Un triángulo heptagonal es un triángulo escaleno obtuso cuyos vértices coinciden con el primer, segundo y cuarto vértices de un heptágono regular (desde un vértice inicial arbitrario). Por lo tanto, sus tres lados coinciden con un lado y con las diagonales adyacentes más cortas y más largas de un heptágono regular. Todos los triángulos heptagonales son similares (tienen la misma forma), por lo que se conocen colectivamente como el triángulo heptagonal. Sus ángulos miden
π
/
7
,
2
π
/
7
,
{\displaystyle \pi /7,2\pi /7,}
y
4
π
/
7
,
{\displaystyle 4\pi /7,}
y es el único triángulo con ángulos en las relaciones 1: 2: 4. El triángulo heptagonal tiene varias propiedades notables.
Un heptágono regular (con lados rojos), sus diagonales más largas (verdes), y sus diagonales más cortas (azules). Cada uno de los catorce triángulos heptagonales congruentes tiene un lado verde, un lado azul, y un lado rojo.
El centro de nueve puntos del triángulo heptagonal es también su primer punto de Brocard .[1] : Propos. 12
El segundo punto de Brocard se encuentra en el círculo de nueve puntos.[2] : p. 19
El circuncentro y los puntos de Fermat de un triángulo heptagonal forman un triángulo equilátero .[1] : Thm. 22
La distancia entre el circuncentro O y el ortocentro H viene dada por[2] : p. 19
O
H
=
R
2
,
{\displaystyle OH=R{\sqrt {2}},}
donde R es el circunradio . La distancia al cuadrado desde el incentro I al ortocentro es[2] : p. 19
I
H
2
=
R
2
+
4
r
2
2
,
{\displaystyle IH^{2}={\frac {R^{2}+4r^{2}}{2}},}
donde r es el inradio .
Las dos tangentes desde el ortocentro hasta el circuncírculo son mutuamente perpendiculares .[2] : p. 19
Lados
Los lados del triángulo heptagonal a < b < c coinciden respectivamente con el lado del heptágono regular, diagonal más corta y diagonal más larga. Satisfacen que[3] : Lemma 1
a
2
=
c
(
c
−
b
)
,
b
2
=
a
(
c
+
a
)
,
c
2
=
b
(
a
+
b
)
,
1
a
=
1
b
+
1
c
{\displaystyle {\begin{aligned}a^{2}&=c(c-b),\\[5pt]b^{2}&=a(c+a),\\[5pt]c^{2}&=b(a+b),\\[5pt]{\frac {1}{a}}&={\frac {1}{b}}+{\frac {1}{c}}\end{aligned}}}
(la última[2] : p. 13 es la ecuación óptica ) y por lo tanto
a
b
+
a
c
=
b
c
,
{\displaystyle ab+ac=bc,}
y[3] : Coro. 2
b
3
+
2
b
2
c
−
b
c
2
−
c
3
=
0
,
{\displaystyle b^{3}+2b^{2}c-bc^{2}-c^{3}=0,}
c
3
−
2
c
2
a
−
c
a
2
+
a
3
=
0
,
{\displaystyle c^{3}-2c^{2}a-ca^{2}+a^{3}=0,}
a
3
−
2
a
2
b
−
a
b
2
+
b
3
=
0.
{\displaystyle a^{3}-2a^{2}b-ab^{2}+b^{3}=0.}
Por lo tanto, -b /c , c /a y a /b satisfacen la ecuación cúbica
t
3
−
2
t
2
−
t
+
1
=
0.
{\displaystyle t^{3}-2t^{2}-t+1=0.}
La relación entre los lados es
b
=
2
cos
(
π
7
)
⋅
a
,
c
=
(
1
+
2
cos
(
2
π
7
)
)
⋅
a
.
{\displaystyle b=2\cos \left({\frac {\pi }{7}}\right)\cdot a,\qquad c=\left(1+2\cos \left({\frac {2\pi }{7}}\right)\right)\cdot a.}
y las raíces de esta ecuación son:
{
t
1
=
1
−
2
cos
(
π
7
)
t
2
=
1
+
2
cos
(
2
π
7
)
t
3
=
4
cos
(
2
π
7
)
cos
(
3
π
7
)
{\displaystyle {\begin{cases}t_{1}=1-2\cos \left({\frac {\pi }{7}}\right)\\t_{2}=1+2\cos \left({\frac {2\pi }{7}}\right)\\t_{3}=4\cos \left({\frac {2\pi }{7}}\right)\cos \left({\frac {3\pi }{7}}\right)\end{cases}}}
También se tiene que[4]
a
2
b
c
,
−
b
2
c
a
,
−
c
2
a
b
{\displaystyle {\frac {a^{2}}{bc}},\quad -{\frac {b^{2}}{ca}},\quad -{\frac {c^{2}}{ab}}}
satisface la ecuación cúbica
t
3
+
4
t
2
+
3
t
−
1
=
0
{\displaystyle t^{3}+4t^{2}+3t-1=0}
y las raíces de esta ecuación son:
{
t
1
=
−
1
−
2
cos
(
π
7
)
t
2
=
−
1
+
2
cos
(
2
π
7
)
t
3
=
4
cos
(
2
π
7
)
cos
(
3
π
7
)
−
2
{\displaystyle {\begin{cases}t_{1}=-1-2\cos \left({\frac {\pi }{7}}\right)\\t_{2}=-1+2\cos \left({\frac {2\pi }{7}}\right)\\t_{3}=4\cos \left({\frac {2\pi }{7}}\right)\cos \left({\frac {3\pi }{7}}\right)-2\end{cases}}}
También se tiene que[4]
a
3
b
c
2
,
−
b
3
c
a
2
,
c
3
a
b
2
{\displaystyle {\frac {a^{3}}{bc^{2}}},\quad -{\frac {b^{3}}{ca^{2}}},\quad {\frac {c^{3}}{ab^{2}}}}
satisface la ecuación cúbica
t
3
−
t
2
−
9
t
+
1
=
0
{\displaystyle t^{3}-t^{2}-9t+1=0}
y las raíces de esta ecuación son:
{
t
1
=
1
−
4
cos
(
π
7
)
t
2
=
1
+
4
cos
(
2
π
7
)
t
3
=
8
cos
(
2
π
7
)
cos
(
3
π
7
)
−
1
{\displaystyle {\begin{cases}t_{1}=1-4\cos \left({\frac {\pi }{7}}\right)\\t_{2}=1+4\cos \left({\frac {2\pi }{7}}\right)\\t_{3}=8\cos \left({\frac {2\pi }{7}}\right)\cos \left({\frac {3\pi }{7}}\right)-1\end{cases}}}
Así mismo, los valores[4]
a
3
b
2
c
,
b
3
c
2
a
,
−
c
3
a
2
b
{\displaystyle {\frac {a^{3}}{b^{2}c}},\quad {\frac {b^{3}}{c^{2}a}},\quad -{\frac {c^{3}}{a^{2}b}}}
satisfacen la ecuación cúbica
t
3
+
5
t
2
−
8
t
+
1
=
0
{\displaystyle t^{3}+5t^{2}-8t+1=0}
y las raíces de esta ecuación son:
{
t
1
=
−
2
[
cos
(
π
7
)
+
2
cos
(
2
π
7
)
+
1
]
t
2
=
6
cos
(
π
7
)
−
2
cos
(
2
π
7
)
−
3
t
3
=
2
[
3
cos
(
2
π
7
)
−
2
cos
(
π
7
)
]
{\displaystyle {\begin{cases}t_{1}=-2\left[\cos \left({\frac {\pi }{7}}\right)+2\cos \left({\frac {2\pi }{7}}\right)+1\right]\\t_{2}=6\cos \left({\frac {\pi }{7}}\right)-2\cos \left({\frac {2\pi }{7}}\right)-3\\t_{3}=2\left[3\cos \left({\frac {2\pi }{7}}\right)-2\cos \left({\frac {\pi }{7}}\right)\right]\end{cases}}}
También se tiene que[2] : p. 14
b
2
−
a
2
=
a
c
,
{\displaystyle b^{2}-a^{2}=ac,}
c
2
−
b
2
=
a
b
,
{\displaystyle c^{2}-b^{2}=ab,}
a
2
−
c
2
=
−
b
c
,
{\displaystyle a^{2}-c^{2}=-bc,}
y[2] : p. 15
b
2
a
2
+
c
2
b
2
+
a
2
c
2
=
5.
