Ángulos primarios en la forma (cos, sen) en la circunferencia goniométrica, múltiplos de 30 y 45 grados Las expresiones algebraicas exactas de valores trigonométricos pueden ser útiles principalmente para obtener soluciones en forma de radicales que permiten simplificar determinados resultados.
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.
[ 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
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°.
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}}\,}
[ 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)}}}
[ 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}}}}}}}}}}}}
[ 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}}}}}}}}}}
[ 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}}}}}}}}}}}
[ 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}}\,}
[ 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}}}}}}}}}
[ 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}}}}}}}}
[ 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}}}}}}}}}
[ 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}}\,}
[ 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)}
[ 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}}}}\,}
[ 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}
[ 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]\,}
[ 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}}\,}
[ 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}}}}\,}
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)\,}
[ 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
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]\,}
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}}}}\,}
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}
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]\,}
[ 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}}}}}
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]\,}
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}}\,}
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}}\,}
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}}}}\,}
[ 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}}}\,}
[ 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\,}
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)}}\,}
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}}\,}
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\,}
[ 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}}.\,}
[ 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°: icosakaitetrágono (24 lados)
Triángulo de 86,25° -3,75° -90°: tetracontakaioctágono (48 lados)
Triángulo de 88,125° -1,875° -90°: eneacontakaihexágono (96 lados)
Triángulo de 89,0625° -0,9375° -90°: hectaeneacontakaidígono (192 lados)
Triángulo de 89,53125° -0,46875° -90°: trihectaoctacontakaitetrágono (384 lados)
...
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°: triacontakaidígono (32 lados)
Triángulo de 87,1875° -2,8125° -90°: hexacontakaitetrágono (64 lados)
Triángulo de 88,09375° -1,40625° -90°: hectadicontakaioctágono (128 lados)
Triángulo de 89,046875° -0,703125° -90°: dihectapentacontakaihexágono (256 lados)
...
5 × 2n
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
45 × 2n [ 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.
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.
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.
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.
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}}.}
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}}.}
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.
Cuerda(36°) = a / b 1 / φ número áureo , a partir del teorema de Ptolomeo 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}}.}
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.
↑ 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 Weisstein, Eric W . «Trigonometry angles» . En Weisstein, Eric W, ed. MathWorld (en inglés) . Wolfram Research . 
![{\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