Slide 2
What we will study: Definition. Important coordinates of the number circle. How to find the coordinate of the number circle? Table of basic coordinates of the number circle. Examples of tasks.
Slide 3
Definition. 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 combined with the point (1;0). Each point on the number circle has its own coordinates x and y 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 on the number circle presented in the figure below:
Slide 5
Let's find the coordinate of point π/4: Point M(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 a system of equations: Having solved this system we get: We found that the coordinates of the point M corresponding to the number π /4 will The coordinates of the points presented on the previous slide are calculated in a similar way.
Slide 6
Slide 7
Coordinates of points on the number circle.
Slide 8
Example Find the coordinate of a point on a number circle: Р(45π/4) Solution: Because. the numbers t and t+2π k (k-integer) correspond to the same point on the number circle then: 45π/4 = (10 + 5/4) π = 10π +5π/4 = 5π/4 + 2π 5 So, the number 45π/4 corresponds to the same point on 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 number circle: Р(-37π/3) Solution: Because. the numbers t and t+2π k (k-integer) correspond to the same point on the number circle then: -37π/3 = -(12 + 1/3) π = -12π –π/3 = -π/3 + 2π (-6) This means that the number -37π/3 corresponds to the same point on the number 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 ordinate y = 1/2 and write down which numbers t they correspond to. Example The straight line y = 1/2 intersects the number circle at points M and P. Point M corresponds to the number π/6 (from the table data), which means, and any number of the form π/6+2π k. Point P corresponds to the number 5π/6, and therefore to any number of the form 5π/6+2 π k. We obtained, as is often said in such cases, two series of values: π/6+2 π k and 5π/6+2 π k Answer: t= π/6+2 π k and t= 5π/6+2 π k Number circle on the coordinate plane.
Slide 11
Example Find points on the number circle with abscissa x≥ and write down which numbers t they correspond to. The straight line x= 1/2 intersects the number circle at points M and P. The inequality x ≥ corresponds to the points of the arc PM. Point M corresponds to the number 3π/4 (from the table data), which means, and any number of the form -3π/4+2π k. Point P corresponds to the number -3π/4, and therefore 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
Number circle on the coordinate plane.
Problems for independent solution. 1) Find the coordinate of a point on the number circle: P(61π/6)? 2) Find the coordinate of a point on the number circle: P(-52π/3) 3) Find points on the number circle with ordinate y = -1/2 and write down which numbers t they correspond to. 4) Find points on the number circle with ordinate y ≥-1/2 and write down which numbers t they correspond to. 5) Find the points on the number circle with the abscissa x≥ and write down which numbers t they correspond to.
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Municipal educational institution secondary school No. 1
KHMAO-Yugra
Lesson development
in 10th grade
on algebra and principles of analysis
Nadezhda Mikhailovna
mathematic teacher
Sovetsky
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 for learning new material. Therefore, the main time of the lesson is devoted to learning new material, and students do most of this work independently.
Types of student activities in the lesson: frontal, independent and individual work.
Since a lot of work needs to be done in a lesson and the results of student activities must be monitored, an interactive whiteboard is used at the stages of updating knowledge and learning new material. For a more visual representation of the overlay of the number circle on the coordinate plane and for reflection on the content of the educational material at the end of the training session, Power Point presentations are also used.
educational
Learn to independently acquire knowledge
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 the 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 your ability to find the coordinates of points on a number circle.
Updating knowledge
In the 9th grade geometry course we studied the following
material:
On a unit semicircle (R = 1), we considered a point M with coordinates X And at
Excerpts from a geometry textbook
Having learned to find the coordinates of a point on the unit circle,
Let’s easily move on to their other names: sines and cosines, i.e.
to the main topic - TRIGONOMETRY
The first task is given on the interactive whiteboard, where students need to place the dots and their corresponding numbers in places on the number circle by dragging them with their finger on the board.
Exercise 1
We got the result:
The second task is given on the interactive board. The answers are closed with a “curtain” and are revealed as they are solved.
Task 2
Result of the task:
Learning new material
Let's take a coordinate system and put 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 to both the number circle and the coordinate plane. Let's consider one of these points, for example, point M (Power Point presentation)
M(t)
Let's plot the coordinates of this point
Let's find the coordinates of the points of interest to us on the unit circle, which we considered earlier with denominators 4, 3, 6 and numerator π.
Find the coordinates of the point on the unit circle corresponding to the number and, accordingly, the angle
Task 3
(Power Point presentation)
Let's depict the radius and coordinates of the point
By the Pythagorean theorem we have X 2+ x 2 = 12
But the angles of the triangle are π/4 = 45° , This means that the triangle is isosceles and x = y
Find the coordinates of a point on 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, which means their legs are equal
In the previous lesson, students received sheets with blanks of number circles and various tables.
Fill out the first table.
Task 5
(interactive board)
First, enter the points of the circle that are multiples of 2 and 4 into the table.
Checking the result:
(interactive board)
Fill in the ordinates and abscissas of these points yourself in the table, taking into account the coordinate signs, 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 obtained, the rest check their answers, then to successfully correct 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 board.
Checking the result:
(interactive board)
Fill out the second table.
Task 7
(interactive board)
First, enter into the table the points of the circle that are multiples of 3 and 6
Checking the result:
(interactive board)
Fill in the ordinates and abscissas of these points yourself in the table
Task 8
Checking the result:
(interactive board)
(Power Point presentation)
Let's conduct a short mathematical dictation followed by self-control.
