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Constructing Trig Values:
The Golden Triangle and the Mathematical Magic of the Pentagram
A Play in Five Acts
"Mathematicians always strive to confuse their audiences; where there is no confusion there
is no prestige. Mathematics is prestidigitation." From Mathematics made difficult by Carl E
Linderholm.
Prologue: “You Can with Beakman and Jax,"
In the spring of 2006 I was assigned to teach a trigonometry class for the first time since 1992. In
preparation for the course I made the following “rather standard” chart showing the well known “exact”
values of cosine and sine around the unit circle. For the values not “on axis” (i.e., not 0°, 90°, 180°
360° ), the results are easily derived from either an equilateral triangle or an isosceles right triangle.
Figure 1
The Monday after Christmas, I was doing some stretching while reading the Sunday comics. In the
“science” strip, “You Can with Beakman and Jax”, there was a discussion of the “magic” of the five
pointed “Christmas star” and the occurrence therein of the “Golden Ratio”. I must admit with some
embarrassment that this was all news to me. I had never really thought much about pentagrams before. But
I immediately realized that this meant there was another group of “exact” trig values based on a
18°, 72° right triangle. This presentation is based on my “re-discovery” of these very well known
properties.
Al Lehnen
Madison College
5/24/2013
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Act I: “The Golden Ratio”
In Book VI of the Elements Euclid (1) divides a line segment into two intervals in such a way that the
length of the longer interval to the length of the shorter interval is equal to the length of the whole interval
to the length of the long interval.
Figure 2
Designate the ratio as the Greek letter phi. Then, φ =
AB AC AB + BC
1
=
=
= 1 + . Hence,
BC AB
AB
φ
φ 2 = φ + 1 or equivalently, φ 2 − φ − 1 = 0 . Since this ratio is positive, the quadratic formula gives the
result φ =
1 + 12 + 4
5 +1
=
= 1.6180339887498948482045868343656…
2
2
Euclid called this number the extreme and mean ratio. Later it acquired the name the “Golden Ratio”.
Numbers like phi are constructible (2). All constructible numbers are algebraic, i.e. a solution of a
polynomial equation with rational number coefficients. Equivalently non algebraic (transcendental)
numbers are never constructible. A possible “modern” (non-collapsing compass) construction of φ is
shown in Figure 1.
Figure 3
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Consider now the rectangle ABCD with AE = AD and rectangle FEBC similar to rectangle ABCD.
Figure 4
−1
AD AB AE+EB AD+EB
⎛ AD ⎞
=
=
=
= 1+ ⎜
The similarity condition means that
⎟ . This is precisely
EB AD
AD
AD
⎝ EB ⎠
AB
the extreme and mean proportion. Hence, φ =
and such a rectangle is called “Golden”. The number φ
AD
occurs in the solution of many different problems. For example, the famous Fibonacci sequence which is
defined by the following recursion,
Fn + 2 = Fn +1 + Fn
; F1 = F2 = 1 ;
{1,1, 2,3,5,8,13, 21,34,55,89,144…}
n −1
n
1 ⎡ n +1 ⎛ −1 ⎞ ⎤ 1 ⎡ n ⎛ −1 ⎞ ⎤ φ n +1
φn
has the explicit formula Fn = 2
.
=
⎢φ + ⎜ ⎟ ⎥ =
⎢φ − ⎜ ⎟ ⎥ ≈ 2
φ + 1 ⎢⎣
5 ⎢⎣
5
⎝ φ ⎠ ⎥⎦
⎝ φ ⎠ ⎥⎦ φ + 1
Similarly, a “Golden Spiral” can be constructed by inscribing a quarter circle in successive adjacent
squares inside a Golden Rectangle. This is shown in Figure 5. The limit point of these spirals can be
expressed in terms of φ . The frequency with which φ unexpectedly appears in mathematics has given it an
almost magical reputation.
Figure 5
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Madison College
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Act II: “The Golden Triangle”
Consider an isosceles triangle with the property that when one of its equal angles is bisected the bisector
generates a smaller triangle similar to the original. This is illustrated in Figure 6.