{\displaystyle {\frac {b^{2}}{a^{2}}}+{\frac {c^{2}}{b^{2}}}+{\frac {a^{2}}{c^{2}}}=5.}
Por otro lado[4]
a
b
−
b
c
+
c
a
=
0
,
{\displaystyle ab-bc+ca=0,}
a
3
b
−
b
3
c
+
c
3
a
=
0
,
{\displaystyle a^{3}b-b^{3}c+c^{3}a=0,}
a
4
b
+
b
4
c
−
c
4
a
=
0
,
{\displaystyle a^{4}b+b^{4}c-c^{4}a=0,}
a
11
b
3
−
b
11
c
3
+
c
11
a
3
=
0.
{\displaystyle a^{11}b^{3}-b^{11}c^{3}+c^{11}a^{3}=0.}
No hay otro par de números (m, n ), tales que m, n > 0 y que m, n <2000, que cumplan [cita requerida ]
a
m
b
n
±
b
m
c
n
±
c
m
a
n
=
0.
{\displaystyle a^{m}b^{n}\pm b^{m}c^{n}\pm c^{m}a^{n}=0.}
Alturas
Las alturas h a , h b y h c satisfacen
h
a
=
h
b
+
h
c
{\displaystyle h_{a}=h_{b}+h_{c}}
[2] : p. 13
y
h
a
2
+
h
b
2
+
h
c
2
=
a
2
+
b
2
+
c
2
2
.
{\displaystyle h_{a}^{2}+h_{b}^{2}+h_{c}^{2}={\frac {a^{2}+b^{2}+c^{2}}{2}}.}
[2] : p. 14
La altura desde el lado b (ángulo opuesto B ) es la mitad de la bisectriz del ángulo interno
w
A
{\displaystyle w_{A}}
de A :[2] : p. 19
2
h
b
=
w
A
.
{\displaystyle 2h_{b}=w_{A}.}
Aquí el ángulo A es el ángulo más pequeño y B es el segundo ángulo más pequeño.
Circunradio, inradio y exinradios
El área del triángulo es[5]
A
=
7
4
R
2
,
{\displaystyle A={\frac {\sqrt {7}}{4}}R^{2},}
donde R es el circunradio del triángulo.
Se tiene que[2] : p. 12
a
2
+
b
2
+
c
2
=
7
R
2
.
{\displaystyle a^{2}+b^{2}+c^{2}=7R^{2}.}
También se tiene que[6]
a
4
+
b
4
+
c
4
=
21
R
4
.
{\displaystyle a^{4}+b^{4}+c^{4}=21R^{4}.}
a
6
+
b
6
+
c
6
=
70
R
6
.
{\displaystyle a^{6}+b^{6}+c^{6}=70R^{6}.}
La relación
r
R
=
2
cos
(
π
7
)
−
3
2
{\displaystyle {\frac {r}{R}}=2\cos \left({\frac {\pi }{7}}\right)-{\frac {3}{2}}}
del inradio respecto al circunradio es la solución positiva de la ecuación cúbica[5]
8
x
3
+
28
x
2
+
14
x
−
7
=
0
{\displaystyle 8x^{3}+28x^{2}+14x-7=0}
siendo las otras dos raíces de esta ecuación
2
cos
(
3
π
7
)
−
3
2
{\displaystyle 2\cos \left({\frac {3\pi }{7}}\right)-{\frac {3}{2}}}
y
2
cos
(
5
π
7
)
−
3
2
{\displaystyle 2\cos \left({\frac {5\pi }{7}}\right)-{\frac {3}{2}}}
.
La relación
r
a
+
r
b
+
r
c
R
=
2
cos
(
π
7
)
+
5
2
{\displaystyle {\frac {r_{a}+r_{b}+r_{c}}{R}}=2\cos \left({\frac {\pi }{7}}\right)+{\frac {5}{2}}}
de la suma de los exinradios respecto al circunradio es la mayor de las raíces de la ecuación cúbica:
8
x
3
−
68
x
2
+
174
x
−
127
=
0
{\displaystyle 8x^{3}-68x^{2}+174x-127=0}
siendo las otras dos raíces de esta ecuación
2
cos
(
3
π
7
)
+
5
2
{\displaystyle 2\cos \left({\frac {3\pi }{7}}\right)+{\frac {5}{2}}}
y
2
cos
(
5
π
7
)
+
5
2
{\displaystyle 2\cos \left({\frac {5\pi }{7}}\right)+{\frac {5}{2}}}
.
La relación
1
r
a
+
1
r
b
+
1
r
c
R
=
4
cos
(
2
π
7
)
{\displaystyle {\frac {{\frac {1}{r_{a}}}+{\frac {1}{r_{b}}}+{\frac {1}{r_{c}}}}{R}}=4\cos \left({\frac {2\pi }{7}}\right)}
de la suma de los inversos de los exinradios respecto al circunradio es la única raíz positiva de la ecuación cúbica:
x
3
+
2
x
2
−
8
x
−
8
=
0
{\displaystyle x^{3}+2x^{2}-8x-8=0}
siendo las otras dos raíces de esta ecuación
4
cos
(
4
π
7
)
{\displaystyle 4\cos \left({\frac {4\pi }{7}}\right)}
y
4
cos
(
6
π
7
)
{\displaystyle 4\cos \left({\frac {6\pi }{7}}\right)}
.
