If `x=ycos((2pi)/(3))=zcos((4pi)/(3))`,then what is xy + yz + zx equal to ?

Options

A. `-1`
B. 0
C. 1
D. 2

Explanation

To solve the problem, we start with the given equations:

1. \( x = y \cos\left(\frac{2\pi}{3}\right) \)
2. \( x = z \cos\left(\frac{4\pi}{3}\right) \)

We need to find the value of \( xy + yz + zx \).

### Step 1: Calculate \( \cos\left(\frac{2\pi}{3}\right) \)

Using the cosine identity:
\[ \cos\left(\frac{2\pi}{3}\right) = \cos\left(\pi - \frac{\pi}{3}\right) = -\cos\left(\frac{\pi}{3}\right) \]
Since \( \cos\left(\frac{\pi}{3}\right) = \frac{1}{2} \):
\[ \cos\left(\frac{2\pi}{3}\right) = -\frac{1}{2} \]

### Step 2: Substitute into the equation for \( x \)

Substituting the value of \( \cos\left(\frac{2\pi}{3}\right) \) into the equation for \( x \):
\[ x = y \left(-\frac{1}{2}\right) \implies x = -\frac{y}{2} \]

### Step 3: Calculate \( \cos\left(\frac{4\pi}{3}\right) \)

Using the cosine identity:
\[ \cos\left(\frac{4\pi}{3}\right) = \cos\left(\pi + \frac{\pi}{3}\right) = -\cos\left(\frac{\pi}{3}\right) \]
Thus:
\[ \cos\left(\frac{4\pi}{3}\right) = -\frac{1}{2} \]

### Step 4: Substitute into the equation for \( x \)

Substituting the value of \( \cos\left(\frac{4\pi}{3}\right) \) into the equation for \( x \):
\[ x = z \left(-\frac{1}{2}\right) \implies x = -\frac{z}{2} \]

### Step 5: Set the equations for \( y \) and \( z \)

From the equations we have:
1. \( x = -\frac{y}{2} \)
2. \( x = -\frac{z}{2} \)

Equating the two expressions for \( x \):
\[ -\frac{y}{2} = -\frac{z}{2} \]
This simplifies to:
\[ y = z \]

### Step 6: Substitute \( y = z \) into \( xy + yz + zx \)

Now we can substitute \( y = z \) into the expression \( xy + yz + zx \):
\[ xy + yz + zx = xy + y^2 + zx \]
Since \( z = y \), we have:
\[ = xy + y^2 + yx = 2xy + y^2 \]

### Step 7: Substitute \( x = -\frac{y}{2} \)

Substituting \( x = -\frac{y}{2} \) into \( 2xy + y^2 \):
\[ 2\left(-\frac{y}{2}\right)y + y^2 = -y^2 + y^2 = 0 \]

### Final Answer

Thus, the value of \( xy + yz + zx \) is:
\[ \boxed{0} \]