If `x=ycos((2pi)/3)=zcos((4pi)/3)`, then xy+yz+zx= (a) -1 (b) 0 (c) 1 (d) 2

Options

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

Answer

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

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

### Step 1: Calculate \(\cos\left(\frac{2\pi}{3}\right)\) and \(\cos\left(\frac{4\pi}{3}\right)\)

We know that:
- \(\cos\left(\frac{2\pi}{3}\right) = -\frac{1}{2}\)
- \(\cos\left(\frac{4\pi}{3}\right) = -\frac{1}{2}\)

### Step 2: Substitute the cosine values into the equations

From the first equation:
\[ x = y \left(-\frac{1}{2}\right) \implies y = -2x \]

From the second equation:
\[ x = z \left(-\frac{1}{2}\right) \implies z = -2x \]

### Step 3: Substitute \(y\) and \(z\) in the expression \(xy + yz + zx\)

Now we need to find \(xy + yz + zx\):
\[ xy = x(-2x) = -2x^2 \]
\[ yz = (-2x)(-2x) = 4x^2 \]
\[ zx = (-2x)x = -2x^2 \]

### Step 4: Combine the terms

Now, we combine all the terms:
\[ xy + yz + zx = -2x^2 + 4x^2 - 2x^2 \]
\[ = (-2x^2 - 2x^2 + 4x^2) = 0 \]

### Final Answer

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