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