taeto 93 Posted March 19, 2020 Share Posted March 19, 2020 Your diagram shows a sensible interpretation of Dima's construction. But it does not seem to say anywhere, nor is it made clear from the diagram, whether the lines are supposed to intersect in a common point. That part may have been something that I inadvertently invented. Link to post Share on other sites

studiot 2222 Posted March 19, 2020 Share Posted March 19, 2020 30 minutes ago, taeto said: Your diagram shows a sensible interpretation of Dima's construction. But it does not seem to say anywhere, nor is it made clear from the diagram, whether the lines are supposed to intersect in a common point. That part may have been something that I inadvertently invented. No I don't think they do all intersect at a common point. Normals certainly cannot. That is why I asked for proof that they were supposed to. Link to post Share on other sites

taeto 93 Posted March 19, 2020 Share Posted March 19, 2020 (edited) Then let us say that the arc has length \(\pi - 2\alpha\) and is placed symmetrically around the \(y\)-axis as in this diagram. Then the arc itself is the segment of the circle between the points \((\cos \alpha,\sin \alpha)\) and \((-\cos \alpha,\sin \alpha)\) in the upper half of \(\mathbb{R}^2.\) And its chord is the horizontal line segment connecting its endpoints as shown. If we take a random point \(p=(\sin t,\cos t)\) in the upper right quadrant, i.e. the point for which its distance to the \(y\)-axis along the arc is \(t,\) with \(0 < t < \pi/2,\) then the line defined by this point is the line \(L\) through \(p\) and the point \(q=(s,\sin \alpha)\) on the chord, such that the ratio \(s/\cos \alpha\) of the distance from the \(y\)-axis to \(q\) equals the ratio \(t/(\pi/2-\alpha)\) of the distance along the arc between the \(y\)-axis and \(p\) itself to the entire piece of arc in the upper right quadrant. So \(s = \frac{\cos \alpha}{\pi/2-\alpha}t\) follows. The equation for \(L\) becomes \(L : y = \frac{\cos t - \sin \alpha}{\sin t -s}(x - s) + \sin \alpha,\) which is checked by setting \(x = s,\sin t.\) Clearly the \(y\)-intercept of \(L\) is the point \((0,-\frac{\cos t - \sin \alpha}{\sin t - s}s+\sin \alpha)\) by setting \(x=0\) in the equation for \(L.\) If this is correct, then the question seems to be whether there exists a value of \(\alpha\) with \(0 < \alpha < \pi/2\) for which this intersection point does not depend on \(t.\) Edited March 19, 2020 by taeto 1 Link to post Share on other sites

DimaMazin 93 Posted March 21, 2020 Author Share Posted March 21, 2020 On 3/19/2020 at 8:54 PM, taeto said: Then let us say that the arc has length π−2α and is placed symmetrically around the y -axis as in this diagram. Then the arc itself is the segment of the circle between the points (cosα,sinα) and (−cosα,sinα) in the upper half of R2. And its chord is the horizontal line segment connecting its endpoints as shown. If we take a random point p=(sint,cost) in the upper right quadrant, i.e. the point for which its distance to the y -axis along the arc is t, with 0<t<π/2, then the line defined by this point is the line L through p and the point q=(s,sinα) on the chord, such that the ratio s/cosα of the distance from the y -axis to q equals the ratio t/(π/2−α) of the distance along the arc between the y -axis and p itself to the entire piece of arc in the upper right quadrant. So s=cosαπ/2−αt follows. The equation for L becomes L:y=cost−sinαsint−s(x−s)+sinα, which is checked by setting x=s,sint. Clearly the y -intercept of L is the point (0,−cost−sinαsint−ss+sinα) by setting x=0 in the equation for L. If this is correct, then the question seems to be whether there exists a value of α with 0<α<π/2 for which this intersection point does not depend on t. Great diagram. Thanks Taeto. You are the best companion. Now I can see the arc can be defined if it exists. But the solve may be very complex . Therefore it will take a lot of time. On 3/19/2020 at 6:21 PM, studiot said: No I don't think they do all intersect at a common point. Normals certainly cannot. That is why I asked for proof that they were supposed to. There is no prize for such proof . Therefore we should firstly define the arc then maybe some prize will appear. Because it is simpler to define the arc than to prove it exists. Link to post Share on other sites

taeto 93 Posted March 21, 2020 Share Posted March 21, 2020 (edited) Dear това́рищ Dima, thank you for your praises! Unfortunately I suspect that such a special arc cannot exist. Referring to my diagram in the preceding post, I have tried to look at the case of the two lines defined by \(t = \pi/6\) and \(t=\pi/3.\) I calculate that they intersect on the \(y\)-axis only if \[\frac{\alpha}{\pi/2} = 1 - \frac{4(\sqrt{3} -1)\cos \alpha}{6(2-\sqrt{3})\sin \alpha +3}.\] (If you prefer to calculate in degrees instead of radians, the only change is to replace \(\pi/2\) by \(90.\) ) But when I plug some values of \(\alpha\) into this, it always seems that the expression on the right side is a larger number than the left side value. Edited March 21, 2020 by taeto Link to post Share on other sites

