4) Find the exact value of \( \sin (\alpha-\beta) \) if \( \cos \alpha=\frac{1}{4}, 0<\alpha<\frac{\pi}{2} \) and \( \sin \beta=\frac{3}{4}, \frac{\pi}{2}<\beta<\pi \). a) \( \frac{\sqrt{105}+3}{4} \) b) \( 0 \frac{-\sqrt{105}-3}{16} \) c) \( 0 \frac{\sqrt{105}+3}{16} \) d) \( 0 \frac{\sqrt{105}-3}{4} \)

To find the exact value of \( \sin (\alpha - \beta) \), we can utilize the sine subtraction formula: \[ \sin(\alpha - \beta) = \sin \alpha \cos \beta - \cos \alpha \sin \beta \] We know \( \cos \alpha = \frac{1}{4} \) and need to find \( \sin \alpha \). Since \( \alpha \) is in the first quadrant, we can use the Pythagorean identity: \[ \sin^2 \alpha + \cos^2 \alpha = 1 \implies \sin^2 \alpha + \left(\frac{1}{4}\right)^2 = 1 \] \[ \sin^2 \alpha + \frac{1}{16} = 1 \implies \sin^2 \alpha = 1 - \frac{1}{16} = \frac{15}{16} \implies \sin \alpha = \frac{\sqrt{15}}{4} \] Next, we find \( \sin \beta \) and \( \cos \beta \). Given \( \sin \beta = \frac{3}{4} \) and knowing \( \beta \) is in the second quadrant, we use the identity again to find \( \cos \beta \): \[ \sin^2 \beta + \cos^2 \beta = 1 \implies \left(\frac{3}{4}\right)^2 + \cos^2 \beta = 1 \] \[ \frac{9}{16} + \cos^2 \beta = 1 \implies \cos^2 \beta = 1 - \frac{9}{16} = \frac{7}{16} \implies \cos \beta = -\frac{\sqrt{7}}{4} \] Now we substitute \( \sin \alpha \), \( \cos \alpha \), \( \sin \beta \), and \( \cos \beta \) into the sine subtraction formula: \[ \sin(\alpha - \beta) = \left(\frac{\sqrt{15}}{4}\right)\left(-\frac{\sqrt{7}}{4}\right) - \left(\frac{1}{4}\right)\left(\frac{3}{4}\right) \] \[ = -\frac{\sqrt{105}}{16} - \frac{3}{16} \] \[ = -\frac{\sqrt{105}+3}{16} \] Thus, the exact value of \( \sin(\alpha - \beta) \) is \[ \boxed{-\frac{\sqrt{105}+3}{16}} \] So, the correct choice is (b) \( 0 \frac{-\sqrt{105}-3}{16} \).