abstract:
A table of the exact values of the trigometric functions is
shown and the values are derived. Further, the exact values
for the angles of the dodecahedron and icosahedron are derived.
The five fold values are derived by repeated use of the cosine
sum of angles and double and triple angle formula resulting in
a quadratic equation. The seven fold problem is shown to
result in a cubic equation which is discussed elsewhere.
angle cos sin
deg radians
0 1 0
3 pi/60
18 pi/10 [5+[5]]/(2[2]) [3-[5]]/(2[2]) =([5]-1)/4
30 [3]/2 1/2
36 pi/5 [3+[5]]/(2[2]) =([5]+1)/4 [5-[5]]/(2[2])
45 1/[2] 1/[2]
54 3pi/10 [5-[5]]/(2[2]) [3+[5]]/(2[2]) =([5]+1)/4
60 1/2 [3]/2
72 2pi/5 [3-[5]]/(2[2]) =([5]-1)/4 [5+[5]]/(2[2])
87
90 0 1
cos angle tan
.9510565 1/10 [5-2[5]]/[5]
.8090169 1/5 [5-2[5]]
.5877852 3/10 [5+2[5]]/[5]
.3090166 2/5 [5+2[5]]
[5]tan(pi/10) = tan(pi/5)
[5]tan(3pi/10) = tan(2pi/5)
( parenthetical tangential digression:
( proof: 5-[5]+3+[5]=8
(
( Phi = ([5]+1)/2 = 2/([5]-1) 1.618033989
( phi = ([5]-1)/2 = 2/([5]+1) 0.6180339
( Phi - phi = 1
(
(
(
( http://www.mcs.surrey.ac.uk/Personal/R.Knott/Fibonacci/phi2DGeomTrig.html#simpletrig
( sin(3pi/10) = Phi/2 = 1/(2phi) 0.8090169
( cos(3pi/10) = [2-phi]/2 0.5877852
( cos(2pi/5) = ([5] - 1) /4 = 0.3090169
(
( sin(3pi/10) = ([5]+1)/4 0.8090169
( cos(3pi/10) = [10-2[5]]/4 0.5877852
(
( http://www.mcs.surrey.ac.uk/Personal/R.Knott/Fibonacci/simpleTrig#2pmphi
( cos(3pi/10) = sin(pi/5) = [2-phi]/2
( sin(3pi/10) = cos(pi/5) = [2+phi]/2 0.8090169
(
(
( cos(pi/5) = [2+phi]/2 = [3+[5]]/(2[2])
( cos(pi/10) = [5+[5]]/(2[2])
( cos(pi/5) = Phi/2
(
( http://www.mcs.surrey.ac.uk/Personal/R.Knott/Fibonacci/trig6.gif
(
( Gerald de Jong
( www.rwgrayprojects.com/OswegoOct2001/Presentation/presentationWeb4.html
resume train of thought
start with well known trigonometric identities:
cos(2A) = cos2(A) - sin2(A) = 2cos2(A) - 1
cos(A/2) = [(1+cos(A))/2]
sin(A/2) = [(1-cos(A))/2]
cos(3A) = 4cos3(A) - 3cos(A)
sin(3A) = 4cos2(A)sin(A) - sin(A)
cos(2A+3A) = cos(2A) cos(3A) - sin(2A) sin(3A)
cos(2A+3A) = cos(2A) cos(3A) - 2sin(A) cos(A) sin(3A)
cos(2A+3A) = (2cos2(A) - 1) (4cos3(A) - 3cos(A))
- 2sin(A) cos(A) (4cos2(A)sin(A) - sin(A))
cos(2A+3A) = (2cos2(A) - 1) (4cos3(A) - 3cos(A))
- 2sin(A) cos(A) (4cos2(A) - 1) sin(A)
cos(2A+3A) = (2cos2(A) - 1) (4cos3(A) - 3cos(A))
- 2 cos(A) (4cos2(A) - 1) (1-cos2(A))
cos(2A+3A) = 8cos5(A) -10cos3(A) + 3cos(A)
- 2 cos(A) (4cos2(A) - 1) (1-cos2(A))
cos(2A+3A) = 8cos5(A) - 10cos3(A) + 3cos(A)
+ 8cos5(A) - 10cos3(A) + 2cos(A)
cos(2A+3A) = 16cos5(A) -20cos3(A) + 5cos(A)
A = pi/5 cos(pi) = -1
16x^5 -20x^3 + 5x + 1 = 0
(4x^3 +2x^2 -3x -1)(4x^2 -2x -1) = 0
