If `"sin" A = "sin"B, "cos"A = "cos"B,` then the value of A im terms of B, is

Options

A. `n pi + 8`
B. `n pi + (-1)^(n)B`
C. `2n pi + B`
D. `2n pi -B`

Answer

To solve the problem, we need to find the value of \( A \) in terms of \( B \) given that \( \sin A = \sin B \) and \( \cos A = \cos B \).

### Step 1: Analyze the given equations
We know that:
1. \( \sin A = \sin B \)
2. \( \cos A = \cos B \)

### Step 2: Use the general solutions for sine and cosine
From the properties of trigonometric functions, we can derive the general solutions for \( A \) based on the equations provided.

#### For \( \sin A = \sin B \):
The general solution for \( A \) in terms of \( B \) is:
\[ A = n\pi + (-1)^n B \]
where \( n \) is any integer.

#### For \( \cos A = \cos B \):
The general solution for \( A \) in terms of \( B \) is:
\[ A = 2n\pi \pm B \]
where \( n \) is any integer.

### Step 3: Find the common solutions
Now, we need to find a common expression for \( A \) from both equations.

1. From \( \sin A = \sin B \):
\[ A = n\pi + (-1)^n B \]
2. From \( \cos A = \cos B \):
\[ A = 2m\pi \pm B \]

### Step 4: Set the two expressions equal
To find a common solution, we can set the two expressions equal to each other.

Let’s consider the case where \( n \) is even (i.e., \( n = 2k \)):
\[ A = 2k\pi + B \quad \text{(from sine)} \]
And for \( m = k \):
\[ A = 2k\pi + B \quad \text{(from cosine)} \]

### Step 5: Conclusion
Thus, the common solution that satisfies both equations is:
\[ A = 2n\pi + B \]
where \( n \) is any integer.

### Final Answer
The value of \( A \) in terms of \( B \) is:
\[ A = 2n\pi + B \]

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