On the unit circle two diametrically opposite ones are marked. Spherical geometry. Lines, segments, distances and angles on a sphere

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Apparently, mankind’s first appeal to what would later be called spherical geometry was the planetary theory of the Greek mathematician Eudoxus (c. 408–355), one of the participants in Plato’s Academy. It was an attempt to explain the motion of the planets around the Earth with the help of four rotating concentric spheres, each of which had a special axis of rotation with the ends fixed on the enclosing sphere, to which, in turn, the stars were “nailed.” In this way, the intricate trajectories of the planets were explained (translated from Greek, “planet” means wandering). It was thanks to this model that ancient Greek scientists were able to quite accurately describe and predict the movements of the planets. This was necessary, for example, in navigation, as well as in many other “earthly” tasks, where it was necessary to take into account that the Earth is not a flat pancake resting on three pillars. Significant contributions to spherical geometry were made by Menelaus of Alexandria (c. 100 AD). His work Spherics became the pinnacle of Greek achievements in this area. IN Sferike spherical triangles are considered - a subject that is not found in Euclid. Menelaus transferred the Euclidean theory of flat triangles to the sphere and, among other things, obtained a condition under which three points on the sides of a spherical triangle or their extensions lie on the same straight line. The corresponding theorem for the plane was already widely known at that time, but it entered the history of geometry precisely as the theorem of Menelaus, and, unlike Ptolemy (c. 150), who had many calculations in his works, Menelaus’ treatise is geometric strictly in the spirit of the Euclidean tradition .

Basic principles of spherical geometry.

Any plane intersecting a sphere produces a circle in cross-section. If the plane passes through the center of the sphere, then the cross section results in a so-called great circle. Through any two points on a sphere, except those that are diametrically opposite, a single large circle can be drawn. (On the globe, an example of a great circle is the equator and all the meridians.) An infinite number of great circles pass through diametrically opposite points. Lesser arc AmB(Fig. 1) of the great circle is the shortest of all lines on the sphere connecting given points. This line is called geodetic. Geodesic lines play the same role on a sphere as straight lines do in planimetry. Many provisions of geometry on the plane are also valid on the sphere, but, unlike the plane, two spherical lines intersect at two diametrically opposite points. Thus, the concept of parallelism simply does not exist in spherical geometry. Another difference is that the spherical line is closed, i.e. moving along it in the same direction, we will return to the starting point; the point does not split the line into two parts. And another surprising fact from the point of view of planimetry is that a triangle on a sphere can have all three right angles.

Lines, segments, distances and angles on a sphere.

Great circles on a sphere are considered to be straight lines. If two points belong to a great circle, then the length of the smaller of the arcs connecting these points is defined as spherical distance between these points, and the arc itself is like a spherical segment. Diametrically opposite points are connected by an infinite number of spherical segments - large semicircles. The length of a spherical segment is determined through the radian measure of the central angle a and the radius of the sphere R(Fig. 2), according to the arc length formula it is equal to R a. Any point WITH spherical segment AB splits it into two, and the sum of their spherical lengths, as in planimetry, is equal to the length of the entire segment, i.e. R AOC+ R OWL= P AOB. For any point D outside the segment AB there is a “spherical triangle inequality”: the sum of the spherical distances from D before A and from D before IN more AB, i.e. R AOD+ R DOB> R AOB, – complete correspondence between spherical and flat geometries. The triangle inequality is one of the fundamental ones in spherical geometry; it follows from it that, as in planimetry, a spherical segment is shorter than any spherical broken line, and therefore any curve on the sphere connecting its ends.

In the same way, many other concepts of planimetry can be transferred to the sphere, in particular those that can be expressed through distances. For example, spherical circle– a set of points on the sphere equidistant from a given point R. It is easy to show that the circle lies in a plane perpendicular to the diameter of the sphere RR` (Fig. 3), i.e. this is an ordinary flat circle with a center on the diameter RR`. But it has two spherical centers: R And R`. These centers are usually called poles. If we turn to the globe, we can see that we are talking about circles such as parallels, and the spherical centers of all parallels are the North and South Poles. If the diameter r of a spherical circle is equal to p/2, then the spherical circle turns into a spherical straight line. (On the globe there is the equator). In this case, such a circle is called polar each of the points R And P`.

One of the most important concepts in geometry is the equality of figures. Figures are considered equal if one can be displayed on top of the other in such a way (by rotation and translation) that the distances are preserved. This is also true for spherical geometry.

