Question

If $ \sin \theta = -\frac{3}{4} $, then compute $ \sin 2\theta $

• A. \( \frac{3\sqrt{7}}{8} \)
• B. \( -\frac{3\sqrt{7}}{8} \)
• C. \( \frac{2\sqrt{3}}{7} \)
• D. \( -\frac{2\sqrt{3}}{7} \)

Hint:

Use \( \sin 2\theta = 2 \sin \theta \cos \theta \) and remember quadrant signs when extracting square roots.

Answer 1

The Correct Option is B

We know: \[ \sin 2\theta = 2 \sin \theta \cos \theta \] Given \( \sin \theta = -\frac{3}{4} \), we find \( \cos \theta \) using: \[ \cos^2 \theta = 1 - \sin^2 \theta = 1 - \left(\frac{9}{16}\right) = \frac{7}{16} \Rightarrow \cos \theta = \pm \frac{\sqrt{7}}{4} \] The sign of \( \cos \theta \) depends on the quadrant. Since \( \sin \theta<0 \), suppose \( \theta \) lies in 3rd or 4th quadrant. Assume \( \cos \theta>0 \), so: \[ \cos \theta = \frac{\sqrt{7}}{4} \] Then: \[ \sin 2\theta = 2 \cdot \left(-\frac{3}{4}\right) \cdot \frac{\sqrt{7}}{4} = -\frac{6\sqrt{7}}{16} = \boxed{ -\frac{3\sqrt{7}}{8} } \]