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

Options

A. `(sqrt2)/(16)`
B. `1/8`
C. `1/6`
D. `(sqrt2)/(32)`

Answer

To find the value of \( \sin \frac{\pi}{16} \sin \frac{3\pi}{16} \sin \frac{5\pi}{16} \sin \frac{7\pi}{16} \), we can follow these steps:

### Step 1: Group the Sine Terms
We can group the sine terms as follows:
\[ \sin \frac{\pi}{16} \sin \frac{7\pi}{16} \quad \text{and} \quad \sin \frac{3\pi}{16} \sin \frac{5\pi}{16} \]

### Step 2: Use the Product-to-Sum Formula
Using the product-to-sum formula:
\[ 2 \sin a \sin b = \cos(a - b) - \cos(a + b) \]
we can apply this to both pairs.

#### For \( \sin \frac{\pi}{16} \sin \frac{7\pi}{16} \):
Let \( a = \frac{\pi}{16} \) and \( b = \frac{7\pi}{16} \):
\[ 2 \sin \frac{\pi}{16} \sin \frac{7\pi}{16} = \cos\left(\frac{7\pi}{16} - \frac{\pi}{16}\right) - \cos\left(\frac{7\pi}{16} + \frac{\pi}{16}\right) \]
Calculating the angles:
\[ = \cos\left(\frac{6\pi}{16}\right) - \cos\left(\frac{8\pi}{16}\right) = \cos\left(\frac{3\pi}{8}\right) - \cos\left(\frac{\pi}{2}\right) \]
Since \( \cos\left(\frac{\pi}{2}\right) = 0 \):
\[ = \cos\left(\frac{3\pi}{8}\right) \]

#### For \( \sin \frac{3\pi}{16} \sin \frac{5\pi}{16} \):
Let \( a = \frac{3\pi}{16} \) and \( b = \frac{5\pi}{16} \):
\[ 2 \sin \frac{3\pi}{16} \sin \frac{5\pi}{16} = \cos\left(\frac{5\pi}{16} - \frac{3\pi}{16}\right) - \cos\left(\frac{5\pi}{16} + \frac{3\pi}{16}\right) \]
Calculating the angles:
\[ = \cos\left(\frac{2\pi}{16}\right) - \cos\left(\frac{8\pi}{16}\right) = \cos\left(\frac{\pi}{8}\right) - \cos\left(\frac{\pi}{2}\right) \]
Again, since \( \cos\left(\frac{\pi}{2}\right) = 0 \):
\[ = \cos\left(\frac{\pi}{8}\right) \]

### Step 3: Combine the Results
Now we have:
\[ \sin \frac{\pi}{16} \sin \frac{7\pi}{16} \sin \frac{3\pi}{16} \sin \frac{5\pi}{16} = \frac{1}{4} \cos\left(\frac{3\pi}{8}\right) \cos\left(\frac{\pi}{8}\right) \]

### Step 4: Use the Product-to-Sum Formula Again
Using the product-to-sum formula again on \( \cos\left(\frac{3\pi}{8}\right) \cos\left(\frac{\pi}{8}\right) \):
\[ 2 \cos a \cos b = \cos(a + b) + \cos(a - b) \]
Let \( a = \frac{3\pi}{8} \) and \( b = \frac{\pi}{8} \):
\[ 2 \cos\left(\frac{3\pi}{8}\right) \cos\left(\frac{\pi}{8}\right) = \cos\left(\frac{4\pi}{8}\right) + \cos\left(\frac{2\pi}{8}\right) = \cos\left(\frac{\pi}{2}\right) + \cos\left(\frac{\pi}{4}\right) \]
Since \( \cos\left(\frac{\pi}{2}\right) = 0 \):
\[ = \cos\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} \]

### Step 5: Final Calculation
Thus, we have:
\[ \sin \frac{\pi}{16} \sin \frac{3\pi}{16} \sin \frac{5\pi}{16} \sin \frac{7\pi}{16} = \frac{1}{8} \cdot \frac{1}{\sqrt{2}} = \frac{1}{8\sqrt{2}} \]
To rationalize the denominator:
\[ = \frac{\sqrt{2}}{16} \]

### Final Answer
The value of \( \sin \frac{\pi}{16} \sin \frac{3\pi}{16} \sin \frac{5\pi}{16} \sin \frac{7\pi}{16} \) is:
\[ \frac{\sqrt{2}}{16} \]