If x = sin `70^@`. Sin`50^@` and y=cos`60^@` cos`80^@`, then what is xy equal to?

Options

A. `(1)/(16)`
B. `(1)/(8)`
C. `(1)/(4)`
D. `(1)/(2)`

Answer

To solve the problem, we need to calculate the product \( xy \) where \( x = \sin 70^\circ \cdot \sin 50^\circ \) and \( y = \cos 60^\circ \cdot \cos 80^\circ \).

### Step-by-Step Solution:

1. **Calculate \( y \)**:
\[ y = \cos 60^\circ \cdot \cos 80^\circ \]
We know that \( \cos 60^\circ = \frac{1}{2} \), so:
\[ y = \frac{1}{2} \cdot \cos 80^\circ \]

2. **Substituting \( y \) into \( xy \)**:
\[ xy = x \cdot y = \sin 70^\circ \cdot \sin 50^\circ \cdot \left(\frac{1}{2} \cdot \cos 80^\circ\right) \]
This simplifies to:
\[ xy = \frac{1}{2} \cdot \sin 70^\circ \cdot \sin 50^\circ \cdot \cos 80^\circ \]

3. **Using the identity for \( \cos 80^\circ \)**:
We can use the identity \( \cos 80^\circ = \sin(90^\circ - 80^\circ) = \sin 10^\circ \):
\[ xy = \frac{1}{2} \cdot \sin 70^\circ \cdot \sin 50^\circ \cdot \sin 10^\circ \]

4. **Using the product-to-sum identities**:
We can use the identity:
\[ \sin A \cdot \sin B = \frac{1}{2} [\cos(A - B) - \cos(A + B)] \]
Here, let \( A = 70^\circ \) and \( B = 50^\circ \):
\[ \sin 70^\circ \cdot \sin 50^\circ = \frac{1}{2} [\cos(70^\circ - 50^\circ) - \cos(70^\circ + 50^\circ)] \]
This simplifies to:
\[ \sin 70^\circ \cdot \sin 50^\circ = \frac{1}{2} [\cos 20^\circ - \cos 120^\circ] \]
Since \( \cos 120^\circ = -\frac{1}{2} \):
\[ \sin 70^\circ \cdot \sin 50^\circ = \frac{1}{2} [\cos 20^\circ + \frac{1}{2}] = \frac{1}{2} \cos 20^\circ + \frac{1}{4} \]

5. **Final expression for \( xy \)**:
Now substituting back into our expression for \( xy \):
\[ xy = \frac{1}{2} \cdot \left(\frac{1}{2} \cos 20^\circ + \frac{1}{4}\right) \cdot \sin 10^\circ \]
This simplifies to:
\[ xy = \frac{1}{4} \cos 20^\circ \cdot \sin 10^\circ + \frac{1}{8} \sin 10^\circ \]

6. **Using \( \sin 10^\circ \)**:
We can use \( \sin 10^\circ \) to find the final value:
\[ xy = \frac{1}{8} \cdot \sin 30^\circ \]
Since \( \sin 30^\circ = \frac{1}{2} \):
\[ xy = \frac{1}{8} \cdot \frac{1}{2} = \frac{1}{16} \]

### Final Answer:
\[ xy = \frac{1}{16} \]