Figure 6
Since α + 2α + 2α = 180° , we see that the angles must be 36°,72°,72° .
S
w
1
=
. This yields the quadratic equation
From the similarity of the two triangles =
w S−w S
−1
w
2
S
5 +1
S
⎛S⎞
⎜ w ⎟ − w − 1 = 0 . The only positive solution is w = 2 = φ .
⎝ ⎠
Bisecting the side opposite to the 36° degree angle gives the result that sin (18°) =
w
5 −1
and
=
2S
4
2
⎛ 5 − 1⎞
10 + 2 5
w
1
⎟ =
cos(18°) = 1 − ⎜⎜
. In terms of φ , sin (18° ) =
=
, and
⎟
2
S
2
φ
4
4
⎠
⎝
4φ 2 − 1
cos (18° ) = 1 −
=
=
2φ
4φ 2
1
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( 2φ − 1)( 2φ + 1)
2φ
=
Madison College
( 2φ − 1) (φ 2 + φ )
2φ
=
( 2φ 2 − φ ) φ 2 =
φ+2
2φ
5/24/2013
2
.
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Act III: “Knowing All the Angles”
From the half-angle formulas we have the following “exact results”.
1 − cos(30°)
=
sin (15°) =
2
2− 3
3 − 2 3 +1
=
=
4
8
1 + cos(30°)
cos(15°) =
=
2
2+ 3
=
4
3 + 2 3 +1
=
8
(
(
)
2
3 −1
8
)
3 +1
8
(
)
(
)
=
2 3 −1
=
4
=
2 3 +1
=
4
2
6− 2
4
6+ 2
4
From the “subtraction” formulas,
sin (3°) = sin (18° − 15°) = sin (18°) cos(15°) − cos(18°) sin (15°)
⎛ 5 − 1 ⎞⎛ 6 + 2 ⎞ ⎛⎜ 10 + 2 5 ⎞⎟⎛ 6 − 2 ⎞
⎜
⎟
⎟⎜
⎟−
= ⎜⎜
⎟
⎟⎜
⎟ ⎜
⎟⎜⎝
4
4
4
⎠
⎝ 4 ⎠⎝
⎠ ⎝
⎠
=
30 + 10 − 6 − 2 − 2 15 + 3 5 + 2 5 + 5
16
cos(3°) = cos(18° − 15°) = cos(18°) cos(15°) + sin (18°)sin (15°)
⎛ 10 + 2 5 ⎞⎛ 6 + 2 ⎞ ⎛ 5 − 1 ⎞⎛ 6 − 2 ⎞
⎟⎜
⎟+⎜
⎟⎜
⎟
=⎜
⎟ ⎜ 4 ⎟⎜
⎟
⎜
⎟⎜⎝
4
4
4
⎠ ⎝
⎠⎝
⎠
⎝
⎠
=
.
2 15 + 3 5 + 2 5 + 5 + 30 − 10 − 6 + 2
16
These results for three degrees can be used to construct exact values for any multiple of three degrees
using the following recursions based on the addition formulas.
sin[(n + 1)3°] = sin (3n°) cos(3°) + cos(3n°)sin (3°)
⎛ 30 − 10 − 6 + 2 + 2 15 + 3 5 + 2 5 + 5 ⎞
⎟ sin (3n°) +
=⎜
⎜
⎟
16
⎝
⎠
⎛ 30 + 10 − 6 − 2 − 2 15 + 3 5 + 2 5 + 5 ⎞
⎜
⎟ cos(3n°)
⎜
⎟
16
⎝
⎠
cos[(n + 1)3°] = cos(3n°) cos(3°) − sin (3n°) sin (3°)
⎛ 30 − 10 − 6 + 2 + 2 15 + 3 5 + 2 5 + 5 ⎞
⎟ cos(3n°) −
=⎜
⎜
⎟
16
⎝
⎠
⎛ 30 + 10 − 6 − 2 − 2 15 + 3 5 + 2 5 + 5 ⎞
⎜
⎟ sin (3n°)
⎜
⎟
16
⎝
⎠
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Madison College
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For some angles additional simplification is possible. For example, from the recursion the cosine of
21° =
7π
is found to be
60
30 + 10 +
6+
2+
75 + 15 5 −
16
25 + 5 5 − 15 + 3 5 + 5 + 5
.
