Let's say Achilles runs ten times faster than the tortoise and is a thousand steps behind it. During the time it takes Achilles to run this distance, the tortoise will crawl a hundred steps in the same direction. When Achilles runs a hundred steps, the tortoise crawls another ten steps, and so on. The process will continue ad infinitum, Achilles will never catch up with the tortoise.
This reasoning became a logical shock for all subsequent generations. Aristotle, Diogenes, Kant, Hegel, Hilbert... They all considered Zeno's aporia in one way or another. The shock was so strong that " ...discussions continue to this day, to reach a common opinion about the essence of paradoxes scientific community so far it has not been possible... we were involved in the study of the issue mathematical analysis, set theory, new physical and philosophical approaches; none of them became a generally accepted solution to the problem..."[Wikipedia, "Zeno's Aporia". Everyone understands that they are being fooled, but no one understands what the deception consists of.
From a mathematical point of view, Zeno in his aporia clearly demonstrated the transition from quantity to . This transition implies application instead of permanent ones. As far as I understand, mathematical apparatus The use of variable units of measurement has either not yet been developed, or it has not been applied to Zeno’s aporia. Applying our usual logic leads us into a trap. We, due to the inertia of thinking, apply constant units of time to the reciprocal value. WITH physical point From a perspective, it looks like time slowing down until it stops completely at the moment when Achilles catches up with the turtle. If time stops, Achilles can no longer outrun the tortoise.
If we turn our usual logic around, everything falls into place. Achilles runs with constant speed. Each subsequent segment of his path is ten times shorter than the previous one. Accordingly, the time spent on overcoming it is ten times less than the previous one. If we apply the concept of “infinity” in this situation, then it would be correct to say “Achilles will catch up with the turtle infinitely quickly.”
How to avoid this logical trap? Stay in constant units measurements of time and do not go to reciprocal quantities. In Zeno's language it looks like this:
In the time it takes Achilles to run a thousand steps, the tortoise will crawl a hundred steps in the same direction. For the next time interval, equal to first, Achilles will run another thousand steps, and the tortoise will crawl a hundred steps. Now Achilles is eight hundred steps ahead of the tortoise.
This approach adequately describes reality without any logical paradoxes. But it is not complete solution Problems. Einstein’s statement about the irresistibility of the speed of light is very similar to Zeno’s aporia “Achilles and the Tortoise”. We still have to study, rethink and solve this problem. And the solution must be sought not in infinitely large numbers, but in units of measurement.
Another interesting aporia of Zeno tells about a flying arrow:
A flying arrow is motionless, since at every moment of time it is at rest, and since it is at rest at every moment of time, it is always at rest.
In this aporia, the logical paradox is overcome very simply - it is enough to clarify that at each moment of time a flying arrow is at rest at different points in space, which, in fact, is motion. Another point needs to be noted here. From one photograph of a car on the road it is impossible to determine either the fact of its movement or the distance to it. To determine whether a car is moving, you need two photographs taken from the same point in different moments time, but distance cannot be determined from them. To determine the distance to the car, you need two photographs taken from different points space at one point in time, but it is impossible to determine the fact of movement from them (naturally, additional data is still needed for calculations, trigonometry will help you). What I want to point out Special attention, is that two points in time and two points in space are different things that should not be confused, because they provide different opportunities for research.
Wednesday, July 4, 2018
The differences between set and multiset are described very well on Wikipedia. Let's see.
As you can see, “there cannot be two identical elements in a set,” but if there are identical elements in a set, such a set is called a “multiset.” Reasonable beings will never understand such absurd logic. This is the level talking parrots and trained monkeys, who have no intelligence from the word “completely”. Mathematicians act as ordinary trainers, preaching to us their absurd ideas.
Once upon a time, the engineers who built the bridge were in a boat under the bridge while testing the bridge. If the bridge collapsed, the mediocre engineer died under the rubble of his creation. If the bridge could withstand the load, the talented engineer built other bridges.
No matter how mathematicians hide behind the phrase “screw me, I’m in the house”, or rather “mathematics studies abstract concepts", there is one umbilical cord that inextricably connects them with reality. This umbilical cord is money. Apply mathematical theory sets to the mathematicians themselves.
We studied mathematics very well and now we are sitting at the cash register, giving out salaries. So a mathematician comes to us for his money. We count out the entire amount to him and lay it out on our table in different piles, into which we put bills of the same denomination. Then we take one bill from each stack and hand it to the mathematician" mathematical set salaries." We explain to the mathematics that he will receive the remaining bills only when he proves that a set without identical elements is not equal to a set with identical elements. This is where the fun begins.
First of all, the logic of the deputies will work: “This can be applied to others, but not to me!” Then they will begin to reassure us that bills of the same denomination have different bill numbers, which means they cannot be considered the same elements. Okay, let's count salaries in coins - there are no numbers on the coins. Here the mathematician will begin to frantically remember physics: on different coins there is different quantities mud, crystal structure and the arrangement of atoms in each coin is unique...
And now I have the most interest Ask: where is the line beyond which the elements of a multiset turn into elements of a set and vice versa? Such a line does not exist - everything is decided by shamans, science is not even close to lying here.
