Let's place the number circle in the coordinate plane so that the center of the circle is aligned with the origin, and its radius is taken as a unit segment. The starting point of the numerical circle A is aligned with the point (1;0). Each point of the numerical circle has its own coordinates x and y in the coordinate plane, and: 1) x > 0, y > 0 in the first quarter; 2) x 0 in the second quarter; 3) x 0, y 0, y > 0 in the first quarter; 2) x 0 in the second quarter; 3) x 0, y 
Find the coordinate of the point π/4: Point M(π/4) is the middle of the first quarter. Let us drop the perpendicular MP from the point M to the line OA and consider the triangle OMP. Since the arc AM is half of the arc AB, then MOP = 45 °. at point M, the abscissa and ordinate are equal: x \u003d y Since the coordinates of the point M (x; y) satisfy the equation of the numerical circle, then to find them, you need to solve the system of equations: Having solved this system, we get: We got that the coordinates of the point M, corresponding to the number π /4 will be In the same way, the coordinates of the points presented on the previous slide are calculated.
Find the coordinate of a point on a numerical circle: Р(45π/4) Solution: the numbers t and t + 2πk (k-integer) correspond to the same point of the number circle then: 45π/4 = (10 + 5/4) π = 10π +5π/4 = 5π/4 + 2π5 4 corresponds to the same point of the numerical circle as the number 5π/4. Looking at the value of the point 5π/4 in the table, we get:
Find the coordinate of a point on a numerical circle: Р(-37π/3) Solution: the numbers t and t + 2πk (k-integer) correspond to the same point of the number circle, then: -37π/3 = -(12 + 1/3) π = -12π –π/3 = -π/3 + 2π( -6) So, the number -37π/3 corresponds to the same point of the numerical circle as the number -π/3, and the number -π/3 corresponds to the same point as 5π/3. Looking at the value of the point 5π/3 in the table, we get:
Find points on the number circle with the ordinate y \u003d 1/2 and write down what numbers t they correspond to. The straight line y \u003d 1/2 intersects the number circle at points M and P. The point M corresponds to the number π / 6 (from the data in the table), which means that any number of the form π / 6 + 2π k. The point P corresponds to the number 5π/6, and hence to any number of the form 5π/6 +2 π k Answer: t= π/6 +2 π k and t= 5π/6 +2 π k
Find points on the number circle with abscissa x and write down what numbers t they correspond to. The straight line x = 1/2 intersects the number circle at the points M and P. The inequality x corresponds to the points of the arc PM. The point M corresponds to the number 3π/4 (from the data in the table) and, therefore, to any number of the form -3π/4 + 2πk. The point P corresponds to the number -3π/4, and hence to any number of the form -3π/4 +2 π k Then we get -3π/4 +2 π k t3π/4 +2 π k Answer: -3π/4 +2 π k t3π/4 +2 π k
1) Find the coordinate of a point on a numerical circle: Р(61π/6)? 2) Find the coordinate of the point of the numerical circle: P (-52π / 3) 3) Find on the numerical circle the points with the ordinate y \u003d -1/2 and write down what numbers t they correspond to. 4) Find points on the number circle with the ordinate y -1/2 and write down what numbers t they correspond to. 5) Find on the number circle the points with the abscissa x and write down what numbers t they correspond to.
If you place a unit number circle on the 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
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)
Additional materials
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Manuals and simulators in the online store "Integral" 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 will we study:
1. Definition.
2. Important coordinates of the numerical circle.
3. How to find the coordinate of a numerical circle?
4. Table of the main coordinates of the numerical circle.
5. Examples of problem solving.
Each point of the number circle has its x and y coordinates in the coordinate plane, and:
1) for $x > 0$, $y > 0$ - in the first quarter;
2) with $x 0$ - in the second quarter;
3) for $x 4) for $x > 0$, $y
For any point $M(x; y)$ of the numerical circle, the following inequalities hold: $-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 of the numerical circle shown in the figure. 
The point $M(\frac(π)(4))$ is the middle of the first quarter. Let us drop the perpendicular MP from the point M to the line OA and consider the triangle OMP. Since the arc AM is half of the arc AB, then $∠MOP=45°$. 
Solution:
$45\frac(π)(4) = (10 + \frac(5)(4)) * π = 10π +5\frac(π)(4) = 5\frac(π)(4) + 2π*5 $.
Hence, the number $45\frac(π)(4)$ corresponds to the same point of 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 a 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 of the numerical circle, then:
$-\frac(37π)(3) = -(12 + \frac(1)(3))*π = -12π –\frac(π)(3) = -\frac(π)(3) + 2π *(-6)$.
Hence, the number $-\frac(37π)(3)$ corresponds to the same point of the number circle as the number $–\frac(π)(3)$, and the number –$\frac(π)(3)$ corresponds to 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 $t$ do they correspond to?
