How to denote numbers with pi on the number circle? Lesson "definition of sine and cosine on the unit circle" Summary and basic formulas.

Lesson and presentation on the topic: "Number circle on the coordinate plane"

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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 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.

When studying trigonometry at school, every student is faced with the very interesting concept of “number circle”. How well the student will learn trigonometry later depends on the school teacher’s ability to explain what it is and why it is needed. Unfortunately, not every teacher can explain this material clearly. As a result, many students are confused even about how to mark points on the number circle. If you read this article to the end, you will learn how to do this without any problems.

So let's get started. Let's draw a circle whose radius is 1. Let's denote the “rightmost” point of this circle with the letter O:

Congratulations, you've just drawn a unit circle. Since the radius of this circle is 1, its length is .

To each real number you can match the length of the trajectory along the number circle from the point O. The direction of movement counterclockwise is taken as a positive direction. For negative – clockwise:

Location of points on the number circle

As we have already noted, the length of the number circle (unit circle) is equal to . Where then will the number be located on this circle? Obviously, from the point O counterclockwise we need to go half the length of the circle, and we will find ourselves at the desired point. Let's denote it by the letter B:

Note that the same point could be reached by walking a semicircle in the negative direction. Then we would plot the number on the unit circle. That is, the numbers correspond to the same point.

Moreover, this same point also corresponds to the numbers , , , and, in general, to an infinite set of numbers that can be written in the form , where , that is, belongs to the set of integers. All this because from the point B you can make a “round-the-world” trip in any direction (add or subtract the circumference) and get to the same point. We get an important conclusion that needs to be understood and remembered.

Each number corresponds to a single point on the number circle. But each point on the number circle corresponds to an infinite number of numbers.

Let us now divide the upper semicircle of the number circle into arcs of equal length by a point C. It is easy to see that the arc length O.C. equal to . Let us now postpone from the point C an arc of the same length in a counterclockwise direction. As a result, we will get to the point B. The result is quite expected, since . Let's lay this arc in the same direction again, but now from the point B. As a result, we will get to the point D, which will already correspond to the number:

Note again that this point corresponds not only to the number, but also, for example, to the number, because this point can be reached by moving away from the point O quarter circle in a clockwise direction (negative direction).

And, in general, we note again that this point corresponds to infinitely many numbers that can be written in the form . But they can also be written in the form . Or, if you prefer, in the form of . All these records are absolutely equivalent, and they can be obtained from one another.

Let us now divide the arc into O.C. half dot M. Now figure out what the length of the arc is OM? That's right, half the arc O.C.. That is . What numbers does the dot correspond to? M on the number circle? I am sure that now you will realize that these numbers can be written as .

But it can be done differently. Let's take . Then we get that . That is, these numbers can be written in the form . The same result could be obtained using the number circle. As I already said, both records are equivalent, and they can be obtained from each other.

Now you can easily give an example of the numbers that the points correspond to N, P And K on the number circle. For example, the numbers , and :

Often it is the minimal positive numbers that are taken to designate the corresponding points on the number circle. Although this is not at all necessary, period N, as you already know, corresponds to an infinite number of other numbers. Including, for example, the number.

If you break the arc O.C. into three equal arcs with points S And L, so that's the point S will lie between the points O And L, then the arc length OS will be equal to , and the arc length OL will be equal to . Using the knowledge you gained in the previous part of the lesson, you can easily figure out how the remaining points on the number circle turned out:

Numbers not multiples of π on the number circle

Let us now ask ourselves the question: where on the number line should we mark the point corresponding to the number 1? To do this, you need to start from the most “right” point of the unit circle O plot an arc whose length would be equal to 1. We can only approximately indicate the location of the desired point. Let's proceed as follows.

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

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

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

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

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

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

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

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

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

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

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

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

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

This is how they are located relative to each other:

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

The different distances on the circle are clearly shown:

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

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

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

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

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

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

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

From this example we can conclude:

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

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

Another conclusion:

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

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

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

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

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

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5. TRIGONOMETRIC FUNCTIONS OF ANY ARGUMENT

§ 20. UNIT CIRCLE

948. What is the relationship between the arc length of a unit circle and its radian measure?

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

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

951. Mark the points on the unit circle and on the number axis that correspond to the numbers:
1) α = π k, k= 0; ±1, ±2, ...;
2) α = π / 2 (2k + 1), k= 0; ± 1; ±2; ...;
3) α = π k / 6 , k= 0; ±1; ±2; ... .
How many such points are there on the number line and how many on the unit circle?

