DimaMazin 93 Posted March 14, 2020 Author Share Posted March 14, 2020 27 minutes ago, studiot said: @DimaMazin I am still waiting for an answer. Shape of arc is feature of arc, but when we know radius we can use length as shape. 1 rad is complex angle because it has complex ratio to Pi. My formulas show some problems with ratio to Pi .And I don't know will the method solve them or not. If even the arc of definition doesn't define anything, I don't understand why it should be unknown? Link to post Share on other sites

studiot 2222 Posted March 14, 2020 Share Posted March 14, 2020 27 minutes ago, DimaMazin said: Shape of arc is feature of arc, but when we know radius we can use length as shape. 1 rad is complex angle because it has complex ratio to Pi. My formulas show some problems with ratio to Pi .And I don't know will the method solve them or not. That did not answer my question. Circular arcs have only one shape, by definition of circular. You still have not clarified what this thread is all about. Please do not use the word complex in mathematical discussion, unless you are actually referring to complex numbers. Use the word complicated perhaps. Link to post Share on other sites

Strange 4273 Posted March 14, 2020 Share Posted March 14, 2020 31 minutes ago, DimaMazin said: 1 rad is complex angle because it has complex ratio to Pi. I'm not sure what you mean by "complex". Obviously not the mathematical sense, as there are no imaginary numbers involved. And not "complicated" because it is just a simple ratio. 33 minutes ago, DimaMazin said: My formulas show some problems with ratio to Pi . I assume that is because they are wrong. 34 minutes ago, DimaMazin said: If even the arc of definition doesn't define anything, I don't understand why it should be unknown? You still have not explained what "arc of definition" means. It is an apparently meaningless phrase, even when translated into your language. Pi is well defined. The trigonometric functions are well defined. The only thing that seems to be undefined is the purpose of this thread. Link to post Share on other sites

Ghideon 465 Posted March 14, 2020 Share Posted March 14, 2020 1 hour ago, taeto said: "non repeating". Thanks, that's a better expression. Link to post Share on other sites

studiot 2222 Posted March 14, 2020 Share Posted March 14, 2020 1 hour ago, Strange said: I'm not sure what you mean by "complex". Obviously not the mathematical sense, as there are no imaginary numbers involved. And not "complicated" because it is just a simple ratio. In Mathematics, the term 'a simplex' is used to refer to (mathematical) objects that have only one part. This is distinguished from 'a complex' that has more than one part, for example complex numbers. To enjoy the benefit of a ratio one need to be working with a set structure that allows division or fractions. Link to post Share on other sites

DimaMazin 93 Posted March 15, 2020 Author Share Posted March 15, 2020 9 hours ago, studiot said: Circular arcs have only one shape, by definition of circular. Let's consider arc 2Pi and arc close to zero. Do you think circle and straight line have similar shapes? Link to post Share on other sites

studiot 2222 Posted March 15, 2020 Share Posted March 15, 2020 4 hours ago, DimaMazin said: Let's consider arc 2Pi and arc close to zero. Do you think circle and straight line have similar shapes? No. I think perhaps there really is a language difficulty here. I don't know how they teach geometry in Russia, but 'similar' is another technical word in Mathematics that has a special meaning, This meaning relates particularly to shape. (basic) geometry distinguishes 'lines' (as straight lines) and curves which are ''line' that are not straight. An arc is part of (a segment of) a curve; a line segment is part of a line. Link to post Share on other sites

taeto 93 Posted March 15, 2020 Share Posted March 15, 2020 5 hours ago, studiot said: I think perhaps there really is a language difficulty here. An arc is part of (a segment of) a curve; a line segment is part of a line. Dima has been courteous enough to explain some of his use of language. He said earlier that he does consider an arc to be a (connected) piece of the unit circle, and that he identifies the essential property of an arc to be its length. He considers arcs "modulo rotation" so to speak. Thus the use of language such as "the arc \(\pi\)" is unambiguous. We know by definition that the entire unit circle is the arc \(2\pi\). Therefore it makes no difference to speak of an arc which is \(1/n\)'th of the whole circle, and an arc of length \(\frac{2\pi}{n},\) for any \(n > 0.\) What gets a bit murky is how then it can make a difference how to "define an arc", when it is just the same as a real number between \(0\) and \(2\pi?\) Is the question really about how to define the various trigonometric functions? What we would answer then is that if \(\alpha \) is a real number in this range, then \(\cos \alpha\) is the \(x\)-coordinate of the left endpoint of the arc \(\alpha\) when it is positioned so that its right endpoint is \( (1,0). \) And so on, to define the functions \(\sin,\) \(\tan,\) \(\cot,\) etc. Or is it about something else entirely? Link to post Share on other sites

