If : `sin(A-B)=cos(A+B)=(1)/(2), "then" : (A,B)`

Options

A. `(15^(@),45^(@))`
B. `(15^(@),30^(@))`
C. `(45^(@),15^(@))`
D. `(90^(@),0^(@))`

Answer

To solve the problem where \( \sin(A - B) = \cos(A + B) = \frac{1}{2} \), we can follow these steps:

### Step 1: Set up the equations
We know that:
\[ \sin(A - B) = \frac{1}{2} \]
\[ \cos(A + B) = \frac{1}{2} \]

### Step 2: Find the angles corresponding to the sine and cosine values
From trigonometric values, we know:
\[ \sin \theta = \frac{1}{2} \implies \theta = 30^\circ \text{ or } 150^\circ \]
Thus, we can write:
\[ A - B = 30^\circ \quad \text{(1)} \]
or
\[ A - B = 150^\circ \quad \text{(2)} \]

For cosine:
\[ \cos \phi = \frac{1}{2} \implies \phi = 60^\circ \text{ or } 300^\circ \]
Thus, we can write:
\[ A + B = 60^\circ \quad \text{(3)} \]
or
\[ A + B = 300^\circ \quad \text{(4)} \]

### Step 3: Solve the equations
We will first consider the pairs of equations from (1) and (3):
1. \( A - B = 30^\circ \)
2. \( A + B = 60^\circ \)

Adding these two equations:
\[ (A - B) + (A + B) = 30^\circ + 60^\circ \]
\[ 2A = 90^\circ \implies A = 45^\circ \]

Now substituting \( A = 45^\circ \) back into equation (3):
\[ 45^\circ + B = 60^\circ \implies B = 60^\circ - 45^\circ = 15^\circ \]

So, we have:
\[ A = 45^\circ, \quad B = 15^\circ \]

### Step 4: Check the other pairs of equations
Now let's check the other possible pairs:
Using (2) and (3):
1. \( A - B = 150^\circ \)
2. \( A + B = 60^\circ \)

Adding these:
\[ (A - B) + (A + B) = 150^\circ + 60^\circ \]
\[ 2A = 210^\circ \implies A = 105^\circ \]

Substituting \( A = 105^\circ \) into equation (3):
\[ 105^\circ + B = 60^\circ \implies B = 60^\circ - 105^\circ = -45^\circ \]
This is not valid for angles A and B in the context of this problem.

Using (1) and (4):
1. \( A - B = 30^\circ \)
2. \( A + B = 300^\circ \)

Adding these:
\[ (A - B) + (A + B) = 30^\circ + 300^\circ \]
\[ 2A = 330^\circ \implies A = 165^\circ \]

Substituting \( A = 165^\circ \) into equation (4):
\[ 165^\circ + B = 300^\circ \implies B = 300^\circ - 165^\circ = 135^\circ \]
This is also not valid for angles A and B in the context of this problem.

### Conclusion
The only valid solution is:
\[ A = 45^\circ, \quad B = 15^\circ \]