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 use the sine difference identity: \[ \sin(\alpha - \beta) = \sin \alpha \cos \beta - \cos \alpha \sin \beta \] We are given \( \cos \alpha = \frac{1}{4} \) and \( \sin \beta = \frac{3}{4} \). We need to find \( \sin \alpha \) and \( \cos \beta \). ### Step 1: Find \( \sin \alpha \) Since \( \cos \alpha = \frac{1}{4} \) and \( 0 < \alpha < \frac{\pi}{2} \), we can use the Pythagorean identity: \[ \sin^2 \alpha + \cos^2 \alpha = 1 \] Substituting \( \cos \alpha \): \[ \sin^2 \alpha + \left(\frac{1}{4}\right)^2 = 1 \] \[ \sin^2 \alpha + \frac{1}{16} = 1 \] \[ \sin^2 \alpha = 1 - \frac{1}{16} = \frac{16}{16} - \frac{1}{16} = \frac{15}{16} \] Taking the positive square root (since \( \alpha \) is in the first quadrant): \[ \sin \alpha = \sqrt{\frac{15}{16}} = \frac{\sqrt{15}}{4} \] ### Step 2: Find \( \cos \beta \) Since \( \sin \beta = \frac{3}{4} \) and \( \frac{\pi}{2} < \beta < \pi \), we can again use the Pythagorean identity: \[ \sin^2 \beta + \cos^2 \beta = 1 \] Substituting \( \sin \beta \): \[ \left(\frac{3}{4}\right)^2 + \cos^2 \beta = 1 \] \[ \frac{9}{16} + \cos^2 \beta = 1 \] \[ \cos^2 \beta = 1 - \frac{9}{16} = \frac{16}{16} - \frac{9}{16} = \frac{7}{16} \] Taking the negative square root (since \( \beta \) is in the second quadrant): \[ \cos \beta = -\sqrt{\frac{7}{16}} = -\frac{\sqrt{7}}{4} \] ### Step 3: Substitute into the sine difference formula Now we can substitute \( \sin \alpha \), \( \cos \alpha \), \( \sin \beta \), and \( \cos \beta \) into the sine difference formula: \[ \sin(\alpha - \beta) = \sin \alpha \cos \beta - \cos \alpha \sin \beta \] Substituting the values we found: \[ \sin(\alpha - \beta) = \left(\frac{\sqrt{15}}{4}\right)\left(-\frac{\sqrt{7}}{4}\right) - \left(\frac{1}{4}\right)\left(\frac{3}{4}\right) \] Calculating each term: \[ = -\frac{\sqrt{15} \sqrt{7}}{16} - \frac{3}{16} \] Combining the terms: \[ = -\frac{\sqrt{105}}{16} - \frac{3}{16} = -\frac{\sqrt{105} + 3}{16} \] ### Final Answer Thus, the exact value of \( \sin(\alpha - \beta) \) is: \[ \boxed{-\frac{\sqrt{105} + 3}{16}} \] This corresponds to option b).