(4)$. Looking at the value of the point $\frac(5π)(4)$ in the table, we get: $P(\frac(45π)(4))=P(-\frac(\sqrt(2))(2);-\frac (\sqrt(2))(2))$.
Example 2.
Find the coordinate of a point on the number circle: $P(-\frac(37π)(3))$.
Solution:
Because the numbers $t$ and $t+2π*k$, where k is an integer, correspond to the same point on the number circle then:
$-\frac(37π)(3) = -(12 + \frac(1)(3))*π = -12π –\frac(π)(3) = -\frac(π)(3) + 2π *(-6)$.
This means that the number $-\frac(37π)(3)$ corresponds to the same point on the number circle as the number $–\frac(π)(3)$, and the number –$\frac(π)(3)$ corresponds the same point as $\frac(5π)(3)$. Looking at the value of the point $\frac(5π)(3)$ in the table, we get:
$P(-\frac(37π)(3))=P(\frac((1))(2);-\frac(\sqrt(3))(2))$.
Example 3.
Find points on the number circle with ordinate $y =\frac(1)(2)$ and write down what numbers $t$ they correspond to?
Solution: 
The straight line $y =\frac(1)(2)$ intersects the number circle at points M and P. Point M corresponds to the number $\frac(π)(6)$ (from the table data). This means any number of the form: $\frac(π)(6)+2π*k$. Point P corresponds to the number $\frac(5π)(6)$, and therefore to any number of the form $\frac(5π)(6) +2 π*k$.
We received, as is often said in such cases, two series of values:
$\frac(π)(6) +2 π*k$ and $\frac(5π)(6) +2π*k$.
Answer: $t=\frac(π)(6) +2 π*k$ and $t=\frac(5π)(6) +2π*k$.
Example 4.
Find points on the number circle with abscissa $x≥-\frac(\sqrt(2))(2)$ and write down which numbers $t$ they correspond to.
Solution:
The straight line $x =-\frac(\sqrt(2))(2)$ intersects the number circle at points M and P. The inequality $x≥-\frac(\sqrt(2))(2)$ corresponds to the points of the arc PM. Point M corresponds to the number $3\frac(π)(4)$ (from the table data). This means any number of the form $-\frac(3π)(4) +2π*k$. Point P corresponds to the number $-\frac(3π)(4)$, and therefore to any number of the form $-\frac(3π)(4) +2π*k$.
Then we get $-\frac(3π)(4) +2 π*k ≤t≤\frac(3π)(4) +2πk$.
Answer: $-\frac(3π)(4) +2 π*k ≤t≤\frac(3π)(4) +2πk$.
Problems to solve independently
1) Find the coordinate of a point on the number circle: $P(\frac(61π)(6))$.2) Find the coordinate of a point on the number circle: $P(-\frac(52π)(3))$.
3) Find points on the number circle with ordinate $y = -\frac(1)(2)$ and write down which numbers $t$ they correspond to.
4) Find points on the number circle with ordinate $y ≥ -\frac(1)(2)$ and write down which numbers $t$ they correspond to.
5) Find points on the number circle with the abscissa $x≥-\frac(\sqrt(3))(2)$ and write down which numbers $t$ they correspond to.
Quite a lot of time is devoted to the number circle in 10th grade. This is due to the significance of this mathematical object for the entire mathematics course.
The correct selection of teaching aids is of great importance for good mastery of the material. The most effective such tools include video tutorials. Recently they have reached the peak of popularity. Therefore, the author did not lag behind the times and developed such a wonderful manual to help mathematics teachers - a video lesson on the topic “Number circle on the coordinate plane.”
This lesson lasts 15:22 minutes. This is practically the maximum time that a teacher can spend on independently explaining material on a topic. Since it takes so much time to explain new material, it is necessary to select the most effective tasks and exercises for consolidation, and also select another lesson where students will solve tasks on this topic.
The lesson begins with an image of a number circle in a coordinate system. The author builds this circle and explains his actions. Then the author names the points of intersection of the number circle with the coordinate axes. The following explains what coordinates the points of the circle will have in different quarters.
After this, the author reminds us what the equation of a circle looks like. And the listeners are presented with two models depicting some points on the circle. Thanks to this, in the next step the author shows how to find the coordinates of the points on the circle corresponding to certain numbers marked on the templates. This produces a table of values for the variables x and y in the equation of a circle.
Next, we propose to consider an example where it is necessary to determine the coordinates of points on a circle. Before starting to solve the example, some remark is introduced that helps in solving it. And then a complete, clearly structured and illustrated solution appears on the screen. There are also tables here that make it easier to understand the essence of the example.
