If `0 le x, y le 2 pi "and cos"x + "cos" y = -2, " then "cos" (x+y)=`

Options

A. 0
B. 1
C. `-1`
D. none of these

Explanation

To solve the problem, we need to find the value of \( \cos(x+y) \) given the condition \( \cos x + \cos y = -2 \) and that \( 0 \leq x, y \leq 2\pi \).

### Step-by-Step Solution:

1. **Understanding the Condition**:
The equation \( \cos x + \cos y = -2 \) implies that both \( \cos x \) and \( \cos y \) must take their minimum possible values. The cosine function ranges from -1 to 1. Therefore, the only way for their sum to equal -2 is if both \( \cos x \) and \( \cos y \) are equal to -1.

**Hint**: Recall that the cosine function achieves its minimum value of -1 at specific angles.

2. **Finding Values of \( x \) and \( y \)**:
Since \( \cos x = -1 \), we find the angle \( x \) in the range \( [0, 2\pi] \):
\[ x = \pi \]
Similarly, for \( \cos y = -1 \):
\[ y = \pi \]

**Hint**: Identify the angles where the cosine function equals -1.

3. **Calculating \( x + y \)**:
Now that we have \( x = \pi \) and \( y = \pi \), we can calculate \( x + y \):
\[ x + y = \pi + \pi = 2\pi \]

**Hint**: Add the values of \( x \) and \( y \) to find the sum.

4. **Finding \( \cos(x+y) \)**:
We need to find \( \cos(x+y) \):
\[ \cos(x+y) = \cos(2\pi) \]
We know that:
\[ \cos(2\pi) = 1 \]

**Hint**: Recall the value of cosine at key angles, particularly at multiples of \( 2\pi \).

5. **Final Answer**:
Therefore, the value of \( \cos(x+y) \) is:
\[ \boxed{1} \]