DimaMazin 93 Posted March 27, 2020 Author Share Posted March 27, 2020 (edited) On 3/21/2020 at 8:04 PM, taeto said: Dear това́рищ Dima, thank you for your praises! Unfortunately I suspect that such a special arc cannot exist. Referring to my diagram in the preceding post, I have tried to look at the case of the two lines defined by t=π/6 and t=π/3. I calculate that they intersect on the y -axis only if απ/2=1−4(3–√−1)cosα6(2−3–√)sinα+3. (If you prefer to calculate in degrees instead of radians, the only change is to replace π/2 by 90. ) But when I plug some values of α into this, it always seems that the expression on the right side is a larger number than the left side value. . Edited March 27, 2020 by DimaMazin IIf x is point with coordinates(s, sint) then what is (x-s)) Link to post Share on other sites

DimaMazin 93 Posted March 28, 2020 Author Share Posted March 28, 2020 Point of definition has coordinates (0 ; y) where y =[ 2t*cos(a)*cos(t) - Pi*sin(t)*sin(a)+2a*sin(t)*sin(a)] / [2t*cos(a) - Pi*sin(t)+2a*sin(t)] Link to post Share on other sites

taeto 93 Posted March 28, 2020 Share Posted March 28, 2020 12 hours ago, DimaMazin said: y =[ 2t*cos(a)*cos(t) - Pi*sin(t)*sin(a)+2a*sin(t)*sin(a)] / [2t*cos(a) - Pi*sin(t)+2a*sin(t)] It seems we agree on this, or at least something very close to it. To simplify more, it could be useful to introduce (in your notation) b = Pi/2 - a. Either way this is a function of t which is not constant. As t grows, the point (0,y) moves upwards on the y-axis. Link to post Share on other sites

DimaMazin 93 Posted April 3, 2020 Author Share Posted April 3, 2020 On 3/28/2020 at 7:08 PM, taeto said: It seems we agree on this, or at least something very close to it. To simplify more, it could be useful to introduce (in your notation) b = Pi/2 - a. Either way this is a function of t which is not constant. As t grows, the point (0,y) moves upwards on the y-axis. No. (0,y) is constant. And we can define everything because: a = [2Pi(t+sin(t))*sin(a)*cos(a) - 4t(1+cos(t))*cos^{2}(a)+2Pi(sin(t) - t*cos(t))*cos(a) - 2Pi^{2}*sin(t)*sin(a)] / [4(t+sin(t))*sin(a)*cos(a)+4(sin(t) - t*cos(t))*cos(a) - 4Pi*sin(t)*sin(a)] Link to post Share on other sites

taeto 93 Posted April 3, 2020 Share Posted April 3, 2020 26 minutes ago, DimaMazin said: No. (0,y) is constant. And we can define everything because: a = [2Pi(t+sin(t))*sin(a)*cos(a) - 4t(1+cos(t))*cos^{2}(a)+2Pi(sin(t) - t*cos(t))*cos(a) - 2Pi^{2}*sin(t)*sin(a)] / [4(t+sin(t))*sin(a)*cos(a)+4(sin(t) - t*cos(t))*cos(a) - 4Pi*sin(t)*sin(a)] If this answers your question, then fine. All I was saying is that if you keep your \(a\) constant, then the point \((0,y)\) depends on the value of \(t.\) Link to post Share on other sites

DimaMazin 93 Posted April 3, 2020 Author Share Posted April 3, 2020 3 minutes ago, taeto said: If this answers your question, then fine. All I was saying is that if you keep your a constant, then the point (0,y) depends on the value of t. Changing t does not change y and a . We can exchange t by Pi/6 and define a . And we can exchange t by Pi/3 and define the same a .Then we can exchange a by new value and get long equation without a . There are only unknown sin(a) and cos(a), therefore they can be defined . Link to post Share on other sites

taeto 93 Posted April 3, 2020 Share Posted April 3, 2020 14 minutes ago, DimaMazin said: Changing t does not change y and a . We can exchange t by Pi/6 and define a . And we can exchange t by Pi/3 and define the same a .Then we can exchange a by new value and get long equation without a . There are only unknown sin(a) and cos(a), therefore they can be defined . I do not understand what you mean. If you set \(t = \pi /6\) in your equation, then we can (if we are lucky) solve the equation and find a solution \(a = a(\pi/6).\) If we set \(t = \pi/3\) we may also find a solution \(a = a(\pi/3)\) to the same equation. It is extremely unlikely that we will discover that \(a(\pi/6) = a(\pi/3)\) holds. I worry when you say that \(\sin(a)\) and \(\cos(a)\) are "unknown" and "can be defined". Once we have \(a,\) then normally they should be completely known and determined by calculation of the function values of \(\sin\) and \(\cos\) with argument \(a.\) Link to post Share on other sites

