DimaMazin

What is the velocity of a photon in a mirror tube when the diameter of this tube is less than the wavelength of the photon? Does it reduce or increase velocity of the photon due to wave strikes?

I have got such complex equation a= ((31/2 - 21/2)cos(a)+(2*21/2 - 3)sin(a)+(3*21/2 -2*61/2)/2))*Pi / ((4*21/2 - 6)sin(a) - 2*61/2+3*21/2)= ={4Pi - 4cos(a) + 4sin(a)cos(a) - 4Pi*sin2(a) - [(4cos(a) - 4Pi - 4sin(a)cos(a)+4Pi*sin2(a))2 - (16 - 16sin2(a))*(Pi2 - 2Pi*cos(a)+4sin(a)cos2(a) - Pi2sin2(a)+2Pi*sin(a)cos(a))]1/2} / {8-8sin2(a)} Theoretically it can be solved, then sin(a) , cos(a) and a will be known. Then we can not only exactly define trigonometric functions but and write them down as numbers. Ofcourse we can describe trigonometric functions which relate to Pi by nambers with exponents and mathematical actions only relative to Pi. But it maybe usefule for solve of Travelling salesman problem. 😛

Can wave motion be caused by quantum entanglement? Why entangled particles on distance cannot wave each other?

What force should be bigger: necessary force for change of polarization of red photon or necessary force for change of polarization of blue photon?

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.

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.

What will be when ( Pi/infinity)*infinity ?

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.

I wrongly was sure. We can do equation for check of your arc with using calculator. Do you want to check every arc?

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.

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)

sin(t)=s[-2y(sin(a)-y)+or- ((sin(a)-y)2+s2 - 4y2s2)1/2] / [2(sin(a) - y)2+s2 ]

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

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 .

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)]

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)]

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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.

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.

Division by equidistant points works too because it is proportional(similar) division of the arc and its chord.

Then divide them for 10001000 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.

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 (x1 ; y1) x1 = a*sin(a/2)/2+cos(a/2) y1 = sin2(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(x2 ; y2) x2 =( 3 - cos(a))/2 y2 = (a - sin(a))/2 If you draw straight line through points (x2 ; y2) 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 k1 for one of the lines and k2 for another of the lines and create equation of the arc of definition: k1 = k2

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))

Therefore shape of circular arc is angle of its segment.

Let's consider arc 2Pi and arc close to zero. Do you think circle and straight line have similar shapes?