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,}
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. Puntos clave
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Relaciones de distancias
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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.
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
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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
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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
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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.}
↑ 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).
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![{\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