The value of `sin""(pi)/(10)sin""(13pi)/(10)` is

Options

A. `(1)/(2)`
B. `-(1)/(2)`
C. `-(1)/(4)`
D. 1

Explanation

To find the value of \( \sin\left(\frac{\pi}{10}\right) \sin\left(\frac{13\pi}{10}\right) \), we can follow these steps:

### Step 1: Rewrite \( \sin\left(\frac{13\pi}{10}\right) \)
We can express \( \frac{13\pi}{10} \) as:
\[ \frac{13\pi}{10} = \pi + \frac{3\pi}{10} \]
Using the sine addition formula, we know:
\[ \sin(\pi + \theta) = -\sin(\theta) \]
Thus,
\[ \sin\left(\frac{13\pi}{10}\right) = \sin\left(\pi + \frac{3\pi}{10}\right) = -\sin\left(\frac{3\pi}{10}\right) \]

### Step 2: Substitute into the original expression
Now substituting this back into our expression, we have:
\[ \sin\left(\frac{\pi}{10}\right) \sin\left(\frac{13\pi}{10}\right) = \sin\left(\frac{\pi}{10}\right) \left(-\sin\left(\frac{3\pi}{10}\right)\right) \]
This simplifies to:
\[ -\sin\left(\frac{\pi}{10}\right) \sin\left(\frac{3\pi}{10}\right) \]

### Step 3: Use the product-to-sum identities
We can use the product-to-sum identities:
\[ \sin A \sin B = \frac{1}{2} \left[ \cos(A - B) - \cos(A + B) \right] \]
Let \( A = \frac{\pi}{10} \) and \( B = \frac{3\pi}{10} \):
\[ -\sin\left(\frac{\pi}{10}\right) \sin\left(\frac{3\pi}{10}\right) = -\frac{1}{2} \left[ \cos\left(\frac{\pi}{10} - \frac{3\pi}{10}\right) - \cos\left(\frac{\pi}{10} + \frac{3\pi}{10}\right) \right] \]
Calculating \( A - B \) and \( A + B \):
\[ A - B = \frac{\pi}{10} - \frac{3\pi}{10} = -\frac{2\pi}{10} = -\frac{\pi}{5} \]
\[ A + B = \frac{\pi}{10} + \frac{3\pi}{10} = \frac{4\pi}{10} = \frac{2\pi}{5} \]
Thus, we have:
\[ -\sin\left(\frac{\pi}{10}\right) \sin\left(\frac{3\pi}{10}\right) = -\frac{1}{2} \left[ \cos\left(-\frac{\pi}{5}\right) - \cos\left(\frac{2\pi}{5}\right) \right] \]
Using the property \( \cos(-x) = \cos(x) \):
\[ = -\frac{1}{2} \left[ \cos\left(\frac{\pi}{5}\right) - \cos\left(\frac{2\pi}{5}\right) \right] \]

### Step 4: Evaluate the cosines
We know from trigonometric identities:
\[ \cos\left(\frac{\pi}{5}\right) = \frac{1 + \sqrt{5}}{4}, \quad \cos\left(\frac{2\pi}{5}\right) = \frac{\sqrt{5} - 1}{4} \]
Substituting these values:
\[ = -\frac{1}{2} \left[ \frac{1 + \sqrt{5}}{4} - \frac{\sqrt{5} - 1}{4} \right] \]
Simplifying:
\[ = -\frac{1}{2} \left[ \frac{1 + \sqrt{5} - \sqrt{5} + 1}{4} \right] = -\frac{1}{2} \left[ \frac{2}{4} \right] = -\frac{1}{2} \cdot \frac{1}{2} = -\frac{1}{4} \]

### Final Answer
Thus, the value of \( \sin\left(\frac{\pi}{10}\right) \sin\left(\frac{13\pi}{10}\right) \) is:
\[ \boxed{-\frac{1}{4}} \]