Then six more examples are considered, which are less time-consuming than the first, but no less important and reflect the main idea of the lesson. Here the solutions are presented in full, with a detailed story and elements of clarity. Namely, the solution contains drawings illustrating the progress of the solution, and a mathematical notation that forms the students’ mathematical literacy.
The teacher may limit himself to the examples discussed in the lesson, but this may not be enough for a high-quality learning of the material. Therefore, choosing tasks to reinforce is simply extremely important.
The lesson can be useful not only for teachers, whose time is constantly limited, but also for students. Especially for those who receive family education or engage in self-education. The materials can be used by those students who missed a lesson on this topic.
TEXT DECODING:
The topic of our lesson is “NUMERIC CIRCLE ON THE COORDINATE PLANE”
We are already familiar with the Cartesian rectangular coordinate system xOy (x o y). In this coordinate system, we will position the number circle so that the center of the circle is aligned with the origin of coordinates, and its radius will be taken as a scale segment.
The starting point A of the number circle is combined with a point with coordinates (1;0), B - with a point (0;1), C - with (-1;0) (minus one, zero), and D - with (0; - 1)(zero, minus one).
(see figure 1)
Since each point on the number circle has its own coordinates in the xOy (x o y) system, then for the points of the first quarter yx is greater than zero and y is greater than zero;
Secondly, the ikx is less than zero and the yk is greater than zero,
for points of the third quarter ikx is less than zero and yk is less than zero,
and for the fourth quarter ikx is greater than zero and yk is less than zero
For any point E (x;y) (with coordinates x, y) of the number circle, the inequalities -1≤ x≤ 1, -1≤y≤1 (x is greater than or equal to minus one, but less than or equal to one; y is greater than or equal to minus one, but less than or equal to one).
Recall that the equation of a circle of radius R with center at the origin has the form x 2 + y 2 = R 2 (x square plus y square equals er square). And for the unit circle R = 1, so we get x 2 + y 2 = 1
(x square plus y square equals one).
Let's find the coordinates of the points on the number circle, which are presented on two layouts (see Fig. 2, 3)
Let point E, which corresponds to
(pi by four) - the middle of the first quarter shown in the figure. From point E we lower the perpendicular EK to the straight line OA and consider the triangle OEK. Angle AOE =45 0, since arc AE is half of arc AB. Therefore, the triangle OEK is an isosceles right triangle, for which OK = EC. This means that the abscissa and ordinate of point E are equal, i.e. x equals game. To find the coordinates of point E, we solve the system of equations: (x is equal to y - the first equation of the system and x square plus y square is equal to one - the second equation of the system). In the second equation of the system, instead of x, we substitute y, we get 2y 2 = 1 (two y square is equal to one), whence y = = (the y is equal to one divided by the root of two is equal to the root of two divided by two) (the ordinate is positive). This means that point E in the rectangular coordinate system has coordinates (,)(root of two divided by two, root of two divided by two).
Reasoning in a similar way, we find the coordinates for the points corresponding to other numbers of the first layout and get: the corresponding point is with coordinates (- ,) (minus root of two divided by two, root of two divided by two); for - (- ,-) (minus root of two divided by two, minus root of two divided by two); for (seven pi over four) (,)(root two divided by two, minus root two divided by two).
Let point D correspond to (Fig. 5). Let us drop the perpendicular from DP(de pe) to OA and consider the triangle ODP. The hypotenuse of this triangle OD is equal to the radius of the unit circle, that is, one, and the angle DOP is equal to thirty degrees, since arc AD = digi AB (a de is equal to one third a be), and arc AB is equal to ninety degrees. Therefore, DP = (de pe is equal to one half O de is equal to one half) Since the leg lying opposite the angle of thirty degrees is equal to half the hypotenuse, that is, y = (y is equal to one half). Applying the Pythagorean theorem, we obtain OR 2 = OD 2 - DP 2 (o pe square equals o de square minus de pe square), but OR = x (o pe equals x). This means x 2 = OD 2 - DP 2 =
this means x 2 = (x square is equal to three-quarters) and x = (x is equal to the root of three times two).
X is positive, because is in the first quarter. We found that point D in a rectangular coordinate system has coordinates (,) root of three divided by two, one half.
Reasoning in a similar way, we will find the coordinates for the points corresponding to other numbers of the second layout and write all the data obtained in the tables:
Let's look at examples.
EXAMPLE 1. Find the coordinates of the points on the number circle: a) C 1 ();
b) C 2 (); c) C 3 (41π); d) C 4 (- 26π). (tse one corresponding to thirty-five pi by four, tse two corresponding to minus forty-nine pi by three, tse three corresponding to forty-one pi, tse four corresponding to minus twenty-six pi).