DimaMazin 93 Posted March 16, 2020 Author Share Posted March 16, 2020 19 hours ago, studiot said: No. I think perhaps there really is a language difficulty here. I don't know how they teach geometry in Russia, but 'similar' is another technical word in Mathematics that has a special meaning, This meaning relates particularly to shape. (basic) geometry distinguishes 'lines' (as straight lines) and curves which are ''line' that are not straight. An arc is part of (a segment of) a curve; a line segment is part of a line. Therefore shape of circular arc is angle of its segment. Link to post Share on other sites

taeto 93 Posted March 16, 2020 Share Posted March 16, 2020 5 hours ago, DimaMazin said: Therefore shape of circular arc is angle of its segment. It answers one question in the original post, is that not right? A little more formally one would state that an angle is a congruence class (instead of "shape") of arcs of the unit circle. Does the original question about definitions of trigonometric functions still need an answer? Link to post Share on other sites

DimaMazin 93 Posted March 17, 2020 Author Share Posted March 17, 2020 On 3/14/2020 at 9:42 PM, Strange said: I assume that is because they are wrong. Yes. There: x = cos(a/2)*(a*cos(a/2) - 2sin(a/2)) / (a - 2sin(a/2)) y = sin(a/2)*(a*cos(a/2) - 2sin(a/2)) / (a - 2sin(a/2)) Link to post Share on other sites

taeto 93 Posted March 17, 2020 Share Posted March 17, 2020 11 hours ago, DimaMazin said: Yes. There: x = cos(a/2)*(a*cos(a/2) - 2sin(a/2)) / (a - 2sin(a/2)) y = sin(a/2)*(a*cos(a/2) - 2sin(a/2)) / (a - 2sin(a/2)) Then for every \(a \neq 0\) you can calculate an \(x\) and a \(y.\) Is there anything wrong with that? Link to post Share on other sites

DimaMazin 93 Posted March 18, 2020 Author Share Posted March 18, 2020 (edited) 13 hours ago, taeto said: Then for every a≠0 you can calculate an x and a y. Is there anything wrong with that? Correct. Divide arc for infinite quantity of equal parts and divide its chord for infinite quantity of equal parts. Draw straight line through nearest point of division of the arc to middle of the arc and through nearest point of division of the chord to middle of the chord, Draw straight line though middle point of the arc and middle point of the chord. Point of cross of the lines has coordinates (x ; y}. Instead of first line you can draw next line: draw tangent in point (cos(a/2) ; sin(a/2)), mark length a/2 from point (cos(a/2) ; sin(a/2)) on the tangent and draw straight line through marked point and point (1: 0). Point of cross of new line and second line has coordinates (x ; y). Marked point has coordinates (x_{1} ; y_{1}) x_{1} = a*sin(a/2)/2+cos(a/2) y_{1} = sin^{2}(a/2) - a*sin(a/2)*cos(a/2)/2 Yet we should draw straight line through nearest point of division the arc to edge and nearest point of division of the chord to edge. Then add half of the chord to point (1 ; 0) on straight line of the chord . On edge of the half of chord draw perpendicular to axis OX and mark length a/2 by point with coordinates(x_{2} ; y_{2}) x_{2} =( 3 - cos(a))/2 y_{2} = (a - sin(a))/2 If you draw straight line through points (x_{2} ; y_{2}) and (1 ; 0) then the line crosses so called second line in point (x ; y). If the arc is the arc of definition then coordinates (x ; y) are the same in both cases.But it is simpler to use equations of the straight lines. Both the lines have point(1;0) therefore in equation y=kx+b b = - k We define k_{1} for one of the lines and k_{2} for another of the lines and create equation of the arc of definition: k_{1} = k_{2} Edited March 18, 2020 by DimaMazin -1 Link to post Share on other sites

Ghideon 465 Posted March 18, 2020 Share Posted March 18, 2020 46 minutes ago, DimaMazin said: Divide arc for infinite quantity of equal parts and divide its chord for infinite quantity of equal parts. ... It may be easier and more efficient if you also provide a picture? If each interested member have to interpret your description there is plenty of room for confusion? 1 Link to post Share on other sites

taeto 93 Posted March 18, 2020 Share Posted March 18, 2020 2 hours ago, DimaMazin said: Correct. Divide arc for infinite quantity of equal parts and divide its chord for infinite quantity of equal parts. Draw straight line through nearest point of division of the arc That is impossible, since there is no nearest point. It would be like asking for the smallest positive number. Link to post Share on other sites

studiot 2222 Posted March 18, 2020 Share Posted March 18, 2020 3 hours ago, Ghideon said: It may be easier and more efficient if you also provide a picture? If each interested member have to interpret your description there is plenty of room for confusion? +1 1 hour ago, taeto said: That is impossible, since there is no nearest point. It would be like asking for the smallest positive number. We're back to this again. It never was properly resolved. On 3/2/2020 at 11:04 PM, studiot said: Here is my attempt to follow your instructions. I have used 168 degrees since this is actually divisible by 6. The arc looks like semicircle, but is actually 168^{o}. Is joining the points as instructed supposed to create straight lines that all meet at one point? How does this help define a sine ? And more particularly the sine of what angle? It is clear you have several members interested in a technical discussion about this so it is in your own interests to engage as fully as you can. Link to post Share on other sites