DimaMazin 93 Posted April 3, 2020 Author Share Posted April 3, 2020 5 minutes ago, taeto said: I do not understand what you mean. If you set t=π/6 in your equation, then we can (if we are lucky) solve the equation and find a solution a=a(π/6). If we set t=π/3 we may also find a solution a=a(π/3) to the same equation. It is extremely unlikely that we will discover that a(π/6)=a(π/3) holds. I worry when you say that sin(a) and cos(a) are "unknown" and "can be defined". Once we have a, then normally they should be completely known and determined by calculation of the function values of sin and cos with argument a. That is wrong because changing t changes sin(t) and cos(t) . There is one equation for sin(a) and cos(a). Second equation we have sin(a) = (1 - cos^{2}(a))^{1/2} Link to post Share on other sites

taeto 93 Posted April 3, 2020 Share Posted April 3, 2020 (edited) 48 minutes ago, DimaMazin said: That is wrong because changing t changes sin(t) and cos(t) . There is one equation for sin(a) and cos(a). Second equation we have sin(a) = (1 - cos^{2}(a))^{1/2} I do not understand any of that. But do you think that what you have now is useful to you, or have you got any additional questions? 48 minutes ago, DimaMazin said: That is wrong because changing t changes sin(t) and cos(t) . Just to make sure. Once you decide on assigning some value to \(t,\) like \(\pi/6, \pi/4, \pi/3\) or any other value, then we can calculate \(\sin t\) and \(\cos t\) just in the same way that we are used to? Edited April 3, 2020 by taeto Link to post Share on other sites

DimaMazin 93 Posted April 7, 2020 Author Share Posted April 7, 2020 On 4/3/2020 at 5:29 PM, taeto said: I do not understand any of that. But do you think that what you have now is useful to you, or have you got any additional questions? Just to make sure. Once you decide on assigning some value to t, like π/6,π/4,π/3 or any other value, then we can calculate sint and cost just in the same way that we are used to? sin(t)=s[-2y(sin(a)-y)+or- ((sin(a)-y)^{2}+s^{2} - 4y^{2}s^{2})^{1/2}] / [2(sin(a) - y)^{2}+s^{2} ] Link to post Share on other sites

taeto 93 Posted April 7, 2020 Share Posted April 7, 2020 2 hours ago, DimaMazin said: sin(t)=s[-2y(sin(a)-y)+or- ((sin(a)-y)^{2}+s^{2} - 4y^{2}s^{2})^{1/2}] / [2(sin(a) - y)^{2}+s^{2} ] Assuming you worked it out correctly, this might be another version of the same identity. I have not checked it. I suppose that you have no additional questions? If we let \(t = \pi/4\) and \(a = \pi/6,\) then \(\sin t = \frac{\sqrt{2}}{2}\) and \(\sin a = \frac{1}{2}.\) Can you solve to find \(y\) for that case? Link to post Share on other sites

DimaMazin 93 Posted April 7, 2020 Author Share Posted April 7, 2020 7 hours ago, taeto said: tsina=12. Can you solve to find y for that case? a is constant . It exactly cannot be equal Pi/6. sin a is constant. You can use t_{1 } or t_{2} instead of t . t , cos t and sin t are variables . And y of point of definition is constant. I don't understand why you don't understand your diagram. When you take t_{1} and t_{2} you get L_{1} and L_{2} . Therefore you can do equation for definition of sin( a ) and cos(a) (using my equation for "a" definition) Link to post Share on other sites

taeto 93 Posted April 7, 2020 Share Posted April 7, 2020 27 minutes ago, DimaMazin said: a is constant . It exactly cannot be equal Pi/6. sin a is constant. You can use t_{1 } or t_{2} instead of t . t , cos t and sin t are variables . And y of point of definition is constant. I don't understand why you don't understand your diagram. When you take t_{1} and t_{2} you get L_{1} and L_{2} . Therefore you can do equation for definition of sin( a ) and cos(a) (using my equation for "a" definition) I understand my own diagram in the way that it was constructed: there is an angle \(t\) that can be chosen freely, and there is an angle \(\alpha,\) or \(a\) as you call it now, which can also be chosen freely, except of course their sum \(t+\alpha\) must be less than \(\pi/2.\) There is nothing there about \(t\) or \(a\) being constant. Then the diagram contains an expression for the resulting value of the \(y\)-intercept of interest. So if you now say that your \(a\) is a constant, and you also say that it is not \(\pi/6\) (which is a constant), then how do you know that \(a\) is not \(\pi/6?\) Is it because you already know what \(a\) is? Can you explain how you know this, and what you think is the actual value of \(a,\) as a fraction of \(\pi\) or in decimal expansion? Link to post Share on other sites

