The video lesson “Definition of sine and cosine on the unit circle” provides visual material for a lesson on the relevant topic. During the lesson, the concepts of sine and cosine for numbers corresponding to points of the unit circle are discussed, many examples are described that form the ability to solve problems where this interpretation of the concepts is used. Convenient and understandable illustrations of solutions, a detailed course of reasoning help to quickly achieve learning goals and increase the effectiveness of the lesson.
The video lesson begins by introducing the topic. At the beginning of the demonstration, the definition of sine and cosine of a number is given. A unit circle with a center at the origin of coordinates is demonstrated on the screen, the points of intersection of the unit circle with the coordinate axes A, B, C, D are marked. A definition is highlighted in the frame, which states that if a point M belonging to the unit circle corresponds to a certain number t, then the abscissa of this point is the cosine of the number t and is denoted cos t, the ordinate of the point is a sine and is denoted sin t. The voicing of the definition is accompanied by an image of point M on the unit circle, indicating its abscissa and ordinate. A short notation is presented using the notation that for M(t)=M(x;y), x= cos t, y= sin t. The restrictions imposed on the value of the cosine and sine of a number are indicated. According to the data reviewed, -1<=cos t<=1 и -1<= sin t<=1.
It is also easy to see from the figure how the sign of the function changes depending on which quarter the point is located in. A table is compiled on the screen in which for each function its sign is indicated depending on the quarter. The sign of cos t is plus in the first and fourth quarters and minus in the second and third quarters. The sin t sign is plus in the first and second quarters, minus in the third and fourth quarters.
Students are reminded of the unit circle equation x 2 + y 2 = 1. It is noted that after substituting instead of the coordinates of the corresponding functions, we obtain cos 2 t+ sin 2 t=1 - the main trigonometric identity. Using the method of finding sin t and cos t using the unit circle, fill in a table of the basic values of sine and cosine for numbers from 0 to 2π in increments of π/4 and for numbers from π/6 to 11π/6 in increments of π/6. These tables are shown on the screen. Using them and the drawing, the teacher can check how well the material has been mastered and how well the students understand the origin of the sin t and cos t values.
An example is considered in which sin t and cos t are calculated for t=41π/4. The solution is illustrated by a figure that shows a unit circle with its center at the origin. The point 41π/4 is marked on it. It is noted that this point coincides with the position of the point π/4. This is proven by representing this fraction as a mixed fraction 41π/4=π/4+2π·5. Using the table of cosine values, we obtain the values cos π/4=√2/2 and sinπ/4=√2/2. From the information obtained it follows that cos 41π/4=√2/2 and sin 41π/4=√2/2.
In the second example, it is necessary to calculate sin t and cos t for t=-25π/3. The screen displays a unit circle with the point t=-25π/3 marked on it. First, to solve the problem, the number -25π/3 is represented as a mixed fraction in order to discover which table value its sin t and cos t will correspond to. After the transformation we get -25π/3=-π/3+2π·(-4). Obviously, t=-25π/3 will coincide on the circle with the point -π/3 or 5π/3. From the table we select the corresponding values of sine and cosine cos 5π/3=1/2 and sin 5π/3=-√3/2. These values will be true for the number in question cos (-25π/3)=1/2 and sin (-25π/3)=-√3/2. The problem is solved.
Example 3 is solved similarly, in which it is necessary to calculate sin t and cos t for t=37π. To solve the example, the number 37π is expanded, isolating π and 2π. In this representation it turns out 37π=π+2π·18. On the unit circle, which is depicted next to the solution, this point is marked at the intersection of the negative part of the ordinate axis and the unit circle - point π. Obviously, the values of the sine and cosine of the number will coincide with the tabulated values of π. From the table we find the values sin π=-1 and cos π=0. Accordingly, these same values are the desired ones, that is, sin 37π=-1 and cos 37π=0.
In example 4 it is required to calculate sin t and cos t at t=-12π. We represent the number as -12π=0+2π·(-6). Accordingly, point -12π coincides with point 0. The cosine and sine values of this point are sin 0=1 and cos 0=0. These values are the required ones sin (-12π)=1 and cos (-12π)=0.
