If $\cos \theta=-\frac{3}{4}$ and is in Quadrant III, then $\sin \theta$ is equivalent to

Step 1 ：Given $\cos \theta = -\frac{3}{4}$, we can plug this value into the Pythagorean identity $\sin^2 \theta + \cos^2 \theta = 1$ and solve for $\sin \theta$.

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

Step 3 ：Since $\theta$ is in Quadrant III, both $\sin \theta$ and $\cos \theta$ are negative. Therefore, $\sin \theta = -\sqrt{\frac{7}{16}}$

Step 4 ：\(\boxed{\sin \theta = -\frac{\sqrt{7}}{4}}\)