DimaMazin 93 Posted April 7, 2020 Author Share Posted April 7, 2020 18 minutes ago, taeto said: So if you now say that your a is a constant, and you also say that it is not π/6 (which is a constant), then how do you know that a is not π/6? Is it because you already know what a is? Can you explain how you know this, and what you think is the actual value of a, as a fraction of π or in decimal expansion? I don't want to repeat my explain about approximate exploring of value of Pi - 2a. We have enough equations for exact definition of a . We should recheck them and define first sin(a) and cos(a) then a and y of point of definition.It will be very complex. Link to post Share on other sites

taeto 93 Posted April 8, 2020 Share Posted April 8, 2020 Why did you tell me that you know that \(a\) is not equal to \(\pi/6?\) I have to understand this first. Link to post Share on other sites

DimaMazin 93 Posted April 9, 2020 Author Share Posted April 9, 2020 On 4/8/2020 at 3:06 AM, taeto said: Why did you tell me that you know that a is not equal to π/6? I have to understand this first. I wrongly was sure. We can do equation for check of your arc with using calculator. Do you want to check every arc? Link to post Share on other sites

taeto 93 Posted April 9, 2020 Share Posted April 9, 2020 Dima, your guess is that there is some value of \(a\) for which the value of \(y\) does not change if we look at all possible value of \(t\), is that right? My guess is that this is not true. Meaning that for any value of \(a\) that we choose, there will be at least two different values of \(t\) that produce different values of \(y\). Do you agree that my description of our guesses is correct? Now, since you are the one who says that there exists a specific value of \(a\) which wins the argument for your case, do you not think that you should have to present this particular value of \(a\)? Are you waiting for someone who is smarter than you to do so? Link to post Share on other sites

DimaMazin 93 Posted April 10, 2020 Author Share Posted April 10, 2020 10 hours ago, taeto said: Dima, your guess is that there is some value of a for which the value of y does not change if we look at all possible value of t , is that right? My guess is that this is not true. Meaning that for any value of a that we choose, there will be at least two different values of t that produce different values of y . Do you agree that my description of our guesses is correct? Now, since you are the one who says that there exists a specific value of a which wins the argument for your case, do you not think that you should have to present this particular value of a ? Are you waiting for someone who is smarter than you to do so? Yes. Let's consider next thing: Arc Pi -2 a is divided for infinite quantity equal parts and its chord is divided for infinite quantity of equal parts.The straight line intersects the nearest points of the divisions to its middle . The point on arc has coordinates([Pi-2a]/infinity ; 1), the point on chord has coordinates (cos(a)/infinity ; sin(a)). Their straight line intersects y axis in point(0; y_{1}) . Increased side (Pi - 2a)/infinity of the triangle in infinite quantity of times on tangent in point(0 ; 1) creates new point on its edge with coordinates ([Pi - 2a] ; 1). Increased side cos(a)/infinity of the smaller triangle in infinite quantity times has edge point(cos(a) ; sin(a)). The new straight line of the new points intersects y axis in the same point (0 ; y_{1}). y_{1} = (2a - Pi*sin^{2}(a)) / (2a*sin(a) - Pi*sin(a)+2a*cos(a)) Some similar thing we can make with edge points of the division. Then their derivative points lie on straight line which intersects y axis in point(0 ; y_{2}) I did not make equation for y_{2}. But feature exists there : y_{1} is lower than y_{2} when arc is bigger than Pi - 2a and y_{1} is upper than y_{2} when arc is smaller than Pi - 2a It can prove that the arc of definition of trigonometric functions exists. Link to post Share on other sites

taeto 93 Posted April 10, 2020 Share Posted April 10, 2020 5 hours ago, DimaMazin said: The point on arc has coordinates([Pi-2a]/infinity ; 1), the point on chord has coordinates (cos(a)/infinity ; sin(a)). What do you mean by dividing by "infinity"? When extending the real numbers by new elements \(\{-\infty, +\infty\},\) and extending the usual arithmetic of real numbers to the new structure, usually you get \(0\) when you divide a real number by \(\infty\). Maybe your "infinity" is not the same as \(\infty\)? Link to post Share on other sites

DimaMazin 93 Posted April 10, 2020 Author Share Posted April 10, 2020 8 minutes ago, taeto said: What do you mean by dividing by "infinity"? When extending the real numbers by new elements {−∞,+∞}, and extending the usual arithmetic of real numbers to the new structure, usually you get 0 when you divide a real number by ∞ . Maybe your "infinity" is not the same as ∞ ? What will be when ( Pi/infinity)*infinity ? Link to post Share on other sites

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