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\( \sin \theta=-\frac{\sqrt{3}}{3}, \frac{3 \pi}{2}<\theta<2 \pi \) Find \( \sin (2 \theta) \)

Ask by Matthews King. in the United States
Mar 14,2025

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Tutor-Verified Answer

Answer

\[ \sin(2\theta) = -\frac{2\sqrt{2}}{3} \]

Solution

We are given \[ \sin \theta = -\frac{\sqrt{3}}{3} \quad \text{with} \quad \frac{3\pi}{2} < \theta < 2\pi. \] Since \(\theta\) is in the fourth quadrant, the cosine is positive. We use the Pythagorean identity: \[ \sin^2\theta + \cos^2\theta = 1. \] Substitute the given value: \[ \left(-\frac{\sqrt{3}}{3}\right)^2 + \cos^2\theta = 1 \quad \Longrightarrow \quad \frac{3}{9} + \cos^2\theta = 1. \] This simplifies to: \[ \frac{1}{3} + \cos^2\theta = 1 \quad \Longrightarrow \quad \cos^2\theta = 1 - \frac{1}{3} = \frac{2}{3}. \] Taking the positive square root (since \(\theta\) is in the fourth quadrant): \[ \cos\theta = \sqrt{\frac{2}{3}}. \] Now, we use the double-angle formula for sine: \[ \sin(2\theta) = 2\sin\theta\cos\theta. \] Substitute the known values: \[ \sin(2\theta) = 2 \left(-\frac{\sqrt{3}}{3}\right)\left(\sqrt{\frac{2}{3}}\right). \] Simplify the expression: \[ \sin(2\theta) = -\frac{2\sqrt{3}}{3}\sqrt{\frac{2}{3}} = -\frac{2\sqrt{3}}{3} \cdot \frac{\sqrt{2}}{\sqrt{3}} = -\frac{2\sqrt{2}}{3}. \] Thus, the solution is: \[ \boxed{\sin(2\theta) = -\frac{2\sqrt{2}}{3}}. \]

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The Deep Dive

To solve for \( \sin(2\theta) \) when \( \sin \theta = -\frac{\sqrt{3}}{3} \) and \( \frac{3 \pi}{2} < \theta < 2 \pi \), we first need to determine \( \theta \). Given that \( \sin \theta = -\frac{\sqrt{3}}{3} \), we know this occurs at angles in the fourth quadrant. The reference angle corresponding to \( \sin^{-1} \left(-\frac{\sqrt{3}}{3}\right) \) is \( \frac{\pi}{6} \). Hence, in the fourth quadrant, the value of \( \theta \) can be expressed as: \[ \theta = 2\pi - \frac{\pi}{6} = \frac{12\pi}{6} - \frac{\pi}{6} = \frac{11\pi}{6} \] Next, we can find \( \sin(2\theta) \) using the double angle identity for sine, which is given by: \[ \sin(2\theta) = 2 \sin(\theta) \cos(\theta) \] We already have \( \sin \theta = -\frac{\sqrt{3}}{3} \). To find \( \cos \theta \), we can use the Pythagorean identity: \[ \sin^2 \theta + \cos^2 \theta = 1 \] Substituting for \( \sin \theta \): \[ \left(-\frac{\sqrt{3}}{3}\right)^2 + \cos^2 \theta = 1 \] \[ \frac{3}{9} + \cos^2 \theta = 1 \] \[ \cos^2 \theta = 1 - \frac{1}{3} = \frac{2}{3} \] Since \( \theta \) is in the fourth quadrant, \( \cos \theta \) is positive: \[ \cos \theta = \sqrt{\frac{2}{3}} = \frac{\sqrt{6}}{3} \] Now substituting back into the double angle formula: \[ \sin(2\theta) = 2 \left(-\frac{\sqrt{3}}{3}\right) \left(\frac{\sqrt{6}}{3}\right) \] \[ = 2 \left(-\frac{\sqrt{18}}{9}\right) = -\frac{2\sqrt{18}}{9} = -\frac{6\sqrt{2}}{9} = -\frac{2\sqrt{2}}{3} \] Thus, the value of \( \sin(2\theta) \) is \[ \boxed{-\frac{2\sqrt{2}}{3}} \]

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