Look here. We select football stadiums with the same field area. The areas of the fields are the same - which means we have a multiset. But if we look at the names of these same stadiums, we get many, because the names are different. As you can see, the same set of elements is both a set and a multiset. Which is correct? And here the mathematician-shaman-sharpist pulls out an ace of trumps from his sleeve and begins to tell us either about a set or a multiset. In any case, he will convince us that he is right.
To understand how modern shamans operate with set theory, tying it to reality, it is enough to answer one question: how do the elements of one set differ from the elements of another set? I'll show you, without any "conceivable as not a single whole" or "not conceivable as a single whole."
Sunday, March 18, 2018
The sum of the digits of a number is a dance of shamans with a tambourine, which has nothing to do with mathematics. Yes, in mathematics lessons we are taught to find the sum of the digits of a number and use it, but that’s why they are shamans, to teach their descendants their skills and wisdom, otherwise shamans will simply die out.
Do you need proof? Open Wikipedia and try to find the page "Sum of digits of a number." She doesn't exist. There is no formula in mathematics that can be used to find the sum of the digits of any number. After all, numbers are graphic symbols, with the help of which we write numbers and in the language of mathematics the task sounds like this: “Find the sum of graphic symbols representing any number.” Mathematicians cannot solve this problem, but shamans can do it easily.
Let's figure out what and how we do in order to find the sum of numbers given number. And so, let us have the number 12345. What needs to be done in order to find the sum of the digits of this number? Let's consider all the steps in order.
1. Write down the number on a piece of paper. What have we done? We have converted the number into a graphical number symbol. This is not a mathematical operation.
2. We cut one resulting picture into several pictures containing individual numbers. Cutting a picture is not a mathematical operation.
3. Convert individual graphic symbols into numbers. This is not a mathematical operation.
4. Add the resulting numbers. Now this is mathematics.
The sum of the digits of the number 12345 is 15. These are the “cutting and sewing courses” taught by shamans that mathematicians use. But that is not all.
From a mathematical point of view, it does not matter in which number system we write a number. So, in different systems In calculus, the sum of the digits of the same number will be different. In mathematics, the number system is indicated as a subscript to the right of the number. WITH a large number 12345 I don’t want to fool my head, let’s look at the number 26 from the article about . Let's write this number in binary, octal, decimal and hexadecimal number systems. We won't look at every step under a microscope; we've already done that. Let's look at the result.
As you can see, in different number systems the sum of the digits of the same number is different. This result has nothing to do with mathematics. It’s the same as if you determined the area of a rectangle in meters and centimeters, you would get completely different results.
Zero looks the same in all number systems and has no sum of digits. This is another argument in favor of the fact that. Question for mathematicians: how is something that is not a number designated in mathematics? What, for mathematicians nothing exists except numbers? I can allow this for shamans, but not for scientists. Reality is not just about numbers.
The result obtained should be considered as proof that number systems are units of measurement for numbers. After all, we cannot compare numbers with different units measurements. If the same actions with different units of measurement of the same quantity lead to different results after comparing them, then this has nothing to do with mathematics.
What is real mathematics? This is when the result mathematical operation does not depend on the size of the number, the unit of measurement used and who performs the action.
Oh! Isn't this the women's restroom?
- Young woman! This is a laboratory for the study of the indephilic holiness of souls during their ascension to heaven! Halo on top and arrow up. What other toilet?
Female... The halo on top and the arrow down are male.
If such a work of design art flashes before your eyes several times a day,
Then it’s not surprising that you suddenly find a strange icon in your car:
Personally, I make an effort to see minus four degrees in a pooping person (one picture) (a composition of several pictures: a minus sign, the number four, a designation of degrees). And I don't think this girl is stupid, no knowledgeable in physics. She just has an arch stereotype of perception graphic images. And mathematicians teach us this all the time. Here's an example.
1A is not “minus four degrees” or “one a”. This is "pooping man" or the number "twenty-six" in hexadecimal notation. Those people who constantly work in this number system automatically perceive a number and a letter as one graphic symbol.
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! If, of course, you are going to find points on number circle.
We denote the numbers \(2π\), \(π\), \(\frac(π)(2)\), \(-\frac(π)(2)\), \(\frac(3π)(2 )\)
As you know from last 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 in last time, but in a negative direction.
Let's put \(-π\). For this let's go the 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 we need from \(0\) to positive side walk 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 \(\frac(π)(4)\), \(\frac(π)(3)\), \(\frac(π)(6)\) ,\(\frac(7π)(6 )\), \(-\frac(4π)(3)\), \(\frac(7π)(4)\)
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.
Let's now 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 \(0\) you 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).
If you place the unit number circle on coordinate plane, then coordinates can be found 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 on it unit circle you can find the coordinates corresponding to these points on 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? For this they build 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 equal to half 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)
Video lesson “Definition of sine and cosine on the unit circle” presents visual material for a lesson on a related 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 it is used this interpretation concepts. 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 correct 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 shown 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 table 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. The screen displays a figure 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 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 these equalities work on the coordinate plane. 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 on the number circle has the xOy system has its own coordinates, you can make 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.