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 use trigonometric identities and properties step by step.

### Step 1: Use the sum-to-product identities
Recall the identity for the sum of cosines:
\[ \cos x + \cos y = 2 \cos\left(\frac{x+y}{2}\right) \cos\left(\frac{x-y}{2}\right) \]
Substituting this into the equation gives:
\[ 2 \cos\left(\frac{x+y}{2}\right) \cos\left(\frac{x-y}{2}\right) - \cos(x+y) = \frac{3}{2} \]

### Step 2: Rewrite \(\cos(x+y)\)
Using the double angle identity for cosine:
\[ \cos(x+y) = \cos^2\left(\frac{x+y}{2}\right) - \sin^2\left(\frac{x+y}{2}\right) = 2\cos^2\left(\frac{x+y}{2}\right) - 1 \]
Substituting this back into the equation gives:
\[ 2 \cos\left(\frac{x+y}{2}\right) \cos\left(\frac{x-y}{2}\right) - (2\cos^2\left(\frac{x+y}{2}\right) - 1) = \frac{3}{2} \]

### Step 3: Simplify the equation
Rearranging the equation:
\[ 2 \cos\left(\frac{x+y}{2}\right) \cos\left(\frac{x-y}{2}\right) - 2\cos^2\left(\frac{x+y}{2}\right) + 1 = \frac{3}{2} \]
This simplifies to:
\[ 2 \cos\left(\frac{x+y}{2}\right) \cos\left(\frac{x-y}{2}\right) - 2\cos^2\left(\frac{x+y}{2}\right) = \frac{1}{2} \]

### Step 4: Factor the equation
Factoring out \(2\):
\[ 2\left(\cos\left(\frac{x+y}{2}\right) \cos\left(\frac{x-y}{2}\right) - \cos^2\left(\frac{x+y}{2}\right)\right) = \frac{1}{2} \]
Dividing both sides by 2:
\[ \cos\left(\frac{x+y}{2}\right) \cos\left(\frac{x-y}{2}\right) - \cos^2\left(\frac{x+y}{2}\right) = \frac{1}{4} \]

### Step 5: Rearranging and solving
Rearranging gives:
\[ \cos\left(\frac{x+y}{2}\right) \left(\cos\left(\frac{x-y}{2}\right) - \cos\left(\frac{x+y}{2}\right)\right) = \frac{1}{4} \]

### Step 6: Analyze the conditions
Since \( \cos\left(\frac{x+y}{2}\right) \) can take values between -1 and 1, we need to analyze the conditions under which this holds true. The maximum value of \( \cos\left(\frac{x+y}{2}\right) \) is 1, which implies:
\[ \cos\left(\frac{x-y}{2}\right) - \cos\left(\frac{x+y}{2}\right) = \frac{1}{4} \]
This indicates that \( \cos\left(\frac{x-y}{2}\right) \) must also be constrained.

### Step 7: Conclusion
To satisfy the equation, \( \cos\left(\frac{x-y}{2}\right) \) must equal \( 1 \), which implies:
\[ \frac{x-y}{2} = 0 \implies x - y = 0 \implies x = y \]

Thus, the final answer is:
\[ \boxed{x = y} \]