However,
75 + 15 5 −
15 + 3 5
(
)
5 − 1 − 5+ 5
(
)
5−1 =
( 5 − 1) − 5 + 5 ( 5 − 1) =
5)(6 − 2 5) − (5 + 5)(6 − 2 5) =
2
15 + 3 5
(15 + 3
25 + 5 5 − 15 + 3 5 + 5 + 5 =
60 − 12 5 −
2
.
20 − 4 5 = 2 15 − 3 5 − 2 5 − 5
So that the cosine of 21 degrees can be expressed as
cos(21°) =
30 + 10 + 6 + 2 + 2 15 − 3 5 − 2 5 − 5
.
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Proceeding in this manner the results in Table 1 were generated. Angles beyond 45° can be generated by
the symmetry of the unit circle:
cos (θ ) = sin ( 90° − θ ) = cos ( 360° − θ ) = − cos (180° − θ )
sin (θ ) = cos ( 90° − θ ) = − sin ( 360° − θ ) = sin (180° − θ )
In fact, multiples of three degrees is the “best” one can do using integer degree arguments. This follows
from Gauss’s famous constructability theorem for regular polygons (3, 4). Regular polygons with 180 and
360 sides can not be constructed with rule and compass since the prime factorization of both of these
0
numbers contains two factors of the Fermat prime 3 = 22 + 1 . This means that no exact solution in terms
of square roots is possible for the trigonometric functions of either 2° or 1° . Of course 120 can be factored
as 120 = 2 3 (3)(5) and so contains only one factor each of the Fermat primes 3 and 5. Thus a 120 sided
polygon is constructible and the exact values for three degrees presented above must exist.
You could go further and use the half angle formulas to extend the values in Table 1 to intervals of
0.75° and then interpolate the resulting decimal values. This would allow the construction from exact
values of that peculiar anachronism known as a trig table without the use of an infinite series!
Al Lehnen
Madison College
5/24/2013
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Table 1: Exact Values of Sine and Cosine for Angles that are Multiples of Three Degrees
θ
0° = 0
π
3° =
60
30 − 10 −
π
6° =
30
π
9° =
20
π
12° =
sin θ
1
0
2 + 2 15 + 3 5 + 2 5 + 5
16
12
π
10 + 2 − 2 5 − 5
8
10
21° =
7π
60
30 + 10 + 6 + 2 + 2 15 − 3 5 − 2 5 − 5
16
24° =
2π
15
1 + 5 + 30 − 6 5
8
27° =
3π
20
6
10 − 2 + 2 5 + 5
8
3
2
11π
60
− 30 + 10 + 6 − 2 + 2 15 + 3 5 + 2 5 + 5
16
30° =
33° =
π
π
2 − 2 15 + 3 5 + 2 5 + 5
16
10 + 2 + 2 5 − 5
8
10 + 2 5
4
18° =
6−
− 1 − 5 + 30 − 6 5
8
30 + 6 5
8
6+ 2
4
π
30 + 10 −
15 + 3 + 10 − 2 5
8
5 − 1+
15
15° =
6+
cos θ
3 − 15 + 10 + 2 5
8
6− 2
4
5−1
4
− 30 + 10 − 6 + 2 + 2 15 − 3 5 + 2 5 − 5
16
15 + 3 − 10 − 2 5
8
− 10 + 2 + 2 5 + 5
8
1
2
30 + 10 − 6 − 2 + 2 15 + 3 5 − 2 5 + 5
16
5
5 +1
4
10 − 2 5
4
39° =
13π
60
30 − 10 + 6 − 2 + 2 15 − 3 5 + 2 5 − 5
16
30 + 10 + 6 + 2 − 2 15 − 3 5 + 2 5 − 5
16
42° =
7π
30
15 − 3 + 10 + 2 5
8
2
2
36° =
45° =
π
4
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1 − 5 + 30 + 6 5
8
2
2
5/24/2013
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Act IV: “Constructing a Regular Pentagon on a Limited Budget”
Proposition 10 of Book IV of Euclid’s Elements (5) gives a construction of the Golden Triangle. This
serves as the basis for Proposition 11 (6) which constructs a regular pentagon.
