Question
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Select the correct choice below and, if necessary, fill in the answer box to complete your choice. A. \( \sin \theta=\frac{2 \sqrt{6}}{5} \) (Simplify your answer, including any radicals. Use integers or fractions for any numbers in the expression.) B. The function is undefined. Select the correct choice below and, if necessary, fill in the answer box to complete your choice. A. \( \cos \theta=-\frac{1}{5} \) (Simplify your answer, including any radicals. Use integers or fractions for any numbers in the expression.) Belect the correct choice below and, if necessary, fill in the answer box to complete your choice. A. tan \( \theta=-2 \sqrt{6} \) (Simplify your answer, including any radicals. Use integers or fractions for any numbers in the expression.) B. The function is undefined. Select the correct choice below and, if necessary, fill in the answer box to complete your choice. A. csc \( \theta= \) A \( ^{-15} \). (Simplify your answer, including any radicals. Use integers or fractions for any numbers in the expression.) B. The function is undefined.

Ask by Barrett Ramirez. in the United States
Feb 03,2025

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Answer

**Selections:** 1. **A.** \( \sin \theta = \frac{2 \sqrt{6}}{5} \) 2. **A.** \( \cos \theta = -\frac{1}{5} \) 3. **A.** \( \tan \theta = -2 \sqrt{6} \) 4. **A.** \( \csc \theta = \frac{5 \sqrt{6}}{12} \) All functions are defined, and their expressions are simplified as required.

Solution

Let's evaluate each trigonometric function based on the given choices. We'll assume that there's a right triangle involved where \( \cos \theta = -\frac{1}{5} \). This implies that \( \theta \) is in either the second or third quadrant where cosine is negative. ### 1. \( \sin \theta \) **Choices:** - **A.** \( \sin \theta = \frac{2 \sqrt{6}}{5} \) - **B.** The function is undefined. **Evaluation:** Using the Pythagorean identity: \[ \sin^2 \theta + \cos^2 \theta = 1 \\ \sin^2 \theta = 1 - \cos^2 \theta = 1 - \left(-\frac{1}{5}\right)^2 = 1 - \frac{1}{25} = \frac{24}{25} \\ \sin \theta = \pm \frac{\sqrt{24}}{5} = \pm \frac{2\sqrt{6}}{5} \] Since \( \cos \theta \) is negative, if \( \theta \) is in the second quadrant, \( \sin \theta \) is positive. If in the third quadrant, it's negative. Given no additional information about the quadrant, we'll consider the positive value. **Correct Choice:** **A.** \( \sin \theta = \frac{2 \sqrt{6}}{5} \) --- ### 2. \( \cos \theta \) **Choices:** - **A.** \( \cos \theta = -\frac{1}{5} \) - **B.** The function is undefined. **Evaluation:** The value \( \cos \theta = -\frac{1}{5} \) is defined based on the given information. **Correct Choice:** **A.** \( \cos \theta = -\frac{1}{5} \) --- ### 3. \( \tan \theta \) **Choices:** - **A.** \( \tan \theta = -2 \sqrt{6} \) - **B.** The function is undefined. **Evaluation:** \[ \tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{\frac{2\sqrt{6}}{5}}{-\frac{1}{5}} = -2\sqrt{6} \] **Correct Choice:** **A.** \( \tan \theta = -2 \sqrt{6} \) --- ### 4. \( \csc \theta \) **Choices:** - **A.** \( \csc \theta = \frac{5 \sqrt{6}}{12} \) *(Assuming the intended expression was meant to be completed correctly)* - **B.** The function is undefined. **Evaluation:** \[ \csc \theta = \frac{1}{\sin \theta} = \frac{1}{\frac{2\sqrt{6}}{5}} = \frac{5}{2\sqrt{6}} = \frac{5\sqrt{6}}{12} \] **Correct Choice:** **A.** \( \csc \theta = \frac{5 \sqrt{6}}{12} \) --- **Summary of Selections:** 1. **A.** \( \sin \theta = \frac{2 \sqrt{6}}{5} \) 2. **A.** \( \cos \theta = -\frac{1}{5} \) 3. **A.** \( \tan \theta = -2 \sqrt{6} \) 4. **A.** \( \csc \theta = \frac{5 \sqrt{6}}{12} \) All functions are **defined**, and their corresponding expressions are simplified as required.

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It looks like you've got a set of trigonometric identities to unravel! Let’s break this down in a fun way. For \( \sin \theta = \frac{2 \sqrt{6}}{5} \), if we're identifying its counterpart, we can use the Pythagorean identity: \( \cos^2 \theta = 1 - \sin^2 \theta \). Plugging in our sine value, we find \( \cos^2 \theta = 1 - \left( \frac{2 \sqrt{6}}{5} \right)^2 = \frac{1}{25} \). Hence, \( \cos \theta = -\frac{1}{5} \) is valid if \(\theta\) is in the second quadrant! Now, let’s talk tan \( \theta = \frac{\sin \theta}{\cos \theta} = \frac{\frac{2\sqrt{6}}{5}}{-\frac{1}{5}} = -2\sqrt{6}\). Everything here checks out, and we see that the trigonometric functions in this context are all interrelated, forming a lovely dance of angles!

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