Question - Trigonometric Function Problem Involving Sine in Quadrants

Given that \(\sin \theta = -\frac{3}{4}\) and \(\theta\) lies in the third quadrant, we know that \(\sin \theta < 0\) and \(\cos \theta < 0\).

To find \( \cos ^2 \theta \), let's first find \( \cos \theta \).

We use the Pythagorean identity \( \sin^2 \theta + \cos^2 \theta = 1 \).

\( \cos^2 \theta = 1 - \sin^2 \theta \)

\( \cos^2 \theta = 1 - \left(-\frac{3}{4}\right)^2 = 1 - \frac{9}{16} = \frac{16}{16} - \frac{9}{16} = \frac{7}{16} \).

The cosine function is negative in the third quadrant, so \(\cos \theta = -\sqrt{\frac{7}{16}} = -\frac{\sqrt{7}}{4}\).

Therefore, \( \cos^2 \theta = \left(-\frac{\sqrt{7}}{4}\right)^2 = \frac{7}{16} \).