x = ([5]+1)/4 = cos(pi/5)
Once one value cos(pi/5)=([5]+1)/4 in the table is derived the
remaining ones can be derived using the half angle formula,
cos(A/2)= [(cos(A) + 1)/2] ,
the double angle formula, cos(2A) = 2cos2(A) - 1 , Pythagoras,
sin2(A) + cos2(A) = 1, and the compliment
equivalences, sin(A) = cos(pi/2 - A) and cos(A) = sin(pi/2 - A)
Pythagoras: sin2(pi/5) = 1-(6+2[5])/16=(10-2[5])/16=(5-[5])/8
sin(pi/5) = [5-[5]]/(2[2])
half angle: cos(pi/10) = [([5]+5)/8]
double angle: cos(2pi/5) = (6+2[5])/8 -1 = ([5]-1)/4
compliment: cos(3pi/10) = sin(pi/5) = [5-[5]]/(2[2])
sin(3pi/10) = cos(pi/5) = ([5]+1)/4
...
(parenthetical tangential digression:
( for pi/7
( cos(2A+5A) = cos(2A)cos(5A) - sin(2A)sin(5A)
( cos(3A+4A) = cos(3A)cos(4A) - sin(3A)sin(4A)
resume train of thought
Consider the icosahedron:
An equilateral triangle is tipped up from a vertex or corner,
until the projection of the angle of the side from
the altitude increases from an angle of 30deg=pi/6
to an angle of 54deg=pi/5.
face of icosahedron
The triangle is tilted up an angle alpha
cos(pi/6) = [3]/2
sin(pi/6) = 1/2
| cos 0 sin | |[3]/2| | [3]/2 cos |
| | | | | |
| 0 1 0 | | 1/2 | = | 1/2 |
| | | | | |
| -sin 0 cos | | 0 | |-[3]/2 sin |
1/([3]cos(alpha)) = tan(pi/5) = [5-2[5]]
cos(alpha) = [5+2[5]] / [15]
the angle between the face and vertex of icosahedron
Consider the dodecahedron:
A pentagon is tipped up from a vertex or corner,
until the projection of the angle of the side from
the altitude increases from an angle of 54deg=pi/5
to an alngle of 60deg=pi/3.
face of dodecahedron
The pentagon is tilted up an angle beta.
cos(3pi/10) = [5-[5]]/(2[2])
sin(3pi/10) = [3+[5]]/(2[2])
| cos 0 sin | |[5-[5]]/(2[2])| | [5-[5]]/(2[2]) cos |
| | | | | |
| 0 1 0 | |[3+[5]]/(2[2])| = | [3+[5]]/(2[2]) |
| | | | | |
| -sin 0 cos | | 0 | |-[5-[5]]/(2[2]) sin |
[3+[5]]/([5-[5]] cos(beta)) = tan(pi/3) = [3]
1/cos(beta) = [3] *( [5-[5]]/[3+[5]] )
= [3] *( [20-8[5]]/2 )
= [3] [5-2[5]]
= [3] * [5-2[5]]
cos(beta) = [5+2[5]] / [15] = 0.794654472
Thus, the angle between the face and vertex of dodecahedron is the same as
that for the icosahedron: the cosine of the angle is also the ratio of the
radius of the inscribed sphere to the radius of the circumscribed sphere.
sin(alpha) = [15/15 - (5+2[5])/15] = [2][5-[5]]/[15] = 0.607061998
cos(alpha) cos(beta) = (5+2[5])/15
cos(alpha) / cos(beta) = 1
cos(alpha) sin(alpha) = [2][15+5[5]]/[15] =[2/5][3+[5]]/[3]
consider the icosahedron; find the dot product of the vector to
the center of two faces; assume one vertex is on the z axis;
the next face would be 2pi/5 away so rotate it the first face
vector so that it is at the next face.
