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Question 13, 1.4.77 HW Score: \( 80 \%, 12 \) of 15 points Pat 2 of 5 Points: 0 of 1 Find the exact value of each of the remaining trigonometric functions of \( \theta \). Rationalize denominators when applicable. \( \cot \theta=-\frac{\sqrt{3}}{7} \), given that \( \theta \) is in quadrant II Select the correct choice below and, if necessary, fill in the answer box to complete your choice. A. \( \sin \theta=\frac{7 \sqrt{13}}{26} \) (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. \( \cos \theta= \) \( \square \) (Simplify your answer, including any radicals. Use integers or fractions for any numbers in the expression.) B. The function is undefined.

Ask by Vaughan Delgado. in the United States
Feb 03,2025

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Tutor-Verified Answer

Answer

- **\( \sin \theta = \frac{7\sqrt{13}}{26} \)** - **\( \cos \theta = -\frac{\sqrt{39}}{26} \)**

Solution

To find the exact values of the remaining trigonometric functions for \( \theta \) given that \( \cot \theta = -\frac{\sqrt{3}}{7} \) and \( \theta \) is in Quadrant II, follow these steps: ### 1. Understanding the Given: - **Quadrant II**: In this quadrant, sine is positive, and cosine and tangent are negative. - **Cotangent Definition**: \[ \cot \theta = \frac{\cos \theta}{\sin \theta} = -\frac{\sqrt{3}}{7} \] ### 2. Setting Up the Relationship: Let’s consider a right triangle where: - Adjacent side (related to cosine) = \( -\sqrt{3} \) (negative in Quadrant II) - Opposite side (related to sine) = \( 7 \) (positive in Quadrant II) ### 3. Calculating the Hypotenuse: \[ \text{Hypotenuse} = \sqrt{(\text{Opposite})^2 + (\text{Adjacent})^2} = \sqrt{7^2 + (\sqrt{3})^2} = \sqrt{49 + 3} = \sqrt{52} = 2\sqrt{13} \] ### 4. Finding \( \sin \theta \) and \( \cos \theta \): - **Sine**: \[ \sin \theta = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{7}{2\sqrt{13}} = \frac{7\sqrt{13}}{26} \] - **Cosine**: \[ \cos \theta = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{-\sqrt{3}}{2\sqrt{13}} = \frac{-\sqrt{39}}{26} \] *(Rationalized the denominator by multiplying numerator and denominator by \( \sqrt{13} \))* ### 5. Conclusion: - **For \( \sin \theta \)**: - **Choice A** is correct. \[ \sin \theta = \frac{7\sqrt{13}}{26} \] - **For \( \cos \theta \)**: - **Choice A** is correct. \[ \cos \theta = \frac{-\sqrt{39}}{26} \] *(Alternatively written as \( -\frac{\sqrt{39}}{26} \))* ### Final Answers: - **\( \sin \theta = \frac{7\sqrt{13}}{26} \)** - **\( \cos \theta = -\frac{\sqrt{39}}{26} \)** **Therefore:** - **Select Choice A for \( \sin \theta \) with \( \sin \theta = \frac{7\sqrt{13}}{26} \).** - **Select Choice A for \( \cos \theta \) and fill in \( \cos \theta = -\frac{\sqrt{39}}{26} \).**

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Bonus Knowledge

When given \(\cot \theta = -\frac{\sqrt{3}}{7}\) and knowing \(\theta\) is in quadrant II, we can find the values of \(\sin \theta\) and \(\cos \theta\). Since \(\cot \theta = \frac{\cos \theta}{\sin \theta}\), we can let \(\cos \theta = -\sqrt{3}k\) and \(\sin \theta = 7k\) where \(k\) is a positive value. Using the Pythagorean identity \(\sin^2 \theta + \cos^2 \theta = 1\), we can solve for \(k\) and thus determine \(\sin \theta\) and \(\cos \theta\). Once you find \(k\) using \(k = \frac{1}{4}\), you'll have: \[ \sin \theta = 7k = \frac{7}{4} \] And since \(\cos \theta\) relates directly to the y-axis, we’ll deduce: \[ \cos \theta = -\frac{\sqrt{3}}{4} \] To wrap up, you can confidently determine the values of both trigonometric functions according to the properties of the Cartesian plane. For additional context, many basic trigonometric identities form the foundation of more complex mathematics, like calculus and physics. Mastering these identities can lead to thrilling discoveries in higher-level math and real-world applications such as modeling sound waves or analyzing harmonic motion, often seen in music and engineering. Also, if you'd like to dive deeper, consider textbooks or online resources that focus on trigonometry applications in real life. You can uncover how architects use these principles to design buildings or how pilots rely on trigonometric calculations for navigation. Happy learning!

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