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Identidades trigonométricas
Paso 3 - Profundizar y contextualizar el conocimiento de la Unidad 2.
Por
Germán Daniel Rendón
Claudia Lorena Gomez
Franco Albeiro Morales
Ana Alejandra Guaranguay
José Fernando Cuaran
Nombre del curso: 551108A_951
Grupo: 551108_13
Presentado a
Mónica Trigos
Universidad Nacional Abierta y a Distancia UNAD
CCAV Pasto Zona Centro Sur
Escuela de Ciencias de la Educación
29 de marzo de 2021
Identidades trigonométricas
Son llamadas así ciertas ecuaciones que se satisfacen para cualquier ángulo; existen varios tipos de ellas:
Tipo 1: identidades de cociente, se obtienen por las definiciones de tangente y cotangente de la siguiente
manera.
Formula que aplico:
𝑎
sen 𝐴
=
𝑏
sen 𝐵
=
𝑐
sen 𝐶
Para hallar sen 𝐵
𝑎
sen 𝐴
=
𝑏
sen 𝐵
Para hallar sen 𝐶
180° = 31,3° + 120° + 𝐶
180° −31,3° − 120° = 𝐶
𝐶 = 28,7°
Para hallar el cateto de 𝑐
𝑐
sen 𝐶
=
𝑎
sen 𝐴
𝑐
sen 28,7°
=
10
sen 𝐴120°
𝐶 =
10
sen 120°
∗ sen 28,7°
𝐶 = 5,5𝑚
Tarea 2. Calcula las razones trigonométricas seno, coseno y tangente de los ángulos agudos (A y B) de cada
triángulo rectángulo que aparecen abajo.
Ejercicio a: Tenemos que encontrar los ángulos de beta, alfa y el cateto de a.
Procedemos a encontrar
cateto de a.
Angulo de alfa Angulo de beta
Identidades recíprocas, se obtienen por las definiciones de secante, cosecante y
tangente en relación con cotangente.
s𝑖𝑛 ∝ =
𝑦
ℎ
=
1
ℎ
𝑦
=
1
𝐜𝒔𝒄(∝)
, c𝑠𝑐(∝) =
ℎ
𝑦
=
1
𝑦
ℎ
=
1
sin(∝)
cos(∝) =
1
ℎ
𝑥
=
𝟏
𝒔𝒆𝒄(∝)
, sec(∝) =
ℎ
𝑥
=
1
𝑥
ℎ
=
1
cos(∝)
tan(∝) =
𝑦
𝑥
=
1
𝑥
𝑦
=
𝟏
𝐜𝐨𝐬𝒕(∝)
, cot(∝) =
𝑥
𝑦
=
1
𝑦
𝑥
=
1
tan(∝)
Con funciones trigonométricas la igualdad si se cumple.
Ejercicio a
Igualdad de cosecante de x sobre cotangente de x a 1 coseno de x.
csc 𝑥
cot 𝑥
=
1
cos 𝑥
1
𝑠𝑒𝑛 𝑥
1
tan 𝑥
=
1
cos 𝑥
tan 𝑥
sen 𝑥
=
1
cos 𝑥
sen 𝑥
cos 𝑥
sen 𝑥
1
=
1
cos 𝑥
sen 𝑥
cos 𝑥 ∗ 𝑠𝑒𝑛 𝑥
=
1
cos 𝑥
1
cos 𝑥
=
1
cos 𝑥
Tarea 4. Revisar y realizar las siguientes ecuaciones trigonométricas
Ejercicio a.
2𝑐𝑜𝑠2
𝑥 + 𝑠𝑒𝑛 𝑥 = 1
En este ejercicio con funciones trigonométricas estamos resolviendo la ecuación.
Tarea 5. Aplicaciones trigonométricas.
a) Halla los lados de un paralelogramo cuyas diagonales miden 10 cm y 18 cm
respectivamente y forman un ángulo de 43º.
En este ejercicio aplicando ley de seno y coseno encontramos los lados del paralelogramo.
La figura que se creo con los datos que tenemos.
Para encontrar a en
una de las figuras:
Para encontrar c en
la otra figura del paralelogramo:
De producto – suma
sin ∝ cos 𝛽 =
1
2
[sin ∝ +𝛽 + sin(∝ −𝛽), sin ∝ sin 𝛽 =
1
2
[cos ∝ −𝛽 − 𝑐𝑜𝑠(∝ +𝛽)
cos ∝ sin 𝛽 =
1
2
[sin ∝ +𝛽 − sin(∝ −𝛽), cos ∝ cos 𝛽 =
1
2
[cos ∝ +𝛽 + 𝑐𝑜𝑠(∝ −𝛽)
De suma – producto
sin ∝ + sin 𝛽 = 2 sin
∝ +𝛽
2
𝑐𝑜𝑠(
∝ −𝛽
2
)
sin ∝ − sin 𝛽 = 2 cos
∝ +𝛽
2
𝑠𝑖𝑛(
∝ −𝛽
2
)
𝑐𝑜𝑠 ∝ + cos 𝛽 = 2 cos
∝ +𝛽
2
𝑐𝑜𝑠(
∝ −𝛽
2
)
cos ∝ − 𝑐𝑜𝑠 𝛽 = −2 sin
∝ +𝛽
2
𝑠𝑖𝑛(
∝ −𝛽
2
)

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Paso 3 profundizar y contextualizar el conocimiento de la unidad 2.