Solution. Let us use the statement obtained earlier: if point D of the number circle corresponds to the number t, then it corresponds to any number of the form t + 2πk(te plus two peaks), where ka is any integer, i.e. kϵZ (ka belongs to z).
a) We get = ∙ π = (8 +) ∙π = + 2π ∙ 4. (thirty-five pi times four equals thirty-five times four, multiplied by pi equals the sum of eight and three quarters, multiplied by pi equals three pi times four plus the product of two pi by four). This means that the number thirty-five pi by four corresponds to the same point on the number circle as the number three pi by four. Using Table 1, we get C 1 () = C 1 (- ;) .
b) Similar to the coordinates C 2: = ∙ π = - (16 + ∙π = + 2π ∙ (- 8). This means that the number
corresponds to the same point on the number circle as the number. And the number corresponds to the same point on the number circle as the number
(show second layout and table 2). For a point we have x = , y =.
c) 41π = 40π + π = π + 2π ∙ 20. This means that the number 41π corresponds to the same point on the number circle as the number π - this is a point with coordinates (-1; 0).
d) - 26π = 0 + 2π ∙ (- 13), that is, the number - 26π corresponds to the same point on the number circle as the number zero - this is a point with coordinates (1;0).
EXAMPLE 2. Find points on the number circle with ordinate y =
Solution. The straight line y = intersects the number circle at two points. One dot corresponds to a number, the second dot corresponds to a number,
Therefore, we get all the points by adding a full revolution 2πk where k shows how many full revolutions the point makes, i.e. we get,
and for any number all numbers of the form + 2πk. Often in such cases they say that they received two series of values: + 2πk, + 2πk.
EXAMPLE 3. Find points on the number circle with abscissa x = and write down which numbers t they correspond to.
Solution. Straight X= intersects the number circle at two points. One dot corresponds to a number (see second layout),
and therefore any number of the form + 2πk. And the second point corresponds to a number, and therefore to any number of the form + 2πk. These two series of values can be covered in one entry: ± + 2πk (plus minus two pi by three plus two pi).
EXAMPLE 4. Find points with ordinate on the number circle at> and write down which numbers t they correspond to.
The straight line y = intersects the number circle at two points M and P. And the inequality y > corresponds to the points of the open arc MR, this means arcs without ends (that is, without u), when moving around the circle counterclockwise, starting from point M and ending at point P. This means that the core of the analytical notation of the arc MR is the inequality< t < (тэ больше, чем пи на три, но меньше двух пи на три) , а сама аналитическая запись дуги имеет вид + 2πk < t < + 2πk(тэ больше, чем пи на три плюс два пи ка, но меньше двух пи на три плюс два пи ка).
EXAMPLE5. Find ordinate points on the number circle at < и записать, каким числам t они соответствуют.
The straight line y = intersects the number circle at two points M and P. And the inequality y< соответствуют точки открытой дуги РМ при движении по окружности против часовой стрелки, начиная с точки Р, а заканчивая в точке М. Значит, ядром аналитической записи дуги РМ является неравенство < t < (тэ больше, чем минус четыре пи на три, но меньше пи на три) , а сама аналитическая запись дуги имеет вид
2πk< t < + 2πk (тэ больше, чем минус четыре пи на три плюс два пи ка, но меньше пи на три плюс два пи ка).
EXAMPLE 6. Find points with abscissa on the number circle X> and write down which numbers t they correspond to.
The straight line x = intersects the number circle at two points M and P. The inequality x > corresponds to the points of the open arc PM when moving along the circle counterclockwise with the beginning at point P, which corresponds, and the end at point M, which corresponds. This means that the core of the analytical notation of the PM arc is the inequality< t <
(te is greater than minus two pi by three, but less than two pi by three), and the analytical notation of the arc itself has the form + 2πk< t < + 2πk (тэ больше, чем минус два пи на три плюс два пи ка, но меньше двух пи на три плюс два пи ка).
EXAMPLE 7. Find points with abscissa on the number circle X < и записать, каким числам t они соответствуют.
The straight line x = intersects the number circle at two points M and P. Inequality x< соответствуют точки открытой дуги МР при движении по окружности против часовой стрелки с началом в точке М, которая соответствует, и концом в точке Р, которая соответствует. Значит, ядром аналитической записи дуги МР является неравенство < t <
(te is more than two pi by three, but less than four pi by three), and the analytical notation of the arc itself has the form + 2πk< t < + 2πk (тэ больше, чем два пи на три плюс два пи ка, но меньше четырех пи на три плюс два пи ка).