Además,[2] : p. 15
1
a
2
+
1
b
2
+
1
c
2
=
2
R
2
.
{\displaystyle {\frac {1}{a^{2}}}+{\frac {1}{b^{2}}}+{\frac {1}{c^{2}}}={\frac {2}{R^{2}}}.}
También se tiene que[6]
1
a
4
+
1
b
4
+
1
c
4
=
2
R
4
.
{\displaystyle {\frac {1}{a^{4}}}+{\frac {1}{b^{4}}}+{\frac {1}{c^{4}}}={\frac {2}{R^{4}}}.}
1
a
6
+
1
b
6
+
1
c
6
=
17
7
R
6
.
{\displaystyle {\frac {1}{a^{6}}}+{\frac {1}{b^{6}}}+{\frac {1}{c^{6}}}={\frac {17}{7R^{6}}}.}
En general para todos los enteros n ,
a
2
n
+
b
2
n
+
c
2
n
=
g
(
n
)
(
2
R
)
2
n
{\displaystyle a^{2n}+b^{2n}+c^{2n}=g(n)(2R)^{2n}}
donde
g
(
−
1
)
=
8
,
g
(
0
)
=
3
,
g
(
1
)
=
7
{\displaystyle g(-1)=8,\quad g(0)=3,\quad g(1)=7}
y
g
(
n
)
=
7
g
(
n
−
1
)
−
14
g
(
n
−
2
)
+
7
g
(
n
−
3
)
.
{\displaystyle g(n)=7g(n-1)-14g(n-2)+7g(n-3).}
Así mismo[6]
2
b
2
−
a
2
=
7
b
R
,
2
c
2
−
b
2
=
7
c
R
,
2
a
2
−
c
2
=
−
7
a
R
.
{\displaystyle 2b^{2}-a^{2}={\sqrt {7}}bR,\quad 2c^{2}-b^{2}={\sqrt {7}}cR,\quad 2a^{2}-c^{2}=-{\sqrt {7}}aR.}
También se tiene que[4]
a
3
c
+
b
3
a
−
c
3
b
=
−
7
R
4
,
{\displaystyle a^{3}c+b^{3}a-c^{3}b=-7R^{4},}
a
4
c
−
b
4
a
+
c
4
b
=
7
7
R
5
,
{\displaystyle a^{4}c-b^{4}a+c^{4}b=7{\sqrt {7}}R^{5},}
a
11
c
3
+
b
11
a
3
−
c
11
b
3
=
−
7
3
17
R
14
.
{\displaystyle a^{11}c^{3}+b^{11}a^{3}-c^{11}b^{3}=-7^{3}17R^{14}.}
El exradio r a correspondiente al lado a es igual al radio de la circunferencia de los nueve puntos del triángulo heptagonal.[2] : p. 15
El triángulo órtico del triángulo heptagonal, con vértices en los pies de las alturas , es similar al triángulo heptagonal, con una relación de similitud de 1: 2. El triángulo heptagonal es el único triángulo obtuso que es similar a su triángulo órtico (el triángulo equilátero es el único agudo con esta propiedad).[2] : pp. 12–13
Las diversas identidades trigonométricas asociadas con el triángulo heptagonal incluyen:[2] : pp. 13–14 [5]
A
=
π
7
,
B
=
2
π
7
,
C
=
4
π
7
.
{\displaystyle A={\frac {\pi }{7}},\quad B={\frac {2\pi }{7}},\quad C={\frac {4\pi }{7}}.}
cos
A
=
b
/
2
a
,
cos
B
=
c
/
2
b
,
cos
C
=
−
a
/
2
c
,
{\displaystyle \cos A=b/2a,\quad \cos B=c/2b,\quad \cos C=-a/2c,}
[4] : Proposition 10
cos
A
cos
B
cos
C
=
−
1
8
,
{\displaystyle \cos A\cos B\cos C=-{\frac {1}{8}},}
cos
2
A
+
cos
2
B
+
cos
2
C
=
5
4
,
{\displaystyle \cos ^{2}A+\cos ^{2}B+\cos ^{2}C={\frac {5}{4}},}
cos
4
A
+
cos
4
B
+
cos
4
C
=
13
16
,
{\displaystyle \cos ^{4}A+\cos ^{4}B+\cos ^{4}C={\frac {13}{16}},}
cot
A
+
cot
B
+
cot
C
=
7
,
{\displaystyle \cot A+\cot B+\cot C={\sqrt {7}},}
cot
2
A
+
cot
2
B
+
cot
2
C
=
5
,
{\displaystyle \cot ^{2}A+\cot ^{2}B+\cot ^{2}C=5,}
csc
2
A
+
csc
2
B
+
csc
2
C
=
8
,
{\displaystyle \csc ^{2}A+\csc ^{2}B+\csc ^{2}C=8,}
csc
4
A
+
csc
4
B
+
csc
4
C
=
32
,
{\displaystyle \csc ^{4}A+\csc ^{4}B+\csc ^{4}C=32,}
sec
2
A
+
sec
2
B
+
sec
2
C
=
24
,
{\displaystyle \sec ^{2}A+\sec ^{2}B+\sec ^{2}C=24,}
sec
4
A
+
sec
4
B
+
sec
4
C
=
416
,
{\displaystyle \sec ^{4}A+\sec ^{4}B+\sec ^{4}C=416,}
sen
A
sen
B
sen
C
=
7
8
,
{\displaystyle \operatorname {sen} A\operatorname {sen} B\operatorname {sen} C={\frac {\sqrt {7}}{8}},}
sen
2
A
sen
2
B
sen
2
C
=
7
64
,
{\displaystyle \operatorname {sen} ^{2}A\operatorname {sen} ^{2}B\operatorname {sen} ^{2}C={\frac {7}{64}},}
sen
2
A
+
sen
2
B
+
sen
2
C
=
7
4
,
{\displaystyle \operatorname {sen} ^{2}A+\operatorname {sen} ^{2}B+\operatorname {sen} ^{2}C={\frac {7}{4}},}
sen
4
A
+
sen
4
B
+
sen
4
C
=
21
16
,
{\displaystyle \operatorname {sen} ^{4}A+\operatorname {sen} ^{4}B+\operatorname {sen} ^{4}C={\frac {21}{16}},}
tan
A
tan
B
tan
C
=
tan
A
+
tan
B
+
tan
C
=
−
7
,
{\displaystyle \tan A\tan B\tan C=\tan A+\tan B+\tan C=-{\sqrt {7}},}
tan
2
A
+
tan
2
B
+
tan
2
C
=
21.
