Bates Ray
06/28/2024 · Primary School
E Asocia cada expresión con su respectivo valor. \( \begin{array}{ll}\text { 111. } \arctan \left(\frac{\sqrt{3}}{3}\right) & \text { a. } 1 \\ \text { 112. } \cot \left(\operatorname{arcsen}\left(-\frac{1}{2}\right)\right) & \text { b. }-\frac{13}{5} \\ \text { 113. } \operatorname{cac}\left(\arctan \left(-\frac{5}{12}\right)\right) & \text { c. }-\sqrt{3} \\ \text { 114. } \tan \left(\arccos \left(\frac{\sqrt{2}}{2}\right)\right) & \text { d. } \frac{5}{3} \\ \text { 115. } \sec \left(\operatorname{arcsen}\left(\frac{4}{5}\right)\right) & \text { c. } \frac{\pi}{6}\end{array} \)
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Vamos a resolver cada expresión y asociarla con su respectivo valor.
1. **\( \arctan \left(\frac{\sqrt{3}}{3}\right) \)**:
- Sabemos que \( \tan\left(\frac{\pi}{6}\right) = \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3} \).
- Por lo tanto, \( \arctan \left(\frac{\sqrt{3}}{3}\right) = \frac{\pi}{6} \).
- **Asociación: 111 - c. \( \frac{\pi}{6} \)**.
2. **\( \cot \left(\operatorname{arcsen}\left(-\frac{1}{2}\right)\right) \)**:
- Si \( y = \operatorname{arcsen}\left(-\frac{1}{2}\right) \), entonces \( \sin(y) = -\frac{1}{2} \).
- Esto implica que \( y = -\frac{\pi}{6} \) (en el rango de \( \arcsen \)).
- Entonces, \( \cos(y) = \sqrt{1 - \sin^2(y)} = \sqrt{1 - \left(-\frac{1}{2}\right)^2} = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2} \).
- Por lo tanto, \( \cot(y) = \frac{\cos(y)}{\sin(y)} = \frac{\frac{\sqrt{3}}{2}}{-\frac{1}{2}} = -\sqrt{3} \).
- **Asociación: 112 - c. \( -\sqrt{3} \)**.
3. **\( \operatorname{cac}\left(\arctan \left(-\frac{5}{12}\right)\right) \)**:
- La función \( \operatorname{cac}(x) \) es la misma que \( \cot(x) \).
- Si \( y = \arctan\left(-\frac{5}{12}\right) \), entonces \( \tan(y) = -\frac{5}{12} \).
- Esto implica que \( \cot(y) = -\frac{12}{5} \).
- **Asociación: 113 - b. \( -\frac{12}{5} \)**.
4. **\( \tan \left(\arccos \left(\frac{\sqrt{2}}{2}\right)\right) \)**:
- Si \( y = \arccos\left(\frac{\sqrt{2}}{2}\right) \), entonces \( y = \frac{\pi}{4} \).
- Por lo tanto, \( \tan(y) = \tan\left(\frac{\pi}{4}\right) = 1 \).
- **Asociación: 114 - a. \( 1 \)**.
5. **\( \sec \left(\operatorname{arcsen}\left(\frac{4}{5}\right)\right) \)**:
- Si \( y = \operatorname{arcsen}\left(\frac{4}{5}\right) \), entonces \( \sin(y) = \frac{4}{5} \).
- Usando el teorema de Pitágoras, \( \cos(y) = \sqrt{1 - \sin^2(y)} = \sqrt{1 - \left(\frac{4}{5}\right)^2} = \sqrt{\frac{9}{25}} = \frac{3}{5} \).
- Por lo tanto, \( \sec(y) = \frac{1}{\cos(y)} = \frac{5}{3} \).
- **Asociación: 115 - d. \( \frac{5}{3} \)**.
Ahora, resumiendo las asociaciones:
- 111 - c. \( \frac{\pi}{6} \)
- 112 - c. \( -\sqrt{3} \)
- 113 - b. \( -\frac{12}{5} \)
- 114 - a. \( 1 \)
- 115 - d. \( \frac{5}{3} \)
Quick Answer
111 - c. \( \frac{\pi}{6} \)
112 - c. \( -\sqrt{3} \)
113 - b. \( -\frac{12}{5} \)
114 - a. \( 1 \)
115 - d. \( \frac{5}{3} \)
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