Welcome to Westonci.ca, where curiosity meets expertise. Ask any question and receive fast, accurate answers from our knowledgeable community. Get detailed and accurate answers to your questions from a community of experts on our comprehensive Q&A platform. Join our platform to connect with experts ready to provide precise answers to your questions in different areas.
Sagot :
To evaluate [tex]\(\text{arccsc}(\sqrt{2})\)[/tex], we need to first recall what the arccsc function represents. The [tex]\(\text{arccsc}(x)\)[/tex] function gives the angle [tex]\(\theta\)[/tex] such that [tex]\(\csc(\theta) = x\)[/tex].
The cosecant function is defined as the reciprocal of the sine function, i.e., [tex]\(\csc(\theta) = \frac{1}{\sin(\theta)}\)[/tex]. Therefore, finding [tex]\(\text{arccsc}(\sqrt{2})\)[/tex] means finding the angle [tex]\(\theta\)[/tex] for which:
[tex]\[ \csc(\theta) = \sqrt{2} \][/tex]
This implies:
[tex]\[ \sin(\theta) = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2} \][/tex]
We now need to determine all angles [tex]\(\theta\)[/tex] within the principal range of the cosecant function, specifically for [tex]\(\theta \in [0, \pi]\)[/tex], where [tex]\(\sin(\theta) = \frac{\sqrt{2}}{2}\)[/tex].
We know that the sine function achieves the value [tex]\(\frac{\sqrt{2}}{2}\)[/tex] at the angles:
[tex]\[ \theta = \frac{\pi}{4} \quad \text{and} \quad \theta = \frac{3\pi}{4} \][/tex]
Therefore, the angles [tex]\(\theta\)[/tex] that satisfy [tex]\(\csc(\theta) = \sqrt{2}\)[/tex] are:
[tex]\[ \theta = \frac{\pi}{4} \quad \text{and} \quad \theta = \frac{3\pi}{4} \][/tex]
Thus, the solution to [tex]\(\text{arccsc}(\sqrt{2})\)[/tex] is:
[tex]\[ \left\{ \frac{\pi}{4}, \frac{3\pi}{4} \right\} \][/tex]
The given options are:
A. [tex]\(\left\{\frac{\pi}{6}, \frac{5 \pi}{6}\right\}\)[/tex]
B. [tex]\(\left\{\frac{\pi}{3} \pm 2 \pi n, \frac{5 \pi}{3} \pm 2 \pi n\right\}\)[/tex]
C. [tex]\(\frac{\pi}{3}\)[/tex]
Since none of these options directly match the set of angles [tex]\(\left\{\frac{\pi}{4}, \frac{3\pi}{4}\right\}\)[/tex], it seems there might be a misunderstanding or an error in the provided answer choices. Therefore, the correct answer based on our evaluation is:
[tex]\[ \left\{\frac{\pi}{4}, \frac{3\pi}{4}\right\} \][/tex]
The cosecant function is defined as the reciprocal of the sine function, i.e., [tex]\(\csc(\theta) = \frac{1}{\sin(\theta)}\)[/tex]. Therefore, finding [tex]\(\text{arccsc}(\sqrt{2})\)[/tex] means finding the angle [tex]\(\theta\)[/tex] for which:
[tex]\[ \csc(\theta) = \sqrt{2} \][/tex]
This implies:
[tex]\[ \sin(\theta) = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2} \][/tex]
We now need to determine all angles [tex]\(\theta\)[/tex] within the principal range of the cosecant function, specifically for [tex]\(\theta \in [0, \pi]\)[/tex], where [tex]\(\sin(\theta) = \frac{\sqrt{2}}{2}\)[/tex].
We know that the sine function achieves the value [tex]\(\frac{\sqrt{2}}{2}\)[/tex] at the angles:
[tex]\[ \theta = \frac{\pi}{4} \quad \text{and} \quad \theta = \frac{3\pi}{4} \][/tex]
Therefore, the angles [tex]\(\theta\)[/tex] that satisfy [tex]\(\csc(\theta) = \sqrt{2}\)[/tex] are:
[tex]\[ \theta = \frac{\pi}{4} \quad \text{and} \quad \theta = \frac{3\pi}{4} \][/tex]
Thus, the solution to [tex]\(\text{arccsc}(\sqrt{2})\)[/tex] is:
[tex]\[ \left\{ \frac{\pi}{4}, \frac{3\pi}{4} \right\} \][/tex]
The given options are:
A. [tex]\(\left\{\frac{\pi}{6}, \frac{5 \pi}{6}\right\}\)[/tex]
B. [tex]\(\left\{\frac{\pi}{3} \pm 2 \pi n, \frac{5 \pi}{3} \pm 2 \pi n\right\}\)[/tex]
C. [tex]\(\frac{\pi}{3}\)[/tex]
Since none of these options directly match the set of angles [tex]\(\left\{\frac{\pi}{4}, \frac{3\pi}{4}\right\}\)[/tex], it seems there might be a misunderstanding or an error in the provided answer choices. Therefore, the correct answer based on our evaluation is:
[tex]\[ \left\{\frac{\pi}{4}, \frac{3\pi}{4}\right\} \][/tex]
We hope you found what you were looking for. Feel free to revisit us for more answers and updated information. Thank you for choosing our platform. We're dedicated to providing the best answers for all your questions. Visit us again. Find reliable answers at Westonci.ca. Visit us again for the latest updates and expert advice.