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.

Puntos clave

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

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 ha, hb y hc 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.

Bisectrices

Se tienen las siguientes propiedades de las bisectrices w A , w B , {\displaystyle w_{A},w_{B},} y w C {\displaystyle w_{C}} de los ángulos 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

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 ra correspondiente al lado a es igual al radio de la circunferencia de los nueve puntos del triángulo heptagonal.[2]​ : p. 15 

Triángulo órtico

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

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.}

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