In the fifth example, you need to solve the equation sin t=√3/2. In solving the equation, the concept of the sine of a number is used. Since it represents the ordinate of the point M(t), it is necessary to find the point with the ordinate √3/2. The figure accompanying the solution shows that the ordinate √3/2 corresponds to two points - the first π/3 and the second 2π/3. Considering the periodicity of the function, we note that t=π/3+2πk and t= 2π/3+2πk for integer k.
In example 6, the equation with cosine is solved - cos t=-1/2. In searching for solutions to the equation, we find points on the unit circle with the abscissa 2π/3. A picture is shown on the screen in which the abscissa -1/2 is marked. It corresponds to two points on the circle - 2π/3 and -2π/3. Taking into account the periodicity of functions, the found solution is written in the form t=2π/3+2πk and t=-2π/3+2πk, where k is an integer.
In example 7 the equation sin t-1=0 is solved. To find a solution, the equation is transformed to the form sin t=1. Sine 1 corresponds to the number π/2. Taking into account the periodicity of the function, the found solution is written in the form t=π/2+2πk, where k is an integer. Similarly, in example 8 the equation cos t+1=0 is solved. Let's transform the equation to the form cos t=-1. The point whose abscissa is -1 corresponds to the number π. This point is marked on the unit circle shown next to the text solution. Accordingly, the solution to this equation is the number t=π+2πk, where k is an integer. It is no more difficult to solve the equation cos t+1=1 in example 9. Transforming the equation, we obtain cos t=0. On the unit circle shown next to the solution, we mark the points -π/2 and -3π/2, at which the cosine takes the value 0. Obviously, the solution to this equation will be a series of values t=π/2+πk, where k is an integer.
In example 10, the values of sin 2 and cos 3 are compared. To make the solution clear, a figure is shown where points 2 and 3 are marked. Knowing that π/2≈1.57, we estimate the distance of the points from it. The figure notes that point 2 is 0.43 away from π/2, while 3 is 1.43 away, so point 2 has a larger abscissa than point 3. This means sin 2>cos 3.
Example 11 describes the calculation of the expression sin 5π/4. Since 5π/4 is π/4+π, then, using reduction formulas, the expression can be converted into the form - sin π/4. From the table we select its value - sin π/4=-√2/2. Similarly, in example 12 the value of the expression cos7π/6 is found. Transforming it to the form cos(π/6+π), we obtain the expression - cos π/6. The table value is cos π/6=-√3/2. This value will be the solution.
Next, it is suggested to remember important equalities that help in solving problems - these are sin(-t)= -sin t and cos (-t)=cos t. In fact, this expression reflects the evenness of the cosine and the oddness of the sine. In the image of the unit circle next to the equalities you can see how coordinate plane the equality data works. Two equalities are also presented that reflect the periodicity of functions, which are important for solving problems sin(t+2πk)= sin t and cos (t+2πk)=cos t. Equalities are demonstrated that reflect the symmetrical arrangement of points on the unit circle sin(t+π)= -sin t and cos (t+π)=-cos t. Next to the equalities, an image is constructed that displays the location of these points on the unit circle. And the last presented equalities sin(t+π/2)= cos t and cos (t+π/2)=- sin t.
The video lesson “Definition of sine and cosine on the unit circle” is recommended for use in a traditional school mathematics lesson to increase its effectiveness and ensure the clarity of the teacher’s explanation. For the same purpose, the material can be used during distance learning. The manual can also be useful for developing appropriate problem-solving skills in students when mastering the material independently.
TEXT DECODING:
"Definition of sine and cosine on the unit circle."
Let's define the sine and cosine of a number
DEFINITION: if a point M of a numerical unit circle corresponds to the number t(te), then the abscissa of the point M is called the cosine of the number t(te) and is designated cost, and the ordinate of the point M is called the sine of the number t(te) and is designated sint(fig).
This means that if M(t) = M (x,y)(em from te is equal to em with coordinates x and y), then x = cost, y= sint (x is equal to the cosine of te, y is equal to the sine of te). Consequently, - 1≤ cost ≤ 1, -1≤ sint ≤1 (cosine te is greater than or equal to minus one, but less than or equal to one; sine te is greater than or equal to minus one, but less than or equal to one). Knowing that each point number circle has its own coordinates in the xOy system, you can create a table of the values of sine and cosine by quarters of a circle, where the cosine value is positive in the first and fourth quarters and, accordingly, negative in the second and third quarters.
