Copyright Carl Janssen 2024 August 24
Calculating sine of average of angles and cosine of average of angles from tangent of average of angles and other proofs for other trigonometric identities
Explaining with a lot of words
It might be technically better to use the term point instead of vertex
This proof is intended to work for real number values and all values plugged in are intended to be real numbers for the way the proof is written but that does not mean whether or not the end result will or will not work if either complex or pure imaginary numbers are used
All angles mentioned in this proof before the second use of the word "solved" refer to angles measured at the origin relative to a horizontal line y = 0 and a second line segment in quadrant 1 of the unit circle
For positive angles between greater than 0 degrees and less than 90 degrees when the angle at the origin is measured. Although this proof will in the end work for any real number angles it is easier to visualize it within quadrant 1
If you take two triangles each on the unit circle with a unit-less radius of 1 and a shared Cartesian origin of (0,0) for one of their vertexes and a second vertex of (1,0) and the third or interesting vertex of (cos(angle), sin(angle))
And convert the interesting vertex of each triangle into Cartesian coordinates
Then if you take the average of the Cartesian coordinates of each interesting vertex with the other interesting vertex for the two angles you will get a new vertex which is not on the unit circle and has a distance other than 1 from the origin
If you draw a line segment from the origin to the new vertex created by the average already mentioned then that line will have a slope that is the same as a triangle with an angle equal to the average of the two original angles and the length of the line segment will not be 1
Thus you can use the coordinates of that new point to calculate the tangent of the average of the two angles for the original triangles on the unit circle even thought that point is not on the unit circle so it's coordinates can not directly be used to calculate the sine or cosine
If you multiply both horizontal or X and vertical or Y Caartesian coordinates of the new vertex already mentioned by the same constant and select the correct constant you can get coordinates which form a line segment with a length of 1 from the origin with the same slope as the tangent of the average of the two original triangles angles that can be used to form a third triangle on the unit circle. With this third triangle the sine and cosine can be calculated. The cosine of the average of the two angles will be the X coordinates of this newest line segment with a length of 1 and the sine will be the Y coordinates of this newest line segment with a length of 1 that has the same slope as that for the average of the two angles in the original triangles.
Explaining using more algebra and less words if you can not understand why what is being done go to the wordy section above or to the diagram if I add it later
Given
tangent(0.5A+0.5B) = ( N * [ 0.5sin(A) + 0.5sin(B) ] ) / ( N * [ 0.5*cos(A) + 0.5*cos(B) ] )
sin(0.5A+0.5B) = N * [ 0.5sin(A) + 0.5sin(B) ]
cos(0.5A+0.5B) = N * [ 0.5cos(A) + 0.5cos(B) ]
Solve for both
sin(0.5A+0.5B)
cos(0.5A+0.5B)
as functions of these four functions
cos(A), cos(B), sin(A), sin(B)
Solution
This value of N when multiplied by the coordinates of ( 0.5cos(A) + 0.5cos(B), 0.5sin(A) + 0.5sin(B) ) creates coordinates of a vertex which can form a line segment with a length of 1 from the origin (0, 0)
This line segment has a slope that is equivalent to the angle of ( 0.5A + 0.5B ). N is calculated using the Pythagorean theorem to calculate the distance from the origin (0,0) to ( 0.5cos(A) + 0.5cos(B), 0.5sin(A) + 0.5sin(B) ) and then taking the reciprocal of that distance
N= 1 / ( [ 0.5sin(A) + 0.5sin(B) ] ^ 2 + [ 0.5cos(A) + 0.5cos(B) ] ^ 2 ) ^ 0.5
sin(0.5A+0.5B) = [ 0.5sin(A) + 0.5sin(B) ] * N
cos(0.5A+0.5B) = [ 0.5sin(A) + 0.5sin(B) ] * N
Solved but needs to be simplified to get other trigonometric identities in simplified form
Solving for the square of the sine of a half angle by letting B = 0
sin(0)=0
cos(0)=1
sin(0.5A+ 0 ) ^ 2 = [ 0.5sin(A) + 0.5sin(0) ] ^ 2 * N ^2
[sin(0.5A) ]^2= [0.5sin(A)]^2 / ( [ 0.5sin(A) + 0.5sin(0) ] ^ 2 + [ 0.5cos(A) + 0.5cos(0) ] ^ 2 )
[sin(0.5A) ]^2= [0.5sin(A)]^2 / ( [ 0.5sin(A) ] ^ 2 + [ 0.5cos(A) + 0.5 ] ^ 2 )
[sin(0.5A) ]^2= [0.5sin(A)]^2 / ( [ 0.5sin(A) ] ^ 2 + [ 0.5cos(A) + 0.5 ] ^ 2 )
[ 1 + cos(A) ] ^ 2 = 1 + 2cos(A) + cos(A)^2 = 2 + 2cos(A) - [sin(A)]^2
[ 0.5cos(A) + 0.5 ] ^ 2 = 0.25 * [ 1 + cos(A) ] ^ 2 = 0.5 + 0.5cos(A) - 0.25[sin(A)]^2
[ 0.5sin(A) ] ^ 2 = 0.25[sin(A)]^2
[sin(0.5A) ]^2= [0.5sin(A)]^2 / ( [ 0.5sin(A) ] ^ 2 + 0.5 + 0.5cos(A) - 0.25[sin(A)]^2 )
(0.5)^2/ 0.5=0.5 = 1 / 2
[sin(0.5A) ]^2= [0.5sin(A)]^2 / [ 0.5 + 0.5cos(A) ] = [sin(A)]^2 / [ 2 + 2cos(A) ]
https://www.geogebra.org/graphing
