The value of `sin ""(pi)/(16) sin ""(3pi)/(16) sin ""(5pi)/(16) sin ""(7pi)/(16)` is

Options

A. `1/16`
B. `(sqrt2)/(16)`
C. `1/8`
D. `(sqrt2)/(8)`

Answer

To solve the problem \( \sin\left(\frac{\pi}{16}\right) \sin\left(\frac{3\pi}{16}\right) \sin\left(\frac{5\pi}{16}\right) \sin\left(\frac{7\pi}{16}\right) \), we can use a known product-to-sum formula.

### Step-by-step Solution:

1. **Recognize the Sine Product**:
We need to evaluate the product of four sine functions. We can pair them to simplify the calculation:
\[ \sin\left(\frac{\pi}{16}\right) \sin\left(\frac{7\pi}{16}\right) \quad \text{and} \quad \sin\left(\frac{3\pi}{16}\right) \sin\left(\frac{5\pi}{16}\right) \]

2. **Use the Identity**:
We can use the identity:
\[ \sin A \sin B = \frac{1}{2} [\cos(A-B) - \cos(A+B)] \]
Applying this to our pairs:
- For \( \sin\left(\frac{\pi}{16}\right) \sin\left(\frac{7\pi}{16}\right) \):
\[ \sin\left(\frac{\pi}{16}\right) \sin\left(\frac{7\pi}{16}\right) = \frac{1}{2} \left[\cos\left(\frac{\pi}{16} - \frac{7\pi}{16}\right) - \cos\left(\frac{\pi}{16} + \frac{7\pi}{16}\right)\right] \]
\[ = \frac{1}{2} \left[\cos\left(-\frac{6\pi}{16}\right) - \cos\left(\frac{8\pi}{16}\right)\right] \]
\[ = \frac{1}{2} \left[\cos\left(\frac{3\pi}{8}\right) - 0\right] = \frac{1}{2} \cos\left(\frac{3\pi}{8}\right) \]

- For \( \sin\left(\frac{3\pi}{16}\right) \sin\left(\frac{5\pi}{16}\right) \):
\[ \sin\left(\frac{3\pi}{16}\right) \sin\left(\frac{5\pi}{16}\right) = \frac{1}{2} \left[\cos\left(\frac{3\pi}{16} - \frac{5\pi}{16}\right) - \cos\left(\frac{3\pi}{16} + \frac{5\pi}{16}\right)\right] \]
\[ = \frac{1}{2} \left[\cos\left(-\frac{2\pi}{16}\right) - \cos\left(\frac{8\pi}{16}\right)\right] \]
\[ = \frac{1}{2} \left[\cos\left(\frac{\pi}{8}\right) - 0\right] = \frac{1}{2} \cos\left(\frac{\pi}{8}\right) \]

3. **Combine the Results**:
Now we can combine the results from both pairs:
\[ \sin\left(\frac{\pi}{16}\right) \sin\left(\frac{7\pi}{16}\right) \sin\left(\frac{3\pi}{16}\right) \sin\left(\frac{5\pi}{16}\right) = \left(\frac{1}{2} \cos\left(\frac{3\pi}{8}\right)\right) \left(\frac{1}{2} \cos\left(\frac{\pi}{8}\right)\right) \]
\[ = \frac{1}{4} \cos\left(\frac{3\pi}{8}\right) \cos\left(\frac{\pi}{8}\right) \]

4. **Final Calculation**:
We can use the product-to-sum formula again on \( \cos\left(\frac{3\pi}{8}\right) \cos\left(\frac{\pi}{8}\right) \):
\[ \cos A \cos B = \frac{1}{2} [\cos(A+B) + \cos(A-B)] \]
Thus,
\[ \cos\left(\frac{3\pi}{8}\right) \cos\left(\frac{\pi}{8}\right) = \frac{1}{2} \left[\cos\left(\frac{4\pi}{8}\right) + \cos\left(\frac{2\pi}{8}\right)\right] \]
\[ = \frac{1}{2} \left[0 + \cos\left(\frac{\pi}{4}\right)\right] = \frac{1}{2} \cdot \frac{\sqrt{2}}{2} = \frac{\sqrt{2}}{4} \]

5. **Final Result**:
Therefore,
\[ \sin\left(\frac{\pi}{16}\right) \sin\left(\frac{3\pi}{16}\right) \sin\left(\frac{5\pi}{16}\right) \sin\left(\frac{7\pi}{16}\right) = \frac{1}{4} \cdot \frac{\sqrt{2}}{4} = \frac{\sqrt{2}}{16} \]

### Final Answer:
\[ \sin\left(\frac{\pi}{16}\right) \sin\left(\frac{3\pi}{16}\right) \sin\left(\frac{5\pi}{16}\right) \sin\left(\frac{7\pi}{16}\right) = \frac{\sqrt{2}}{16} \]