DimaMazin

Unit circle has r=1. a is angle in radians. You have confused in your inattentiveness . (61/2 - 21/2)/4 = (2-31/2)1/2/2 21/2(31/2-1)/2=(2-31/2)1/2 (31/2-1)2/2=2-31/2 3-2*31/2+1=4-2*31/2

For example one circular sector of unit circle has area a/2, other circular sector has area Pi/4 - a/2 then a/2=sin(a)*cosin(a)/2 +(area of circular segment of angle 2a)/2 and Pi/4 - a/2 = sin(a)*cosin(a)/2+(area of circular segment of angle Pi - 2a)/2 Then area of the twq equal triangles is sin*cosin

(61/2 - 21/2)/4 = (2 - 31/2)1/2/2 I have taken sin * cos=1/4 then I have got sine=(2 - 31/2)1/2/2 or (2 + 31/2)1/2/2 Calculator has shown arcsin=15 degrees but sine 15 degrees=(61/2 - 21/2)/4 I don't understand how they are connected.

Rather that is wrong because t is not obligated to be proportional to ( Pi-2a)/2 at level 1.

The idea is wrong because next formula must work but it is not working t/(1-y)=sin t/(cos t -y) y=(t cost-sint)/(t-sint)

For example two photons with the same fraquency are traveling together. They have horizontal polarizations. Then fraquency is time to do left amlitude or right amlitude. When one photon makes left amplitude then another photon makes right amlitude for support of momentum law.

How do you define what photon cannot travel in thin waveguide becouse it is thin for the photon?Why do you think that source frame is more important for such definition? Why polarity of photon wave cannot be rotating? If it can be rotating then what is energy of the rotation?

The acceleration increases negative velocity of the waveguide relative to source of the photon. Negative velocity creates redshift of the wavelength relative to the waveguide.

The acceleration of the waveguide can turn the blue photon into red photon relative to the waveguide only when it does not affect the photon.

If cutoff exists for red photon, but blue photon can be there. Then acceleration of the very long waveguide from the source of the blue photon can turn the blue photon into red photon . Then how the photon can disappear in the waveguide?

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)]