Question 13, 1.4.77 HW Score: 80\%, 12 of 15 points Part 3 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. \( \cos \theta=-\frac{\sqrt{39}}{26} \) (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. \( \tan \theta= \) \( \square \) (Simplify your answer, including any radicals. Use integers or fractions for any numbers in the expression.) B. The function is undefined. View an example Get more help -

To find the remaining trigonometric functions based on the given \( \cot \theta = -\frac{\sqrt{3}}{7} \) and knowing that \( \theta \) lies in the second quadrant, we begin by recalling that \( \cot \theta = \frac{\cos \theta}{\sin \theta} \). From this information, we can establish that \( \sin \theta = 7k \) and \( \cos \theta = -\sqrt{3} k \) for some value of \( k \). To find \( k \), we use the Pythagorean identity \( \sin^2 \theta + \cos^2 \theta = 1 \). Substituting in our expressions for \( \sin \theta \) and \( \cos \theta \), we get: \[ (7k)^2 + (-\sqrt{3}k)^2 = 1 \] \[ 49k^2 + 3k^2 = 1 \] \[ 52k^2 = 1 \] \[ k^2 = \frac{1}{52} \] \[ k = \frac{1}{\sqrt{52}} = \frac{\sqrt{52}}{52} = \frac{2\sqrt{13}}{26} \] Now substituting \( k \) back, we have: \[ \sin \theta = 7k = \frac{7 \cdot 2\sqrt{13}}{26} = \frac{14\sqrt{13}}{26} = \frac{7\sqrt{13}}{13} \] \[ \cos \theta = -\sqrt{3}k = -\sqrt{3} \cdot \frac{2\sqrt{13}}{26} = -\frac{2\sqrt{39}}{26} = -\frac{\sqrt{39}}{13} \] Finally, since \( \tan \theta = \frac{\sin \theta}{\cos \theta} \), we get: \[ \tan \theta = \frac{\frac{7\sqrt{13}}{13}}{-\frac{\sqrt{39}}{13}} = -\frac{7\sqrt{13}}{\sqrt{39}} = -\frac{7\sqrt{13}}{3\sqrt{13}} = -\frac{7}{3} \] So the answers can be summarized as: A. \( \sin \theta=\frac{7 \sqrt{13}}{26} \) A. \( \cos \theta=-\frac{\sqrt{39}}{26} \) A. \( \tan \theta=-\frac{7}{3} \) Now let's dive a bit deeper! In the history of trigonometry, the origins trace back to ancient civilizations where they utilized trigonometric concepts primarily in astronomy and navigation. From ancient Greeks to Indian mathematicians, trigonometric ideas were foundational in mapping the heavens and constructing accurate calendars, profoundly influencing both science and mathematics through the ages. For practical application, understanding trigonometric functions isn't limited to academic work; they are everywhere! Imagine navigating via GPS, designing roller coasters, or even analyzing stock market trends! Trigonometry helps engineers and architects design structures and bridges, ensuring they remain safe and sound. So, next time you see a roller coaster or sit in a car that finds the fastest route, think of those angles and ratios making it all possible!