In a triangle `A B C ,` if `A-B=120^0a n dsinA/2sinB/2sinC/2=1/(32),` then the value of `8cosC` is_______

Explanation

To solve the problem, we need to find the value of \(8 \cos C\) given that \(A - B = 120^\circ\) and \(\frac{\sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}}{1} = \frac{1}{32}\).

### Step-by-Step Solution:

1. **Rearranging the Given Equation**:
\[ \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2} = \frac{1}{32} \]
Multiply both sides by 2:
\[ 2 \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2} = \frac{1}{16} \]

2. **Using the Sine Product Identity**:
The product \(2 \sin \frac{A}{2} \sin \frac{B}{2}\) can be expressed using the cosine difference identity:
\[ 2 \sin \frac{A}{2} \sin \frac{B}{2} = \cos\left(\frac{A - B}{2}\right) - \cos\left(\frac{A + B}{2}\right) \]
Given \(A - B = 120^\circ\), we have:
\[ \frac{A - B}{2} = 60^\circ \]
Therefore,
\[ 2 \sin \frac{A}{2} \sin \frac{B}{2} = \cos(60^\circ) - \cos\left(\frac{A + B}{2}\right) \]
Since \(\cos(60^\circ) = \frac{1}{2}\), we can write:
\[ 2 \sin \frac{A}{2} \sin \frac{B}{2} = \frac{1}{2} - \cos\left(\frac{A + B}{2}\right) \]

3. **Substituting Back**:
Now substituting back into our equation:
\[ \left(\frac{1}{2} - \cos\left(\frac{A + B}{2}\right)\right) \sin \frac{C}{2} = \frac{1}{16} \]

4. **Expressing \(C\)**:
Since \(A + B + C = 180^\circ\), we have:
\[ C = 180^\circ - (A + B) \]
Thus,
\[ \frac{A + B}{2} = 90^\circ - \frac{C}{2} \]

5. **Finding \(\sin \frac{C}{2}\)**:
Let \(x = \sin \frac{C}{2}\). Then we can express the equation as:
\[ \left(\frac{1}{2} - \cos(90^\circ - \frac{C}{2})\right)x = \frac{1}{16} \]
Since \(\cos(90^\circ - \theta) = \sin(\theta)\):
\[ \left(\frac{1}{2} - \sin\left(\frac{C}{2}\right)\right)x = \frac{1}{16} \]

6. **Solving the Quadratic Equation**:
Rearranging gives us:
\[ x^2 - \frac{1}{16} = 0 \]
This leads to:
\[ 16x^2 - 1 = 0 \quad \Rightarrow \quad x^2 = \frac{1}{16} \quad \Rightarrow \quad x = \frac{1}{4} \]

7. **Finding \(\cos C\)**:
Using the identity:
\[ \cos C = 1 - 2 \sin^2\left(\frac{C}{2}\right) = 1 - 2 \left(\frac{1}{4}\right)^2 = 1 - 2 \cdot \frac{1}{16} = 1 - \frac{1}{8} = \frac{7}{8} \]

8. **Calculating \(8 \cos C\)**:
Finally, we find:
\[ 8 \cos C = 8 \cdot \frac{7}{8} = 7 \]

### Final Answer:
The value of \(8 \cos C\) is \(7\).