Find `sin20^(@)sin40^(@)sin60^(@)sin80^(@)`.

Answer

To find the value of \( \sin 20^\circ \sin 40^\circ \sin 60^\circ \sin 80^\circ \), we can use a trigonometric identity that relates the product of sines to a simpler expression.

### Step-by-Step Solution:

1. **Identify the Sine Product Formula**:
We can use the formula:
\[ \sin A \sin B \sin C = \frac{1}{4} \left( \sin (A + B + C) + \sin (A + B - C) + \sin (A - B + C) + \sin (-A + B + C) \right) \]
However, a more suitable formula for our case is:
\[ \sin \theta \sin (60^\circ - \theta) \sin (60^\circ + \theta) = \frac{1}{4} \sin (3\theta) \]

2. **Set Up the Angles**:
Let's set \( \theta = 20^\circ \). Then:
- \( 60^\circ - \theta = 40^\circ \)
- \( 60^\circ + \theta = 80^\circ \)

Thus, we can rewrite the product:
\[ \sin 20^\circ \sin 40^\circ \sin 80^\circ = \frac{1}{4} \sin (3 \times 20^\circ) \]

3. **Calculate \( \sin (3 \times 20^\circ) \)**:
\[ 3 \times 20^\circ = 60^\circ \]
Therefore:
\[ \sin (3 \times 20^\circ) = \sin 60^\circ \]

4. **Substitute \( \sin 60^\circ \)**:
We know that:
\[ \sin 60^\circ = \frac{\sqrt{3}}{2} \]
Now substituting back:
\[ \sin 20^\circ \sin 40^\circ \sin 80^\circ = \frac{1}{4} \cdot \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{8} \]

5. **Include \( \sin 60^\circ \)**:
Now we need to include \( \sin 60^\circ \) in our original product:
\[ \sin 20^\circ \sin 40^\circ \sin 60^\circ \sin 80^\circ = \left( \sin 20^\circ \sin 40^\circ \sin 80^\circ \right) \sin 60^\circ \]
Thus:
\[ = \frac{\sqrt{3}}{8} \cdot \frac{\sqrt{3}}{2} = \frac{3}{16} \]

### Final Answer:
\[ \sin 20^\circ \sin 40^\circ \sin 60^\circ \sin 80^\circ = \frac{3}{16} \]