$ABCD$ is a square and $M$ and $N$ are the midpoints of $BC$ and $CD$ respectively. Then $\sin \theta=$
[asy]
draw((0,0)--(2,0)--(2,2)--(0,2)--cycle);
draw((0,0)--(2,1));
draw((0,0)--(1,2));
label("A", (0,0), SW);
label("B", (0,2), NW);
label("C", (2,2), NE);
label("D", (2,0), SE);
label("M", (1,2), N);
label("N", (2,1), E);
label("$\theta$", (.5,.5), SW);
[/asy]
$ \textbf{(A)}\ \frac{\sqrt{5}}{5} \qquad\textbf{(B)}\ \frac{3}{5} \qquad\textbf{(C)}\ \frac{\sqrt{10}}{5} \qquad\textbf{(D)}\ \frac{4}{5} \qquad\textbf{(E)}\ \text{none of these} $
1. **Assign Coordinates:**
   Let's place the square \(ABCD\) on the coordinate plane with \(A = (0,0)\), \(B = (0,1)\), \(C = (1,1)\), and \(D = (1,0)\). The side length of the square is 1.

2. **Find Midpoints:**
   - \(M\) is the midpoint of \(BC\), so \(M = \left(\frac{0+1}{2}, \frac{1+1}{2}\right) = \left(\frac{1}{2}, 1\right)\).
   - \(N\) is the midpoint of \(CD\), so \(N = \left(\frac{1+1}{2}, \frac{1+0}{2}\right) = \left(1, \frac{1}{2}\right)\).

3. **Calculate Lengths:**
   - The length of \(AM\) is calculated using the distance formula:
     \[
     AM = \sqrt{\left(\frac{1}{2} - 0\right)^2 + \left(1 - 0\right)^2} = \sqrt{\left(\frac{1}{2}\right)^2 + 1^2} = \sqrt{\frac{1}{4} + 1} = \sqrt{\frac{5}{4}} = \frac{\sqrt{5}}{2}
     \]
   - Similarly, the length of \(AN\) is:
     \[
     AN = \sqrt{\left(1 - 0\right)^2 + \left(\frac{1}{2} - 0\right)^2} = \sqrt{1^2 + \left(\frac{1}{2}\right)^2} = \sqrt{1 + \frac{1}{4}} = \sqrt{\frac{5}{4}} = \frac{\sqrt{5}}{2}
     \]

4. **Area of Triangle \(AMN\):**
   - The area of \(\triangle AMN\) can be calculated using the determinant formula for the area of a triangle with vertices \((x_1, y_1)\), \((x_2, y_2)\), and \((x_3, y_3)\):
     \[
     \text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right|
     \]
     Substituting \(A = (0,0)\), \(M = \left(\frac{1}{2}, 1\right)\), and \(N = \left(1, \frac{1}{2}\right)\):
     \[
     \text{Area} = \frac{1}{2} \left| 0(1 - \frac{1}{2}) + \frac{1}{2}(\frac{1}{2} - 0) + 1(0 - 1) \right| = \frac{1}{2} \left| 0 + \frac{1}{4} - 1 \right| = \frac{1}{2} \left| -\frac{3}{4} \right| = \frac{3}{8}
     \]

5. **Using the Area Formula:**
   - The area of \(\triangle AMN\) can also be expressed using the formula:
     \[
     \text{Area} = \frac{1}{2} \cdot AM \cdot AN \cdot \sin \theta
     \]
     Substituting the known values:
     \[
     \frac{3}{8} = \frac{1}{2} \cdot \left(\frac{\sqrt{5}}{2}\right)^2 \cdot \sin \theta
     \]
     Simplifying:
     \[
     \frac{3}{8} = \frac{1}{2} \cdot \frac{5}{4} \cdot \sin \theta = \frac{5}{8} \cdot \sin \theta
     \]
     Solving for \(\sin \theta\):
     \[
     \sin \theta = \frac{3}{5}
     \]

Conclusion:
\[
\boxed{\frac{3}{5}}
\]