952. How are the points corresponding to numbers located on the number axis and on the unit circle:
1) A And - A; 2) A And A±π; 3) A+ π and A- π; 4) A And A+ 2π k, k= 0; ±1; ±2; ...?

953. What is the fundamental difference between the representation of numbers by points on the number axis and their representation by points on the unit circle?

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

2) In each case write general formula numbers corresponding to the indicated points of the unit circle.

955. Knowing that A is one of the numbers corresponding to a given point on the unit circle, find:
1) all numbers corresponding to a given point;
2) all numbers corresponding to a point on the unit circle symmetrical to the given one:
a) relative to the x-axis; b) relative to the ordinate axis; c) relative to the origin.
Solve the problem by accepting A = 0; π / 2 ; 1 ; 2 ; π / 6; - π / 4 .

956. Find the condition that the numbers satisfy A, corresponding:
1) points of the 1st quarter of the unit circle;
2) points of the 2nd quarter of the unit circle;
3) points of the 3rd quarter of the unit circle;
4) points of the 4th quarter of the unit circle.

957. Vertex A of a regular octagon ABCDEFKL inscribed in a unit circle has coordinates (1; 0) (Fig. 39).

1) Determine the coordinates of the remaining vertices of the octagon.
2) Create a general formula for arcs of the unit circle ending:
a) at points A, C, E and K; b) at points B, D, F and L; c) at points A, B, C, D, E, F, K and L.

958. 1) Construct a point on the unit circle whose ordinate is 0.5. How many points on the unit circle have a given ordinate? How are these points located relative to the ordinate axis?

2) Measure with a protractor (with an accuracy of 1°) the smallest arc in absolute value, the end of which has an ordinate of 0.5, and draw up a general formula for arcs of the unit circle ending at points with an ordinate of 0.5.

959. Solve problem 958, taking the ordinate at equal to:

1) - 0,5; 2) 0 4; 3) 0,5√3 .

960. 1) Construct a point on the unit circle whose abscissa is 0.5. How many points on the unit circle have a given abscissa? How are these points located relative to the x-axis?

2) Measure with a protractor (with an accuracy of 1°) the smallest positive arc, the end of which has an abscissa equal to 0.5, and draw up a general formula for unit circle arcs ending at points with an abscissa of 0.5.

961. Solve problem 960, taking the abscissa X equal to:

1) - 2 / 3 ; 2) 0,4; 3) 0,5√2 .

962. Determine the coordinates of the ends of the arcs of the unit circle given by the formula ( k= 0; ±1; ±2; ...):

1) α = 30°(2 k+ 1); 2) α = π k / 3 .

963. Express the following series of angles ( k= 0; ±1; ±2; ...):

1) α 1 = 180° k+ 120° and α 2 = 180° k+ 30°;

2) α 1 = π k + π / 6 and α 2 = π k - π / 3 ;

3) α 1 = 90° k and α 2 = 45° (2 k + 1);

4) α 1 = π k and α 2 = π / 3 (3k± 1);

5) α 1 = 120° k± 15° and α 2 = 120° k± 45°;

6) α 1 = π k; α2 = 2π k ± π / 3 and α 3 = 2l k± 2π / 3 ;

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

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

964. Eliminate duplicate angles in the following formulas ( k= 0-±1; ±2; ...):

1) α 1 = 90° k and α 2 = 60° k+ 30°;

2) α 1 = π k / 2 and α 2 = π k / 5 ;

3) α 1 = 1 / 4 π k and α 2 = 1 / 2 π k± 1/4 π;

4) α 1 = π (2 k+ 1) - π / 6 and α 2 = 2 / 5 π k+ 1 / 30 π;

5) α 1 = 72° k+ 36° and α 2 = 120° k+ 60°.