Force on charges stacked in evenly spaced series at a single moment in time

 

Net force on point charges stacked in evenly spaced series at a single moment in time

It is at a single moment in time when all of the charges would be an equal distance from the next nearest charge as measured in a single dimension after that point of time some of the charges might move to a new position

In a one dimensional tube uncharged tube of length L going from left to right

The distance between the two farthest point charges from each other is L

N is an integer greater than or equal to two equal to the number of point charges

The total charge of all point charges is N

The charge of each point charge is Q/N

The distance between each point charge and the nearest point charge is D

D = L / ( N -1 )

Fi is the Individual force a point charge on the farthest right experiences from another point charge listed with an index of i where the charge farthest on the left has i = 0 the charge second farthest to the right has i = n - 2

Fi = [ L - i * L / ( n - 1 ) ] ^ 2

F = sum of all Fi from i = 0 to i = n - 2

Incomplete next steps

Look up notation I used in place of standard calculus symbols when letters and numbers and askey symbols are used instead of hand drawings from previous articles

Take limit as n approaches infinity to calculate F for a unlimmitted number of point charges using an integral

Compare that for all other variables having the same value but n equalling two

Do in two and then three dimensions next for the point charge in a single corner

Note :  trig functions might be useable to estimate distance between charge as a function of angle but might give wrong results if used as an integral of force with respect to angle from one angle to another angle to calculate net total force experienced by a charge in a corner of a square or cube, so it is probably better to use indexes i, j and k and do double or triple integral using caartesian coordinates instead of single or double integral using polar or spherical coordinates because the number of point charges in each angular direction maybe different 

Copyright Carl Janssen 2022 Jan 12

Tangent of the average of two angles and other trigonometric identities derived from a combination of it and the pythagorean theoreom

Tangent of the average of two angles and other trigonometric identities derived from a combination of it and the pythagorean theoreom

The tangent converts an angle into a slope

Take two points in polar coordinates of (1, A) and (1, B) on the unit circle

convert those points into caartesian coordinates and get (cosA, sinA) and (cosB, sinB)

A midpoint that is the average of their caartesian coordinates exists in between those two points having coordinates of 

( 0.5*cos(A)+0.5*cos(B) , 0.5*sin(A) + 0.5*sin(B) ) caartesian

It has a radius in polar coordinates of

( [ 0.5*cos(A)+0.5*cos(B)  ] ^ 2+ [ 0.5*sin(A) + 0.5*sin(B) ] ^ 2) ^ 0.5

The midpoints angle in polar coordinates must be the average of the polar coordinates' angles of the two points it is in between in this specific case.  The midpoint of any two polar coordinates with the same radius away from a common origin ( pole 0 ) is located at the average of their angles but maybe with a different radius when expressed in polar coordinates.

This midpoint is located at

( [ 0.5*cos(A)+0.5*cos(B)  ] ^ 2+ [ 0.5*sin(A) + 0.5*sin(B) ] ^ 2) ^ 0.5, (A+B)/2 )    polar 

the slope from

 (0,0) to the midpoint 

(0,0) to (0.5*cos(A)+0.5*cos(B) , 0.5*sin(A) + 0.5*sin(B) ) caartesian

(0,0) to ( [ 0.5*cos(A)+0.5*cos(B)  ] ^ 2+ [ 0.5*sin(A) + 0.5*sin(B) ] ^ 2) ^ 0.5, (A+B)/2 )    polar  

Is the same as the slope from (0,0) to a point on the unit circle with the same angle in polar coordinates as the midpoint

(0,0) to ( 1, (A+B)/2 ) in polar coordinates

N = 1 / ( [ 0.5*cos(A)+0.5*cos(B)  ] ^ 2+ [ 0.5*sin(A) + 0.5*sin(B) ] ^ 2) ^ 0.5

The point on the unit circle (1, (A+B)/2 ) polar 

is the same as

( N*[0.5*cos(A)+0.5*cos(B)],  N*[0.5*sin(A) + 0.5*sin(B)] ) caartesian

Since that point is on the unit circle the following is therefore true

cos([A+B]/2) = N*[0.5*cos(A)+0.5*cos(B)] =

= [0.5*cos(A)+0.5*cos(B)] / ( [ 0.5*cos(A)+0.5*cos(B)  ] ^ 2+ [ 0.5*sin(A) + 0.5*sin(B) ] ^ 2) ^ 0.5

sin([A+B]/2) = N*[0.5*sin(A)+0.5*sin(B)] =

= [0.5*sin(A)+0.5*sin(B)] / ( [ 0.5*cos(A)+0.5*cos(B)  ] ^ 2+ [ 0.5*sin(A) + 0.5*sin(B) ] ^ 2) ^ 0.5