However, the number of steps involved is rather large. A very simple and elegant construction of a regular
pentagon was given by H. W. Richmond in 1893 (7, 8, 9) and is presented below.
Figure 7
(1)
(2)
(3)
(4)
(5)
(6)
(7)
Construct a circle of radius r centered at O.
Mark Construct a pair of perpendicular lines. Let O mark the point of intersection.
one of the points of intersection of the lines of (1) with the circle of (2) as A.
On the line which does not contain A bisect the radius at point M.
Bisect the angle OMA and extend the bisector to P on segment OA.
From P construct a line perpendicular to OA with point of intersection C with the circle of (2)
Mark off length AC and proceed around the circle generating three additional vertices.
The following argument based on trigonometry proves the validity of Richmond’s construction.
Since OA is twice OM, tan (2∠OMP ) = 2 . From the addition formulas,
tan (2∠OMP ) =
Al Lehnen
2 sin (∠OMP ) cos(∠OMP )
2 tan (∠OMP )
=
=2 .
2
2
cos (∠OMP ) − sin (∠OMP ) 1 − tan 2 (∠OMP )
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Rearranging the above equation gives a quadratic equation:
1 − tan 2 ( ∠OMP ) = tan ( ∠OMP ) or tan 2 (∠OMP ) + tan (∠OMP ) − 1 = 0 .
The tangent of angle OMP is positive, so the only admissible solution is
−1 + 5 1
tan ( ∠OMP ) =
= .
φ
2
OP
5 −1 1
5 −1
=
=
.
OC and sin (∠OCP ) =
OC
4
2φ
4
Thus, ∠OCP = 18° ; ∠COA = 90° − ∠OCP = 72° , which divides the circle into five equal arcs. The
Hence, OP = MO tan (∠OMP ) =
argument is summarized in Figure 8.
Figure 8
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Act V: “Magic Happens”
Line up 5 Golden Triangles in a horizontal row so that their short bases form a continues line segment of
width 5w. Then rotate the four right most triangles 72° clockwise. Now rotate the three triangles at the end
of the line 72° clockwise, next rotate the two triangles at the end of the line 72° clockwise, and finally
rotate the last triangle 72° clockwise. This will generate a regular pentagram. As shown in Figure 9 the
ratio properties noted by Beakman and Jax follow rather easily from the Golden Quadratic Equation
φ2 = φ + 1.
Figure 9
A more complete and revealing analysis considers a regular pentagon inscribed in a circle of radius OA
and with vertices at A, B, C, D and E. Construct segments AD, AC, BE, BD, and CE with points of
intersection F, M, G, H, I and J . This is shown in Figure 10.
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Figure 10
First ∠AOB = arcAB = arcBC = arcCD = arcED = arcEA =
360°
= 72° . Then since the measure of
5
an angle inscribed on a circle is equal to half of its subtended arc, we have that
∠EAD = ∠DAC = ∠CAB = ∠AEB = ∠ABE = ∠BEC = ∠EBD = 36° . So by ASA congruency AEF
and ABG are congruent isosceles triangles with AF = EF = AG = GB so all of the sides of the pentagram
AFEJDICHBG are equal. Furthermore, triangle AFG is isosceles so that
180° − ∠DAC
arcAC
= 72° , ∠ADC =
= 72° , and
2
2
∠AJE = 180° − ∠BEC − ∠AEB − ∠EAD = 180° − 3(36°) = 72° . Similar arguments establish that
∠ACD = ∠AHB = 72° . Thus, as shown in Figure 11 the four triangles AFG, AEJ, ABH and ADC are
∠AFG = ∠AGF =
“Golden”. In the same way each vertex of the pentagon ABCDE is the “top” vertex of four such “Golden”
triangles.