1) Find the coordinates of the points of the unit circle:
Option 2
1 option
2) Find the abscissa of the points of the unit circle:
1) Find the coordinates of the points on the unit circle
Option 2
1 option
2) Find the abscissa of the points on the unit circle
check yourself
3) Find the ordinates of the points of the unit circle:
For yourself, you can mark “5” for 4 completed examples,
“4” for 3 examples and mark “3” for 2 examples
Summing up the lesson
1) In the future, to find the values of sine, cosine, tangent and cotangent of points and angles, it is necessary to learn from the completed tables the values of the coordinates of points belonging to the first quarter because further we will learn to express the coordinate values of all other points through the values of the points of the first quarter;
2) Prepare theoretical questions for testing.
Homework:
Lesson summary
The grade is given to the students who worked most actively in the lesson. The work of all students is not graded, since errors are corrected immediately during the lesson. The dictation was conducted for self-control; there is insufficient volume for assessment.
Analytical geometry provides uniform techniques for solving geometric problems. To do this, all given and sought-after points and lines are assigned to one coordinate system.
In a coordinate system, each point can be characterized by its coordinates, and each line – by an equation with two unknowns, the graph of which this line is. Thus, the geometric problem is reduced to an algebraic one, where all calculation methods are well developed.
A circle is a geometric locus of points with one specific property (each point on the circle is equidistant from one point, called the center). The equation of a circle must reflect this property and satisfy this condition.
The geometric interpretation of the equation of a circle is the line of a circle.
If you place a circle in a coordinate system, then all points on the circle satisfy one condition - the distance from them to the center of the circle must be the same and equal to the circle.
Circle with center at a point A and radius R place it in the coordinate plane.
If the center coordinates (a;b) , and the coordinates of any point on the circle (x;y) , then the equation of the circle has the form:
If the square of the radius of a circle is equal to the sum of the squares of the differences between the corresponding coordinates of any point on the circle and its center, then this equation is the equation of a circle in a plane coordinate system.
If the center of the circle coincides with the origin, then the square of the radius of the circle is equal to the sum of the squares of the coordinates of any point on the circle. In this case, the equation of the circle takes the form:
Consequently, any geometric figure as a locus of points is determined by an equation connecting the coordinates of its points. Conversely, the equation relating the coordinates X And at , define a line as the geometric locus of points on the plane whose coordinates satisfy this equation.
Examples of solving problems about the equation of a circle
Task. Write an equation for a given circle
Write an equation for a circle with center at point O (2;-3) and radius 4.Solution.
Let us turn to the formula for the equation of a circle:
R 2 = (x-a) 2 + (y-b) 2
Let's substitute the values into the formula.
Circle radius R = 4
Coordinates of the center of the circle (according to the condition)
a = 2
b = -3
We get:
(x - 2 ) 2 + (y - (-3 )) 2 = 4 2
or
(x - 2) 2 + (y + 3) 2 = 16.
Task. Does a point belong to the equation of a circle?
Check if a point belongs to A(2;3) equation of a circle (x - 2) 2 +(y+3) 2 = 16 .Solution.
If a point belongs to a circle, then its coordinates satisfy the equation of the circle.
To check whether a point with given coordinates belongs to a circle, substitute the coordinates of the point into the equation of the given circle.
In equation ( x - 2) 2 + (y + 3) 2 = 16
Let us substitute, according to the condition, the coordinates of point A(2;3), that is
x = 2
y=3
Let's check the truth of the resulting equality
(x - 2) 2 + (y + 3) 2 = 16
(2
- 2) 2 + (3
+ 3) 2 = 16
0 + 36 = 16 equality is false
So the given point do not belong given equation of a circle.
If you place the unit number circle on the coordinate plane, then you can find the coordinates for its points. The number circle is positioned so that its center coincides with the origin of the plane, i.e., point O (0; 0).
Usually on the unit number circle the points corresponding to the origin of the circle are marked
- quarters - 0 or 2π, π/2, π, (2π)/3,
- middle quarters - π/4, (3π)/4, (5π)/4, (7π)/4,
- thirds of quarters - π/6, π/3, (2π)/3, (5π)/6, (7π)/6, (4π)/3, (5π)/3, (11π)/6.
On the coordinate plane, with the above location of the unit circle on it, you can find the coordinates corresponding to these points of the circle.
The coordinates of the ends of the quarters are very easy to find. At point 0 of the circle, the x coordinate is 1, and the y coordinate is 0. We can denote it as 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 semi-axis: C (π) = C (-1; 0).
End of third quarter: D ((2π)/3) = D (0; -1).
But how to find the coordinates of the midpoints of the quarters? To do this, construct a right triangle. Its hypotenuse is a segment from the center of the circle (or 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, draw a perpendicular from a point on the circle to any axis. Let it be towards the x axis. The result is a right triangle, the lengths of the legs of which are the x and y coordinates of the point on the circle.
A quarter circle is 90º. And half a quarter is 45º. Since the hypotenuse is drawn to the midpoint of the quadrant, the angle between the hypotenuse and the leg extending from the origin is 45º. But the sum of the angles of any triangle is 180º. Consequently, the angle between the hypotenuse and the other leg also remains 45º. This results in 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 = √½ = 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 the other quarters, only the signs will change, and the modules of the values will remain the same, since the right triangle will only be turned 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 a circle, a right triangle is also constructed. 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 a leg lying opposite an angle of 30º is equal to half the hypotenuse. This means that we have found the y coordinate, it is equal to ½.
Knowing the lengths of the hypotenuse and one of the legs, using the Pythagorean theorem we find the other leg:
x 2 + (½) 2 = 1 2
x 2 = 1 - ¼ = 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 be equal to ½, and y, respectively, √3/2: T 2 (π/3) = T 2 (½; √3/2).
For other points of the third quarters, the signs and order of the coordinate values will change. All points that are closer to the x axis will have a modulus x coordinate value equal to √3/2. Those points that are closer to the y axis will have a modulus 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)