Since the tangent is the slope of an angle the following is therefore true

tan([A+B]/2) = N*[0.5*sin(A) + 0.5*sin(B)] / (N*[0.5*cos(A)+0.5*cos(B)])

tan([A+B]/2) = [0.5*sin(A) + 0.5*sin(B)] / [0.5*cos(A)+0.5*cos(B)]

tan([A+B]/2) = [sin(A) + sin(B)] / [cos(A) + cos(B)]

The following simplifications can be made for half and double angle trigonometric values

sin(0) = 0

cos(0) = 1

tan([A+0]/2) = [sin(A) + sin(0)] / [cos(A)+0.5*cos(0)]

tan(A/2) = sin(A) / [ 1 + cos(A) ]

[tan(A/2)]^2 = [ sin(A) ] ^ 2 / [ 1 + cos(A) ] ) ^2

[sin(theta)]^2 = 1 - [cos(theta)]^2 = [1-cos(theta)]*[1+cos(theta)] 

[ sin(A) ] ^ 2 / [ 1 + cos(A) ] ) ^2 = [1-cos(A)]*[1+cos(A)] / [1+cos(A)]^2

[ sin(A) ] ^2 / [ 1 + cos(A) ] ) ^2 = [1-cos(A)] / [1+cos(A)]

[tan(A/2)]^2 = [1-cos(A)] / [1+cos(A)]

tangent of half angle confirmed 

https://en.m.wikipedia.org/wiki/List_of_trigonometric_identities#Half-angle_formulae

let x = cos(A/2) and solve using quadratic equation

(1 - [cos(A/2)]^2 ) / [cos(A/2)]^2 = [1-cos(A)] / [1+cos(A)]

( 1 - x^2 ) / x^2 = [1-cos(A)] / [1+cos(A)]

1 - x ^2 = ( x^2 ) *  [1-cos(A)] / [1+cos(A)]

0 = ( x ^2 ) + ( x^2 ) * [1-cos(A)] / [1+cos(A)] - 1

0 = ( x ^2 ) * [1 + cos(A) + 1 - cos(A)] / [1+cos(A)]   - 1

0 =  ( x ^2 ) * 2 / [1+cos(A)]   - 1

a = 2 / 1 + cos(A) , b = 0 , c = -1

x= [ - b +- ( b^2 - 4*a*c) ] / ( 2 *a )

x = [- 0 +- ( 0 - 4 * 2 /  [ 1 + cos(A)] * -1) ^ 0.5 ] / 2 * [2 / 1 + cos(A)]

x = +- ( 8 / [1 + cos(A)] ^ 0.5 ) /  ( 4 / [ 1 + cos(A) ] )

cos(A/2) =  +- ( 8 ^ 0.5 ) * ( 1 / [ 1 + cos(A) ] ^ - 0.5 ) / 4

8 ^ 0.5 = 2 * 2 ^ 0.5

4 = 2 * 2 ^ 1

(8^0.5) / 4 = 2 ^ -0.5

cos(A/2) = +- ( [ 1 + cos(A) ] ^ 0.5 ) / ( 2 ^ 0.5)

[ sin(A/2) ] ^ 2 =  1 - [ cos (A/2) ] ^2

[ sin(A/2) ] ^ 2 = 1 - [ 1 + cos(A) ] / 2 = 0.5 - 0.5 * cos(A) = [ 1 - cos(A) ] / 2

sin(A/2) =  +- ( [ 1 - cos(A) ] / 2 ) ^ 0.5

confirmed

https://web.archive.org/web/20210912185657/https://www.intmath.com/analytic-trigonometry/4-half-angle-formulas.php

