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. 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,}
4
π
/
7
,
{\displaystyle 4\pi /7,}
Puntos clave [ editar ] 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
Relaciones de distancias [ editar ] 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 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}}}
[ 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.}
Las alturas h a h b h c
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}}
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.
Bisectrices [ editar ] Se tienen las siguientes propiedades de las bisectrices
w
A
,
w
B
,
{\displaystyle w_{A},w_{B},}
w
C
{\displaystyle w_{C}}
A, B y C respectivamente:[ 2] : p. 16
w
A
=
b
+
c
,
{\displaystyle w_{A}=b+c,}
w
B
=
c
−
a
,
{\displaystyle w_{B}=c-a,}
w
C
=
b
−
a
.
{\displaystyle w_{C}=b-a.}
Circunradio, inradio y exinradios [ editar ] 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}}}
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}}}
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}}}
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}}}
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)}
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)}
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 a es igual al radio de la circunferencia de los nueve puntos del triángulo heptagonal.[ 2] : p. 15
Triángulo órtico [ editar ] 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
Propiedades trigonométricas [ editar ] 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}},}
sen
2
4
π
7
,
{\displaystyle \operatorname {sen} ^{2}{\frac {4\pi }{7}},}
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}},}
[ 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.}
Referencias [ editar ]
↑ a b Paul Yiu, "Heptagonal Triangles and Their Companions", Forum Geometricorum http://forumgeom.fau.edu/FG2009volume9/FG200912.pdf
↑ a b c d e f g h i j k l m n ñ o p Leon Bankoff and Jack Garfunkel, "The heptagonal triangle", Mathematics Magazine
↑ a b Abdilkadir Altintas, "Some Collinearities in the Heptagonal Triangle", Forum Geometricorum http://forumgeom.fau.edu/FG2016volume16/FG201630.pdf
↑ a b c d e f g h i Wang, Kai. “Heptagonal Triangle and Trigonometric Identities”, Forum Geometricorum 19, 2019, 29–38.
↑ a b c Weisstein, Eric W. "Heptagonal Triangle." From MathWorld--A Wolfram Web Resource. http://mathworld.wolfram.com/HeptagonalTriangle.html
↑ a b c d e Wang, Kai. https://www.researchgate.net/publication/327825153_Trigonometric_Properties_For_Heptagonal_Triangle
↑ Gleason, Andrew Mattei (March 1988). «Angle trisection, the heptagon, and the triskaidecagon» . The American Mathematical Monthly 95 (3): 185-194. doi :10.2307/2323624 el original el 19 de diciembre de 2015. ↑ Victor H. Moll, An elementary trigonometric equation, https://arxiv.org/abs/0709.3755 , 2007
↑ a b Kai Wang,
https://www.researchgate.net/publication/336813631_Topics_of_Ramanujan_type_identities_for_PI7
↑ Roman Witula and Damian Slota, New Ramanujan-Type Formulas and Quasi-Fibonacci Numbers of Order 7, Journal of Integer Sequences, Vol. 10 (2007).
![{\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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/63107bd555aab31d4dbd45b3bea63fcb044aeb43)
![{\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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b4adbd5db36fdb12f6fcfbde16edc40037bf88b6)
![{\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}})}}=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/944fa265cc4a9e01179032bc93e04f3bad74375c)
![{\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)}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c8a93053ce8ad0eb9ddcb3cc47d03aeb44597eb1)
![{\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}})}}}=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f48096e10b7da3bcd8754842b04dd84755e91613)
![{\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)}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ee8a1e5f7a4a7646390f002f725c8937e8282894)
![{\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}})}}=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/64d8d1922a51de41472b406b267ec8d1edb9e0f4)
![{\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)}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/aba2a19e87ed078ce46a84894486a90ae81b9a1f)
![{\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}})}}}=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ad5bf445b03e9e6b31f71598805dccbf20a7e195)
![{\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)}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ba0013568e9972e6aae1aec3e8c85493e626fdcd)
![{\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}}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b314c223d0be63a0f72ff996b5fd8fcc12d7e7f0)
![{\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}}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b166c9b5596c43166f06dae47417e000f8f47807)
![{\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}})}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/195af596d79a11379a62885b6934e92f9e028386)
![{\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}})}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9b69e7f053c8c3e7301b9043df9d030c894ffe8b)
![{\displaystyle {\sqrt[{3}]{\tan({\frac {2\pi }{7}})}}+{\sqrt[{3}]{\tan({\frac {4\pi }{7}})}}+{\sqrt[{3}]{\tan({\frac {8\pi }{7}})}}=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9a52f81bb6d9f9c7d2edf6a15127b6509d173c2a)
![{\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)}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d40e8336542726b021dbc22dfc8df6f1f0360223)
![{\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}})}}}=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/93fae279c3c18761c02980f1de1f56add9b85ac2)
![{\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)}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/fc97fe95f2898ee704610d68aa4b316420511852)
![{\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}})}}=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5fb410a07939f6d720b679c008c5e43729fc9998)
![{\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)}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8c44c9c7171b3fe668d604b31cdd4770b74662fd)
![{\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}})}}}=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2ae013c02c7bfa040e9e92343e56a699900fcce0)
![{\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)}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d7f12b27533fca50aee78833799b2de718f13068)
![{\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}}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e5c104a4280b7a96ff7a0675b28b99ab5ae46262)
![{\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.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/63365b601c7fdc95ac2ecf405bf3f8fe5a016364)
![{\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)}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/307026a941c7352dfed95f0c14aaa465015817fd)
![{\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.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/fc578e6203860ba7806ed05aab4c5ae03fc9c07f)
![{\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.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/932f7f96c30f5d525cfc095ad31c67c2ab91b322)
![{\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.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6bd72754057556b76216a587321ce081449f8094)
![{\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.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b714fc2f8791a0274bb594df329dffafe897c617)
![{\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.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0e9412d68dbce7ac0c38e9580d731a5281295f43)
Datos: Q24883336