{\displaystyle \tan ^{2}A+\tan ^{2}B+\tan ^{2}C=21.}
La ecuación cúbica
64
y
3
−
112
y
2
+
56
y
−
7
=
0
{\displaystyle 64y^{3}-112y^{2}+56y-7=0}
tiene soluciones[2] : p. 14
sen
2
π
7
,
sen
2
2
π
7
,
{\displaystyle \operatorname {sen} ^{2}{\frac {\pi }{7}},\operatorname {sen} ^{2}{\frac {2\pi }{7}},}
y
sen
2
4
π
7
,
{\displaystyle \operatorname {sen} ^{2}{\frac {4\pi }{7}},}
que son los senos al cuadrado de los ángulos del triángulo.
La solución positiva de la ecuación cúbica
x
3
+
x
2
−
2
x
−
1
=
0
{\displaystyle x^{3}+x^{2}-2x-1=0}
es igual
2
cos
2
π
7
,
{\displaystyle 2\cos {\frac {2\pi }{7}},}
que es el doble del coseno de uno de los ángulos del triángulo.[7] : p. 186–187
Sen (2π/7), sen (4π/7) y sen (8π/7) son las raíces de[4]
x
3
−
7
2
x
2
+
7
8
=
0.
{\displaystyle x^{3}-{\frac {\sqrt {7}}{2}}x^{2}+{\frac {\sqrt {7}}{8}}=0.}
También se tiene que:[6]
sen
A
−
sen
B
−
sen
C
=
−
7
2
,
{\displaystyle \operatorname {sen} A-\operatorname {sen} B-\operatorname {sen} C=-{\frac {\sqrt {7}}{2}},}
sen
A
sen
B
−
sen
B
sen
C
+
sen
C
sen
A
=
0
,
{\displaystyle \operatorname {sen} A\operatorname {sen} B-\operatorname {sen} B\operatorname {sen} C+\operatorname {sen} C\operatorname {sen} A=0,}
sen
A
sen
B
sen
C
=
7
8
.
{\displaystyle \operatorname {sen} A\operatorname {sen} B\operatorname {sen} C={\frac {\sqrt {7}}{8}}.}
−
sen
A
,
sen
B
,
sen
C
son las raíces de
x
3
−
7
2
x
2
+
7
8
=
0.
{\displaystyle -\operatorname {sen} A,\operatorname {sen} B,\operatorname {sen} C{\text{ son las raíces de }}x^{3}-{\frac {\sqrt {7}}{2}}x^{2}+{\frac {\sqrt {7}}{8}}=0.}
Para un entero n , sea
S
(
n
)
=
(
−
sen
A
)
n
+
sen
n
B
+
sen
n
C
.
{\displaystyle S(n)=(-\operatorname {sen} {A})^{n}+\operatorname {sen} ^{n}{B}+\operatorname {sen} ^{n}{C}.}
Para n = 0, ..., 20,
S
(
n
)
=
3
,
7
2
,
7
2
2
,
7
2
,
7
⋅
3
2
4
,
7
7
2
4
,
7
⋅
5
2
5
,
7
2
7
2
7
,
7
2
⋅
5
2
8
,
7
⋅
25
7
2
9
,
7
2
⋅
9
2
9
,
7
2
⋅
13
7
2
11
,
{\displaystyle S(n)=3,{\frac {\sqrt {7}}{2}},{\frac {7}{2^{2}}},{\frac {\sqrt {7}}{2}},{\frac {7\cdot 3}{2^{4}}},{\frac {7{\sqrt {7}}}{2^{4}}},{\frac {7\cdot 5}{2^{5}}},{\frac {7^{2}{\sqrt {7}}}{2^{7}}},{\frac {7^{2}\cdot 5}{2^{8}}},{\frac {7\cdot 25{\sqrt {7}}}{2^{9}}},{\frac {7^{2}\cdot 9}{2^{9}}},{\frac {7^{2}\cdot 13{\sqrt {7}}}{2^{11}}},}
7
2
⋅
33
2
11
,
7
2
⋅
3
7
2
9
,
7
4
⋅
5
2
14
,
7
2
⋅
179
7
2
15
,
7
3
⋅
131
2
16
,
7
3
⋅
3
7
2
12
,
7
3
⋅
493
2
18
,
7
3
⋅
181
7
2
18
,
7
5
⋅
19
2
19
.
{\displaystyle {\frac {7^{2}\cdot 33}{2^{11}}},{\frac {7^{2}\cdot 3{\sqrt {7}}}{2^{9}}},{\frac {7^{4}\cdot 5}{2^{14}}},{\frac {7^{2}\cdot 179{\sqrt {7}}}{2^{15}}},{\frac {7^{3}\cdot 131}{2^{16}}},{\frac {7^{3}\cdot 3{\sqrt {7}}}{2^{12}}},{\frac {7^{3}\cdot 493}{2^{18}}},{\frac {7^{3}\cdot 181{\sqrt {7}}}{2^{18}}},{\frac {7^{5}\cdot 19}{2^{19}}}.}
Para n = 0, -1,, ..-20,
S
(
n
)
=
3
,
0
,
2
3
,
−
2
3
⋅
3
7
7
,
2
5
,
−
2
5
⋅
5
7
7
,
2
6
⋅
17
7
,
−
2
7
7
,
2
9
⋅
11
7
,
−
2
10
⋅
33
7
7
2
,
2
10
⋅
29
7
,
−
2
14
⋅
11
7
7
2
,
2
12
⋅
269
7
2
,
{\displaystyle S(n)=3,0,2^{3},-{\frac {2^{3}\cdot 3{\sqrt {7}}}{7}},2^{5},-{\frac {2^{5}\cdot 5{\sqrt {7}}}{7}},{\frac {2^{6}\cdot 17}{7}},-2^{7}{\sqrt {7}},{\frac {2^{9}\cdot 11}{7}},-{\frac {2^{10}\cdot 33{\sqrt {7}}}{7^{2}}},{\frac {2^{10}\cdot 29}{7}},-{\frac {2^{14}\cdot 11{\sqrt {7}}}{7^{2}}},{\frac {2^{12}\cdot 269}{7^{2}}},}
−
2
13
⋅
117
7
7
2
,
2
14
⋅
51
7
,
−
2
21
⋅
17
7
7
3
,
2
17
⋅
237
7
2
,
−
2
17
⋅
1445
7
7
3
,
2
19
⋅
2203
7
3
,
−
2
19
⋅
1919
7
7
3
,
2
20
⋅
5851
7
3
.
{\displaystyle -{\frac {2^{13}\cdot 117{\sqrt {7}}}{7^{2}}},{\frac {2^{14}\cdot 51}{7}},-{\frac {2^{21}\cdot 17{\sqrt {7}}}{7^{3}}},{\frac {2^{17}\cdot 237}{7^{2}}},-{\frac {2^{17}\cdot 1445{\sqrt {7}}}{7^{3}}},{\frac {2^{19}\cdot 2203}{7^{3}}},-{\frac {2^{19}\cdot 1919{\sqrt {7}}}{7^{3}}},{\frac {2^{20}\cdot 5851}{7^{3}}}.}
−
cos
A
,
cos
B
,
cos
C
son las raíces de
x
3
+
1
2
x
2
−
1
2
x
−
1
8
=
0.
