If `sin(x-y)=cos(x+y)=1/2,` the value of x and y lyingbetween `0^(@)and 90^(@)` are given by

Options

A. `x=15^(@),y=25^(@)`
B. `x=65^(@),y=15^(@)`
C. `x=45^(@),y=45^(@)`
D. `x=45^(@),y=15^(@)`

Answer

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

### Step 1: Set up the equations from the given conditions
From the problem, we have:
1. \( \sin(x - y) = \frac{1}{2} \)
2. \( \cos(x + y) = \frac{1}{2} \)

### Step 2: Solve for \( x - y \)
The sine function equals \( \frac{1}{2} \) at specific angles. In the interval \( 0^\circ \) to \( 90^\circ \), we have:
\[ x - y = 30^\circ \quad \text{(or } \frac{\pi}{6} \text{ radians)} \]

### Step 3: Solve for \( x + y \)
The cosine function equals \( \frac{1}{2} \) at specific angles. In the interval \( 0^\circ \) to \( 90^\circ \), we have:
\[ x + y = 60^\circ \quad \text{(or } \frac{\pi}{3} \text{ radians)} \]

### Step 4: Set up a system of equations
Now we have a system of equations:
1. \( x - y = 30^\circ \)
2. \( x + y = 60^\circ \)

### Step 5: Add the equations
Adding the two equations:
\[ (x - y) + (x + y) = 30^\circ + 60^\circ \]
This simplifies to:
\[ 2x = 90^\circ \]
Thus, we find:
\[ x = 45^\circ \]

### Step 6: Substitute to find \( y \)
Now substitute \( x = 45^\circ \) back into one of the original equations to find \( y \). Using \( x + y = 60^\circ \):
\[ 45^\circ + y = 60^\circ \]
Solving for \( y \):
\[ y = 60^\circ - 45^\circ = 15^\circ \]

### Final Answer
Thus, the values of \( x \) and \( y \) are:
\[ x = 45^\circ, \quad y = 15^\circ \]