If `sinA=sinBandcosA=cosB` then

Options

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

Answer

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

### Step 1: Set up the equations
Given:
\[ \sin A = \sin B \]
\[ \cos A = \cos B \]

### Step 2: Use the sine and cosine subtraction formulas
From the sine equation, we can rewrite it as:
\[ \sin A - \sin B = 0 \]
Using the sine subtraction formula:
\[ \sin A - \sin B = 2 \sin\left(\frac{A - B}{2}\right) \cos\left(\frac{A + B}{2}\right) = 0 \]

### Step 3: Analyze the sine equation
For the product to be zero, either:
1. \( \sin\left(\frac{A - B}{2}\right) = 0 \) or
2. \( \cos\left(\frac{A + B}{2}\right) = 0 \)

### Step 4: Use the cosine equation
Similarly, for the cosine equation:
\[ \cos A - \cos B = 0 \]
Using the cosine subtraction formula:
\[ \cos A - \cos B = -2 \sin\left(\frac{A + B}{2}\right) \sin\left(\frac{A - B}{2}\right) = 0 \]

### Step 5: Analyze the cosine equation
For this product to be zero, either:
1. \( \sin\left(\frac{A + B}{2}\right) = 0 \) or
2. \( \sin\left(\frac{A - B}{2}\right) = 0 \)

### Step 6: Combine the results
From both equations, we see that \( \sin\left(\frac{A - B}{2}\right) = 0 \) is a common factor. This implies:
\[ \frac{A - B}{2} = n\pi \quad \text{for some integer } n \]
Thus,
\[ A - B = 2n\pi \]
This means:
\[ A = B + 2n\pi \]

### Step 7: Conclusion
Since both sine and cosine functions are periodic, we conclude that:
\[ A = B + 2n\pi \quad \text{or} \quad A = \pi - B + 2n\pi \]