Strange 4273 Posted March 18, 2020 Share Posted March 18, 2020 5 hours ago, DimaMazin said: Draw straight line through nearest point of division of the arc to middle of the arc and through nearest point of division of the chord to middle of the chord, Draw straight line though middle point of the arc and middle point of the chord. Go on then. Show us the diagram. Link to post Share on other sites

DimaMazin 93 Posted March 18, 2020 Author Share Posted March 18, 2020 2 hours ago, taeto said: That is impossible, since there is no nearest point. It would be like asking for the smallest positive number. Then divide them for 1000^{1000} equal parts. We don't need see it because we can use similar increasing of cathetuses of triangles (they increased are a/2 and chord/2) . I told about them. Link to post Share on other sites

taeto 93 Posted March 18, 2020 Share Posted March 18, 2020 Thanks Dima, that is helpful; Спасибо! I stand to be corrected. But is the question anything like this: There is a special measure \(\alpha\) of a circular arc, with \(\alpha < \pi\), for which if you divide the arc of measure \(\alpha\) into \(N\) equidistant points, and you divide the chord between the two ends of the arc into \(N\) equidistant points, for any \(N > 0,\) and the line \(\ell_n\) is defined as the line that passes through the \(n\)'th point on the arc and the \(n\)'th point on the chord, for each \(n=1,2,\ldots,N-1,\) then the lines \(\ell_n\) all intersect in a common point? Maybe I just misunderstand the figure that got posted by studiot a couple of times now. Link to post Share on other sites

DimaMazin 93 Posted March 19, 2020 Author Share Posted March 19, 2020 12 hours ago, taeto said: Thanks Dima, that is helpful; Спасибо! I stand to be corrected. But is the question anything like this: There is a special measure α of a circular arc, with α<π , for which if you divide the arc of measure α into N equidistant points, and you divide the chord between the two ends of the arc into N equidistant points, for any N>0, and the line ℓn is defined as the line that passes through the n 'th point on the arc and the n 'th point on the chord, for each n=1,2,…,N−1, then the lines ℓn all intersect in a common point? Maybe I just misunderstand the figure that got posted by studiot a couple of times now. Division by equidistant points works too because it is proportional(similar) division of the arc and its chord. Link to post Share on other sites

DimaMazin 93 Posted March 19, 2020 Author Share Posted March 19, 2020 If points of division are dividing the arc and its chord so: left part of the arc / left part of its chord = right part of the arc / right part of its chord then their straight line crosses point of definition. Link to post Share on other sites

Strange 4273 Posted March 19, 2020 Share Posted March 19, 2020 3 hours ago, DimaMazin said: If points of division are dividing the arc and its chord so: left part of the arc / left part of its chord = right part of the arc / right part of its chord then their straight line crosses point of definition. SHOW US THE DIAGRAM Link to post Share on other sites

taeto 93 Posted March 19, 2020 Share Posted March 19, 2020 (edited) If my interpretation is correct, then it is not possible that all the lines will intersect in a single point. If they did, the common intersection must lie on the line \(L\) defined by the midpoint \(m\) of the arc and the midpoint of its chord, which is the symmetry axis of the whole setup. If you compute how far down along \(L\) you find the intersection with another line \(\ell\), then you get an expression that contains nonzero terms with \(\cos t\) and \(\sin t,\) where \(t\) is the distance along the arc from \(m\) to the point on the arc that defines \(\ell.\) As well as it contains other rational nonzero terms that contain terms like \(t\) in their denominator. It is not difficult to see that if you differentiate w.r.t. \(t,\) then the derivative still contains such terms. There is no way that they may cancel each other out, hence the derivative cannot vanish. This means that the intersection points move around on \(L\) as \(t\) varies, in particular they are different for different pairs of lines. For the argument to be completely convincing, it would be better to calculate the precise expressions. But they get pretty complicated. Edited March 19, 2020 by taeto Link to post Share on other sites

Phi for All 5935 Posted March 19, 2020 Share Posted March 19, 2020 On 3/18/2020 at 1:13 AM, DimaMazin said: Draw straight line through nearest point of division of the arc to middle of the arc and through nearest point of division of the chord to middle of the chord, Draw straight line though middle point of the arc and middle point of the chord. Point of cross of the lines has coordinates (x ; y}. Instead of first line you can draw next line: draw tangent in point (cos(a/2) ; sin(a/2)), mark length a/2 from point (cos(a/2) ; sin(a/2)) on the tangent and draw straight line through marked point and point (1: 0). Point of cross of new line and second line has coordinates (x ; y). Marked point has coordinates (x_{1} ; y_{1}) ! Moderator Note A picture is worth a thousand words in any language. Please draw this and post it. It should help the discussion immensely. Link to post Share on other sites

studiot 2222 Posted March 19, 2020 Share Posted March 19, 2020 (edited) On 3/18/2020 at 2:18 PM, taeto said: Maybe I just misunderstand the figure that got posted by studiot a couple of times now. I am sorry if my diagram confused you. All I did was cut and paste the drawing instructions given by Dima and attempt to use them to draw a diagram. That diagram was the result. I then asked if my diagram correctly showed his intentions but was not answered. Edited March 19, 2020 by studiot Link to post Share on other sites

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