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

Answer

To solve the problem, we need to find the values of angles A and B given the equations:

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

Let's solve this step by step.

### Step 1: Simplify the equations
From the first equation:
\[ 2 \cos(A + B) = 1 \]
Dividing both sides by 2 gives:
\[ \cos(A + B) = \frac{1}{2} \]

From the second equation:
\[ 2 \sin(A - B) = 1 \]
Dividing both sides by 2 gives:
\[ \sin(A - B) = \frac{1}{2} \]

### Step 2: Identify standard angles
We know that:
\[ \cos(60^\circ) = \frac{1}{2} \]
Thus, we can write:
\[ A + B = 60^\circ \quad \text{(Equation 1)} \]

Also, we know that:
\[ \sin(30^\circ) = \frac{1}{2} \]
Thus, we can write:
\[ A - B = 30^\circ \quad \text{(Equation 2)} \]

### Step 3: Solve the system of equations
Now we have a system of two equations:
1. \( A + B = 60^\circ \)
2. \( A - B = 30^\circ \)

We can solve these equations by adding them together:
\[ (A + B) + (A - B) = 60^\circ + 30^\circ \]
This simplifies to:
\[ 2A = 90^\circ \]
Dividing both sides by 2 gives:
\[ A = 45^\circ \]

### Step 4: Substitute to find B
Now we can substitute \( A = 45^\circ \) back into Equation 1 to find B:
\[ 45^\circ + B = 60^\circ \]
Subtracting \( 45^\circ \) from both sides gives:
\[ B = 60^\circ - 45^\circ = 15^\circ \]

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

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