  # DimaMazin

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• ### dimreepr

1. I mistaken there y1=[sin(a)*(Pi - 2a) - 2cos(a)] / [Pi-2cos(a)-2a] y2= [2sin(a)*cos(a)+2a - Pi ] / [2cos(a) - sin(a)*(Pi - 2a)] Now everyone can define is any arc less or more than the arc of definition.
2. Let's consider that stupid rule instead of true science because it will appear again and again . (Pi/million)*million=Pi (Pi/billion)*billion=Pi There is no trend to reduce result therefore so called scientists should prove that billion is not nearer to infinity than million.
3. What will be when ( Pi/infinity)*infinity ?
4. 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; y1) . 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 ; y1). y1 = (2a - Pi*sin2(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 ; y2) I did not make equation for y2. But feature exists there : y1 is lower than y2 when arc is bigger than Pi - 2a and y1 is upper than y2 when arc is smaller than Pi - 2a It can prove that the arc of definition of trigonometric functions exists.
5. I wrongly was sure. We can do equation for check of your arc with using calculator. Do you want to check every arc?
6. 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.
7. a is constant . It exactly cannot be equal Pi/6. sin a is constant. You can use t1 or t2 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 t1 and t2 you get L1 and L2 . Therefore you can do equation for definition of sin( a ) and cos(a) (using my equation for "a" definition)
8. sin(t)=s[-2y(sin(a)-y)+or- ((sin(a)-y)2+s2 - 4y2s2)1/2] / [2(sin(a) - y)2+s2 ]
9. 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 - cos2(a))1/2
10. 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 .
11. No. (0,y) is constant. And we can define everything because: a = [2Pi(t+sin(t))*sin(a)*cos(a) - 4t(1+cos(t))*cos2(a)+2Pi(sin(t) - t*cos(t))*cos(a) - 2Pi2*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)]
12. 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)]
13. 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. 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.
14. 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.
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