Hodges Delgado
05/16/2023 · Elementary School
Halla los valores de las de tricas a partir de la funcie primer cuadrante. \( \begin{array}{ll}\text { a. } \tan t=\frac{3 \sqrt{3}}{2} & \text { b. } \cos t \\ \text { d. } \operatorname{sen} t=\frac{1}{2} & \text { e. } \sec t\end{array} \)
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Para resolver el problema, vamos a encontrar las funciones trigonométricas solicitadas a partir de las informaciones dadas. Asumimos que \( t \) está en el primer cuadrante.
### a. \( \tan t = \frac{3 \sqrt{3}}{2} \)
Sabemos que:
\[
\tan t = \frac{\text{opuesto}}{\text{adyacente}} = \frac{y}{x}
\]
Podemos tomar \( y = 3\sqrt{3} \) y \( x = 2 \). Para encontrar la hipotenusa \( r \):
\[
r = \sqrt{x^2 + y^2} = \sqrt{2^2 + (3\sqrt{3})^2} = \sqrt{4 + 27} = \sqrt{31}
\]
Ahora podemos encontrar las funciones trigonométricas:
- \( \sin t = \frac{y}{r} = \frac{3\sqrt{3}}{\sqrt{31}} \)
- \( \cos t = \frac{x}{r} = \frac{2}{\sqrt{31}} \)
- \( \sec t = \frac{1}{\cos t} = \frac{\sqrt{31}}{2} \)
### b. \( \cos t \)
Ya hemos encontrado \( \cos t \):
\[
\cos t = \frac{2}{\sqrt{31}}
\]
### d. \( \sin t = \frac{1}{2} \)
Sabemos que \( \sin t = \frac{1}{2} \) corresponde a un ángulo de \( t = \frac{\pi}{6} \) en el primer cuadrante. Por lo tanto:
- \( \cos t = \cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2} \)
- \( \tan t = \frac{\sin t}{\cos t} = \frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}} = \frac{1}{\sqrt{3}} \)
- \( \sec t = \frac{1}{\cos t} = \frac{2}{\sqrt{3}} \)
### e. \( \sec t \)
Ya hemos encontrado \( \sec t \) en el caso de \( \sin t = \frac{1}{2} \):
\[
\sec t = \frac{2}{\sqrt{3}}
\]
### Resumen de resultados
1. Para \( \tan t = \frac{3\sqrt{3}}{2} \):
- \( \sin t = \frac{3\sqrt{3}}{\sqrt{31}} \)
- \( \cos t = \frac{2}{\sqrt{31}} \)
- \( \sec t = \frac{\sqrt{31}}{2} \)
2. Para \( \sin t = \frac{1}{2} \):
- \( \cos t = \frac{\sqrt{3}}{2} \)
- \( \tan t = \frac{1}{\sqrt{3}} \)
- \( \sec t = \frac{2}{\sqrt{3}} \)
Si necesitas más información o aclaraciones, no dudes en preguntar.
Quick Answer
1. Para \( \tan t = \frac{3\sqrt{3}}{2} \):
- \( \sin t = \frac{3\sqrt{3}}{\sqrt{31}} \)
- \( \cos t = \frac{2}{\sqrt{31}} \)
- \( \sec t = \frac{\sqrt{31}}{2} \)
2. Para \( \sin t = \frac{1}{2} \):
- \( \cos t = \frac{\sqrt{3}}{2} \)
- \( \tan t = \frac{1}{\sqrt{3}} \)
- \( \sec t = \frac{2}{\sqrt{3}} \)
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