If `cosx + cosy + cosz = 0` and `sinx + siny + sinz =0`, then show that
`cos (x-y) + cos(y - z) + cos(z - x) = - 3/2.`

Options

A. `cos x cos ycosz=1`
B. `cos x+cosy+cos z=0`
C. `sinx+siny+sinz=1`
D. `cos 3x+cos3y+cos3z=12 cosx cosy cosz`

Answer

To solve the problem, we need to show that if \( \cos x + \cos y + \cos z = 0 \) and \( \sin x + \sin y + \sin z = 0 \), then \( \cos(x-y) + \cos(y-z) + \cos(z-x) = -\frac{3}{2} \).

### Step-by-Step Solution:

1. **Start with Given Equations**:
\[ \cos x + \cos y + \cos z = 0 \quad \text{(1)} \]
\[ \sin x + \sin y + \sin z = 0 \quad \text{(2)} \]

2. **Rearranging the Equations**:
From equation (1), we can express \( \cos z \):
\[ \cos z = -(\cos x + \cos y) \quad \text{(3)} \]
From equation (2), we can express \( \sin z \):
\[ \sin z = -(\sin x + \sin y) \quad \text{(4)} \]

3. **Square and Add the Equations**:
We square both sides of equations (3) and (4) and add them:
\[ \cos^2 z + \sin^2 z = \cos^2 x + \cos^2 y + \sin^2 x + \sin^2 y + 2(\cos x \cos y + \sin x \sin y) \]

Using the identity \( \cos^2 \theta + \sin^2 \theta = 1 \):
\[ 1 = (\cos^2 x + \sin^2 x) + (\cos^2 y + \sin^2 y) + 2(\cos x \cos y + \sin x \sin y) \]
\[ 1 = 1 + 1 + 2(\cos x \cos y + \sin x \sin y) \]

4. **Simplifying the Equation**:
Thus, we have:
\[ 1 = 2 + 2(\cos x \cos y + \sin x \sin y) \]
Rearranging gives:
\[ 2(\cos x \cos y + \sin x \sin y) = -1 \]
\[ \cos x \cos y + \sin x \sin y = -\frac{1}{2} \]

5. **Using the Cosine of Angle Difference**:
Recall the cosine of the difference of angles:
\[ \cos(x-y) = \cos x \cos y + \sin x \sin y \]
Therefore:
\[ \cos(x-y) = -\frac{1}{2} \quad \text{(5)} \]

6. **Applying the Same Logic for Other Angles**:
By symmetry, we can apply the same reasoning to find:
\[ \cos(y-z) = -\frac{1}{2} \quad \text{(6)} \]
\[ \cos(z-x) = -\frac{1}{2} \quad \text{(7)} \]

7. **Adding the Results**:
Now, we add equations (5), (6), and (7):
\[ \cos(x-y) + \cos(y-z) + \cos(z-x) = -\frac{1}{2} - \frac{1}{2} - \frac{1}{2} = -\frac{3}{2} \]

8. **Conclusion**:
Thus, we have shown that:
\[ \cos(x-y) + \cos(y-z) + \cos(z-x) = -\frac{3}{2} \]