Solution: 
The line $y =\frac(1)(2)$ intersects the number circle at the points M and P. The point M corresponds to the number $\frac(π)(6)$ (from the data in the table). Hence, any number of the form: $\frac(π)(6)+2π*k$. The point P corresponds to the number $\frac(5π)(6)$, and hence to any number of the form $\frac(5π)(6) +2 π*k$.
We got, as they often say 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 line $x =-\frac(\sqrt(2))(2)$ intersects the number circle at the points M and P. The inequality $x≥-\frac(\sqrt(2))(2)$ corresponds to the points of the arc PM. The point M corresponds to the number $3\frac(π)(4)$ (from the data in the table). Hence, any number of the form $-\frac(3π)(4) +2π*k$. The point P corresponds to the number $-\frac(3π)(4)$, and hence 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: $-\frac(3π)(4) +2 π*k ≤t≤\frac(3π)(4) +2πk$.
Municipal educational institution secondary school No. 1
KhMAO-Yugra
Lesson Development
in 10 "b" class
in algebra and the beginnings of analysis
Nadezhda Mikhailovna
mathematic teacher
Soviet
Topic: TRIGONOMETRY
Trigonometric functions
Trigonometric equations
Trigonometric transformations
Number circle on
coordinate plane
The subject is taught using block-modular technology.
This lesson is one of the lessons of learning new material. Therefore, the main time of the lesson is devoted to the study of new material, and students do most of this work on their own.
Types of students' activities in the lesson: frontal, independent and individual work.
Since it is necessary to do a lot of work in the lesson and be sure to control the results of student activities, an interactive whiteboard is used at the stages of updating knowledge and learning new material. For a more visual representation of the imposition of a number circle on the coordinate plane and for reflecting the content of the educational material, Power Point presentations are also used at the end of the lesson.
cognitive
Learn to acquire knowledge on your own
nurturing
Cultivate composure, responsibility, diligence
developing
Learn to analyze, compare, build analogies
Lesson plan:
1) Organizational moment, topic, purpose of lesson 2 min.
2) Updating knowledge 4 min.
3) Learning new material 30 min.
4) Reflection 3 min.
5) Summary of lesson 1 min.
Organizing time
Number circle
coordinate plane
consider a number circle on the coordinate plane; together find the coordinates of two points; then independently compile tables of coordinate values of other main points of the circle;
test the ability to find the coordinates of points on a numerical circle.
Knowledge update
In the 9th grade geometry course, we studied the following
material:
On the unit semicircle (R = 1) we considered the point M with coordinates X and at
Excerpts from the geometry textbook
Having learned to find the coordinates of a point on a unit circle,
we can easily move on to their other names: sines and cosines, i.e.
to the main topic - TRIGONOMETRY
The first task is given on an interactive whiteboard, where students need to put the dots and their corresponding numbers in place on the number circle by dragging them with their finger across the board.
Exercise 1
Got the result:
The second task is given on the interactive whiteboard. The answers are closed by a “curtain”, they open as they are solved.
Task 2
The result of the task:
Learning new material
Let's take a coordinate system and impose a number circle on it so that their centers coincide, and the horizontal radius of the circle coincides with the positive direction of the OX axis (Power Point presentation)
As a result, we have points that belong simultaneously to the numerical circle and the coordinate plane. Consider one of these points, for example, point M (Power Point presentation)
M(t)
Draw the coordinates of this point
Let's find the coordinates of the points of the unit circle that are of interest to us, which were considered earlier with the denominators 4, 3, 6 and the numerator π.
Find the coordinates of the point of the unit circle corresponding to the number, respectively, and the angle
Task 3
(power point presentation)
Draw the radius and coordinates of a point
By the Pythagorean theorem, we have X 2+ x 2 = 12
But the angles of the triangle in π/4 = 45° , so the triangle is isosceles and x = y
Find the coordinates of the point of the unit circle corresponding to the numbers (angles)
Task 4
(power point presentation)
Means at= 1/2
According to the Pythagorean theorem
Triangles are equal in hypotenuse
and an acute angle, so their legs are equal
In the previous lesson, students received sheets with blanks for number circles and various tables.
Complete the first table.
Task 5
(interactive board)
First, enter the circle points in the table that are multiples of 2 and 4
Checking the result:
(interactive board)
Fill in independently in the table the ordinates and abscissas of these points, taking into account the signs of the coordinates, depending on which quarter the point is located in, using the lengths of the segments obtained above for the coordinates of the points.
Task 6
One of the students names the results, the rest check with their answers, then for successful correction of the results (since these tables will be used later in the work to develop skills and deepen knowledge on the topic), a correctly completed table is shown on the interactive whiteboard.
Checking the result:
(interactive board)
Complete the second table.
