If `sin(x+y)=cos(x-y)`, then find the value of `cos^(2)x`.

Options

A. 5
B. `-(1)/(2)`
C. `(1)/(2)`
D. 3

Answer

To solve the equation \( \sin(x+y) = \cos(x-y) \) and find the value of \( \cos^2 x \), we can follow these steps:

### Step 1: Use the co-function identity
We know that:
\[ \sin(90^\circ - \theta) = \cos(\theta) \]
So, we can rewrite \( \cos(x-y) \) as:
\[ \cos(x-y) = \sin(90^\circ - (x-y)) = \sin(90^\circ - x + y) \]

### Step 2: Set the equation
Now we have:
\[ \sin(x+y) = \sin(90^\circ - x + y) \]

### Step 3: Apply the sine identity
Since \( \sin A = \sin B \) implies that:
\[ A = B + n \cdot 180^\circ \quad \text{or} \quad A = 180^\circ - B + n \cdot 180^\circ \quad (n \in \mathbb{Z}) \]
We can set:
1. \( x + y = 90^\circ - x + y + n \cdot 180^\circ \)
2. \( x + y = 180^\circ - (90^\circ - x + y) + n \cdot 180^\circ \)

### Step 4: Simplify the first case
From the first case:
\[ x + y = 90^\circ - x + y + n \cdot 180^\circ \]
Subtract \( y \) from both sides:
\[ x = 90^\circ - x + n \cdot 180^\circ \]
Adding \( x \) to both sides gives:
\[ 2x = 90^\circ + n \cdot 180^\circ \]
Thus:
\[ x = 45^\circ + \frac{n \cdot 180^\circ}{2} \]

### Step 5: Consider the second case
From the second case:
\[ x + y = 180^\circ - (90^\circ - x + y) + n \cdot 180^\circ \]
This simplifies to:
\[ x + y = 90^\circ + x - y + n \cdot 180^\circ \]
Subtract \( x \) and \( y \) from both sides:
\[ 0 = 90^\circ - 2y + n \cdot 180^\circ \]
This gives:
\[ 2y = 90^\circ + n \cdot 180^\circ \]
So:
\[ y = 45^\circ + \frac{n \cdot 180^\circ}{2} \]

### Step 6: Find \( \cos^2 x \)
From our earlier result, we can take \( n = 0 \) for simplicity:
\[ x = 45^\circ \]
Now we can find \( \cos^2 x \):
\[ \cos^2 x = \cos^2(45^\circ) = \left(\frac{1}{\sqrt{2}}\right)^2 = \frac{1}{2} \]

### Final Answer
Thus, the value of \( \cos^2 x \) is:
\[ \cos^2 x = \frac{1}{2} \]

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