46. If \( \sin x=\frac{1}{4}, x \) in quadrant II, then find \( \sin \frac{x}{2}, \cos \frac{x}{2}, \tan \frac{x}{2} \)

Given that \( \sin x = \frac{1}{4} \) and \( x \) is in the second quadrant, we can find \( \sin \frac{x}{2} \), \( \cos \frac{x}{2} \), and \( \tan \frac{x}{2} \) using half-angle identities. ### Step 1: Determine \( \cos x \) Since \( x \) is in the second quadrant, \( \cos x \) is negative. Using the Pythagorean identity: \[ \cos^2 x = 1 - \sin^2 x = 1 - \left(\frac{1}{4}\right)^2 = 1 - \frac{1}{16} = \frac{15}{16} \] \[ \cos x = -\sqrt{\frac{15}{16}} = -\frac{\sqrt{15}}{4} \] ### Step 2: Find \( \sin \frac{x}{2} \) Using the half-angle formula: \[ \sin \frac{x}{2} = \sqrt{\frac{1 - \cos x}{2}} = \sqrt{\frac{1 - \left(-\frac{\sqrt{15}}{4}\right)}{2}} = \sqrt{\frac{1 + \frac{\sqrt{15}}{4}}{2}} = \frac{\sqrt{4 + \sqrt{15}}}{2\sqrt{2}} \] ### Step 3: Find \( \cos \frac{x}{2} \) Using the half-angle formula: \[ \cos \frac{x}{2} = \sqrt{\frac{1 + \cos x}{2}} = \sqrt{\frac{1 + \left(-\frac{\sqrt{15}}{4}\right)}{2}} = \sqrt{\frac{1 - \frac{\sqrt{15}}{4}}{2}} = \frac{\sqrt{4 - \sqrt{15}}}{2\sqrt{2}} \] ### Step 4: Find \( \tan \frac{x}{2} \) Using the tangent half-angle formula: \[ \tan \frac{x}{2} = \frac{\sin x}{1 + \cos x} = \frac{\frac{1}{4}}{1 - \frac{\sqrt{15}}{4}} = \frac{1}{4 - \sqrt{15}} \] Rationalizing the denominator: \[ \tan \frac{x}{2} = \frac{1}{4 - \sqrt{15}} \times \frac{4 + \sqrt{15}}{4 + \sqrt{15}} = \frac{4 + \sqrt{15}}{16 - 15} = 4 + \sqrt{15} \] ### Final Answers \[ \sin \frac{x}{2} = \frac{\sqrt{4 + \sqrt{15}}}{2\,\sqrt{2}} \] \[ \cos \frac{x}{2} = \frac{\sqrt{4 - \sqrt{15}}}{2\,\sqrt{2}} \] \[ \tan \frac{x}{2} = 4 + \sqrt{15} \] **Answer Summary:** After simplifying, the half-angles are equal to √(4 + √15) divided by twice √2, √(4 – √15) divided by twice √2, and 4 + √15. Thus, sin(x⁄2) = √(4 + √15)/(2√2), cos(x⁄2) = √(4 − √15)/(2√2), and tan(x⁄2) = 4 + √15.