{\displaystyle -\cos A,\cos B,\cos C{\text{ son las raíces de }}x^{3}+{\frac {1}{2}}x^{2}-{\frac {1}{2}}x-{\frac {1}{8}}=0.}
Para cualquier entero n
C
(
n
)
=
(
−
cos
A
)
n
+
cos
n
B
+
cos
n
C
.
{\displaystyle C(n)=(-\cos {A})^{n}+\cos ^{n}{B}+\cos ^{n}{C}.}
Para n = 0, 1, ... 10,
C
(
n
)
=
3
,
−
1
2
,
5
4
,
−
1
2
,
13
16
,
−
1
2
,
19
32
,
−
57
128
,
117
256
,
−
193
512
,
185
512
,
.
.
.
{\displaystyle C(n)=3,-{\frac {1}{2}},{\frac {5}{4}},-{\frac {1}{2}},{\frac {13}{16}},-{\frac {1}{2}},{\frac {19}{32}},-{\frac {57}{128}},{\frac {117}{256}},-{\frac {193}{512}},{\frac {185}{512}},...}
C
(
−
n
)
=
3
,
−
4
,
24
,
−
88
,
416
,
−
1824
,
8256
,
−
36992
,
166400
,
−
747520
,
3359744
,
.
.
.
{\displaystyle C(-n)=3,-4,24,-88,416,-1824,8256,-36992,166400,-747520,3359744,...}
tan
A
,
tan
B
,
tan
C
son las raíces de
x
3
+
7
x
2
−
7
x
+
7
=
0.
{\displaystyle \tan A,\tan B,\tan C{\text{ son las raíces de }}x^{3}+{\sqrt {7}}x^{2}-7x+{\sqrt {7}}=0.}
tan
2
A
,
tan
2
B
,
tan
2
C
son las raíces de
x
3
−
21
x
2
+
35
x
−
7
=
0.
{\displaystyle \tan ^{2}A,\tan ^{2}B,\tan ^{2}C{\text{ son las raíces de }}x^{3}-21x^{2}+35x-7=0.}
Para un entero n , sea
T
(
n
)
=
tan
n
A
+
tan
n
B
+
tan
n
C
.
{\displaystyle T(n)=\tan ^{n}{A}+\tan ^{n}{B}+\tan ^{n}{C}.}
Para n = 0, 1, ... 10,
T
(
n
)
=
3
,
−
7
,
7
⋅
3
,
−
31
7
,
7
⋅
53
,
−
7
⋅
87
7
,
7
⋅
1011
,
−
7
2
⋅
239
7
,
7
2
⋅
2771
,
−
7
⋅
32119
7
,
7
2
⋅
53189
,
{\displaystyle T(n)=3,-{\sqrt {7}},7\cdot 3,-31{\sqrt {7}},7\cdot 53,-7\cdot 87{\sqrt {7}},7\cdot 1011,-7^{2}\cdot 239{\sqrt {7}},7^{2}\cdot 2771,-7\cdot 32119{\sqrt {7}},7^{2}\cdot 53189,}
T
(
−
n
)
=
3
,
7
,
5
,
25
7
7
,
19
,
103
7
7
,
563
7
,
7
⋅
9
7
,
2421
7
,
13297
7
7
2
,
10435
7
,
.
.
.
{\displaystyle T(-n)=3,{\sqrt {7}},5,{\frac {25{\sqrt {7}}}{7}},19,{\frac {103{\sqrt {7}}}{7}},{\frac {563}{7}},7\cdot 9{\sqrt {7}},{\frac {2421}{7}},{\frac {13297{\sqrt {7}}}{7^{2}}},{\frac {10435}{7}},...}
También se tiene que[6] [8]
tan
A
−
4
sen
B
=
−
7
,
{\displaystyle \tan A-4\operatorname {sen} B=-{\sqrt {7}},}
tan
B
−
4
sen
C
=
−
7
,
{\displaystyle \tan B-4\operatorname {sen} C=-{\sqrt {7}},}
tan
C
+
4
sen
A
=
−
7
.
{\displaystyle \tan C+4\operatorname {sen} A=-{\sqrt {7}}.}
Así mismo[4]
cot
2
A
=
1
−
2
tan
C
7
,
{\displaystyle \cot ^{2}A=1-{\frac {2\tan C}{\sqrt {7}}},}
cot
2
B
=
1
−
2
tan
A
7
,
{\displaystyle \cot ^{2}B=1-{\frac {2\tan A}{\sqrt {7}}},}
cot
2
C
=
1
−
2
tan
B
7
.