cos(2pi/5) = ([5]-1)/4 = [6-2[5]]/4
|([5]-1)/4 -[(5+[5])/8 0||[2][5-[5]]/[15]| |[2][5-[5]]/[15]|
| || | | |
|-[(5+[5])/8 ([5]-1)/4 0|| 0 | dot| 0 |
| || | | |
| 0 0 1|| [5+2[5]]/[15] | | [5+2[5]]/[15] |
|[40-16[5]]/(2[2])| |[2][5-[5]]|
| | | |
=(1/15)| [20]/2 | dot | 0 |
| | | |
| [5+2[5]] | | |
= (1/15)([2][35-15[5] + 5+2[5])
= (1/15)([2](-5+3[5]) + 5+2[5]) = [5]/3
Thus cos(icosahedron_face_normals) = [5]/3
cos(41.8103149deg=0.232279527pi=0.729727656)=0.745355993
and sin(icosahedron_face_normals) = 2/3
likewise with the dodecahedron; find the dot product of the vector to
the center of two faces; assume one vertex is on the z axis;
the next face would be 2pi/3 away so rotate it the first face
vector so that it is at the next face.
| -1/2 -[3]/2 0 | |[2][5-[5]]/[15]| |[2][5-[5]]/[15]|
| | | | | |
|[3]/2 -1/2 0 | | 0 | dot| 0 |
| | | | | |
| 0 0 2/2 | | [5+2[5]]/[15] | | [5+2[5]]/[15] |
| -q | | q |
| | | |
= (1/30)| [3]q | dot | 0 |
| | | |
| 2p | | p |
= (2p^2-q^2)/30 = 6[5]/30 = 1/[5]
where p = [5+2[5]]
q = [2][5-[5]]
2p^2 - q^2 = 10 + 4[5] - 10 + 2[5] = 6[5]
Thus cos(dodecahedron_face_normals) = [5]/5
and sin(dodecahedron_face_normals) = (2/5)[5]
Let R be the circumscribed radius, E the radius of the
center of the edge, and r the inscribed radius.
chord or edge from center for dodecahedron
vertex to vertex angle is arccos([5]/3)
cos(arccos([5]/3)/2)
Ed = [(1+[5]/3)/2] R = [3+[5]]/[6] R = 0.934172359
face from center
face to face angle arccos([5]/5)
cos(arccos([5]/5)/2)
r = [(1+[5]/5)/2] Ed = [5+[5]]/[10] Ed = 0.850650808
the length of the edge of the dodecahedron is
Ld = 2 sin(arccos([5]/3)/2)
= 2[(1-[5]/3)/2] = [2/3][3-[5]] = 0.71364418
of the circumscribed sphere radius
or [2/3][3-[5]] / [5+2[5]]/[15] = [10][25-11[5]] / [5]
= [2][25-11[5]] = 0.898055953
of the inscribed sphere radius
As a final check for the dodecahedron the ratio of the length
of an edge,[2/3][3-[5]], to the radius to the edge, Ed = [3+[5]]/[6],
should be twice the tangent of 1/2 the angle between vertices,
arccos([5]/3).
2 [1- [5]/3]/[1+ [5]/3] = [2/3][3-[5]] / [3+[5]]/[6]
2 [14/9 - (2/3)[5]] /[4/9] = [4] [14-6[5]/ [4]
[14 - 6[5]] = [14-6[5]]
[7-3[5]] = [7-3[5]]
chord or edge from center for icosahedron
vertex to vertex angle arccos([5]/5)
cos(arccos([5]/5)/2)
Ei = [(1+[5]/5)/2] = [5+[5]]/[10] R = 0.850650808
face from center
face to face angle arccos([5]/3)
[(1+[5]/3)/2]
cos(arccos([5]/3)/2)
ri = [(1+[5]/3)/2] Ei = [3+[5]]/[6] Ei =0.934172359
the length of the edge of the icosahedron is
Li = 2 sin(arccos([5]/5)/2)
= 2[(1-[5]/5)/2] = [2/5][5-[5]] = 1.051462224
of the circumscribed sphere radius
or [2/5][5-[5]] / [5+2[5]]/[15] = [6] [35-15[5]] / [5]
= [6] [7-3[5]] = 0. = 1.323169076
of the inscribed sphere radius
As a final check for the icosahedron the ratio of the length
of an edge,[2/5][5-[5]], to the radius to the edge, Ei = [5+[5]]/[10],
should be twice the tangent of 1/2 the angle between vertices,
arccos([5]/5).
