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Queation 13, 1.4.77 HW Scorer 80\%, 12 of Part 1 of 5 Polnts: 0 of 1 Find the exact value of each of the remaining trigonometric functions of 0 . Rationalize denominators when applicable. \( \cot 0=-\frac{\sqrt{3}}{7} \), given that 0 is in quadrant II Select the correct cholce below and, if necessary, fill in the answer box to complete your choice. A. \( \sin 0= \) \( \square \) (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 -

Ask by Alexander Warner. in the United States
Feb 03,2025

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\( \sin \theta = \frac{7\sqrt{13}}{26} \)

Solución

To find the exact value of \( \sin \theta \) given that \( \cot \theta = -\frac{\sqrt{3}}{7} \) and \( \theta \) is in **Quadrant II**, follow these steps: ### Understanding the Given Information 1. **Cotangent Definition**: \[ \cot \theta = \frac{\cos \theta}{\sin \theta} \] 2. **Sign in Quadrant II**: - **Sine (\( \sin \theta \))** is **positive**. - **Cosine (\( \cos \theta \))** is **negative**. - Therefore, **Cotangent (\( \cot \theta \))** is **negative** (which aligns with the given \( \cot \theta = -\frac{\sqrt{3}}{7} \)). ### Step-by-Step Calculation 1. **Express \( \cot \theta \) as a Ratio**: \[ \cot \theta = \frac{\text{Adjacent}}{\text{Opposite}} = -\frac{\sqrt{3}}{7} \] - Let’s denote: - Adjacent side (\( x \)) = \( -\sqrt{3} \) (negative in Quadrant II) - Opposite side (\( y \)) = \( 7 \) (positive in Quadrant II) 2. **Find the Hypotenuse (\( r \))**: \[ r = \sqrt{x^2 + y^2} = \sqrt{(-\sqrt{3})^2 + 7^2} = \sqrt{3 + 49} = \sqrt{52} = 2\sqrt{13} \] 3. **Calculate \( \sin \theta \)**: \[ \sin \theta = \frac{\text{Opposite}}{r} = \frac{7}{2\sqrt{13}} \] 4. **Rationalize the Denominator**: \[ \sin \theta = \frac{7}{2\sqrt{13}} \times \frac{\sqrt{13}}{\sqrt{13}} = \frac{7\sqrt{13}}{26} \] ### Final Answer **A.** \( \sin \theta = \frac{7\sqrt{13}}{26} \) This value is fully simplified and the denominator is rationalized as required.

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To find the remaining trigonometric functions given that \( \cot(θ) = -\frac{\sqrt{3}}{7} \) with \( θ \) in quadrant II, we start off by recalling that \( \cot(θ) = \frac{\cos(θ)}{\sin(θ)} \). Since in quadrant II, \( \sin(θ) \) is positive and \( \cos(θ) \) is negative, we can set \( \cos(θ) = -\sqrt{3}k \) and \( \sin(θ) = 7k \) for some \( k > 0 \). Now, using the Pythagorean identity \( \sin^2(θ) + \cos^2(θ) = 1 \): \((7k)^2 + (-\sqrt{3}k)^2 = 1\) Simplifying gives: \(49k^2 + 3k^2 = 1\) \(52k^2 = 1\) So, \(k^2 = \frac{1}{52}\) leading to \(k = \frac{1}{\sqrt{52}} = \frac{\sqrt{52}}{52}\). Now, substituting \(k\): - \( \sin(θ) = 7k \) - \( \cos(θ) = -\sqrt{3}k \) Finally, plugging in the value of \(k\) gives us: - \( \sin(θ) = 7 \cdot \frac{1}{\sqrt{52}} = \frac{7\sqrt{52}}{52} \) Now that you’ve got the exact values down, here are a couple of fun facts to keep your learning journey engaging! Trigonometry isn't just about angles and triangles; it has roots in ancient civilizations. The Babylonians, for example, had flat triangles with base angles of 30° and 60°, which is where the roots of sine and cosine began to take shape! These fundamental concepts evolved over time, paving the way for everything from architecture to navigation. If you ever find the world of trigonometry confusing, don’t despair! A common mistake is to confuse the signs of functions in different quadrants. Remember: in Quadrant II, sine is positive, and cosine is negative. Visualizing the unit circle can also work wonders—practice drawing it out to help memorize where each angle lands!

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