If `sin(A-B)=sinAcosB-cosAsinB`, then `sin15^(@)` will be

Options

A. `(sqrt(3)+1)/(2sqrt(2))`
B. `(sqrt(3))/(2sqrt(2))`
C. `(sqrt(3)-1)/(-sqrt(2))`
D. `(sqrt(3)-1)/(2sqrt(2))`

Answer

To find the value of \( \sin 15^\circ \) using the formula \( \sin(A - B) = \sin A \cos B - \cos A \sin B \), we can set \( A = 45^\circ \) and \( B = 30^\circ \). This is because \( 15^\circ = 45^\circ - 30^\circ \).

### Step-by-Step Solution:

1. **Identify A and B**:
\[ A = 45^\circ, \quad B = 30^\circ \]

2. **Use the sine difference formula**:
\[ \sin(15^\circ) = \sin(45^\circ - 30^\circ) = \sin 45^\circ \cos 30^\circ - \cos 45^\circ \sin 30^\circ \]

3. **Substitute known values**:
- \( \sin 45^\circ = \frac{1}{\sqrt{2}} \)
- \( \cos 30^\circ = \frac{\sqrt{3}}{2} \)
- \( \cos 45^\circ = \frac{1}{\sqrt{2}} \)
- \( \sin 30^\circ = \frac{1}{2} \)

So, substituting these values gives:
\[ \sin(15^\circ) = \left(\frac{1}{\sqrt{2}}\right) \left(\frac{\sqrt{3}}{2}\right) - \left(\frac{1}{\sqrt{2}}\right) \left(\frac{1}{2}\right) \]

4. **Simplify the expression**:
\[ \sin(15^\circ) = \frac{\sqrt{3}}{2\sqrt{2}} - \frac{1}{2\sqrt{2}} \]

5. **Combine the fractions**:
\[ \sin(15^\circ) = \frac{\sqrt{3} - 1}{2\sqrt{2}} \]

### Final Result:
Thus, the value of \( \sin 15^\circ \) is:
\[ \sin 15^\circ = \frac{\sqrt{3} - 1}{2\sqrt{2}} \]