Angles on a sphere are defined as follows. When two spherical lines intersect a And b Four spherical bigons are formed on the sphere, just as two intersecting lines on a plane divide it into four plane angles (Fig. 4). Each of the diagons corresponds to a dihedral angle formed by the diametrical planes containing a And b. And the angle between spherical straight lines is equal to the smaller of the angles of the diagons they form.

We also note that angle P ABC, formed on a sphere by two arcs of a great circle, is measured by angle P A`B.C.` between tangents to the corresponding arcs at a point IN(Fig. 5) or a dihedral angle formed by diametrical planes containing spherical segments AB And Sun.

In the same way as in stereometry, each point on the sphere is associated with a ray drawn from the center of the sphere to this point, and any figure on the sphere is associated with the union of all rays intersecting it. Thus, a spherical straight line corresponds to the diametrical plane containing it, a spherical segment corresponds to a plane angle, a digon corresponds to a dihedral angle, and a spherical circle corresponds to a conical surface whose axis passes through the poles of the circle.

A polyhedral angle with a vertex at the center of the sphere intersects the sphere along a spherical polygon (Fig. 6). This is an area on a sphere bounded by a broken line of spherical segments. The links of the broken line are the sides of a spherical polygon. Their lengths are equal to the values of the corresponding plane angles of the polyhedral angle, and the value of the angle at any vertex A equal to the dihedral angle at the edge OA.

Spherical triangle.

Among all spherical polygons, the spherical triangle is of greatest interest. Three large circles, intersecting in pairs at two points, form eight spherical triangles on the sphere. Knowing the elements (sides and angles) of one of them, it is possible to determine the elements of all the others, so we consider the relationships between the elements of one of them, the one whose all sides are less than half of the great circle. The sides of a triangle are measured by the plane angles of the trihedral angle OABC, the angles of the triangle are dihedral angles of the same trihedral angle (Fig. 7).

Many properties of a spherical triangle (and they are also properties of trihedral angles) almost completely repeat the properties of an ordinary triangle. Among them is the triangle inequality, which, in the language of trihedral angles, states that any plane angle of a trihedral angle is less than the sum of the other two. Or, for example, three signs of equality of triangles. All planimetric consequences of the mentioned theorems, together with their proofs, remain valid on the sphere. Thus, the set of points equidistant from the ends of the segment will also be on the sphere perpendicular to it, a straight line passing through its middle, from which it follows that the bisectors are perpendicular to the sides of a spherical triangle ABC have a common point, or rather, two diametrically opposed common points R And R`, which are the poles of its only circumscribed circle (Fig. 8). In stereometry, this means that a cone can be described around any trihedral angle. It is easy to transfer to the sphere the theorem that the bisectors of a triangle intersect at the center of its incircle.

Theorems on the intersection of heights and medians also remain true, but their usual proofs in planimetry directly or indirectly use parallelism, which does not exist on a sphere, and therefore it is easier to prove them again, in the language of stereometry. Rice. Figure 9 illustrates the proof of the spherical median theorem: planes containing the medians of a spherical triangle ABC, intersect a plane triangle with the same vertices along its usual medians, therefore, they all contain the radius of the sphere passing through the intersection point of the plane medians. The end of the radius will be the common point of the three “spherical” medians.

The properties of spherical triangles differ in many ways from the properties of triangles on a plane. Thus, to the known three cases of equality of rectilinear triangles, a fourth is added: two triangles ABC And А`В`С` are equal if three angles P are equal, respectively A= P A`, R IN= P IN`, R WITH= P WITH`. Thus, there are no similar triangles on the sphere; moreover, in spherical geometry there is no very concept of similarity, because There are no transformations that change all distances by the same (not equal to 1) number of times. These features are associated with a violation of the Euclidean axiom of parallel lines and are also inherent in Lobachevsky’s geometry. Triangles that have equal elements and different orientations are called symmetrical, such as triangles AC`WITH And VSS` (Fig. 10).

The sum of the angles of any spherical triangle is always greater than 180°. Difference P A+P IN+P WITH - p = d (measured in radians) is a positive quantity and is called spherical excess of a given spherical triangle. Area of a spherical triangle: S = R 2 d where R is the radius of the sphere, and d is the spherical excess. This formula was first published by the Dutchman A. Girard in 1629 and named after him.

