1. The problem asks us to find the exact coordinates of a point on the unit circle where $\sin \theta = \frac{1}{4}$. The unit circle has radius 1. 2. Recall the Pythagorean identity for any angle $\theta$: $$\sin^2 \theta + \cos^2 \theta = 1$$ This means the sum of the squares of sine and cosine of the angle is always 1. 3. Given $\sin \theta = \frac{1}{4}$, substitute into the identity: $$\left(\frac{1}{4}\right)^2 + \cos^2 \theta = 1$$ $$\frac{1}{16} + \cos^2 \theta = 1$$ 4. Solve for $\cos^2 \theta$: $$\cos^2 \theta = 1 - \frac{1}{16} = \frac{16}{16} - \frac{1}{16} = \frac{15}{16}$$ 5. Take the square root to find $\cos \theta$: $$\cos \theta = \pm \sqrt{\frac{15}{16}} = \pm \frac{\sqrt{15}}{4}$$ 6. Therefore, the coordinates of the point on the unit circle are: $$\left(\cos \theta, \sin \theta\right) = \left(\pm \frac{\sqrt{15}}{4}, \frac{1}{4}\right)$$ 7. The sign of $\cos \theta$ depends on the quadrant where $\theta$ lies. Since $\sin \theta$ is positive, $\theta$ is in either the first or second quadrant. In the first quadrant, $\cos \theta$ is positive; in the second quadrant, it is negative. Final answer: $$\boxed{\left(\frac{\sqrt{15}}{4}, \frac{1}{4}\right) \text{ or } \left(-\frac{\sqrt{15}}{4}, \frac{1}{4}\right)}$$