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Find the exact value of each of the remaining trigonometric functions of \( \theta \). Rationalize denominators when applicable. Select the correct choice below and, if necessary, fill in the answer box to complete your choice. 14 of 20 point(s) possible A. sin \( \theta=-\frac{\sqrt{5}}{6} \), given that \( \theta \) is in quadrant iv (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, Select the correct choice below and, if necessary, fill in the answer box to complete your choice.

Ask by Matthews Chen. in the United States
Mar 13,2025

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Answer

- \(\sin\theta = -\frac{\sqrt{5}}{6}\) - \(\cos\theta = \frac{\sqrt{31}}{6}\) - \(\tan\theta = -\frac{\sqrt{155}}{31}\)

Solution

We are given \[ \sin\theta=-\frac{\sqrt{5}}{6} \] with \(\theta\) in quadrant IV. In quadrant IV the cosine function is positive, so we proceed as follows. 1. Use the Pythagorean identity: \[ \sin^2\theta + \cos^2\theta = 1. \] Substitute \(\sin\theta\): \[ \left(-\frac{\sqrt{5}}{6}\right)^2 + \cos^2\theta = 1 \quad \Longrightarrow \quad \frac{5}{36} + \cos^2\theta = 1. \] 2. Solve for \(\cos^2\theta\): \[ \cos^2\theta = 1 - \frac{5}{36} = \frac{36}{36} - \frac{5}{36} = \frac{31}{36}. \] Since \(\theta\) is in quadrant IV, take the positive square root: \[ \cos\theta = \frac{\sqrt{31}}{6}. \] 3. Find \(\tan\theta\) using the definition: \[ \tan\theta=\frac{\sin\theta}{\cos\theta} = \frac{-\frac{\sqrt{5}}{6}}{\frac{\sqrt{31}}{6}} = -\frac{\sqrt{5}}{\sqrt{31}}. \] To rationalize the denominator, multiply numerator and denominator by \(\sqrt{31}\): \[ \tan\theta = -\frac{\sqrt{5}\,\sqrt{31}}{31} = -\frac{\sqrt{155}}{31}. \] Thus, the exact values are: - For part A: \(\sin\theta = -\frac{\sqrt{5}}{6}\). - For the cosine, choose (A): \(\cos\theta = \frac{\sqrt{31}}{6}\). - For the tangent (if required), we have: \(\tan\theta = -\frac{\sqrt{155}}{31}\).

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To find the remaining trigonometric functions based on the provided sine value of \( \theta \) in the fourth quadrant, we start by using the sine value given: \( \sin \theta = -\frac{\sqrt{5}}{6} \). First, we can find \( \cos \theta \) using the Pythagorean identity: \[ \sin^2 \theta + \cos^2 \theta = 1 \] Substituting the value of sine, we get: \[ \left(-\frac{\sqrt{5}}{6}\right)^2 + \cos^2 \theta = 1 \] \[ \frac{5}{36} + \cos^2 \theta = 1 \] \[ \cos^2 \theta = 1 - \frac{5}{36} = \frac{36}{36} - \frac{5}{36} = \frac{31}{36} \] Now taking the positive square root (since cosine is positive in the fourth quadrant): \[ \cos \theta = \sqrt{\frac{31}{36}} = \frac{\sqrt{31}}{6} \] Next, we can find the other trigonometric functions. The value of \( \tan \theta \) can be found using: \[ \tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{-\frac{\sqrt{5}}{6}}{\frac{\sqrt{31}}{6}} = -\frac{\sqrt{5}}{\sqrt{31}} = -\frac{\sqrt{5 \cdot 31}}{31} = -\frac{\sqrt{155}}{31} \] Finally, we can mention the remaining functions: - \( \csc \theta = \frac{1}{\sin \theta} = -\frac{6}{\sqrt{5}} = -\frac{6\sqrt{5}}{5} \) (rationalizing), - \( \sec \theta = \frac{1}{\cos \theta} = \frac{6}{\sqrt{31}} = \frac{6\sqrt{31}}{31} \) (rationalizing), - \( \cot \theta = \frac{1}{\tan \theta} = -\frac{\sqrt{31}}{\sqrt{5}} = -\frac{\sqrt{31 \cdot 5}}{5} = -\frac{\sqrt{155}}{5} \) (rationalizing). So, for the answers: - \( \cos \theta = \frac{\sqrt{31}}{6} \), - \( \tan \theta = -\frac{\sqrt{155}}{31} \). Now you can fill in the blanks accordingly!

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