If `alpha and beta` be between `0 and (pi)/(2)and if cos(alpha+beta)=(12)/(13) and sin(alpha-beta)=3/5,` then sin 2 `alpha` is equal to

Options

A. `64//65`
B. `56//65`
C. 0
D. `16//15`

Answer

To solve the problem, we need to find the value of \( \sin 2\alpha \) given that \( \cos(\alpha + \beta) = \frac{12}{13} \) and \( \sin(\alpha - \beta) = \frac{3}{5} \).

### Step-by-step Solution:

1. **Understanding the Given Values**:
- We have \( \cos(\alpha + \beta) = \frac{12}{13} \).
- We have \( \sin(\alpha - \beta) = \frac{3}{5} \).

2. **Finding \( \sin(\alpha + \beta) \)**:
- Since \( \cos^2(\alpha + \beta) + \sin^2(\alpha + \beta) = 1 \):
\[ \sin^2(\alpha + \beta) = 1 - \cos^2(\alpha + \beta) = 1 - \left(\frac{12}{13}\right)^2 = 1 - \frac{144}{169} = \frac{25}{169} \]
- Therefore, \( \sin(\alpha + \beta) = \sqrt{\frac{25}{169}} = \frac{5}{13} \) (since \( \alpha + \beta \) is in the first quadrant).

3. **Finding \( \cos(\alpha - \beta) \)**:
- Again using \( \sin^2(\alpha - \beta) + \cos^2(\alpha - \beta) = 1 \):
\[ \cos^2(\alpha - \beta) = 1 - \sin^2(\alpha - \beta) = 1 - \left(\frac{3}{5}\right)^2 = 1 - \frac{9}{25} = \frac{16}{25} \]
- Therefore, \( \cos(\alpha - \beta) = \sqrt{\frac{16}{25}} = \frac{4}{5} \) (since \( \alpha - \beta \) is also in the first quadrant).

4. **Using the Sum and Difference Formulas**:
- We know:
\[ \sin 2\alpha = \sin(\alpha + \beta + \alpha - \beta) = \sin(\alpha + \beta)\cos(\alpha - \beta) + \cos(\alpha + \beta)\sin(\alpha - \beta) \]
- Substituting the known values:
\[ \sin 2\alpha = \left(\frac{5}{13}\right)\left(\frac{4}{5}\right) + \left(\frac{12}{13}\right)\left(\frac{3}{5}\right) \]

5. **Calculating the Expression**:
- Simplifying:
\[ \sin 2\alpha = \frac{5 \cdot 4}{13 \cdot 5} + \frac{12 \cdot 3}{13 \cdot 5} = \frac{20}{65} + \frac{36}{65} = \frac{56}{65} \]

### Final Answer:
Thus, \( \sin 2\alpha = \frac{56}{65} \).

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