If `2 cos(A + B) = 2 sin(A - B) = 1`, find the values of A and B.

Answer

To solve the problem, we start with the equations given:

1. \( 2 \cos(A + B) = 1 \)
2. \( 2 \sin(A - B) = 1 \)

### Step 1: Simplify the equations

From the first equation, we can simplify it as follows:
\[ 2 \cos(A + B) = 1 \implies \cos(A + B) = \frac{1}{2} \]

From the second equation, we simplify it similarly:
\[ 2 \sin(A - B) = 1 \implies \sin(A - B) = \frac{1}{2} \]

### Step 2: Find angles from trigonometric values

We know that:
- \( \cos(60^\circ) = \frac{1}{2} \)
- \( \sin(30^\circ) = \frac{1}{2} \)

From this, we can set up our equations based on the angles:
1. \( A + B = 60^\circ \) (from \( \cos(A + B) = \frac{1}{2} \))
2. \( A - B = 30^\circ \) (from \( \sin(A - B) = \frac{1}{2} \))

### Step 3: Solve the system of equations

Now we have a system of two equations:
1. \( A + B = 60^\circ \) (Equation 1)
2. \( A - B = 30^\circ \) (Equation 2)

To solve for \( A \) and \( B \), we can add both equations:
\[ (A + B) + (A - B) = 60^\circ + 30^\circ \]
This simplifies to:
\[ 2A = 90^\circ \implies A = 45^\circ \]

Now, substitute \( A = 45^\circ \) back into Equation 1 to find \( B \):
\[ 45^\circ + B = 60^\circ \]
Solving for \( B \):
\[ B = 60^\circ - 45^\circ = 15^\circ \]

### Final Values
Thus, the values of \( A \) and \( B \) are:
\[ A = 45^\circ, \quad B = 15^\circ \]