N = 1 / ( [ 0.5*cos(A)+0.5*cos(B)  ] ^ 2+ [ 0.5*sin(A) + 0.5*sin(B) ] ^ 2) ^ 0.5

when B = 0  then N = 1 / ( [ 0.5*cos(A)+0.5*1  ] ^ 2+ [ 0.5*sin(A) + 0.5*0 ] ^ 2) ^ 0.5

( [ 0.5*cos(A)+0.5*1  ] ^ 2+ [ 0.5*sin(A) + 0.5*0 ] ^ 2) = 0.25 * ( [1 + cos(A) ] ^ 2 + [ sin(A) ] ^2 )

( [1 + cos(A) ] ^ 2 + [ sin(A) ] ^2 ) = [ cos(A) ] ^ 2 + [ sin(A) ] ^ 2 + 2 * [ cos(A) ] + 1

( [1 + cos(A) ] ^ 2 + [ sin(A) ] ^2 ) = 2 + 2 * cos(A)

 ( [ 0.5*cos(A)+0.5*1  ] ^ 2+ [ 0.5*sin(A) + 0.5*0 ] ^ 2) = 0.5 + 0.5 * cos(A) 

cos(A/2) = [ 0.5*cos(A) + 0.5*1 ] / ( [ 0.5*cos(A)+0.5*1  ] ^ 2+ [ 0.5*sin(A) + 0.5*0 ] ^ 2) ^ 0.5  

[ cos(A/2) ] ^ 2 =  ( [ 0.5*cos(A) + 0.5* ] ^ 2 ) / [ 0.5 + 0.5 * cos(A) ]

[ cos(A/2) ] ^ 2 = 0.5 + 0.5 * cos(A) = [ 1 + cos(A) ] / 2

cos(A/2) =  +- ( [ 1 + cos(A) ] ^ 0.5 ) / ( 2 ^ 0.5)

[ cos(A/2) ] ^ 2 = [ 1 + cos(A) ] / 2

[ cos(A) ] ^ 2 = [ 1 + cos( 2 * A ) ] / 2

cos( 2 * A ) = ( 2 * [ cos(A) ] ^ 2 ) - 1 

confirmed

https://web.archive.org/web/20210421123600/https://www.intmath.com/analytic-trigonometry/3-double-angle-formulas.php

cos([A+B]/2) = [ 0.5*cos(A)+0.5*cos(B) ] / ( [ 0.5*cos(A)+0.5*cos(B)  ] ^ 2+ [ 0.5*sin(A) + 0.5*sin(B) ] ^ 2) ^ 0.5

cos(A+B) = ( 2 * [ cos( [A+B] / 2) ] ^ 2 ) - 1

cos(A+B) = [ 2 * ( [ 0.5*cos(A)+0.5*cos(B) ] ^ 2 ) / ( [ 0.5*cos(A)+0.5*cos(B)  ] ^ 2+ [ 0.5*sin(A) + 0.5*sin(B) ] ^ 2) ] - 1

[ tan(A/2) ] ^ 2 = [ ( sin(A) ] ^ 2 / [ 1 + cos(A) ] ) ^2

[ tan(A) ] ^ 2 = [ ( sin( 2*A ) ] ^2 / [ 1 + cos( 2*A) ] ) ^2

[ sin( 2*A ) ] ^ 2 = [ tan(A) ] ^2 * [ 1 + cos( 2*A) ] ) ^2

[ sin( 2*A) ] ^ 2 = [ tan(A) ] ^ 2 * ( 1 + [ ( 2 * [ cos(A) ] ^ 2 ) - 1 ] ) ^2

[ sin( 2*A) ] ^ 2 = 4 * [ tan(A) ] ^ 2 * [cos(A)] ^ 4 

[ sin( 2*A) ] ^ 2=  4 * [ sin (A) ] ^ 2 * [ cos(A) ] ^ 2 

sin( 2 * A) = 2 * [ sin (A) ] * [ cos (A) ]

confirmed

https://web.archive.org/web/20210421123600/https://www.intmath.com/analytic-trigonometry/3-double-angle-formulas.php

sin(A/2) = [ 0.5 * sin(A) + 0.5*0 ] / ( [ 0.5*cos(A)+0.5*1  ] ^ 2+ [ 0.5*sin(A) + 0.5*0 ] ^ 2) ^ 0.5 