{\displaystyle \cot ^{2}C=1-{\frac {2\tan B}{\sqrt {7}}}.}
También se tiene que[4]
cos
A
=
−
1
2
+
4
7
sen
3
C
,
{\displaystyle \cos A=-{\frac {1}{2}}+{\frac {4}{\sqrt {7}}}\operatorname {sen} ^{3}C,}
cos
2
A
=
3
4
+
2
7
sen
3
A
,
{\displaystyle \cos ^{2}A={\frac {3}{4}}+{\frac {2}{\sqrt {7}}}\operatorname {sen} ^{3}A,}
cot
A
=
3
7
+
4
7
cos
B
,
{\displaystyle \cot A={\frac {3}{\sqrt {7}}}+{\frac {4}{\sqrt {7}}}\cos B,}
cot
2
A
=
3
+
8
7
sen
A
,
{\displaystyle \cot ^{2}A=3+{\frac {8}{\sqrt {7}}}\operatorname {sen} A,}
cot
A
=
7
+
8
7
sen
2
B
,
{\displaystyle \cot A={\sqrt {7}}+{\frac {8}{\sqrt {7}}}\operatorname {sen} ^{2}B,}
csc
3
A
=
−
6
7
+
2
7
tan
2
C
,
{\displaystyle \csc ^{3}A=-{\frac {6}{\sqrt {7}}}+{\frac {2}{\sqrt {7}}}\tan ^{2}C,}
sec
A
=
2
+
4
cos
C
,
{\displaystyle \sec A=2+4\cos C,}
sec
A
=
6
−
8
sen
2
B
,
{\displaystyle \sec A=6-8\operatorname {sen} ^{2}B,}
sec
A
=
4
−
16
7
sen
3
B
,
{\displaystyle \sec A=4-{\frac {16}{\sqrt {7}}}\operatorname {sen} ^{3}B,}
sen
2
A
=
1
2
+
1
2
cos
B
,
{\displaystyle \operatorname {sen} ^{2}A={\frac {1}{2}}+{\frac {1}{2}}\cos B,}
sen
3
A
=
−
7
8
+
7
4
cos
B
,
{\displaystyle \operatorname {sen} ^{3}A=-{\frac {\sqrt {7}}{8}}+{\frac {\sqrt {7}}{4}}\cos B,}
También se tiene que[9]
sen
3
B
sen
C
−
sen
3
C
sen
A
−
sen
3
A
sen
B
=
0
,
{\displaystyle \operatorname {sen} ^{3}B\operatorname {sen} C-\operatorname {sen} ^{3}C\operatorname {sen} A-\operatorname {sen} ^{3}A\operatorname {sen} B=0,}
sen
B
sen
3
C
−
sen
C
sen
3
A
−
sen
A
sen
3
B
=
7
2
4
,
{\displaystyle \operatorname {sen} B\operatorname {sen} ^{3}C-\operatorname {sen} C\operatorname {sen} ^{3}A-\operatorname {sen} A\operatorname {sen} ^{3}B={\frac {7}{2^{4}}},}
sen
4
B
sen
C
−
sen
4
C
sen
A
+
sen
4
A
sen
B
=
0
,
{\displaystyle \operatorname {sen} ^{4}B\operatorname {sen} C-\operatorname {sen} ^{4}C\operatorname {sen} A+\operatorname {sen} ^{4}A\operatorname {sen} B=0,}
sen
B
sen
4
C
+
sen
C
sen
4
A
−
sen
A
sen
4
B
=
7
7
2
5
,
{\displaystyle \operatorname {sen} B\operatorname {sen} ^{4}C+\operatorname {sen} C\operatorname {sen} ^{4}A-\operatorname {sen} A\operatorname {sen} ^{4}B={\frac {7{\sqrt {7}}}{2^{5}}},}
sen
11
B
sen
3
C
−
sen
11
C
sen
3
A
−
sen
11
A
sen
3
B
=
0
,
{\displaystyle \operatorname {sen} ^{11}B\operatorname {sen} ^{3}C-\operatorname {sen} ^{11}C\operatorname {sen} ^{3}A-\operatorname {sen} ^{11}A\operatorname {sen} ^{3}B=0,}
sen
3
B
sen
11
C
−
sen
3
C
sen
11
A
−
sen
3
A
sen
11
B
=
7
3
⋅
17
2
14
.
{\displaystyle \operatorname {sen} ^{3}B\operatorname {sen} ^{11}C-\operatorname {sen} ^{3}C\operatorname {sen} ^{11}A-\operatorname {sen} ^{3}A\operatorname {sen} ^{11}B={\frac {7^{3}\cdot 17}{2^{14}}}.}
También se cumplen identidades de tipo Ramanujan,[10]
2
sen
(
2
π
7
)
3
+
2
sen
(
4
π
7
)
3
+
2
sen
(
8
π
7
)
3
=
{\displaystyle {\sqrt[{3}]{2\operatorname {sen}({\frac {2\pi }{7}})}}+{\sqrt[{3}]{2\operatorname {sen}({\frac {4\pi }{7}})}}+{\sqrt[{3}]{2\operatorname {sen}({\frac {8\pi }{7}})}}=}
.......
(
−
7
18
)
−
7
3
+
6
+
3
(
5
−
3
7
3
3
+
4
−
3
7
3
3
)
3
{\displaystyle {\text{.......}}\left(-{\sqrt[{18}]{7}}\right){\sqrt[{3}]{-{\sqrt[{3}]{7}}+6+3\left({\sqrt[{3}]{5-3{\sqrt[{3}]{7}}}}+{\sqrt[{3}]{4-3{\sqrt[{3}]{7}}}}\right)}}}
1
2
sen
(
2
π
7
)
3
+
1
2
sen
(
4
π
7
)
3
+
1
2
sen
(
8
π
7
)
3
=
{\displaystyle {\frac {1}{\sqrt[{3}]{2\operatorname {sen}({\frac {2\pi }{7}})}}}+{\frac {1}{\sqrt[{3}]{2\operatorname {sen}({\frac {4\pi }{7}})}}}+{\frac {1}{\sqrt[{3}]{2\operatorname {sen}({\frac {8\pi }{7}})}}}=}
.......
(
−
1
7
18
)
6
+
3
(
5
−
3
7
3
3
+
4
−
3
7
3
3
)
3
{\displaystyle {\text{.......}}\left(-{\frac {1}{\sqrt[{18}]{7}}}\right){\sqrt[{3}]{6+3\left({\sqrt[{3}]{5-3{\sqrt[{3}]{7}}}}+{\sqrt[{3}]{4-3{\sqrt[{3}]{7}}}}\right)}}}
4
sen
2
(
2
π
7
)
3
+
4
sen
2
(
4
π
7
)
3
+
4
sen
2
(
8
π
7
)
3
=
{\displaystyle {\sqrt[{3}]{4\operatorname {sen} ^{2}({\frac {2\pi }{7}})}}+{\sqrt[{3}]{4\operatorname {sen} ^{2}({\frac {4\pi }{7}})}}+{\sqrt[{3}]{4\operatorname {sen} ^{2}({\frac {8\pi }{7}})}}=}
.......
(
49
18
)
49
3
+
6
+
3
(
12
+
3
(
49
3
+
2
7
3
)
3
+
11
+
3
(
49
3
+
2
7
3
)
3
)
3
{\displaystyle {\text{.......}}\left({\sqrt[{18}]{49}}\right){\sqrt[{3}]{{\sqrt[{3}]{49}}+6+3\left({\sqrt[{3}]{12+3({\sqrt[{3}]{49}}+2{\sqrt[{3}]{7}})}}+{\sqrt[{3}]{11+3({\sqrt[{3}]{49}}+2{\sqrt[{3}]{7}})}}\right)}}}
1
4
sen
2
(
2
π
7
)
3
+
1
4
sen
2
(
4
π
7
)
3
+
1
4
sen
2
(
8
π
7
)
3
=
{\displaystyle {\frac {1}{\sqrt[{3}]{4\operatorname {sen} ^{2}({\frac {2\pi }{7}})}}}+{\frac {1}{\sqrt[{3}]{4\operatorname {sen} ^{2}({\frac {4\pi }{7}})}}}+{\frac {1}{\sqrt[{3}]{4\operatorname {sen} ^{2}({\frac {8\pi }{7}})}}}=}
.......