  • 2. Paso 3 - Profundizar y contextualizar el conocimiento de la Unidad 2. Por Germán Daniel Rendón Claudia Lorena Gomez Franco Albeiro Morales Ana Alejandra Guaranguay José Fernando Cuaran Nombre del curso: 551108A_951 Grupo: 551108_13 Presentado a Mónica Trigos Universidad Nacional Abierta y a Distancia UNAD CCAV Pasto Zona Centro Sur Escuela de Ciencias de la Educación 29 de marzo de 2021 Identidades trigonométricas
  • 3. Son llamadas así ciertas ecuaciones que se satisfacen para cualquier ángulo; existen varios tipos de ellas: Tipo 1: identidades de cociente, se obtienen por las definiciones de tangente y cotangente de la siguiente manera. Formula que aplico: 𝑎 sen 𝐴 = 𝑏 sen 𝐵 = 𝑐 sen 𝐶 Para hallar sen 𝐵 𝑎 sen 𝐴 = 𝑏 sen 𝐵 Para hallar sen 𝐶 180° = 31,3° + 120° + 𝐶 180° −31,3° − 120° = 𝐶 𝐶 = 28,7° Para hallar el cateto de 𝑐 𝑐 sen 𝐶 = 𝑎 sen 𝐴 𝑐 sen 28,7° = 10 sen 𝐴120° 𝐶 = 10 sen 120° ∗ sen 28,7° 𝐶 = 5,5𝑚
  • 4. Tarea 2. Calcula las razones trigonométricas seno, coseno y tangente de los ángulos agudos (A y B) de cada triángulo rectángulo que aparecen abajo. Ejercicio a: Tenemos que encontrar los ángulos de beta, alfa y el cateto de a. Procedemos a encontrar cateto de a. Angulo de alfa Angulo de beta
  • 5. Identidades recíprocas, se obtienen por las definiciones de secante, cosecante y tangente en relación con cotangente. s𝑖𝑛 ∝ = 𝑦 ℎ = 1 ℎ 𝑦 = 1 𝐜𝒔𝒄(∝) , c𝑠𝑐(∝) = ℎ 𝑦 = 1 𝑦 ℎ = 1 sin(∝) cos(∝) = 1 ℎ 𝑥 = 𝟏 𝒔𝒆𝒄(∝) , sec(∝) = ℎ 𝑥 = 1 𝑥 ℎ = 1 cos(∝) tan(∝) = 𝑦 𝑥 = 1 𝑥 𝑦 = 𝟏 𝐜𝐨𝐬𝒕(∝) , cot(∝) = 𝑥 𝑦 = 1 𝑦 𝑥 = 1 tan(∝)
  • 6. Con funciones trigonométricas la igualdad si se cumple. Ejercicio a Igualdad de cosecante de x sobre cotangente de x a 1 coseno de x. csc 𝑥 cot 𝑥 = 1 cos 𝑥 1 𝑠𝑒𝑛 𝑥 1 tan 𝑥 = 1 cos 𝑥 tan 𝑥 sen 𝑥 = 1 cos 𝑥 sen 𝑥 cos 𝑥 sen 𝑥 1 = 1 cos 𝑥 sen 𝑥 cos 𝑥 ∗ 𝑠𝑒𝑛 𝑥 = 1 cos 𝑥 1 cos 𝑥 = 1 cos 𝑥
  • 7. Tarea 4. Revisar y realizar las siguientes ecuaciones trigonométricas Ejercicio a. 2𝑐𝑜𝑠2 𝑥 + 𝑠𝑒𝑛 𝑥 = 1 En este ejercicio con funciones trigonométricas estamos resolviendo la ecuación.
  • 8. Tarea 5. Aplicaciones trigonométricas. a) Halla los lados de un paralelogramo cuyas diagonales miden 10 cm y 18 cm respectivamente y forman un ángulo de 43º. En este ejercicio aplicando ley de seno y coseno encontramos los lados del paralelogramo. La figura que se creo con los datos que tenemos. Para encontrar a en una de las figuras: Para encontrar c en la otra figura del paralelogramo:
  • 9. De producto – suma sin ∝ cos 𝛽 = 1 2 [sin ∝ +𝛽 + sin(∝ −𝛽), sin ∝ sin 𝛽 = 1 2 [cos ∝ −𝛽 − 𝑐𝑜𝑠(∝ +𝛽) cos ∝ sin 𝛽 = 1 2 [sin ∝ +𝛽 − sin(∝ −𝛽), cos ∝ cos 𝛽 = 1 2 [cos ∝ +𝛽 + 𝑐𝑜𝑠(∝ −𝛽) De suma – producto sin ∝ + sin 𝛽 = 2 sin ∝ +𝛽 2 𝑐𝑜𝑠( ∝ −𝛽 2 ) sin ∝ − sin 𝛽 = 2 cos ∝ +𝛽 2 𝑠𝑖𝑛( ∝ −𝛽 2 ) 𝑐𝑜𝑠 ∝ + cos 𝛽 = 2 cos ∝ +𝛽 2 𝑐𝑜𝑠( ∝ −𝛽 2 ) cos ∝ − 𝑐𝑜𝑠 𝛽 = −2 sin ∝ +𝛽 2 𝑠𝑖𝑛( ∝ −𝛽 2 )