[ sin(A/2) ] ^ 2 =  ( [ 0.5 * sin(A) ] ^2 )  /  0.5 + 0.5 * cos(A) 

[ sin(A/2) ] ^ 2 = 0.5 * ( [ sin(A) ] ^2 ) / [ ( 1 + cos((A) ]

[ sin(theta) ] ^ 2 = 1 - [ cos(theta) ] ^ 2 = [ 1 - cos(theta) ] * [ 1 + cos(theta) ]

[ sin(A/2) ] ^ 2 = 0.5 *  [ 1 - cos(A/2) ]

sin(A/2) =  +- ( [ 1 - cos(A) ] / 2 ) ^ 0.5

[ cos(theta) ] ^ 2 = 1 - [ sin(theta) ] ^ 2

1 = ( 1 / [ cos(theta) ] ^ 2 ) - [ tan(theta) ] ^2

1 + [ tan(theta) ] ^2 = [ cos(theta) ] ^ -2

[ cos(theta) ] ^ 2 = 1 / (1 + [ tan(theta) ] ^2)

cos(theta) = +- 1 / (1 + [ tan(theta) ] ^2) ^ 0.5

[ sin(theta) ] ^ 2 = [ tan(theta) ] ^ 2 * [ cos(theta) ] ^2

[ sin(theta) ] ^ 2 = ( [ tan(theta) ] ^ 2 ) / (1 + [ tan(theta) ] ^2)

cos( [A+B] / 2 ) = +- 1 / ( 1 + [ tan( [A+B] / 2 ) ] ^2) ^ 0.5

cos( [A+B] / 2 ) = +- 1 / ( 1 + ( [ sin(A) + sin(B) ] / [ cos(A) + cos(B) ] ) ^2) ^ 0.5

[ sin( [A+B] / 2 ) ] ^ 2 =  [ tan( [A+B] / 2 ) ] * cos( [A+B] / 2 ) 

sin( [A+B] / 2 ) = +- ( [ sin(A) + sin(B) ] / [ cos(A) + cos(B) ] ) / ( 1 + ( [ sin(A) + sin(B) ] / [ cos(A) + cos(B) ] ) ^2) ^ 0.5

Using sin, cos and tan functions to calculate each other

[tan(theta)]^2 = [sin(theta)/cos(theta)]^2

[sin(theta)/cos(theta)]^2 = (1 - [cos(theta)]^2 ) / [cos(theta)]^2

[tan(theta)]^2 = (1 - [cos(theta)]^2 ) / [cos(theta)]^2

[tan(theta)]^2 = [sin(theta)]^2 / (1 - [sin(theta)]^2 )

Nothing in high school or college geometry is used to do this that has not been taught in prerequisite classes for high school or college geometry other than the definition of trigonometric functions the pythagorean theoreom and converting polar to caartesian coordinates or caartesian to polar coordinates all of which can be taught without prerequisite knowledge from high school or college geometry.  See my essay on why high school or college geometry should be removed and students should go straight to trigonometry without it which explains the cult like nature of high school or college geometry class.

http://teachingthenarrowway.blogspot.com/2021/12/why-you-should-be-able-to-skip-high.html

http://web.archive.org/web/*/http://teachingthenarrowway.blogspot.com/2021/12/why-you-should-be-able-to-skip-high.html

Copyright Carl Janssen 2022

Incomplete work on the following

work on half and double angles, added angles and average angles for functions other than tan(0.5A + 0.5B)

I originally planned to get half an double angles for each trigonometric identity in terms of itself then planned to compare with results published elsewhere but apparently that is not commonly put that way in a table.  Such as for example 

sin of half an angle as a function of sin of a whole angle

cos of half an angle as a function of cos of a whole angle

tan of half an angle as a function of tan of a whole angle

"Each trigonometric function in terms of each of the other five" is not for half and double angles 

https://en.m.wikipedia.org/wiki/List_of_trigonometric_identities

http://web.archive.org/web/20211227144328/https://en.m.wikipedia.org/wiki/List_of_trigonometric_identities