(
1
49
18
)
2
7
3
+
6
+
3
(
12
+
3
(
49
3
+
2
7
3
)
3
+
11
+
3
(
49
3
+
2
7
3
)
3
)
3
{\displaystyle {\text{.......}}\left({\frac {1}{\sqrt[{18}]{49}}}\right){\sqrt[{3}]{2{\sqrt[{3}]{7}}+6+3\left({\sqrt[{3}]{12+3({\sqrt[{3}]{49}}+2{\sqrt[{3}]{7}})}}+{\sqrt[{3}]{11+3({\sqrt[{3}]{49}}+2{\sqrt[{3}]{7}})}}\right)}}}
2
cos
(
2
π
7
)
3
+
2
cos
(
4
π
7
)
3
+
2
cos
(
8
π
7
)
3
=
5
−
3
7
3
3
{\displaystyle {\sqrt[{3}]{2\cos({\frac {2\pi }{7}})}}+{\sqrt[{3}]{2\cos({\frac {4\pi }{7}})}}+{\sqrt[{3}]{2\cos({\frac {8\pi }{7}})}}={\sqrt[{3}]{5-3{\sqrt[{3}]{7}}}}}
1
2
cos
(
2
π
7
)
3
+
1
2
cos
(
4
π
7
)
3
+
1
2
cos
(
8
π
7
)
3
=
4
−
3
7
3
3
{\displaystyle {\frac {1}{\sqrt[{3}]{2\cos({\frac {2\pi }{7}})}}}+{\frac {1}{\sqrt[{3}]{2\cos({\frac {4\pi }{7}})}}}+{\frac {1}{\sqrt[{3}]{2\cos({\frac {8\pi }{7}})}}}={\sqrt[{3}]{4-3{\sqrt[{3}]{7}}}}}
4
cos
2
(
2
π
7
)
3
+
4
cos
2
(
4
π
7
)
3
+
4
cos
2
(
8
π
7
)
3
=
11
+
3
(
2
7
3
+
49
3
)
3
{\displaystyle {\sqrt[{3}]{4\cos ^{2}({\frac {2\pi }{7}})}}+{\sqrt[{3}]{4\cos ^{2}({\frac {4\pi }{7}})}}+{\sqrt[{3}]{4\cos ^{2}({\frac {8\pi }{7}})}}={\sqrt[{3}]{11+3(2{\sqrt[{3}]{7}}+{\sqrt[{3}]{49}})}}}
1
4
cos
2
(
2
π
7
)
3
+
1
4
cos
2
(
4
π
7
)
3
+
1
4
cos
2
(
8
π
7
)
3
=
12
+
3
(
2
7
3
+
49
3
)
3
{\displaystyle {\frac {1}{\sqrt[{3}]{4\cos ^{2}({\frac {2\pi }{7}})}}}+{\frac {1}{\sqrt[{3}]{4\cos ^{2}({\frac {4\pi }{7}})}}}+{\frac {1}{\sqrt[{3}]{4\cos ^{2}({\frac {8\pi }{7}})}}}={\sqrt[{3}]{12+3(2{\sqrt[{3}]{7}}+{\sqrt[{3}]{49}})}}}
tan
(
2
π
7
)
3
+
tan
(
4
π
7
)
3
+
tan
(
8
π
7
)
3
=
{\displaystyle {\sqrt[{3}]{\tan({\frac {2\pi }{7}})}}+{\sqrt[{3}]{\tan({\frac {4\pi }{7}})}}+{\sqrt[{3}]{\tan({\frac {8\pi }{7}})}}=}
.......
(
−
7
18
)
7
3
+
6
+
3
(
5
+
3
(
7
3
−
49
3
)
3
+
−
3
+
3
(
7
3
−
49
3
)
3
)
3
{\displaystyle {\text{.......}}\left(-{\sqrt[{18}]{7}}\right){\sqrt[{3}]{{\sqrt[{3}]{7}}+6+3\left({\sqrt[{3}]{5+3({\sqrt[{3}]{7}}-{\sqrt[{3}]{49}})}}+{\sqrt[{3}]{-3+3({\sqrt[{3}]{7}}-{\sqrt[{3}]{49}})}}\right)}}}
1
tan
(
2
π
7
)
3
+
1
tan
(
4
π
7
)
3
+
1
tan
(
8
π
7
)
3
=
{\displaystyle {\frac {1}{\sqrt[{3}]{\tan({\frac {2\pi }{7}})}}}+{\frac {1}{\sqrt[{3}]{\tan({\frac {4\pi }{7}})}}}+{\frac {1}{\sqrt[{3}]{\tan({\frac {8\pi }{7}})}}}=}
.......
(
−
1
7
18
)
−
49
3
+
6
+
3
(
5
+
3
(
7
3
−
49
3
)
3
+
−
3
+
3
(
7
3
−
49
3
)
3
)
3
{\displaystyle {\text{.......}}\left(-{\frac {1}{\sqrt[{18}]{7}}}\right){\sqrt[{3}]{-{\sqrt[{3}]{49}}+6+3\left({\sqrt[{3}]{5+3({\sqrt[{3}]{7}}-{\sqrt[{3}]{49}})}}+{\sqrt[{3}]{-3+3({\sqrt[{3}]{7}}-{\sqrt[{3}]{49}})}}\right)}}}
tan
2
(
2
π
7
)
3
+
tan
2
(
4
π
7
)
3
+
tan
2
(
8
π
7
)
3
=
{\displaystyle {\sqrt[{3}]{\tan ^{2}({\frac {2\pi }{7}})}}+{\sqrt[{3}]{\tan ^{2}({\frac {4\pi }{7}})}}+{\sqrt[{3}]{\tan ^{2}({\frac {8\pi }{7}})}}=}
.......
(
49
18
)
3
49
3
+
6
+
3
(
89
+
3
(
3
49
3
+
5
7
3
)
3
+
25
+
3
(
3
49
3
+
5
7
3
)
3
)
3
{\displaystyle {\text{.......}}\left({\sqrt[{18}]{49}}\right){\sqrt[{3}]{3{\sqrt[{3}]{49}}+6+3\left({\sqrt[{3}]{89+3(3{\sqrt[{3}]{49}}+5{\sqrt[{3}]{7}})}}+{\sqrt[{3}]{25+3(3{\sqrt[{3}]{49}}+5{\sqrt[{3}]{7}})}}\right)}}}
1
tan
2
(
2
π
7
)
3
+
1
tan
2
(
4
π
7
)
3
+
1
tan
2
(
8
π
7
)
3
=
{\displaystyle {\frac {1}{\sqrt[{3}]{\tan ^{2}({\frac {2\pi }{7}})}}}+{\frac {1}{\sqrt[{3}]{\tan ^{2}({\frac {4\pi }{7}})}}}+{\frac {1}{\sqrt[{3}]{\tan ^{2}({\frac {8\pi }{7}})}}}=}
.......
