The greatest value of
`cos^(2)((pi)/(3)-x)-cos^(2)((pi)/(3)+x)` is

Options

A. `-(sqrt(3))/(2)`
B. `(1)/(2)`
C. `(sqrt(3))/(2)`
D. `(3)/(2)`

Answer

To find the greatest value of the expression \( \cos^2\left(\frac{\pi}{3} - x\right) - \cos^2\left(\frac{\pi}{3} + x\right) \), we can follow these steps:

### Step 1: Use the identity for the difference of squares
We can use the trigonometric identity:
\[ \cos^2 A - \cos^2 B = \sin(A - B) \sin(A + B) \]
Let \( A = \frac{\pi}{3} - x \) and \( B = \frac{\pi}{3} + x \).

### Step 2: Calculate \( A - B \) and \( A + B \)
Calculating \( A - B \):
\[ A - B = \left(\frac{\pi}{3} - x\right) - \left(\frac{\pi}{3} + x\right) = -2x \]
Calculating \( A + B \):
\[ A + B = \left(\frac{\pi}{3} - x\right) + \left(\frac{\pi}{3} + x\right) = \frac{2\pi}{3} \]

### Step 3: Substitute into the identity
Now substituting back into the identity:
\[ \cos^2\left(\frac{\pi}{3} - x\right) - \cos^2\left(\frac{\pi}{3} + x\right) = \sin(-2x) \sin\left(\frac{2\pi}{3}\right) \]
Since \( \sin(-\theta) = -\sin(\theta) \), we have:
\[ \sin(-2x) = -\sin(2x) \]
Thus, the expression becomes:
\[ -\sin(2x) \sin\left(\frac{2\pi}{3}\right) \]

### Step 4: Calculate \( \sin\left(\frac{2\pi}{3}\right) \)
We know:
\[ \sin\left(\frac{2\pi}{3}\right) = \sin\left(\pi - \frac{\pi}{3}\right) = \sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2} \]
So we can rewrite our expression:
\[ -\sin(2x) \cdot \frac{\sqrt{3}}{2} \]

### Step 5: Find the maximum value
The maximum value of \( -\sin(2x) \) occurs when \( \sin(2x) \) is at its minimum, which is -1. Therefore:
\[ \text{Maximum value} = -(-1) \cdot \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{2} \]

### Conclusion
Thus, the greatest value of the expression \( \cos^2\left(\frac{\pi}{3} - x\right) - \cos^2\left(\frac{\pi}{3} + x\right) \) is:
\[ \frac{\sqrt{3}}{2} \]