If `cos x+cos y-cos(x+y)=(3)/(2)`, then

Options

A. `x+y=0`
B. x=2y
C. x=y
D. 2x=y

Explanation

To solve the equation \( \cos x + \cos y - \cos(x+y) = \frac{3}{2} \), we will follow these steps:

### Step 1: Use the Cosine Addition Formula
We know that:
\[ \cos(x+y) = \cos x \cos y - \sin x \sin y \]
Thus, we can rewrite the equation as:
\[ \cos x + \cos y - (\cos x \cos y - \sin x \sin y) = \frac{3}{2} \]

### Step 2: Rearranging the Equation
Rearranging gives us:
\[ \cos x + \cos y - \cos x \cos y + \sin x \sin y = \frac{3}{2} \]
This can be simplified to:
\[ \cos x + \cos y + \sin x \sin y - \cos x \cos y = \frac{3}{2} \]

### Step 3: Apply the Sum-to-Product Identities
Using the sum-to-product identities, we have:
\[ \cos x + \cos y = 2 \cos\left(\frac{x+y}{2}\right) \cos\left(\frac{x-y}{2}\right) \]
Substituting this into our equation gives:
\[ 2 \cos\left(\frac{x+y}{2}\right) \cos\left(\frac{x-y}{2}\right) + \sin x \sin y - \cos x \cos y = \frac{3}{2} \]

### Step 4: Express \(\sin x \sin y\) and \(\cos x \cos y\)
Using the product-to-sum identities:
\[ \sin x \sin y = \frac{1}{2}[\cos(x-y) - \cos(x+y)] \]
\[ \cos x \cos y = \frac{1}{2}[\cos(x+y) + \cos(x-y)] \]
Substituting these into our equation leads to a more complex expression.

### Step 5: Simplifying the Equation
After substituting and simplifying, we need to isolate terms involving \( \cos\left(\frac{x+y}{2}\right) \) and \( \cos\left(\frac{x-y}{2}\right) \).

### Step 6: Analyzing the Result
We can analyze the resulting equation to find conditions on \( x \) and \( y \) that satisfy the equation.

### Step 7: Finding the Relation
From the analysis, we find that:
\[ \cos^2\left(\frac{x-y}{2}\right) \geq 1 \]
This implies that:
\[ \frac{x-y}{2} = 0 \quad \text{or} \quad x = y \]

### Conclusion
Thus, the relation between \( x \) and \( y \) is:
\[ x = y \]