Prove that `sin 6^(@) sin 42^(@) sin 66^(@) sin 78^(@) = (1)/(16)`.

Answer

To prove that \( \sin 6^\circ \sin 42^\circ \sin 66^\circ \sin 78^\circ = \frac{1}{16} \), we will follow a systematic approach using trigonometric identities.

### Step-by-Step Solution:

1. **Start with the Left-Hand Side (LHS)**:
\[ \text{LHS} = \sin 6^\circ \sin 42^\circ \sin 66^\circ \sin 78^\circ \]

2. **Multiply LHS by 1 (in the form of \( \frac{1}{4} \cdot 4 \))**:
\[ \text{LHS} = \frac{1}{4} \cdot 4 \cdot \sin 6^\circ \sin 42^\circ \sin 66^\circ \sin 78^\circ \]

3. **Rearranging the terms**:
\[ \text{LHS} = \frac{1}{4} \left( 2 \sin 6^\circ \sin 66^\circ \right) \left( 2 \sin 42^\circ \sin 78^\circ \right) \]

4. **Using the identity \( 2 \sin A \sin B = \cos(A - B) - \cos(A + B) \)**:
- For \( 2 \sin 6^\circ \sin 66^\circ \):
\[ 2 \sin 6^\circ \sin 66^\circ = \cos(66^\circ - 6^\circ) - \cos(66^\circ + 6^\circ) = \cos 60^\circ - \cos 72^\circ \]
- For \( 2 \sin 42^\circ \sin 78^\circ \):
\[ 2 \sin 42^\circ \sin 78^\circ = \cos(78^\circ - 42^\circ) - \cos(78^\circ + 42^\circ) = \cos 36^\circ - \cos 120^\circ \]

5. **Substituting back into LHS**:
\[ \text{LHS} = \frac{1}{4} \left( \cos 60^\circ - \cos 72^\circ \right) \left( \cos 36^\circ - \cos 120^\circ \right) \]

6. **Using known values of cosine**:
- \( \cos 60^\circ = \frac{1}{2} \)
- \( \cos 72^\circ = \frac{\sqrt{5}-1}{4} \)
- \( \cos 36^\circ = \frac{\sqrt{5}+1}{4} \)
- \( \cos 120^\circ = -\frac{1}{2} \)

7. **Substituting these values**:
\[ \text{LHS} = \frac{1}{4} \left( \frac{1}{2} - \frac{\sqrt{5}-1}{4} \right) \left( \frac{\sqrt{5}+1}{4} + \frac{1}{2} \right) \]

8. **Simplifying the expressions**:
- First term:
\[ \frac{1}{2} - \frac{\sqrt{5}-1}{4} = \frac{2}{4} - \frac{\sqrt{5}-1}{4} = \frac{2 - \sqrt{5} + 1}{4} = \frac{3 - \sqrt{5}}{4} \]
- Second term:
\[ \frac{\sqrt{5}+1}{4} + \frac{1}{2} = \frac{\sqrt{5}+1}{4} + \frac{2}{4} = \frac{\sqrt{5}+3}{4} \]

9. **Final multiplication**:
\[ \text{LHS} = \frac{1}{4} \cdot \frac{3 - \sqrt{5}}{4} \cdot \frac{\sqrt{5}+3}{4} \]
\[ = \frac{1}{4} \cdot \frac{(3 - \sqrt{5})(\sqrt{5} + 3)}{16} \]
\[ = \frac{1}{4} \cdot \frac{9 - 5}{16} = \frac{1}{4} \cdot \frac{4}{16} = \frac{1}{16} \]

10. **Conclusion**:
\[ \text{LHS} = \frac{1}{16} \]
Thus, we have proved that:
\[ \sin 6^\circ \sin 42^\circ \sin 66^\circ \sin 78^\circ = \frac{1}{16} \]