The sine value is positive in the first and second quarters and, accordingly, negative in the third and fourth quarters. (show on drawing)
Since the equation of the number circle has the form x 2 + y 2 = 1 (x square plus y square equals one), then we get the equality:
(cosine squared te plus sine squared te equals one).
Based on the tables that we compiled when determining the coordinates of points on the numerical circle, we will compile tables for the coordinates of points on the numerical circle for the values of cost and sint.
Let's look at examples.
EXAMPLE 1. Calculate cos t and sin t if t = (te equals forty-one pi over four).
Solution. The number t = corresponds to the same point on the number circle as the number, since = ∙π = (10 +) ∙π = + 2π ∙ 5 (forty-one pi times four is equal to the sum of pi times four and the product of two pi times five). And for point t = according to the table the value of cosines 1 we have cos = and sin =. Hence,
EXAMPLE 2. Calculate cos t and sin t, if t = (te equals minus twenty-five pi over three).
SOLUTION: The number t = corresponds to the same point on the number circle as the number, since = ∙ π = - (8 +)∙π = + 2π ∙ (- 4) (minus twenty-five pi over three is equal to the sum of minus pi over three and the product of two pi times minus four). And the number corresponds to the same point on the number circle as the number. And for point t = according to Table 2 we have cos = and sin =. Therefore, cos () = and sin () =.
EXAMPLE 3. Calculate cos t and sin t if t = 37π; (te equals thirty-seven pi).
SOLUTION: 37π = 36π + π = π + 2π ∙ 18. This means that the number 37π corresponds to the same point on the number circle as the number π. And for the point t = π, according to Table 1, we have cos π = -1, sin π = 0. This means cos37π = -1, sin37π = 0.
EXAMPLE 4. Calculate cos t and sin t if t = -12π (equal to minus twelve pi).
SOLUTION: - 12π = 0 + 2π ∙ (- 6), that is, the number - 12π corresponds to the same point on the number circle as the number zero. And for the point t = 0, according to Table 1, we have cos 0 = 1, sin 0 =0. This means cos(-12π) =1, sin(-12π) =0.
EXAMPLE 5. Solve the equation sin t = .
Solution. Considering that sin t is the ordinate of the point M(t) (em from te) of the number circle, we will find points with the ordinate on the number circle and write down which numbers t they correspond to. One point corresponds to a number, and therefore to any number of the form + 2πk. The second point corresponds to a number, and therefore to any number of the form + 2πk. Answer: t = + 2πk, where kϵZ (ka belongs to zet), t= + 2πk, where kϵZ (ka belongs to zet).
EXAMPLE 6. Solve the equation cos t = .
Solution. Considering that cos t is the abscissa of the point M(t) (em from te) of the number circle, we will find the points with the abscissa on the number circle and write down which numbers t they correspond to. One point corresponds to a number, and therefore to any number of the form + 2πk. And the second point corresponds to the number or, and therefore to any number of the form + 2πk or + 2πk.
Answer: t = + 2πk, t=+ 2πk (or ± + 2πk (plus minus two pi by three plus two pi ka), where kϵZ (ka belongs to zet).
EXAMPLE 7. Solve the equation cos t = .
Solution. Similar to the previous example, you need to find points with an abscissa on the number circle and write down which numbers t they correspond to.
The figure shows that two points E and S have an abscissa, but we cannot yet say which numbers they correspond to. We will return to this issue later.
EXAMPLE 8. Solve the equation sin t = - 0.3.
Solution. On the number circle we find points with ordinate - 0.3 and write down which numbers t they correspond to.
The ordinate - 0.3 has two points P and H, but we cannot yet say what numbers they correspond to. We will also return to this issue later.
EXAMPLE 9. Solve the equation sin t -1 =0
Solution. Let's move minus one to the right side of the equation, we get sine te equals one (sin t = 1). On the number circle we need to find a point whose ordinate is equal to one. This point corresponds to a number, and therefore to all numbers of the form + 2πk (pi times two plus two peaks).
