If `sin alpha=sinbeta and cos alpha=cosbeta,` then

Options

A. `sin""(alpha+beta)/(2)=0`
B. `cos""(alpha+beta)/(2)=0`
C. `sin""(alpha-beta)/(2)=0`
D. `cos""((alpha-beta)/(2))=0`

Answer

To solve the problem where \( \sin \alpha = \sin \beta \) and \( \cos \alpha = \cos \beta \), we can follow these steps:

### Step-by-Step Solution:

1. **Start with the given equations:**
\[ \sin \alpha = \sin \beta \quad \text{and} \quad \cos \alpha = \cos \beta \]

2. **Rearrange the sine equation:**
\[ \sin \alpha - \sin \beta = 0 \]
Using the sine difference identity, we can rewrite this as:
\[ 2 \sin\left(\frac{\alpha + \beta}{2}\right) \cos\left(\frac{\alpha - \beta}{2}\right) = 0 \]
This gives us two possible cases:
- \( \sin\left(\frac{\alpha + \beta}{2}\right) = 0 \)
- \( \cos\left(\frac{\alpha - \beta}{2}\right) = 0 \)

3. **Rearrange the cosine equation:**
\[ \cos \alpha - \cos \beta = 0 \]
Using the cosine difference identity, we can rewrite this as:
\[ 2 \sin\left(\frac{\alpha + \beta}{2}\right) \sin\left(\frac{\alpha - \beta}{2}\right) = 0 \]
This gives us two possible cases:
- \( \sin\left(\frac{\alpha + \beta}{2}\right) = 0 \)
- \( \sin\left(\frac{\alpha - \beta}{2}\right) = 0 \)

4. **Analyze the equations:**
From both sets of equations, we see that:
- If \( \sin\left(\frac{\alpha - \beta}{2}\right) = 0 \), then \( \frac{\alpha - \beta}{2} = n\pi \) for some integer \( n \), which implies \( \alpha - \beta = 2n\pi \) or \( \alpha = \beta + 2n\pi \).
- If \( \sin\left(\frac{\alpha + \beta}{2}\right) = 0 \), then \( \frac{\alpha + \beta}{2} = m\pi \) for some integer \( m \), which implies \( \alpha + \beta = 2m\pi \).

5. **Conclusion:**
Since both conditions lead to the conclusion that \( \alpha \) and \( \beta \) must be equal (considering the periodic nature of sine and cosine), we can conclude:
\[ \alpha = \beta + 2n\pi \quad \text{or} \quad \alpha = \beta \]

### Final Result:
Thus, if \( \sin \alpha = \sin \beta \) and \( \cos \alpha = \cos \beta \), then:
\[ \alpha = \beta + 2n\pi \quad \text{(for some integer } n\text{)} \]