If `cosx+cosy+cosz=sinx+siny+sinz=0" then "cos(x-y)=` (a) 0 (b) `-1/2` (c) 2 (d) 1

Options

A. 0
B. `-(1)/(2)`
C. 2
D. 1

Answer

To solve the problem, we need to find the value of \( \cos(x - y) \) given that:

\[ \cos x + \cos y + \cos z = 0 \]
\[ \sin x + \sin y + \sin z = 0 \]

### Step-by-Step Solution:

1. **Rearranging the Equations**:
From the first equation, we can express \( \cos z \):
\[ \cos z = -(\cos x + \cos y) \]
From the second equation, we can express \( \sin z \):
\[ \sin z = -(\sin x + \sin y) \]

2. **Squaring and Adding the Equations**:
We square both equations and add them:
\[ (\cos x + \cos y)^2 + (\sin x + \sin y)^2 = \cos^2 z + \sin^2 z \]

3. **Expanding the Left Side**:
Using the identity \( (a + b)^2 = a^2 + b^2 + 2ab \):
\[ \cos^2 x + \cos^2 y + 2\cos x \cos y + \sin^2 x + \sin^2 y + 2\sin x \sin y \]

4. **Using the Pythagorean Identity**:
We know that \( \cos^2 \theta + \sin^2 \theta = 1 \):
\[ (1 + 1) + 2(\cos x \cos y + \sin x \sin y) = \cos^2 z + \sin^2 z \]
Thus, we have:
\[ 2 + 2(\cos x \cos y + \sin x \sin y) = 1 \]

5. **Simplifying the Equation**:
Rearranging gives:
\[ 2(\cos x \cos y + \sin x \sin y) = 1 - 2 \]
\[ 2(\cos x \cos y + \sin x \sin y) = -1 \]
Dividing by 2:
\[ \cos x \cos y + \sin x \sin y = -\frac{1}{2} \]

6. **Using the Cosine of Difference Identity**:
Recall that:
\[ \cos(x - y) = \cos x \cos y + \sin x \sin y \]
Therefore:
\[ \cos(x - y) = -\frac{1}{2} \]

### Conclusion:
The value of \( \cos(x - y) \) is \( -\frac{1}{2} \).

### Final Answer:
(b) \( -\frac{1}{2} \)