1. **Stating the problem:** We have a point $P$ moving in the $Oxy$-plane with parametric equations: $$x(t) = \frac{1}{2} \sin(t), \quad y(t) = \sin\left(t + \frac{1}{3} \pi\right)$$ We need to find the maximum distance of $P$ from the origin, i.e., maximize $d(t) = \sqrt{x(t)^2 + y(t)^2}$. 2. **Formula and rules:** The distance from the origin is given by $$d(t) = \sqrt{x(t)^2 + y(t)^2}$$ To find the maximum distance, we maximize $d(t)^2$ to avoid the square root: $$d(t)^2 = x(t)^2 + y(t)^2$$ 3. **Calculate $d(t)^2$:** $$d(t)^2 = \left(\frac{1}{2} \sin(t)\right)^2 + \sin^2\left(t + \frac{1}{3} \pi\right) = \frac{1}{4} \sin^2(t) + \sin^2\left(t + \frac{\pi}{3}\right)$$ 4. **Use trigonometric identity:** Recall that $$\sin^2(\alpha) = \frac{1 - \cos(2\alpha)}{2}$$ Apply this: $$d(t)^2 = \frac{1}{4} \cdot \frac{1 - \cos(2t)}{2} + \frac{1 - \cos\left(2t + \frac{2\pi}{3}\right)}{2} = \frac{1}{8}(1 - \cos(2t)) + \frac{1}{2} \left(1 - \cos\left(2t + \frac{2\pi}{3}\right)\right)$$ 5. **Simplify:** $$d(t)^2 = \frac{1}{8} - \frac{1}{8} \cos(2t) + \frac{1}{2} - \frac{1}{2} \cos\left(2t + \frac{2\pi}{3}\right) = \frac{5}{8} - \frac{1}{8} \cos(2t) - \frac{1}{2} \cos\left(2t + \frac{2\pi}{3}\right)$$ 6. **Maximize $d(t)^2$:** Use the cosine addition formula: $$\cos\left(2t + \frac{2\pi}{3}\right) = \cos(2t) \cos\left(\frac{2\pi}{3}\right) - \sin(2t) \sin\left(\frac{2\pi}{3}\right) = -\frac{1}{2} \cos(2t) - \frac{\sqrt{3}}{2} \sin(2t)$$ Substitute back: $$d(t)^2 = \frac{5}{8} - \frac{1}{8} \cos(2t) - \frac{1}{2} \left(-\frac{1}{2} \cos(2t) - \frac{\sqrt{3}}{2} \sin(2t)\right) = \frac{5}{8} - \frac{1}{8} \cos(2t) + \frac{1}{4} \cos(2t) + \frac{\sqrt{3}}{4} \sin(2t)$$ Simplify terms: $$d(t)^2 = \frac{5}{8} + \frac{1}{8} \cos(2t) + \frac{\sqrt{3}}{4} \sin(2t)$$ 7. **Rewrite as a single cosine:** Let $$A = \frac{1}{8}, \quad B = \frac{\sqrt{3}}{4}$$ Then $$d(t)^2 = \frac{5}{8} + A \cos(2t) + B \sin(2t) = \frac{5}{8} + R \cos(2t - \phi)$$ where $$R = \sqrt{A^2 + B^2} = \sqrt{\left(\frac{1}{8}\right)^2 + \left(\frac{\sqrt{3}}{4}\right)^2} = \sqrt{\frac{1}{64} + \frac{3}{16}} = \sqrt{\frac{1}{64} + \frac{12}{64}} = \sqrt{\frac{13}{64}} = \frac{\sqrt{13}}{8}$$ and $$\phi = \arctan\left(\frac{B}{A}\right) = \arctan\left(\frac{\frac{\sqrt{3}}{4}}{\frac{1}{8}}\right) = \arctan(2\sqrt{3})$$ 8. **Maximum value of $d(t)^2$:** $$\max d(t)^2 = \frac{5}{8} + R = \frac{5}{8} + \frac{\sqrt{13}}{8} = \frac{5 + \sqrt{13}}{8}$$ 9. **Maximum distance:** $$d_{max} = \sqrt{\max d(t)^2} = \sqrt{\frac{5 + \sqrt{13}}{8}} \approx 0.97$$ **Final answer:** $$\boxed{0.97}$$