2 [1 - [5]/5]/[1 + [5]/5] = [2/5][5-[5]]/([5+[5]]/[10])
2 [30/25 -(2/5)[5]] /[4/5] = 2 [30-10[5]] / [20]
2 [(3/2) -(1/2)[5]] = [3-[5]] [2]
2 [3-[5]/5]/[2] = [3-[5]] [2]
for both the icosahedron and dodecahedron the
the ratio of the radius of the inscribed/circumscribed sphere is
product of both r/E and E/R :
[1+[5]/5][1+[5]/3]/2 = [20/15 + 8[5]/15]/2
= [20+8[5]]/[60] = [5+2[5]]/[15] = 0.794654472,
the same as the cosine of the vertex to face center angle
at /wiki/Exact_trigonometric_constantw#Uses_for_constants
(c)2009 Wm.C.Corwin www.ConcurrentInverse.com
optional
The symetry of the dodecahedron and icosahedron are equivalent if the
angles to the vertices and faces are interchanged. The angle of faces
or vertices can be found by rotating the face vector about a vertex
vector by 2pi/5 in the case of the icosahedron or pi/6 in the case of
the dodecahedron.
center of pentagon face perpendicular to z axix
rotation about y axis of alpha=beta
gives x',y,z' reference frame with z' axis going to corner
a rotation of 2pi/3 about the z' axis will be tranformed to the x,y,z frame
that rotation will take one face normal to another
cos(beta) = [5+2[5]]/[15] cos(2pi/3) = 1/2
sin(beta) = [2][5-[5]]/[15] sin(2pi/3) = -[3]/2
the three fold rotation T is transformed by the five fold rotation F
T = (1/F') T' F'
| cos(beta) 0 sin(beta) | | 1/2 -[3] 0 | | ... |
| | | | | |
| 0 1 0 | | [3] 1/2 0 | | ... |
| | | | | |
|-sin(beta) 0 cos(beta) | | 0 0 1 | | ...|
| [5+2[5]] 0 [2][5-[5]]| |1 -[3] 0 | | [5+2[5]] 0 -[2][5-[5]]|
| | | | | |
1/30| 0 [15] 0 | |[3] 1 0 | | 0 [15] 0 |
| | | | | |
|-[2][5-[5]] 0 [5+2[5]] | |0 0 2 | | [2][5-[5]] 0 [5+2[5]] |
p = [5+2[5]]
q = [2][5-[5]]
2p^2 - q^2 = 10 + 4[5] - 10 + 2[5] = 6[5]
p^2 + q^2 = 15
pq = [30+10[5]] = [10][3+[5]]
2p^2+q^2 = 10+4[5]+10-2[5] = 20+2[5]
| p -[3]p 2q | | p 0 -q |
| | | |
1/30 |[45] [15] 0 | | 0 [15] 0 |
| | | |
| q -[3]q 2p | | -q 0 p |
|p^2+2q^2 -[45]p pq?-6[5] | | ... -[5] |
| | | |
T = 1/30 |[45]p 15 -[45]q?6[15]|= (1/5)| ... [15]|
| | | |
|-pq -[45]q 2p^2-q^2=6[5] | | [5] |
The normal vector to the other face would be
|0| | [10][3+[5]]?|
T |0| = 1/30 |-[90][5-[5]]?|
|1| | 6[5] |
| sin(d) cos(2pi/3) | |(2/5)[5] (-1/2) | | -[5]/5 |
= | sin(d) sin(2pi/3) | = |(2/5)[5] [3]/2 | = | [15]/5 |
| cos(d) | | 1/[5] | | [5]/5 |
tan(2pi/5) = [5+2[5]] =
=
sin(d) = [1-1/[5]] = 2/[5] = 2[5]/5
but 6[5] / 30 = 1/[5]
dodecahedron angle between face normals arccos(1/[5])
sin(arccos(1/[5]) = [1-1/5] = 2/[5] = 2[5]/5
cos(2pi/5) = ([5]-1)/4
sin(2pi/5) = [5+[5]]/(2[2])
icosahedron angle between face normals arccos([5]/3)
(c) 2009 Wm.C.Corwin