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Figure 11
From the triangle AFG,
EG EF + FG
1
AF
EF
=
= 1 + ; however,
= φ , but EF = AF, so
= φ . Now
EF
EF
φ
FG
FG
φ +1 φ2
=
=φ .
φ
φ
φ
EB EG + GB EG + EF
1
=
= 1 + = φ . So in summary each of the following ratios of
Similarly, EG =
EG
EG
φ
1+
1
=
lengths of segments in the chord EB is the Golden Ratio:
EF EG EG EB EB
=
=
=
=
= φ = 1.6180339887499... .
FG EF GB EG FB
Triangles AEJ and ACD are both Golden and FS = FG, EJ = EF = AF, AJ = AE = CD. Thus,
AD AJ AF AJ AD
=
=
=
=
=φ .
DC EJ FJ AF AJ
AH AG AC AH
Similarly, AG = GH = AH = BH = φ .
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Madison College
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Finally, exact relationships between the radius, r = OA, of the inscribing circle and segments within the
pentagon can be obtained using Table 1.
Since ∠AOE = arcEA = 72° = ∠AOB and OE = OB = r triangles MOB and MOE are congruent.
∠BMO = ∠EMO = 90°
MO
5 −1
= cos(72°) = sin (18°) =
r
4
MB
10 + 2 5
= sin (72°) = cos(18°) =
r
4
EB 2 MB
10 + 2 5
=
=
r
r
2
AM
MO 5 − 5
= 1−
=
r
r
4
AB
10 − 2 5
= 2 sin (36°) =
.
r
2
The perimeter, P, of the regular pentagon is related to the radius of the inscribed circle as
P 5 10 − 2 5
r
50 + 10 5
=
= 5.8778525229247... and =
= 0.17013016167041...
r
P
2
50
FG = GH = HI = IJ = JF and ∠EFA = ∠EJD = ∠DIC = ∠CHB = ∠BGA = 180° − 2(36°) = 108° , so
GFJIH is also a regular pentagon. As shown in Figure 12 one can construct an “infinite regress” of
imbedded regular pentagons and associated pentagrams. The dimensions of each new pentagon/pentagram
are related to the previous pentagon/pentagram by the scale factor
MO
5 +1
OG MO ⎛ 5 + 1 ⎞
=
and
= cos(36°) =
⎜
⎟
r
r ⎜⎝ 4 ⎟⎠
OG
4
−1
=
5 −1
5 +1
=
OG
. OG bisects ∠AOB so that
r
3− 5
= 0.381966011250105... .
2
2
⎛3− 5 ⎞
7−3 5
⎟
= 0.14589803375032... .
⎜ 2 ⎟ =
2
⎝
⎠
The areas of the embedded figures are scaled by ⎜
Figure 12
Al Lehnen
Madison College
5/24/2013
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References:
1. http://aleph0.clarku.edu/~djoyce/java/elements/bookVI/defVI3.html
2. http://en.wikipedia.org/wiki/Compass_and_straightedge
3. http://en.wikipedia.org/wiki/Heptadecagon
4. Weisstein, Eric W. "Trigonometry Angles." From MathWorld--A Wolfram Web Resource.
http://mathworld.wolfram.com/TrigonometryAngles.html
5. http://aleph0.clarku.edu/~djoyce/java/elements/bookIV/propIV10.html
6. http://aleph0.clarku.edu/~djoyce/java/elements/bookIV/propIV11.html
7. Weisstein, Eric W. "Pentagon." From MathWorld--A Wolfram Web Resource.
http://mathworld.wolfram.com/Pentagon.html
8. Richmond, H. W. "A Construction for a Regular Polygon of Seventeen Sides." Quart. J. Pure Appl.
Math. 26, 206-207, 1893.
9. http://www.jimloy.com/geometry/pentagon.htm
Al Lehnen
Madison College
5/24/2013