If we consider a diagon with angle a, then at 226 = 2p/ n (n – integer) the sphere can be cut exactly into P copies of such a diagon, and the area of the sphere is 4 nR 2 = 4p at R= 1, so the area of the diagon is 4p/ n= 2a. This formula is also true for a = 2p t/n and therefore true for all a. If we continue the sides of a spherical triangle ABC and express the area of the sphere through the areas of the resulting bigons with angles A,IN,WITH and its own area, then we can arrive at the above Girard formula.

Coordinates on the sphere.

Each point on the sphere is completely determined by specifying two numbers; these numbers ( coordinates) are determined as follows (Fig. 11). Some large circle is fixed QQ` (equator), one of the two points of intersection of the diameter of the sphere PP`, perpendicular to the equatorial plane, with the surface of a sphere, for example R (pole), and one of the large semicircles PAP` coming out of the pole ( first meridian). Large semicircles coming out of P, called meridians, small circles parallel to the equator, such as LL`, – parallels. As one of the point coordinates M on the sphere the angle q is taken = POM (point height), as the second – angle j = AON between the first meridian and the meridian passing through the point M (longitude points, counted counterclockwise).

In geography (on the globe), it is customary to use the Greenwich meridian as the first meridian, passing through the main hall of the Greenwich Observatory (Greenwich is a London borough), it divides the Earth into the Eastern and Western hemispheres, respectively, and longitude is eastern or western and measured from 0 to 180° in both directions from Greenwich. And instead of the height of a point in geography, it is customary to use latitude at, i.e. corner NOM = 90° – q, measured from the equator. Because Since the equator divides the Earth into the Northern and Southern Hemispheres, the latitude is either northern or southern and varies from 0 to 90°.

Marina Fedosova

I once witnessed a conversation between two applicants:

– When should you add 2πn, and when should you add πn? I just can't remember!

– And I have the same problem.

I just wanted to tell them: “You don’t need to memorize, but understand!”

This article is addressed primarily to high school students and, I hope, will help them solve the simplest trigonometric equations with “understanding”:

Number circle

Along with the concept of a number line, there is also the concept of a number circle. As we know, in a rectangular coordinate system, a circle with a center at the point (0;0) and radius 1 is called a unit circle. Let’s imagine the number line as a thin thread and wind it around this circle: we will attach the origin (point 0) to the “right” point of the unit circle, we will wrap the positive semi-axis counterclockwise, and the negative semi-axis in the direction (Fig. 1). Such a unit circle is called a numerical circle.

Properties of the number circle

Each real number lies on one point on the number circle.
There are infinitely many real numbers at every point on the number circle. Since the length of the unit circle is 2π, the difference between any two numbers at one point on the circle is equal to one of the numbers ±2π; ±4π ; ±6π ; ...

Let's conclude: knowing one of the numbers of point A, we can find all the numbers of point A.

Let's draw the diameter of the AC (Fig. 2). Since x_0 is one of the numbers of point A, then the numbers x_0±π ; x_0±3π; x_0±5π; ... and only they will be the numbers of point C. Let's choose one of these numbers, say, x_0+π, and use it to write down all the numbers of point C: x_C=x_0+π+2πk ,k∈Z. Note that the numbers at points A and C can be combined into one formula: x_(A ; C)=x_0+πk ,k∈Z (for k = 0; ±2; ±4; ... we obtain the numbers of point A, and for k = ±1; ±3; ±5; … – numbers of point C).

Let's conclude: knowing one of the numbers at one of the points A or C of the diameter AC, we can find all the numbers at these points.

Two opposite numbers are located on points of the circle that are symmetrical with respect to the abscissa axis.

Let's draw a vertical chord AB (Fig. 2). Since points A and B are symmetrical about the Ox axis, the number -x_0 is located at point B and, therefore, all numbers of point B are given by the formula: x_B=-x_0+2πk ,k∈Z. We write the numbers at points A and B using one formula: x_(A ; B)=±x_0+2πk ,k∈Z. Let us conclude: knowing one of the numbers at one of the points A or B of the vertical chord AB, we can find all the numbers at these points. Let's consider the horizontal chord AD and find the numbers of point D (Fig. 2). Since BD is a diameter and the number -x_0 belongs to point B, then -x_0 + π is one of the numbers of point D and, therefore, all the numbers of this point are given by the formula x_D=-x_0+π+2πk ,k∈Z. The numbers at points A and D can be written using one formula: x_(A ; D)=(-1)^k∙x_0+πk ,k∈Z . (for k= 0; ±2; ±4; … we get the numbers of point A, and for k = ±1; ±3; ±5; … – the numbers of point D).