Task 7
(interactive board)
First, enter the circle points in the table that are multiples of 3 and 6
Checking the result:
(interactive board)
Fill in independently in the table the ordinates and abscissas of these points
Task 8
Checking the result:
(interactive board)
(power point presentation)
We will conduct a small mathematical dictation with subsequent self-control.
1) Find the coordinates of the points of the unit circle:
Option 2
1 option
2) Find the abscissas of the points of the unit circle:
1) Find the coordinates of the points of the unit circle
Option 2
1 option
2) Find the abscissas of the points of the unit circle
test yourself
3) Find the ordinates of the points of the unit circle:
For yourself, you can put a mark "5" for 4 completed examples,
"4" for 3 examples and "3" for 2 examples
Summing up the lesson
1) In the future, to find the values of the sine, cosine, tangent and cotangent of points and angles, it is necessary to learn from the completed tables the coordinates of the points belonging to the first quarter, because further we will learn to express the values of the coordinates of all other points through the values of the points of the first quarter;
2) Prepare theoretical questions for the test.
Homework:
Lesson summary
The grade is given to the most active students in the lesson. The work of all students is not evaluated, as errors are corrected immediately during the lesson. The dictation was carried out for self-control, for evaluation there is not enough volume.
slide 2
What we will study: Definition. Important coordinates of the number circle. How to find the coordinate of a number circle? Table of basic coordinates of the numerical circle. Task examples.
slide 3
Definition. Let's place the number circle in the coordinate plane so that the center of the circle is aligned with the origin, and its radius is taken as a unit segment. The starting point of the numerical circle A is aligned with the point (1;0). Each point of the numerical circle has its own x and y coordinates in the coordinate plane, and: x > 0, y > 0 in the first quarter; x 0 in the second quarter; x 0, y
slide 4
It is important for us to learn how to find the coordinates of the points of the numerical circle shown in the figure below:
slide 5
Find the coordinate of the point π/4: Point M(π/4) is the middle of the first quarter. Let us drop the perpendicular MP from the point M to the line OA and consider the triangle OMP. at point M, the abscissa and ordinate are equal: x \u003d y Since the coordinates of the point M (x; y) satisfy the equation of the numerical circle, then to find them, you need to solve the system of equations: Having solved this system, we get: We got that the coordinates of the point M, corresponding to the number π /4 will be In the same way, the coordinates of the points presented on the previous slide are calculated.
slide 6
Slide 7
Coordinates of points on a numerical circle.
Slide 8
Example Find the coordinate of a point on a numerical circle: Р(45π/4) Solution: Since. the numbers t and t + 2π k (k-integer) correspond to the same point of the number circle then: 45π/4 = (10 + 5/4) π = 10π +5π/4 = 5π/4 + 2π 5 45π/4 corresponds to the same point of the number circle as the number 5π/4. Looking at the value of the point 5π/4 in the table, we get:
Slide 9
Example Find the coordinate of a point on a numerical circle: Р(-37π/3) Solution: the numbers t and t + 2π k (k-integer) correspond to the same point of the numerical circle then: -37π/3 = -(12 + 1/3) π = -12π –π/3 = -π/3 + 2π (-6) So, the number -37π/3 corresponds to the same point of the numerical circle as the number –π/3, and the number –π/3 corresponds to the same point as 5π/3. Looking at the value of the point 5π/3 in the table, we get:
Slide 10
Find points on the number circle with the ordinate y \u003d 1/2 and write down what numbers t they correspond to. Example The straight line y \u003d 1/2 intersects the number circle at points M and P. The point M corresponds to the number π / 6 (from the table data), which means, and any number of the form π / 6 + 2π k. The point P corresponds to the number 5π/6, and hence to any number of the form 5π/6+2 π k Answer: t= π/6+2 π k and t= 5π/6+2 π k The number circle on the coordinate plane.
slide 11
Example Find points on the number circle with abscissa x≥ and write down what numbers t they correspond to. The straight line x= 1/2 intersects the number circle at the points M and P. The inequality x ≥ corresponds to the points of the arc PM. The point M corresponds to the number 3π/4 (from the data in the table), which means that any number of the form -3π/4+2π k. The point Р corresponds to the number -3π/4, and hence to any number of the form – -3π/4+2 π k Then we get -3π/4+2 π k≤t≤3π/4+2 π k Answer: -3π/ 4+2 π k≤t≤3π/4+2 π k Number circle on the coordinate plane.
slide 12
Tasks for independent solution. 1) Find the coordinate of a point on a numerical circle: Р(61π/6)? 2) Find the coordinate of the point of the numerical circle: P (-52π / 3) 3) Find on the numerical circle the points with the ordinate y \u003d -1/2 and write down what numbers t they correspond to. 4) Find points on the number circle with ordinate y ≥-1/2 and write down what numbers t they correspond to. 5) Find on the number circle points with abscissa x≥ and write down what numbers t they correspond to.
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