If `sin(x-y)=(1)/(2)` and `cos(x+y)=(1)/(2)` , then value of `x` is

Options

A. `15^(@)`
B. `30^(@)`
C. `45^(@)`
D. `60^(@)`

Answer

To solve the problem, we need to find the value of \( x \) given the equations:

1. \( \sin(x - y) = \frac{1}{2} \)
2. \( \cos(x + y) = \frac{1}{2} \)

### Step 1: Solve for \( x - y \)

From the first equation, we know that:

\[ \sin(x - y) = \frac{1}{2} \]

The angle whose sine is \( \frac{1}{2} \) is \( 30^\circ \) (or \( \frac{\pi}{6} \) radians). Therefore, we can write:

\[ x - y = 30^\circ \quad \text{(Equation 1)} \]

### Step 2: Solve for \( x + y \)

From the second equation, we have:

\[ \cos(x + y) = \frac{1}{2} \]

The angle whose cosine is \( \frac{1}{2} \) is \( 60^\circ \) (or \( \frac{\pi}{3} \) radians). Thus, we can write:

\[ x + y = 60^\circ \quad \text{(Equation 2)} \]

### Step 3: Solve the system of equations

Now we have a system of two equations:

1. \( x - y = 30^\circ \)
2. \( x + y = 60^\circ \)

We can solve these equations simultaneously.

### Step 4: Add the equations

Adding both equations:

\[ (x - y) + (x + y) = 30^\circ + 60^\circ \]

This simplifies to:

\[ 2x = 90^\circ \]

### Step 5: Solve for \( x \)

Dividing both sides by 2 gives:

\[ x = 45^\circ \]

### Step 6: Find \( y \)

Now we can find \( y \) using either of the original equations. Let's use Equation 1:

\[ x - y = 30^\circ \]

Substituting \( x = 45^\circ \):

\[ 45^\circ - y = 30^\circ \]

Rearranging gives:

\[ y = 45^\circ - 30^\circ = 15^\circ \]

### Conclusion

The value of \( x \) is:

\[ \boxed{45^\circ} \]