Let's conclude: knowing one of the numbers at one of the points A or D of the horizontal chord AD, we can find all the numbers at these points.

Sixteen main points of the number circle

In practice, solving most of the simplest trigonometric equations involves sixteen points on a circle (Fig. 3). What are these dots? Red, blue and green dots divide the circle into 12 equal parts. Since the length of the semicircle is π, then the length of the arc A1A2 is π/2, the length of the arc A1B1 is π/6, and the length of the arc A1C1 is π/3.

Now we can indicate one number at a time:

π/3 on C1 and

The vertices of the orange square are the midpoints of the arcs of each quarter, therefore, the length of the arc A1D1 is equal to π/4 and, therefore, π/4 is one of the numbers of point D1. Using the properties of the number circle, we can use formulas to write down all the numbers on all marked points of our circle. The coordinates of these points are also marked in the figure (we will omit the description of their acquisition).

Having mastered the above, we now have sufficient preparation to solve special cases (for nine values of the number a) simplest equations.

Solve equations

1)sinx=1⁄(2).

– What is required of us?

– Find all those numbers x whose sine is equal to 1/2.

Let's remember the definition of sine: sinx – ordinate of the point on the number circle on which the number x is located. We have two points on the circle whose ordinate is equal to 1/2. These are the ends of the horizontal chord B1B2. This means that the requirement “solve the equation sinx=1⁄2” is equivalent to the requirement “find all the numbers at point B1 and all the numbers at point B2.”

2)sinx=-√3⁄2 .

We need to find all the numbers at points C4 and C3.

3) sinx=1. On the circle we have only one point with ordinate 1 - point A2 and, therefore, we need to find only all the numbers of this point.

Answer: x=π/2+2πk, k∈Z.

4)sinx=-1 .

Only point A_4 has an ordinate of -1. All the numbers of this point will be the horses of the equation.

Answer: x=-π/2+2πk, k∈Z.

5) sinx=0 .

On the circle we have two points with ordinate 0 - points A1 and A3. You can indicate the numbers at each of the points separately, but given that these points are diametrically opposite, it is better to combine them into one formula: x=πk,k∈Z.

Answer: x=πk ,k∈Z .

6)cosx=√2⁄2 .

Let's remember the definition of cosine: cosx is the abscissa of the point on the number circle on which the number x is located. On the circle we have two points with the abscissa √2⁄2 - the ends of the horizontal chord D1D4. We need to find all the numbers on these points. Let's write them down, combining them into one formula.

Answer: x=±π/4+2πk, k∈Z.

7) cosx=-1⁄2 .

We need to find the numbers at points C_2 and C_3.

Answer: x=±2π/3+2πk , k∈Z .

10) cosx=0 .

Only points A2 and A4 have an abscissa of 0, which means that all the numbers at each of these points will be solutions to the equation.
.

The solutions to the equation of the system are the numbers at points B_3 and B_4. To the cosx inequality<0 удовлетворяют только числа b_3
Answer: x=-5π/6+2πk, k∈Z.

Note that for any admissible value of x, the second factor is positive and, therefore, the equation is equivalent to the system

The solutions to the system equation are the number of points D_2 and D_3. The numbers of point D_2 do not satisfy the inequality sinx≤0.5, but the numbers of point D_3 do.

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Question: On a circle, diametrically opposite points A and B and a different point C are chosen. The tangent drawn to the circle at point A and the line BC intersect at point D. Prove that the tangent drawn to the circle at point C bisects the segment A.D. The incircle of triangle ABC touches sides AB and BC at points M and N respectively. A line passes through the midpoint of AC parallel to the line. MN intersects lines BA and BC at points D and E, respectively. Prove that AD=CE.

On the circle, diametrically opposite points A and B and a different point C are chosen. The tangent drawn to the circle at point A and the straight line BC intersect at point D. Prove that the tangent drawn to the circle at point C bisects the segment AD. The incircle of triangle ABC touches sides AB and BC at points M and N respectively. A line passes through the midpoint of AC parallel to the line. MN intersects lines BA and BC at points D and E, respectively. Prove that AD=CE.

On the unit circle two diametrically opposite ones are marked. Spherical geometry. Lines, segments, distances and angles on a sphere

Basic principles of spherical geometry.

Lines, segments, distances and angles on a sphere.

Spherical triangle.

Coordinates on the sphere.

Answers: