Las expresiones algebraicas exactas de valores trigonométricos pueden ser útiles principalmente para obtener soluciones en forma de radicales que permiten simplificar determinados resultados.
Ángulos primarios en la forma (cos, sen) en la circunferencia goniométrica, múltiplos de 30 y 45 grados Todos los números trigonométricos (senos o cosenos de submúltiplos racionales de 360°) son números algebraicos (es decir, soluciones de ecuaciones algebraicas con coeficientes enteros). Además, pueden expresarse en términos de radicales de números complejos ; pero no todos ellos pueden expresarse en términos de radicales reales . Cuando lo son, se pueden expresar más específicamente en términos de raíces cuadradas.
Todos los valores de los senos, cosenos y tangentes de los ángulos en incrementos de 3° se pueden expresar en términos de raíces cuadradas, usando identidades (como las del ángulo mitad , las del ángulo doble y las de las sumas y restas de ángulos ) y usando los valores conocidos para 0°, 30°, 36°, y 45°. Para un ángulo de un número entero de grados que no es múltiplo de 3° (π / 60 radianes ), los valores de seno, coseno y tangente no se pueden expresar en términos de radicales reales.
Según el teorema de Niven , los únicos valores racionales de la función seno para los que el argumento es un número racional de grados son 0, 1 / 2 1 / 2
Según el teorema de Baker , si el valor de un seno, un coseno o una tangente es algebraico, entonces el ángulo es un número racional de grados o un número trascendente de grados. Es decir, si el ángulo es un número de grados algebraico, pero no racional, todas las funciones trigonométricas tienen valores trascendentes.
Alcance de este artículo
editar
La lista de este artículo está incompleta en varios sentidos. Primero, las funciones trigonométricas de todos los ángulos que son múltiplos enteros de los dados también pueden expresarse en radicales, pero algunos se omiten aquí.
En segundo lugar, siempre es posible aplicar la fórmula del ángulo mitad para encontrar una expresión en radicales para una función trigonométrica de la mitad de cualquier ángulo de la lista, aplicando sucesivamente este procedimiento las veces que se desee.
En tercer lugar, existen expresiones en radicales reales para una función trigonométrica de un múltiplo racional de π si y solo si el denominador del múltiplo racional completamente reducido es una potencia de 2 por sí mismo o el producto de una potencia de 2 con el producto de números de Fermat distintos, de los cuales los conocidos son 3, 5, 17, 257 y 65537.
En cuarto lugar, este artículo solo se ocupa de los valores de las funciones trigonométricas cuando la expresión puede obtenerse en radicales reales , es decir, en raíces de números reales. Muchos otros valores de funciones trigonométricas se pueden expresar, por ejemplo, en raíces cúbicas de números complejos que no se pueden reescribir en términos de raíces de números reales. Por ejemplo, los valores de la función trigonométrica de cualquier ángulo que sea un tercio de un ángulo θ considerado en este artículo se pueden expresar en raíces cúbicas y raíces cuadradas usando fórmula de la ecuación cúbica para resolver
4
cos
3
θ
3
−
3
cos
θ
3
=
cos
θ
,
{\displaystyle 4\cos ^{3}{\frac {\theta }{3}}-3\cos {\frac {\theta }{3}}=\cos \theta ,}
pero, en general, la solución para el coseno de un tercio de un ángulo implica usar la raíz cúbica de un número complejo (en el denominado casus irreducibilis
En la práctica, todos los valores de senos, cosenos y tangentes que no se encuentran en este artículo se aproximan utilizando las técnicas descritas en el artículo dedicado a las tablas trigonométricas
Algunos ángulos
editar
Tabla trigonométrica exacta para múltiplos de 3 grados Los valores fuera del rango angular [0°, 45°] se deducen trivialmente de los siguientes valores, utilizando simples relaciones de reflexión y simetría . (Véase Lista de identidades trigonométricas )
En las entradas siguientes, cuando un cierto número de grados está relacionado con un polígono regular, la relación es que el número de grados en cada ángulo del polígono es (n - 2) veces el número indicado de grados (donde n es el número de lados). Esto se debe a que la suma de los ángulos de cualquier n -gono es 180° × (n - 2) y, por lo tanto, la medida de cada ángulo de cualquier n -gono regular es 180° × (n - 2) ÷ n . Así, por ejemplo, la entrada "45°: cuadrado" significa que, con n = 4, 180° ÷ n = 45°, y el número de grados en cada ángulo de un cuadrado hay (n - 2) × 45° = 90°.
0°: fundamental
editar
sen
0
=
0
{\displaystyle \operatorname {sen} 0=0\,}
cos
0
=
1
{\displaystyle \cos 0=1\,}
tan
0
=
0
{\displaystyle \tan 0=0\,}
cot
0
está indefinida
{\displaystyle \cot 0{\text{ está indefinida}}\,}
sec
0
=
1
{\displaystyle \sec 0=1\,}
csc
0
está indefinida
{\displaystyle \csc 0{\text{ está indefinida}}\,}
1.5°: hecatonicoságono regular (polígono de 120 lados)
editar
sen
(
π
120
)
=
sen
(
1.5
∘
)
=
(
2
+
2
)
(
15
+
3
−
10
−
2
5
)
−
(
2
−
2
)
(
30
−
6
5
+
5
+
1
)
16
{\displaystyle \operatorname {sen} \left({\frac {\pi }{120}}\right)=\operatorname {sen} \left(1.5^{\circ }\right)={\frac {\left({\sqrt {2+{\sqrt {2}}}}\right)\left({\sqrt {15}}+{\sqrt {3}}-{\sqrt {10-2{\sqrt {5}}}}\right)-\left({\sqrt {2-{\sqrt {2}}}}\right)\left({\sqrt {30-6{\sqrt {5}}}}+{\sqrt {5}}+1\right)}{16}}}
cos
(
π
120
)
=
cos
(
1.5
∘
)
=
(
2
+
2
)
(
30
−
6
5
+
5
+
1
)
+
(
2
−
2
)
(
15
+
3
−
10
−
2
5
)
16
{\displaystyle \cos \left({\frac {\pi }{120}}\right)=\cos \left(1.5^{\circ }\right)={\frac {\left({\sqrt {2+{\sqrt {2}}}}\right)\left({\sqrt {30-6{\sqrt {5}}}}+{\sqrt {5}}+1\right)+\left({\sqrt {2-{\sqrt {2}}}}\right)\left({\sqrt {15}}+{\sqrt {3}}-{\sqrt {10-2{\sqrt {5}}}}\right)}{16}}}
tan
(
π
120
)
=
tan
(
1.5
∘
)
=
(
2
+
2
)
(
15
+
3
−
10
−
2
5
)
−
(
2
−
2
)
(
30
−
6
5
+
5
+
1
)
(
2
+
2
)
(
30
−
6
5
+
5
+
1
)
+
(
2
−
2
)
(
15
+
3
−
10
−
2
5
)
{\displaystyle \tan \left({\frac {\pi }{120}}\right)=\tan \left(1.5^{\circ }\right)={\frac {\left({\sqrt {2+{\sqrt {2}}}}\right)\left({\sqrt {15}}+{\sqrt {3}}-{\sqrt {10-2{\sqrt {5}}}}\right)-\left({\sqrt {2-{\sqrt {2}}}}\right)\left({\sqrt {30-6{\sqrt {5}}}}+{\sqrt {5}}+1\right)}{\left({\sqrt {2+{\sqrt {2}}}}\right)\left({\sqrt {30-6{\sqrt {5}}}}+{\sqrt {5}}+1\right)+\left({\sqrt {2-{\sqrt {2}}}}\right)\left({\sqrt {15}}+{\sqrt {3}}-{\sqrt {10-2{\sqrt {5}}}}\right)}}}
cot
(
π
120
)
=
cot
(
1.5
∘
)
=
(
2
+
2
)
(
30
−
6
5
+
5
+
1
)
+
(
2
−
2
)
(
15
+
3
−
10
−
2
5
)
(
2
+
2
)
(
15
+
3
−
10
−
2
5
)
−
(
2
−
2
)
(
30
−
6
5
+
5
+
1
)
{\displaystyle \cot \left({\frac {\pi }{120}}\right)=\cot \left(1.5^{\circ }\right)={\frac {\left({\sqrt {2+{\sqrt {2}}}}\right)\left({\sqrt {30-6{\sqrt {5}}}}+{\sqrt {5}}+1\right)+\left({\sqrt {2-{\sqrt {2}}}}\right)\left({\sqrt {15}}+{\sqrt {3}}-{\sqrt {10-2{\sqrt {5}}}}\right)}{\left({\sqrt {2+{\sqrt {2}}}}\right)\left({\sqrt {15}}+{\sqrt {3}}-{\sqrt {10-2{\sqrt {5}}}}\right)-\left({\sqrt {2-{\sqrt {2}}}}\right)\left({\sqrt {30-6{\sqrt {5}}}}+{\sqrt {5}}+1\right)}}}
sec
(
π
120
)
=
sec
(
1.5
∘
)
=
16
(
2
+
2
)
(
30
−
6
5
+
5
+
1
)
+
(
2
−
2
)
(
15
+
3
−
10
−
2
5
)
{\displaystyle \sec \left({\frac {\pi }{120}}\right)=\sec \left(1.5^{\circ }\right)={\frac {16}{\left({\sqrt {2+{\sqrt {2}}}}\right)\left({\sqrt {30-6{\sqrt {5}}}}+{\sqrt {5}}+1\right)+\left({\sqrt {2-{\sqrt {2}}}}\right)\left({\sqrt {15}}+{\sqrt {3}}-{\sqrt {10-2{\sqrt {5}}}}\right)}}}
csc
(
π
120
)
=
csc
(
1.5
∘
)
=
16
(
2
+
2
)
(
15
+
3
−
10
−
2
5
)
−
(
2
−
2
)
(
30
−
6
5
+
5
+
1
)
{\displaystyle \csc \left({\frac {\pi }{120}}\right)=\csc \left(1.5^{\circ }\right)={\frac {16}{\left({\sqrt {2+{\sqrt {2}}}}\right)\left({\sqrt {15}}+{\sqrt {3}}-{\sqrt {10-2{\sqrt {5}}}}\right)-\left({\sqrt {2-{\sqrt {2}}}}\right)\left({\sqrt {30-6{\sqrt {5}}}}+{\sqrt {5}}+1\right)}}}
1.875°: eneacontahexágono regular (polígono de 96 lados)
editar
sen
(
π
96
)
=
sen
(
1.875
∘
)
=
2
−
2
+
2
+
2
+
3
2
{\displaystyle \operatorname {sen} \left({\frac {\pi }{96}}\right)=\operatorname {sen} \left(1.875^{\circ }\right)={\frac {\sqrt {2-{\sqrt {2+{\sqrt {2+{\sqrt {2+{\sqrt {3}}}}}}}}}}{2}}}
cos
(
π
96
)
=
cos
(
1.875
∘
)
=
2
+
2
+
2
+
2
+
3
2
{\displaystyle \cos \left({\frac {\pi }{96}}\right)=\cos \left(1.875^{\circ }\right)={\frac {\sqrt {2+{\sqrt {2+{\sqrt {2+{\sqrt {2+{\sqrt {3}}}}}}}}}}{2}}}
tan
(
π
96
)
=
tan
(
1.875
∘
)
=
2
−
2
+
2
+
2
+
3
2
+
2
+
2
+
2
+
3
{\displaystyle \tan \left({\frac {\pi }{96}}\right)=\tan \left(1.875^{\circ }\right)={\frac {\sqrt {2-{\sqrt {2+{\sqrt {2+{\sqrt {2+{\sqrt {3}}}}}}}}}}{\sqrt {2+{\sqrt {2+{\sqrt {2+{\sqrt {2+{\sqrt {3}}}}}}}}}}}}
cot
(
π
96
)
=
cot
(
1.875
∘
)
=
2
+
2
+
2
+
2
+
3
2
−
2
+
2
+
2
+
3
{\displaystyle \cot \left({\frac {\pi }{96}}\right)=\cot \left(1.875^{\circ }\right)={\frac {\sqrt {2+{\sqrt {2+{\sqrt {2+{\sqrt {2+{\sqrt {3}}}}}}}}}}{\sqrt {2-{\sqrt {2+{\sqrt {2+{\sqrt {2+{\sqrt {3}}}}}}}}}}}}
sec
(
π
96
)
=
sec
(
1.875
∘
)
=
2
2
+
2
+
2
+
2
+
3
{\displaystyle \sec \left({\frac {\pi }{96}}\right)=\sec \left(1.875^{\circ }\right)={\frac {2}{\sqrt {2+{\sqrt {2+{\sqrt {2+{\sqrt {2+{\sqrt {3}}}}}}}}}}}}
csc
(
π
96
)
=
csc
(
1.875
∘
)
=
2
2
−
2
+
2
+
2
+
3
{\displaystyle \csc \left({\frac {\pi }{96}}\right)=\csc \left(1.875^{\circ }\right)={\frac {2}{\sqrt {2-{\sqrt {2+{\sqrt {2+{\sqrt {2+{\sqrt {3}}}}}}}}}}}}
2.25°: octacontágono regular (polígono de 80 lados)
editar
sen
(
π
80
)
=
sen
(
2.25
∘
)
=
1
2
2
−
2
+
2
+
5
+
5
2
{\displaystyle \operatorname {sen} \left({\frac {\pi }{80}}\right)=\operatorname {sen} \left(2.25^{\circ }\right)={\frac {1}{2}}{\sqrt {2-{\sqrt {2+{\sqrt {2+{\sqrt {\frac {5+{\sqrt {5}}}{2}}}}}}}}}}
cos
(
π
80
)
=
cos
(
2.25
∘
)
=
1
2
2
+
2
+
2
+
5
+
5
2
{\displaystyle \cos \left({\frac {\pi }{80}}\right)=\cos \left(2.25^{\circ }\right)={\frac {1}{2}}{\sqrt {2+{\sqrt {2+{\sqrt {2+{\sqrt {\frac {5+{\sqrt {5}}}{2}}}}}}}}}}
2.8125°: hexacontatetrágono regular (polígono de 64 lados)
editar
sen
(
π
64
)
=
sen
(
2.8125
∘
)
=
1
2
2
−
2
+
2
+
2
+
2
{\displaystyle \operatorname {sen} \left({\frac {\pi }{64}}\right)=\operatorname {sen} \left(2.8125^{\circ }\right)={\frac {1}{2}}{\sqrt {2-{\sqrt {2+{\sqrt {2+{\sqrt {2+{\sqrt {2}}}}}}}}}}}
cos
(
π
64
)
=
cos
(
2.8125
∘
)
=
1
2
2
+
2
+
2
+
2
+
2
{\displaystyle \cos \left({\frac {\pi }{64}}\right)=\cos \left(2.8125^{\circ }\right)={\frac {1}{2}}{\sqrt {2+{\sqrt {2+{\sqrt {2+{\sqrt {2+{\sqrt {2}}}}}}}}}}}
3°: hexacontágono regular (polígono de 60 lados)
editar
sen
(
π
60
)
=
sen
(
3
∘
)
=
2
(
1
−
3
)
5
+
5
+
(
10
−
2
)
(
3
+
1
)
16
{\displaystyle \operatorname {sen} \left({\frac {\pi }{60}}\right)=\operatorname {sen} \left(3^{\circ }\right)={\frac {2\left(1-{\sqrt {3}}\right){\sqrt {5+{\sqrt {5}}}}+\left({\sqrt {10}}-{\sqrt {2}}\right)\left({\sqrt {3}}+1\right)}{16}}\,}
cos
(
π
60
)
=
cos
(
3
∘
)
=
2
(
1
+
3
)
5
+
5
+
(
10
−
2
)
(
3
−
1
)
16
{\displaystyle \cos \left({\frac {\pi }{60}}\right)=\cos \left(3^{\circ }\right)={\frac {2\left(1+{\sqrt {3}}\right){\sqrt {5+{\sqrt {5}}}}+\left({\sqrt {10}}-{\sqrt {2}}\right)\left({\sqrt {3}}-1\right)}{16}}\,}
tan
(
π
60
)
=
tan
(
3
∘
)
=
[
(
2
−
3
)
(
3
+
5
)
−
2
]
[
2
−
10
−
2
5
]
4
{\displaystyle \tan \left({\frac {\pi }{60}}\right)=\tan \left(3^{\circ }\right)={\frac {\left[\left(2-{\sqrt {3}}\right)\left(3+{\sqrt {5}}\right)-2\right]\left[2-{\sqrt {10-2{\sqrt {5}}}}\right]}{4}}\,}
cot
(
π
60
)
=
cot
(
3
∘
)
=
[
(
2
+
3
)
(
3
+
5
)
−
2
]
[
2
+
10
−
2
5
]
4
{\displaystyle \cot \left({\frac {\pi }{60}}\right)=\cot \left(3^{\circ }\right)={\frac {\left[\left(2+{\sqrt {3}}\right)\left(3+{\sqrt {5}}\right)-2\right]\left[2+{\sqrt {10-2{\sqrt {5}}}}\right]}{4}}\,}
3,75°: tetracontaoctágono regular (polígono de 48 lados)
editar
sen
(
π
48
)
=
sen
(
3.75
∘
)
=
1
2
2
−
2
+
2
+
3
{\displaystyle \operatorname {sen} \left({\frac {\pi }{48}}\right)=\operatorname {sen} \left(3.75^{\circ }\right)={\frac {1}{2}}{\sqrt {2-{\sqrt {2+{\sqrt {2+{\sqrt {3}}}}}}}}}
cos
(
π
48
)
=
cos
(
3.75
∘
)
=
1
2
2
+
2
+
2
+
3
{\displaystyle \cos \left({\frac {\pi }{48}}\right)=\cos \left(3.75^{\circ }\right)={\frac {1}{2}}{\sqrt {2+{\sqrt {2+{\sqrt {2+{\sqrt {3}}}}}}}}}
4.5°: tetracontágono regular (polígono de 40 lados)
editar
sen
(
π
40
)
=
sen
(
4.5
∘
)
=
1
2
2
−
2
+
5
+
5
2
{\displaystyle \operatorname {sen} \left({\frac {\pi }{40}}\right)=\operatorname {sen} \left(4.5^{\circ }\right)={\frac {1}{2}}{\sqrt {2-{\sqrt {2+{\sqrt {\frac {5+{\sqrt {5}}}{2}}}}}}}}
cos
(
π
40
)
=
cos
(
4.5
∘
)
=
1
2
2
+
2
+
5
+
5
2
{\displaystyle \cos \left({\frac {\pi }{40}}\right)=\cos \left(4.5^{\circ }\right)={\frac {1}{2}}{\sqrt {2+{\sqrt {2+{\sqrt {\frac {5+{\sqrt {5}}}{2}}}}}}}}
5.625°: triacontadígono regular (polígono de 32 lados)
editar
sen
(
π
32
)
=
sen
(
5.625
∘
)
=
1
2
2
−
2
+
2
+
2
{\displaystyle \operatorname {sen} \left({\frac {\pi }{32}}\right)=\operatorname {sen} \left(5.625^{\circ }\right)={\frac {1}{2}}{\sqrt {2-{\sqrt {2+{\sqrt {2+{\sqrt {2}}}}}}}}}
cos
(
π
32
)
=
cos
(
5.625
∘
)
=
1
2
2
+
2
+
2
+
2
{\displaystyle \cos \left({\frac {\pi }{32}}\right)=\cos \left(5.625^{\circ }\right)={\frac {1}{2}}{\sqrt {2+{\sqrt {2+{\sqrt {2+{\sqrt {2}}}}}}}}}
6°: triacontágono regular (polígono de 30 lados)
editar
sen
π
30
=
sen
6
∘
=
30
−
180
−
5
−
1
8
{\displaystyle \operatorname {sen} {\frac {\pi }{30}}=\operatorname {sen} 6^{\circ }={\frac {{\sqrt {30-{\sqrt {180}}}}-{\sqrt {5}}-1}{8}}\,}
cos
π
30
=
cos
6
∘
=
10
−
20
+
3
+
15
8
{\displaystyle \cos {\frac {\pi }{30}}=\cos 6^{\circ }={\frac {{\sqrt {10-{\sqrt {20}}}}+{\sqrt {3}}+{\sqrt {15}}}{8}}\,}
tan
π
30
=
tan
6
∘
=
10
−
20
+
3
−
15
2
{\displaystyle \tan {\frac {\pi }{30}}=\tan 6^{\circ }={\frac {{\sqrt {10-{\sqrt {20}}}}+{\sqrt {3}}-{\sqrt {15}}}{2}}\,}
cot
π
30
=
cot
6
∘
=
27
+
15
+
50
+
2420
2
{\displaystyle \cot {\frac {\pi }{30}}=\cot 6^{\circ }={\frac {{\sqrt {27}}+{\sqrt {15}}+{\sqrt {50+{\sqrt {2420}}}}}{2}}\,}
7,5°: icositetrágono regular (polígono de 24 lados)
editar
sen
(
π
24
)
=
sen
(
7.5
∘
)
=
1
2
2
−
2
+
3
=
1
4
8
−
2
6
−
2
2
{\displaystyle \operatorname {sen} \left({\frac {\pi }{24}}\right)=\operatorname {sen} \left(7.5^{\circ }\right)={\frac {1}{2}}{\sqrt {2-{\sqrt {2+{\sqrt {3}}}}}}={\frac {1}{4}}{\sqrt {8-2{\sqrt {6}}-2{\sqrt {2}}}}}
cos
(
π
24
)
=
cos
(
7.5
∘
)
=
1
2
2
+
2
+
3
=
1
4
8
+
2
6
+
2
2
{\displaystyle \cos \left({\frac {\pi }{24}}\right)=\cos \left(7.5^{\circ }\right)={\frac {1}{2}}{\sqrt {2+{\sqrt {2+{\sqrt {3}}}}}}={\frac {1}{4}}{\sqrt {8+2{\sqrt {6}}+2{\sqrt {2}}}}}
tan
(
π
24
)
=
tan
(
7.5
∘
)
=
6
−
3
+
2
−
2
=
(
2
−
1
)
(
3
−
2
)
{\displaystyle \tan \left({\frac {\pi }{24}}\right)=\tan \left(7.5^{\circ }\right)={\sqrt {6}}-{\sqrt {3}}+{\sqrt {2}}-2\ =\left({\sqrt {2}}-1\right)\left({\sqrt {3}}-{\sqrt {2}}\right)}
cot
(
π
24
)
=
cot
(
7.5
∘
)
=
6
+
3
+
2
+
2
=
(
2
+
1
)
(
3
+
2
)
{\displaystyle \cot \left({\frac {\pi }{24}}\right)=\cot \left(7.5^{\circ }\right)={\sqrt {6}}+{\sqrt {3}}+{\sqrt {2}}+2\ =\left({\sqrt {2}}+1\right)\left({\sqrt {3}}+{\sqrt {2}}\right)}
9°: icoságono regular (polígono de 20 lados)
editar
sen
π
20
=
sen
9
∘
=
1
2
2
−
5
+
5
2
{\displaystyle \operatorname {sen} {\frac {\pi }{20}}=\operatorname {sen} 9^{\circ }={\frac {1}{2}}{\sqrt {2-{\sqrt {\frac {5+{\sqrt {5}}}{2}}}}}}
cos
π
20
=
cos
9
∘
=
1
2
2
+
5
+
5
2
{\displaystyle \cos {\frac {\pi }{20}}=\cos 9^{\circ }={\frac {1}{2}}{\sqrt {2+{\sqrt {\frac {5+{\sqrt {5}}}{2}}}}}}
tan
π
20
=
tan
9
∘
=
5
+
1
−
5
+
2
5
{\displaystyle \tan {\frac {\pi }{20}}=\tan 9^{\circ }={\sqrt {5}}+1-{\sqrt {5+2{\sqrt {5}}}}\,}
cot
π
20
=
cot
9
∘
=
5
+
1
+
5
+
2
5
{\displaystyle \cot {\frac {\pi }{20}}=\cot 9^{\circ }={\sqrt {5}}+1+{\sqrt {5+2{\sqrt {5}}}}\,}
11.25°: hexadecágono regular (polígono de 16 lados)
editar
sen
π
16
=
sen
11.25
∘
=
1
2
2
−
2
+
2
{\displaystyle \operatorname {sen} {\frac {\pi }{16}}=\operatorname {sen} 11.25^{\circ }={\frac {1}{2}}{\sqrt {2-{\sqrt {2+{\sqrt {2}}}}}}}
cos
π
16
=
cos
11.25
∘
=
1
2
2
+
2
+
2
{\displaystyle \cos {\frac {\pi }{16}}=\cos 11.25^{\circ }={\frac {1}{2}}{\sqrt {2+{\sqrt {2+{\sqrt {2}}}}}}}
tan
π
16
=
tan
11.25
∘
=
4
+
2
2
−
2
−
1
{\displaystyle \tan {\frac {\pi }{16}}=\tan 11.25^{\circ }={\sqrt {4+2{\sqrt {2}}}}-{\sqrt {2}}-1}
cot
π
16
=
cot
11.25
∘
=
4
+
2
2
+
2
+
1
{\displaystyle \cot {\frac {\pi }{16}}=\cot 11.25^{\circ }={\sqrt {4+2{\sqrt {2}}}}+{\sqrt {2}}+1}
12°: pentadecágono regular (polígono de 15 lados)
editar
sen
π
15
=
sen
12
∘
=
1
8
[
2
(
5
+
5
)
+
3
−
15
]
{\displaystyle \operatorname {sen} {\frac {\pi }{15}}=\operatorname {sen} 12^{\circ }={\tfrac {1}{8}}\left[{\sqrt {2\left(5+{\sqrt {5}}\right)}}+{\sqrt {3}}-{\sqrt {15}}\right]\,}
cos
π
15
=
cos
12
∘
=
1
8
[
6
(
5
+
5
)
+
5
−
1
]
{\displaystyle \cos {\frac {\pi }{15}}=\cos 12^{\circ }={\tfrac {1}{8}}\left[{\sqrt {6\left(5+{\sqrt {5}}\right)}}+{\sqrt {5}}-1\right]\,}
tan
π
15
=
tan
12
∘
=
1
2
[
3
3
−
15
−
2
(
25
−
11
5
)
]
{\displaystyle \tan {\frac {\pi }{15}}=\tan 12^{\circ }={\tfrac {1}{2}}\left[3{\sqrt {3}}-{\sqrt {15}}-{\sqrt {2\left(25-11{\sqrt {5}}\right)}}\,\right]\,}
cot
π
15
=
cot
12
∘
=
1
2
[
15
+
3
+
2
(
5
+
5
)
]
{\displaystyle \cot {\frac {\pi }{15}}=\cot 12^{\circ }={\tfrac {1}{2}}\left[{\sqrt {15}}+{\sqrt {3}}+{\sqrt {2\left(5+{\sqrt {5}}\right)}}\,\right]\,}
15°: dodecágono regular (polígono de 12 lados)
editar
sen
π
12
=
sen
15
∘
=
1
4
(
6
−
2
)
=
1
2
2
−
3
{\displaystyle \operatorname {sen} {\frac {\pi }{12}}=\operatorname {sen} 15^{\circ }={\frac {1}{4}}\left({\sqrt {6}}-{\sqrt {2}}\right)={\frac {1}{2}}{\sqrt {2-{\sqrt {3}}}}}
cos
π
12
=
cos
15
∘
=
1
4
(
6
+
2
)
=
1
2
2
+
3
{\displaystyle \cos {\frac {\pi }{12}}=\cos 15^{\circ }={\frac {1}{4}}\left({\sqrt {6}}+{\sqrt {2}}\right)={\frac {1}{2}}{\sqrt {2+{\sqrt {3}}}}}
tan
π
12
=
tan
15
∘
=
2
−
3
{\displaystyle \tan {\frac {\pi }{12}}=\tan 15^{\circ }=2-{\sqrt {3}}\,}
cot
π
12
=
cot
15
∘
=
2
+
3
{\displaystyle \cot {\frac {\pi }{12}}=\cot 15^{\circ }=2+{\sqrt {3}}\,}
18°: decágono regular (polígono de 10 lados)[ 1]
editar
sen
π
10
=
sen
18
∘
=
1
4
(
5
−
1
)
=
1
1
+
5
{\displaystyle \operatorname {sen} {\frac {\pi }{10}}=\operatorname {sen} 18^{\circ }={\tfrac {1}{4}}\left({\sqrt {5}}-1\right)={\frac {1}{1+{\sqrt {5}}}}\,}
cos
π
10
=
cos
18
∘
=
1
4
2
(
5
+
5
)
{\displaystyle \cos {\frac {\pi }{10}}=\cos 18^{\circ }={\tfrac {1}{4}}{\sqrt {2\left(5+{\sqrt {5}}\right)}}\,}
tan
π
10
=
tan
18
∘
=
1
5
5
(
5
−
2
5
)
{\displaystyle \tan {\frac {\pi }{10}}=\tan 18^{\circ }={\tfrac {1}{5}}{\sqrt {5\left(5-2{\sqrt {5}}\right)}}\,}
cot
π
10
=
cot
18
∘
=
5
+
2
5
{\displaystyle \cot {\frac {\pi }{10}}=\cot 18^{\circ }={\sqrt {5+2{\sqrt {5}}}}\,}
21°: suma 9° + 12°
editar
sen
7
π
60
=
sen
21
∘
=
1
16
(
2
(
3
+
1
)
5
−
5
−
(
6
−
2
)
(
1
+
5
)
)
{\displaystyle \operatorname {sen} {\frac {7\pi }{60}}=\operatorname {sen} 21^{\circ }={\frac {1}{16}}\left(2\left({\sqrt {3}}+1\right){\sqrt {5-{\sqrt {5}}}}-\left({\sqrt {6}}-{\sqrt {2}}\right)\left(1+{\sqrt {5}}\right)\right)\,}
cos
7
π
60
=
cos
21
∘
=
1
16
(
2
(
3
−
1
)
5
−
5
+
(
6
+
2
)
(
1
+
5
)
)
{\displaystyle \cos {\frac {7\pi }{60}}=\cos 21^{\circ }={\frac {1}{16}}\left(2\left({\sqrt {3}}-1\right){\sqrt {5-{\sqrt {5}}}}+\left({\sqrt {6}}+{\sqrt {2}}\right)\left(1+{\sqrt {5}}\right)\right)\,}
tan
7
π
60
=
tan
21
∘
=
1
4
(
2
−
(
2
+
3
)
(
3
−
5
)
)
(
2
−
2
(
5
+
5
)
)
{\displaystyle \tan {\frac {7\pi }{60}}=\tan 21^{\circ }={\frac {1}{4}}\left(2-\left(2+{\sqrt {3}}\right)\left(3-{\sqrt {5}}\right)\right)\left(2-{\sqrt {2\left(5+{\sqrt {5}}\right)}}\right)\,}
cot
7
π
60
=
cot
21
∘
=
1
4
(
2
−
(
2
−
3
)
(
3
−
5
)
)
(
2
+
2
(
5
+
5
)
)
{\displaystyle \cot {\frac {7\pi }{60}}=\cot 21^{\circ }={\frac {1}{4}}\left(2-\left(2-{\sqrt {3}}\right)\left(3-{\sqrt {5}}\right)\right)\left(2+{\sqrt {2\left(5+{\sqrt {5}}\right)}}\right)\,}
22.5°: octágono regular
editar
sen
π
8
=
sen
22.5
∘
=
1
2
2
−
2
,
{\displaystyle \operatorname {sen} {\frac {\pi }{8}}=\operatorname {sen} 22.5^{\circ }={\frac {1}{2}}{\sqrt {2-{\sqrt {2}}}},}
cos
π
8
=
cos
22.5
∘
=
1
2
2
+
2
{\displaystyle \cos {\frac {\pi }{8}}=\cos 22.5^{\circ }={\frac {1}{2}}{\sqrt {2+{\sqrt {2}}}}\,}
tan
π
8
=
tan
22.5
∘
=
2
−
1
{\displaystyle \tan {\frac {\pi }{8}}=\tan 22.5^{\circ }={\sqrt {2}}-1\,}
cot
π
8
=
cot
22.5
∘
=
2
+
1
=
δ
S
{\displaystyle \cot {\frac {\pi }{8}}=\cot 22.5^{\circ }={\sqrt {2}}+1=\delta _{S}\,}
número plateado 24°: suma 12° + 12°
editar
sen
2
π
15
=
sen
24
∘
=
1
8
[
15
+
3
−
2
(
5
−
5
)
]
{\displaystyle \operatorname {sen} {\frac {2\pi }{15}}=\operatorname {sen} 24^{\circ }={\tfrac {1}{8}}\left[{\sqrt {15}}+{\sqrt {3}}-{\sqrt {2\left(5-{\sqrt {5}}\right)}}\right]\,}
cos
2
π
15
=
cos
24
∘
=
1
8
(
6
(
5
−
5
)
+
5
+
1
)
{\displaystyle \cos {\frac {2\pi }{15}}=\cos 24^{\circ }={\tfrac {1}{8}}\left({\sqrt {6\left(5-{\sqrt {5}}\right)}}+{\sqrt {5}}+1\right)\,}
tan
2
π
15
=
tan
24
∘
=
1
2
[
50
+
22
5
−
3
3
−
15
]
{\displaystyle \tan {\frac {2\pi }{15}}=\tan 24^{\circ }={\tfrac {1}{2}}\left[{\sqrt {50+22{\sqrt {5}}}}-3{\sqrt {3}}-{\sqrt {15}}\right]\,}
cot
2
π
15
=
cot
24
∘
=
1
2
[
15
−
3
+
2
(
5
−
5
)
]
{\displaystyle \cot {\frac {2\pi }{15}}=\cot 24^{\circ }={\tfrac {1}{2}}\left[{\sqrt {15}}-{\sqrt {3}}+{\sqrt {2\left(5-{\sqrt {5}}\right)}}\right]\,}
27°: suma 12° + 15°
editar
sen
3
π
20
=
sen
27
∘
=
1
8
[
2
5
+
5
−
2
(
5
−
1
)
]
{\displaystyle \operatorname {sen} {\frac {3\pi }{20}}=\operatorname {sen} 27^{\circ }={\tfrac {1}{8}}\left[2{\sqrt {5+{\sqrt {5}}}}-{\sqrt {2}}\;\left({\sqrt {5}}-1\right)\right]\,}
cos
3
π
20
=
cos
27
∘
=
1
8
[
2
5
+
5
+
2
(
5
−
1
)
]
{\displaystyle \cos {\frac {3\pi }{20}}=\cos 27^{\circ }={\tfrac {1}{8}}\left[2{\sqrt {5+{\sqrt {5}}}}+{\sqrt {2}}\;\left({\sqrt {5}}-1\right)\right]\,}
tan
3
π
20
=
tan
27
∘
=
5
−
1
−
5
−
2
5
{\displaystyle \tan {\frac {3\pi }{20}}=\tan 27^{\circ }={\sqrt {5}}-1-{\sqrt {5-2{\sqrt {5}}}}\,}
cot
3
π
20
=
cot
27
∘
=
5
−
1
+
5
−
2
5
{\displaystyle \cot {\frac {3\pi }{20}}=\cot 27^{\circ }={\sqrt {5}}-1+{\sqrt {5-2{\sqrt {5}}}}\,}
30°: hexágono regular
editar
sen
π
6
=
sen
30
∘
=
1
2
{\displaystyle \operatorname {sen} {\frac {\pi }{6}}=\operatorname {sen} 30^{\circ }={\frac {1}{2}}\,}
cos
π
6
=
cos
30
∘
=
3
2
{\displaystyle \cos {\frac {\pi }{6}}=\cos 30^{\circ }={\frac {\sqrt {3}}{2}}\,}
tan
π
6
=
tan
30
∘
=
3
3
=
1
3
{\displaystyle \tan {\frac {\pi }{6}}=\tan 30^{\circ }={\frac {\sqrt {3}}{3}}={\frac {1}{\sqrt {3}}}\,}
cot
π
6
=
cot
30
∘
=
3
{\displaystyle \cot {\frac {\pi }{6}}=\cot 30^{\circ }={\sqrt {3}}\,}
sec
π
6
=
sec
30
∘
=
2
3
3
=
2
3
{\displaystyle \sec {\frac {\pi }{6}}=\sec 30^{\circ }={\frac {2{\sqrt {3}}}{3}}={\frac {2}{\sqrt {3}}}\,}
csc
π
6
=
csc
30
∘
=
2
{\displaystyle \csc {\frac {\pi }{6}}=\csc 30^{\circ }=2}
33°: suma 15° + 18°
editar
sen
11
π
60
=
sen
33
∘
=
1
16
[
2
(
3
−
1
)
5
+
5
+
2
(
1
+
3
)
(
5
−
1
)
]
{\displaystyle \operatorname {sen} {\frac {11\pi }{60}}=\operatorname {sen} 33^{\circ }={\tfrac {1}{16}}\left[2\left({\sqrt {3}}-1\right){\sqrt {5+{\sqrt {5}}}}+{\sqrt {2}}\left(1+{\sqrt {3}}\right)\left({\sqrt {5}}-1\right)\right]\,}
cos
11
π
60
=
cos
33
∘
=
1
16
[
2
(
3
+
1
)
5
+
5
+
2
(
1
−
3
)
(
5
−
1
)
]
{\displaystyle \cos {\frac {11\pi }{60}}=\cos 33^{\circ }={\tfrac {1}{16}}\left[2\left({\sqrt {3}}+1\right){\sqrt {5+{\sqrt {5}}}}+{\sqrt {2}}\left(1-{\sqrt {3}}\right)\left({\sqrt {5}}-1\right)\right]\,}
tan
11
π
60
=
tan
33
∘
=
1
4
[
2
−
(
2
−
3
)
(
3
+
5
)
]
[
2
+
2
(
5
−
5
)
]
{\displaystyle \tan {\frac {11\pi }{60}}=\tan 33^{\circ }={\tfrac {1}{4}}\left[2-\left(2-{\sqrt {3}}\right)\left(3+{\sqrt {5}}\right)\right]\left[2+{\sqrt {2\left(5-{\sqrt {5}}\right)}}\,\right]\,}
cot
11
π
60
=
cot
33
∘
=
1
4
[
2
−
(
2
+
3
)
(
3
+
5
)
]
[
2
−
2
(
5
−
5
)
]
{\displaystyle \cot {\frac {11\pi }{60}}=\cot 33^{\circ }={\tfrac {1}{4}}\left[2-\left(2+{\sqrt {3}}\right)\left(3+{\sqrt {5}}\right)\right]\left[2-{\sqrt {2\left(5-{\sqrt {5}}\right)}}\,\right]\,}
36°: pentágono regular
editar
[ 1]
sen
π
5
=
sen
36
∘
=
1
4
10
−
2
5
{\displaystyle \operatorname {sen} {\frac {\pi }{5}}=\operatorname {sen} 36^{\circ }={\frac {1}{4}}{\sqrt {10-2{\sqrt {5}}}}}
cos
π
5
=
cos
36
∘
=
1
+
5
4
=
φ
2
,
{\displaystyle \cos {\frac {\pi }{5}}=\cos 36^{\circ }={\frac {1+{\sqrt {5}}}{4}}={\frac {\varphi }{2}},}
φ número áureo ;
tan
π
5
=
tan
36
∘
=
5
−
2
5
{\displaystyle \tan {\frac {\pi }{5}}=\tan 36^{\circ }={\sqrt {5-2{\sqrt {5}}}}\,}
cot
π
5
=
cot
36
∘
=
1
5
25
+
10
5
{\displaystyle \cot {\frac {\pi }{5}}=\cot 36^{\circ }={\frac {1}{5}}{\sqrt {25+10{\sqrt {5}}}}}
39°: suma 18° + 21°
editar
sen
13
π
60
=
sen
39
∘
=
1
16
[
2
(
1
−
3
)
5
−
5
+
2
(
3
+
1
)
(
5
+
1
)
]
{\displaystyle \operatorname {sen} {\frac {13\pi }{60}}=\operatorname {sen} 39^{\circ }={\tfrac {1}{16}}\left[2\left(1-{\sqrt {3}}\right){\sqrt {5-{\sqrt {5}}}}+{\sqrt {2}}\left({\sqrt {3}}+1\right)\left({\sqrt {5}}+1\right)\right]\,}
cos
13
π
60
=
cos
39
∘
=
1
16
[
2
(
1
+
3
)
5
−
5
+
2
(
3
−
1
)
(
5
+
1
)
]
{\displaystyle \cos {\frac {13\pi }{60}}=\cos 39^{\circ }={\tfrac {1}{16}}\left[2\left(1+{\sqrt {3}}\right){\sqrt {5-{\sqrt {5}}}}+{\sqrt {2}}\left({\sqrt {3}}-1\right)\left({\sqrt {5}}+1\right)\right]\,}
tan
13
π
60
=
tan
39
∘
=
1
4
[
(
2
−
3
)
(
3
−
5
)
−
2
]
[
2
−
2
(
5
+
5
)
]
{\displaystyle \tan {\frac {13\pi }{60}}=\tan 39^{\circ }={\tfrac {1}{4}}\left[\left(2-{\sqrt {3}}\right)\left(3-{\sqrt {5}}\right)-2\right]\left[2-{\sqrt {2\left(5+{\sqrt {5}}\right)}}\,\right]\,}
cot
13
π
60
=
cot
39
∘
=
1
4
[
(
2
+
3
)
(
3
−
5
)
−
2
]
[
2
+
2
(
5
+
5
)
]
{\displaystyle \cot {\frac {13\pi }{60}}=\cot 39^{\circ }={\tfrac {1}{4}}\left[\left(2+{\sqrt {3}}\right)\left(3-{\sqrt {5}}\right)-2\right]\left[2+{\sqrt {2\left(5+{\sqrt {5}}\right)}}\,\right]\,}
42°: suma 21° + 21°
editar
sen
7
π
30
=
sen
42
∘
=
30
+
6
5
−
5
+
1
8
{\displaystyle \operatorname {sen} {\frac {7\pi }{30}}=\operatorname {sen} 42^{\circ }={\frac {{\sqrt {30+6{\sqrt {5}}}}-{\sqrt {5}}+1}{8}}\,}
cos
7
π
30
=
cos
42
∘
=
15
−
3
+
10
+
2
5
8
{\displaystyle \cos {\frac {7\pi }{30}}=\cos 42^{\circ }={\frac {{\sqrt {15}}-{\sqrt {3}}+{\sqrt {10+2{\sqrt {5}}}}}{8}}\,}
tan
7
π
30
=
tan
42
∘
=
15
+
3
−
10
+
2
5
2
{\displaystyle \tan {\frac {7\pi }{30}}=\tan 42^{\circ }={\frac {{\sqrt {15}}+{\sqrt {3}}-{\sqrt {10+2{\sqrt {5}}}}}{2}}\,}
cot
7
π
30
=
cot
42
∘
=
50
−
22
5
+
3
3
−
15
2
{\displaystyle \cot {\frac {7\pi }{30}}=\cot 42^{\circ }={\frac {{\sqrt {50-22{\sqrt {5}}}}+3{\sqrt {3}}-{\sqrt {15}}}{2}}\,}
45°: cuadrado
editar
sen
π
4
=
sen
45
∘
=
2
2
=
1
2
{\displaystyle \operatorname {sen} {\frac {\pi }{4}}=\operatorname {sen} 45^{\circ }={\frac {\sqrt {2}}{2}}={\frac {1}{\sqrt {2}}}\,}
cos
π
4
=
cos
45
∘
=
2
2
=
1
2
{\displaystyle \cos {\frac {\pi }{4}}=\cos 45^{\circ }={\frac {\sqrt {2}}{2}}={\frac {1}{\sqrt {2}}}\,}
tan
π
4
=
tan
45
∘
=
1
{\displaystyle \tan {\frac {\pi }{4}}=\tan 45^{\circ }=1\,}
cot
π
4
=
cot
45
∘
=
1
{\displaystyle \cot {\frac {\pi }{4}}=\cot 45^{\circ }=1\,}
sec
π
4
=
sec
45
∘
=
2
{\displaystyle \sec {\frac {\pi }{4}}=\sec 45^{\circ }={\sqrt {2}}\,}
csc
π
4
=
csc
45
∘
=
2
{\displaystyle \csc {\frac {\pi }{4}}=\csc 45^{\circ }={\sqrt {2}}\,}
54°: suma 27° + 27°
editar
sen
3
π
10
=
sen
54
∘
=
5
+
1
4
{\displaystyle \operatorname {sen} {\frac {3\pi }{10}}=\operatorname {sen} 54^{\circ }={\frac {{\sqrt {5}}+1}{4}}\,\!}
cos
3
π
10
=
cos
54
∘
=
10
−
2
5
4
{\displaystyle \cos {\frac {3\pi }{10}}=\cos 54^{\circ }={\frac {\sqrt {10-2{\sqrt {5}}}}{4}}}
tan
3
π
10
=
tan
54
∘
=
25
+
10
5
5
{\displaystyle \tan {\frac {3\pi }{10}}=\tan 54^{\circ }={\frac {\sqrt {25+10{\sqrt {5}}}}{5}}\,}
cot
3
π
10
=
cot
54
∘
=
5
−
2
5
{\displaystyle \cot {\frac {3\pi }{10}}=\cot 54^{\circ }={\sqrt {5-2{\sqrt {5}}}}\,}
60°: triángulo equilátero
editar
sen
π
3
=
sen
60
∘
=
3
2
{\displaystyle \operatorname {sen} {\frac {\pi }{3}}=\operatorname {sen} 60^{\circ }={\frac {\sqrt {3}}{2}}\,}
cos
π
3
=
cos
60
∘
=
1
2
{\displaystyle \cos {\frac {\pi }{3}}=\cos 60^{\circ }={\frac {1}{2}}\,}
tan
π
3
=
tan
60
∘
=
3
{\displaystyle \tan {\frac {\pi }{3}}=\tan 60^{\circ }={\sqrt {3}}\,}
cot
π
3
=
cot
60
∘
=
3
3
=
1
3
{\displaystyle \cot {\frac {\pi }{3}}=\cot 60^{\circ }={\frac {\sqrt {3}}{3}}={\frac {1}{\sqrt {3}}}\,}
sec
π
3
=
sec
60
∘
=
2
{\displaystyle \sec {\frac {\pi }{3}}=\sec 60^{\circ }=2}
csc
π
3
=
csc
60
∘
=
2
3
3
=
2
3
{\displaystyle \csc {\frac {\pi }{3}}=\csc 60^{\circ }={\frac {2{\sqrt {3}}}{3}}={\frac {2}{\sqrt {3}}}\,}
67,5°: suma 7,5° + 60°
editar
sen
3
π
8
=
sen
67.5
∘
=
1
2
2
+
2
{\displaystyle \operatorname {sen} {\frac {3\pi }{8}}=\operatorname {sen} 67.5^{\circ }={\tfrac {1}{2}}{\sqrt {2+{\sqrt {2}}}}\,}
cos
3
π
8
=
cos
67.5
∘
=
1
2
2
−
2
{\displaystyle \cos {\frac {3\pi }{8}}=\cos 67.5^{\circ }={\tfrac {1}{2}}{\sqrt {2-{\sqrt {2}}}}\,}
tan
3
π
8
=
tan
67.5
∘
=
2
+
1
{\displaystyle \tan {\frac {3\pi }{8}}=\tan 67.5^{\circ }={\sqrt {2}}+1\,}
cot
3
π
8
=
cot
67.5
∘
=
2
−
1
{\displaystyle \cot {\frac {3\pi }{8}}=\cot 67.5^{\circ }={\sqrt {2}}-1\,}
72°: suma 36° + 36°
editar
sen
2
π
5
=
sen
72
∘
=
1
4
2
(
5
+
5
)
{\displaystyle \operatorname {sen} {\frac {2\pi }{5}}=\operatorname {sen} 72^{\circ }={\tfrac {1}{4}}{\sqrt {2\left(5+{\sqrt {5}}\right)}}\,}
cos
2
π
5
=
cos
72
∘
=
1
4
(
5
−
1
)
{\displaystyle \cos {\frac {2\pi }{5}}=\cos 72^{\circ }={\tfrac {1}{4}}\left({\sqrt {5}}-1\right)\,}
tan
2
π
5
=
tan
72
∘
=
5
+
2
5
{\displaystyle \tan {\frac {2\pi }{5}}=\tan 72^{\circ }={\sqrt {5+2{\sqrt {5}}}}\,}
cot
2
π
5
=
cot
72
∘
=
1
5
5
(
5
−
2
5
)
{\displaystyle \cot {\frac {2\pi }{5}}=\cot 72^{\circ }={\tfrac {1}{5}}{\sqrt {5\left(5-2{\sqrt {5}}\right)}}\,}
75°: suma 30° + 45°
editar
sen
5
π
12
=
sen
75
∘
=
1
4
(
6
+
2
)
{\displaystyle \operatorname {sen} {\frac {5\pi }{12}}=\operatorname {sen} 75^{\circ }={\tfrac {1}{4}}\left({\sqrt {6}}+{\sqrt {2}}\right)\,}
cos
5
π
12
=
cos
75
∘
=
1
4
(
6
−
2
)
{\displaystyle \cos {\frac {5\pi }{12}}=\cos 75^{\circ }={\tfrac {1}{4}}\left({\sqrt {6}}-{\sqrt {2}}\right)\,}
tan
5
π
12
=
tan
75
∘
=
2
+
3
{\displaystyle \tan {\frac {5\pi }{12}}=\tan 75^{\circ }=2+{\sqrt {3}}\,}
cot
5
π
12
=
cot
75
∘
=
2
−
3
{\displaystyle \cot {\frac {5\pi }{12}}=\cot 75^{\circ }=2-{\sqrt {3}}\,}
90°: fundamental
editar
sen
π
2
=
sen
90
∘
=
1
{\displaystyle \operatorname {sen} {\frac {\pi }{2}}=\operatorname {sen} 90^{\circ }=1\,}
cos
π
2
=
cos
90
∘
=
0
{\displaystyle \cos {\frac {\pi }{2}}=\cos 90^{\circ }=0\,}
tan
π
2
=
tan
90
∘
está indefinida
{\displaystyle \tan {\frac {\pi }{2}}=\tan 90^{\circ }{\text{ está indefinida}}\,}
cot
π
2
=
cot
90
∘
=
0
{\displaystyle \cot {\frac {\pi }{2}}=\cot 90^{\circ }=0\,}
sec
π
2
=
sec
90
∘
está indefinida
{\displaystyle \sec {\frac {\pi }{2}}=\sec 90^{\circ }{\text{ está indefinida}}\,}
csc
π
2
=
csc
90
∘
=
1
{\displaystyle \csc {\frac {\pi }{2}}=\csc 90^{\circ }=1\,}
Lista de constantes trigonométricas de 2π/n
editar
Para las raíces cúbicas de números no reales que aparecen en esta tabla, se tiene que tomar el valor principal , que es la raíz cúbica con la parte real más grande; esta mayor parte real es siempre positiva. Por lo tanto, las sumas de las raíces cúbicas que aparecen en la tabla son todos números reales positivos.
SENO:
n
sen
(
2
π
n
)
1
0
2
0
3
1
2
3
4
1
5
1
4
(
10
+
2
5
)
6
1
2
3
7
8
1
2
2
9
i
2
(
−
1
−
−
3
2
3
−
−
1
+
−
3
2
3
)
10
1
4
(
10
−
2
5
)
11
12
1
2
13
14
1
24
3
(
112
−
14336
+
−
5549064192
3
−
14336
−
−
5549064192
3
)
15
1
8
(
15
+
3
−
10
−
2
5
)
16
1
2
(
2
−
2
)
17
1
4
8
−
2
(
15
+
17
+
34
−
2
17
−
2
17
+
3
17
−
170
+
38
17
)
18
i
4
(
4
−
4
−
3
3
−
4
+
4
−
3
3
)
20
1
4
(
5
−
1
)
24
1
4
(
6
−
2
)
{\displaystyle {\begin{array}{r|l|l|l}n&\operatorname {sen} \left({\frac {2\pi }{n}}\right)\\\hline 1&0\\\hline 2&0\\\hline 3&{\frac {1}{2}}{\sqrt {3}}\\\hline 4&1\\\hline 5&{\frac {1}{4}}\left({\sqrt {10+2{\sqrt {5}}}}\right)\\\hline 6&{\frac {1}{2}}{\sqrt {3}}\\\hline 7&\\\hline 8&{\frac {1}{2}}{\sqrt {2}}\\\hline 9&{\frac {i}{2}}\left({\sqrt[{3}]{\frac {-1-{\sqrt {-3}}}{2}}}-{\sqrt[{3}]{\frac {-1+{\sqrt {-3}}}{2}}}\right)\\\hline 10&{\frac {1}{4}}\left({\sqrt {10-2{\sqrt {5}}}}\right)\\\hline 11&\\\hline 12&{\frac {1}{2}}\\\hline 13&\\\hline 14&{\frac {1}{24}}{\sqrt {3\left(112-{\sqrt[{3}]{14336+{\sqrt {-5549064192}}}}-{\sqrt[{3}]{14336-{\sqrt {-5549064192}}}}\right)}}\\\hline 15&{\frac {1}{8}}\left({\sqrt {15}}+{\sqrt {3}}-{\sqrt {10-2{\sqrt {5}}}}\right)\\\hline 16&{\frac {1}{2}}\left({\sqrt {2-{\sqrt {2}}}}\right)\\\hline 17&{\frac {1}{4}}{\sqrt {8-{\sqrt {2\left(15+{\sqrt {17}}+{\sqrt {34-2{\sqrt {17}}}}-2{\sqrt {17+3{\sqrt {17}}-{\sqrt {170+38{\sqrt {17}}}}}}\right)}}}}\\\hline 18&{\frac {i}{4}}\left({\sqrt[{3}]{4-4{\sqrt {-3}}}}-{\sqrt[{3}]{4+4{\sqrt {-3}}}}\right)\\\hline 20&{\frac {1}{4}}\left({\sqrt {5}}-1\right)\\\hline 24&{\frac {1}{4}}\left({\sqrt {6}}-{\sqrt {2}}\right)\end{array}}}
COSENO:
n
cos
(
2
π
n
)
1
1
2
−
1
3
−
1
2
4
0
5
1
4
(
5
−
1
)
6
1
2
7
1
6
(
−
1
+
7
+
21
−
3
2
3
+
7
−
21
−
3
2
3
)
8
1
2
2
9
1
2
(
−
1
+
−
3
2
3
+
−
1
−
−
3
2
3
)
10
1
4
(
5
+
1
)
11
12
1
2
3
13
1
12
(
104
−
20
13
+
12
−
39
3
+
104
−
20
13
−
12
−
39
3
+
13
−
1
)
14
1
24
3
(
80
+
14336
+
−
5549064192
3
+
14336
−
−
5549064192
3
)
15
1
8
(
1
+
5
+
30
−
6
5
)
16
1
2
(
2
+
2
)
17
1
16
(
−
1
+
17
+
34
−
2
17
+
2
17
+
3
17
−
34
−
2
17
−
2
34
+
2
17
)
18
1
4
(
4
+
4
−
3
3
+
4
−
4
−
3
3
)
20
1
4
(
10
+
2
5
)
24
1
4
(
6
+
2
)
{\displaystyle {\begin{array}{r|l|l|l}n&\cos \left({\frac {2\pi }{n}}\right)\\\hline 1&1\\\hline 2&-1\\\hline 3&-{\frac {1}{2}}\\\hline 4&0\\\hline 5&{\frac {1}{4}}\left({\sqrt {5}}-1\right)\\\hline 6&{\frac {1}{2}}\\\hline 7&{\frac {1}{6}}\left(-1+{\sqrt[{3}]{\frac {7+21{\sqrt {-3}}}{2}}}+{\sqrt[{3}]{\frac {7-21{\sqrt {-3}}}{2}}}\right)\\\hline 8&{\frac {1}{2}}{\sqrt {2}}\\\hline 9&{\frac {1}{2}}\left({\sqrt[{3}]{\frac {-1+{\sqrt {-3}}}{2}}}+{\sqrt[{3}]{\frac {-1-{\sqrt {-3}}}{2}}}\right)\\\hline 10&{\frac {1}{4}}\left({\sqrt {5}}+1\right)\\\hline 11&\\\hline 12&{\frac {1}{2}}{\sqrt {3}}\\\hline 13&{\frac {1}{12}}\left({\sqrt[{3}]{104-20{\sqrt {13}}+12{\sqrt {-39}}}}+{\sqrt[{3}]{104-20{\sqrt {13}}-12{\sqrt {-39}}}}+{\sqrt {13}}-1\right)\\\hline 14&{\frac {1}{24}}{\sqrt {3\left(80+{\sqrt[{3}]{14336+{\sqrt {-5549064192}}}}+{\sqrt[{3}]{14336-{\sqrt {-5549064192}}}}\right)}}\\\hline 15&{\frac {1}{8}}\left(1+{\sqrt {5}}+{\sqrt {30-6{\sqrt {5}}}}\right)\\\hline 16&{\frac {1}{2}}\left({\sqrt {2+{\sqrt {2}}}}\right)\\\hline 17&{\frac {1}{16}}\left(-1+{\sqrt {17}}+{\sqrt {34-2{\sqrt {17}}}}+2{\sqrt {17+3{\sqrt {17}}-{\sqrt {34-2{\sqrt {17}}}}-2{\sqrt {34+2{\sqrt {17}}}}}}\right)\\\hline 18&{\frac {1}{4}}\left({\sqrt[{3}]{4+4{\sqrt {-3}}}}+{\sqrt[{3}]{4-4{\sqrt {-3}}}}\right)\\\hline 20&{\frac {1}{4}}\left({\sqrt {10+2{\sqrt {5}}}}\right)\\\hline 24&{\frac {1}{4}}\left({\sqrt {6}}+{\sqrt {2}}\right)\end{array}}}
TANGENTE:
n
tan
(
2
π
n
)
1
0
2
0
3
−
3
4
±
∞
5
5
+
2
5
6
3
7
8
1
9
10
5
−
2
5
11
12
1
3
3
13
14
112
−
14336
+
−
5549064192
3
−
14336
−
−
5549064192
3
80
+
14336
+
−
5549064192
3
+
14336
−
−
5549064192
3
15
1
2
(
−
3
3
−
15
+
50
+
22
5
)
16
2
−
1
17
18
20
1
5
(
25
−
10
5
)
24
2
−
3
{\displaystyle {\begin{array}{r|l|l|l}n&\tan \left({\frac {2\pi }{n}}\right)\\\hline 1&0\\\hline 2&0\\\hline 3&-{\sqrt {3}}\\\hline 4&\pm \infty \\\hline 5&{\sqrt {5+2{\sqrt {5}}}}\\\hline 6&{\sqrt {3}}\\\hline 7&\\\hline 8&1\\\hline 9&\\\hline 10&{\sqrt {5-2{\sqrt {5}}}}\\\hline 11&\\\hline 12&{\frac {1}{3}}{\sqrt {3}}\\\hline 13&\\\hline 14&{\sqrt {\frac {112-{\sqrt[{3}]{14336+{\sqrt {-5549064192}}}}-{\sqrt[{3}]{14336-{\sqrt {-5549064192}}}}}{80+{\sqrt[{3}]{14336+{\sqrt {-5549064192}}}}+{\sqrt[{3}]{14336-{\sqrt {-5549064192}}}}}}}\\\hline 15&{\frac {1}{2}}\left(-3{\sqrt {3}}-{\sqrt {15}}+{\sqrt {50+22{\sqrt {5}}}}\right)\\\hline 16&{\sqrt {2}}-1\\\hline 17&\\\hline 18&\\\hline 20&{\frac {1}{5}}\left({\sqrt {25-10{\sqrt {5}}}}\right)\\\hline 24&2-{\sqrt {3}}\end{array}}}
Usos de las constantes
editar
Como ejemplo del uso de estas constantes, considere el volumen de un dodecaedro regular , donde a es la longitud de una arista:
V
=
5
a
3
cos
36
∘
tan
2
36
∘
.
{\displaystyle V={\frac {5a^{3}\cos 36^{\circ }}{\tan ^{2}{36^{\circ }}}}.}
Utilizando
cos
36
∘
=
5
+
1
4
,
{\displaystyle \cos 36^{\circ }={\frac {{\sqrt {5}}+1}{4}},\,}
tan
36
∘
=
5
−
2
5
,
{\displaystyle \tan 36^{\circ }={\sqrt {5-2{\sqrt {5}}}},\,}
esto se puede simplificar a:
V
=
a
3
(
15
+
7
5
)
4
.
{\displaystyle V={\frac {a^{3}\left(15+7{\sqrt {5}}\right)}{4}}.\,}
Deducción de triángulos
editar
Polígono regular (de "n"-lados) y su triángulo rectángulo fundamental. Ángulos: a = 180° / n b =90(1 - 2 / n La deducción de los valores de seno, coseno y tangente en formas radiales se basa en el construibilidad de los triángulos rectángulos.
Aquí, los triángulos rectángulos formados a partir de secciones de simetría de polígonos regulares se utilizan para calcular sus razones trigonométricas fundamentales. Cada triángulo rectángulo representa tres puntos en un polígono regular: un vértice, un centro de arista que contiene ese vértice y el centro del polígono. Un n -gono se puede dividir en 2n triángulos rectángulos con ángulos de 180 / n 180 / n n = 3, 4, 5,…
La construcción de polígonos de 3, 4, 5 y 15 lados es la base, y las bisectrices de los ángulos permiten deducir también múltiplos de dos.
Construibles
Polígonos regulares de 3 × 2n n = 0, 1, 2, 3, ...
Triángulo de 30° -60° -90°: triángulo equilátero (3 lados)
Triángulo de 60° -30° -90°: hexágono (6 lados)
Triángulo de 75° -15° -90°: dodecágono (12 lados)
Triángulo de 82,5° -7,5° -90°: icositetrágono (24 lados)
Triángulo de 86,25° -3,75° -90°: tetracontaoctágono (48 lados)
Triángulo de 88,125° -1,875° -90°: eneacontaexágono (96 lados)
Triángulo de 89,0625° -0,9375° -90°: 192-gono
Triángulo de 89,53125° -0,46875° -90°: 384-gono
...
4 × 2n Triángulo de 45° -45° -90°: cuadrado (4 lados)
Triángulo de 67,5° -22,5° -90°: octógono (8 lados)
Triángulo de 78,75° -11,25° -90°: hexadecágono (16 lados)
Triángulo de 84,375° -5,625° -90°: triacontadígono (32 lados)
Triángulo de 87,1875° -2,8125° -90°: hexacontatetrágono (64 lados)
Triángulo de 88,09375° -1,40625° -90°: 128-gono
Triángulo de 89,046875° -0,703125° -90°: 256-gono
...
5 × 2n Triángulo de 54° -36° -90°: pentágono (5 lados)
Triángulo de 72° -18° -90°: decágono (10 lados)
Triángulo de 81° -9° -90°: isodecágono (20 lados)
Triángulo de 85,5° -4,5° -90°: tetracontágono (40 lados)
Triángulo de 87,75° -2,25° -90°: octacontágono (80 lados)
Triángulo de 88,875° -1,125° -90°: 160-gono
Triángulo de 89,4375° -0,5625° -90°: 320-gono
...
15 × 2n
... También hay polígonos regulares construibles superiores: 17 , 51, 85, 255, 257 , 353, 449, 641, 1409, 2547, ..., 65535, 65537 , 69481, 73697, ..., 4294967295). No construibles (con ángulos enteros o de medio grado) - No son posibles expresiones radicales finitas que involucren números reales para estas relaciones de aristas de triángulo, por lo tanto, sus múltiplos de dos tampoco son posibles.
9 × 2n Triángulo de 70° -20° -90°: eneágono (9 lados)
Triángulo de 80° -10° -90°: octodecágono (18 lados)
Triángulo de 85° -5° -90°: triacontahexágono (36 lados)
Triángulo de 87,5° -2,5° -90°: heptacontágono (72 lados)
...
45 × 2n Triángulo de 86° -4° -90°: tetracontapentágono (45 lados)
Triángulo de 88° -2° -90°: eneacontágono (90 lados)
Triángulo de 89° -1° -90°: 180 gono
Triángulo de 89,5° -0,5° -90°: 360-gono
... Valores trigonométricos calculados para seno y coseno
editar
Valores triviales
editar
En formato de grados, sen y cos de 0, 30, 45, 60 y 90 se pueden calcular a partir de sus triángulos rectángulos, utilizando el teorema de Pitágoras.
En formato radianes, sen y cos de π/2n se pueden expresar en formato radical aplicando de forma recursiva lo siguiente:
2
cos
θ
=
2
+
2
cos
2
θ
=
2
+
2
+
2
cos
4
θ
=
2
+
2
+
2
+
2
cos
8
θ
{\displaystyle 2\cos \theta ={\sqrt {2+2\cos 2\theta }}={\sqrt {2+{\sqrt {2+2\cos 4\theta }}}}={\sqrt {2+{\sqrt {2+{\sqrt {2+2\cos 8\theta }}}}}}}
2
sen
θ
=
2
−
2
cos
2
θ
=
2
−
2
+
2
cos
4
θ
=
2
−
2
+
2
+
2
cos
8
θ
{\displaystyle 2\operatorname {sen} \theta ={\sqrt {2-2\cos 2\theta }}={\sqrt {2-{\sqrt {2+2\cos 4\theta }}}}={\sqrt {2-{\sqrt {2+{\sqrt {2+2\cos 8\theta }}}}}}}
Por ejemplo:
cos
π
2
1
=
0
2
{\displaystyle \cos {\frac {\pi }{2^{1}}}={\frac {0}{2}}}
cos
π
2
2
=
2
+
0
2
{\displaystyle \cos {\frac {\pi }{2^{2}}}={\frac {\sqrt {2+0}}{2}}}
sen
π
2
2
=
2
−
0
2
{\displaystyle \operatorname {sen} {\frac {\pi }{2^{2}}}={\frac {\sqrt {2-0}}{2}}}
cos
π
2
3
=
2
+
2
2
{\displaystyle \cos {\frac {\pi }{2^{3}}}={\frac {\sqrt {2+{\sqrt {2}}}}{2}}}
sen
π
2
3
=
2
−
2
2
{\displaystyle \operatorname {sen} {\frac {\pi }{2^{3}}}={\frac {\sqrt {2-{\sqrt {2}}}}{2}}}
cos
π
2
4
=
2
+
2
+
2
2
{\displaystyle \cos {\frac {\pi }{2^{4}}}={\frac {\sqrt {2+{\sqrt {2+{\sqrt {2}}}}}}{2}}}
sen
π
2
4
=
2
−
2
+
2
2
{\displaystyle \operatorname {sen} {\frac {\pi }{2^{4}}}={\frac {\sqrt {2-{\sqrt {2+{\sqrt {2}}}}}}{2}}}
cos
π
2
5
=
2
+
2
+
2
+
2
2
{\displaystyle \cos {\frac {\pi }{2^{5}}}={\frac {\sqrt {2+{\sqrt {2+{\sqrt {2+{\sqrt {2}}}}}}}}{2}}}
sen
π
2
5
=
2
−
2
+
2
+
2
2
{\displaystyle \operatorname {sen} {\frac {\pi }{2^{5}}}={\frac {\sqrt {2-{\sqrt {2+{\sqrt {2+{\sqrt {2}}}}}}}}{2}}}
cos
π
2
6
=
2
+
2
+
2
+
2
+
2
2
{\displaystyle \cos {\frac {\pi }{2^{6}}}={\frac {\sqrt {2+{\sqrt {2+{\sqrt {2+{\sqrt {2+{\sqrt {2}}}}}}}}}}{2}}}
sen
π
2
6
=
2
−
2
+
2
+
2
+
2
2
{\displaystyle \operatorname {sen} {\frac {\pi }{2^{6}}}={\frac {\sqrt {2-{\sqrt {2+{\sqrt {2+{\sqrt {2+{\sqrt {2}}}}}}}}}}{2}}}
y así sucesivamente.
Forma radical, sen y cos de π / (3 × 2n )
editar
cos
2
π
3
=
−
1
2
{\displaystyle \cos {\frac {2\pi }{3}}={\frac {-1}{2}}}
cos
π
3
×
2
0
=
2
−
1
2
{\displaystyle \cos {\frac {\pi }{3\times 2^{0}}}={\frac {\sqrt {2-1}}{2}}}
sen
π
3
×
2
0
=
2
+
1
2
{\displaystyle \operatorname {sen} {\frac {\pi }{3\times 2^{0}}}={\frac {\sqrt {2+1}}{2}}}
cos
π
3
×
2
1
=
2
+
1
2
{\displaystyle \cos {\frac {\pi }{3\times 2^{1}}}={\frac {\sqrt {2+1}}{2}}}
sen
π
3
×
2
1
=
2
−
1
2
{\displaystyle \operatorname {sen} {\frac {\pi }{3\times 2^{1}}}={\frac {\sqrt {2-1}}{2}}}
cos
π
3
×
2
2
=
2
+
3
2
{\displaystyle \cos {\frac {\pi }{3\times 2^{2}}}={\frac {\sqrt {2+{\sqrt {3}}}}{2}}}
sen
π
3
×
2
2
=
2
−
3
2
{\displaystyle \operatorname {sen} {\frac {\pi }{3\times 2^{2}}}={\frac {\sqrt {2-{\sqrt {3}}}}{2}}}
cos
π
3
×
2
3
=
2
+
2
+
3
2
{\displaystyle \cos {\frac {\pi }{3\times 2^{3}}}={\frac {\sqrt {2+{\sqrt {2+{\sqrt {3}}}}}}{2}}}
sen
π
3
×
2
3
=
2
−
2
+
3
2
{\displaystyle \operatorname {sen} {\frac {\pi }{3\times 2^{3}}}={\frac {\sqrt {2-{\sqrt {2+{\sqrt {3}}}}}}{2}}}
cos
π
3
×
2
4
=
2
+
2
+
2
+
3
2
{\displaystyle \cos {\frac {\pi }{3\times 2^{4}}}={\frac {\sqrt {2+{\sqrt {2+{\sqrt {2+{\sqrt {3}}}}}}}}{2}}}
sen
π
3
×
2
4
=
2
−
2
+
2
+
3
2
{\displaystyle \operatorname {sen} {\frac {\pi }{3\times 2^{4}}}={\frac {\sqrt {2-{\sqrt {2+{\sqrt {2+{\sqrt {3}}}}}}}}{2}}}
cos
π
3
×
2
5
=
2
+
2
+
2
+
2
+
3
2
{\displaystyle \cos {\frac {\pi }{3\times 2^{5}}}={\frac {\sqrt {2+{\sqrt {2+{\sqrt {2+{\sqrt {2+{\sqrt {3}}}}}}}}}}{2}}}
sen
π
3
×
2
5
=
2
−
2
+
2
+
2
+
3
2
{\displaystyle \operatorname {sen} {\frac {\pi }{3\times 2^{5}}}={\frac {\sqrt {2-{\sqrt {2+{\sqrt {2+{\sqrt {2+{\sqrt {3}}}}}}}}}}{2}}}
y así sucesivamente.
Forma radical, sen y cos de π / (5 × 2n )
editar
cos
2
π
5
=
5
−
1
4
{\displaystyle \cos {\frac {2\pi }{5}}={\frac {{\sqrt {5}}-1}{4}}}
cos
π
5
×
2
0
=
5
+
1
4
{\displaystyle \cos {\frac {\pi }{5\times 2^{0}}}={\frac {{\sqrt {5}}+1}{4}}}
2
+
2
cos
π
5
=
2
+
1.25
+
0.5
{\displaystyle 2+2\cos {\frac {\pi }{5}}=2+{\sqrt {1.25}}+0.5}
cos
π
5
×
2
1
=
2.5
+
1.25
2
{\displaystyle \cos {\frac {\pi }{5\times 2^{1}}}={\frac {\sqrt {2.5+{\sqrt {1.25}}}}{2}}}
sen
π
5
×
2
1
=
1.5
−
1.25
2
{\displaystyle \operatorname {sen} {\frac {\pi }{5\times 2^{1}}}={\frac {\sqrt {1.5-{\sqrt {1.25}}}}{2}}}
cos
π
5
×
2
2
=
2
+
2.5
+
1.25
2
{\displaystyle \cos {\frac {\pi }{5\times 2^{2}}}={\frac {\sqrt {2+{\sqrt {2.5+{\sqrt {1.25}}}}}}{2}}}
sen
π
5
×
2
2
=
2
−
2.5
+
1.25
2
{\displaystyle \operatorname {sen} {\frac {\pi }{5\times 2^{2}}}={\frac {\sqrt {2-{\sqrt {2.5+{\sqrt {1.25}}}}}}{2}}}
cos
π
5
×
2
3
=
2
+
2
+
2.5
+
1.25
2
{\displaystyle \cos {\frac {\pi }{5\times 2^{3}}}={\frac {\sqrt {2+{\sqrt {2+{\sqrt {2.5+{\sqrt {1.25}}}}}}}}{2}}}
sen
π
5
×
2
3
=
2
−
2
+
2.5
+
1.25
2
{\displaystyle \operatorname {sen} {\frac {\pi }{5\times 2^{3}}}={\frac {\sqrt {2-{\sqrt {2+{\sqrt {2.5+{\sqrt {1.25}}}}}}}}{2}}}
cos
π
5
×
2
4
=
2
+
2
+
2
+
2.5
+
1.25
2
{\displaystyle \cos {\frac {\pi }{5\times 2^{4}}}={\frac {\sqrt {2+{\sqrt {2+{\sqrt {2+{\sqrt {2.5+{\sqrt {1.25}}}}}}}}}}{2}}}
sen
π
5
×
2
4
=
2
−
2
+
2
+
2.5
+
1.25
2
{\displaystyle \operatorname {sen} {\frac {\pi }{5\times 2^{4}}}={\frac {\sqrt {2-{\sqrt {2+{\sqrt {2+{\sqrt {2.5+{\sqrt {1.25}}}}}}}}}}{2}}}
cos
π
5
×
2
5
=
2
+
2
+
2
+
2
+
2.5
+
1.25
2
{\displaystyle \cos {\frac {\pi }{5\times 2^{5}}}={\frac {\sqrt {2+{\sqrt {2+{\sqrt {2+{\sqrt {2+{\sqrt {2.5+{\sqrt {1.25}}}}}}}}}}}}{2}}}
sen
π
5
×
2
5
=
2
−
2
+
2
+
2
+
2.5
+
1.25
2
{\displaystyle \operatorname {sen} {\frac {\pi }{5\times 2^{5}}}={\frac {\sqrt {2-{\sqrt {2+{\sqrt {2+{\sqrt {2+{\sqrt {2.5+{\sqrt {1.25}}}}}}}}}}}}{2}}}
y así sucesivamente.
Forma radical, sen y cos de π / (5 × 3 × 2n
editar
cos
π
15
×
2
0
=
0.703125
+
1.875
+
0.3125
−
0.25
2
{\displaystyle \cos {\frac {\pi }{15\times 2^{0}}}={\frac {{\sqrt {{\sqrt {0.703125}}+1.875}}+{\sqrt {0.3125}}-0.25}{2}}}
cos
π
15
×
2
1
=
0.703125
+
1.875
+
0.3125
+
1.75
2
{\displaystyle \cos {\frac {\pi }{15\times 2^{1}}}={\frac {\sqrt {{\sqrt {{\sqrt {0.703125}}+1.875}}+{\sqrt {0.3125}}+1.75}}{2}}}
sen
π
15
×
2
1
=
2.25
−
0.703125
+
1.875
−
0.3125
2
{\displaystyle \operatorname {sen} {\frac {\pi }{15\times 2^{1}}}={\frac {\sqrt {2.25-{\sqrt {{\sqrt {0.703125}}+1.875}}-{\sqrt {0.3125}}}}{2}}}
cos
π
15
×
2
2
=
2
+
0.703125
+
1.875
+
0.3125
+
1.75
2
{\displaystyle \cos {\frac {\pi }{15\times 2^{2}}}={\frac {\sqrt {2+{\sqrt {{\sqrt {{\sqrt {0.703125}}+1.875}}+{\sqrt {0.3125}}+1.75}}}}{2}}}
sen
π
15
×
2
2
=
2
−
0.703125
+
1.875
+
0.3125
+
1.75
2
{\displaystyle \operatorname {sen} {\frac {\pi }{15\times 2^{2}}}={\frac {\sqrt {2-{\sqrt {{\sqrt {{\sqrt {0.703125}}+1.875}}+{\sqrt {0.3125}}+1.75}}}}{2}}}
cos
π
15
×
2
3
=
2
+
2
+
0.703125
+
1.875
+
0.3125
+
1.75
2
{\displaystyle \cos {\frac {\pi }{15\times 2^{3}}}={\frac {\sqrt {2+{\sqrt {2+{\sqrt {{\sqrt {{\sqrt {0.703125}}+1.875}}+{\sqrt {0.3125}}+1.75}}}}}}{2}}}
sen
π
15
×
2
3
=
2
−
2
+
0.703125
+
1.875
+
0.3125
+
1.75
2
{\displaystyle \operatorname {sen} {\frac {\pi }{15\times 2^{3}}}={\frac {\sqrt {2-{\sqrt {2+{\sqrt {{\sqrt {{\sqrt {0.703125}}+1.875}}+{\sqrt {0.3125}}+1.75}}}}}}{2}}}
cos
π
15
×
2
4
=
2
+
2
+
2
+
0.703125
+
1.875
+
0.3125
+
1.75
2
{\displaystyle \cos {\frac {\pi }{15\times 2^{4}}}={\frac {\sqrt {2+{\sqrt {2+{\sqrt {2+{\sqrt {{\sqrt {{\sqrt {0.703125}}+1.875}}+{\sqrt {0.3125}}+1.75}}}}}}}}{2}}}
sen
π
15
×
2
4
=
2
−
2
+
2
+
0.703125
+
1.875
+
0.3125
+
1.75
2
{\displaystyle \operatorname {sen} {\frac {\pi }{15\times 2^{4}}}={\frac {\sqrt {2-{\sqrt {2+{\sqrt {2+{\sqrt {{\sqrt {{\sqrt {0.703125}}+1.875}}+{\sqrt {0.3125}}+1.75}}}}}}}}{2}}}
cos
π
15
×
2
5
=
2
+
2
+
2
+
2
+
0.703125
+
1.875
+
0.3125
+
1.75
2
{\displaystyle \cos {\frac {\pi }{15\times 2^{5}}}={\frac {\sqrt {2+{\sqrt {2+{\sqrt {2+{\sqrt {2+{\sqrt {{\sqrt {{\sqrt {0.703125}}+1.875}}+{\sqrt {0.3125}}+1.75}}}}}}}}}}{2}}}
sen
π
15
×
2
5
=
2
−
2
+
2
+
2
+
0.703125
+
1.875
+
0.3125
+
1.75
2
{\displaystyle \operatorname {sen} {\frac {\pi }{15\times 2^{5}}}={\frac {\sqrt {2-{\sqrt {2+{\sqrt {2+{\sqrt {2+{\sqrt {{\sqrt {{\sqrt {0.703125}}+1.875}}+{\sqrt {0.3125}}+1.75}}}}}}}}}}{2}}}
y así sucesivamente.
Forma radical, sen y cos de π / (17 × 2n )
editar
Si
M
=
2
(
17
+
17
)
{\displaystyle M=2(17+{\sqrt {17}})}
N
=
2
(
17
−
17
)
{\displaystyle N=2(17-{\sqrt {17}})}
cos
π
17
=
M
−
4
+
2
(
N
+
2
(
2
M
−
N
+
17
N
−
N
−
8
M
)
)
8
.
{\displaystyle \cos {\frac {\pi }{17}}={\frac {\sqrt {M-4+2({\sqrt {N}}+{\sqrt {2(2M-N+{\sqrt {17N}}-{\sqrt {N}}-8{\sqrt {M}})}})}}{8}}.}
Por lo tanto, aplicando inducción:
cos
π
17
×
2
0
=
30
+
2
17
+
136
−
8
17
+
272
+
48
17
+
8
34
−
2
17
×
(
17
−
1
)
−
64
34
+
2
17
8
;
{\displaystyle \cos {\frac {\pi }{17\times 2^{0}}}={\frac {\sqrt {30+2{\sqrt {17}}+{\sqrt {136-8{\sqrt {17}}}}+{\sqrt {272+48{\sqrt {17}}+8{\sqrt {34-2{\sqrt {17}}}}\times ({\sqrt {17}}-1)-64{\sqrt {34+2{\sqrt {17}}}}}}}}{8}};}
cos
π
17
×
2
n
+
1
=
2
+
2
cos
π
17
×
2
n
2
{\displaystyle \cos {\frac {\pi }{17\times 2^{n+1}}}={\frac {\sqrt {2+2\cos {\frac {\pi }{17\times 2^{n}}}}}{2}}}
sen
π
17
×
2
n
+
1
=
2
−
2
cos
π
17
×
2
n
2
.
{\displaystyle \operatorname {sen} {\frac {\pi }{17\times 2^{n+1}}}={\frac {\sqrt {2-2\cos {\frac {\pi }{17\times 2^{n}}}}}{2}}.}
Forma radical, sen y cos de π / (257 × 2n ) π / (65537 × 2n )
editar
La inducción anterior se puede aplicar de la misma manera a todos los Número de Fermat restantes (F3 = 223 + 1 = 28 + 1 = 257 y F4 = 224 + 1 = 216 + 1 = 65537 ), los factores de π cuyas expresiones radicales cos y sen se sabe que existen pero son muy largas para expresarlas aquí.
cos
π
257
×
2
n
+
1
=
2
+
2
cos
π
257
×
2
n
2
{\displaystyle \cos {\frac {\pi }{257\times 2^{n+1}}}={\frac {\sqrt {2+2\cos {\frac {\pi }{257\times 2^{n}}}}}{2}}}
sen
π
257
×
2
n
+
1
=
2
−
2
cos
π
257
×
2
n
2
;
{\displaystyle \operatorname {sen} {\frac {\pi }{257\times 2^{n+1}}}={\frac {\sqrt {2-2\cos {\frac {\pi }{257\times 2^{n}}}}}{2}};}
cos
π
65537
×
2
n
+
1
=
2
+
2
cos
π
65537
×
2
n
2
{\displaystyle \cos {\frac {\pi }{65537\times 2^{n+1}}}={\frac {\sqrt {2+2\cos {\frac {\pi }{65537\times 2^{n}}}}}{2}}}
sen
π
65537
×
2
n
+
1
=
2
−
2
cos
π
65537
×
2
n
2
.
{\displaystyle \operatorname {sen} {\frac {\pi }{65537\times 2^{n+1}}}={\frac {\sqrt {2-2\cos {\frac {\pi }{65537\times 2^{n}}}}}{2}}.}
Forma radical, sen y cos de π / (255 × 2n ) π / (65535 × 2n ) π / (4294967295 × 2n )
editar
D = 232 - 1 = 4,294,967,295 es el denominador entero impar más grande para el cual se sabe que existen formas radicales para sen (π / D) y cos (π / D).
Usando los valores de forma radical de las secciones anteriores, y aplicando cos (A-B) = cosA cosB + senA senB, seguido de inducción, se obtiene -
cos
π
255
×
2
0
=
2
+
2
cos
(
π
15
−
π
17
)
2
{\displaystyle \cos {\frac {\pi }{255\times 2^{0}}}={\frac {\sqrt {2+2\cos({\frac {\pi }{15}}-{\frac {\pi }{17}})}}{2}}}
sen
π
255
×
2
0
=
2
−
2
cos
(
π
15
−
π
17
)
2
;
{\displaystyle \operatorname {sen} {\frac {\pi }{255\times 2^{0}}}={\frac {\sqrt {2-2\cos({\frac {\pi }{15}}-{\frac {\pi }{17}})}}{2}};}
cos
π
255
×
2
n
+
1
=
2
+
2
cos
π
255
×
2
n
2
{\displaystyle \cos {\frac {\pi }{255\times 2^{n+1}}}={\frac {\sqrt {2+2\cos {\frac {\pi }{255\times 2^{n}}}}}{2}}}
sen
π
255
×
2
n
+
1
=
2
−
2
cos
π
255
×
2
n
2
;
{\displaystyle \operatorname {sen} {\frac {\pi }{255\times 2^{n+1}}}={\frac {\sqrt {2-2\cos {\frac {\pi }{255\times 2^{n}}}}}{2}};}
Por lo tanto, utilizando los valores de forma radical de las secciones anteriores y aplicando cos (A-B) = cosA cosB + senA senB, seguido de inducción, se obtiene -
cos
π
65535
×
2
0
=
2
+
2
cos
(
π
255
−
π
257
)
2
{\displaystyle \cos {\frac {\pi }{65535\times 2^{0}}}={\frac {\sqrt {2+2\cos({\frac {\pi }{255}}-{\frac {\pi }{257}})}}{2}}}
sen
π
65535
×
2
0
=
2
−
2
cos
(
π
255
−
π
257
)
2
;
{\displaystyle \operatorname {sen} {\frac {\pi }{65535\times 2^{0}}}={\frac {\sqrt {2-2\cos({\frac {\pi }{255}}-{\frac {\pi }{257}})}}{2}};}
cos
π
65535
×
2
n
+
1
=
2
+
2
cos
π
65535
×
2
n
2
{\displaystyle \cos {\frac {\pi }{65535\times 2^{n+1}}}={\frac {\sqrt {2+2\cos {\frac {\pi }{65535\times 2^{n}}}}}{2}}}
sen
π
65535
×
2
n
+
1
=
2
−
2
cos
π
65535
×
2
n
2
.
{\displaystyle \operatorname {sen} {\frac {\pi }{65535\times 2^{n+1}}}={\frac {\sqrt {2-2\cos {\frac {\pi }{65535\times 2^{n}}}}}{2}}.}
Finalmente, usando los valores de forma radical de las secciones anteriores, y aplicando cos (A-B) = cosA cosB + senA senB, seguido de inducción, se obtiene -
cos
π
4294967295
×
2
0
=
2
+
2
cos
(
π
65535
−
π
65537
)
2
{\displaystyle \cos {\frac {\pi }{4294967295\times 2^{0}}}={\frac {\sqrt {2+2\cos({\frac {\pi }{65535}}-{\frac {\pi }{65537}})}}{2}}}
sen
π
4294967295
×
2
0
=
2
−
2
cos
(
π
65535
−
π
65537
)
2
;
{\displaystyle \operatorname {sen} {\frac {\pi }{4294967295\times 2^{0}}}={\frac {\sqrt {2-2\cos({\frac {\pi }{65535}}-{\frac {\pi }{65537}})}}{2}};}
cos
π
4294967295
×
2
n
+
1
=
2
+
2
cos
π
4294967295
×
2
n
2
{\displaystyle \cos {\frac {\pi }{4294967295\times 2^{n+1}}}={\frac {\sqrt {2+2\cos {\frac {\pi }{4294967295\times 2^{n}}}}}{2}}}
sen
π
4294967295
×
2
n
+
1
=
2
−
2
cos
π
4294967295
×
2
n
2
.
{\displaystyle \operatorname {sen} {\frac {\pi }{4294967295\times 2^{n+1}}}={\frac {\sqrt {2-2\cos {\frac {\pi }{4294967295\times 2^{n}}}}}{2}}.}
La expansión de la última forma radical es muy grande, por lo que se expresa en la forma más simple anterior.
n × π / (5 × 2m
editar
Cuerda(36°) = a / b 1 / φ número áureo , a partir del teorema de Ptolomeo Método geométrico
editar
Aplicando el teorema de Ptolomeo al cuadrilátero cíclico ABCD definido por cuatro vértices sucesivos del pentágono, se deduce que:
crd
36
∘
=
crd
(
∠
A
D
B
)
=
a
b
=
2
1
+
5
=
5
−
1
2
{\displaystyle \operatorname {crd} 36^{\circ }=\operatorname {crd} (\angle \mathrm {ADB} )={\frac {a}{b}}={\frac {2}{1+{\sqrt {5}}}}={\frac {{\sqrt {5}}-1}{2}}}
que es el 1 / φ número áureo . crd es la función cuerda ,
crd
θ
=
2
sen
θ
2
.
{\displaystyle \operatorname {crd} \ {\theta }=2\operatorname {sen} {\frac {\theta }{2}}.\,}
(Consúltese también la tabla de cuerdas de Ptolomeo )
Así
sen
18
∘
=
1
1
+
5
=
5
−
1
4
.
{\displaystyle \operatorname {sen} 18^{\circ }={\frac {1}{1+{\sqrt {5}}}}={\frac {{\sqrt {5}}-1}{4}}.}
(Alternativamente, sin usar el teorema de Ptolomeo, rotúlese como X la intersección de AC y BD, y obsérvese considerando los ángulos que el triángulo AXB es isósceles , entonces AX = AB = a . Los triángulos AXD y CXB son semejantes , porque AD es paralelo a BC. Entonces XC = a · (a / b a + a 2 / b b . Al resolver esta expresión, se obtiene a / b 1 / φ
De forma similar
crd
108
∘
=
crd
(
∠
A
B
C
)
=
b
a
=
1
+
5
2
,
{\displaystyle \operatorname {crd} \ 108^{\circ }=\operatorname {crd} (\angle \mathrm {ABC} )={\frac {b}{a}}={\frac {1+{\sqrt {5}}}{2}},}
y entonces
sen
54
∘
=
cos
36
∘
=
1
+
5
4
.
{\displaystyle \operatorname {sen} 54^{\circ }=\cos 36^{\circ }={\frac {1+{\sqrt {5}}}{4}}.}
Método algebraico
editar
Si θ es 18° o -54°, entonces 2θ y 3θ suman 5θ = 90° o -270°, por lo tanto sen 2θ es igual a cos 3θ.
(
2
sen
θ
)
cos
θ
=
sen
2
θ
=
cos
3
θ
=
4
cos
3
θ
−
3
cos
θ
=
(
4
cos
2
θ
−
3
)
cos
θ
=
(
1
−
4
sen
2
θ
)
cos
θ
{\displaystyle (2\operatorname {sen} \theta )\cos \theta =\operatorname {sen} 2\theta =\cos 3\theta =4\cos ^{3}\theta -3\cos \theta =(4\cos ^{2}\theta -3)\cos \theta =(1-4\operatorname {sen} ^{2}\theta )\cos \theta }
Entonces,
4
sen
2
θ
+
2
sen
θ
−
1
=
0
{\displaystyle 4\operatorname {sen} ^{2}\theta +2\operatorname {sen} \theta -1=0}
sen
θ
=
sen
(
18
∘
,
−
54
∘
)
=
−
1
±
5
4
.
{\displaystyle \operatorname {sen} \theta =\operatorname {sen}(18^{\circ },-54^{\circ })={\frac {-1\pm {\sqrt {5}}}{4}}.}
Por lo tanto,
sen
(
18
∘
)
=
cos
(
72
∘
)
=
5
−
1
4
{\displaystyle \operatorname {sen}(18^{\circ })=\cos(72^{\circ })={\frac {{\sqrt {5}}-1}{4}}}
sen
(
54
∘
)
=
cos
(
36
∘
)
=
5
+
1
4
{\displaystyle \operatorname {sen}(54^{\circ })=\cos(36^{\circ })={\frac {{\sqrt {5}}+1}{4}}}
sen
(
36
∘
)
=
cos
(
54
∘
)
=
10
−
2
5
4
{\displaystyle \operatorname {sen}(36^{\circ })=\cos(54^{\circ })={\frac {\sqrt {10-2{\sqrt {5}}}}{4}}}
sen
(
72
∘
)
=
cos
(
18
∘
)
=
10
+
2
5
4
.
{\displaystyle \operatorname {sen}(72^{\circ })=\cos(18^{\circ })={\frac {\sqrt {10+2{\sqrt {5}}}}{4}}.}
Alternativamente, las fórmulas de ángulos múltiplos para funciones de 5x , donde x ∈ {18, 36, 54, 72, 90} y 5x ∈ {90, 180, 270, 360, 450}, se puede resolver para las funciones de x , ya que se conocen los valores de la función de 5x . Las fórmulas de múltiplos de ángulos son:
sen
5
x
=
16
sen
5
x
−
20
sen
3
x
+
5
sen
x
,
{\displaystyle \operatorname {sen} 5x=16\operatorname {sen} ^{5}x-20\operatorname {sen} ^{3}x+5\operatorname {sen} x,\,}
cos
5
x
=
16
cos
5
x
−
20
cos
3
x
+
5
cos
x
.
{\displaystyle \cos 5x=16\cos ^{5}x-20\cos ^{3}x+5\cos x.\,}
Cuando sen 5x = 0 o cos 5x = 0, dejamos que y = sen x o y = cos xy que resolviendo para y da:
16
y
5
−
20
y
3
+
5
y
=
0.
{\displaystyle 16y^{5}-20y^{3}+5y=0.\,}
Una solución es cero, y la ecuación de cuarto grado resultante se puede resolver como una cuadrática en y 2 . Cuando sen 5x = 1 o cos 5x = 1, dejamos de nuevo y = sen x o y = cos xy resuelva para y :
16
y
5
−
20
y
3
+
5
y
−
1
=
0
,
{\displaystyle 16y^{5}-20y^{3}+5y-1=0,\,}
que permite obtener:
(
y
−
1
)
(
4
y
2
+
2
y
−
1
)
2
=
0.
{\displaystyle (y-1)\left(4y^{2}+2y-1\right)^{2}=0.\,}
9° es 45 - 36, y 27° es 45 - 18; entonces se usan las fórmulas de la resta para seno y coseno. 6° es 36 - 30, 12° es 30 - 18, 24° es 54 - 30 y 42° es 60 - 18; entonces se usan las fórmulas de la resta para seno y coseno. 3° es 18 - 15, 21° es 36 - 15, 33° es 18 + 15 y 39° es 54 - 15, por lo que se usan las fórmulas de la resta (o la suma) para seno y coseno. Véase también
editar
↑ a b Bradie, Brian (Sep 2002). «Exact values for the sine and cosine of multiples of 18°: A geometric approach». The College Mathematics Journal 33 (4): 318-319. JSTOR 1559057 doi :10.2307/1559057 Bibliografía
editar
Enlaces externos
editar
![{\displaystyle \tan \left({\frac {\pi }{60}}\right)=\tan \left(3^{\circ }\right)={\frac {\left[\left(2-{\sqrt {3}}\right)\left(3+{\sqrt {5}}\right)-2\right]\left[2-{\sqrt {10-2{\sqrt {5}}}}\right]}{4}}\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3def0c4623027bd006e5772927044dc5f2c82f41)
![{\displaystyle \cot \left({\frac {\pi }{60}}\right)=\cot \left(3^{\circ }\right)={\frac {\left[\left(2+{\sqrt {3}}\right)\left(3+{\sqrt {5}}\right)-2\right]\left[2+{\sqrt {10-2{\sqrt {5}}}}\right]}{4}}\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5604f62a7ebebb43eb6e64baa6f88872c7e9d069)
![{\displaystyle \operatorname {sen} {\frac {\pi }{15}}=\operatorname {sen} 12^{\circ }={\tfrac {1}{8}}\left[{\sqrt {2\left(5+{\sqrt {5}}\right)}}+{\sqrt {3}}-{\sqrt {15}}\right]\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/501b094d25f92fb3934140508580a664eb5956a8)
![{\displaystyle \cos {\frac {\pi }{15}}=\cos 12^{\circ }={\tfrac {1}{8}}\left[{\sqrt {6\left(5+{\sqrt {5}}\right)}}+{\sqrt {5}}-1\right]\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/fb2ac6239ea26bcfc4e7eefcc375f7f67b6a0acb)
![{\displaystyle \tan {\frac {\pi }{15}}=\tan 12^{\circ }={\tfrac {1}{2}}\left[3{\sqrt {3}}-{\sqrt {15}}-{\sqrt {2\left(25-11{\sqrt {5}}\right)}}\,\right]\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/72b57f4f65cc3f783c5b77b860a2297a8a8cbab4)
![{\displaystyle \cot {\frac {\pi }{15}}=\cot 12^{\circ }={\tfrac {1}{2}}\left[{\sqrt {15}}+{\sqrt {3}}+{\sqrt {2\left(5+{\sqrt {5}}\right)}}\,\right]\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/fe18dfa13f93c50f52d019bc3ea48fa0ca9bf472)
![{\displaystyle \operatorname {sen} {\frac {2\pi }{15}}=\operatorname {sen} 24^{\circ }={\tfrac {1}{8}}\left[{\sqrt {15}}+{\sqrt {3}}-{\sqrt {2\left(5-{\sqrt {5}}\right)}}\right]\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4b608583ac90c7e11cfd7189b647446771fcf620)
![{\displaystyle \tan {\frac {2\pi }{15}}=\tan 24^{\circ }={\tfrac {1}{2}}\left[{\sqrt {50+22{\sqrt {5}}}}-3{\sqrt {3}}-{\sqrt {15}}\right]\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6fa351918ed8323175584ed4751bfacf5f61e67e)
![{\displaystyle \cot {\frac {2\pi }{15}}=\cot 24^{\circ }={\tfrac {1}{2}}\left[{\sqrt {15}}-{\sqrt {3}}+{\sqrt {2\left(5-{\sqrt {5}}\right)}}\right]\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/679f693d22aa397b2aa744452edc6d64496e09fd)
![{\displaystyle \operatorname {sen} {\frac {3\pi }{20}}=\operatorname {sen} 27^{\circ }={\tfrac {1}{8}}\left[2{\sqrt {5+{\sqrt {5}}}}-{\sqrt {2}}\;\left({\sqrt {5}}-1\right)\right]\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b298e1d507c0d9f81a5abb002fed56d421a64dc0)
![{\displaystyle \cos {\frac {3\pi }{20}}=\cos 27^{\circ }={\tfrac {1}{8}}\left[2{\sqrt {5+{\sqrt {5}}}}+{\sqrt {2}}\;\left({\sqrt {5}}-1\right)\right]\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/96ed3130d2995af89ffd7fed92ca75b79e11b580)
![{\displaystyle \operatorname {sen} {\frac {11\pi }{60}}=\operatorname {sen} 33^{\circ }={\tfrac {1}{16}}\left[2\left({\sqrt {3}}-1\right){\sqrt {5+{\sqrt {5}}}}+{\sqrt {2}}\left(1+{\sqrt {3}}\right)\left({\sqrt {5}}-1\right)\right]\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/acca62ba3594054a7b46bc9a0929fe2d64f834fc)
![{\displaystyle \cos {\frac {11\pi }{60}}=\cos 33^{\circ }={\tfrac {1}{16}}\left[2\left({\sqrt {3}}+1\right){\sqrt {5+{\sqrt {5}}}}+{\sqrt {2}}\left(1-{\sqrt {3}}\right)\left({\sqrt {5}}-1\right)\right]\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/89d8b1c326975404f0f829d7b88fcfd06abfee06)
![{\displaystyle \tan {\frac {11\pi }{60}}=\tan 33^{\circ }={\tfrac {1}{4}}\left[2-\left(2-{\sqrt {3}}\right)\left(3+{\sqrt {5}}\right)\right]\left[2+{\sqrt {2\left(5-{\sqrt {5}}\right)}}\,\right]\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/fdb93b95e09248b579de718f6c6670c64b38a785)
![{\displaystyle \cot {\frac {11\pi }{60}}=\cot 33^{\circ }={\tfrac {1}{4}}\left[2-\left(2+{\sqrt {3}}\right)\left(3+{\sqrt {5}}\right)\right]\left[2-{\sqrt {2\left(5-{\sqrt {5}}\right)}}\,\right]\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/637555b8bd052954d2c6cff5ec40071d29d1ffeb)
![{\displaystyle \operatorname {sen} {\frac {13\pi }{60}}=\operatorname {sen} 39^{\circ }={\tfrac {1}{16}}\left[2\left(1-{\sqrt {3}}\right){\sqrt {5-{\sqrt {5}}}}+{\sqrt {2}}\left({\sqrt {3}}+1\right)\left({\sqrt {5}}+1\right)\right]\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2f93a9e7cf6d20a08841d7aca3f639b33680aedd)
![{\displaystyle \cos {\frac {13\pi }{60}}=\cos 39^{\circ }={\tfrac {1}{16}}\left[2\left(1+{\sqrt {3}}\right){\sqrt {5-{\sqrt {5}}}}+{\sqrt {2}}\left({\sqrt {3}}-1\right)\left({\sqrt {5}}+1\right)\right]\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d84363bc38d47561a10590df7ae08c6353619596)
![{\displaystyle \tan {\frac {13\pi }{60}}=\tan 39^{\circ }={\tfrac {1}{4}}\left[\left(2-{\sqrt {3}}\right)\left(3-{\sqrt {5}}\right)-2\right]\left[2-{\sqrt {2\left(5+{\sqrt {5}}\right)}}\,\right]\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b7739885ff94d55132c58be5a2ca59b17eb44037)
![{\displaystyle \cot {\frac {13\pi }{60}}=\cot 39^{\circ }={\tfrac {1}{4}}\left[\left(2+{\sqrt {3}}\right)\left(3-{\sqrt {5}}\right)-2\right]\left[2+{\sqrt {2\left(5+{\sqrt {5}}\right)}}\,\right]\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2a960ac308ab19d558f2f1c6399a1913bfe1f5ca)
![{\displaystyle {\begin{array}{r|l|l|l}n&\operatorname {sen} \left({\frac {2\pi }{n}}\right)\\\hline 1&0\\\hline 2&0\\\hline 3&{\frac {1}{2}}{\sqrt {3}}\\\hline 4&1\\\hline 5&{\frac {1}{4}}\left({\sqrt {10+2{\sqrt {5}}}}\right)\\\hline 6&{\frac {1}{2}}{\sqrt {3}}\\\hline 7&\\\hline 8&{\frac {1}{2}}{\sqrt {2}}\\\hline 9&{\frac {i}{2}}\left({\sqrt[{3}]{\frac {-1-{\sqrt {-3}}}{2}}}-{\sqrt[{3}]{\frac {-1+{\sqrt {-3}}}{2}}}\right)\\\hline 10&{\frac {1}{4}}\left({\sqrt {10-2{\sqrt {5}}}}\right)\\\hline 11&\\\hline 12&{\frac {1}{2}}\\\hline 13&\\\hline 14&{\frac {1}{24}}{\sqrt {3\left(112-{\sqrt[{3}]{14336+{\sqrt {-5549064192}}}}-{\sqrt[{3}]{14336-{\sqrt {-5549064192}}}}\right)}}\\\hline 15&{\frac {1}{8}}\left({\sqrt {15}}+{\sqrt {3}}-{\sqrt {10-2{\sqrt {5}}}}\right)\\\hline 16&{\frac {1}{2}}\left({\sqrt {2-{\sqrt {2}}}}\right)\\\hline 17&{\frac {1}{4}}{\sqrt {8-{\sqrt {2\left(15+{\sqrt {17}}+{\sqrt {34-2{\sqrt {17}}}}-2{\sqrt {17+3{\sqrt {17}}-{\sqrt {170+38{\sqrt {17}}}}}}\right)}}}}\\\hline 18&{\frac {i}{4}}\left({\sqrt[{3}]{4-4{\sqrt {-3}}}}-{\sqrt[{3}]{4+4{\sqrt {-3}}}}\right)\\\hline 20&{\frac {1}{4}}\left({\sqrt {5}}-1\right)\\\hline 24&{\frac {1}{4}}\left({\sqrt {6}}-{\sqrt {2}}\right)\end{array}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c9d705dc002fd87aec636b3981c5af6dc81b35cb)
![{\displaystyle {\begin{array}{r|l|l|l}n&\cos \left({\frac {2\pi }{n}}\right)\\\hline 1&1\\\hline 2&-1\\\hline 3&-{\frac {1}{2}}\\\hline 4&0\\\hline 5&{\frac {1}{4}}\left({\sqrt {5}}-1\right)\\\hline 6&{\frac {1}{2}}\\\hline 7&{\frac {1}{6}}\left(-1+{\sqrt[{3}]{\frac {7+21{\sqrt {-3}}}{2}}}+{\sqrt[{3}]{\frac {7-21{\sqrt {-3}}}{2}}}\right)\\\hline 8&{\frac {1}{2}}{\sqrt {2}}\\\hline 9&{\frac {1}{2}}\left({\sqrt[{3}]{\frac {-1+{\sqrt {-3}}}{2}}}+{\sqrt[{3}]{\frac {-1-{\sqrt {-3}}}{2}}}\right)\\\hline 10&{\frac {1}{4}}\left({\sqrt {5}}+1\right)\\\hline 11&\\\hline 12&{\frac {1}{2}}{\sqrt {3}}\\\hline 13&{\frac {1}{12}}\left({\sqrt[{3}]{104-20{\sqrt {13}}+12{\sqrt {-39}}}}+{\sqrt[{3}]{104-20{\sqrt {13}}-12{\sqrt {-39}}}}+{\sqrt {13}}-1\right)\\\hline 14&{\frac {1}{24}}{\sqrt {3\left(80+{\sqrt[{3}]{14336+{\sqrt {-5549064192}}}}+{\sqrt[{3}]{14336-{\sqrt {-5549064192}}}}\right)}}\\\hline 15&{\frac {1}{8}}\left(1+{\sqrt {5}}+{\sqrt {30-6{\sqrt {5}}}}\right)\\\hline 16&{\frac {1}{2}}\left({\sqrt {2+{\sqrt {2}}}}\right)\\\hline 17&{\frac {1}{16}}\left(-1+{\sqrt {17}}+{\sqrt {34-2{\sqrt {17}}}}+2{\sqrt {17+3{\sqrt {17}}-{\sqrt {34-2{\sqrt {17}}}}-2{\sqrt {34+2{\sqrt {17}}}}}}\right)\\\hline 18&{\frac {1}{4}}\left({\sqrt[{3}]{4+4{\sqrt {-3}}}}+{\sqrt[{3}]{4-4{\sqrt {-3}}}}\right)\\\hline 20&{\frac {1}{4}}\left({\sqrt {10+2{\sqrt {5}}}}\right)\\\hline 24&{\frac {1}{4}}\left({\sqrt {6}}+{\sqrt {2}}\right)\end{array}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a62131084211d314bc7750aab3da6b4485a1d290)
![{\displaystyle {\begin{array}{r|l|l|l}n&\tan \left({\frac {2\pi }{n}}\right)\\\hline 1&0\\\hline 2&0\\\hline 3&-{\sqrt {3}}\\\hline 4&\pm \infty \\\hline 5&{\sqrt {5+2{\sqrt {5}}}}\\\hline 6&{\sqrt {3}}\\\hline 7&\\\hline 8&1\\\hline 9&\\\hline 10&{\sqrt {5-2{\sqrt {5}}}}\\\hline 11&\\\hline 12&{\frac {1}{3}}{\sqrt {3}}\\\hline 13&\\\hline 14&{\sqrt {\frac {112-{\sqrt[{3}]{14336+{\sqrt {-5549064192}}}}-{\sqrt[{3}]{14336-{\sqrt {-5549064192}}}}}{80+{\sqrt[{3}]{14336+{\sqrt {-5549064192}}}}+{\sqrt[{3}]{14336-{\sqrt {-5549064192}}}}}}}\\\hline 15&{\frac {1}{2}}\left(-3{\sqrt {3}}-{\sqrt {15}}+{\sqrt {50+22{\sqrt {5}}}}\right)\\\hline 16&{\sqrt {2}}-1\\\hline 17&\\\hline 18&\\\hline 20&{\frac {1}{5}}\left({\sqrt {25-10{\sqrt {5}}}}\right)\\\hline 24&2-{\sqrt {3}}\end{array}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/31a80b8d620055fc090dd2e55484610e289741ea)
Datos: Q2533727