(
1
49
18
)
5
7
3
+
6
+
3
(
89
+
3
(
3
49
3
+
5
7
3
)
3
+
25
+
3
(
3
49
3
+
5
7
3
)
3
)
3
{\displaystyle {\text{.......}}\left({\frac {1}{\sqrt[{18}]{49}}}\right){\sqrt[{3}]{5{\sqrt[{3}]{7}}+6+3\left({\sqrt[{3}]{89+3(3{\sqrt[{3}]{49}}+5{\sqrt[{3}]{7}})}}+{\sqrt[{3}]{25+3(3{\sqrt[{3}]{49}}+5{\sqrt[{3}]{7}})}}\right)}}}
También se tiene que[9]
cos
(
2
π
7
)
/
cos
(
4
π
7
)
3
+
cos
(
4
π
7
)
/
cos
(
8
π
7
)
3
+
cos
(
8
π
7
)
/
cos
(
2
π
7
)
3
=
−
7
3
.
{\displaystyle {\sqrt[{3}]{\cos({\frac {2\pi }{7}})/\cos({\frac {4\pi }{7}})}}+{\sqrt[{3}]{\cos({\frac {4\pi }{7}})/\cos({\frac {8\pi }{7}})}}+{\sqrt[{3}]{\cos({\frac {8\pi }{7}})/\cos({\frac {2\pi }{7}})}}=-{\sqrt[{3}]{7}}.}
cos
(
4
π
7
)
/
cos
(
2
π
7
)
3
+
cos
(
8
π
7
)
/
cos
(
4
π
7
)
3
+
cos
(
2
π
7
)
/
cos
(
8
π
7
)
3
=
0.
{\displaystyle {\sqrt[{3}]{\cos({\frac {4\pi }{7}})/\cos({\frac {2\pi }{7}})}}+{\sqrt[{3}]{\cos({\frac {8\pi }{7}})/\cos({\frac {4\pi }{7}})}}+{\sqrt[{3}]{\cos({\frac {2\pi }{7}})/\cos({\frac {8\pi }{7}})}}=0.}
2
sen
(
2
π
7
3
+
2
sen
(
4
π
7
3
+
2
sen
(
8
π
7
3
=
(
−
7
18
)
−
7
3
+
6
+
3
(
5
−
3
7
3
3
+
4
−
3
7
3
3
)
3
{\displaystyle {\sqrt[{3}]{2\operatorname {sen}({2\pi }{7}}}+{\sqrt[{3}]{2\operatorname {sen}({4\pi }{7}}}+{\sqrt[{3}]{2\operatorname {sen}({8\pi }{7}}}=\left(-{\sqrt[{18}]{7}}\right){\sqrt[{3}]{-{\sqrt[{3}]{7}}+6+3\left({\sqrt[{3}]{5-3{\sqrt[{3}]{7}}}}+{\sqrt[{3}]{4-3{\sqrt[{3}]{7}}}}\right)}}}
cos
4
(
4
π
7
)
/
cos
(
2
π
7
)
3
+
cos
4
(
8
π
7
)
/
cos
(
4
π
7
)
3
+
cos
4
(
2
π
7
)
/
cos
(
8
π
7
)
3
=
−
49
3
/
2.
{\displaystyle {\sqrt[{3}]{\cos ^{4}({\frac {4\pi }{7}})/\cos({\frac {2\pi }{7}})}}+{\sqrt[{3}]{\cos ^{4}({\frac {8\pi }{7}})/\cos({\frac {4\pi }{7}})}}+{\sqrt[{3}]{\cos ^{4}({\frac {2\pi }{7}})/\cos({\frac {8\pi }{7}})}}=-{\sqrt[{3}]{49}}/2.}
cos
5
(
2
π
7
)
/
cos
2
(
4
π
7
)
3
+
cos
5
(
4
π
7
)
/
cos
2
(
8
π
7
)
3
+
cos
5
(
8
π
7
)
/
cos
2
(
2
π
7
)
3
=
0.
{\displaystyle {\sqrt[{3}]{\cos ^{5}({\frac {2\pi }{7}})/\cos ^{2}({\frac {4\pi }{7}})}}+{\sqrt[{3}]{\cos ^{5}({\frac {4\pi }{7}})/\cos ^{2}({\frac {8\pi }{7}})}}+{\sqrt[{3}]{\cos ^{5}({\frac {8\pi }{7}})/\cos ^{2}({\frac {2\pi }{7}})}}=0.}
cos
5
(
4
π
7
)
/
cos
2
(
2
π
7
)
3
+
cos
5
(
8
π
7
)
/
cos
2
(
4
π
7
)
3
+
cos
5
(
2
π
7
)
/
cos
2
(
9
π
7
)
3
=
−
3
∗
7
3
/
2.
{\displaystyle {\sqrt[{3}]{\cos ^{5}({\frac {4\pi }{7}})/\cos ^{2}({\frac {2\pi }{7}})}}+{\sqrt[{3}]{\cos ^{5}({\frac {8\pi }{7}})/\cos ^{2}({\frac {4\pi }{7}})}}+{\sqrt[{3}]{\cos ^{5}({\frac {2\pi }{7}})/\cos ^{2}({\frac {9\pi }{7}})}}=-3*{\sqrt[{3}]{7}}/2.}
cos
14
(
2
π
7
)
/
cos
5
(
4
π
7
)
3
+
cos
14
(
4
π
7
)
/
cos
5
(
8
π
7
)
3
+
cos
14
(
8
π
7
)
/
cos
5
(
2
π
7
3
=
0.
{\displaystyle {\sqrt[{3}]{\cos ^{14}({\frac {2\pi }{7}})/\cos ^{5}({\frac {4\pi }{7}})}}+{\sqrt[{3}]{\cos ^{14}({\frac {4\pi }{7}})/\cos ^{5}({\frac {8\pi }{7}})}}+{\sqrt[{3}]{\cos ^{14}({\frac {8\pi }{7}})/\cos ^{5}({\frac {2\pi }{7}}}}=0.}
cos
14
(
4
π
7
)
/
cos
5
(
2
π
7
)
3
+
cos
14
(
8
π
7
)
/
cos
5
(
4
π
7
)
3
+
cos
14
(
2
π
7
)
/
cos
5
(
8
π
7
)
3
=
−
61
∗
7
3
/
8.
{\displaystyle {\sqrt[{3}]{\cos ^{14}({\frac {4\pi }{7}})/\cos ^{5}({\frac {2\pi }{7}})}}+{\sqrt[{3}]{\cos ^{14}({\frac {8\pi }{7}})/\cos ^{5}({\frac {4\pi }{7}})}}+{\sqrt[{3}]{\cos ^{14}({\frac {2\pi }{7}})/\cos ^{5}({\frac {8\pi }{7}})}}=-61*{\sqrt[{3}]{7}}/8.}
Leon Bankoff and Jack Garfunkel, "The heptagonal triangle", Mathematics Magazine 46 (1), January 1973, 7–19.
Wang, Kai. “Heptagonal Triangle and Trigonometric Identities”, Forum Geometricorum 19, 2019, 29–38.
Roman Witula and Damian Slota, New Ramanujan-Type Formulas and Quasi-Fibonacci Numbers of Order 7, Journal of Integer Sequences, Vol. 10 (2007).