Answer: t = + 2πk, kϵZ(ka belongs to zet).
EXAMPLE 10. Solve the equation cos t + 1 = 0.
Let's move one to the right side of the equation, we get the cosine te equals minus one (cos t = - 1). The abscissa minus one has a point on the number circle, which corresponds to the number π, and this means all numbers of the form π+2πk. Answer: t = π+ 2πk, kϵZ.
EXAMPLE 11. Solve the equation cos t + 1 = 1.
Let's move the unit to the right side of the equation, we get the cosine te equals zero (cos t = 0). Abscissa zero has points B and D (Figure 1), which correspond to numbers, etc. These numbers can be written as + πk. Answer: t = + πk, kϵZ.
EXAMPLE 12. Which of the two numbers is greater, cos 2 or cos 3? (cosine of two or cosine of three)
Solution. Let's reformulate the question differently: points 2 and 3 are marked on the number circle. Which of them has a larger abscissa?
On the number circle, mark points 2 and 3. Remember that. This means that point 2 is removed from the circle by approximately 0.43 (zero point forty-three hundredths) (2 -≈ 2 - 1.57 = 0.43), and point 3 by 1.43 (one point forty three hundredths). Therefore, point 2 is closer to point than point 3, so it has a larger abscissa (we took into account that both abscissas are negative).
Answer: cos 2 > cos 3.
EXAMPLE 13. Calculate sin (sine five pi times four)
Solution. sin(+ π) = - sin = (sine five pi over four equals the sum of pi over four and pi equals minus sine pi over four equals minus root two over two).
EXAMPLE 14. Calculate cos (cosine of seven pi by six).
cos(+ π) = - cos =. (we represented seven pi over six as the sum of pi over six and pi and applied the third equality).
For sine and cosine we get some important formulas.
1. For any value of t the following equalities are true:
sin (-t) = -sin t
cos (-t) = cos t
The sine of minus te is equal to minus sine of te
The cosine of minu te is equal to the cosine of te.
The figure shows that points E and L, symmetrical with respect to the abscissa axis, have the same abscissa, this means
cos(-t) = cost, but the ordinates are equal in magnitude and opposite in sign (this means sin(- t) = - sint.
2. For any value of t the following equalities are valid:
sin (t+2πk) = sin t
cos (t+2πk) = cos t
The sine of te plus two pi is equal to the sine of te
The cosine of te plus two pi is equal to the cosine of te
This is true, since the numbers t and t+2πk correspond to the same point.
3. For any value of t the following equalities are valid:
sin (t+π) = -sin t
cos (t+π) = -cos t
The sine of te plus pi is equal to minus the sine of te
cosine of te plus pi equals minus cosine of te
Let the number t correspond to point E of the number circle, then the number t+π corresponds to point L, which is symmetrical to point E relative to the origin. The figure shows that at these points the abscissa and ordinate are equal in magnitude and opposite in sign. This means,
cos(t +π)= - cost;
sin(t +π)= - sint.
4. For any value of t the following equalities are valid:
sin(t+) = cos t
cos(t+) = -sin t
Sine te plus pi by two equals cosine te
Cosine te plus pi by two is equal to minus sine te.
>> Number circle
While studying the algebra course for grades 7-9, we have so far dealt with algebraic functions, i.e. functions specified analytically by expressions, in the recording of which we used algebraic operations over numbers and variables (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.
For introduction trigonometric functions we will need a new one 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's a 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. Of course you understand the convention 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, anyone can negative number match a point on the circle: 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, to find the corresponding points on the circle, you need to walk along the circle given length, moving from point A in a 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 get to the 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. Please note 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 given points assign other “names”.
Have you noticed that in all the 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 with a path 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? Slightly 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 to a single number. 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 make seven (| k | = | -71 = 7) rounds of the circle in the negative direction. But if we are at point M(1), then, having also performed | to | full circles around the circle, we will again find ourselves at point M.
A.G. Mordkovich Algebra 10th grade
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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 combined 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 $∠MOP=45°$. So triangle OMP is 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 decided this system, we get: $y = x =\frac(\sqrt(2))(2)$.
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.
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)
